Xtraedge April 2010

XtraEdge for IIT-JEE 1 APRIL 2010

Dear Students,

Motivate Yourself One of the greatest virtues of human beings is their ability to think and act accordingly. The emergence of the techno savvy human from the tree swinging ape has really been a long journey. This transition has taken a span of countless centuries and lots of thinking caps have been involved. Inquisitiveness and aspiration to come out with the best have been the pillars for man's quest for development. Self-motivation is the sheer force, which pulled him apart and distinguished him from his primitive ancestors. Many times, in our life, when we are reviving old memories we get into a phase of nostalgia. We feel that we could have done better than what we had achieved. But thinking back won’t rewind the tireless worker called time. All we can do is promise ourselves that we will give our very best in the future. But do we really keep up to our mental commitments? I can guess that 90% answers are in the negative. This is because of that creepy careless attitude which is slowly, but surely entering into our mind. We easily forget the pains of yesterday to relish the joys of today. This is the only time in our life, when we can control our fate, by controlling our mind. So it is time to pull up our socks and really motivate ourselves so that we can give our best shot in the future. Self-motivation is the need of the hour. Only we can control and restrict ourselves. It’s up to us, how we use our mental capabilities to the best of our abilities. Here are some Funda's for self-motivation. Don't just read them digest each one of them and apply them and I bet it will make a better YOU. • The ultimate motivator is defeat. Once you are defeated, you

have nowhere to go except the top. • Then only thing stopping you is yourself. • There is no guarantee that tomorrow will come. So do it today. • Intentions don't count, but action's do. • Don't let who you are, stunt what you want to be. • Success is the greatest motivator. • Your goals must be clear, but the guidelines must be flexible. Try to include these one liners in your scrapbook or on your favorite poster. You will be sub-consciously tuned to achieve what you want. Also do keep in mind that nothing can control your destiny but you! With Best Wishes for Your Future.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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To fly, we have to have resistance Volume - 5 Issue - 10

April, 2010 (Monthly Magazine)

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Volume-5 Issue-10 April, 2010 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Key Concepts & Problem Solving strategy for IIT-JEE.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics, Chemistry & Maths

Much more IIT-JEE News.

Xtra Edge Test Series for JEE – 2010 & 2011

Success Tips for the Month

• "The way to succeed is to double your error rate."

• "Success is the ability to go from failure to failure without losing your enthusiasm."

• "Success is the maximum utilization of the ability that you have."

• We are all motivated by a keen desire for praise, and the better a man is, the more he is inspired to glory.

• Along with success comes a reputation for wisdom.

• They can, because they think they can.

• Nothing can stop the man with the right mental attitude from achieving his goal; nothing on earth can help the man with the wrong mental attitude.

• Keep steadily before you the fact that all true success depends at last upon yourself.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 • President buries 'Time Capsule' on IIT Kanpur campus • IITs admission criteria set for an overhaul

IITian ON THE PATH OF SUCCESS 6 Mr. Krishnamurthy Rengarajan

KNOW IIT-JEE 7 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 52 Mock Test IIT-JEE Paper-1 & Paper-2 Mock Test AIEEE Mock Test BIT SAT SOLUTIONS 92

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set# 12] Students’ Forum Physics Fundamentals Calorimetry, K.T.G., Heat Transfer Atomic Structure, X-Ray & Radio Activity CATALYST CHEMISTRY 30

Key Concept Aromatic Hydrocarbon Solubility Product Understanding: Organic Chemistry

DICEY MATHS 38

Mathematical Challenges Students’ Forum Key Concept Calculus Algebra

Study Time........

Test Time ..........


President buries 'Time Capsule' on IIT Kanpur campus

Kanpur: President Pratibha Patil recently buried a 'Time Capsule' on the Indian Institute of Technology, Kanpur (IIT-K) campus on the occasion of its golden jubilee celebrations and also unveiled a nanosatellite developed by the institute. The capsule, which is made of a special metal, contains pen-drives, chips, images and several other documents related to the landmark achieve-ments of the IIT-K. Lauding the nanosatellite Jugnu's development team, Patil said it projects the complex nature of tasks that the students there were equipped to handle. Congratulating IIT-K students and faculty, Patil said that the institute has come a long way in its 50 years of its existence, and also called upon the institute's students and faculty members to develop such devices that can harness energy in efficient ways with minimal negative impact on the environment. "It (IIT-K) had made an impact on technical education within the country, while its students through their innovations, have played an important role in India, as well as around the world," the President said. Jugnu, developed by a team of 50 IIT students, will help in collection of information related to floods, drought and other natural calamities.

IITs admission criteria set for an overhaul New Delhi: The admission criteria for admission to the premier engineering institutes of the country - the Indian Institutes of Technology (IITs), is all set for an overhaul.

According to the recommendation of a panel set up under IIT Kaharagpur's Director, Mr. Damodar Acharya, the Joint Entrance Examination (JEE) for undergraduates and Graduate Aptitude Test in Engineering (GATE) for postgraduates may soon not be the only criteria for their admission. The committee proposes to consider the Class XII marks as well.

Presently, the eligibility for a student to sit for a JEE is 60% in class XII after which a cut-off is decided every year for admission into an IIT. Following the consensus of the last meeting of the IIT Council to give more weightage to the school leaving examination, the committee proposes to mull over the issue of factoring in class XII examination result in the cut-off for admission to the IITs.

The IIT Directors in a meeting with the Human Resource Development (HRD) Minister, Kapil Sibal, in Manesar, also expressed their discontent over the existing pattern for GATE as they felt that not enough students with research orientation were being picked up through this exam.

Sibal said that each IIT should submit a proposal within a month on one area of expertise in higher

research. Although the admission criteria is becoming stringent with the proposal of IIT-Kharagpur Director's recommendation, the HRD minister retreated with a firm 'No' to IIT-Kanpur Director's proposal for a fee hike. According to the fee-hike proposal, which the IIT-Kanpur's Director framed, it suggested to hike fees by eight times. At present, the B.Tech students pay Rs.50,000 per year as their fee which the committee proposed to increase up to Rs.4 lakh per annum over the period of 10 years with a Rs.35,000 mark-up every year.

"The proposal on fee hike should be discussed as the government was planning to set up a Higher Education Funding Corporation which would address poor students," said a senior official.

During the meeting, Sibal also asked the IITs to come up with their plans for the future endeavors in a specialized area in which they want to emerge as global giants by 2020 in four weeks time.

Nanotech is used to treat cancer - IIT-B & Docs In Mumbai it could have been India’s second nano success if it passes the muster. Only this nano creation is revealed in the healthcare field and it was possible because of partnership between oncologists and scientists of the Indian Institute of Technology-Bombay.

The joint effort of the IIT -B and doctors from Tata Memorial


Hospital in Parel and Apollo Hospital in Hyderabad has the possibility to transform treatment of retinoblastoma—a rare cancer of the retina that mainly affects children under two years of age. They have developed a nano-particle that could conquer the child killer.

Shirin Thakur, Guntur-based teenager was suffering from recurrent retinoblastoma since she was two years old. Last week, she took the third injection of a special mixture into the tissues around her left eye which was made of nano-particles of carboplatin which is commonly used to treat retinoblastoma. While she was standing with her doctor in the Apollo hospital, Hyderabad said, “I have been suffering from attacks of retinoblastoma in my left eye since I was two. Even in the US, they told me there is no hope but to remove my eye.’’ Now, she has “fuzzy’’ vision in the nearly blind eye. “My vision gets better every day.’’

PM was asked to fine-tune JEE by Maths Wizard PM Manmohan singh was given suggestions on the improvement of IIT-JEE pattern by Maths Wizard Anand Kumar. He pleaded that poor aspirants should be given at least three attempts for IIT-JEE exam, as they are usually the late starters.

Every year Anand’s Super-30 offers free accommodation to 30 poor students and provides coaching to them free of cost in order to help them to crack JEE. This noble initiative, has of late reported 100% success rate with all its 30 students making it to IIT’S, without seeking any financial support from government or non-government sources.

In his 15-minute interaction with the PM in New Delhi, Anand enlightened him of his efforts usher in a new awakening in Bihar by sending 182 poor students to IITs within the past seven years. He further informed the PM that he has decided to increase the student intake to 60 from 30. Very soon he would open schools for talented poor children so as to provide them the right momentum at the secondary school level itself. He added that government should run coaching programmes on the pattern of Super-30 for talented students from the countryside and it should not be confined to IIT only.

Audibly elated Anand told TOI over the phone that Pm patted him on his back and also instructed PMO to look into his suggestions.

He said that many problems asked in JEE are of Maths Olympiad level so students from villages find it difficult to solve such problems even if they have sound knowledge of the subject at +2 level. He pleaded that the exam should be designed in such a way that conceptual or analytical problems of the +2 level should be asked in JEE."

IIT Rajasthan moves to state varsity campus Jaipur: The Indian Institute of Technology, Rajasthan (IIT-R) has been temporarily shifted to the engineering faculty building of Jai Narain Vyas University according to a Memorandum of Understanding (MoU) signed in Jodhpur over the weekend. Students of IIT-R earlier had to rely on IIT-Kanpur and other institutions to attend classes.

IIT-Rajasthan founder-director P. K. Kalra and JNV University Registrar Nirmala Meena, in the

presence of Chief Minister Ashok Gehlot who was on a visit to Jodhpur, signed the MoU.

The CM also inaugurated the work to extend the building. Gehlot pointed out in an official press release that the establishment of IIT in Jodhpur was recommended and headed by the noted economist V.S. Vyas after he studied the facilities and educational standard of various towns in the State.

"The final decision to nod for an IIT in the city was accepted at the central level," added Gehlot. He further announced that the construction of a new building in Rajasthan would require some time, and for the time being the classes would be held in JNV University while those in the extended portion of the building would begin from May his year.

Following special efforts made by the state government Mr. Gehlot said that the union government has also nodded to use the newly constructed A.S.K. Hostel in the university for IIT students.

"The establishment of an IIT in the state would definitely boost its higher education and will prove to be a milestone," said Gehlot.

According to the MoU, it approves the functioning of IIT-Rajasthan only on temporary basis for two years in the engineering faculty building.

State Technical Education Minister Mahendrajit Singh Malviya and JNV University Vice-Chancellor Naveen Mathur signed the MoU as witnesses. Among others, Jodhpur MP Chandresh Kumari, Barmer-Jaisalmer MP Harish Chaudhary, Jodhpur Mayor Rameshwar Dadich, Deputy Mayor Niyaz Mohammed and Collector Naveen Mahajan were present on the occasion.


Knowledge is indeed wealth. Who better exemplifies it than Krishnamurthy Rengarajan,IIT-B gold medallist (B Tech dual-degree course). Krishnamurthy's story is that of hard work, sheer grit and determination. His undying passion for learning and excellence has paid off. Coming from a lower middle class background, Krishnamurthy has made his parents proud when he passed with flying colours. His father, who works as a typist at Bharatiya Vidya Bhavan is overwhelmed by his son's achievement. Rengarajan, who hails from Tamil Nadu, came to Mumbai 28 years ago and settled down in a distant Mumbai suburb of Dombivli. Though the family went through a lot of hardships initially, he made sure that his children were well educated. "My son always wanted to join the IIT. When people asked him what if you don't get through the entrance examinations, he used to say, `there is no question of me not clearing the test," says his proud father. And, of course, he did top all the five years at IIT, a result of sheer hard work and brilliance, says his mother, barely able to control her excitement. "I am very happy for him," says Radha Rengarajan. Krishnamurthy did his schooling at the Kidland School in Dombivli and pre-degree from V G Vaze College at Mulund. His favourite subject being mathematics it was obvious that he would pursue a degree in engineering. He won the Rakesh Mathur award of Rs 1 lakh (Rs 100,000) during his third year and other scholarships throughout the four years. Here's what Krishnamurthy had to say on his IIT experience. My IIT experience The five years I spent at IIT were the best in my life. I will cherish each and every moment here. I loved everything here: the professors are the best one can ever get, the facilities to study and the extra-curricular activities are excellent. I made best of friends and thoroughly enjoyed my college life. I don't think I will ever get this experience anywhere else.

On studies Before joining IIT, I used to study for 7 to 8 hours daily. After joining IIT, I used to spend about a couple of hours. I did not go for anything coaching classes. I learnt through Brilliant Tutorial correspondence course and my preparation began after I finished my 10th standard. Why IIT IIT is one of the premier institutes in India. I always wanted to get good higher education, so I opted for IIT. My mantra for success There is no short cut to success. One has to work very hard, put in a lot of effort, should have a problem-solving mentality and a right approach to every problem. My parents always stood by me, their support has been invaluable and am overwhelmed. Advice to IIT aspirants Work hard. You have to spend a lot of time preparing as exams are getting more and more competitive. You must also have problem-solving skills. Interests Solving math puzzles, reading books. I used to play cricket, but now I don't get the time. Next move Money is the least important thing for Krishnamurthy. So no jobs for the time being. "I have been selected for the scholarship programme at Stanford University for a PhD in operations research. I would like to research on optimising computer networks and operation systems. Quality research is available abroad. After the PhD programme I would like to join any academia of good repute and continue my research activities. Among corporates, I admire Google. It is the one company that reflects perfection, hard work and efficiency." Will you come back to India? Of course, I will. The brain drain phenomenon is dying out. It's the time for reverse brain.

Mr. Krishnamurthy Rengarajan IIT-B Gold Medallist

Success StoryThis article contains story of a person who is successful after graduation from different IIT's


PHYSICS

1. A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure in solution). The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surface are frictionless. The ball is given a gentle push (towards the right see figure in solution) The angle made by the radius vector of the ball with the upward vertical is denoted by θ [IIT-2002]

(a) Express the total normal reaction force exerted by the sphere on the ball as a function of angle θ.

(b) Let NA and NB denote the magnitudes of the normal reaction forces on the ball exerted by the sphere A and B, respectively. Sketch the variations of NA and NB as functions of cos θ in the range 0 ≤ θ ≤ π by drawing two separate graphs in your answer book, taking cos θ on the horizontal axes.

Sol. The ball is moving in a circular motion. The necessary centripetal force is provided by

(mg cos θ – N). Therefore

V mgsinθ

mgcosθ mg

R θ θ

C d/2 NA

D A

B

mg cos θ – NA =

+

2dR

mv2 …(i)

According to energy conservation

21 mv2 = mg

+

2dR (1 – cos θ) …(ii)

From (i) and (ii) NA = mg (3 cos θ – 2) …(iii) The above equation shows that as θ increases NA

decreases. At a particular value of θ, NA will become zero and the ball will lose contact with sphere A. This condition can be found by putting NA = 0 in eq. (iii) 0 = mg (3 cos θ – 2)

∴ θ = cos–1

32

The graph between NA and cos θ From eq (iii) when θ = 0, NA = mg. 2mg

NA

mg

cos θ 1

2mg

cos θ–1

5mg

NB

When θ = cos–1

32 ; NA = 0

The graph is a straight line as shown.

When θ > cos–1

32

NB – (– mg cos θ) =

2dR

mv2

+

⇒ NB + mg cos θ =

+

2dR

mv2 …(iv)

Using energy conservation

2mv21 = mg

θ

+−

+ cos

2dR

2dR

+

2dR

mv2 = 2 mg [1 – cos θ] …(v)

From (iv) and (v) we get NB + mg cos θ = 2 mg – 2mg cos θ NB – mg (2 – 3 cos θ)

When cos θ = 32 , NB = 0

When cos θ = – 1, NB = 5 mg Therefore the graph is as shown.

2. A cylindrical block of length 0.4 m and area of cross section 0.04 m2 is placed coaxially on a a thin metal disc of mass 0.4 Kg and the same cross section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the materical of all cylinder is 10 watt/kg. K, how long will it take for the temperature of the disc to increases to 350 K? Assume, for purpose of calculation, the thermal conductivity of the disc to be very high and

KNOW IIT-JEE By Previous Exam Questions


the system to be thermally insulated except for the upper face of the cylinder. [IIT-1992]

Sol. Initially the temperature at the open end is 400 K and the temperature at the metal disc-cylinder interface is 300 K.

Metal disc Insulation

Cylinder

H

Openend

θ

As heat passes through the cylinder and reaches the

metal disc, the temperature of metal disc rises. Since the conductivity (thermal) of metal disc is very high so temperature of the whole disc will rise and along with that the temperature of the other end of the cylinder (metal disc-cylinder interface) also rises simultaneously. Let at any instant of time, the temperature of the metal disc-cylinder interface is θ. At this instant the rate of heat crossing the cylinder.

dtdQ =

l

)–400(KA θ ...(i)

where K = thermal conductivity A = area of cross section of cylinder l = length of cylinder The same amount of heat is received by the metal

disc. Therefore

dtdQ = mc

dtdθ ...(ii)

m = mass of disc c = specified heat of metal disc From (i) and (ii)

mcdtdθ =

l

)–400(KA θ

θ−θ

400d ×

KAmcl = dt

On integrating At t = 0, θ = 300

∫t

0dt =

KAmcl

∫ θ−θ350

300 )400(d At t = t, θ = 350

⇒ t = KAmcl− [log (400 – θ) 350

300]

= 04.010

303.24.06004.0×

×××− log10 300400350400

−−

= 166.38 sec.

3. A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic fields E and B are present. E and B are parallel to each other. At time t = 0 the velocity v0 of the particle is perpendicular to E. (Assume that its speed is always <<c, the speed of light in vaccum.) Find the velocity v of the particle at time t. You must express your answer in terms of t, q, m, the vectors v0, E and B and their magnitudes v0, E and B. [IIT-1998]

Sol. Because the forces due to parallel electric and magnetic fields on a charged particle moving perpendicular to the fields will be at right angles to each other (electric force being along the direction of →E while magnetic force perpendicular to the plane containing v and B ) so magnetic force will not affect the motion of charged particle in the direction of electric field and vice-versa. So the problem is equivalent to superposition of two independent motions as shown in the adjoining figures.

Y

X

Zv0 Fm

k

j

i

B E Fe

So for motion of the particle under electric field

alone,

ay = mqE i.e.,

dtdvy =

mqE

or ∫y

0ydv = ∫

y

0 mqE dt i.e., vy =

mqE t ...(1)

While at the same instant, the charged particle under the action of magnetic field will describe a circle in the x-z plane with

r = qB

mv0 i.e., ω = r

v0 = mqB

Xv0

v0cos θ

v0

θ

θ

Z

v0 sin θ

So angular position of the particle at time t in the x-z

plane will be given by

θ = ωt = mqB t

and therefore in accordance with figure

vx = v0 cosθ = v0 cos ωt = v0 cos

t

mqB ...(2)

and vz = v0 sinθ = v0 sin ωt = v0sin

t

mqB ...(3)

So in the light of equations (1), (2) and (3), we get

→v = i vx + j vy + k vz

= i v0 cos

t

mqB + j

tmqE + v0 sin

tmqB

k


But because here, i = 0

0

vv→

, j = EE→

= BB→

and k = BvBv

0

0 ×

So, →0v =

→

0

0

vv

v0 cos

mqBt +

→

EE

mqE t

+ BvBv

0

0 × v0sin

mqBt

4. A long solenoid of radius 'a' and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d(d<<R) and length L. A variable current i = i0 sin ωt flows through the coil. If the resitivity of the material of cylindrical shell is ρ, find the induced current in the shell. [IIT-2005]

Sol. The magnetic field in the solenoid is given by B = µ0ni

a

L

d R

⇒ B = µ0 n i0 sin ωt [Q i = i0 sin ωt given] The magnetic flux linked with the solenoid

φ = →B .

→A = BA cos 90º

= (µ0 n i0 sin ωt) (πa2) ∴ The rate of change of magnetic flux through the

solenoid

dtdφ = π µ0 n a2 i0 ω cos ωt

The same rate of change of flux is linked with the cylindrical shell. By the principle of electromagetic induction, the induced emf produced in the cylinderical shell is

ITOP VIEW

e = –dtdφ = – πµ0 n a2 i0 ω cos ωt ...(i)

The resistance offered by the cylindrical shell to the flow of induced current I will be

R = ρAl

Here, l = 2πR,

A = L × d

∴ R = ρLd

R2π ...(ii)

The induced current I will be

I = R

|e| = R2

Ld]tcosinaµ[ 02

0

π×ρ×ωωπ

I = R2

Ldtcosinaµ 02

0

ρ×ωωπ

5. A beam of light has three wavelength 4144 Å, 4972 Å and 6216 Å with a total intensity of 3.6 × 10–3 Wm–2 equally distributed amongst the three wavelengths. The beam falls normally on an area 1.0 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photo electrons liberated in two seconds. [IIT-1989]

Sol. Work function = 2.3 eV = 2.3 × 1.6 × 10–19J = 3.68 × 10–19 J Energy of a photon of wave length (4144 Å)

= λhc = 10

834

104144103106.6

−

−

××××

= 4.78 × 10–3 × 10–16 = 4.78 × 10–19 J Energy of a photon of wave length 4972 Å

= 10

834

104972103106.6

−

−

×××× = 3.98 × 10–19J

Energy of a photon of wavelength 6216 Å

= 10

834

106216103106.6

−

−

×××× = 3.18 × 10–19 J

This means that light of wavelength 4144 Å and 4972Å are capable of ejection of electrons from the metal surface

The energy incident on 1 cm2 area of metal per second = 3.6 × 10–3 × 10–4 = 3.6 × 10–7 J

The energy/s of wavelength 4972 Å = 1.2 × 10–7 J No of photon incident of wavelength 4972 Å

E = λhcn ∴ n =

hcEλ

⇒ n = 834

107

103106.6104972102.1

××××××

−

−−

= 301.33 × 109

= 3.01 × 1011 Similarly number of photon incident of wavelength

4144 Å.

n = 834

107

103106.6104144102.1

××××××

−

−−

= 2.51 × 1011

⇒ Total number of photons capable of ejection of electrons per second

= 3.01 × 1011 + 2.51 × 1011 = 5.52 × 1011 ∴ Total number of photoelectrons ejected in two

seconds ≈ 11 × 1011.


CHEMISTRY

6. From the following data, form the reaction between A and B. [IIT-1994]

Initial rate (mol L–1s–1) [A]

mol L–1 [B]

mol L–1 300 K 320 K 2.5 ×10–4 3.0 ×10–5 5.0 ×10–4 2.0 × 10–3

5.0 × 10–4 6.0 × 10–4 4.0 × 10–3 – 1.0 × 10–3 6.0 × 10–5 1.6 × 10–2 –

Calculate (a) the order of reaction with respect to A and with

respect to B, (b) the rate constant at 300 K, (c) the energy of activation, (d) the pre exponential factor.

Sol. Rate of reaction = k[A]l [B]m where l and m are the order of reaction with respect

to A and B respectively. From the given data, we obtain following expressions :

5.0 × 10–4 = k[2.5 × 10–4]l [3.0 × 10–5]m ...(i) 4.0 × 10–3 = k[5.0 × 10–4] l [6.0 × 10–5]m ...(ii) 1.6 × 10–2 = k[1.0 × 10–3]l [6.0 × 10–5]m ..(iii) From eq. (ii) and eq. (iii), we get

2

3

106.1100.4

−

−

×× =

l

××

−

−

3

4

100.1100.5

or 0.25 = (0.5)l or (0.5)2 = (0.5) l or l = 2 From eq. (i) and eq. (ii), we get

3

4

100.1100.5

−

−

×× =

m

5

52

4

4

100.6100.3

100.5105.2

××

××

−

−

−

−

or 81 =

41 ×

m

21

or 21 =

m

21

or m = 1 (b) At T1 = 300 K,

k1 = 12 ]B[]A[reactionofRate =

]100.3[]105.2[100.5

524

4

−−

−

×××

= 2.67 × 108 L2 mol–2 s–1 (c) At T2 = 320 K,

k2 = 12 ]B[]A[reactionofRate

= ]100.3[]105.2[

100.2524

3

−−

−

×××

= 1.067 × 109 L2 mol–2 s–1

We know, 2.303 log 1

2

kk =

−

21

12a

TTTT

RE

or 2.303 log 8

9

1067.210067.1

×× =

×−

300320300320

314.8Ea

or 2.303 × 0.6017 =

×30032020

314.8Ea

or Ea = 20

300320314.86017.0303.2 ××××

= 55.3 kJ mol–1 (d) According to Arrhenius equation, k = RT/EaAe−

or 2.303 log k = 2.303 log A – RTEa

At 300 K,

2.303 log (2.67 × 108) = 2.303 log A – 300314.8

103.55 3

××

or 2.303 × 8.4265 = 2.303 log A – 22.17

or logA = 303.2

17.224062.19 + = 303.25762.41 = 18.0531

A = Antilog 18.0531 = 1.13 × 1018 s–1

7. 1 g of a mixture containing equal number of moles of carbonates of two alkali metals, required 44.4 ml of 0.5 N HCl for complete reaction. The atomic weight of one metal is 7, find the atomic weight of other metal. Also calculate the amount of sulphate formed on quantitative conversion of 1.0 g of the mixture in two sulphates. [IIT-1972]

Sol. Let, Mass of one alkali metal carbonate M2CO3 = xg Then, mass of other alkali metal carbonate M2ĆO3

= (1 – x)g Step 1.

Equivalent mass of M2CO3 = 2

massMolecular = 2

74

= 37 (at mass of M = 7)

Meq. of M2CO3 = 37x ×1000

Moles of M2CO3 = 74x

Step 2

Equivalent mass of M2ĆO3 = 2

massMolecular

= 2

60m2 +

( m = atomic mass of M´)

Meq. of M2ĆO3 = 60m2

)x1(2+− × 1000

Moles of M2ĆO3 = 60m2

x1+

−

Step 3. Meq. of HCl = NHCl × VHCl = 0.5 × 44.4 Step 4. According to the question, Moles of M2CO3 = Moles of M2ĆO3

or 74x =

60m2x1

+− ...(i)


And Meq. of M2CO3 + Meq . of M2ĆO3 = Meq. of HCl

or 37x × 1000 +

60m2)x1(2

+− × 1000

= 0.5 × 44.4 ...(ii) Solving eq. (i) and (ii), we get m = 23 and x = 0.41 ∴ Mass of M2CO3 = x = 0.41 g and Mass of M2ĆO3 = 1 – x = 0.59 g Step 5.

Equivalent mass of M2SO4 = 2

massMolecular

= 2

110 = 55

Meq. of M2SO4 = 55

W42SOM × 1000

But, Meq. of M2SO4 = Meq. of M2CO3

∴ 55

W42SOM × 1000 =

3741.0 × 1000

or 4SOM2W = 0.6095 g

Step 6.

Equivalent mass of M2´SO4 = 2

massMolecular

= 2

142 = 71

Meq. of M2´SO4 = 71

W42 SO´M × 1000

But, Meq. of M2´SO4 = Meq. of M2ĆO3

∴ 71

W42 SO´M × 1000 =

10659.02× × 1000

or 42 SO´MW = 0.7904 g

∴ Total mass of sulphates = 42SOMW +

42 SO´MW = 0.6095 + 0.7904 = 1.3999 g

8. 0.9 g of a solid organic compound (molecular mass 90), containing carbon, hydrogen and oxygen, was heated with oxygen corresponding to a volume of 224 ml at STP. After combustion the total volume of the gases was 560 ml at STP. On treatment with potassium hydroxide, the volume decreased to 112 ml. Determine the molecular formula of the compound. [IIT-1972]

Sol. Given that, Mass of solid organic compound = 0.9 g Molecular mass of organic compound = 90 ∴ No. of moles of organic compound available

= 90

9.0 = 0.01

Volume of O2 taken = 224 ml Volume of O2 used = 224 – 112 = 112 ml 22400 ml O2 = 1 mol.

∴ 112 ml O2 = 22400112 = 0.005 mol

∴ At STP, no. of moles of O2 used = 22400112

= 0.005 mol Volume of CO2 obtained = 560 – 112 = 448 ml

∴ At STP, no. of moles of CO2 used = 22400

448

= 0.02 mol 0.01 mol organic compound yields = 0.02 mol CO2. ∴ 1 mol organic compound yields = 2 mol CO2 or 2 mol C ∴ The molecular formula of organic compound is

C2HyOz. The reaction is :

C2HyOz + 21 O2 → 2CO2 +

2y H2O

Equating no. of oxygen atoms,

z + 1 = 4 + 2y

or z = 3 + 2y

Molecular mass of C2HyOz = 2 × 12 + y × 1 + z × 16

Hence, 2 × 12 + y × 1 +

+

2y3 × 16 = 90

or y = 2

and z = 3 + 2y = 4

∴ The molecular formula of organic compound is C2H2O4.

9. (a) Write the intermediate steps for each of the

following reactions. (i) C6H5CHOHC ≡ CH →

+OH3 C6H5CH=CHCHO

(ii)

OH

H+

O CH3

(b) Show the steps to carry out the following transformations :

(i) Ethylbenzene → benzene (ii) Ethylbenzene → 2-phenylpropionic acid [IIT-1998] Sol. (a) (i)

C6H5CH(OH)C ≡ CH H+ C6H5CH – C ≡ CH

OH2+ –H2O

C6H5CH = C = CH C6H5CH – C ≡ CH ⊕ ⊕

OH–

C6H5CH = C = CHOH C6H5CH = CHCHO Tautomerism


(ii)

OH H+ CH3

OH

CH3CH3 O O

H⊕

⊕

(b) (i)

C2H5

alk. KMnO4

Ethyl benzene

COOH

NaOH

Benzoic acid

–H2O

COONa

Sodium benzoate

NaOH

+CaO

Benzene

+ Na2CaO3

(ii)

CH2CH3

Br2/hv

CHBrCH3

Mg

Ether

CH3CH – MgBr

CO2

CH3CHCOOMgBrCH3CHCOOH

H2O/H+

Mg + Br

OH 2-Phenylpropionic acid

10. Interpret the non-linear shape of H2S molecule and non-planar shape of PCl3 using valence shell electron pair repulsion (VSEPR) theory. (Atomic numbers : H = 1, P = 15, S = 16, Cl = 17.) [IIT-1998]

Sol. In H2S, no. of hybrid orbitals = 21 (6 + 2 – 0 + 0) = 4

Hence here sulphur is sp3 hybridised, so 16S = 1s2, 2s22p6,

44 344 21ionhybridisatsp

1z

1y

2x

2

3

p3p3p3s3

or

S

H HH H

S

Due to repulsion between lp - lp; the geometry of

H2S is distorted from tetrahedral to V-shape.

In PCl3, no. of hybrid orbitals = 21 [5 + 3 – 0 + 0] = 4

Hence, here P shows sp3-hybridisation 15P = 1s2, 2s22p6,

44 344 21ionhybridisatsp

1z

1y

1x

2

3

p3p3p3s3

P

or

P

Cl Cl

ClClCl

Cl

Thus due to repulsion between lp – bp, geometry is

distorted from tetrahedral to pyramidal.T

MATHEMATICS

11. 7 relatives of a man comprises 4 ladies and 3 gentlemen; his wife has also 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives ? [IIT-1985]

Sol. The possible cases are : Case I : A man invites 3 ladies and women invites 3

gentlemen ⇒ 4C3.4C3 = 16 Case II : A man invites (2 ladies, 1 gentleman) and

women invites (2 gentlemen, 1 lady) ⇒ (4C2.3C1).(3C1.4C2) = 324 Case III : A man invites (1 lady, 2 gentlemen) and

women invites (2 ladies, 1 gentleman) ⇒ (4C1.3C2).(3C2.4C1) = 144 Case IV : A man invites (3 gentlemen) and women

invites (3 ladies) ⇒ 3C3.3C3 = 1 ∴ Total number of ways = 16 + 324 + 144 + 1 = 485 12. Let n be a positive integer and (1 + x + x2)n = a0 + a1x + ..... + a2nx2n Show that a0

2 – a12 + ...... + a2n

2 = an. [IIT-1994] Sol. (1 + x + x2)n = a0 + a1x + .... + a2nx2n ...(1) Replacing x by –1/x. we obtain

n

2x1

x11

+− = a0 –

xa1 + 2

2

xa – 3

3

xa

+...+ n2n2

xa ..(2)

Now, a02 – a1

2 + a22 – a3

2 + ... + a2n2 = coefficient of

the term independent of x in [a0 + a1x + a2x2 + ... + a2nx2n]

+−+− n2

n2221

0 xa...

xa

xaa

= coefficient of the term independent of x in

(1 + x + x2)nn

2x1

x11

+−

But, R.H.S. = (1 + x + x2)nn

2x1

x11

+−

= n2

n2n2

x)1xx()xx1( +−++


= n2

n222

x]x)1x[( −+

= n2

n242

x)xxx21( −++

= n2

n42

x)xx21( ++

Thus, a02 – a1

2 + a22 – a3

2 + ... a2n2

= coefficient of the term independent of x in

n2x1 (1 + x2 + x4)n

= coefficient of x2n in (1 + x2 + x4)n = coefficient of tn in (1 + t + t2)n = an 13. Solve for x the following equation : log(2x + 3) (6x2 + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9) [IIT-1987] Sol. log(2x + 3) (6x2 + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9) ⇒ log(2x + 3)(2x + 3) . (3x + 7) = 4 – log(3x + 7)(2x + 3)2 ⇒ 1 + log(2x + 3) )(3x + 7) = 4 – 2log(3x + 7) (2x + 3) Put log(2x + 3) (3x + 7) = y

⇒ y + y2 – 3 = 0 ⇒ y2 – 3y + 2 = 0

⇒ (y – 1) (y – 2) = 0 ⇒ y = 1 or y = 2 ⇒ log(2x + 3) (3x + 7) = 1 or log(2x + 3)(3x + 7) = 2 ⇒ 3x + 7 = 2x + 3 or (3x + 7) = (2x + 3)2 ⇒ x = – 4 or 3x + 7 = 4x2 + 12x + 9 4x2 + 9x + 2 = 0 4x2 + 8x + x + 2 = 0 (4x + 1) (x + 2) = 0 x = – 2, –1/4 ∴ x = – 2, –4, –1/4 But, log exists only when, 6x2 + 23x + 21 > 0, 4x2 + 12x + 9 > 0, 2x + 3 > 0 and 3x + 7 > 0 ⇒ x > –3/2 ∴ x = –1/4 is the only solution. 14. Let f[(x + y)/2] = f(x) + f(y) / 2 for all real x and y,

if f´(0) exists and equals –1 and f(0) = 1, find f(2). [ΙΙΤ−1992]

Sol. f

+

2yx =

2)y(f)x(f + ∀ x, y ∈ R (given)

Putting y = 0, we get

f

2x =

2)0(f)x(f + =

21 [1 + f(x)] [Q f(0) = 1]

⇒ 2f(x/2) = f(x) + 1 ⇒ f(x) = 2f(x/2) – 1 ∀ x, y ∈ R ...(1) Since f´(0) = –1, we get

⇒ h

)0(f)h0(flim0h

−+→

= – 1 ⇒ h

1)h(flim0h

−→

Now, let x ∈ R then applying formula of differentiability.

f´(x) = h

)x(f)hx(flim0h

−+→

= h

)x(f2

h2x2flim

0h

−

+

→

= h

)x(f2

)h2(f)x2(f

lim0h

−+

→

= h

)x(f12h2f21

2x2f2

21

lim0h

−

−

+−

→

[using equation (1)]

= h

)x(f1)h(f21–)x(f221

lim0h

−−+

→

= h

1)h(flim0h

−→

= –1

Therefore f´(x) = – ∀ x ∈ R ⇒ ∫ )x´(f dx = ∫ −1 dx

⇒ f(x) = – x + k where k is a constant. But f(0) = 1, therefore f(0) = – 0 + k ⇒ 1 = k ⇒ f(x) = 1 – x ∀ x ∈ R ⇒ f(2) = – 1 15. If (a + bx)ey/x = x, then prove that

x32

2

dxyd =

2

ydxdyx

− [IIT-1983]

Sol. (a + bx).ey/x = x ...(1) Differentiating both sides, we get

(a + bx).ey/x.

−

2x

ydxdyx

+ bey/x = 1

⇒ x. 2x

ydxdyx

−

+ beb/x = 1, (using (1))

or dxdy –

xy + bey/x = 1,

Again differentiation both sides,

2

2

dxyd – 2x

ydxdyx −

+ bey/x.

−

2x

ydxdyx

= 0 ..(2)

from (2), 2

2

dxyd – 2x

ydxdyx −

−

xy

dxdy = 0

⇒ x32

2

dxyd =

2

ydxdyx

−


1. Four infinite thin current carrying sheets are placed in YZ plane. The 2D view of the arrangement is as shown in fig. Direction of current has also been shown in the figure. The linear current density. i.e. current per unit width in the four sheets are I, 2I, 3I and 4I, respectively.

I Y

X

II III IV

a a aa The magnetic field as a function of x is best

represented by

(A)

B

X µ0I

a

2µ0I5µ0I

2a 3a 4a 5a

(B)

B

X µ0I

a

3µ0I

–µ0I 2a 3a 4a 5a

(C)

B

X +µ0I

a

4µ0I

–µ0I 2a 3a 4a 5a

(D)

B

X +µ0I

a

2µ0I

–µ0I2a 3a 4a 5a

2. Match the column

Column – I Column – II

(A) a charge particle is (P) Velocity of the moving in uniform particle may be electric and magnetic constant fields in gravity free space (B) a charge particle is (Q) Path of the particle moving in uniform may be straight line electric, magnetic and gravitational fields (C) a charge particle is (R) Path of the particle moving in uniform may be circular magnetic and gravitational fields (where electric field is zero) (D) A charge particle is (S) Path of the particle moving in only may be helical uniform electric field (T) None

3. Magnetic flux in a circular coil of resistance 10Ω changes with time as shown in fig. Cross indicates a direction perpendicular to paper inwards.

Match the following :

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Dev Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions given in same issue

Set #12


φ(Magnetic flux)

10

2 t(s)–10

68 10 14 16

× × × × × ×× × × × × ×× × × × × ×× × × × × × × × × × × × × × × × × ×

Column – I Column – II

(A) At 1s, induced current is (P) Clockwise (B) At 5s, induced current is (Q) Anticlockwise (C) At 9s, induced current is (R) Zero (D) At 15s, induced current is (S) 2A (T) None 4. A conducting rod of length l is moved at constant

velocity v0 on two parallel, conducting, smooth, fixed rails, which are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct ?

⊗v0 R ⊗B

(A) The thermal power dissipated in the resistor is

equal to the rate of work done by an external person pulling the rod

(B) If applied external force is doubled, then a part of the external power increases the velocity of the rod

(C) Lenz’s law is not satisfied if the rod is accelerated by an external force

(D) If resistance R is doubled, then power required to maintain the constant velocity v0 becomes half

5. The x-z plane separates two media A & B of

refractive indices µ1 = 1.5 & µ2 = 2. A ray of light travels from A to B. Its directions in the two media

are given by unit vectors ,jbia1

∧∧→+=µ .jbic2

∧∧→+=µ

Then

(A) 34

ca

= (B) 43

ca

=

(C) 34

db

= (D) 43

db

=

6. Two converging lenses of the same focal length f are separated by distance 2f. The axis of the second lens is inclined at angle º60=θ with respect to the axis of the first lens. A parallel paraxial beam of light is incident from left side of the lens. Then

2f

60º

(A) Final image after all possible refraction will

formed at optical centre of first lens (B) Final image after all possible refraction will

formed at optical centre of second lens (C) Final image after all possible refraction will

formed at distance f from second lens

(D) Final image after all possible refraction will formed at distance f from first lens

7. If Cv for an ideal gas is given by Cv = (3 + 2T)R, where T is absolute temperature of gas, then the equation of adiabatic process for this gas is

(A) VT2 = constant

(B) VT3e-2T = C

(C) VT2e2T = constant

(D) VT3e2T = constant

8. The pressure of one mole of ideal gas varies according to the law P = P0 – αV2 where P0 & α are positive constant constants. The highest temperature that gas may attain -

(A) 2/1

00

3P

R3P2

α

(B) 2/1

00

3P

R2P3

α

(C) 2/1

00

3P

RP

α

(D) 2/1

00 PRP

α


1.[C] The given circuit as an R-L-C series circuit when frequency of the source varies the impedance of the R-L-C series circuit varies and correspondingly the current in the circuit get varied

Impedance variation and current variation are shown in figure.

R = Z minimum

f1 f=fr f2 f

∆f = f2-f1

I

I0

f1 f=fr f2 f

∆f = f2-f1

2I

I 0=

At frequency f1 XC > XL Power factor – leading nature of circuit is capacitance

At frequency f2 XL > XC Power factor – leading nature of circuit is inducting

At frequency f1 and frequency f2 impedance Z= R.2

Because of the fact –

As )i........(RV

ZhmVI0 ==

ZV

2I0 ==±

⇒ ZV

22/V

=

⇒

2Rand

ZV

2RV

=

f1 < f < f2

2.[A] 12 fff −=∆ = B and width of R-L-C series circuit

L/R.

21π

=

3[C] At frequency f1 current

.Amp2102

2010200.

21

RV.

21

2I0 ====±

Watt less current

φ=µ sinII

.Amp10

21.210I ==µ

º45As =φ because 2

12R

RZ/Rcos ===φ

4[A] At frequency current .Amp2010200

RV

ZmVI ====

Potential difference across capacitor

VC = IC.XC = 1.XC = 20.XC

Charge on capacitor QC = C.VC = C. 20XC

= C. (2v) C

1ω

π

=π

=π

=ω

=51

)50(220

f22020

Coulomb)5( 1−π=

cb)5( 1−π=

5.[C] Longest wavelength

m59.0

60058.0350

fvv s

max =××

=+

=λ

6.[B] fvv

vfs

max −=

Hz607

58.0350350

=×−

=

7.[A] 345.5/346.0 × 600 = 599 Hz

8.[A] 2

2

22

2

dtyd

v1

dxyd&

dtdy

v1

dxdy

+=−=

Solution Physics Challenging Problems

Set # 11

8 Questions were Publ ished in March Issue


1.[C] Magnetic field due to infinite current carrying

sheet is given by ,2

JB 0µ

= where J is linear current

density.

2J0µ

II III

(a)

2J0µ

(b)

2J0µ

2J0µ

Fig. (a) and (b) represent the direction of magnetic field due to current carrying sheets. For x < a,

2)J4(

2)J3(

2)J2(J

2J

B 0000ttanresul

µ+

µ−

µ−

µ=

For a < x < 2a,

J2

)J4(2

)J3(2

)J2(2

JB 0

0000ttanresul µ−=

µ+

µ−

µ−

µ=

For 2a < x < 3a,

02

)J4(2

)J3(2

)J2(2

JB 0000

ttanresul =µ

−µ

−µ

+µ

=

So, the required curve is B

XO a 2a 3a 4a 5a

2. A → P,Q,S B → P,Q,R,S C → P,Q,R,S D → Q

(A) Velocity of the particle may be constant, if forces of electric and magnetic fields balance each other. Then, path of particle will be straight line. Also, path of particle may be helical if magnetic and electric fields are in same direction. But path of particle cannot be circular. Path can be circular if only magnetic field is present, or if some other forces is present which can cancel the effect of electric field.

(B) Here, all the possibilities are possible depending upon the combinations of the three fields.

(C) This situation is similar to part (i) (D) In a uniform electric field, path can be only

3. A → Q B → R C → P D → Q (A) At t = 1s, flux is increasing in the inward

direction, hence induced e.m.f. will be in anticlockwise direction.

(B) At t = 5s, there is no change in flux, so induced e.m.f. is zero

(C) At t = 9s, flux is increasing in upward direction hence induced e.m.f. will be in clockwise direction.

(D) At t = 15s, flux is decreasing in upward direction, so induced e.m.f. will be in anticlockwise direction.

4.[A, B, D] Rate of work done by external agent is

de/dt = BIL.dx/dt = BILv and thermal power dissipated in resistor = eI = (BvL) I clearly both are equal, hence (A). If applied external force is doubled, the rod will experience a net force and hence acceleration. As a result velocity increase, hence (B). Since, I = e/R On doubling R, current and hence required power become half. Since, P = BILv Hence (D)

5.[A]

xz

n1sini = n2sinr

j

)j(2)j(5.1 21

∧→∧→×µ=×µ

]j)jdic[(2j)jbia(5.1∧∧∧∧∧∧

×+=×+

∧∧

= kc2ka5.1

34

5.120

ca

==

Solution Physics Challenging Problems

Set # 12

8 Questions Published in this Issue


6.[A]

I2

+ –

f

f cos60º

I1

f sin60ºf

x u = -f cos60º f = +f

º60cosf

1v1

f1

−−=

f2

v1

f1

+=

v1

f2

f1

=−

v = -f

º60cosxf

=

xº60cos

f=

x = 2f ∴ final image will formed at optical centre of first lens.

7.[C] Cv = (3 + 2T)R dQ = dU + PdV adiabatic process dQ = 0 0 = Rn (3 + 2T)dT + PdV

dVV

nRTdT)T23(Rn0 ++=

∫ ∫

+

=− dTT

T23V

dV

– log V = 3 logT + 2T + C – logV – logT3 = 2T + C log VT3 = 2T + C VT3 = e2T VT3e-2T = C

8.[A] 20 VPP α−=

PV = RT

20 VP

VRT

α−=

RV

RVP

T3

0 α−=

0dVdT

=

0RV3

RP 2

0 =α

−

α

=3P

V 0 Now put V in T.

• Saturn’s rings are made up of particles of ice, dust and rock. Some particles are as small as grains of sand while others are much larger than skyscrapers.

• Jupiter is larger than 1,000 Earths.

• The Great Red Spot on Jupiter is a hurricane-like storm system that was first detected in the early 1600’s.

• Comet Hale-Bopp is putting out approximately 250 tons of gas and dust per second. This is about 50 times more than most comets produce.

• The Sun looks 1600 times fainter from Pluto than it does from the Earth.

• There is a supermassive black hole right in the middle of the Milky Way galaxy that is 4 million times the mass of the Sun.

• Halley’s Comet appears about every 76 years.

• The orbits of most asteroids lie partially between the orbits of Mars and Jupiter.

• Asteroids and comets are believed to be ancient remnants of the formation of our Solar System (More than 4 billion years ago!).

• Comets are bodies of ice, rock and organic compounds that can be several miles in diameter.

• The most dangerous asteroids, those capable of causing major regional or global disasters, usually impact the Earth only once every 100,000 years on average.

• Some large asteroids even have their own moon.

• Near-Earth asteriods have orbits that cross the Earth’s orbit. These could potentially impact the Earth.

• There are over 20 million observable meteors per day.

• Only one or two meteorites per day reach the surface of Earth.

• The largest found meteorite was found in Hoba, Namibia. It weighed 60 tons.


1. A beam of length L, breadth b and thickness d when loaded by a weight Mg in the middle, a depression e is produced in it. By measuring this depression e, the value of Young's modulus of the material of beam can be calculated by using the expression

Y = edb4

LgM3

3

Following are the values of different physical quantities obtained in one set of observations on this experiment :

M = 1000 gms, L = 200 cm, b = 2.54 cm, d = 0.620 cm, e = 0.1764 cm. If M is measured by spring balance, L by metre scale,

b by vernier calipers, d by screw gauge and e by spherometer, then what will be the maximum possible percentage errors in Y ?

Sol. Given that Y = edb4

LgM3

3

Taking log on both sides of above equation, we get log Y = log M + log g + 3 log L – log 4 – log b – 3 log d – log e Differentiating above equations, we have :

YY∆ =

MM∆ + 3

LL∆ –

bb∆ – 3

dd∆ –

ee∆

In order to calculate maximum possible error, we shall convert negative sign into positive sign.

∴ YY∆ =

MM∆ + 3

LL∆ +

bb∆ + 3

dd∆ +

ee∆

Now, least counts of the different measuring instruments used in the experiment are as under :

Least count of spring balance = 5 gm i.e. ∆M = 5gm Least count of metre scale = 0.1 cm i.e. ∆L = 0.1 cm Least count of vernier callipers = 0.01 cm i.e. ∆b = 0.001 cm Least count of screw gauge = 0.001 cm i.e. ∆d = 0.001 cm Least count of spherometer = 0.005 cm i.e ∆e = 0.005 cm

∴ YY∆ =

10005 +

2001.03× +

54.201.0 +

62.0001.03× +

1764.0005.0

= 0.005 + 0.0015 + 0.00393 + 0.00484 + 0.02834 = 0.0436 or 4.36% Hence the maximum possible percentage error is

4.36%.

2. A small glass ball is released from rest from the top of a smooth incline plane of constant base b. find the angle of inclination of the plane for minimum time of motion of the glass ball.

A

BCb

θ

Sol. Let the angle of inclination be θ. If, the glass ball

reaches the bottom B of the inclined plane after a time, say t, the equation of motion along the plane is given as

AB = (VA)t + 21 (g sin θ)t2

A

B Cb

θ g

L = b sec θ

g sin θ

Since the glass ball is released from rest VA = 0,

hence AB = 21 (g sin θ)t2 ...(1)

Putting AB = (BC) sec θ = b sec θ, in equation (1), we obtain

t = θθcossing

b2 = θ2sing

b4

For t to be minimum, sin 2θ is maximum ∴ sin 2θ|max = 1

or, 2θ = 2π or, θ =

4π

3. Two particles, both of mass m, attract each other with

the force )r(Frr

= – rr 2α

where α is a positive constant. At a certain moment (t = 0), the distance between the particles is R, and their velocities are

−==

xv2vxvv

02

01

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumPHYSICSS


Assuming the two-particle system reaches a minimum of kinetic energy at a certain moment and at a certain finite distance between the particles (in the laboratory frame), find the distance between the particles at that moment and the value of that minimal kinetic energy.

Sol. For the sake of convenience, we will first solve the problem in the frame of the centre of mass. Then, we will transform the results into the laboratory frame. We will determine the velocity of the center of mass by :

cmvr

= xm2

mv2mv 00 − = – xv

21

0 ...(1)

This velocity remains constant since there are no external forces acting on the whole system. In the velocity of the centre of mass, the velocities of the particles are :

−=−=

=−=

xv23vvu

xv23vvu

0cm22

0cm11

rrr

rrr

...(2)

By definition of a centre of mass frame, the total momentum of the particles is zero. The kinetic energy in this system at t = 0 is :

K´ = 21 mu1

2 + 21 mu2

2 = 49 mv0

2 ...(3)

Therefore, the total energy in the center of mass frame at t = 0 is

E´ = K´ + U´ = 49 mv0

2 – Rα ...(4)

where we define R ≡ r(t = 0). Note that since the force is conservative, we have F

r = – ∇ u. The scalar

function is u = – α/r. The advantage of using the center of mass frame is

evident when one inspects the moment of arrival at a minimal distance, t0. At that moment, in this system, the two particles stop and reverse their directions. The kinetic energy, therefore, vanishes at t0 in the center of mass frame, or,

K´(t0) = 'minK = 0 ...(5)

Hence, E´(t0) = –minRα = E´ ...(6)

Plugging in the value of E´, we find :

Rmax = Rmv94

R420−α

α ...(7)

We now transform the centre of mass frame to the laboratory frame. Since Rmax is the relative distance between the two particles, it is unchanged by the transformation. Recall that distance is an invariant quantity of the transformations of displacement and / or rotation. Therefore,

Kmin – maxRα = E(0) = E =

21 mv0

2 + 21 m(2v0)2 –

Rα

....(8)

The kinetic energy in the laboratory frame is, therefore :

Kmin = 25 mv0

2 – Rα +

R4Rmv94 2

0

α−α

α

=

−

49

25 mv0

2 = 41 mv0

2 ...(9)

Another way of finding the minimal kinetic energy is by using the following formula :

K = K´ + 2cmMv

21 ...(10)

where K´ is the kinetic energy in the centre of mass frame, K is the energy in the laboratory frame, and M is the total mass of the system. In our case,

K = 0 + 2cmMv)m2(

21 = 2

0Mv41 ...(11)

Note : Generally, in transforming from system S to system S´ with relative velocity V

r, the kinetic

energy is transformed as :

K´ = K – M2

p2 +

2

MpV

2M

−

rr ..(12)

where M is the total mass and pr

is the total momentum in S. K´ is minimal in the center of mass

frame if we choose Vr

= cmvr

= Mpr

. We obtain :

K´ = K = M2

p2 ...(13)

4. A hot body is being cooled in air according to

Newton's law of cooling, the rate of fall of temperature being K times the difference of its temperature with respect to that of surroundings. Calculate the time after which the body will lose half the maximum heat it can lose. The time is to be counted from the instant t = 0.

Sol. According to Newton's law of cooling, we have

dtdθ = – K(θ – θ0)

where θ0 is the temperature of the surrounding and θ is the temperature of the body at time t. Suppose θ = θ1 at time t = 0.

Then, ∫θ

θ θ−θθ

1 0

d = –K ∫t

0dt or, log

01

0

θ−θθ−θ

= – Kt

or, θ – θ0 = (θ1 – θ0)e–Kt ...(1) The body continues to lose heat till its temperature

becomes equal to that of the surroundings. The loss of heat in this entire period is dQm = ms(θ1 – θ0).

This is the maximum heat the body can lose. If the body loses half this heat, the decreases in its temperature will be

ms2

dQm = 2

01 θ−θ


If the body loses this heat in time t1, the temperature at t1 will be

θ1 – 2

01 θ−θ =

201 θ+θ

Putting these values of time and temperature in (1) :

2

01 θ+θ – θ0 = (θ1 – θ0) 1Kte−

or, 1Kte− = 21 or t1 =

K2log

5. In a certain region surrounding the origin of the

coordinates, →B = 5 × 10–4

→k T and

→E = k V/m. A

proton enters the fields at the origin with an initial

velocity →

0v = 2.5 × 105 i m/s. Describe the proton's motion and give its position after three complete revolutions.

Sol. The z-component of the force →F is a constant

electrical force. It produces a constant acceleration along z-axis given as

Z = 21 at2 =

21

meE t2

BE z

x

y The other component of the force F is a magnetic

force which provides the necessary centripetal force to keep the proton in a circular path of radius r(say). The period of revolution of the proton

T = 0vr2π As Fcp =

rmv2

0 = ev0B

∴ v0 = m

eBr Hence, T = eB

m2π

Since the particle (proton) moves in circular path having a period of revolution T in x, y plane and moves along z-axis with a constant acceleration a = eE/m, the path of the proton is a helix.

After three revolutions, putting t = 3T, we obtain

z =

meE

21 (3T)2 =

meET

29 2

Putting T = 2πm/eB, we get

z = 2

2

eBEm18π = 246

272

)105(106.11066.15)7/22(18

−−

−

××××××

= 37m

Interesting Science Facts

• The Universe contains over 100 billion galaxies. • Wounds infested with maggots heal quickly and

without spread of gangrene or other infection. • More germs are transferred shaking hands than

kissing. • The longest glacier in Antarctica, the Almbert

glacier, is 250 miles long and 40 miles wide. • The fastest speed a falling raindrop can hit you is

18mph. • A salmon-rich, low cholesterol diet means that

Inuits rarely suffer from heart disease. • Inbreeding causes 3 out of every 10 Dalmation

dogs to suffer from hearing disability. • The world’s smallest winged insect, the Tanzanian

parasitic wasp, is smaller than the eye of a housefly.

• If the Sun were the size of a beach ball then Jupiter would be the size of a golf ball and the Earth would be as small as a pea.

• It would take over an hour for a heavy object to sink 6.7 miles down to the deepest part of the ocean.

• There are more living organisms on the skin of each human than there are humans on the surface of the earth.

• The grey whale migrates 12,500 miles from the Artic to Mexico and back every year.

• Quasars emit more energy than 100 giant galaxies. • Quasars are the most distant objects in the

Universe. • The Saturn V rocket which carried man to the

Moon develops power equivalent to fifty 747 jumbo jets.

• Koalas sleep an average of 22 hours a day, two hours more than the sloth.

• Light would take .13 seconds to travel around the Earth.

• Neutron stars are so dense that a teaspoonful would weigh more than all the people on Earth.

• One in every 2000 babies is born with a tooth. • Every hour the Universe expands by a billion

miles in all directions. • Somewhere in the flicker of a badly tuned TV set

is the background radiation from the Big Bang. • The temperature in Antarctica plummets as low

as -35 degrees Celsius. • Space debris travels through space at over 18,000

mph.


Calorimetry : The specific heat capacity of a material is the amount

of heat required to raise the temperature of 1 kg of it by 1 K. This leads to the relation

Q = ms θ where Q = heat supplied, m = mass, θ = rise in

temperature. The relative specific heat capacity of a material is the

ratio of its specific heat capacity to the specific heat capacity of water (4200 J kg–1K–1).

Heat capacity or thermal capacity of a body is the amount of heat required to raise its temperature by 1 K. [Unit : J K–1]

Thus heat capacity = Q/θ = ms

Also dtdθ =

ms1 ×

dtdQ

i.e., the rate of heating (or cooling) of a body depends inversely on its heat capacity.

The water equivalent of a body is that mass of water which has the same heat capacity as the body itself. [Unit : g or kg] This is given by

W = ws

sm×

where m = mass of body, s = specific heat capacity of the body, sw = specific heat capacity of water.

Principle of Calorimetry : The heat lost by one system = the heat gained by another system. Or, the net heat lost or gainsed by an isolated system is zero.

It system with masses m1, m2, ...., specific heat capacities s1, s2, ...., and initial temperatures θ1, θ2, .... are mixed and attain an equilibrium temperature θ then

θ = ´ms

msΣ

θΣ , for equal masses θ = ss

ΣθΣ

Newton's law of cooling : The rate of loss of heat from a body in an

environment of constant temperature is proportional to the difference between its temperature and that of the surroundings.

If θ = temperature of the surroundings then

– msdtdθ = C´(θ – θ0)

where C´ is a constant that depends on the nature and extent of the surface exposed. Simplifying

dtdθ = –C(θ – θ0) where C =

ms´C = constant

Kinetic theory of gases :

The pressure of an ideal gas is given by p = 31 µnC2

where µ = mass of each molecule, n = number of molecules per unit volume and C is the root square speed of molecules.

p = 31

ρC2 or pV = 31 mC2

where ρ is the density of the gas and m = mass of the gas.

Root Mean Square Speed of Molecules : This is defined as

C = N

C...CCC 2N

23

22

21 ++++

where N = total number of molecules. It can be obtained through these relations

C = ρp3 =

MRT3

Total Energy of an ideal gas (E) : This is equal to the sum of the kinetic energies of all

the molecules. It is assumed that the molecules do not have any potential energy. This follows from the assumption that these molecules do not exert any force on each other.

E = 21 mC2 =

23

Mm RT =

23 pV

Thus, the energy per unit mass of gas = 21 C2

The energy per unit volume = 23 p

The energy per mole = 23 pV =

23 RT

Calorimetry, K.T.G., Heat transfer

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Perfect gas equation : From the kinetic theory of gases the equation of an

ideal gas is pV = RT for a mole

and pV = Mm RT for any mass m

Avogadro number (N) and Boltzmann constant (k) : The number of entities in a mole of a substance is

called the Avogadro number. Its value is 6.023 × 1023 mol–1.

The value of the universal gas constant per molecular is called Boltzmann constant (k). Its value is 1.38 × 10–23 J K–1.

Degrees of Freedom : Principle of equipartition of energy : The number of ways in which energy may be stored

by a system is called its degrees of freedom. Principle of Equipartition of Energy : This

principle states that the total energy of a gas in thermal equilibrium is divided equally among its degrees of freedom and that the energy per degree of freedom is kT/2 where T is the temperature of the gas. For a monoatomic atom the number of degrees of freedom is 3, for a diatomic atom it is 5, for a polyatomic atom it is 6.

Hence the energy of a mole of a monoatomic gas is

µ = N

× kT

213 =

23 RT

Which is the same as that given by the kinetic theory. For a mole of diatomic gas µ

= N

× kT

215 =

25 RT

For a mole of polyatomic gas µ

= N

× kT

216 = 3RT

When the irrational degrees of freedom are also taken into account, the number of degrees of freedom

= 6n – 6 for non-linear molecules = 6n – 5 for linear molecules where n = number of atoms in a molecule. Kinetic Temperature : The kinetic temperature of a moving particle is the

temperature of an ideal gas in thermal equilibrium whose rms velocity equals the velocity of the given particle.

Maxwellian distribution of velocities : In a perfect gas all the molecules do not have the

same velocity, rather velocities are distributed among them. Maxwell enunciated a law of distribution of velocities among the molecules of a perfect gas. According to this law, the number of molecules with

velocities between c and c + dc per unit volume is

dn = 4πna3 2bce− c2 dc where

b = kT2m and a =

kT2mπ

and the number of molecules with the velocity c per unit volume is

nc = 4πna3 2bce− c2 The plot of nc and c is shown in the figure. The

velocity possessed by the maximum number of molecules is called the most probable velocity

α c Crms α = m/kT2 The mean velocity

c = πm/kT8 and vrms = πm/kT3

Conduction : The transfer of heat through solids occurs mainly by

conduction, in which each particle passes on thermal energy to the neighboring particle but does not move from its position. Very little conduction occurs in liquids and gases.

A

QQ

θ1 θ2

d Consider a slab of area A and thickness d, whose

opposite faces are at temperature θ1 and θ2 (θ1 > θ2). Let Q heat be conducted through the slab in time t.

Then Q = λA

θ−θd

21 t

where λ = thermal conductivity of the material. This has a fixed value for a particular material, being

large for good conductors (e.g., Cu, Ag) and low for insulators (e.g., glass, wood).

Heat Current : The quantity Q/t gives the heat flow per unit time, and is called the heat current.

In the steady state, the heat current must be the same across every cross-section. This is a very useful principle, and can be applied also to layers or slabs in contact.

tQ = – λA

dxdθ where the quantity

dxdθ =

d21 θ−θ is

called the temperature gradient.


Unit of λ : Different units are used, e.g., cal cm s ºC–1, cal m–1 s–1 ºC–1, jm´1 s–1 ºC–1. Convection : It is a process by which heat is conveyed by the

actual movement of particles. Particles closest to the source receive heat by conduction through the wall of the vessel. They rise up-wards and are replaced by colder particles from the sides. Thus, a circulation of particles is set up – hot particles constitute the upward current and cold particles, the side and downward current.

The transfer of heat by convection occurs only in fluids, and is the main mode of heat transfer in them. Most fluids are very poor conductors.

Radiation : Thermal Radiation : Thermal radiations are

electromagnetic waves of long wavelengths. Black Body : Bodies which absorb the whole of the

incident radiation and emit radiations of all wavelengths are called black bodies.

It is difficult to realize a perfect black body in practice. However, a cavity whose interior walls are dull black does behave like a black body.

Absorption : Every surface absorbs a part or all of the radiation falling on it. The degree of absorption depends on the nature and colour of the surface. Dull, black surfaces are the best absorbers. Polished, white surfaces absorb the least. The coefficient of absorption for a surface is

aλ = incidentradiationabsorbedradiation

The suffix λ denotes the wavelength of the radiation being considered, Clearly, aλ = 1 for a black body, for all values of λ.

Emission : Each surface emits radiation (radiates) continuously. The emissive power (eλ) is defined as the radiation emitted normally per second per unit solid angle per unit area, in the wave-length range λ and λ + dλ. Clearly, the emissive power of a black body (denoted by Eλ) is the maximum.

Kirchhoff's Law : According to this law, for the same conditions of temperature and wavelength, the ratio eλ /aλ is the same for all surfaces and is equal to Eλ. This simply means that good absorbers are good emitters. Hence, a black body is the best emitter, and a polished white body, the poorest emitter.

Prevost's Theory of Exchanges : All bodies emit radiations irrespective of their temperatures. They emit radiations to their environments and receive radiations from their environments simultaneously. In the equilibrium state the exchange between a body and the environment of energy continues in equal amounts.

Stefan-Boltzmann Law : If a black body at an absolute temperature T be surrounded by another black body at an absolute temperature T0, the rate of loss of radiant energy per unit area is

E = σ(T4 – T04)

where σ is a constant called Stefan constant and its value is 5.6697 × 10–8 W m–2 K–4

The total energy radiated by a black body at an absolute temperature T is given by

E = σT4 × surface area × time Note : Remember that rate of generation of heat by

electricity is given by H = I2 R or R

V2or VI Js–1 or W.

1. An earthenware vessel loses 1 g of water per second due to evaporation. The water equivalent of the vessel is 0.5 kg and the vessel contains 9.5 kg of water. Find the time required for the water in the vessel to cool to 28ºC from 30ºC. Neglect radiation losses. Latent heat of vaporization of water in this range of temperature is 540 cal g–1.

Sol. Here water at the surface is evaporated at the cost of the water in the vessel losing heat.

Heat lost by the water in the vessel = (9.5 + 0.5) × 1000 × (30 – 20) = 105 cal Let t be the required time in seconds. Heat gained by the water at the surface = (t × 10–3) × 540 × 103

(Q L = 540 cal g–1 = 540 × 103 cal kg–1)

∴ 105 = 540t or t = 185 s = 3 min 5s

2. 15 gm of nitrogen is enclosed in a vessel at temperature T = 300 K. Find the amount of heat required to double the root mean square velocity of these molecules.

Sol. The kinetic energy of each molecule with mass m is given by

21 m 2

rmsv = 23 kT ...(1)

If we want to increase the r.m.s. speed to η times, then the temperature has to be raised to T´. Then,

2rmsmv

21 =

23 kT´ or

21 mη2 2

rmsv = 23 kT´ ...(2)

From eqs. (1) and (2), T´ = η2T ...(3)

Solved Examples


The internal energy of n molecules at temperature T is given by

U = 25 nRT

Similarly, U´ = 25 nRT´

∴ Change in internal energy ∆U = 25 nR[T´ – T]

or ∆U = 25 nRT[η2 – 1]

= 25

Mm RT[η2 – 1]

= 25

2815 (8.31) (300) [4 – 1] = 104 J

3. 10 gm of oxygen at a pressure 3 × 105 N/m2 and

temperature 10ºC is heated at constant pressure and after heating it occupies a volume of 10 litres (a) find the amount of heat received by the gas and (b) the energy of thermal motion of gas molecules before heating.

Sol. (a) The states of the gas before and after heating are

PV1 = µM RT1 and PV2 =

µM RT2

Solving these equations for T2, we have

T2 = MR

PµV2 = )1031.8)(1010()103)(1010(32

33

53

×××××

−

−

= 1156 K

Now T2 – T1 = 1156 – 283 = 873 K The amount of heat received by the gas is given by

∆Q = µM CP(T2 – T1)

= 32

8731008.29)1010( 33 ××× −

= 7.9 × 103 J (b) The energy of the gas before heating

E1 = µM ×

2i × RT1

where i = number of degrees of freedom = 5 (for oxygen)

= 322

)283)(1031.8(5)1010( 33

×××× −−

= 1.8 × 103 J

4. A slab of stone of area 3600 sq cm and thickness 10 cm is exposed on the lower surface of steam 100ºC. A block of ice at 0ºC rests on upper surface of the slab. In one hour 4800 gm of ice is melted. Calculate the thermal conductivity of the stone.

Sol. The quantity of heat Q passing across the stone is given by

Q = d

t)TT(KA 21 −

Here A = 3600 sq. cm = 0.36 m2 d = 10 cm = 0.10 m, (T1 – T2) = 100 – 0 = 100ºC and

t = 1 hour = 3600 sec.

∴ Q = 10.0

360010036.0K ××× kilo-calories ...(1)

Now heat gained by the ice in one hour = mass of the ice × latent heat of ice = 4.8 × 80 kilo calories ...(2) From eqs. (1) and (2)

4.8 × 80 = 10.0

360010036.0K ×××

or K = 360010036.010.0808.4

××××

= 3 × 10–4 kilo cal m–1(ºC)–1s–1 5. A flat bottomed metal tank of water is dragged along

a horizontal floor at the rate of 20m/sec. The tank is of mass 20 kg and contains 1000 kg of water and all the heat produced in the dragging is conducted to the water through the bottom plate of the tank. If the bottom plate has an effective area of conduction 1 m2 and the thickness 5 cm and the temperature of water in the tank remains constant at 50ºC, calculate the temperature of the bottom surface of the tank, given the coefficient of friction between the tank and the floor is 0.343 and K for the material of the tank is 25 cal m–1 s–1 K–1.

Sol. Frictional force = µ m g = 0.343 × (1000 + 20) × 9.81 = 3432 N The rate of dragging, i.e., the distance travelled in

one second = 20 m. ∴ Work done per second = (3432 × 20) Nm/sec. This work done appears as heat at the bottom plate of

the tank. Hence

H = 18.4

203432× cal/sec

But H = d

)TT(KA 21 − (Q t = 1 sec)

Now 18.4

203432× = 05.0

)TT(125 21 −××

∴ T1 – T2 = 12518.405.0203432

×××× = 32.84

Temp. of bottom surface T1 = 50 + 32.84 = 82.84ºC


Atomic Structure :

According to Neil Bohr's hypothesis is the angular momentum of an electron is quantised.

mvr = n

π2h or L = n

π2h

2πr = nλ

vn = Znmr2h

π =

137

c × nz ms–1

rn =

π 22

2

mke4h

Zn 2

= 0.529Z

n 2Å where k =

041πε

fn =

hrke2

× n1 =

n1058.6 15× Hz

K.E. = 21

rZke 2

; P.E. = rke2− × Z; T.E. = –

r2ke2

× Z

T.E. = 2

2

nZ6.13− ev/atom where –13.6

= Ionisation energy

⇒ +T.E. = 2

.E.P+ = – K.E.

Note : If dielectric medium is present then εr has to be taken into consideration.

cv =

λ1 = v =

−

ε 22

21

320

24

n1

n1

ch8zme

= RZ2

− 2

221 n

1n1 =

hp =

hmv

n = ∞ n = 7 n = 6 n = 5 n = 4

n = 3

n = 2

n = 1

Kδ

Kγ

Kβ

Lγ

Lβ

Lα

Balmer (Visible)

Limiting line of Lyman series

Lyman Series (U.V. rays)

Paschen (I.R.)

Brackett (I.R.)

Pfund (I.R.)

–0.85 eV

–1.5 eV

–3.4 eV

–13.6 eV

The maximum number of electrons that can be accommodated in an orbit is 2n2.

X-rays : When fast moving electron strikes a hard metal,

X-rays are produced. When the number of electrons striking the target metal increases, the intensity of X-rays increases. When the accelerating voltage/kinetic energy of electron increases λmin decreases. X-rays have the following properties :

(a) Radiations of short wavelength (0.01 Å – 10Å); high pentrating power; having a speed of 3 × 108 m/s in vacuum.

λmin λ

Inte

nsity

Continuous spectrum (Varies & depends on accelerating voltage)

Kβ

Kα

Lγ Lβ

Lα

Characteristic spectrum (fixed for a target material)

(b) λmin = eVhc =

E.Khc =

V12400 Å

(c) λ1 = R(Z – b)2

− 2n

11

b = 1 for k-line transfer of electron

(d) Moseley law ν = a(z – b)

R = R0A1/3 where R0 = 1.2 × 10–15 m R = radius of nucleus of mass number A. * Nucleus density is of the order of 1017 kg/m3

Isomers are nuclides which have identical atomic number and mass number but differ in their energy states.

Nucleon

energybindingNuclear = Amc2∆

where ∆m = mass defect

= A

c]Mm)ZA(Zm[ 2np −−+

Atomic Structure, X-Ray & Radio Activity

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


* The binding energy per nucleon is small for small nuclei.

* For 2 < A < 20, there are well defined maxima which indicate that these nuclei are more stable.

* For 30 < A < 120 the average B.E./A is 8.5 MeV / nucleon with a peak value of 8.8 MeV for Iron.

* For A > 120, there is a gradual decreases in B.E./nucleon.

* More the B.E./A, more is the stability.

Radioactivity :

β particles are electrons emitted from the nucleus. (n → p + β)

(a) N = N0e–λt

(b) dtdN− = λN where

dtdN = activity level

(c) N = N0

n

21

= N0

2/1Tt

21

⇒ A = A0

n

21

where A = activity level

(d) T1/2 = λ693.0

(e) τ = λ1

(f) τ = 1.4 T1/2

(g) t = λ303.2 log10 N

N0 = λ303.2 log10 A

A0

= λ303.2 log

mm0

(h) If a radioactive element decays by simultaneous

emission of two particle then dtdN− = λ1N + λ2N

The following parameters remain conserved during a nuclear reaction

(a) linear momentum (b) Angular momentum (c) Number of nucleons (d) Charge (e) The energy released in a nuclear reaction X + P → Y + Z + Q

Q = [mx + mp) – (my + mz)]c2 = ∆m × c2 Q = ∆m × 931 MeV

(f) In a nuclear fusion reaction small nuclei fuse to give big nuclei whereas in a nuclear fusion reaction a big nuclei breaks down.

Thermal neutrons produce fission in fissile nuclei. Fast moving neutrons, when collide with atoms of comparable masses, transfer their kinetic energy to colliding particle and slow down.

According to Doppler's effect of light λλ∆ =

cv

Power, P = tE =

tnhν =

tnhcλ

η = putInputout

1. The energy of an excited hydrogen atom is –3.4 eV.

Calculate the angular momentum of the electron according to Bohr theory.

Sol. The energy of the electron in the nth orbit is

En = – 2n6.13 eV

Here, – 2n6.13 = –3.4

or n = 2

Angular momentum = π2

nh = 14.321063.62 34

××× −

= 2.11 × 10–34 Js.

2. The wavelength of the first member of the Balmer series in the hydrogen spectrum is 6563 Å. Calculate the wavelength of the first member of the Lyman series.

Sol. For the first member of the Balmer series

λ1 = R

− 22 3

121 =

36R5 ...(1)

For the first member of the Lyman series

´

1λ

= R

− 22 2

111 =

4R3 ...(2)

Dividing Eq. (1) by Eq. (2)

λλ´ =

33645

×× =

275

or λ´ = 275

λ = 275 × 6563 = 1215 Å

Solved Examples


3. Hydrogen atom in its ground state is excited by means of a monochromatic radiation of wavelength 970.6 Å. How many different wavelengths are possible in the resulting emission spectrum ? Find the longest wavelength amongst these.

Sol. Energy the radiation quantum

E = hv = λhc = 1910

834

106.1106.970103106.6

−−

−

××××××

= 12.75 eV Energy of the excited sate En = – 13.6 + 12.75 = – 0.85 eV

Now, we know that En = – 2n6.13

or n2 = –nE6.13 =

85.06.13

−− = 16

or n = 4 The number of possible transition in going to the

ground state and hence the number of different wavelengths in the spectrum will be six as shown in the figure.

n 4 3

2

1 The longest wavelength corresponds to minimum

energy difference, i.e., for the transition 4 → 3.

Now E3 = – 236.13 = – 1.51 eV

max

hcλ

= E4 – E3

or λmax = 19

834

106.1)85.051.1(103106.6

−

−

××−×××

= 18.75 × 10–7m = 18750 Å

4. X-rays are produced in an X-ray tube by electrons

accelerated through a potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.

Sol. Initial kinetic energy of the electron = 50.0 keV Energy of the photon produced in the first collision,

E1 = 50.0 – 25.0 = 25.0 keV Wavelength of this photon

λ1 = 1E

hc = 319

834

105.12106.1103106.6×××

×××−

−

= 0.99 × 10–10 m = 0.99 Å Kinetic energy of the electron after third collision = 0 Energy of the photon produced in the third collision , E3 = 12.5 – 0 = 12.5 keV This is same as E2. Therefore, wavelength of this

photon, λ3 = λ2 = 0.99 Å

5. In an experiment on two radioactive isotopes of an elements (which do not decay into each other), their mass ratio at a given instant was found to be 3. The rapidly decaying isotopes has larger mass and an activity of 1.0 µCi initially. The half lives of the two isotopes are known to be 12 hours and 16 hours. What would be the activity of each isotope and their mass ratio after two days ?

Sol. We have, after two days, i.e., 48 hours,

N1 = 4

01 2

1N

= 0

1N /16

N2 = 3

02 2

1N

= 0

2N /8

Mass ratio = 2

1

NN = 0

2

01

NN .

168 =

16283× =

23

Now, 01A = λ1

01N = 1.0 µCi

After two days,

A1 = λ1N1 = λ101N /16 = 0

1A /16 = (1/16)µCi

A2 = λ2N2 = λ202N /8

But 1

2

λλ =

2

1

TT =

1612 =

43

or λ2 = 43

λ1

A2 =

λ14

3 ×

0

1N31 ×

81

= 321

λ101N =

321 0

1A

= (1/32) µCi


Halogenation of Benzene : Benzene does not react with bromine or chlorine

unless a Lewis acid is present in the mixture. (As a consequence, benzene does not decolorize a solution of bromine in carbon tetrachloride.) When Lewis acids are present, however, benzene reacts readily with bromine or chlorine, and the reactions give bromobenzene and chlorobenzene, respectively, in good yields :

+ Cl2

FeCl3

25ºC

Cl + HCl

Chlorobenzene (90%)

+ Br2FeBr3

heat

Br+ HBr

Bromobenzene (75%) The Lewis acids most commonly used to effect

chlorination and bromination reactions are FeCl3, FeBr3, and AlCl3, all in the anhydrous form.

A mechanism for the reaction : Electrophillic Aromatic Bromination : Step 1

Br – Br : + FeBr3 → :Br – Br – FeBr3

→ :Br+ + :Br – FeBr3

–

–

+

Bromine combines with FeBr3 to form a complex that dissociates

to form a positive bromine ion and FeBr4–

Step 2

+ Br:

slow

HBr:

HBr:

HBr:

Arenium ion The positive bromine ion attacks benzene to

form an arenium ion

+

+

+ +

Step 3

HBr: + H – Br: + FeBr3

Br:

A proton is removed from the arenium ionto become bromobenzene

+

–:Br – FeBr3

The function of the Lewis acid can be seen in step 1. The ferric bromide reacts with bromine to produce a positive bromine ion, Br+ (and FeBr4

–). In step 2 this Br+ ion attacks the benzene ring to produce an arenium ion. Then, finally in step 3 a proton is removed from the arenium ion by FeBr4

–. This results in the formation of bromobenzene and hydrogen bromide the products of the reaction. At the same time this step regenerates the catalyst, FeBr3.

Nucleophilic Aromatic Substitution through an Elimination – Addition Mechanism : Benzyne Although aryl halides such as chlorobenzene and

bromobenzene do not react with most nucleophiles under ordinary circumstances, they do react under highly forcing conditions. Chlorobenzene can be converted to phenol by heating it with aqueous sodium hydroxide in a pressurized reactor at 350ºC .

+ NaOH

Cl

350ºCH2O

ONa

PhenolH3O+

OH

Bromobenzene reacts with the very powerful base,

NH2– , in liquid ammonia :

+ K :NH2

Br

+ – -33ºCNH3

+ KBr

NH2

Aniline These reactions take place through an elimination –

addition mechanism that involves the formation of an interesting intermediate called benzyne (or dehydrobenzene). We can illustrate this mechanism with the reaction of bromobenzene and amide ion.

In the first step, the amide ion initiates an elimination by abstracting one of the ortho protons because they are the most acidic. The negative charge that develops on the ortho carbon is stabilized by the inductive effect of the bromine. The anion then loses a bromide ion. This elimination produces the highly unstable, and thus highly reactive, benzyne. Benzyne then reacts with any available nucleophile (in this case, an amide ion) by a two-step addition reaction to produce aniline.

Organic Chemistry

Fundamentals

AROMATIC HYDROCARBON

KEY CONCEPT


The Benzyne Elimination – Addition Mechanism :

Br(–NH3)

Benzyne

(–Br–)

H

Br

(or dehydrobenzene)

:NH2

–

NH2

–

:NH3

NH2

H+ :NH2

:NH2 –

–

–

Addition

Elimination

Evidence for an elimination-addition mechanism : When 14C-labeled (C*) chlorobenzene is treated with

amide ion in liquid ammonia, the aniline that is produced has the label equally divided between the 1 and 2 positions.

Cl K+NH2

– * NH2–

NH3

NH2*

NH2

*(50%)

(50%)AdditionElimination

*

When the ortho derivative 1 is treated with sodium

amide, the only organic product obtained is m-(trifluoromethyl) aniline :

Cl

NaNH2

CF3

NH3 NH2

CF3

m-(Trifluoromethy)aniline1

This result can also be explained by an elimination –

addition mechanism. The first step produces the benzyne 2 :

Cl

NaNH2

CF3

NH3 + Cl–

CF3

1 2

This benzyne then adds an amide ion in the way that produces the more stable carbanion 3 rather than the less stable carbanion 4 :

CF3

:NH2

NH2

NH2

4

CF3

CF3

–

NH2 +:NH2

CF3

–

–

:NH3

2

3

Less stable carbanion

More stable carbanion (The negative charge is closer to the

electronegative trifluoromethyl group)

–

Carbanion 3 then accepts a proton from ammonia to

form m-(trifluoromethyl) aniline. Carbanion 3 is more stable than 4 because the carbon

atom bearing the negative charge is closer to the highly electronegative trifluoromethyl group. The trifluoromethyl group stabilizes the negative charge through its inductive effect. (Resonance effects are not important here because the sp2 orbital that contains the electron pair does not overlap with the π orbitals of the aromatic system.)

The Birch Reduction : Benzene can be reduced to 1, 4-cyclohexadiene by

treating it with an alkali metal (sodium, Lithium, or potassium) in a mixture of liquid ammonia and an alcohol.

Na

NH3, EtOHBenzene 1, 4-cyclohexadiene

A Mechanism for the Reaction : Brich Reduction : Na

etc.– –

Benzene Benzene radical anion

The first electron transfer produces a delocalized benzene radical anion.

EtOH

etc.

Cyclohexadienyl radical

H

H

H

H

Protonation produces a cyclohexadienyl radical(also a delocalized species)

Na

etc.

Cyclohexadienyl anion

H

HH

H

–

–

EtOHH

H

HH

1,4-CyclohexadieneTransfer of another electron leads to the formation of a delocalized

cyclohexadienyl anion, and protonation of this produces the 1,4-cyclohexadiene.


Solubility : The amount of a solute, dissolved in a given volume

of a solvent (water) in 100 mL or in 1L to form a saturated solution at a given temperature is termed as the solubility of the solute.

Solubility Product : Salts like AgI, AgCl, PbI2, BaSO4, PbSO4 etc. are

ordinarily considered insoluble but they do possess some solubility. These are sparingly soluble salts. A saturated solution of sparingly soluble salt contains a very small amount of the dissolved salt. It is assumed that whole of the dissolved salt is present in the form of ions, i.e., it is completely dissociated. Consider a sparingly soluble salt like AgCl, the following equilibrium occurs between the undissolved solid salt and the silver and chloride ions in the saturated solution.

AgCl (s) Ag+ (aq) + Cl–(aq) Applying the law of mass action to the ionic

equilibrium,

K = )]s(AgCl[

]Cl][Ag[ −+

or K[AgCl(s) = [Ag+] [Cl–] The concentration of solid AgCl in the solid state i.e.

[AgCl(s)] is constant at a particular temperature, no matter how much solid is present in contact with the solution. It follows that

[AgCl(s)] = K´ = constant Hence, [Ag+] [Cl–] = KK´ = Ksp (constant) Ksp is termed as the solubility product. It is defined as

the product of the concentration of ions in a saturated solution of a salt at a given temperature. Consider, in general, the salt of the type AxBy which is dissociated as :

AxBy x Ay+ + y Bx– Applying law of mass action,

]BA[

]B[]A[

yx

yxxy −+

= K

when the solution is saturated, [Ax By] = K´ (constant) or [Ay+]x [Bx–]y = K[AxBy] = KK´ = Ksp (constant) Thus, solubility product is defined as the product of

concentrations of the ions raised to a power equal to the number of times the ions occur in the equation

representing the dissociation of the salt at a given temperature when the solution is saturated.

Solubility product is not the ionic product under all conditions but only when the solution is saturated. Ionic product has a broad meaning since it is applicable to all types of solutions, may be saturated or unsaturated.

Relationship between solubility and solubility product : The equilibrium for a saturated solution of any

sparingly soluble salt be expressed as : Ax By x Ay+ + y Bx– Thus, solubility product, Ksp = [Ay+]x [Bx–]y . Let 's' mole per litre be the solubility of the salt, then Ax By x Ay+ + y Bx– xs ys So KSP = [xs]x [ys]y = xx.yy(s)x+y Since the solubility of a salt varies with temperature,

the numerical value of Ksp for a salt changes with temperature; values usually recorded at 25ºC.

Common Ion Effect : The suppression of the degree of dissociation of a

weak acid or a weak base by the addition of a strong electrolyte containing a common ion. The common ion effect play an important role in the qualitative analysis.

Application of solubility product in qualitative analysis: Precipitation of sulphides of group II. Sulphides of

group II are precipitated by passing H2S gas through the solution of these cations in presence of dil HCl. H2S being a weak electrolyte ionizes only sligthtly, while HCl being a strong electrolyte is almost completely ionized.

H2S 2H+ + S2– ; HCl → H+ + Cl– Thus, the common ion effect takes place. As a result,

the degree of dissociation of H2S decreases sufficiently and the concentration of S2– ions in the solution becomes very small. But with this low concentration of second group and the sulphide ions exceeds the very low solubility products of their corresponding sulphides. Therefore, the cations of group II get precipitated as their insoluble sulphides.

On the other hand, the sulphides of the cations of the other groups (III, IV, V and Mg) are not precipitated under these conditions because their solubility products are quite high.

Physical Chemistry

Fundamentals

SOLUBILITY PRODUCT

KEY CONCEPT


Precipitation of the hydroxides of group III : Hydroxides of group III are precipitated by adding an excess of solid NH4Cl to the solutions of these cations followed by the addition of excess of NH4OH. Being a weak electrolyte, NH4OH is only slightly ionised, whereas NH4Cl, being a strong electrolyte, ionizes almost completely to give at large concentration of NH4

+ ions. NH4OH NH4

+ + OH–; NH4Cl → NH4+ + Cl–

Due to the common ion effect, the degree of dissociation of NH4OH gets suppressed and hence the concentration of OH– ions in solution decreases appreciably. But even with this low conc. of OH– ions, the ionic products of the cations of group III and OH–ions exceed the low values of the solubility products of their corresponding metal hydroxides. As a result, the cations of group III get precipitated as their insoluble hydroxides.

On the other hand, cations of groups IV, V and Mg, which require a large conc. of OH– ions due to their high solubility products will not be precipitated.

Precipitation of sulphides of group IV. The sulphides of group IV are precipitated by passing H2S through ammoniacal solution of these cations.

Both H2S and NH4OH, being weak electrolytes, ionize only slightly as :

H2S 2H+ + S2–

NH4OH +4NH + OH–

The H+ ions and OH– ions combine to produce practically unionised molecules of water

H+ + OH– → H2O As a result, the above dissociation equilibrium

reactions get shifted in the forward direction, so that the concentration of S2– ions goes on increasing. Ultimately, the ionic product of the cations of group IV and S2– ions exceed the solubility products of their corresponding metal sulphides and hence get precipitated.

Precipitation of carbonates of group V : The carbonates of group V are precipitated by adding (NH4)2CO3 solution to the solution of these cations in the presence of NH4Cl and NH4OH. (NH4)2CO3, being a weak electrolyte ionises only slightly to give a small concentration of CO3

2– ions. (NH4)2CO3 2NH4

+ + CO32–

On the other hand, NH4Cl being a strong electrolyte, ionises almost completely to give a large concentration of +

4NH ions. Due to the common ion effect, the dissociation of (NH4)2CO3 is suppressed and hence the concentration of CO3

2– ions in the solution decreases considerably. But even with this low concentration of CO3

2– ions, the ionic products of these cations and CO3

2– ions exceed the low values of the solubility products of their corresponding metal carbonates and thus get precipitated.

However, under these conditions, Mg salts do not get precipitated as MgCO3 since its solubility product is comparatively high and thus requires a high concentration of CO3

2– ions for precipitation. The carbonates of Na+, K+ and +

4NH ions are also not precipitated because they are quite soluble.

The necessity of adding NH4OH arises due to the fact that (NH4)2CO3 solution usually contains a large amount of NH4HCO3. Thus, the cations of group V will form not only insoluble carbonates but soluble bicarbonates as well. As a result, the precipitation will not be complete. In order to convert NH4HCO3 to (NH4)2CO3, NH4OH is always added.

NH4HCO3 + NH4OH → (NH4)2CO3 + H2O Preferential precipitation of Salts : A solution contains more than one ion capable of

forming a precipitate with another ion which is added to the solution. For example, in a solution containing Cl–, Br–, and I– ions, if Ag+ ions are added, then out of the three, the least soluble silver salt is precipitated first. If the addition of Ag+ ions is continued, eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stoichiometry of the precipitated salts is the same, then the salt with the minimum solubility product (and hence also the minimum solubility) will precipitate first followed by the salt of next higher solubility product and so on.

If the stoichiometry of the precipitated salts is not the same, then, from the solubility product data alone, we cannot predict which ion will precipitate first. Take, for example, a solution containing Cl– and CrO4

2–. Both these ions form precipitates with Ag+. Through the solubility product product of AgCl is larger than that of Ag2CrO4, yet it is AgCl (lesser soluble) which precipitates first when Ag+ ions are added to the solution. In order to predict which ion (Cl– or CrO4

2–) precipitates first, we have to calculate the concentration of Ag+ ions needed to start the precipitation through the solubility product data and the given concentration of Cl– or CrO4

2–. Since square root is involved in the expression for computing Ag+ for silver chromate, the quantity of Ag+ needed to start the precipitation of CrO4

2– is larger than that for Cl–. Hence, as AgNO3 is added to the solution, the minimum of the two concentrations of Ag+ to start the precipitation will be reached first and thus the corresponding ion (Cl– in this case) will be precipitated in preference to the other. During the course of precipitating, concentration of Cl– decreases and the corresponding concentration of Ag+ to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO4

2–. At this stage, practically the whole of Cl– ions have been precipitated. The addition of more of AgNO3 causes the precipitation of both the ions together.


1. An organic compound (A), C4H9Cl, on reacting with aqueous KOH gives (B) and on reaction with alcoholic KOH gives (C) which is also formed on passing vapours of (B) over heated copper. The compound (C) readily decolourise bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to give (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg++ and H2SO4. Identify (A) to (H) with proper reasoning.

Sol.

C4H9Cl

(A) (Alkyl halide)

C4H8 (C)

(Alkene)

Alc. KOH

∆;– HCl

C4H9OH (B)

(Alcohol)

Aq. KOH

∆; –KCl

Cu

∆; –H2O

We know that p-alcohol on heating with Cu gives

aldehyde while s-alcohol under similar conditions gives ketone. Thus, (B) is a t-alcohol because it, on heating with Cu gives an alkene (C). Since a t-alcohol is obtained by the hydrolysis of a t-alkyl halide, hence (A) is t-butyl chloride.

Thus, (A) is

3

33

CH|

CHCCH|Cl

−− and (B), is

3

33

CH|

CHCCH|OH

−−

The alkene (C) on ozonolysis gives (D) and (E), hence (C) is not symmetrical alkene. In these compound (E) gives Cannizaro's reaction with NaOH. So, (E) is an aldehyde which does not contain α - H atom. Hence it is HCHO. Compound (D) can also be prepared by the hydration of propyne in the presence of acidic solution and Hg++

CH3 – C ≡ CH + H2O

Hg++

H+ CH3 – C = CH2

OH

CH3 – C – CH3

O

(D)

Hence (D) is acetone and (E) is formaldehyde. Therefore, alkene (C) is 2-methyl propene.

CH3 – C = CH2

CH3

(C) (D) reacts with hydroxyl amine (NH2OH) to form

oxime (F).

C = O + H2 NOH

CH3

(D)CH3

–H2O C = NOHCH3

CH3(F)

Thus, (B) is

3

33

CH|

CHCCH|OH

−− and (A) is

3

33

CH|

CHCCH|Cl

−−

Reactions :

CH3 – C – CH3

Cl

CH3

Aq. KOH ∆; –KCl

CH3 – C – CH3

OH

CH3

Cu/

300º

C

–H2O

CH3 – C = CH2 + H2O

CH3

(A) (B)

(C)

Alc. KOH/∆ –KCl; –H2O

CH3 – C = CH2

CH3

(C)

CH3 – C = CH2(I) O3

(II) H2O/Zn C = O + H –C – H

CH3

CH3

O

(E)(D)

(C)CH3

∆

–H2OC = O + H2NOH

CH3

CH3 (F)(D)C = NOH

CH3

CH3

2HCHO + NaOH → CH3OH + HCOONa (E) (G) (H)

CH3 – C ≡ CH + H2OHg++

CH3 – C – CH3

O

(D)H+

UNDERSTANDINGOrganic Chemistry


2. An organic compound (A) C7H15Cl on treatment with alcoholic KOH gives a hydrocarbon (B) C7H14. (B) on treatment with O3 and subsequent hydrolysis gives acetone and butyraldehyde. What are (A) and (B) ? Explain the reactions.

Sol. In numerical, following data are given :

halide Alkyl)A(

157 ClHC∆

→ KOH .Alc

Alkene)B(147HC → 3O

Ozonide3147 OHC

∆

→ Zn/OH2 CH3COCH3 + CH3CH2CH2CHO

The alkene contains seven carbon atoms. The position of C = C double bond can be located as follows :

C = O + O = C.CH2CH2CH3 CH3

CH3

H

–2[O]

CH3 – C = CH.CH2CH2CH3

CH3 Thus, alkene (B) is 2-methyl hexene-2 The ozonolysis reaction is as follows :


CH3 (B)

O3

CH3 – C CH – CH2CH2CH3

O O

CH3

O

H2O/Zn CH3 – C = O + O = C – CH2CH2CH3

CH3 H

∆

Since alkene (B) is produced by the removal of one

mol of HCl from alkyl halide (A) and thus (A) can be either (I) or (II).

CH3 – C – CH2CH2CH2CH3

Cl

CH3

(I) or

CH3 – CH – CH – CH2CH2CH3

CH3

(II)

Cl The dehydro halogenation reaction by(I) or (II) yeilds


Cl

CH3

KOH alc.

(I). CH3 – C = CH.CH2CH2CH3

CH3 Main product (Saytzeff's rule)


CH3

KOH Alc.

(II) CH3 – C = CH.CH2CH2CH3

CH3 Main product (Saytzeff's rule)

Cl

Thus, both (I) and (II) give main product as 2-methyl hexene-2, hence (A) is either (I) or (II).


Cl

CH3

2-chloro-2-methyl hexane

(A)


ClCH3

3-chloro-2-methyl hexane

or


CH32-methyl hexene-2

(B)

3. An unsaturated hydrocarbon (A), C6H10 readily gives

(B) on treatment with NaNH2 in liquid NH3. When (B) is allowed to react with 1-chloro propane, a compound (C) is obtained. On partial hydrogenation in the presence of Lindlar catalyst (C) gives (D), C9H18. On ozonolysis (D) gives 2, 2-dimethyl propanal and butanal. Give structures of (A), (B), (C) and (D) with proper reasoning.

Sol. The structure of compound (D) can be obtained by joining the products of ozonolysis.

CH3 – C – CH = O + O = CH.CH2CH2CH3

CH3

CH3

2,2-dimethyl propanal

–2[O]

Butanal

CH3 – C – CH = CH.CH2CH2CH3

CH3

CH3 2,2-dimethyl heptene-3 (D)

Ozonolysis equation of (D) is :

CH3 – C – CH = CHCH2CH2CH3

CH3

CH3

(D)

(I) O3

CH3 – C – CHO + CH3CH2CH2CHO

CH3

CH3

(II) H2O/Zn

Alkene (D) is obtained by the partial hydrogenation

of (C), thus (C) contains a – C≡C – triple bond at C3.

CH3 – C – C ≡ C – CH2CH2CH3

CH3

CH3

(C)

H2


CH3

CH3

Lindlar catalyst

(D)


The starting compound (A) reacts with NaNH2 in presence of liquid NH3. It means it contains one –C≡CH at the terminal carbon, and, therefore gives a mono sodium derivative.

)A(106HC

3

2NH

NaNH → C4H9 – )B(

C ≡ C.Na

Compound (B) reacts with 1-chloro propane to give compound (C) as follows :

C4H9 – C ≡ C – Na + Cl – CH2CH2CH3 ∆

C4H9 – C≡C – CH2CH2CH3

–NaCl

(C)

(B) 1-chloro propane

But, (C) is CH3 – C – C ≡ C – CH2CH2CH3

CH3

CH3

Now, putting the value of C4H9 as a t-butyl radical, we have :

CH3 – C – C≡C – H

CH3

CH3 (A)

NaNH2 CH3 – C – C≡CNa + NH3

CH3

CH3 (B)

Hence,

CH3 – C – C≡CH

CH3

CH3 (3,3-dimethyl butyne-1)

(A)

CH3 – C – C≡CNa

CH3

CH3

(B)

CH3 – C – C≡C – CH2CH2CH3

CH3

CH3

(C)


CH3

CH3

(D)

4. A hydrocarbon (A) [C = 90.56%, V.D. = 53] was

subjected to vigrous oxidation to give a dibasic acid (B). 0.10 g of (B) required 24.10 ml of 0.05 N NaOH for complete neutralization. When (B) was heated strongly with soda-lime it gave benzene. Identify (A) and (B) with proper reasoning and also give their structures.

Sol. Determination of empirical formula of (A) :

Element % Atomic wt.

Relative no. of atoms Simplest ratio

C 90.56 12 12

56.90 = 7.55 55.755.7 = 1 or 4

H 9.44 1 144.9 = 9.44 55.7

44.9 = 1.25

or 5

The empirical formula of (A) = C4H5 Empirical formula weight = 48 + 5 = 53 Molecular weight = V.D. × 2 = 53 × 2 = 106

Hence, n = wt.Empirical wt.Molecular =

53106 = 2

Molecular formula = 2 × C4H5 = C8H10 The given equation may be outlined as follows :

C8H10

(A)

Vigrous oxidation6[O] C6H4

(B)

COOH

COOH+ 2H2O

Meq. of dicarboxylic acid = Meq. of NaOH

E

000,11.0 × = 24.1 × 0.05

Equivalent of acid = 83 Molecular wt. = Basicity × Equivalent weight = 2 × 83 = 166 Since (B) on heating with soda-lime gives benzene,

the C6H4 represents to benzene nucleus having two side chains, thus (B) is a benzene dicarboxylic acid.

There are three benzene dicarboxylic acids.

COOHCOOH

Phthalic acid

COOH

COOH Isophthalic acid

COOH

COOHTerphthalic acid

All the above three acids are obtained by the oxidation of respectively xylenes.

CH3

CH3

o-xylene

COOH COOH

6[O]+ 2H2O

CH3

CH3m-xylene

COOH

COOH

6[O]+ 2H2O

CH3

CH3p-xylene

COOH

COOH

6[O]+ 2H2O


All the above three acids on heating with soda-lime yields only benzene.

COOH COOH

COOH COOH , ,

COOH

COOH

NaOH + CaO∆ + 2CO2

Of the three acids, one which on heating gives an anhydride, is o-isomer.

COOH CO CO COOH

O ∆

–H2O

One acid which on nitration gives a mono nitro compound is p-dicarboxylic acid.

COOH

COOH

NO2 HNO3

∆; H2SO4

COOH

COOH

One acid which on nitration gives three mono nitro compounds will be the m-isomer.

COOH

COOH NO2

HNO3

H2SO4

COOH

COOH

NO2

COOH

COOH

NO2

COOH

COOH

5. Two moles of an anhydrous ester (A) are condensed in presence of sodium ethoxide to give a β-keto ester (B) and ethanol. On heating in an acidic solution compound (B) gives ethanol and a β-keto acid (C). (C) on decarboxylation gives (D) of molecular formula C3H6O. Compound (D) reacts with sodamide to give a sodium salt (E), which on heating with CH3I gives (F), C4H8O, which reacts with phenyl hydrazine but not with Fehling reagent. (F) on heating with I2 and NaOH gives yellow precipitate of CHI3 and sodium propionate. Compound (D) also gives iodoform, but sodium salt of acetic acid. The sodium salt of acetic acid on acidification gives acetic acid which on heating with C2H5OH in presence of conc. H2SO4 gives the original ester (A). What are (A) to (F) ?

Sol. (i) Acetic acid on heating with C2H5OH gives original compound (A).

CH3COOH + C2H5OH ∆ → 42SOH

)A(523 HCOOCCH

+ H2O

(ii) CH3COOC2H5 (A) on heating with C2H5ONa undergoes Claisen condensation to give (B), which is aceto acetic ester.

CH3CO OC2H5 + H CH2COOC2H5 C2H5ONaReflux

(A)+ C2H5OH + CH3COCH2COOC2H5

(B) (iii) (B) on heating in acidic solution gives (C) and

ethyl alcohol.

)B(

5223 HCOOCCOCHCH + HOH →+H

)C(

23 COOHCOCHCH + C2H5OH

(iv) (C) on decarboxylation gives acetone (D).

)C(

23 COOHCOCHCH 2CO−

∆→ )D(

33COCHCH

(v) (D) reacts with NaNH2 to form sodium salt (E), which on heating with CH3I gives butanone (F).

)D(

33COCHCH + NaNH2 3NH−

∆→ )E(

23 NaCOCHCH

NaI–ICH3 →

)F(323 CHCOCHCH

(vi) )F(

323 CHCOCHCH + 3I2 + 4NaOH →∆

CHI3 + CH3CH2COONa + 3NaI + 3H2O

(vii) )D(

33COCHCH + 3I2 + 4NaOH →∆

CHI3 + CH3COONa + 3NaI + 3H2O

CH3COONa →HCl CH3COOH + NaCl

Thus, (A) CH3COOC2H5

(B) CH3COCH2COOC2H5

(C) CH3COCH2COOH

(D) CH3COCH3

(E) CH3COCH2Na

(F) CH3COCH2CH3


1. Prove that, if n is a positive integer,

∫ −a

0

nx dxxe =

n !

++++− −

!na...

!2aa1e1

n2a

Also, deduce the value of ∫∞ −

0

nx dxxe

2. Let A ≡ (6, 5), B ≡ (2, –3) and C ≡ (–2, 1) be the

vertices of a triangle. Find the point P in the interior of the triangle such that ∆PBC is an equilateral triangle.

3. If Sn = nC1 + 2.nC2 + 3.nC3 + ....... + n nCn then find

∑=

n

1nnS . Also prove that

nC1 . (nC2)2 . (nC3)3.... (nCn)n ≤ 21n Cn

1n2

+

+.

4. Let z1, z2, z3 be three distinct complex numbers

satisfying |z1 – 1| = |z2 – 1| = |z3 – 1|; Let A, B and C be the

points represented in the argand plane corresponding to z1, z2 and z3 respectively. Prove that z1 + z2 + z3 = 3 if and only if ∆ABC is an equilateral triangle.

5. Let A ≡ (r, 0) be a point on the circle x2 + y2 = r2 and

D be a given point inside the circle. If BC be any arbitrary chord of the circle thorugh point D. Prove that the locus of the centroid of triangle ABC is a circle whose radius is less than r/3.

6. A number is chosen at random from the set 1, 2, 3, …......, 2006. What is the probability that it has no prime factor in common with 10 ! ?

7. Two vertices of a triangle are a – j3i + and j5i2 +

and its orthocenter is at ( j2i + ). Find the position

vector of third vertex. 8. Show that an equilateral triangle is a triangle of

maximum area for a given perimeter and a triangle of minimum perimeter for a given area.

9. The bottom of a tank with a capacity of 300 litres is

covered with a mixture of salt and some insoluble substance. Assuming that the rate at which the salt dissolves is proportional to the difference between the concentration at the given time and the concentration of a saturated solution (1 kg of salt per 3 litres of water) and that the given quantity of pure water disolves 1/3 kg of salt in 1 minute. Find the quantity of salt in solution at the expiration of one hour.

10. An isosceles triangle with its base parallel to the

major axis of the ellipse 9

x2+

3y2

= 1 is

circumscribed with all the three sides touching the ellipse. The least possible area of the triangle is.

`tàxÅtà|vtÄ VtÄÄxÇzxá This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Shailendra Maheshwari Joint Director Academics, Career Point, KotaSolut ions published in this issue

12Set


1. Let z = x + iy so given 1 ≤ x or 1 ≤ r cos θ , if z = ∑eiθ Now,

iyx1iyx1

++−− = 22 y)x1(

)iyx1)(iyx1(++

−+−−

Real part 22

222

y)x1(y)x1(

++−− ≤ 0 as x ≥ 1 given and

imaginary part

22 y)x1(y)x1(y)x1(

+++−−− = 22 y)x1(

y2++

− ≤ 0

as y ≥ 0 given

so z1z1

+− ∩ 0 is true.

2. As ∠ POQ = 90º P

C

Q

O θ 90–θ

so CP = (OC) . tan (90 – θ) = (OC) cot θ & CQ = (OC) tan θ so CP. CQ = (OC)2 = r2 when r is the radius of circle.

3. f(0) = c f(1) = a + b+ c & f(–1) = a – b + c

solving these, a = 21 [f(1) + f(–1) – 2f(0)]

b = 21 [f(1) – f(–1)] & c = f(0)

so f(x) = 2

)1x(x + f(1) + (1 – x2) f(0) + 2

)1x(x − f(–1)

2 | f(x) | ≤ |x| |x + 1| + 2 |1 – x2| + |x| |x – 1|; as |f(1)|, |f(0)|, |f(–1)| ≤ 1. 2 |f(x)| ≤ |x| (x + 1) + 2(1 – x2) + |x| (1 – x) as x ∈ [–1, 1]

so 2 |f(x)| ≤ 2(|x| + 1 – x2) ≤ 2 . 45

so |f(x)| ≤ 45

Now, as g(x) = x2 f

x1

= 21 (1 + x) f(1) + (x2 – 1) f(0) +

21 (1 – x) f(–1)

so 2 | g(x) | ≤ |x + 1| + 2 |1 – x2| + |1 – x| 2 | g(x) | ≤ x + 1 + 2(1 – x2) + 1 – x as x ∈ [–1, 1] 2 |g(x)| ≤ –2x2 + 4 ≤ 4 |g(x)| ≤ 2

4.

−− ý

ya2

y21a1

22

= 22 yaa

1

−+ .

22 ya2

ý.y2

−

− – y1 y´

a1 = y´

−

−−+−

− y1

ya)yaa(

y

yaa

y222222

= 2222

22222222

ya)yaa(ay

ya)yaa(aay)yaa(yý

−−+

−−+−−−+

y(a + 22 ya − ) 22 ya −

= y´[y2a + y2 22 ya − – y2a – a2 22 ya − – a(a2 – y2)]

y (a + 22 ya − ) 22 ya − = –y´(a2 – y2) (a + 22 ya − )

y´ = – 22 ya

y

−

y´M = 22a β−

β−

y = xa 22 β−

β− ...(1)

y = β ...(2) locus of intersection of these two lines is

y = 22 ya

y

−

− x

a2 – y2 = x2 x2 + y2 = a2

MATHEMATICAL CHALLENGES SOLUTION FOR MARCH ISSUE (SET # 11)


5. ∑= ++

−n

0r

rn

r

)1r()2r(2.C)2(

= )2n)(1n(2

1++ ∑

=

−n

0r

r)2( n+2Cr+2 (–2)2

= )2n)(1n(2

1++ ∑

=

+−n

0r

2r)2( n+2Cr+2

= )2n)(1n(2

1++

[(1 – 2)n + 2 –1 + 2(n + 2)]

= )2n)(1n(2

1++

[(–1) n + 2 + 2n + 3]

= 2n

1+

if n is odd

= 2n

1+

if n is even

6. In = 1

0

11n

1nxtanx

+

−+

– ∫ ++

+1

0 2

1n

)x1)(1n(x dx ...(1)

= )1n(4 +

π –

+−+

+ ∫−

dxx1

)11x(x1n

1 1

0 2

21n

= )1n(4 +

π – ∫ −

+

1

0

1n dxx1n

1 + ∫ ++

−1

0 2

1ndx

x1x

1n1

= )1n(4 +

π – 1

0

n

nx

1n1

+ + ∫ ++

−1

0 2

1ndx

x1x

1n1 ...(2)

from (1) ∫ +

+1

0 2

1ndx

x1x =

4π – (n + 1) In

so ∫ +

−1

0 2

1n

x1x dx =

4π – (n – 1)In–2

use it in (2)

In = )1n(4 +

π – n)1n(

1+

+ 1n

1+

−−

π−2nI)1n(

4

(n + 1) In + (n – 1) In – 2 = 2π –

n1

7. diff. partially w. r.t. x

f ´

+

3yx .

31 =

3)x´(f

Let x = 0 f ´(y/3) = f ´ (0) so f ´(x) = f ´(0) hence f ´(x) is a constant so f ´ (x) = 2 (as given) so f(x) = 2x + c; since f(0) = 2 so f(x) = 2x + 2 Hence f(2) = 6

8,9,10. g (f(x)); z → z1 g(f(x)) = 1 ; x ∈ even integer = 0 ; otherwise Hence g(f(x)) is many one onto, periodic and even

function Now, f(g(x)) = p ; if x is perfect square = 1 ; otherwise Hence f(g(x)) = p has infinitely many roots.

Do you know

• The largest telescope in the world is currently being constructed in northern Chile. The telescope will utilize four - 26 ft. 8 in. (8.13 meters) mirrors which will gather as much light as a single 52 ft. 6 in. (16 meters) mirror.

• The Hubble Space Telescope weighs 12 tons (10,896 kilograms), is 43 feet (13.1 meters) long, and cost $2.1 billion to originally build.

• The longest living cells in the body are brain cells which can live an entire lifetime.

• The largest flying animal was the pterosaur which lived 70 million years ago. This reptile had a wing span of 36-39 feet (11-11.9 meters) and weighed 190-250 pounds (86-113.5 kilograms).

• The Atlantic Giant Squid's eye can be as large as 15.75 inches (40 centimeters) wide.

• Armadillos, opossums, and sloth's spend about 80% of their lives sleeping.

• The starfish species, Porcellanaster ivanovi, has been found to live in water as deep as 24,881 feet (7,584 meters).

• The tentacles of the giant Arctic jellyfish can reach 120 feet (36.6 meters) in length.

• The greatest tide change on earth occurs in the Bay of Fundy. The difference between low tide and high tide can be as great as 54 ft. 6 in. (16.6 meters).

• The highest temperature produced in a laboratory was 920,000,000 F (511,000,000 C) at the Tokamak Fusion Test Reactor in Princeton, NJ, USA.

• The fastest computer in the world is the CRAY Y-MP C90 supercomputer. It has two gigabytes of central memory and 16 parallel central processor units.


1. In = ∫ −a0

nx dxxe = ( )a0nx xe− + n ∫ −−a0

1nx xe dx = – e–a an + nIn – 1

In = –e–a an + n[–e–a an–1 + (n–1)In–2] = –e–a[an + nan–1 +n(n – 1)an–2 + n(n – 1)(n – 2)an–3 + ..... + n(n – 1)..... 2a] + n I0

= ( )

+++−

+−−−

−− a2

a....1n

an

aeen21nn

aa0

x

In =

++++− −

na.....

2aa1e1n

n2a

Now, I = ∫∞ −0

nx xe = ∞→a

Lt In = n

2. Mid pt. M of BC = (0, –1)

A(6, 5)

B(2, –3)

M

C (–2, 1) 60º

45º

P

slope of BC = 2213

+−− = –1

so slope of altitude of ∆PBC is = 1. length BC = 1616 + = 24

Now altitude PM = 24 sin 60º

= 24 . 23 = 62

eqn. of PM line is

21

0x − =

21

1y + = r (as its slope is 1)

x = 2r & y =

2r – 1

for req. pt. P take r = 62 .

So pt. P ≡

−1

262,

262 = )132,32( −

3. Sn = ∑=

n

0rr . nCr = n . 2n–1

so S = ∑=

n

1n1S = ∑

=

−n

1n

1n2.n

S = 1 + 2.21 + 3.22 + 4.23 + .... + n . 2n–1 2S = 2 + 2.22 + 3.23 + ..... + (n – 1). 2n–1 + n.2n (1 – 2)S = (1 + 2 + 22 + 23 + .... + 2n–1) – n . 2n

= 1 . 1212n

−− – n . 2n

S = n . 2n – 2n + 1 = (n – 1) 2n + 1 Now A.M. ≥ G. M.

n.....321

C.n....C.3C.2C nn

3n

2n

1n

++++++++

≥

( ) ( ) ( ) n...211

nn

n33

n22

n1

n C......C.C.C +++

2)1n(n

1n

2)1n(n

2.n

+

−

+ ≥ nC1 (nC2)2 ..... (nCn)n

so nC1 . (nC2)2 ..... (nCn)n ≤ 2

1n Cn

1n2

+

+

4. Let P be a point represented by 1. so as |z1 – 1| = |z2 – 1| = |z3 – 1| so P is the

circumcentre of ABC. Its centroid is 3

zzz 321 ++

If the ∆ABC is equilateral then circumcentre = centroid

so 1 = 3

zzz 321 ++

so z1 + z2 + z3 = 3 Now if z1 + z2 + z3 = 3 then centroid of ∆ABC is 1

which is point P and P is already the circumcentre of ∆ABC. So now if they are same then ∆ABC is equilateral.

5. let the centroid of ∆ABC be (h, k) then 3h = r cos α + r cos β + r

rh3 – 1 = 2 cos

2β+α cos

2β−α ...(1)

& rk3 = sin α + sin β

= 2 cos 2

β−α sin 2

β+α ...(2)

MATHEMATICAL CHALLENGES SOLUTION FOR THIS ISSUE (SET # 12)


line BC, x cos2

β+α + y sin2

β+α = r cos2

β−α

Let point D be (a, b)

D C(α)

A(r, 0)

B(β)

then a cos 2

β+α + b sin 2

β+α = r cos 2

β−α

....(3)

Multiply (3) by cos 2

β−α

a cos2

β−α cos2

β+α + b sin2

β+α cos2

β−α

= r cos2

2β−α

use (1) & (2)

−1

rh3

2a + b

r2k3 = r cos2

2β−α ...(4)

square & add (1) & (2)

4 cos2

2β−α =

2

1rh3

− + 2

2

rk9 ...(5)

from (4) & (5)

2

1rh3

41

− + 2

2

r4k9 =

−1

rh3

r2a + 2r2

k3b

so req. locus is (3x – r)2 + 9y2 = 2a(3x – r) + 6.b.y 9x2 + 9y2 – 6rx + r2 = 6ax – 2ar + 6.b.y

x2 + y2 – 32 (r + a)x –

32 by +

9r2

+ 9ar2 = 0

22

b31y)ar(

31x

−+

+−

= 9

bar2r)ar( 222 +−−+ = 9

ba 22 +

It is a circle and radius is 3

ba 22 +

Since point D is interior of circle so a2 + b2 < r2, so radius of this circle is less than r/3. 6. Prime factors dividing 10 are 2, 3, 5, 7. As required

the number chosen should not be divisible by 2 or 3 or 5 or 7. Define, the events as

A : divisible by 2 B : divisible by 3 C : divisible by 5 D : divisible by 7

A ∪ B ∪ C ∪ D = A + B + C + D – A ∩ B – A ∩ C – A ∩ D – B ∩ C – B ∩ D – C ∩ D + A ∩ B ∩ C

+ A ∩ C ∩ D + B ∩ C ∩ D + A ∩ B ∩ D – A ∩ B ∩ C ∩ D

P(A ∪ B ∪ C ∪ D)

= 2006

1 [ 1003 + 668 + 401 + 286 – 334 – 200

– 143 – 133 – 95 – 57 + 66 + 28 + 19 + 47 – 9]

= 20061547

7. Line BC

Aa

H

DB Cb = –i + 3j c = 2i + 5j

→r = – j3i + + )j2i3(t + ...(1) any pt. D on it = (3t – 1) i + (3 + 2t) j As HD ⊥ BC, so ((3t – 1 – 1) i + (3 + 2t – 2) j). (3i + 2j) = 0 3(3t – 2) + 2(2t + 1) = 0

13 t – 4 = 0 ⇒ t = 134

so point D = –13i +

13j47

Now line HD ⇒ →r = i + 2j + s´

−−+− j2i

13j47

13i

= i + 2j + s(–14i + 21j)

⇒ →r = i + 2j + λ(–2i + 3j) Any point A on it = (1 – 2λ)i + (2 + 3λ)j Now as AC ⊥ BH so [ (1 – 2λ – 2)i + (2 + 3λ – 5)j] . [2i – j] = 0 2 (–1 – 2λ) – (3λ – 3) = 0

⇒ –7λ = 2 – 3 ⇒ λ = 71

so pt. A = 7i5 +

7j17

8. If A is the area of the triangle with sides a, b, and c,

then A2 = s(s – a) (s – b) (s – c); where 2s = a + b + c using AM – GM inequality for s – a, s – b, s – c, we

have

A2 ≤ s3

3)cs()bs()as(

−+−+−


A2 ≤ s 3

3s2s3

− = 3

4

3s

A ≤ 33

s2

Let 2s = p, then A ≤ 312

p2

Amax = 312

p2,

As condition of equality holds iff s – a = s – b = s – c which happen if a = b = c so Amax = ; for a = b = c

Now again p ≥ A312

pmin. = A312 ; and again equality holds if a = b = c. 9. Let the amount of salt dissolved at any time t is x kg.

So concentration is 300

x

so dtdx = k

−

300x

31 = k

−

300x100

x100

dx−

= 300k dt

– ln (100 – x) = 300k t + C

at t = 0, x = 0 so C = –ln 100

so 300kt = ln 100 – ln (100 – x) = ln

x100100

−

at t = 1 min., x = 1/3

so 300k = ln

31100

100

−

so 300k = ln

3299100

so 300k = ln

299300

so k = 300 ln 299300

so ln 299300 . t = ln

x100100

−

so t

299300

=

x100100

−

so 100 – x = 100 . t

300299

so x = 100

−

t

3002991

at t = 60 min.

x = 100

−

60

3002991 kg

10. Let P (3 cos θ, 3 sin θ)

line BC : y = – 3 ; line AC : A

B C

P

3x cos θ +

3y sin θ = 1

pt. C

−

θθ+ 3,

cos)sin1(3

pt. A (0, 3 cosec θ )

Area A = 21 . 2 .

θθ+

cos)sin1(3 . ( 3 cosec θ + 3 )

= θθθ+

cossin)sin1(33 2

= θ

θ+2sin

)sin1(36 2

θddA =

θθθ+−θθθ+

2sin)2cos)sin1(2cos2sin)sin1(2(36

2

2

= θ

θθ−θ−θθθ+2sin

)2cossin2coscos2)(sinsin1(3122

= θ

θ−θ−θθ+2sin

)2cos)2)(sin(sin1(3122

= θ

θ+−θθ+2sin

)sin21)(sinsin1(3122

2

= θ

+θ−θθ+2sin

)1)(sin1sin2)(sin1(3122

= θ

−θθ+2sin

)1sin2()sin1(3122

2

Amin at θ = 6π

so Amin =

23

21136

2

+

= 12 . 49

= 27sq. units. = 0027 Ans.


1. Let f : R → R and f(x) = g(x) + h(x), where g(x) is a

polynomial and h(x) is a continuous and differentiable bounded function on both sides, then f(x) is onto if g(x) of odd degree and f(x) is into if g(x) is of even degree. Then check whether f(x) is one one, many one, onto or into.

(i) f(x) = a1x + a3x3 + a5x5 + .... + a2n + 1x2n + 1 – cot–1x where 0 < a1 < a2 < a3 ....... < a2n + 1

(ii) f(x) = 1xx

2x)1x)(1x(x2

44

++

++++

Sol. (i) f(x) = odd degree polynomial + bounded function cot–1x also f´(x) > 0

⇒ y = f(x) will be one one and onto

(ii) f(x) = 1xx

)1xx)(1x(2

24

+++++ = x4 + 1 +

1xx1

2 ++

= even degree polynomial

+ bounded function ∈

34,0

also f´(x) = 0 has at least one root real which is not repeated since f´(x) is a polynomial of degree 7.

⇒ f(x) = 0 has at least one point of extrema. ⇒ many one & Into 2. A rectangle ABCD of dimensions r and 2r is folded

along the diagonal BD such that planes ABD and CBD are perpendicular to each other. Let the position of the vertex A remains unchanged and C0 is the new position of C, then find the distance of C0 from A and shortest distance between the edges AB & C0D.

(2r, r)

C

B (2r, 0)(0, 0) A

D (0, r)

N

Sol. Let the rectangle ABCD lies on the plane xy. After

folding the rectangle along the BD co-ordinates of points in 3-D are-

A : (0, 0, 0), B : (2r, 0, 0), C : (2r, r, 0), D(0, r, 0)

and N :

0,

5r,

5r2 and C0

5r2,

5r,

5r2 ,

Now AC0 = 585 r

and shortest distance = |DCAB|

)DCAB.(AC

0

00

×

×=

3r5 unit.

3. Let f(x) is a polynomial one-one function such that

f(x).f(y) + 2 = f(x) + f(y) + f(xy) ∀ x, y ∈ R–0,

f(1) ≠ 1, f´(1) = 3. Let g(x) = 4x (f(x) + 3) – ∫

x

0)x(f dx,

then prove that g(x) is an identity for all given x ∈ R –0.

Sol. putting x = y = 1 in given condition we get f(1)2 + 2 = f(1) + f(1) + f(1) ⇒ f(1)2 – 3f(1) + 2 = 0 ⇒ f(1) = 1 or 2 ⇒ f(1) = 2 Now put y = 1/x,

f(x) . f

x1 + 2 = f(x) + f

x1 + f(1)

⇒ f(x) = 1 ± xn According to given conditions, f(x) = 1 + x3

Now, g(x) = 4x [1 + x3 + 3] – ∫

x

0(1 + x3) dx = 0

⇒ g(x) = 0 for ∀ x ∈ R –0 4. Let three normals are drawn to the parabola y2 = 4ax

at three points P, Q and R, from a fixed point A. Two circles S1 and S2 are drawn on AP and AQ as diameter. If slope of the common chord of the circles S1 and S2 be m1 and the slope of the tangent to the parabola at the point R be m2, then prove that m1 . m2 = 2.

Sol. Let A(h, k) be a fixed point at3 + (2a – h)t – k = 0 Q three normals are drawn from (h, k) Let feet of normals P,Q and R are three points with

parameters t1, t2 and t3. Common chord of S1 and S2 = S1 – S2 = (t1 + t2)x +

2y – h(t1 + t2) – 2k = 0 Tangent to the parabola at R = t3y = x + at3

2

m1 . m2 = –

+

2tt 21 .

3t1 =

21

(Q t1 + t2 + t3 = 0 for co-normal points).

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' Forum

MATHS


5. Let z1, z2 and z3 are unimodular complex numbers then find the greatest value of |z1 – z2|2 + |z2 – z3|2 + |z3 – z1|2.

Sol. |z1 – z2|2 + |z2 – z3|2 + |z3 – z1|2 = 2[|z1|2 + |z2|2 + |z3|2] – [z1 2z + 1z z2 + 1z z3 + 3z z1 + z2 3z + 2z z3] = 6 – [z1 2z + z2 1z + z1 3z + z3 1z +z2 3z + z3 2z ] ...(1) Now |z1 + z2 + z3|2 ≥ 0 ⇒ z1 2z + z2 3z + z2 1z + z3 2z + z1 3z + 1z z3 ≥ –3 ...(2) From (1) & (2) maxm value of |z1 – z2|2 + |z2 – z3|2 + |z3 – z1|2 = 6 – (–3) = 9 6. Consider following two infinite series in real θ and r

C = 1 + r cos θ + !2

2cosr 22 θ + !3

3cosr3 θ + ....

S = r sin θ + !2

2sinr2 θ + !3

3sinr3 θ + .....

If a remains constant and r varies the prove that

(i) CdrdC + S

drdS = (C2 + S2) cos θ

(ii) 2

drdC

+

2

drdS

= C2 + S2

Sol. We have C + iS =

θiree ...(1)

C – iS = θ−iree ...(2)

Now,

C2 + S2 = 2

reie

θ =

2sinricosr e.e θθ = e2r cos θ ...(3)

Differentiating (1) w.r.t.r, we get

drdC + i

drdS = eiθ .

θie.re ....(4)

∴ 2

drdC

+

2

drdS

=

θθ ie.ri e.e = e2r cos θ

= C2 + S2 (Form 3) multiply (2) and (4)

+

drdSi

drdC (C – iS) = eiθ .

θie.re .θ−ie.re

= eiθ ]e[ )ee(r ii θ−θ + = eiθ . e2r cosθ Now equating real parts in both sides

CdrdC + S

drdS = (C2 + S2) cosθ

Hence proved.

Science Facts

Skin Deep Storage

Chip implants that keep track of personal information seem like a novelty but do they have a more useful future?

These days, some people are following their pets and getting tagged. Radio frequency identification (RFID) chips are the size of a grain of rice and can be loaded up with personal information like passwords and implanted under the skin. Instead of having to remember a login code, an RFID reader can be set up to automatically detect it and grant you access to a range of things from your computer to your front door. It seems like it could be useful to people with exceptionally poor memories, but right now these chips are being snapped up by technology geeks like Amal Graafstra. The 29-year-old businessman from Vancouver, Canada, is one of the first people to have an RFID implant and so far is happy with the results. "I just don't want to be without access to the things that I need to get access to. In his chip, he has stored a unique identification number which can be used to log him into various electronic devices. It didn't cost him an arm and a leg either: he got the whole set-up on the internet for about $50 (£30), including the $2 cost of the chip itself. The procedure to implant the chip is quite simple and painless. Amal's chip was implanted under the skin of his left hand while he was under a local anesthetic. It is possible to inject the chips using a large enough needle, but in Amal's case the chip was inserted by simply cutting through his skin with a scalpel. Other than complaining of sensitivity in the area of the implant, Amal said that it doesn't hurt and he expects that eventually the chip will be completely unobtrusive. A hand implanted with an RFID chip and the chip reader. The chip is made of silicon and is digitally encoded with information.. A RFID reader, which is installed in a computer or an electronic device like a reader by a front door, emits a radio signal of a particular frequency, just like radio stations each broadcast on their own frequency. The chip acts passively when it is within 3 inches of the reader: the right incoming radio signal induces just enough energy in the antenna of the chip for a circuit in the chip to power up and produce a response. The reader can then access the information on the chip and pass it on to the computer or device that requires it.


Function, Limits, Continuity & Differentiability :

If the domain of the function is in one quadrant then the trigonometrical functions are always one-one.

If trigonometrical function changes its sign in two consecutive quadrants then it is one-one but if it does not change the sign then it is many one.

In three consecutive quadrants tigonometrical functions are always many one.

Any continuous function f(x), which has at least one local maximum, is many-one.

Any polynomial function f : R → R is onto if degree of f is odd and into if degree of f is even.

An into function can be made onto by redefining the cordomain as the range of f is even.

An into function can be made onto by redefining the codomain as the range of the original function.

If f(x) is periodic with period T then )x(f

1 is also

periodic with same period T.

If f(x) is periodic with period T, )x(f is also periodic with same period T.

Period of x – [x] is 1. Period of algebraic functions x , x2, x3 + 5 etc. does't exist.

If )x(flimax→

does not exist, then we can not remove

this discontinuity. So this become a non-removable discontinuity or essential discontinuity.

If f is continuous at x = c and g is discontinuous at x = c, then

(a) f + g and f – g are discontinuous (b) f.g may be continuous

If f and g are discontinuous at x = c, then f + g, f – g and fg may still be continuous.

Point functions (domain and range consists one value only) is not a continuous function.

If a function is differentiable at a point, then it is continuous also at that point.

i.e., Differentiability ⇒ Continuity, but the converse need not be true.

If a function 'f' is not differentiable but is continuous at x = a, it geometrically implies a sharp corner or kink at x = a.

If f(x) and g(x) both are not differentiable at x = a then the product function f(x).g(x) can still be differentiable at x = a.

If f(x) is differentiable at x = a and g(x) is not differentiable at x = a then the sum function f(x) + g(x) is also not differentiable at x = a.

If f(x) and g(x) both are not differentiable at x = a, then the sum function may be a differentiable function.

Differentiation and Applications of Derivatives :

dxdy is

dxd (y) in which

dxd is simply a symbol of

operation and not 'd' divided by dx. If f´(x0) = ∞, the function is said to have an infinite

derivative at the point x0. In this case the line tangent to the curve of y = f(x) at the point x0 is perpendicular to the x-axis.

Of all rectangles of a given perimeter, the square has the largest area.

All rectangles of a given area, the squares has the least perimeter.

A cone of maximum volume that can be inscribed in

a sphere of a given radius r, is of height 3r4 .

A right circular cylinder of maximum volume that can be inscribed in a square of radius r, is of height

3r2 .

If at any point P(x1, y1) on the curve y = f(x), the tangent makes equal angle with the axes, then at the

point P, ψ = 4π or

43π . Hence, at P tan ψ =

dxdy = ±1

Indefinite Integral : If F1(x) and F2(x) are two antiderivatives of a

function f(x) on an interval [a, b], then the difference between them is a constant.

The signum function has an antiderivative on any interval which does not contain the point x = 0, and does not possess an anti=derivative on any interval which contains the point.

The antiderivative of every odd function is an even function and vice-versa.

CALCULUS Mathematics Fundamentals M

AT

HS


If In = ∫ axn e.x dx, then In = aex axn

– an In–1

If In = ∫ dx)x(log , then In = x log x – x

If In = ∫ xlog1 dx, then

In = log(logx) + logx + )!2.(2)x(log 2

+ )!3.(3)x(log 3

+ ...

If In = ∫ dx)x(log n ; then In = x(logx)n – n.In–1

Successive integration by parts can be performed when one of the functions is xn (n is positive integer) which will be successively differentiated and the other is either of the following sin ax, cos ax, e–ax, (x +a)m which will be successively integrated.

Chain rule :

∫ dxv.u = uv1 – u´v2 + u"v3 – u"'v4 + ....

+ (–1)n – 1un–1vn + (–1)n ∫ dxv.u n

where un stands for nth differential coefficient of u and vn stands for nth integral of v.

∫ axxe sin(bx + c)dx = 22

ax

baxe

+[a sin(bx + c) – b

cos(bx + c)] – 222

ax

)ba(e+

[(a2 – b2)sin (bx + c) – 2ab

cos (bx + c)] + k

∫ axxe cos(bx + c)dx = 22

ax

bae.x+

[a cos(bx + c) – b

sin(bx + c)] – 222

ax

)ba(e+

[(a2 – b2)cos (bx + c) – 2ab

sin (bx + c)] + k

∫ axxe sin(bx + c)dx

= 22

x

b)a(loga

+[(loga)sin(bx + c) – b cos(bx + c)] + k

∫ axxe cos(bx + c)dx

= 22

x

b)a(loga

+[(loga)cos(bx + c) + b sin(bx + c)] + k

∫ ++

xsindxcoscxsinbxcosa dx

= 22 dcbdac

++ x + 22 dc

bcad+− log |c cos x + d sinx| + k.

Reduction formulae for I(n,m) = ∫ xcosxsin

m

n dx is

I(n,m) = 1m

1−

.xcosxsin

1m

1n

−

−

– )1–m()1n( − .I(n–2, m – 2)

Definite Integral and Area Under Curves :

The number f(c) = ∫−

b

adx)x(f

)ab(1 is called the

mean value of the function f(x) on the interval [a, b]. If m and M are the smallest and greatest values of a

function f(x) on an interval [a, b], then m(b – a) ≤

∫b

adx)x(f ≤ M(b – a).

If f2(x) and g2(x) are integrable on [a, b], then

∫b

adx)x(g)x(f ≤

2/1b

a

2 dx)x(f

∫

2/1b

a

2 dx)x(g

∫

Change of variables : If the function f(x) is continuous on [a, b] and the function x = φ(t) is continuously differentiable on the interval [t1, t2] and a = φ(t1), b = φ(t2), then

∫b

adx)x(f = ∫ φφ

2

1

t

tdt)t´()t((f

Let a function f(x, α) be continuous for a ≤ x ≤ b and c ≤ α ≤ d. Then for any α ∈[c, d], if I(α) =

∫ αb

adx),x(f , then I´(α) = ∫ α

b

adx),x´(f , where

I´(α) is the derivative of I(α) w.r.t. α and f´(x, α) is the derivative of f(x, α) w.r.t α, keeping x constant.

∫b

adx)x´(f = (b – a) ∫ +−

1

0dt]at)ab[(f

∫ −++

b

a )xba(f)x(fdx)x(f =

21 (b – a)

The area of the region bounded by y2 = 4ax, x2 = 4by

is 3ab16 sq. unit.

The area of the region bounded by y2 = 4ax and y =

mx is 3

2

m3a8 sq. unit.

The area of the region bounded by y2 = 4ax and its

latus-rectum is 3a8 2

sq. unit.

The area of the region bounded by one arch of sin(ax) or cos (ax) and x-axis is 2/sq. unit.

Area of the ellipse (x2/a2) + (y2/b2) = 1 is πab sq. unit. Area of region bounded by the curve y = sin x, x-axis

and the line x = 0 and x = 2π is 4 sq. unit.


Complex Numbers :

|z| ≥ |Re(z)| ≥ Re(z) and |z| ≥ |1m(z) | ≥ 1m(z)

|z|

z is always a unimodular complex number if z ≠ 0

|Re(z) | + |1m(z) | ≤ 2 |z|

If z1z + = a, the greatest and least values of |z| are

respectively 2

4aa 2 ++ and 2

4aa 2 ++−

|z1 + 22

21 zz − | + |z2 – 2

221 zz − | = |z1 + z2| + |z1 – z2|

If z1 = z2 ⇔ |z1| = |z2| or arg z1 = arg z2 |z1 + z2| = |z1 – z2| ⇔ arg (z1) – arg(z2) = π/2. If |z1| ≤ 1, |z2| ≤ 1 then

(i) |z1 + z2|2 ≤ (|z1| – |z2|)2 + (arg (z1) – arg (z2))2 (ii) |z1 + z2|2 ≥ (|z1| + |z2|)2 – (arg (z1) – arg(z2))2

|z1 + z2|2 = |z1|2 + |z2|2 + 2||z2| cos(θ1 – θ2). |z1 – z2|2 = |z1|2 + |z2|2 – 2|z1||z2| cos(θ1 – θ2). If z1 and z2 are two complex numbers then

|z1 z2| = r1r2; arg(z1z2) = θ1 + θ2 and 2

1

zz =

2

1

rr ,

arg

2

1

zz = θ1 – θ2 and where |z1| = r1, |z2| = r2,

arg(z1) = θ1 and arg(z2) = θ2. The area of the triangle whose vertices are z, iz and

z + iz is 21 |z|2.

The area of the triangle with vertices z, wz and

z + wz is 43 |z2|.

If z1, z2, z3 be the vertices of an equilateral triangle and z0 be the circumcentre, then

21z + 2

2z + 23z = 3 2

0z . If z1, z2, z3 ....... zn be the vertices of a regular

polygon of n sides and z0 be its centroid, then 2

1z + 22z + .......+ 2

nz = n 20z .

If z1, z2, z3 be the vertices of a triangle, then the triangle is equilateral if

(z1 – z2)2 + (z2 – z3)2 + (z3 – z1)2 = 0 or 2

1z + 22z + 2

3z = z1z2 + z2z3 + z3z1

or 21 zz

1−

+ 32 zz

1−

+ 13 zz

1−

= 0

If z1, z2 z3 are the vertices of an isosceles triangle, right angled at z2 then 2

1z + 22z + 2

3z = 2z2(z1 + z2) If z1, z2, z3 are the vertices of a right-angled isosceles

triangle, then (z1 – z2)2 = 2(z1 – z3)(z3 – z2). If z1, z2, z3 be the affixes of the vertices A, B, C

respectively of a triangle ABC, then its orthocentre is

CseccBsecbAseca

z)Csecc(z)B(secbz)A(seca 321

++++

For any a, b ∈ R

(i) iba + + iba − = aba2 22 ++

(ii) iba + – iba − = aba2 22 −+

The sum and product of two complex numbers are real simultaneously if and only if they are conjugate to each other.

If ω and ω2 are the complex cube roots of unity, then (i) (aω + bω2)(aω2 + bω) = a2 + b2 – ab (ii) (a + b (aω + bω2)(aω2 + b2ω) = a3 + b3 (iii) (a + bω + cω2)(a + bω2 + cω) = a2 + b2 + c2 – ab – bc – ca (iv) (a + b + c) (a + bω + cω2) (a + bω2 + cω) = a3 + b3 + c3 – 3abc If three points z1, z2, z3 connected by relation

az1 + bz2 + cz3 = 0 where a + b + c = 0, then the three points are collinear.

If three complex numbers are in A.P. then they lie on a straight line in the complex plane.

Progression : If Tk and Tp of any A.P. are given, then formula for

obtaining Tn is knTT kn

−− =

kpTT kp

−

−.

If pTp = qTq of an A.P., then Tp + q = 0. If pth term of an A.P. is q and the qth term is p, then

Tp+q = 0 and Tn = p + q – n. If the pth term of an A.P. is 1/q and the qth term is 1/p,

then its pqth term is 1.

ALGEBRA Mathematics Fundamentals M

AT

HS


The common difference of an A.P. is given by d = S2 – 2S1 where S2 is the sum of first two terms and S1 is the sum of first term or the first term.

If sum of n terms Sn is given then general term Tn = Sn – Sn–1, where Sn–1 is sum of (n – 1) terms of A.P.

If for an A.P. sum of p term is q and sum of q terms is p, then sum of (p + q) terms is –(p + q).

If for an A.P., sum of p term is equal to sum of q terms, then sum of (p + q) terms is zero.

If the pth term of an A.P. is 1/q and qth term is 1/p,

then sum of pq terms is given by Spq = 21 (pq + 1).

Sum of n A.M.'s between a and b is equal to n times the single A.M. between a and b.

i.e. A1 + A2 + A3 + ...... + An = n

+

2ba .

If Tk and Tp of any G.P. are given, then formula for

obtaining Tn is kn

1

k

n

TT −

=

kp1

k

p

TT −

Product of n G.M.'s between a and b is equal to nth power of single geometric mean between a and b

i.e. G1G2G3 .... Gn = ( ab )n. The product of n geometric means between a and 1/a

is 1.

If n G.M.'s inserted between a and b then r = 1n

1

ab +

Quadratic Equations and Inequations : An equation of degree n has n roots, real or

imaginary. If f(α) = 0 and f´(α) = 0, then α is a repeated root of

the quadratic equation f(x) = 0 and f(x) = a(x – α)2. In fact α = –b/2a.

If α is repeated common root of two quadratic equations f(x) = 0 and φ(x) = 0, then α is also a common root of the equations f´(x) = 0 and φ´(x) = 0.

In the equation ax2 + bx + = 0 [a, b, c ∈R], if a + b + c = 0 then the roots are 1, c/a and if a – b + c = 0, then the roots are –1 and – c/a.

If one root of the quadratic equation ax2 + bx + c = 0 is equal to the nth power of the other, then

( ) 1n1

nac + + ( ) 1n1

nca + + b = 0. If one root is k times the other root of the quadratic

equation ax2 + bx + c = 0, then k

)1k( 2+ =acb2

.

If an equation has only one change of sign, it has one +ve root and no more.

Permutations and Combinations : nC0 = nCn = 1, nC1 = n nCr + nCr–1 = n+1Cr nCx = nCy ⇔ x = y or x + y = n n. n-1Cr–1 = (n – r + 1)nCr–1 If n is even then the greatest value of nCr is nCn/2.

If n is odd then the greatest value of nCr is 2

C 1nn

+ or

2C 1n

n− .

nCr = rn .n–1Cr–1.

Number of selection of zero or more things out of n different things is, nC0 + nC1 + bC2 + ... + nCn = 2n.

nC0 + nC2 + nC4 + .... = nC1 + nC3 + nC5 + .... = 2n–1. Gap method : Suppose 5 moles A, B, C, D, E are

arranged in a row as × A × B × C × D × E ×. There will be six gaps between these five. Four in between and two at either end. Now if three females P, Q, R are to be arranged so that no two are together we shall use gap method i.e., arrange them in between these 6 gaps. Hence the answer will be 6P3.

Together : Suppose we have to arrange 5 persons in a row which can be done in 5! = 120 ways. But if two particular persons are to be together always, then we tie these two particular persons with a string. Thus we have 5 – 2 + 1 (1 corresponding to these two together) = 3 + 1 = 4 units, which can be arranged in 4! ways. Now we loosen the string and these two particular can be arranged in 2! ways. Thus total arrangements = 24 × 2 = 48.

If we are given n different digits (a, a2, a3 ..... an) then sum of the digits in the unit place of all numbers formed without repetition is (n – 1)!(a1 + a2 + a3 + .... + an). Sum of the total numbers in this case can be obtatined by applying the formula (n – 1)!(a1 + a2 + a3 + ..... + an). (1111 ......... n times).

Binomial Theorem & Mathematical Induction : The number of terms in the expansion of (x + y)n are

(n + 1). If the coefficients of pth, qth terms in the expansion of

(1 – x)n are equal, then p + q = n + 2. For finding the greatest term in the expansion of

(x + y)n. we rewrite the expansion in this form

(x + y)n = xnn

xy1

+ . Greatest term in (x + y)n = xn.

Greatest term in n

xy1

+ .

There are infinite number of terms in the expansion of (1 +x)n, when n is a negative integer or a fraction.

The number of term in the expansion of (x1 + x2 + .... + x2)n = n+r-1Cr–1.


Learning the MIghTy way

When you hear... you forget. When you see... you remember. When you do... you understand. – Confucius

This sounds very much the way we learnt things in our childhood. Our parents did not tell us how to walk, but made us walk. They did not show us how to ride a bicycle, they made us ride it. And that’s how learning goes naturally: deriving inspiration by doing things, ‘Inspired Learning’.

In this era of information overload, educational institutions hardly devote time and resources to make students follow this natural path of learning. An exception to this is Manipal Institute of Technology (MIT), where the ‘been there, done that’ spirit has helped students make a mark worldwide.

While in other colleges, students were seeing how a car is built, students at MIT designed and manufactured a high performance car which made its way into the ‘Formula Student 2009’ (Student edition of Formula One), at Silverstone, UK. MIT got 65th position among teams from 126 universities of 23 countries. Several other feathers of innovation have been added to MIT’s achievement hat within the last 2 years like: 1st position in “Train Blazer- 2008”; 1st position in “GE Edison Innovation Challenge -2008”; 1st and 2nd position in “Schneider Electric Innovation Challenge” 2008; 1st prize for an Innovative proposal on “Solar Power embedded Street Lamp” and many more.

The foundation of this hub of thinking and innovation was laid 52 years ago, and since then is being strengthened by constructive effort and hard work. The quality of technical education is evident in the research activities going on in the campus: 6 patents filed in the current year, 9 grants received for various activities and, several papers, books and awards added to its research portfolio. A major landmark is ‘Manipal University Technology Business Incubator’, which is being established with a huge funding from the Department of Science and Technology, Government of India.

While focusing on innovation, MIT has taken care of students by getting them absorbed into the best industries, through quality industry projects and collaborations with universities and companies. ‘Practice School’ is one such concept, in sync with the idea of ‘Inspired learning’. This was introduced at MIT in 2005, through which students are trained to effectively link the theory learnt in classrooms with the practice in the industry. This training helps reduce the cost for a company, on in-house training of students who are their prospective employees. Due to such efforts placement statistics have been moving up since 2005 reaching 95% in 2008. Even in the recession hit 2008-09, 56 companies turned up offering jobs to 948 students, with the best remuneration reaching 10.75 lakhs per annum. MIT now stands 7th among the private Engineering colleges in India, and 22nd overall.

A student, Gaurav Sinha (4th year, Electrical and Electronics) quotes “MIT has provided me with excellent opportunities to develop to my full potential. The environment is conducive for our growth and along with the state-of-art infrastructure facilities available helps bring out the best in each of us. The platform and exposure that I got at MIT has groomed my personality and prepared me to face any challenge in future”. Already having produced strong personalities of industry like Rajeev Chandrasekhar (Chairman & CEO of Jupiter Capital; FICCI President; Founder & former-CEO of BPL mobile), Ravi Bapna (Executive Director of CITNE & Professor, Indian School of Business) , Amit Behl (Director, Intel India), etc. the excellent infrastructure coupled with innovative learning and course content, holds promise for a much brighter future for the country.

– MIT, Manipal

XtraEdge for IIT-JEE APRIL 2010 51


CHEMISTRY

SECTION – I Straight Objective Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. A bis-aldol dimerization of 1-phenyl-1,2-propanedione (C6H5COCOCH3) gives which of the following ?

(A)

O

O

C6H5

C6H5 (B)

OC6H5

C6H5

(C) O C6H5

C6H5 (D)

OC6H5

O C6H5

2. An alkene (A), C16H16 on ozonolysis gives only one product B(C8H8O). Compound (B) on reaction with NH2OH, H2SO4, ∆ gives N-methyl benzamide. The compound 'A', is –

(A) CH3 C = C

H CH3 H

(B)

C = C CH3 CH3

(C) CH2–CH = CH–CH2

(D) CH = CH

CH3

CH3

3. Identify product D in the following reaction sequence:

H3C–

3

3

CH|

—C|

CH

CH2CH2OHHeat,OH

H,OCrK2

722 →+

A →2SOCl B

→NH)CH( 23 COH.II

ether,LiAlH.I

2

4 → D

(A) H3C–

3

3

CH|

—C|

CH

CH2C ≡ N

(B) H3C–

3

3

CH|

—C|

CH

CH2 23

23

)CH(CHN|

)CH(N

(C) H3C –

3

3

CH|

—C|

CH

CH2 C||O

N(CH3)2

(D) H3C–

3

3

CH|

—C|

CH

CH2CH2N(CH3)2

MOCK TEST FOR IIT-JEE

PAPER - I

Time : 3 Hours Total Marks : 240

Instructions : • This question paper contains 60 questions in Chemistry (20), Mathematics (20) & Physics (20). • In section -I (8 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (4 Ques) of each paper +4 marks will be given for correct answer –1 mark for wrong answer. • In section -III contains 2 groups of questions (2 × 3 = 6 Ques.) of each paper +4 marks will be given for each

correct answer & –1 mark for wrong answer. • In section -IV (2 Ques.) of each paper +8(2×4) marks will be given for correct answer & No Negative marking for

wrong answer.


4. Consider the following reactions : Reaction I :

H3C

O

O

O

CF3 1

23r

OHCHCH → H3C

O

O +

F3C

O

OH

Reaction II :

H3C

O

O

O

CF3 2

23r

OHCHCH →

H3C

O

OH +

F3C

O

O Which of the following is a correct comparison of the

rate of reactions I and II ? (A) rI = rII (B) r1 > rII (C) r1 < rII (D) None

5. A container containing HCl(g) is fitted with a funnel and a long capillary tube as shown below. Capillary is immersed in the water. When few drops of water is introduced into

the container through the funnel then – A

HCl(g)

(A) there is a pressure drop in the container and liquid

level in the capillary rises (B) there is a pressure drop in the container because

HCl(g) effuse out more rapidly than the air effuse in

(C) HCl(g) undergoes spontaneous dissociation to H2(g) & Cl2(g), hence number of moles decreases, resulting pressure drop inside the container. Water level in the capillary rises

(D) HCl(g) spontaneously mixed with water through the capillary therefore, water level in the capillary remain same

6. As per Boyle's law V ∝ 1/P at constant temperature,

As per charles law V ∝ T at constant pressure. Therefore, by combining, one concluded that T ∝ 1/P hence, PT = constant

(A) PT = constant is correct, because volume remain same in both the laws

(B) PT = constant is incorrect, because volume remain same at the constant temperature and at the constant pressure

(C) PT = constant is correct, because volume at constant temperaute and volume at constant pressure are not same

(D) PT = constant is incorrect, because volume at constant temperature and volume at constant pressure for the same amount of gas are different

7. Which of the following statements is correct ? (A) (n – 1) d subshell has lower energy than ns

subshell (B) (n – 1) d subshell has higher energy than ns

subshell (C) (n + 1) d subshell has lower energy than nf

subshell (D) nf subshell has lower energy than (n + 2) s

subshell

8. (A), (B) and (C) are elements in the third short period. Oxide of (A) is ionic , that of (B) is amphoteric and of (C) a giant molecule. (A), (B) and (C) have atomic numbers in the order of -

(A) (A) < (B) < (C) (B) (C) < (B) < (A) (C) (A) < (C) < (B) (D) (B) < (A) < (C)

SECTION – II Multiple Correct Answers Type

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

9.

HH

+ Ph3C⊕BF4– →

⊕

H

BF4– + Ph3 CH

which of the following statements is/are correct -

(A) the cation in reactant side is approximately

1011 times more stable than product side

(B) cation in reaction side is non planer

(C) it is acid base reaction

(D) reaction must be exothermic


10. Which of the following is/are correct ? (A) The efficiency of a solid catalyst depends

upon its surface area (B) Catalyst operates by providing alternate path

for the reaction that involves lower activation energy

(C) Catalyst lowers the activation energy of forward reaction only without affecting the activation energy of backward reaction

(D) Catalyst does not affect the overall enthalpy change of the reaction

11. For the three elements P, Q & R, ionization

enthalpy (IE) and electron gain enthalpies (∆eg H) are given in the following table -

Element IE in kJ/mol ∆eg H in kJ/mol P 1680 –340 Q 1100 –120 R 500 –20

(A) P is the highest electronegative element among P, Q and R

(B) R is the least electronegative among P, Q & R (C) Electro negativity of P is approximately equal to 4 (D) R may be chlorine 12. Which of the following is/are correct ?

(A) For the incompressible liquid TdP

dH

is

approximately equal to volume of liquid

(B) For ideal gas TdP

dH

is equal to zero

(C) For real gas if TdV

dE

= 0 then not

necessarily TdP

dH

is equal to zero

(D) None of these

SECTION – III

Comprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.

Paragraph # 1 (Ques. 13 to 15) A pleasant smelling optically active compound,

monoester 'F' has molecular weight 186. It doesn't react with Br2 in CCl4. Hydrolysis of 'F' gives two optically active compounds 'G', which is soluble in NaOH and 'H'. H gives a positive iodoform test, but on warming with conc. H2SO4 gives I with no disastereomers. When the Ag+ salt of 'G' is reacted with Br2, racemic 'J' is formed. Optically active J is formed when 'H' is treated with tosyl chloride (TsCl), and then with NaBr.

13. The pleasant smelling optically active compound, F is -

(A) (CH3)2CH–

3CH|

–CHC||O

O–

3CH|

–CH CH(CH3)2

(B) (CH3)3C–CH2 C||O

–O–

3CH|

–CH CH(CH3)2

(C) CH3CH2CH2

3CH|

–CH C||O

–O–

3CH|

–CH CH2CH2CH3

(D) CH3CH2

3CH|

–CH CH2– 2COCH||O

–

3

2

CH|

CHCH –CH3

14. How would be the structure of F if I exists as

diastereomers ?

(A) (CH3)2CH

3CH|

CHCO||O

3

23

CH|

)CH(CHCH

(B) (CH3)3CCH2 C||O

O

3

23

CH|

)CH(CHCH

(C) CH3CH2CH2

3CH|CHCO

||O

3

322

CH|

CHCHCHCH

(D) CH3CH2

3

2

HC|

CHCH C||O

OCH2

3

32

CH|

CHCHCH


15. What would be the structure of F if H gives negative idoform test ?

(A) (CH3)2

3

2

CH|

COCHCH||O

3

23

CH|

)CH(CHCH

(B) CH3(CH2)3CH2

3CH|

COCHCH||O

CH(CH3)2

(C) CH3CH2CH2

33

2

CHCH||

CHCOCHCH||O

CH2CH3

(D) CH3CH2

3

2

CH|

COCHCH||O

CH2

3

2

CH|

CHCH CH3

Paragraph # 2 (Ques. 16 to 18) To the 100 ml of 10–2 (M) aqueous solution of HCl

0.1 (M) HA (Ka = 10–2) is added in such a way so that the final pH of the solution become 1.7

Given log 2 = 0.3 16. What was the pH of 10–2 (M) aqueous solution of

HCl ? (A) pH = 2 (B) pH < 2 (C) pH > 2 (D) pH = 1.7 17. What volume of 0.1 (M) HA was required to add in

aqueous HCl to reduced the final pH equal to 1.7 ? (A) 175 ml (B) 100 ml (C) 104 ml (D) 75 ml 18. Which of the following solution is isohydric with

0.1(M) aqueous solution of HA ? (A) 0.01(M) aqueous solution of HB (Ka = 10–2) (B) 0.01(M) aqueous solution of HC (Ka=2×10–1) (C) 0.01(M) aqueous solution of HNO3 (D) 1(M) aqueous solution of HD (Ka = 10–3)

SECTION – IV

Matrix – Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example :

If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.

q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

19. Match the following : Column-I Column-II (A)CH3CH2CH2NH2 (p)Treatment of NaNO2, HCl gives N-nitroso compound (B) CH3CH2NHCH3 (q)Treatment of NaNO2, HCl gives diazonium chloride

(C)CH3–

3CH|N − CH3 (r) Treatment of

CH3I (excess) followed by AgOH; heat gives out alkene

(D) NH2 (s) Treatment of HCl,

heat gives dealkylation (t) Treatment with CHCl3/

Alc.KOH gives isocyanide 20. Match the following : Column-I Column-II

(A) C6H5CH2CHO →− )eq1(Br,OH 2 (p) Redox products

(B) CH3CHO )Cº50(

OH

− →

− (q) Enantiomeric

products

(C) CH3CH2–CH = O )Cº25(

OH →−

(r) H

DKK

α

α → < 1(Primary

isotopic effect)

(D) CH3–

3CH|

–CH

O||

H–C →−OH.Conc (s) Diastereomeric

products

(t) Disproportionation

MATHEMATICS




1. Let F denote the set of all onto functions from A = a1, a2, a3, a4 to B = x, y, z. A function f is chosen at random from F. The probability that f –1 (x) consists of exactly two elements is

(A) 2/3 (B) 1/3 (C) 1/6 (D) 0

2. A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters TA are visible . The probability that the letter has come from CALCUTTA is

(A) 4/11 (B) 1/3 (C) 5/12 (D) None

3. A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelops at random, the probability that not all letters are placed in correct envelopes is

(A) 1/24 (B) 11/24 (C) 5/8 (D) 23/24

4. The value of x for which the matrix

A =

− 121010702

is inverse of

B =

−−

−

x2x4x010x7x14x

is

(A) 21 (B)

31 (C)

41 (D)

51

5. The largest term in the expansion of (3 + 2x)50, where x = 1/5, is

(A) 5th (B) 6th (C) 8th (D) 9th

6. If z2z − = 2, then the greatest value of | z | is

(A) 1 + 2 (B) 2 + 2

(C) 3 +1 (D) 5 + 1

7. The value of ∫ −−16

1

1 dx1xtan is

(A) 3

16π + 32 (B) π34 – 32

(C) π34 + 32 (D) π

316 – 32

8. The solution of (y + x + 5)dy = (y – x + 1) dx is

(A) log ((y + 3)2 + (x + 2)2) + tan–1 2x3y

++ = C

(B) log ((y + 3)2 + (x – 2)2) + tan–1 2x3y

−− = C

(C) log ((y + 3)2 + (x + 2)2) + 2 tan–1 2x3y

++ = C

(D) log ((y + 3)2 + (x + 2)2) – 2 tan–1 2x3y

++ = C

SECTION – II

Multiple Correct Answers Type

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 9. If n > 1, then (1 + x)n – nx – 1 is divisible by - (A) x2 (B) x3 (C) x4 (D) x5

10. If y = 1A4sin11A4sin1

−+

+− , then one of the values of y is -

(A) – tan A (B) cot A

(C) tan

+

π A4

(D) – cot

+

π A4

11. Equation of a common tangent to the circles x2 + y2 – 6x = 0 and x2 + y2 + 2x = 0 is - (A) x = 1 (B) x = 0 (C) x + y3 + 3 = 0 (D) x – y3 + 3 = 0

12. The points of extremum of ∫ ++−2x

0 t

2

e24t5t are -

(A) x = – 2 (B) x = 1 (C) x = 0 (D) x = – 1

SECTION – III Comprehension Type


Paragraph # 1 (Ques. 13 to 15) Consider the region S0 which is enclosed by the curve

y ≥ 2x1− and max. |x|, |y| ≤ 4. If slope of a family of lines is defined as m(t) = cos t where point (t, 2t + 0.4) lies inside the region S0. Any member of this family of lines is called L1= 0 if it passes through (π, max t) and L2 = 0 if it passes through the (π, minm t).

13. Area of region of S0 is - (A) 8 + π/2 sq. units (B) 8 – π/2 sq. units (C) 8 sq. units (D) 8 + π sq. units


14. If line lies inside the region S0, then - (A) t ∈ (0, 0.28) (B) t ∈ (0, 1) (C) t ∈ (0.28, 1) (D) t ∈ [.28, 1]

15. L1 = 0 having maximum slope, is -

(A) π−

−x

1y = cos 1 (B) π−

−x

1y = cos (0.28)

(C) 1x

y−

π− = cos (1) (D) None of these

Paragraph # 2 (Ques. 16 to 18) Two circles S1 = 0 and S2 = 0 are touching to each

other externally at point T, with centre C1, C2 and radii r1 and r2 respectively.

If P and Q be the points of contact of a direct common tangent to the two circles and PQ meets the line joining C1, C2 in S. Tangent at common point T is intersecting to the tangent PQ at R point and to other direct tangent at V point. Let S1= x2 + y2 – 6x = 0 and S2= x2 + y2 + 2x = 0.

16. Angle between the two direct tangents is– (A) 90º (B) 30º (C) 60º (D) None of these

17. Direct tangents are–

(A) y = 3 x + 3 , y = – 3 x + 3

(B) y = 3

x – 3 , y = 3x− – 3

(C) y = 3

x + 3 , y = 3x− – 3

(D) None of these 18. A circle S = 0 of radius 1 units rolls on the outside of

the circle S2 = 0, touching it externally, locus of the centre of this outer circle is –

(A) Circle (B) Ellipse (C) Parabola (D) None of these

SECTION – IV

Matrix – Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.

q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

19. Match the following-:

Column- I Column- II

(A) The sum of the series

∑=

10

0rr

20 C is

(p) – 10C5

(B) The coefficient of x53 in

∑=

−−100

0r

rr100r

100 2)3x(C is

(q) 100C4

(C) (10C0)2 –(10C1)2 +……. …..–(10C9)2 + (10C10)2 equals

(r) 219+21 20C10

(D) The value of

95C4 + ∑=

−5

1j3

j100 C is

(s) – 100C53

(t) 100C47

20. Match the following -

Column -I

Column -II

(A) If the lines 1

2x − = 1

3y − = λ− 4z

and λ−1x =

24y − =

15z −

intersect at (α, β, γ) then λ =

(p) 0

(B) If ∞→x

lim 4x

++

−π −

2x1xtan

41 =

y2 + 4y + 5 then y =

(q) –1

(C) If chord x + y + 1= 0 of parabola y2 = ax subtends 90º at (0, 0) then a =

(r) –2

(D) If a = i + j + k , a . b = 1

and a × b = j – k , then | b | is equal to

(s) 1

(t) –3


PHYSICS



1. ABCD is a smooth horizontal fixed plane on which mass m1 = 0.1 kg is moving in a circular path of radius r = 1 m. It is connected by an ideal string which is passing through a smooth hole and connects

of mass m2 = 1/ 2 kg at the other end as shown. m2 also moves in a horizontal circle of same radius of 1

m with a speed of 10 m/s. If g = 10 m/s2 then the speed of m1 is-

m2

D

m1

A B

C

(A) 10 m/s (B) 10 m/s

(C) 101 m/s (D) None of these

2. A L shaped rod whose one rod is horizontal and other is vertical is rotating about a vertical axis as shown with angular speed ω. The sleeve shown in figure has mass m and friction coefficient between rod and sleeve is µ. The minimum angular speed ω for which sleeve cannot sleep on rod is –

m

ω

sleeve

l

(A) lµ

=ωg (B)

l

gµ=ω

(C) gµ

=ωl (D) None of these

3. Two solid spherical balls of radius r1 & r2

(r2 < r1), of density σ are tied up with a string and released in a viscous liquid of lesser density ρ and coefficient of viscosity η, with the string just taut as shown. The terminal velocity of spheres is -

r1

r2

(A) )(gr92 2

2 ρ−ση

(B) )(gr92 2

1 ρ−ση

(C) ηρ−σ

++ g)(

rr)rr(

92

21

32

31 (D)

ηρ−σ

−− g)(

rr)rr(

92

21

32

31

4. A block of mass m is attached to an ideal spring and system lies in vertical plane as shown. Initially the supporting plane is placed so that spring remains in its natural length then the plane is moved very slowly downwards. The graph showing variation of normal reaction applied by mass on supporting plane with distance travelled by block is –

Supporting plane

M

(A)

mg

x

N

(B)

mg x

N

(C)

mg

x

N

(D) None of these

5. A massless container is filled with liquid of density ρ. It contains two holes as shown in figure. Container rests on ground. Area of the two holes are A each. Container is filled with liquid upto height H. Then –

H3H/4H/4

(A)Torque produced by normal force between container & ground about center of gravity is

2AgH2ρ into the plane of paper

(B) Torque produced by friction about center of gravity is ρAgH2 out of the plane of paper

(C) Net torque produce by thrust force & friction

force about center of gravity is 4

AgH2ρ into the

plane of paper (D) Torque produced by normal force between

container and ground about centre of gravity is zero


6.

A B

Two containers A & B contain ideal gases helium and oxygen respectively. Volume of both containers are equal and pressure is also equal. Container A has twice the number of molecules than container B then if vA & vB represent the rms speed of gases in containers A & B respectively, then -

(A) B

A

vv = 2 (B)

B

A

vv = 4

(C) B

A

vv = 2 (D)

B

A

vv = 8

7. A capacitor is composed of three parallel conducting plates. All three plates are of same area A. The first pair of plates are kept a distance d1 apart and the space between them is filled with a medium of a dielectric ε1. The corresponding data for the second pair are d2 & ε2 respectively. What is the surface charge density on the middle plate ?

d1 d2

ε1 ε2

V0

(A)

ε+

εε

2

2

1

10 ddV (B)

ε+

εε−

2

2

1

10 ddV

(C)

ε+

εε

2

2

1

10 ddV2 (D)

ε+

εε−

2

2

1

10 ddV2

8. The mirror of length 2l makes 10 revolutions per minute about the axis crossing its mid point O and perpendicular to the plane of the figure There is a light source in point A and an observer at point B of the circle of radius R drawn around centre O (∠AOB = 90º)

What is the proportion l

R if the observer B first sees

the light source when the angle of mirror ψ = 15º ?

A

R Bl

l

O

ψ

(A) 2 (B)2

1 (C) 22 (D) 22

1

SECTION – II



9. A body moves in a circular path of radius R with deceleration so that at any moment of time its tangential and normal acceleration are equal in magnitude. At the initial moment t = 0, the velocity of body is v0 then the velocity of body at any time will be –

(A) v =

+

Rtv1

v

0

0 at time t

(B) v = RS

0ev−

after it has moved S meter

(C) v = v0e–SR after it has moved S meter (D) None of these

10. A cylinder block of length L = 1m is in two

immiscible liquids. Part of block inside liquid(1) is

41 m and in liquid (2) is

41 m. Area of cross-section

of block is A. Densities of liquid (1) & (2) are ρ and 2ρ respectively –

Liquid (1) ρ.

2ρ. Liquid (2)


(A) Density of block is 3ρ/4

(B) Force exerted by liquid (1) on block is ρAg/4 (C) Block is depressed so that it is just completely

immersed in liquid (1) and released. A initial acceleration of block is 4/3 g

(D) In case (C) force exerted by liquid (2) on block is 3/2 ρAg

11. R = 10Ω & E = 13 V and voltmeter & Ammeter are ideal then -

V

A

c 6V

b 3Ω

8V

a

R

E

(A) Reading of Ammeter is 2.4 A (B) Reading of Ammeter is 8.4 A (C) Reading of voltmeter is 8.4 V (D) Reading of voltmeter is 27 V 12. A parallel plate air capacitor is connected to a

battery. If plates of the capacitor are pulled further apart, then which of the following statements are correct -

(A) Strength of electric field inside the capacitor remain unchanged, if battery is disconnected before pulling the plate.

(B) During the process, work is done by an external force applied to pull the plates whether battery is disconnected or it remain connected.

(C) Potential energy in the capacitor decreases if the battery remains connected during pulling plates apart.

(D) None of the above

SECTION – III Comprehension Type


Paragraph # 1 (Ques. 13 to 15) In the shown arrangement, both springs are relaxed.

The coefficient of friction between m2 and m1 is µ. There is no friction between m1 and surface. If the blocks are displaced slightly they perform SHM together

m2 m1

k2 k1

13. If the small displacement of blocks is x then

acceleration of m2 is-

(A) 2

2

mxk

(B) 2

21

mx)kk( +

(C) xmmkk

21

21

++ (D) None of these

14. The condition in which frictional force on m2 acts in the direction of its displacement from mean position is –

(A) 2

1

2

1

kk

mm

> (B) 2

1

1

2

kk

mm

>

(C) 2

1

2

1

kk

mm

= (D) None of these

15. If the condition obtained in Q.15 is met, then the maximum amplitude of oscillation is –

(A) 1221

212

kmkm)mm(gm

−+µ (B)

2211

212

kmkm)mm(gm

−+µ

(C) 1221

212

kmkm)mm(gm

++µ (D) None of these

Paragraph # 2 (Ques. 16 to 18) A conducting rod PQ of mass M rotates without

friction on a horizontal plane about Ο on circular rails of diameter 'l'. The centre O and the periphery are connected by resistance R. The system is located in a uniform magnetic field perpendicular to the plane of the loop. At t = 0, PQ starts rotating clockwise with angular velocity ω0. Neglect the resistance of the rails and rod, as well as self inductance.


B

O ω0

Q

R P

⊗

16. Magnitude of current as a function of time

(A) t2

0 eR2

B α−ω l (B) t22

0 eR16

B α−ω l

(C) t2

0 eR8

B α−ω l (D) t22

0 eR8

B α−ω l

Where α = RM8B3 22l

17. Total charge flow through resistance till rod PQ stop rotating .

(A) B8M0ω (B)

B3M0ω (C)

B6M0ω (D)

B9M0ω

18. Heat generated in the circuit by t = ∞

(A) 24

M 20

2ωl (B) 8

M 20

2ωl

(C) 3

M 20

2ωl (D) 32

M 20

2ωl

SECTION – IV Matrix – Match Type


q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

19. A uniform solid cube is floating in a liquid as shown in the figure, with part x inside the liquid. Some changes in parameters are mentioned in Column I. Assuming no other changes, match the following -

x

Column I Column II (A) If density of the liquid (p) Increase decreases, x will (B) If height of the cube is (q) Decreases increased keeping base area and density same, x will (C) If the whole system is (r) Remain same accelerated upward, then x will (D) If the cube is replaced (s) May increase by another cube of same or decrease size but lesser density, x will (t) none 20. A block of mass m = 1 kg is at rest with respect to a

rough wedge as shown in figure.

µm

a

θ The wedge starts moving up from rest with an

acceleration of a = 2m/s2 and the block remains at res with respect to wedge then in 4 sec. of motion of wedge work done on block (assume angle of inclination of wedge is θ = 30º and g = 10 m/s2) –

Column I Column II

(A) By gravity (p) 144 J (in magnitude)

(B) By normal reaction (q) 32 J

(C) By friction (r) 160 J

(D) By all the forces (s) 48 J

(t) none


CHEMISTRY



1.

NH2

NOBr Product

The main product is –

(A)

Br

+ Enantiomer

(B)

Br

+ Enantiomer

(C) Br

+ Enantiomer

(D) NO

+ Enantiomer

2. There are two isomeric carboxylic acids– 'A' and 'B'

C9H8O2. reacts with H2/Pd giving compounds, C9H10O2. 'A' gives a resolvable product and 'B' gives a non-resolvable product. Both isomers could by oxidised to PhCOOH.

The correct structures of 'A' and 'B' are, respectively–

(A) CH2=CH COOH ;

CH=CHCOOH

(B) COOHCH2=C ; CH=CH–COOH

(C) CH=CHCOOH ; CH2=CH COOH

(D) CH=CH–COOH ;

COOHC=CH2

3. If H-He undergoes dissociation, which of the follwoing product most expected to occur –

2HHe products (A) He2 + 2H (B) H2 + 2He (C) D2 + He2 (D) 2H + 2He

4. 100 ml solution (I) of buffer containing 0.1(M) HA and 0.2 (M) A–, is mixed with another solution (II) of 100 ml containing 0.2(M) HA and 0.3(M) A– .After mixing what is the pH of resulting solution ?( Given pKa of HA = 5)

(A) 5 – log 5/3 (B) 5 + log 5/3 (C) 5 + log 2/5 (D) 5 – log 5/2



MOCK TEST FOR IIT-JEE

PAPER - II


Instructions : • This question paper contains 57 questions in Chemistry (19,) Mathematics (19) & Physics (19). • In section -I (4 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (5 Ques) of each paper +4 marks will be given for correct answer & –1 mark for wrong answer • In section -III (2 Ques.) of each paper +8(2×4) marks will be given for correct answer. No Negative marking for

wrong answer. • In section -IV (8 Ques.) of each paper +4 marks will be given for correct answer & –1 mark for wrong answer.


5. Dopamine of a drug used in the treatment of parkinson's disease.

CH2–CH

Dopamin

NH2 COOH HO

HO

Which of the following statements about this compound are correct ?

(A) It can exist in optically active forms. (B) One mole will react with three moles of sodium

hydroxide to form a salt (C) It can exist as a Zwitter ion in the aqueous

solution (D) It gives nitroso compound on treatment with

HNO2.

6. Which of the following method(s) would be useful for preparing ketones ?

(A) Friedel-Crafts reaction of an acyl chloride with benzene (AlCl3 catalyst)

(B) Reaction of methyllithium with the lithium salt of carboxylic acid, followed by hydrolysis

(C) Reaction of R2CuLi with an acyl choride in ether at low temperature

(D) Reaction of Grignard reagents with acyl chloride in ether followed by hydroylsis

7. In the given table types of H bonds and some H bond energies are given and other H bond energies are not given. You are to perdit the unknown H-bond energies.

Types of H-bonds H-bond energies in KJ/mol (I) F – H …….. O – F – H …….. F 30 (II) O – H …….. O – O – H …….. F 15 (III) F – H …….. F– – (IV) N – H …….. N – Correct prediction are – (A) H-bond energy for (I) may be 20 kJ/mol (B) H-bond energy for (II) may be 25 kJ/mol (C) H-bond energy for (III) may be 113 kJ/mol (D) H-bond energy for (IV) may be 12 kJ/mol

8. During conductance measurement of an electrolyte based on the wheatstone bridge principle alternating current is used because direct current produces –

(A) polymerisation (B) Ionisation (C) Electrolysis resulting in the chnge of

concentration and in consequencae the resistance (D) Polarisation at the electrodes resulting in the

change of resistance

9. Consider the cell : Ag | AgCl (s) | KCl (1M) | Hg2Cl2(s) | Hg (l) | Pt. The

cell potential : (A) increases on increasing concentration of Cl– ions (B) decreases on decreasing concentration of Cl– ions (C) is independent of concentration of Cl– ions

(D) is independent of amounts of AgCl (s) and Hg2Cl2 (s)

SECTION – III

Matrix - Match Type


q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

10. Match the following Column I Column II (A) HCOOH (p) Decarboxylation on heating (B) CH3COOH (q) Reation with Br2 (C) OH

COOH (r) Cu2+(alkaline)→Cu2O

(D) PhCH2COOH (s) Decarbonylation or decarboxylation on treatment with conc. H2SO4 (t) Reaction with I2+NaOH

11. Match the following Column I Column II (A) XeO4 (p) Non polar (B) XeF4 (q) Having no lone pair of electron in the central atom (C) SO3 (r) Planar (D) BF3 (s) Having lone pair of electrons in the central atom (t) Tetrahedral


SECTION – IV Integer answer type

This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. 3 ampere current was passed through an aqueous

solution of an unknown salt of Pd for one hour, 2.977g of Pdn+ was deposited at cathode. Find n. (At. wt. of Pd = 106.4)

13. Consider the following reaction sequence

O glycol/KOH

NHNH 22 → W →OH/O 23 X

∆ → 2)OH(Ca

Y →COOOHHC 56 Z

How many carbons are present in the final product Z? 14. For the reaction in the plant cells 6CO2(g) + 12H2O (l) → C6H12O6(s) + 6O2(g) + 6H2O(l) ∆rGº = 3000 kJ/mol ATP → ADP + PO4

3– ∆rGº = – 30 kJ/mol Glucose is stored in the plant cell as starch,

(C6H10O5)n. To produce 162 g of starch how many moles of ATP are minimum required ? Give your answer after divide actual answer by 100.

15. The relative lowering of the vapour pressure of an

aqueous solution containing a non-volatile solute is 0.0125. What is the molality of the solution. Give your answer after multiplying actual answer by 10.

16. A polyvalent metal weighing 0.1g and having atomic

weight 51 reacted with dil. H2SO4 to give 43.9 ml of hydrogen at STP. The solution containing the metal in this lower oxidation state(X), was found to require 58.8 mL of 0.1 N KMnO4 for complete oxidation. What is the higher oxidation state (Y) of the metal ?

17. The stopcock, connecting the two bulbs of volumes 5 litres and 10 litres containing an ideal gas at 9 atm and 6 atm respectively, is opened. What is the final pressure (in atm) in the two bulbs if the temperature remained the same ?

18. Sum of lone pairs present in XeOF4 and XeO3 is ….. . 19. Total number of isomers are possible in [Co(en)2Cl2]

and [Co(C2O4)2(NH3)2]– is.

MATHEMATICS



1. Let ∑=

=n

1r

4 )n(fr , then ∑=

−n

1r

4)1r2( is equal to

(A) f(2n) – 16 f(n) (B) f (2n) – 7f (n) (C) f(2n – 1) – 8 f(n) (D) None of these

2. If fr(α) =

α

+α

22 rsini

rcos ×

α

+α

22 r2sini

r2cos ….

α

+α

rsini

rcos

then nnflim

∞→(π) equals

(A) –1 (B) 1 (C) – i (D) i

3. If α, β, γ, δ are four complex numbers such that δγ is

real and αδ – βγ ≠ 0, then z = tt

δ+γβ+α ,

t ∈ R represents a (A) circle (B) parabola (C) ellipse (D) straight line 4. The inflection points on the graph of function

y = ∫ −−x

0

2 dt)2t)(1t( are

(A) x = –1 (B) x = 3/2 (C) x = 4/3 (D) x =1




5. If A and B are two invertible matrices of the same order, then adj (AB) is equal to -

(A) adj (B) adj (A) (B) |B| |A| B–1A–1 (C) |B| |A| A–1B–1 (D) |A| |B| (AB)–1

6. If A (α , β) =

αα−αα

βe000cossin0sincos

, then

(A) A (α, β)′ = A (– α, β) (B) A (α, β)–1 = A (–α, –β) (C) Adj (A (α, β)) = e–β A (–α, –β) (D) A (α, β)′ = A (α, – β) 7. For a positive integer n, if the expansion of

n4

2 xx5

+ has a term independent of x, then n can

be (A) 18 (B) 21 (C) 27 (D) 99 8. If k is odd then kCr is maximum for r equal to

(A) 21 (k – 1) (B)

21 (k + 1)

(C)k – 1 (D) k 9. Let an =

43421timesn

)1...111( , then

(A) a912 is not prime (B) a951 is not prime (C) a480 is not prime (D) a91 is not prime

SECTION – III Matrix - Match Type


q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

10. Match the following Column-1 Column-II Equation General Solutions

(A) 2 sin θ – 3 = 0 (p) nπ + (–1)n 3π

(B) 2 sin 2θ + 3 (q) 2nπ –3π

= 2 sin θ + 2 3 cos θ

(C) sin 2θ + cos 2θ + 4 sin θ (r) 2nπ +3π

= 1 + 4 cos θ

(D) cos2θ = 41 (s) nπ –

4π

(t)

nπ +

4π

11. Match the following :

Column- I Column- II

(A) If three unequal numbers a, b, c are in A.P. and b – c, c – b, a are in G.P., then

abc3cba 333 ++ is equal to

(p) 1/3

(B) Let x be the arithmetic mean and y,z be two geometric means between any two positive numbers, then

xyzzy 33 + is equal to

(q) 1

(C) If a, b, c be three positive number which form three successive terms of a G.P. and c> 4b –3a, then the common ratio of the G.P. can be equal to

(r) 2

(D) ∞→n

lim tan

∑

=

−n

1r2

1

r21tan

is equal to

(s) 3

(t) 1/2




0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. If z ≠ 0 and 2 + cos θ + i sin θ = 3/z, then find the value of 2 (z + z ) – |z|2.

13. Find the sum of all the integral roots of (log5 x)2 + log5x (5/x) = 1.

14. Four different integers form an increasing A.P. such that one of them is the square of the remaining numbers. Find the largest numbers.

15. If n is a positive integer, and E = 2n + 1C1 + 2n + 1C2 + … + 2n + 1Cn – 2n + 1C2n + 1 – 2n + 1C2n – … – 2n + 1Cn + 1. Find |E|

16. Find the number of values of t for which the system of equations

(a + 2t)x + by + by + cz = 0 bx + (c + 2t)y + az = 0 cx + ay + (b + 2t) z = 0 has non-trivial solutions.

17. A function f(x) is defined for x > 0 and satisfies f(x2) = x3 for all x > 0. Then the value of f ′(4) is ___.

18. If A is the area formed by the positive x-axis, and the normal and tangent to the circle x2 + y2 = 4 at (1, 3 ) then A/ 3 is equal to _____.

19. The area bounded by the curves x = y2 and x = 3 – 2y2 is ______.

PHYSICS


This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. A thin conducting plate is inserted in half way between

the plates of a parallel plates capacitor of capacitance C.

Conducting plate

What does the value of capacitance, if both the plate of capacitor is shortened ?

(A) C (B) 2C (C) 3C (D) 4C 2. White light is incident normally on a glass surface

(n = 1.52) that is coated with a film of mg F2 (n = 1.38). For what minimum thickness of the film will yellow light of wavelength 550 nm (in air) be missing in the reflected light ?

mgF2 glass

(A) 99.6 nm (B) 49.8 nm (C) 19.6 nm (D) 10.6 nm 3. A circuit element is placed in a closed box. At time

t = 0, a constant current generator supplying a current of I amp is connected across the box. Potential diff. across the box varies according to graph as shown in the figure. The element in the box is -

3 2

8Volts(V)

Time t(sec)

(A) a resistance of 2Ω (B) a battery of emf 6V (C) an inductance of 2H (D) a capacitance 4. Consider a usual set-up of Young's double slit

experiment with slits of equal intensity as shown in the figure. Take 'O' as origin and the Y axis as indicated. If average intensity between y1 = λD/4d and y2 = λD/4d equals n times the intensity of maximum, then n equal is (take average over phase difference) -

S2

d

S1

O

y

D

(A)

π+

2121 (B) 2

π+

21

(C)

π+

21 (D)

π−

2121

SECTION – II




5. A long round conductor of cross-sectional area A is made of a material whose resistivity depends on the radial distance r from the axis of the conductor as ρ

= 2rα , α is a constant. The total resistance per unit

length of the conductor is R and the electric field strength in the conductor due to which a current I flows in it is E.

(A) R = 2A2πα

(B) R = 2A4πα

(C) E = 2AI2πα (D) E = 2A

I4πα

6. Two infinite plates carry j ampere of current out of the age per unit width of the plate as shown. BP and BQ represent magnitude of field at points P and Q respectively.

P

Q

(A) BP = 0 (B) BP = µ0j (C) BQ = 0 (D) BQ = µ0j

7. A bar magnet M is allowed to fall towards a fixed conducting ring C. If g is the acceleration due to gravity, v is the velocity of the magnet at t = 2 s and s is the distance traveled by it in the same time then,

M

C

3g

(A) v > 2g (B) v < 2g (C) s > 2g (D) s < 2g 8. In the network shown, the capacitor C is initially

uncharged. The time constant of the circuit is τ and the charge on C at time t after the switch S is closed is q.

C S

R3

R2

R1

(A) τ = CR1 (B) τ = C

+

+32

321 RR

RRR

(C) q = )e1(RR

CVR /t

32

2 τ−−+

(D) q = )e1(RR

CVR /t

21

1 τ−−+

9. Binding energy per nucleon vs mass number curve for nuclei is shown in the figure. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is -

0

Bin

ding

Ene

rgy/

N

ucle

on in

MeV

5.0

7.58.08.5

30 60 90 120Mass number of nuclei

Z

Y X

W

(A) Y → 2Z (B) W → X + Z (C) W → 2Y (D) X → Y + Z

SECTION – III Matrix - Match Type


q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

10. Capillary rise and shape of droplets on a plate due to surface tension are shown in column II.

Column I Column II

(A) Adhesive forces is (p)

B

A

greater than cohesive forces

(B) Cohesive forces is (q)

A

B

greater than adhesive forces


(C) Pressure at A > pressure (r) A mercury drop at B is pressed between two parallel plates of glass

AB

(D) Pressure at B > Pressure (s) AB

at A (t) none 11. Column I Column II (A) The coefficient of (p ) with decrease volume expansion at in pressure constant pressure is (B) Mean free path of (q) at all temperatures molecule increases (C) An ideal gas obeys (r) Same for all gases Boyle’s and Charle’s Law (D) A real gas behaves (s) At high temperature as an ideal gas at low pressure and (t) none



0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. A ball A is falling vertically downwards with

velocity v1. It strikes elastically with a wedge moving horizontally with velocity v2 as shown in figure. Find

the value of 2

1

vv , so that the ball bounces back in

vertically upward direction relative to the wedge.

v2

A

v1

30º

13. Portion AB of the wedge shown in figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. If AB = BC, then find the value of ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches to point C.

A

D

B

C

14. In what minimum time after its motion begins will a

particle oscillating according to the law x = 7sin21

πt

move from the position of equilibrium to the position x = – 7/2 units ?

15. A satellite is revolving in a circular equatorial orbit of radius R = 2 × 104 km from east to west. Calculate the interval (in hrs) after which it will appear at the same equatorial town. Given that the radius of the earth = 6400 km and g (acceleration due to gravity) = 10 ms–2.

16. Three capillary tubes of same radius 1 cm but of lengths 1m, 2m and 3m are fitted horizontally to the bottom of a long cylinder containing a liquid at constant pressure and flowing through these tubes. What is the length of a single tube which can replace the three capillaries.

17. Equations of a stationary and a traveling waves are as follows : y1 = a sin kx cos ωt and y2 = a sin (ωt – kx)

The phase difference between two points x1 = k3π and

x2 = k23π are φ1 and φ2 respectively for the two waves.

Then find the value of 2

1

φφ .

18. A coil, a capacitor and an AC source of voltage 24 V (rms) are connected in series. By varying the frequency of the source, a maximum rms current of 6A is observed. If this coil is connected to a battery of emf 12 V and internal resistance 4 Ω, then find the value of current flow through it.

19. A potential difference of 103 V is applied across an

X-ray tube. Calculate the value of ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced.


PHYSICS

1. An P-N-P transistor circuit is arranged as shown. It is a –

RL = 10 K V

V

PN P

(A) common base amplifier circuit (B) common-emitter amplifier circuit (C) common-collector circuit

(D) None 2. A tuning fork and an air column in resonance tube

whose temperature is 51°C produces 4 beats in 1 second when sounded together. When the temperature of the air column decreases, the number of beats per second decreases. When the temperature remains 16°C, only 1 beat per second is produced. Then the frequency of the tuning fork is -

(A) 55 Hz (B) 50 Hz

(C) 68 Hz (D) none 3. For wave propagation wrong statement is - (A) The wave intensity remains constant for a plane

wave

(B) The wave intensity decreases as the inverse of the distance from the source for a spherical wave

(C) The wave intensity decreases as the inverse square of the distance from the source for spherical wave

(D) Total intensity of the spherical wave over the spherical surface centered at the source remains constant at all times

4. Statement-1: Temperature of the body is lowered considerably if we put wet clothes.

Statement -2: Specific heat of water is high. (A) Both statement -1 and statement -2 are true and statement -2 is a correct explanation of the statement -1

(B) Both statement -1 and statement -2 are true but statement -2 is not a correct explanation of the statement -1 (C) Both statement -1 and statement -2 are false (D) statement -1 is false but the statement -2 is true.

5. An observer starts moving with uniform acceleration a toward a stationary sound source emitting a whistle of frequency n. As the observer approaches source, the apparent frequency n' heard by the observer varies with time as –

(A)

time

n'

(B)

time

n'

(C)

time

n'

(D)

time

n'

6. The main scale of a spectrometer is divided into 720 divisions in all. If the vernier scale consists of 30 divisions, the least count of the instrument is (30 division of vernier scale coincide with 29 division of main scale) -

(A) 0.1° (B) 1'' (C) 1' (D) 0.1''

SYLLABUS : Physics : Full syllabus Chemistry : Full syllabus Mathematics : Full syllabus


Instructions : • Part A – Physics (144 Marks) – Questions No. 1 to 2 and 9 to 30 consist FOUR (4) marks each and Question No.

3 to 8 consist EIGHT (8) marks each for each correct response. Part B – Chemistry (144 Marks) – Questions No. 31 to 39 and 46 to 60 consist FOUR (4) marks each and

Question No. 40 to 45 consist EIGHT (8) marks each for each correct response. Part C – Mathematics (144 Marks) – Questions No.61 to 82 and 89 to 90 consist FOUR (4) marks each and

Question No. 83 to 88 consist EIGHT (8) marks each for each correct response • For each incorrect response, ¼ (one fourth) of the weightage marks allotted of the would be deducted.

MOCK TEST - AIEEE PATTERN


7. Experimental verification of Newton's law of cooling is valid for -

(A) large temperature difference i.e. 30°C to 85°C between hot liquid and surrounding

(B) very large temperature difference i.e. 5°C to 95°C between hot liquid and surrounding

(C) small temperature difference i.e. 30°C to 35°C between hot liquid and surrounding

(D) any temperature difference

8. While studying the dissipation of energy of a simple pendulum by plotting a graph between square of amplitude and time which of the following apparatus is not essential ?

(A) Ticker timer (B) Meter scale (C) Vernier calliper (D) Stop watch

9. An object is weighed on a balance whose pans are not equal in masses when placed in the left pan, the object appears to weigh 10.30g but when placed in the right pan, it appears to weigh 12.62g. What is the correct mass of the object ?

(A) 10.30 g (B) 12.62 g (C) 11.46 g (D) Can not find

10. When jockey is put at two ends of the potentiometer wire, the galvanometer gives diflections in opposite directions. It means that apparatus can -

(A) not give a null point

(B) give a null point (C) be faulty

(D) be used after making some changes in the circuit

11. A student performs an experiment to determine the Young's modulus of a wire exactly 2cm long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young's modulus obtained from the reading is -

(A) (2.0 ± 0.3) × 1011 N/m2

(B) (2.0 ± 0.2) × 1011 N/m2 (C) (2.0 ± 0.1) × 1011 N/m2

(D) (2.0 ± 0.05) × 1011 N/m2

12. A student measures the focal length of a convex lens by putting an object pin at a distance 'u' from the lens and measuring the distance 'v' of the image pin. The graph between 'u' and 'v' plotted by the student should look like -

(A)

v(cm)

O u(cm)

(B) O

v(cm)

u(cm)

(C)

v(cm)

O u(cm)

(D)

v(cm)

Ou(cm)

13. Two concave mirror each of focal length f. A point source is placed at a point midway between two mirror. The minimum value of d for which only one image of s is formed –

d

s

(A) f (B) 2f

(C) 3f (D) 4f 14. In YDSE, if the intensity of central maxima is IO then

the y-coordinate of point where the intensity is 2

IO ,

d = 0.1 mm, D = 1m, λ = 5000A°

d

P

y

O

D

(A) 1.5 mm (B) 2 mm (C) 1.75 mm (D) 1.25 mm 15. Two charges of –4µC and +4µC are placed at points

A (1,0,4) and B (2, –1,5) located in an electric field →E = 0.20 i V/cm. The torque acting on the dipole

is-

(A) 8 × 10–5 N-m (B) 8/ 2 × 10–5 N-m

(C) 8 2 × 10–5 N-m (D) 22 × 10–5 N-m 16. Three concentric spherical metallic shells A, B and C

of radii a, b and c (a < b < c) have surface charge densities σ, – σ and σ respectively. If the shells A and C are at same potential, then the correct relation between a, b and c is -

(A) a + c = b (B) b + c = a (C) a – b = c (D) a + b = c


17. Five identical plates of equal area A are placed parallel to and at equal distance d from each other as shown in figure. The effective capacity of the system between the terminals A and B is -

B

A

(A) 53

dAo∈

(B) 45

dAo∈

(C) 35

dAo∈

(D) 54

dAo∈

18. Read the following statements carefully : Y : The resistivity of semiconductor decreases with

increase of temperature. Z : In a conducting solid, the rate of collision

between free electrons and ions increases with increase of temperature.

(A) Both Y and Z are true and Z is correct explaination of Y

(B) Both Y and Z are true but Z is not correct explaination of Y

(C) Y is true but Z is false

(D) Y is false but Z is true

19. A 1m long metallic wire is broken into two unequal parts A and B. The part A is uniformly extended into another wire C. The length of C is twice the length of A and resistance of C is equal to that of B. The ratio of resistances of parts A and C is -

(A) 4 (B) 41

(C) 2 (D) 21

20. A 600 cm long potentiometer wire is connected to a circuit as shown in figure. The resistance of potentiometer wire is 15r. The distance from point A at which the jockey should touch the wire to get zero deflection in the galvanometer is –

G

E

A B E/2 r

r

J

(A) 320 cm (B) 230 cm (C) 160 cm (D) 460 cm 21. A rectangular loop of metallic wire is of length a and

breadth b and carries a current i. The magnetic field at the centre of the loop is -

(A) π

µ4

i0 ab

ba8 22 + (B)

πµ4

i0ab

ba4 22 +

(C) π

µ4

i0ab

ba2 22 + (D) π

µ4

i0ab

ba 22 +

22. A short magnet produces a deflection of 30° when placed at certain distance in tanA position of magnetometer. If another short magnet of double the length and thrice the pole strength is placed at the same distance in tanB position of the magnetometer, the deflection produced will be -

(A) 60° (B) 30°

(C) 45° (D) None of these

23. A solenoid has 2000 turns wound over a length of 0.30 m. Its area of cross-section is 1.2 × 10–3 m2

. Around its central section a coil of 300 turns is wound. If an initial current of 2A in the solenoid is reversed in 0.25 sec, the emf induced in the coil is equal to -

(A) 6 × 10–4 Volt (B) 4.8 × 10–2 Volt (C) 6 × 10–2 Volt (D) 48 kV

24. A 100 volt AC source of frequency 500 Hz is connected to a L–C-R circuit with L = 8.1 mH, C = 12.5 µF and R = 10 Ω, all connected in series. The potential difference across the resistance is -

(A) 100 V (B) 200 V

(C) 300 V (D) 400 V

25. Which one is correct ? (A) Resultant of two vectors of unequal magnitude

can be zero

(B) Resultant of three non-coplanar vectors of equal magnitude can be zero

(C) Resultant of three coplanar vectors is always zero (D) Minimum number of non-coplanar vectors whose

resultant can be zero is four.

26. A stone thrown with the velocity V0 = 14 m/s at an angle 45° to the horizontal, dropped to the ground at a distance 'S' from the point where it was thrown. From what height should the stone be thrown in horizontal direction with the same initial velocity so that it fall at the same spot -

(A) 14.2 m (B) 16.9 m (C) 10.0 m (D) 9.6 m

27. A small body of mass 'm' is attached to one end of a light inelastic string of length l. The other end of the string is fixed. The string is held initially taut and horizontal and then body is released. The centripetal acceleration of the body and the tension in the string when the string reaches vertical position will be -

(A) g, mg (B) 2g, 3 mg (C) 3g, 2mg (D) 3g, 3 mg


28. Assertion : A rocket moves forward by pushing the surrounding air backwards.

Reason : It derives the necessary thrust to move forward, according to Newton's third law of motion.

(A) Both Assertion and Reason are true and Reason is a correct explanation of the Assertion

(B) Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion

(C) Both Assertion and Reason are false

(D) Assertion is false but the Reason is true

29. While slipping on rough spherical surface of radius

'R', block A of mass 'm' comes with velocity gR4.1 at bottom B. Work done in slipping the block from 'B' to 'C' is –

m

A

B

C

(A) 4

mgR (B) mgR

(C) 1.3 mgR (D) 45 mgR

30. A 2000 kg rocket in free space expels 0.5 kg of gas per second at exhaust velocity 400 ms–1 for 5 seconds. What is the increase in speed of rocket in this time -

(A) 2000 ms–1 (B) 200 ms–1

(C) 0.5 ms–1 (D) zero

CHEMISTRY

31. Which of the following can act as a both Bronsted acid & Bronsted base -

(A) Na2CO3 (B) OH– (C) HCO3

– (D) NH3 32. In which compound the oxidation No. of Oxygen

is +21 -

(A) OF2 (B)O2F2 (C) O2[PtF6] (D) KO2 33. The favourable conditions for a spontaneous

reactions are - (A) T ∆S > ∆H, ∆H = ⊕ , ∆S = ⊕ (B) T ∆S > ∆H, ∆H = ⊕ , ∆S = Θ (C) T ∆S = ∆H, ∆H = Θ , ∆S = Θ (D) T ∆S = ∆H, ∆H = ⊕ , ∆S = ⊕

34. The substance not likely to contains CaCO3 is - (A) Dolomite (B) A marble statue (C) Calcined Gypsum (D) Sea shells

35. On mixing 10ml of acetone with 50ml of CHCl3, the total volume of the solutions-

(A) < 60ml (B) > 60ml (C) = 60ml (D) unpredictable

36. On addition of He gas at constant volume to the reaction N2 + 3H2 2NH3 at equilibrium-

(A) The reaction stops (B) Forward reactions is favoured (C) Reaction remains unaffected (D) Backward reactions is favoured

37. The half life of a reaction is 24 hours . If we start with 10gm of reactant, How many grams of it will reaction after 96 hours, (I order reaction)

(A) 0.625gm (B) 6.25gm (C) 1.25gm (D) 0.125gm

38. The current is passed in Ag2SO4 aqueous solution & 1.6 gm O2 is obtained. The amount of Ag deposited will be- [Ag = 108gm]

(A) 107.8g (B) 1.6g (C) 0.8g (D) 21.6g

39. In decinormal solution CH3COOH is ionized to the extent of 1.3% find the pH of solution.

(A) 3.89 (B) 2.89 (C) 4.89 (D) 5.89 40. A FCC element (atomic wt.= 60) has a cell edge of

400pm. Its density is- (A) 6.23 g/cm3 (B) 6.43 g/cm3 (C) 6.53 g/cm3 (D) 6.63 g/cm3 41. Which set of quantum No. is not possible- n l m s (A) 2 0 0 +1/2 (B) 4 2 –3 –1/2 (C) 3 2 –2 +1/2 (D) 2 1 0 +1/2 42. Give simplest formula of compound which

containing 6gm C, 3.01×1023 atom O and 2 mole H atoms-

(A) CH2O (B) CH4O (C) CHO (D) CH3O

43. The IUPAC Name of

(A) 1,2-dimethyl Cyclohexene (B) 2,3-dimethyl Cyclohexene (C) 1,2-dimethyl Cyclohex-2-ene (D) 5,6-dimethyl Cyclohex-1-ene 44. Which of the following reagent can make distinction

between Pri. and Sec. amines ? (A) NH3 (B) NaNO2/HCl (C) HCl (D) All


45. Toluene reacts with Cl2 in the presence of light to give -

(A) Benzyl chloride (B) Benzoyl chloride (C) p-chlorotoluene (D) o- chlorotoluene 46. Which compound is formed when excess of KCN is

added to an aqueous solution of copper sulphate (A) Cu (CN)2 (B) K2 [Ca(CN)4] (C) K [Cu(CN)2] (D) K3 [Cu(CN)4] 47. The Blue Print Process involves the use of- (A) Indigo dyes (B) Iron compound (C) Vat dyes (D) some other compounds 48. The ionic radii of N3–, O2–, F– and Na+ follow the

order- (A) N3– > O2– > F– > Na+ (B) N3– > Na+ > O2– > F– (C) Na+ > O2– > N3– > F– (D) O2– > F– > Na+ > N3– 49. Reaction xy2 xy + y (g) (g) (g) Initial pressure of xy2 is 600 mm Hg & total pressure

at equilibrium is 800 mm Hg. Kp of reaction is - (A) 50 (B) 100 (C) 166.6 (D) 400 50. Cell: Zn|Zn+2|| Cu+2| Cu If the correct reactions of Zn+2 & Cu+2 ions are

doubled, the emf of the cells: (A) doubled (B) halved (C) same (D) zero 51. What is the pH of buffer solution containing 12g

CH3COOH & 16.4g CH3COONa in 500ml of solution (Ka for CH3COOH = 1.8×10–5).

(A) 4.7447 (B) 4.4774 (C) 4.4477 (D) None 52. How many moles of ferrous oxalate are completely

oxidized by 1 mole KMnO4 in acidic medium- (A) 3/5 (B) 5/3 (C) 1/5 (D) 5 53. In an irreversible process taking place at constant-T

& P and which only pressure-volume work is being done, than (dG) and (dS), satisfy the criteria-

(A) (dS)V, E > 0, (dG)T,P < 0 (B) (dS)V, E = 0, (dG)T,P = 0 (C) (dS)V, E = 0, (dG)T,P > 0 (D) (dS)V, E < 0, (dG)T,P < 0 54. 3,3-dimethylbutan-2-ol, on reaction with Conc.

H2SO4 at 443K will give…… as major product- (A) 3,3-dimethyl but-1-ene (B) 2,3-dimethyl but-2-ene (C) 2,2-dimethyl but-2-ene (D) 2,2-dimethyl-1- butene

55. Select the true statement about benzene from amongst the following-

(A) Because unsaturation benzene easily undergoes addition reaction

(B) There are two types of C-C bonds in benzene molecule

(C) There is a cyclic delocalisation of π es– in benzene

(D) Monosubstitution of benzene group gives three isomeric substances

56. OH → dustZn B3

3

AlCl

ClCH → − K

→ 4KMnO.alk D. Identity ‘D’ ?

(A) CH3 (B)

(C) CHO

(D) COOH

57. In Lassaigne’s test, the organic compound is first fused with sodium metal. The sodium metal is used because (A) The melting point of sodium metal is low (B) Sodium metal reacts with elements present

in organic compounds to form inorganic compounds

(C) All sodium salts are soluble in water (D) All the above 58. Concentrated hydrochloric acid when kept in open air

sometimes produces a cloud at white fumes the explanation for it is that-

(A) Oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas

(B) Strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke

(C) Due to strong affinity for water concentrated hydrochloric acid pulls moisture of air towards itself. This moisture forms droplets of water and hence the cloud.

(D) Concentrated hydrochloric acid emits strongly smelling HCl gas all the time

59. Consider the following complex [Co(NH3)5CO3]ClO4 The coordination number, oxidation number, number

of d-electrons and number of unpaired electrons on the metal respectively-

(A) 6, 3, 6, 0 (B) 7, 2, 7, 1 (C) 7, 1, 6, 4 (D) 6, 2, 7, 3


60. Point out the incorrect statement about resonance (A) Resonance structure should have equal energy (B) In Resonance structure, the constituent atoms

should be in the same position (C) In Resonance structure there should not be same

number of electron pairs (D) Resonance structure should differ only in the

location of electrons around the constituent atoms

MATHEMATICS

61. If the angles of elevation of an aeroplane from two points 1 km apart be 60º and 30º, then the height of the aeroplane is -

(A) 500m (B) 3

500 m

(C) 3

2000 m (D) None

62. Suppose a population A has 100 observations 101, 102, ..........200 and another population B has 100 observations 151, 152, ......... 250. If VA and VB

represents the variances of the two population

respectively then B

A

VV is -

(A) 49 (B)

94

(C) 32 (D) 1

63. If vertices of a triangle are (1, 0), (2, b) & (c2, – 3)

then centroid of triangle (A) can lie on y axis (B) always lie on x axis (C) lie on x axis if a + b = 3

(D) lie on y axis if c = 3 only

64. The value of k in order that f(x) = sin x – cos x – kx + b decreases for all real

values is given by : (A) k < 1 (B) k ≥ 1

(C) k > 2 (D) k < 2

65. 0xlim→ 2

x

xxcos–e

2

is equal to -

(A) 3/2 (B) 1/2 (C) 2/3 (D) None

66. If ∆ =

333

222

111

cbacbacba

, then

33333

22222

11111

cbc3b2acbc3b2acbc3b2a

−+−+−+

is equal to -

(A) ∆ (B) 2∆ (C) – 3∆ (D) 0 67. In a ∆ABC, angle A is greater than angle B. If the

measures of angles A and B satisfy the equation 3sin

x – 4 sin3x – k = 0, 0 < k < 1, then the measure of angle C is -

(A) 3π (B)

2π (C)

32π (D)

65π

68. If the p, q, r have truth values. F, F, T then the statement (p ↔ q) ∨ ~ r → (p ∧ r) will be (A) T (B) F (C) T, F (D) None

69. The statement (p ∨ q) ↔ (q ∧ ~ p) is a (A) Tautology (B) Contradiction (C) Neither tautology nor contradiction (D) None of these

70. If nth term of sequence 212 ,

1371 ,

911 ,

2320 , .... is

175

then value of n is - (A) 20 (B) 10 (C) 5 (D) 13 71. The ratio in which plane 2x – k = 0 divides the line

joining (–2, 4, 7) & (3, – 5, 8) is 9 : 1 then k equal to- (A) 4 (B) 5 (C) 6 (D) 7 72. If |a| = 2, |b| = 5 and |a × b| = 8 then |a – b| is equal to: (A) 12 (B) 15 (C) 17 (D) 5 73. The extremities of a line segment of length 6 move in

two fixed perpendicular lines. If locus of a point P which divides this line segment in ratio 1 : 2 is an ellipse then eccentricity of this ellipse is -

(A) 21 (B)

21 (C)

23 (D)

43

74. The value of p such that the vertex of parabola

y = x2 + 2px + 13 is 4 units above x-axis & lies in first quadrant is :

(A) 3 (B) 4 (C) ± 3 (D) – 3


75. If the lines represented by x2 + 2λx + 2y2 = 0 & lines represented by (1 + λ)x2 – 8xy + y2 = 0 are equally inclined then λ equals :

(A) – 2 (B) + 2 (C) ± 2 (D) ± 4 76. locus of centre of a variable circle tx2 + ty2 + 2(t2 + 1)x – 2(t2 – 1)y + t = 0 is a : (A) Straight line (B) Parabola (C) Ellipse (D) Hyperbola

77. If ∫−

4

1)x(f dx = 4 and ))x(f3(

4

2∫ − dx = 7 then the

value of ∫4

2)x(f dx is -

(A) 2 (B) – 3 (C) – 5 (D) None

78. Bisector of angle between lines 2x + y – 6 = 0 & 4x – 2y + 7 = 0 which contains origin is -

(A) acute angle bisector ; x = 5/8 (B) acute angle bisector ; y = 19/4 (C) obtuse angle bisector ; x = 5/8 (D) obtuse angle bisector ; y = 19/4

79. The value of [ ]∫2

1

1–))x(g(f f 'g(x). g'(x) dx where

g(1) = g(2) is equal to - (A) 1 (B) 2 (C) 0 (D) None

80. If f(x) = ∫ ++

2

22

x1xsinx sec2x dx and f(0) = 0 then

f(1) = (A) 1 – π/4 (B) π/4 – 1 (C) tan1 – π/4 (D) None of these

81. The domain of the function f(x) = ]x[x

xsec 1–

− is -

(A) R (B) R – (– 1, 1)I (C) R – I (D) R – [0, 1) 82. Let f(x) = sin–1x + sec–1x, then - (A) Df = [– 1, 1] (B) Rf = –π/2, π/2 (C) Rf = π/2 (D) None of these 83. If z1, z2, z3 represents the vertices of an equilateral

triangle such that |z1| = |z2| = |z3| then - (A) z1 + z2 = z3 (B) z1 + z2 + z3 = 0 (C) z1z2 = 1/z3 (D) z1 – z2 = z3 – z2

84. In a class of 100 students there are 70 boys whose average marks in a subject are 75.If the average marks of the complete class is 72, then what is the average of the girls.

(A) 73 (B) 65 (C) 68 (D) 74 85. A letter is taken at random from the letters of word

'STATISTICS' and a another letter is taken at random from letters of word 'ASSISTANT'. The probability that they are the same letter is -

(A) 1/45 (B) 13/90 (C) 19/0 (D) 5/18 86. If A = x : x ∈ I ; – 2 ≤ x ≤ 2 Β = x : x ∈ I ; 0 ≤ x ≤ 3 C = x : x ∈ N ; 1 ≤ x ≤ 2 and D = (x, y) ∈ N × N; x + y = 8 then - (A) n(A ∪ (B ∪ C)) = 5 (B) n(D) = 6 (C) n(B ∪ C) = 5 (D) None of these

87. Solution of sec2 ydxdy + 2x tan y = x3 is -

(A) tan y = 2xce− + (x2 – 1)

(B) tan y = 2xce− + (x2 – 1)

(C) tan y = 2xce− – (x2 – 1)

(D) None of these 88. The equation of common tangent to the curves y2 = 8x and xy = –1 is - (A) 3y = 9x + 2 (B) y = 2x + 1 (C) 2y = x + 8 (D) y = x + 2 89. Let f be twice differentiable function such that f "(x) = – f(x) and f '(x) = g(x) h(x) = (f(x))2 + (g(x))2 . If h(5) = 11 then h(10) is equal to (A) 22 (B) 11 (C) 0 (D) None 90. We are required to from different words with the help

of letter of the word INTEGER. Let m1, be the number of words in which I and N are never together and m2 be the number of words which begin with

I and end with R. Then 2

1

mm is given by -

(A) 30 (B) 1/30 (C) 6 (D) 42


PHYSICS 1. If the amplitude of a damped oscillator becomes half

in 2 minutes, the amplitude of oscillation w.r.t. initial one after 6 minutes is

(A) 271 (B)

81 (C)

181 (D)

641

2. An infinite number of spring having force constants as k, 2k, 4k, 8k ..... ∞ and respectively are connected in series; then equivalent spring constant is

(A) k (B) 2k (C) k/2 (D) ∞

3. A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m when the particle passes through the mean position. Its kinetic energy is 8 × 10–3 J. The equation of motion of this particle when the initial phase of oscillation is 45º can be given by

(A) 0.1 cos

π

+4

t4 (B) 0.1 sin

π

+4

t4

(C) 0.4 sin

π

+4

t (D) 0.2 sin

+

π t22

4. A mass m is moving with constant velocity along a line parallel to x-axis away from the origin. It angular momentum with respect to origin.

(A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing

5. A vessel containing oil (density = 0.8g/cm3) over mercury (density = 13.6 g/cm3) has a homogeneous sphere floating with half of its volume immersed in mercury and other half in oil. The density of material of sphere in g/cm3 is

(A) 3.3 (B) 6.4 (C) 7.2 (D) 2.8 6. Two trains move towards each other with the same

speed, speed of sound is 340 ms–1. If the pitch of the tone of the whistle of one is heard on the other changes by 9/8 times then the speed of each train is

V V (A) 2 ms–1 (B) 2000 ms–1 (C) 20 ms–1 (D) 200 ms–1

7. A sound level I differ by 4 dB from another sound of intensity 10 nW cm–2. The absolute value of intensity of sound level I in Wm–2 is

(A) 2.5 × 10–4 (B) 5.2 × 10–4 (C) 2.5 × 10–2 (D) 5.2 × 10–2

8. An ideal gas is taken through the cycle A → B → C → A as shown. If the net heat supplied to the gas in the cycle 5J, the work done by the gas in the process C → A

B

A

C2

1

10 P(N/m2)

V(m3)

(A) – 5 J (B) – 15 J (C) – 10 J (D) –20 J

9. There are n electrons of charge e on a drop of oil of density ρ. It is in equilibrium in an electric field E. Then radius of drop is

(A) 2/1

g4neE2

πρ

(B) 2/1

gneE

ρ

(C) 3/1

g4neE3

πρ

(D) 3/1

gneE2

πρ

10. Two identical cells of emf 1.5 V and internal resistance 1 Ω are in series. A third cell of similar parameters is connected in parallel to the

MOCK TEST – BIT-SAT

Time : 3 Hours Total Marks : 450 Instructions :

• This question paper contains 150 questions in Physics (40) Chemistry (40), Mathematics (45), Logical Reasoning (10) & English (15). There is Negative Marking

• Each question has four option & out of them, ONLY ONE is the correct answer. There is – ve marking.

• +3 Marks for each correct & – 1 Mark for the incorrect answer.


combination. The terminal voltage of the cells A, B, C are

A

1.5 V 1.5 V

1Ω B 1Ω

1.5 V 1Ω

C (A) 1, 1, 2 (B) 1.5, 1.5, 1.5 (C) 1.5, 0, 0 (D) 2, 1, 1

11. A wire has resistance of R ohm at T kelvin. At what temperature the resistance of wire be 2R ohm when temperature coefficient of resistance is α per degree centigrade.

(A) α

+α−2

1)T273( (B) α

−α−2

1)T273(

(C) α

−α− 1)T273( (D) α

+α− 12)T273(

12. Two cells each of same emf but of internal resistance r1 and r2 are joined to form a series circuit through an external resistance R. Value of R in term of r1 and r2 for which cell 1 has zero p.d. across it is

1

R

E E

r1 r2 2

(A) R = r1 – r2 (B) R = r1 + r2

(C) 21

21

rrrr

+ = R (D)

21

21

rrrr + = R

13. A current i flows in the network shown. Resulting mangnetic induction at point p is

2a

aA

B

a

a a

a

aF P

E D

C 2a

(A) a4iµ0

π (B) –

a8iµ2 0

π

(C) –aiµ

28 0

π (D)

aiµ

82 0

π

14. An alpha particle and a proton have same velocity when they enter a uniform magnetic field. The period of rotation of proton will be

(A) double of that of α particle (B) four times that of α particle (C) one half time that of α particle (D) same as that of α particle 15. A coil of inductance 8.4 mH and resistance 6 Ω is

connected to a 12 V battery. The current in the coil is 1A at approximate time

(A) 500 s (B) 20 s (C) 35 ms (D) 1 ms 16. A fish rising vertically up towards the surface with

speed 3ms–1 observe a bird diving vertically down towards it with speed 9 m/s. The actual velocity of bird is

y

y´

(A) 4.5 ms–1 (B) 5.4 ms–1

(C) 3.0 ms–1 (D) 3.4 ms–1

17. A concave lens of glass, refractive index 1.5 has both surface of same radius of curvature R. on immersion in a medium of refractive index 1.75, it will behave as a

(A) convergent lens of focal length 3.5 R (B) convergent lens of focal length 3R (C) divergent lens of focal length 3.5 R (D) divergent lens of focal length 3 R 18. I is the intensity due to source of light at any point P

on the screen if light reaches the point P via two different paths (a) direct (b) after reflection from a plane mirror then path difference between two paths is 3λ/2, the intensity at P is

(A) I (B) zero (C) 2I (D) 4I 19. The surface of some material is radiated, in turn, by

waves of λ = 3.4 × 10–7 m and λ = 5.4 × 10–7 m respectively. The ratio of stopping potential in two cases is 2 : 1, the work function is

(A) 2.05 eV (B) 1.05 eV (C) 3.05 eV (D) None 20. A X-ray tube has a working voltage of 40 × 103 V.

The continuous spectrum limit of the emitted x-rays is (A) 0.17 Å (B) 0.13 Å (C) 0.13 Å (D) 0.31 Å 21. The number of alpha and beta deca 88Ra222

experiences before turning into stable Pb206 isotope is (A) 4, 2 (B) 2, 4 (C) 1, 3 (D) 6, 10


22. The displacement of interfering light waves are y1 = 4 sin ωt and y2 = 3 sin (ωt + π/2). The amplitude of resultant wave is

(A) 5 (B) 7 (C) 1 (D) 0 23. A beam of light of wavelength 600 nm from a

distance source falls on a single slit 1 mm wide and resulting diffraction pattern is observed on a screen 2m away. Distance between first dark fringe on either side of the central bright fringe.

(A) 1.2 mm (B) 3.2 mm (C) 2.4 mm (D) 4.2 mm 24. The intensity of light from one source is double of

the other coherent source in a double slit experiment. The ratio of destructive to constructive interference in the obtained pattern is

(A) 34 (B) 1/34 (C) 17 (D) 1/17 25. Two radioactive material of half life T are produced

at different instants. Their activities area found to be A1 and A2 respectively when A2 < A1. Their age difference is

(A) 0.44 T log1

2

AA (B) 1.44 T log

2

1

AA

(C) 4.44 T log1

2

AA (D) 5.44 T log

1

2

AA

26. Three concentric conducting spherical shell x, y and z

have radii a, b and c respectively such that c > b > a, their surface charge density are σ, –σ and σ respectively. Then potential Vx is given by

C

b

a

y z

x

(A)

+−

εσ cb

ca 2

0 (B)

0εσ [a – b + c]

(C) 0ε

σ [a + b + c] (D) –0ε

σ [a + b – c]

27. A certain physical quantity is calculated from the

formula 3π (a2 – b2)h, where h, a and b are all lengths.

The quantity being calculated is (A) velocity (B) length (C) area (D) volume

28. The potential energy of a particle varies with distance

x from a fixed origin as U = Bx

xA2 +

, where A and B

are dimensional constant then dimensional formula for AB is

(A) M L7/2 T–2 (B) M L11/2 T–2

(C) M2 L9/2 T–2 (D) M L13/2 T–3 29 A particle leaves the origin at t = 0 and moves in the

+ve x axis direction. Velocity of the particle at any

instant is given by v = u

−

´tt1 . If u = 10 m/s and

t´ = 5 sec. Find the x coordinate of the particle at an instant of 10s.

t´ = 5 sec

t

v

u

(A) 0 (B) 10 m (C) 20 m (D) –10 m

30. An aero-plane drops a parachutist. After covering a distance of 40 m, he opens the parachute and retards at 2 ms–2. If he reaches the ground with a speed of 2ms–1, he remains in the air for about

(A) 16 s (B) 3 s (C) 13 s (D) 10 s

31. A tank moves uniformly along x-axis. It fires a shot from origin at an angle of 30º with horizontal while moving along positive x-axis & the second shot is also fired similarly that the tank moved along negative x-axis. If the respectively range of the shots are 250 m and 200 m along x-axis, the velocity of the tank.

30º

(A) 9.4 m/s (B) 4.9 m/s (C) 3.9 m/s (D) 5.9 m/s


32. A large number of particles are moving with same magnitude of velocity v but having random directions. The average relative velocity between any two particles average over all the pairs is

(A) 4π v (B)

2π v (C)

π3 v (D)

π4 v

33. A body is moving with uniform speed v on an horizontal circle from A as shown in the fig. Change in the velocity in the first quarter revolution is

v1

A

O

E

S

W

(A) v2 north (B) 2 v south west

(C) 2 v north-west (D) 2v west

34. A hollow vertical drum of radius r and height H has a small particle in contact with smooth inner surface of the upper rim at point P. The particle is given a horizontal speed u tangential to the rim. It leaves the lower rim at Q vertically below P. Taking n as an integer for number of revolution we get

Q

H

P u

(A) n = H

r2π (B) g/H2

r2π

(C) n = g/H2u

r2π (D) n = g/H2

r2uπ

35. A balloon is descending at a constant acceleration a. The mass of the balloon is M. When a mass m is released from the balloon is starts rising with acceleration a. Assuming that volume does not change when the mass is released, what is the value of m.

(A) )ga(

a2+

M (B)

+

a2ga M

(C) M)ga(

a2+

(D) ga

Ma+

36. Two blocks of masses 2 kg and 5 kg are at rest on ground. The masses are connected by a string passing over a frictionless pulley which in under the influence of a constant upward force F = 50 N. The accelerations of 5 kg and 2 kg masses are

5 kg 2 kg

50 N

(A) 0, 2.5 ms–2 (B) 0, 0 (C) 2.5 m/s2, 2.5 m/s2 (D) 1 m/s2, 2.5 m/s2

37. In the shown system m1 > m2. Thread QR is holding the system. If this thread is cut, then just after cutting

m1

m2

RQ

(A) Acceleration of mass m1 is zero and that of m2 is

directed upward (B) Acceleration of mass m2 is zero and that of m2 is

directed downward (C) Acceleration of both the block will be same (D) Acceleration of system is given by

+−

21

21

mmmm kg, when k is the spring factor

38. A car of mass M accelerates starting from rest.

Velocity of the car is given by v = 2/1

Mpt2

, where p

is the constant power supplied by the engine. The position of car as a function of time is given as

(A) 2/1

M9p8

t3/2 (B)

2/1

M8p9

t3/2

(C) 2/1

M9p8

t2/3 (D)

M8p9 t3

39. Six identical uniform rods PQ, QR, RS and ST, TV, UP each weighing w are freely joined at their ends to form a hexagon. The rod PQ is fixed in a horizontal position and middle points of PQ and ST are connected by a vertical string. The tension in string is

P Q

U R

T S (A) W (B) 3W (C) 2W (D) 4W


40. A body of mass 2 kg is being dragged with a uniform velocity of 2 ms–1 on a horizontal plane. The coefficient of friction between the body and the surface is 0.2. Work in 5 sec. is

(A) 39.2 J (B) 9.32 J (C) 23.9 J (D) 93.2 J

CHEMISTRY

1. 100 kg of iron ore (Fe2O3) containing 20% impurities on reduction with CO give iron equal to -

(A) 112 kg (B) 80 kg (C) 100 kg (D) 56 kg

2. Given : The mass of electron is 9.11 × 10–31 kg, Planck’s constant is 6.626 × 10–34 Js, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is -

(A) 5.79 × 107 ms–1 (B) 5.79 × 108 ms–1 (C) 5.79 × 105 ms–1 (D) 5.79 × 106 ms–1

3. The van der Waal equation for 0.5 mol of real gas is -

(A)

+ 2V4

aP

−

2bV = RT

(B)

+ 2V4

aP (2V – b) = RT

(C)

+ 2V4

aP = )b2–V(2

RT

(D)

+ 2V4

aP = )bV2(

RT2−

4. One mole of N2O4 is enclosed in a 5L container. At equilibrium, the container has 0.5 mol of N2O4. The equilibrium constant for the decomposition of N2O4

[N2O4 (g) 2NO2(g)] is-

(A) 1 (B) 52 (C) 3 (D)

51

5. Which one is the strongest Bronsted Lowry base out of the following -

(A) ClO– (B) −2ClO (C) −

3ClO (D) −4ClO

6. The pH of a solution obtained by mixing 50 mL of 0.4 M HCl with 50 ml of 0.2 N NaOH is -

(A) – log 2 (B) – log 2 × 10–1 (C) 1.0 (D) 2.0

7. Oxidation number of sulphur in H2SO5 is- (A) +2 (B) + 4 (C) +8 (D) +6

8. Equivalent mass of FeC2O4 in the reaction FeC2O4 → Fe3+ + CO2 is - (M.wt of FeC2O4 = M) (A) M (B) M/2 (C) M/3 (D) 2M/3

9. The fraction of total volume occupied by the atoms in a simple cube is -

(A) 4π (B) 2

8π

(C) 2 6π (D)

6π

10. In the diagram given below the value X is -

CuCuCu V50.0V15.02 → → ++++

E° = X Volt (A) 0.325 V (B) 0.65 V (C) – 0.35 V (D) – 0.65 V 11. For a first order reaction, A → B, t1/2 = 1 hr. What

fraction of the initial conc. of A reacts in 4 hrs ?

(A) 1615 (B)

161 (C)

87 (D)

81

12. An azeotropic solution of two liquids has boiling point lower than that of either of them if it -

(A) shows a –ve deviation from Raoult’s Law (B) shows a +ve deviation from Raoult’s Law (C) shows no deviation from Raoult’s Law (D) is saturated

13. In multi-molecular colloidal solution atoms or molecules are held together by -

(A) Hydrogen bonding (B) Strong attraction forces (C) Van der Waal’s forces (D) Strong electrical forces

14. Given : C + 2S → CS2 ; ∆H° = + 117 kJ C + O2 → CO2 ; ∆H° = –393 kJ S + O2 → SO2 ; ∆H° = – 297 kJ The value of ∆Hcombustion of CS2 in kJ mol–1 is (A) – 1104 (B) + 1104 (C) + 807 (D) – 807

15. Aspirin is chemically - (A) Methyl salicylate (B) Ethyl salicylate (C) Acetyl salicylic acid (D) o-hydroxy benzoic acid

16. Aniline when diazotized in cold and then treated with dimethyl aniline gives a colored product. It’s structure would be -

(A) (CH3)2N N = N

(B) (CH3)2N NH

(C) CH3NH N = N NHCH3

(D) CH3 N = N NH2


17. Ethyl ester Excess

MgBrCH3 → P. The product P will be -

(A) H3C

OH H3C

CH3

(B)

H3C

OH H5C2

C2H5

(C)

OH

C2H5 C2H5

C2H5

(D)

OH

C2H5H5C2

H7C3 18. When m-chlorobenzaldehyde is treated with 50%

KOH solution, the product(s) obtained is -

(A)

OH OH

CH – CH OH OH

(B)

OH OH

COO– +

CH2OH

(C)

Cl Cl

COO–

+ CH2OH

(D)

Cl Cl

CH – CH OH OH

19. Phenol is less acidic than - (A) Acetic acid (B) p-Nitrophenol (C) Both (A) & (B) (D) None of these

20.

OH

+ C2H5I OHHC.Anhy

HCO

52

52 → Product

In the above reaction product is - (A) C6H5OC2H5 (B) C2H5OC2H5 (C) C6H5OC6H5 (D) C6H5I 21. When acetylene is passed through dilute H2SO4

containing Hg2+ ions, the product formed is - (A) Acetone (B) Acetic acid (C) Acetaldehyde (D) Formaldehyde 22. Among the following statements on the nitration of

aromatic compounds, the false one is - (A) The rate of nitration of benzene is almost the same

as that of hexadeuterobenzene (B) The rate of nitration of toluene is greater than that

of benzene (C) The rate of nitration of benzene is greater than that

of hexadeuterobenzene (D) Nitration is an electrophilic substitution reaction

23. Which one is electrophilic addition - (A) CH3 – CH3 + Cl2 → C2H5Cl + HCl (B) CH3CH = O + HCN → (CH3)2C(OH)CN (C) (CH3)2C = O + HCN → (CH3)2C(OH)CN (D) CH3 = CH2 + Br2 → CH2BrCH2Br 24. Which one of the following conformation of

cyclohexane is chiral - (A) Twist boat (B) Rigid (C) Chair (D) Boat 25. The dipole moment is the highest for - (A) Trans-2-butene (B) 1,3- dimethyl benzene (C) Acetophenone (D) Ethanol 26. IUPAC name of the following compound -

N

CH3

CH3 C O

(A) N, N-dimethylcyclo propanecarboxamide (B) N-methylcyclopropanamide (C) Cyclopropanamide (D) None of the above 27. When a mixture of solid NaCl, solid K2Cr2O7 is

heated with conc. H2SO4 orange red vapours are obtained of the compound –

(A) Chromous Chloride (B) Chromyl Chloride (C) Chromic Chloride (D) Chromic sulphate

28. Which of the following will give a pair of enantiomorphs -

(A) [Co(NH3)4Cl2] NO2 (B) [Cr(NH3)6] [Co(CN)6] (C) [Co(en)2Cl2]Cl (D) [Pt(NH3)4] [Pt Cl6] 29. In following reaction

−4yMnO + xH+ + −2

42OC → yMn++ + 2CO2 + 2x H2O

x and y are (A) 2 and 16 (B) 16 and 2 (C) 8 and 16 (D) 5 and 2 30. A reduction in atomic size with increase in atomic

number is a characteristic of element of - (A) High atomic mass (B) d-block (C) f – block (D) Radioactive series 31. Which statement is not correct for nitrogen – (A) It has a small size (B) It does not readily react with O2 (C) It is a typical non-metal (D) d-orbitals available for bonding


32. Which one of the following is not an amphoteric substance -

(A) HNO3 (B) −3HCO (C) H2O (D) NH3

33. Which reaction cannot be used for the production of halogen acid –

(A) 2KBr + H2SO4 → K2SO4 +2HBr (B) NaHSO4 + NaCl → Na2SO4 + HCl (C) NaCl + H2SO4 → NaHSO4 + HCl (D) CaF2 + H2SO4 → CaSO4 + 2HF 34. B(OH)3 + NaOH NaBO2 + Na[B (OH)4] + H2O How can this reaction is made to proceed in forward

direction - (A) Addition of cis 1, 2-diol (B) Addition of borax (C) Addition of trans 1, 2-diol (D) Addition of Na2HPO4

35. Sodium thiosulphate is prepared by - (A) Reducing Na2SO4 solution with H2S (B) Boiling Na2SO3 solution with S in alkaline

medium (C) Neutralising H2S2O3 solution with NaOH (D) Boiling Na2SO3 solution with S in acidic medium

36. The critical temperature of water is higher than that of O2 because H2O molecule has -

(A) Fewer electrons than oxygen (B) Two covalent bond (C) V-shape (D) Dipole moment 37. Zone refining is a technique used primarily for which

one of the following process - (A) Alloying (B) Tempring (C) Sintering (D) Purification 38. Which one of the following elements has the highest

ionization energy – (A) [Ne] 3s2 3p1 (B) [Ne] 3s2 3p2 (C) [Ne] 3s2 3p3 (D) [Ar] 3d10 4s2 4p2

39. The correct order of dipole moment is - (A) CH4 < NF3 < NH3 < H2O (B) NF3 < CH4 < NH3 < H2O (C) NH3 < NF3 < CH4 < H2O (D) H2O < NH3 < NF3 < CH4 40. If Nx is the number of bonding orbitals of an atom

and Ny is the no. of the antibonding orbitals, then the molecule/atom will be stable if -

(A) Nx > Ny (B) Nx = Ny (C) Nx < Ny (D) Nx ≤ Ny

MATHEMATICS

1. Consider the sequence (angles are measured in radians) sin log102 , sin log103 , sin log104 ….then -

(A) all the terms of this sequence are positive (B) all the terms of this sequence are negative (C) 1001th term is negative (D) 10001th term is negative

2. The order relation between x, sin–1 x & tan–1x x ∈(0 ,1) is -

(A) tan–1x < x < sin–1x (B) sin–1x < tan–1x < x (C) x < sin–1x < tan–1x (D) None

3. The smallest positive valve of x satisfying the equation log2 cos x + log2 (1 – tan x ) + log2(1 + tan x) – log2sin x = 1 is -

(A) π/8 (B) π/6 (C) π/4 (D) π/6 4. A pole stands at a point A on the boundary of a

circular park of radius r and subtends an angle α at another point B on boundary. If arc AB subtends an angle α at the centre of the path, the height of the pole is -

(A) r sin α/2 tan α (B) 2r sin α/2 tan α (C) 2r sin α/2 cot α (D) None of these 5. The base of a triangle lies along the line x = a and is

of length 2a. The area of the triangle is a2. If the third vertex lies on the line -

(A) x = 0 (B) x = – a (C) x = 2a, or x = 0 (D) x = 0 or x = – 2a 6. If y = mx bisects an angle between the lines

ax2 – 2hxy + by2 = 0 then m

1–m2 =

(A) h

a–b (B) h

b–b (C) h

ba + (D) None

7. If the circle x2 + y2 + 2gx + 2fy + c = 0 passes through all the four quadrant then -

(A) g = – b (B) C > 0 (C) C < 0 (D) None

8. The equation of the circle which has two normals (x–1) (y – 2) = 0 and a tangent 3x + 4y = 6 is

(A) x2 + y2 – 2x – 4y + 4 = 0 (B) x2 + y2 + 2x – 4y + 5 = 0 (C) x2 + y2 = 5 (D) (x –3)2 + (y – 4)2 = 5

9. Circles drawn on the diameter as focal distance of any point lying on the parabola x2 – 4x + 6y + 10 = 0 will touch a fixed line whose equation is

(A) y = 2 (B) y = –1 (C) x + y = 2 (D) x – y = 2


10. The foci of a hyperbola coincide with the foci of the

ellipse 25x2

+ 9y2

= 1. If eccentricity of the

hyperbola is 2, then its equation is (A) x2 – 3y2 – 12 = 0 (B) 3x2 – y2 – 12 = 0 (C) x2 – y2 – 4 = 0 (D) None of these

11. →α and

→β are two mutually perpendicular unit vector

a→α + a

→β + c(

→α ×

→β ),

→α + (

→α ×

→β ) and c

→α + c

→β

+ b (→α ×

→β ) are coplaner then c is

(A) A.M. of a & b (B) G.M. of a & b (C) H.M. of a & b (D) None of these 12. The point of contact of the spheres x2 + y2 + z2 + 2x – 4y – 4z – 7 = 0 x2 + y2 + z2 + 2x – 4y – 16z + 65 = 0 (A) (1, 2, 6) (B) (1, 2, –6) (C) (1, –2, 6) (D) (–1, 2, 6) 13. If f(x) = 3 – 4x2 – 4x + 8–1 then range of f(x) is (A) (–∞, 1) ∪ (3, ∞) (B) (2, 3) (C) [2, 3] (D) None of these 14. If x > 0 and g is a bounded function then

1e

)x(ge)x(flim nx

nx

n ++

∞→ is

(A) 0 (B) f(x) (C) g(x) (D) None 15. If a1 = 1 and an = n(1 + an–1) ∀ n ≥ 2 than

+

+

+

∞→ n21n a11...

a11

a11lim =

(A) 1 (B) e (C) 1/e (D) None 16. Let f(x) = |2 sgn 2x| + 2 then f(x) has (A) removable discontinuity (B) infinte discontinuity (C) No discontinuity (D) essential discontinuity

17. If f(x) = cos

−

π 3x]x[2

, 1 < x < 2 and [.] = G.I.F.

then f´

π32

is

(A) 0 (B) 3(π/2)2/3 (C) –3(π/2)2/3 (D) None of these 18. If yex = cos x then, y4/ y = (A) –1 (B) 2 (C) –4 (D) None

19. Let f & g be differentiable function satisfying g´(a) = 2, g(a) = b and fog = I (Identity function), then f´(b) is equal to

(A) 1/2 (B) 2 (C) 2/3 (D) None

20. Tangents are drawn from origin to the curve y = sin x points of contact lie on the curve

(A) x2 + y2 = x2y2 (B) x2 – y2 = xy (C) x2 – y2 = x2y2 (D) None of these 21. Two positive numbers whose sum is 16 and sum of

whose cubes is maximum are given by (A) 8, 8 (B) no such number exist (C) 0, 16 (D) None of these

22. Let f(x) = 2x1 , g(x) =

x1 on [a, b], 0 < a < b. Let

)a(g)b(g)a(f)b(f

−− =

)c´(g)c´(f for same a < c < b then c is

(A) A.M. of a & b (B) G. M. of a & b (C) H.M. of a & b (D) None of these

23. ∫ + xcos1 2 . sin 2x cos 2x dx =

(A) 52 (1 + cos2x)3/2(3 – 2cos2x)2 + c

(B) 52 (1 + cos2x)3/2(3 – 2 cos2x) + c

(C) 52 (1 + cos2x)3/2(3 + 2 cos2x) + c

(D) None of these

24. ∫

+− ....

4x

2x1

42dx

(A) sin x (B) – sin x (C) cos x (D) None

25.

∫

∫

π

π

−

π→ π−

−

2

2

x

4/

x

2/

tcos

2x dt)2/t(

dt)12(lim =

(A) π

2loge (B) π22nl (C)

π2n2l (D) None

26. ∞→n

lim2n/2

2n11

+

2n/4

2

2

n21

+

2n/6

2

2

n31

+

....

2n/n2

2

2

nn1

+ =

(A) 4/e (B) 3/e (C) 2/e (D) None


27. The area bounded by the curves y = 2x2 & y = x

|x|

and x = 0 is equal to

(A) 32 (B)

622 (C)

62 (D) None

28. Order and degree of the differential equation "y = (y´+ 3)1/3 are respectively (A) 2, 2 (B) 2, 3 (C) 3, 2 (D) None

29. If x18 = y21 = z28 then 3, 3 logyx, 3 logz y, 7 logxz are in

(A) A.P. (B) G.P. (C) H.P. (D) None

30. If log2x + log2y ≥ 6 then least possible value of x + y is

(A) 32 (B) 16 (C) 8 (D) None

31. No. of real roots of the equation x3 + x2 + 10x + sin x = 0 is (A) 1 (B) 2 (C) 3 (D) ∞

32. The roots of the equation ax2 + bx + c = 0, a ∈R+ are two consecutive odd positive integers then

(A) |b| ≤ 4a (B) |b| ≥ 4a (C) |b| ≥ 2a (D) None of these

33. The sum of the terms of an infinitely decreasing G.P. is equal to the greatest value of the function f(x) = x3 + 3x – 9 on the interval [–2, 3] and the difference between the first two terms is f´(0) then sum of first terms is

(A) 19 or – 37 (B) 19 (C) –37 (D) None of these

34. If the complex number z1 = a + i, z2 = 1 + ib, z3 = 0 form an equilateral triangle (a, b are real number between 0 & 1) then :

(A) a = 3 – 1, b = 2/3

(B) a = 2 – 3 , b = 32 −

(C) a = 21 , b =

43

(D) None of these

35. ∑=

−n

0r

r)1( nCr

∞+++ .....

27

23

21

r3

r

r2

r

r is equal to

(A) 12

1n −

(B) 12

3n −

(C) 12

2n −

(D) None

36. The coefficient of x3y4z in the expansion of (1 + x + y – z)9 is (A) 2 . 9C7 . 7C4 (B) – 2 . 9C2 . 7C3

(C) 9C7 . 7C4 (D) None of these

37. If x1

ex

− = B0 + B1x + B2x2 + .... then Bn – Bn–1 = ?

(A) n1 (B)

n1 (C)

1n1−

(D) None

38. The number of point (x, y, z) in space whose each coordinate is a negative integer such that x + y + z + 12 = 0 is

(A) 55 (B) 110 (C) 75 (D) None 39. Six boys and six girls sit along a line alternatively

with probability P1 & along a circle (again alternatively) with probability P2 then P1/P2 is equal to

(A) 1 (B) 1/5 (C) 6 (D) None 40. If f(x) is a polynomial satisfying

f(x) = 21

−

x1f1

)x(fx1f)x(f

and f(2) = 17

then the value of f(5) is (A) 624 (B) –124 (C) 626 (D) 126

41. If A =

20x1

is idempotent then x =

(A) 0 (B) 2 (C) no such x exist (D) None of these 42. Let R be a relation on the set of integers given by

a R b if a = 2k b for some integer k then R is (A) an equivalence relation (B) reflexive and symmetric but not transitive (C) reflexive and transitive but not symmetric (D) symmetric and transitive but not reflxive

43. Minimum value of a

cb + + b

ac + + c

ba + , (for real

+ve numbers a, b, c) is (A) 1 (B) 2 (C) 4 (D) 6 44. From mean value theorem f(b) – f(a) = (b – a) f´(x1);

a < x1 < b if f(x) = x1 then x1 =

(A) ab (B) 2

ba + (C) ba

ab2+

(D) baab

+−

45. If f(x) = ∫ dxxcot4 + 31 cot3x – cot x and f

π

2=

2π

then f(x) is (A) π – x (B) x – π (C) x (D) None


LOGICAL REASONING

1. Fill in the blank spaces. 11, 12, 17, 18, 23, 24, (?) (A) 12 (B) 29 (C) 30 (D) 35

2. Choose the best alternative. Dum-Dum : Calcutta : : Palam : ? (A) Kerala (B) Delhi (C) Madras (D) Bombay

3. Pick the odd one out – (A) Wheat (B) Paddy (C) Towar (D) Mustard

4. Which of the following figures (A), (B), (C) and (D) when folded along the lines, will produce the given figure (X) ?

(X)

(A)

(B)

(C)

(D)

5. In each of the following questions, choose the set of figures which follows the given rule.

Rule : The series becomes complex as it proceeds :

(A)

(B)

(C)

(D)

6. In following question below, you are given a figure (X) followed by four figures (A), (B), (C) and (D) such that (X) is embedded in one of them. Trace out the correct alternative.

(X)

(A)

(B)

(C)

(D)

7. In the following question, find out which of the answer figures (A), (B), (C) and (D) completes the figure-matrix ?

?

(A)

(B)

(C)

(D)

8. The question that follow contain a set of three figures X, Y and Z showing a sequence of a piece of paper. Fig. (Z) shows the manner in which the folded paper has been cut. These three figures are followed by four answer figures from which you have to choose a figure which would most closely resemble the unfolded form of fig. (Z).

X Y Z

(A) (B)

(C) (D)

9. In the following question, complete the missing portion of the given pattern by selecting from the given alternatives (A), (B), (C) and (D).

(X)

?

(A)

(B)

(C)

(D)

10. In the following question, find out which of the answer figures (A), (B), (C) and (D) complete the figure - matrix ?


?

(A)

(B)

(C)

(D)

ENGLISH

1. Choose the one which best expresses the meaningful concept :

The state's duty is to . . . . the safety of its Citizens. (A) assure (B) ensure (C) insure (D) accept 2. Choose the one which best expresses the meaningful

concept : The company went . . . . in the 1990's recession. (A) burst (B) bust (C) bursted (D) busted 3. Choose the one which best expresses the meaningful

concept : What can we . . . . from this evidence, Watson ? (A) deduce (B) deduct (C) reduce (D) conduce 4. Choose the one which best expresses the meaningful

concept in opposite meaning : Zenith : (A) Sky (B) Firmament (C) nadir (D) naive 5. Pick up the correct Synonym for the following word Voracious : (A) Hungry (B) Wild (C) Quick (D) Angry 6. One who travels from place to place : (A) Journey man (B) Tramp (C) Itinerant (D) Mendicant 7. Choose the one which best expresses the meaning of

the given idiom/proverb : To fly off the candle : (A) To dislocate (B) To lose one's temper (C) To take off (D) To be indifferent 8. Fill in the blanks with one of the options given

below: Gandhi Ji . . . . . . smoking in his youth.

(A) took to (B) took for (C) took in (D) took up

9. Select the one which best expresses the same sentence in Passive or Active Voice.

Get the box broken. (A) Get someone to break the box. (B) They have broken the box (C) Have the broken box (D) Break the box

10. Choose the one which best expresses the correct answer in the speech :

He said, "How shabby you are looking!" (A) He asked how shabby I was looking (B) He exclaimed with disgust that I was looking

very shabby (C) He exclaimed with sorrow that they were

looking much shabby (D) He told that I was looking much shabby

11. Pick out the mis-spelt word – (A) Neigh (B) Rein (C) Neice (D) Neither

12. Find out which part of the sentence has an error :

)a(wonderI /

)b(bookthewithdonehashewhat /

)c(himlendI /

)d(ErrorNo

(A) Wonder (B) What he has done with the book (C) I lend him (D) No Error

13. Pick out the most appropriate pair to fill in the blanks in the same order, to make the sentence meaningfully complete :

She was . . . . . because all her plans had gone . . . . . (A) distraught, awry (B) Frustrated, Magnificently (C) Elated, wild (D) Dejected, splendidly 14. Pick out the most effective word from the given

words to fill in the blanks to make the sentence meaningfully complete :

Most of the issues discussed in the meeting were trivial and only a few were :

(A) Interesting (B) Practical (C) Complex (D) significant 15. Pick out the most appropriate pair to fill in the

blanks in the same order, to make the sentence meaningfully complete :

The . . . . . of glory lead but to the . . . . . (A) Paths, grave (B) Ways, happiness (C) Acts, Prosperity (D) Achievements, Suffering


CHEMISTRY

1.[A]

C6H5 – C||O

C||O

CH3 H3C

O||C O

||C

–C6H5 Orientation for aldol condensation

2.[B] A →Ozonolysis B ∆

→ 422 SOH,OHNH oxime

formation followed by Beckmann rearrangement.

O||C –NH–CH3

N-methyl benzamide

So, B is

3CH|

OC = and

A is

33 CHCH||CC =

3.[D] (CH3)3CCH2CH2OH

Heat,OHH,OCrK

2

722 →+

(CH3)3)A(

2CCH COOH

→ 2SOCl (CH3)3 CCH2COCl(B)

(CH3)2NH

(CH3)3)C(

2CCH C||O

N(CH3)2

(CH3)3)D(

CC H2CH2N(CH3)2

OHether,LiAlH

2

4 ←

4.[C] Attack of nucleophile is a rate determining step

F3C

O

O

O

I II

CH3

I is more electron deficient and facilitates a faster attack.

5.[A] As water introduces, water dissolves HCl(g) and a press drop is produced Liquid level in the capillary rises.

6.[D] 7.[B] The energy of AOs depends on the (n + l) values n + l value of (n – 1) d = n + 1 ; n + l value of ns = n (n + l) value of (n + 1)d = n + 3 ; n + l value of nf

= (n + 3) But due to lower value of principle quantum no.

energy of nf < (n + 1) d ∴ energy of (n + 2) s < nf. 8.[A] A, B and C are magnesium ,aluminium and

silicon. Magnesium form ionic oxide, MgO ; Aluminium forms amphoteric oxide ,Al2O3 and silicon forms a giant molecule SiO2.

9.[ B,C,D] Due to resonance cyclohexatriene cation is

aromatic which causes it's stability. 10.[ A,B,D] Catalyst lower the activation energy of

forward reaction & backward direction keeping the same enthalpy of reaction.

11.[ A,B,C] Greater the value of (IE – ∆eg H) greater is the

electronegativity E.N. of P = (1680 + 340)/(4.18) (125) ~– 4 12.[A,B,C] H = E + PV

TdP

dH

=

TdPdE

+ P

TdPdV

+ V

For liquid, TdP

dH

=

TdVdE

TdPdV

+

P TdP

dV

+ V

For incompressible liquid, TdP

dV

~– 0.

∴ TdP

dH

~– V

SOLUTION FOR MOCK TEST IIT-JEE (PAPER - I)


For ideal gas, TdP

dH

= 0

For the real gas, if TdV

dE

= 0,

then TdP

dH

= P

TdPdV

+ V ≠ 0

13.[A] F(Monoester) Molecular weight = 186 No Br2 reaction ⇒ Saturated Two oxygen ⇒ 2 × 16 = 32 No. of CH2 = (186 – 32)/14 = 11 Hence, molecular formula of saturated monoester

F, is C11H22O2 Hydrolysis

G H Optically active, soluble in NaOH ⇒ Acid We have, Ag+ salt →2Br ± J Hunsdiecker reaction; radical intermediate, so racemic mixture (J contains one carbon less than G)

Optically active, (Alcohols are not soluble in NaOH) +ve iodoform test, it suggests HO–

3CH|

R–CH On warning with H2SO4 (dehydration)

I (no diastereomers means no geometrical isomers) It suggests same alkyl group on one of the doubly bonded carbon atoms

H NaBr.IITsCl.I optically active ⇒ J

It suggests G contains one more carbon atom than H.

Hence, molecular formula of ester F, is

C5H11– C||O

–O–C5H11

C5H11– C||O

–OH(G) C5H11OH(H)

⇒ HO – CH CH3

* – –CH|

CH3

CH3 →∆,SOH 42

C=C

H

H3C (I)

No cis-trans isomers

CH3

CH3

H NaBr

TsCl → CH3–

)J(33

*

CHCH||

Br–HC–CH

So, G is CH3–

333 CHCHCH|||

OHCCHCH||O

−−−

)J(

3333

3Br*

3

CHCHCHCH||||

BrCHCHCHOAg

O||CHCCHCH 2

±

−−−→−−−−

F = CH3–

3333

3

CHCHCHCH||||

CHCHCHOCCHCH||O

−−−−−−

14.[C] I exists as diastereomers and H is optically active. So, H is HO–

3CH|CH − CH2CH2–CH3

H

H3CC = C

H

CH2CH3

H2SO4 ∆

, and

G is CH3CH2CH2–

3CH|

OHCHC||O

−

15.[D] Since H is optically active and gives negative iodoform test, so H is

HO–CH2–

3CH|

–CH CH2–CH2 and

G is CH3CH2

3

2

CH|CHCH C

||O

–OH

16.[A]

17.[D] Final pH = 1.7 or – log [H+] = 1.7 or log [H+] = – 2 + 0.3 or [H+] = 2 × 10–2 Let v ml of 0.1 (M) HA solution is mixed with

100 ml of 10–2 (M) HCl. In the mixed solution,

[HA] = )v100(

10v 4

+× −

× 103 =)v100(

v1.0+

(M) and

[HCl]=)v100(

0110 33

+×−

= )v100(

1+

HA H+ + A–

)v100(v1.0+

(1 – α) )v100(

v1.0+

α )v100(

v1.0+

α

+)v100(

1+

∴ [H+] = )v100(

)1v1.0(+

+α = 2 × 10–2


or 0.1 v α + 1 = 2 + 2 × 10–2 × v

Ka =

)v100()1(v1.0

)v00.1(v1.0

)v100()1v1.0(

+α−

+α

×+

+α

=)1( α−

α × 2 × 10–2

∴ )1( α−

α × 2 × 10–2 = 10–2 or α−

α1

= 21

or 2α = 1 – α or α = 1/3

∴ 0.1 v × 31 + 1 = 2 + 2 × 10–2 × v

or 30v –

50v = 1 or

5030v20

× = 1

or v = 2

150 = 75 ml

18.[D] For isohydric solution, Ka1C1 = Ka2C2 Column Matching 19. [A] → r,s,t; [B] → p,r,s; [C] → s; [D] → q,t

For CH3CH2CH2NH2 →HCl/NaNO2

productthenot;ermediateintsimply

isionDiazoniumNCHCHCH 2223+

CH3–CH2–CH2–OH

N – CH3 + HO CH3

CH3

CH3

on heating doesn't give alkene.

20. [A] → q,r; [B] → p,r; [C] → r,s; [D] → p,t

C6H5CH2–CHO →−

2Br/OH C6H5–

)(Br|

CH

±

–CH = O

CH3CHO →−

−OH

Cº250CH3–

)(CH|

OH

±–CH2–CH = O

CH3–CH2CH = O Cº25OH →

−CH3CH2–CH=

3CH|

CHOC −

CH3–

3CH|

CH − CH = O →−OH.Conc

CH3–

3CH|CH − C

||O

–O– + CH3–

3CH|CH − CH2OH

MATHEMATICS

1.[A] Let us first count the number of elements in F.

Total number of functions from A to B is 34 = 81. The number of functions which do not contain

x(y) [z] in its range is 24. ∴ the number of functions which contain exactly

two elements in the range is 3 . 24 = 48. The number of functions which contain exactly

one element in its range is 3. Thus, the number of onto functions from A to B is

81 – 48 + 3 = 36 [using principle of inclusion exclusion] n (F) = 36. Let f ∈ F. We now count the number of ways in

which f –1(x) consists of single element. We can choose preimage of x in 4 ways. The

remaining 3 elements can be mapped onto y, z is 23 – 2 = 6 ways.

∴ f –1 (x) will consists of exactly one element in 4 × 6 = 24 ways.

Thus, the probability of the required event is 24/36 = 2/3

2.[A] Let E1 denote the event that the letter came from

TATANAGAR and E2 the event that the letter came from CALCUTTA. Let A denote the event that the two consecutive alphabets visible on the envelope are TA. We have P(E1) = 1/2, P(E2) = 1/2, P(A / E1) = 2/8, P (A / E2) = 1/7. Therefore, by Bayes' theorem we have

P(E2 / A) = )E/A(P)E(P)E/A(P)E(P

)E/A(P)E(P

2211

22

+

= 114

3.[D] Required probability = 1 – P (all the letters are

put in correct envelops) The number of the ways of putting the letters in

the envelops = 4P4 = 4! The number of ways of putting letters in correct

envelops = 1

∴ Required probability = 1 – 241 =

2423

4.[D] We have

AB =

− x52x10001000x5

=

100010001

⇒ x = 1/5

5.[B] Greatest term in the expansion of (x + y)n is

kth term where k =

++

yxy)1n(


In the present case

k =

++

x23)x2)(150( =

+ 5/23)5/2)(51( =

17102 = 6

Thus, 6th term is the largest term.

6.[D] We have | z | = z4

z4z +− ≤

z4z − +

|z|4

= 2 + |z|

4

⇒ | z |2 ≤ 2 | z | + 4 ⇒ (| z | – 1)2 ≤ 5 ⇒ | z | – 1 ≤ 5 ⇒ | z | ≤ 5 + 1

Also, for z = 5 + 1

z4z − = 2

Therefore, the greatest value of | z | is 5 + 1.

7.[D] Integrating by parts, the given integral is equal to

x tan–1 16

11x − ∫

−−

16

1 1xx4

1x

x dx

= π3

16 – ∫−

16

1 1x

dx41

= π3

16 – ∫+

3

0

2dt

t)t1(t4

41 ( x = 1 + t2)

= π3

16 – ( )33 + = 3

16π – 2 3

8.[C] The intersection of y – x + 1 = 0 and y + x + 5 = 0 is (– 2, –3). Put x = X – 2, y = Y – 3. The given

equation reduces to dXdY =

XYXY

+− . This is a

homogeneous equation, so putting Y = υX, we get

XdXdυ =

112

+υ+υ

−

⇒

+υ−

+υυ

−1

11 22 dυ =

XdX

⇒ –21 log (υ2 + 1) – tan–1 υ = log | X | + C

⇒ log (Y2 + X2) + 2 tan–1 XY = C

⇒ log ((y + 3)2 + (x + 2)2) + 2 tan–1 2x3y

++ = C

9.[A] We have (1 + x)n – nx – 1 = C0 + C1 x + C2 x2 + … + Cn xn – nx – 1 = x2 [C2 + C3x + … + Cn xn–2] [Q C1 = n, C0 = 1] Thus, (1 + x)n – nx – 1 is divisible by x2.

10.[A,B,C,D] y = 1)A2sinA2(cos

1)A2sinA2(cos2

2

−−

+−

⇒ y = 1)A2sinA2(cos1)A2sinA2(cos

−−±+−±

which gives us four values of y, say y1, y2, y3 and y4. We have

y1 = 1A2sinA2cos1A2sinA2cos

−−+− =

A2sin)1A2(cosA2sin)A2cos1(

+−−+

= AcosAsin2Asin2

AcosAsin2Acos22

2

+−−

= )AsinA(cosAsin)AsinA(cosAcos

−− = cot A

y2 =1)A2sinA2(cos1)A2sinA2(cos

−+−+−− =

A2sin)A2cos1(A2sin)A2cos1(

−+−+−

=AcosAsin2Acos2

AcosAsin2Asin22

2

−−

+ = – tan A

y3 =1)A2sinA2(cos

1)A2sinA2(cos−+−

+− =A2sin)A2cos1(

A2sin)A2cos1(−+−

−+

=AcosAsin2Acos2

AcosAsin2Acos22

2

−−− = –

AsinAcosAsinAcos

+−

= –Atan1Atan1

+− = – tan

−

π A4

= – cot

+

π A4

y4 =1)A2sinA2(cos1)A2sinA2(cos

−++−− =

A2sin)A2cos1(A2sin)A2cos1(

+−−+−

=AcosAsin2Asin2

AcosAsin2Asin22

2

+−+ =

AsinAcosAsinAcos

−+

=Atan1Atan1

−+ = tan

+

π A4

.

11.[B, C, D] Equations of the given circles can be written as (x – 3)2 + y2 = 32 (1) and (x + 1)2 + y2 = 12 (2) Equation of any tangent to circle (2) is (x + 1) cos θ + y sin θ = 1 (3) This will be a tangent to circle (1) also if

θ+θ

−θ+22 sincos

1cos)13( = ± 3 ⇒ 4 cos θ – 1 = ± 3

That is, cos θ = 1 or cos θ = –21 . When cos θ = 1,

we have sin θ = 0, and the equation of the common tangent (3) becomes

x + 1 = 1 or x = 0 (4) When cos θ = –1/2, we have sin θ = ± 2/3 , and

the equations of the common tangents are

–21 (x + 1) ±

23 y = 1 ⇒ x – y3 + 3 = 0 (5)

and x + y3 + 3 = 0 (6)


12.[A, B, C, D]

Let F(x) = ∫ ++−2x

0 t

2

e24t5t dt

⇒ F′(x) = 2x

24

e2

4x5x

+

+− .2x

So from F′(x) = 0, we get x = 0 or

x2 =2

16255 −± =2

35 ± = 4, 1

Hence x = 0, ±2, ±1. 13.[B]

D C

–1 1 x = 4 x = –4

y = –4

y = +4

A B

Shaded region is S0. Area of S0

= 4 × 2 – 21 π (1)2 = 8 – π/2

14.[D] y ∈ [0, 4], x ∈ [–1, 1]

m (t) = cost

lines y = 2x + 0.4 lies inside the region so

⇒ t ∈ [0, 1]

t2 + (2t + 0.4)2 – 1 ≥ 0 ⇒ t ∈ [0.28, 1]

15.[B] (Slope)max. = (cos t)max = cos (0.28) and point is (π, 1)

π−

−x

1y = cos (0.28)

16.[C] SCSC

2

1 =2

1

rr =

13

x =13

)3(1)1(3−−− =

26− = – 3

17.[D] tangents = y = ±3

x + RT

tan30º =3

RT ⇒ RT = 3

18.[A]

•2

C1

C2 60º 1

(h + 1)2 + k2 = (1 + 2)2 (circle)

Column Matching 19. [A] → r; [B] → p,r; [C] → s; [D] → r

(A) ∑

=

10

0rr

20 C = 20C0 + 20C1 +……+ 20C10

But, 20C0 + 20C1 +……+ 20C20 = 220 Also, 20C20 = 1 = 20C0, 20C19 = 20C1, 20C18 = 20C2 etc. ∴ given sum = (20C0 + 20C1 +……+ 20C20) – (20C11 +…..+ 20C20) 220 + 20C10– (20C10 + 20C9 + ……+ 20C0) ∴ 2 (20C0 + 20C1 +…..+ 20C10) = 220 + 20C10

(B) ∑=

100

0rr

100 C (x –3)100–r 2r

= ((x–3) +2)100 = (x –1)100 = (1 –x)100

∑=

−100

0r

rr

100 )x(C = ∑=

−−100

0r

rr

100rr xC)1()1(

∴ Coeff. of x53 = (–1)53 100C53 = – 100C53

(C) We have (1+ x)10 = 10C0 + 10C1 x +10C2x2 +……+ 10C10 x10

....(1) Also (1–x)10 = 10C0 – 10C1x + 10C2x2 +……. …..+ 10C10x10 ....(2) Multiplying, we get (1 –x2)10 = (10C0 + 10C1 x + 10C2x2 +…… ….+ 10C10x10) × (10C0 –10C1x + 10C2x2+… …...+ 10C10 x10)

Equating the coefficients of x10, we get 10C5 (–1)5 =10C0

10C10 –10C1 10C9 + 10C2 10C8 +… ….+ 10C10 10C0 ⇒ – 10C5 = (10C0)2 – (10C1)2 + (10C2)2 +…… …+ (10C10)2


(D) 95C4 + ∑=

−5

0j3

j100 C

= 95C4 + 99C3 + 98C3 + 97C3 + 96C3 +95C3 = (95C4 + 95C3) +96C3 + 97C3 + 98C3+ 99C3 = (96C4 + 96C3) + 97C3 +98C3 +99C3 = (97C4 +97C3) +98C3 + 99C3 = (98C4 + 98C3) + 99C3 = 99C4 + 99C3 = 100C4

20. [A] → p,q; [B] → q,t; [C] → q; [D] → s (A) Given lines intersect if

12

11544312

λλ−−−

= 0

⇒ λ = 0, – 1

(B) ∞→x

lim 4x

++

+

++

−−

2x1x1

2x1x1

tan 1 =2 = y2 + 4y + 5

⇒ y = –1, – 3 (C) y2 – ax (– x – y) = 0 ⇒ for perpendicular lines a + b = 0 ⇒ 1 + a = 0 ⇒ a = – 1 (D) ( a × b ) × a = ( j – k ) × a ⇒ ( a . a ) b – ( a . b ) a = ( j – k ) × a on solving,

we get b = i

PHYSICS

1.[B] T = m1r1ω12

also T = m222

22

2 rg ω+

∴ m1r1ω12 = m2 2

222

2 )r(g ω+

or, 0.1 × ω12 =

21 22 )10()10( +

0.1 ω12 = 1

ω1 = 10 rad/s v1 = r1ω1 = 10 m/s 2.[A] As f = µN = mg

or, µmlω2 = mg ⇒ ω = lµ

g

3.[C] At terminal velocity net force is zero.

6πη(r1 + r2) VT +34

π (r13 + r2

3) ρg =34

π (r13 + r2

3) σg

4.[C] As supporting plane is lowered slowly ∴ N = mg – kx

5.[D] Net force acting on container due to liquid coming out from the holes is given by

F = ρA

×−×

4Hg2

4H3g2 = ρgAH towads left

∴ F = f = ρgAH towards right.

Now, τF = ρgAH × 2H into the plane of paper.

τf = ρgAH × 2H out of plane of papers

∴ τF = τf hence τN = 0

6.[C] TA = Rn

VP

A

AA and TB = Rn

VP

B

BB

Given, PA = PB , VA = VB and nA = 2nB

∴ TA = 2

TB

Now, A

B

B

A

B

A

MM

TT

VV

×= = 2

7.[A] Equivalent circuit

2, 2'

V0

ε2 C2V0

C1V0

ε1

1 1'2

2'3 3'

Middle plate

11' 33'

Total charge on 2 & 2' plate = Vd

Ad

A

2

02

1

01

εε+

εε

σ = ε0V

ε+

ε

2

2

1

1

dd

8.[A]

A

30B

15

30ºC

15º

O

AI

30º 120º

45ºC

BR O

l

Sine rule


º45sin

R = º30sin

l

l

R = 2º30sinº45sin

=

9.[A,B] As aT = aN

∴ ds

vdv = Rv2− which can also

be written as dtdv = –

Rv2

Integrating the above equations answer is obtained.

10.[A,C] (A) ρBLAg = ρ × 4L Ag + 2ρ ×

4L Ag

(B) FB = ρ × 4L Ag + 2ρ ×

4L3 Ag

FB = 47 ρLAg; a =

mmgFB −

11.[B,D]

c 6V

b 3Ω

8V

a

10

13V

•

11

← eq. ckt.

Using Kirchoff's Law Solve the circuit.

12.[A,B,C]

V

CV –CV

If battery is disconnected and plate are pulled

apart, then charge will remain constant

E = 0A2

Q∈

× 2 = 0

Q∈Α

∴ E remain same (A) is correct work is done against attractive force

→ ←−+ Fext.Felc by Fext. (B) is correct.

U = 21 CV2

V = constant [as battery is connected]

C = d

A0∈

as d increase C decrease ∴ U decrease option (C) is correct. 13.[C] As springs are in parallel

∴ a = )mm(x)kk(

massF

21

21net

++

=

14.[A] Frictional force on m2 will act in direction of

displacement if k2x > m2 a

15.[A] as k2A – µm2g = m2amax

∴ k2A – µm2g = m2

++

21

21

mmkk A

Solve to get answer. ]

16.[C] Induced emf across OP = 2

2B

21

ωl =

8B 2lω

(i) current = R8

B 2lω … (i)

Torque on the rod = 2 ∫2/

0

dxxBil

= 4

Bi 2l …(ii)

∴ R32

Bdtd

12M 422 ll ω

−=ω

× [substituting τ = I α]

– RM8B3d 22

0

l=

ωω

∫ω

ω∫t

0

dt

Solving this eq. & eq. (i)

i = R8

B 20lω e–αt

17.[B] θ = ∫∞

0

dti

18.[A] Heat generated = 20I

21

ω

Column Matching 19. [A] → p; [B] → p; [C] → r; [D] → q As cube is floating ρsALg = ρLAxg

∴ x =

ρρ

L

s L

20. [A] → r; [B] → p; [C] → s; [D] → q

S = 21 × 2 × 16 = 16 m

| Wg | = mg S = WN = m(g + a) cos2θ. S Wf = m(g + a) sin2θ . S


CHEMISTRY

1.[C]

NH2

NOBr

NH2 +

Br–

Br

+ Enantiomer

2.[B]

CH2=C

COOH

Ph H2/Pd

CH3–CH COOH

Ph (resolvable)

'A'

'B'

CH=CH–COOH H2/Pd

Optically inacitve

CH2–CH2–COOH

3.[B] Distribution of electrons in the MO's in He2 is σ1s

2 σ∗1s2 . He2 is unstable

Distribution of electrons in the MOS in H2 is σ1s2

. H2 is stable. 4.[B] After mixing total moles of A– = 100 × 0.2 × 10–3 + 100 × 0.3 × 10–3

= 100 × 10–3 × 0.5 moles After mixing total moles of HA = 100 × 0.1 × 10–3 + 100 × 0.2 × 10–3

= 100 × 0.3 × 10–3 moles

After mixing resulting pH = 5 + log35

5.[A,B,C]

CH2–CH*

NH2

COOH HO HO

It contains 2 phenolic hydrogens and a carboxylic acidic hydrogens

CH2–HC NH2

COO– HO HO

+

Zwitter ion

6.[A,B,C]

Grignard reagent reacting with acyl halide usually gives 3º alcohol.

7.[B,C,D] The order of H-bond energies F – H …….. F– > F – H …….. O > F – H …….. F >

O – H …….. O > O – H …….. F > N – H …….. N

8.[C,D]

9.[C,D] RHE reaction : Hg2Cl2(s) + 2e → 2Hg(l) + 2Cl– LHE reaction : 2Ag(s) + 2Cl– → 2AgCl(s) + 2e Net reaction : Hg2Cl2 (s) + 2Ag(s) → 2Hg (l) + 2AgCl(s) In case of same concentration of Cl– ions in the

two half cells, Ecell is independent on the concentration of Cl–. Other substances are either pure solids or liquids, which have unit activities irrespective of their amounts.

Column Matching : 10. [A] → p, r, s ; [B] → q;

[C] → p, q,,s ; [D] → q

11. [A] → p, q,t ; [B] → p, r, s; [C] → p, q, r; [D] → p, q, r

Numerical Response type questions : 12. [4]

EW =

96500Q or

n/4.106977.2 =

96500606013 ×××

∴ n = 4

13. [8]

O

)reductionKishnerWolff(

glycol/KOH

NHNH 22

−

→

(W)

)Ozonolysis(

OH/O 23 →

O

OHO

HO(X)

SOLUTION FOR MOCK TEST

PAPER - II IIT-JEE (PAPER - II)


∆

→ 2)OH(Ca O

(Y)

)oxidation

VilligerBaeyer(

COOOHHC 56

− →

(Z)

O

O

14. [1] Molecular weight of starch = 162n ∴ moles of C6H12O6 to be produced = 1 mol

∴ moles of ATP required =30

3000 = 100 mol

15. [7]

0

0

PPP − = x1 = mole fraction of solute

x1 = 0.0125

−1

x1

1=

18m1000

×

∴ m = 0.70

16. [5] Eq. of metal = Eq. of hydrogen

X/511.0 =

112009.43 ⇒ X = 2

Now eq. of metal = eq. of KMnO4

2Y/51

1.0−

=1000

8.581.0 ×

∴ Y = 5

17. [7] For the first bulb p1v1 = n1RT & second bulb p2v2 = n2RT or p1v1 + p2v2 = (n1 + n2) RT …(1) Suppose equilibrium pressure at each bulb = p atm Then p(v1 + v2) = (n1 + n2)RT …(2) From eq. (1) & (2) p1v1 + p2v2 = p(v1 + v2) 9 × 5 + 6 × 10 = p × 15 p = 105/15 = 7 18. [2]

Xe

• •

O O

O

XeO3

Xe

• • F

O FF

F

XeOF4

19. [6] Each has two geometrical isomers and two optical isomers (shown by cis-isomer).

MATHEMATICS

1. [A] We have

∑ ∑ ∑= = =

−=−n

1r

n2

1r

n

1r

444 )r2(r)1r2( = f(2n) – 16f(n)

2. [D] Using De Moivre's theorem fr (α) =

2r/ie α 2r/i2e α …. r/ie α

= )r......21)(r/i( 2e +++α

= ]2/)1r(r)[r/i( 2e +α = )r/11)(2/i(e +α

∴ )(flim nnπ

∞→=

∞→nlim )n/11(2/ie +π

= eiπ/2 = cos

π

2 + i sin

π

2 = i

3. [D] z = tt

δ+γβ+α ⇒ ( γ + δt) z = α + βt

⇒ (δz – β)t = α – γz

⇒ t = β−δ

γ−αz

z [Q αδ – βγ ≠ 0 ]

As t is real, β−δ

γ−αz

z = β−δ

γ−αz

z

⇒ (α – γz)( zδ – β ) = ( α – zγ )(δz – β)

⇒ ( γ δ – γ δ )z z +(γ β – α δ)z + (α δ – β γ ) z

= (α β – α β) ...(1)

Since δγ is real,

δγ =

δγ or γ δ – δ γ = 0

Therefore (1) can be written as a z + a z = c ...(2) where a = i(α δ – β γ ) and c = i( α β – α β ) Note that a ≠ 0 for if a = 0 then

a δ – β γ = 0 ⇒ βα =

δγ =

δγ [Q

δγ is real]

⇒ αδ – βγ = 0, which is against hypothesis. Also, note that c = i( α β – α β ) is a purely real

number.

Thus, z = tt

δ+γβ+α represents a straight line.

4. [C] dxdy = (x – 1)(x – 2)2 so 2

2

dxyd = (x – 2) × (3x – 4).

The points of inflection are given by 2

2

dxyd = 0

So x = 2, x = 4/3 are points of inflection.


5.[A,B,D] We have adj A = |A| A–1 adj (AB) = |AB| (AB)–1 = |A| |B| (AB)–1 = |A| |B| (B–1 A–1) = ( |B| B–1) (|A| A–1) = (adj B) (adj A) 6.[A,B,C] We have

A(α, β)′ =

ααα−α

βe000cossin0sincos

=

α−α−−α−α−

βe000)cos()sin(0)sin()cos(

= A (–α, β) Also, A (α, β) A( – α, –β)

=

αα−αα

βe000cossin0sincos

ααα−α

β−e000cossin0sincos

= I

⇒ A(α, β)–1 = A(–α, –β) Next, adj A(α, β) = |A (α, β)| A (α, β)–1 = eβ A(– α, – β).

7.[A,B,C,D]

Let (r + 1) th term of n

42 x

x5

+ be

independent of x. We have

Tr + 1 = nCr

rn

2x5 −

(x4)r

= nCr 5n – r x6r – 2n For this term to be independent of x, 6r – 2n = 0

or n= 3r. As each of 18, 21, 27 and 99 is divisible by 3, each of this can be a possible value of n.

8.[A,B] Let k = 2n + 1, then 2n + 1Cr is maximum when

r = n. Also 2n+1Cn = 2n + 1Cn+1 Thus, kCr is maximum when

r = 21 (k – 1) or r =

21 (k + 1).

9.[A,B,C,D] As a912, a951 and a480 are divisible by 3, none of

them is prime. For a91, we have

a91 = 43421

times91

)9....99(91 =

91 (1091 – 1)

= 91 [(107)13 – 1]

=

−−110

1)10(7

137

−−1101107

= [(107)12 + (107)11 + ….+ 107 + 1] × [106 + 105 + ….. + 10 + 1] = a91 is not prime.

Column Matching : 10. [A] → p, r; [B] → p,q, r; [C] → t ; [D] → p, q, r

sin θ = 23 = sin

3π

⇒ θ = nπ + (–1)n 3π = 2nπ +

3π

4 sin θ cos θ – 2 sinθ – 2 3 cos θ + 3 = 0

⇒ (2 sin θ – 3 ) (2 cos θ – 1) = 0

⇒ sin θ = 23

⇒ θ = nπ + (–1)n 3π , cosθ =

21

⇒ θ = 2nπ ±3π

(C) sin 2θ + cos 2θ + 4 sin θ = 1 + 4 cos θ ⇒ 2 sin θ cos θ + 1 – 2 sin2θ + 4 sin θ = 1+ 4cos θ ⇒ 2 sin θ (cos θ– sin θ) – 4(cosθ– sinθ) = 0 ⇒ (2 sin θ –4) (cos θ – sin θ) = 0 ⇒ sin θ = 2 or sin θ = cos θ ⇒ tan θ = 1 ⇒ θ = nπ + π/4, n ∈ I

(D) cos2 θ = 1/4= cos2 3π

⇒ θ = 2nπ ±3π

11. [A] → r ; [B] → r ; [C] → p ; [D] → q (A)We have b – a = c – b and (c – b)2 = a(b – a) ⇒ (b – a)2 = a(b – a) ⇒ b = 2a and so c = 3a.

Thus a : b : c = 1 : 2 : 3

(B) If the numbers are a and b, then x = 2

ba + and

b = ar3 ⇒ r = 31

ab

Now,

xyz

zy 33 +

=

)ar()ar(xrara2

6333 +

=

x)r1(a 3+

=

2baba

++ = 2

(C) c > 4b –3a ⇒ ar2 + > 4ar –3a ⇒ r2 – 4r + 3 > 0 ⇒ r < 1 or r > 3, But the terms

are positive so r ∈ (0, 1) ∪ (3, ∞)

(D) tan–1

2r21 = tan–1

2r42

= tan–1 )1r2()1r2(1)1r2()1r2(

−++−−+

= tan–1 (2r +1) –tan–1(2r–1)

∴

∑

=

−2

n

1r

1

r21tan (2n+1) – tan–1(1)

= tan–1 (2n + 1) – 4π


∴ tan

∑

=

−2

n

1r

1

r21tan =

1).1n2(11)1n2(

++−+ =

1nn+

∴ ∞→n

lim tan

∑

=

−2

n

1r

1

r21tan =

∞→nlim

1nn+

= 1

Numerical Response type questions : 12. [3] We have

2|z|9 = (2 + cos θ)2 + sin2 θ = 5 + 4 cos θ (1)

and

z3 +

z3 = 4 + 2 cos θ (2)

Eliminating θ from (1) and (2), we get

2|z|9 – 6

+

z1

z1 = – 3

⇒ 3 = 2( z + z) – |z|2 13. [6] Clearly x > 0 and x ≠ 1/5

log5x (5/x) =xlog5logxlog5log

55

55

+−

Putting log5x = t, then equation becomes

t2 +t1t1

+− = 1

⇔ t3 + t2 – 2t = 0 ⇔ t(t – 1) (t + 2) = 0 ⇔ t = 0, 1, – 2 So integral roots are 1 and 5.

14. [2] Let the A.P. be a – d, a, a + d, a + 2d. Note that a

and d must be integers. Also as this is an increasing a + 2d is the largest. We have

a + 2d = (a – d)2 + a2 + (a + d)2 = 3a2 + 2d2 ⇒ 3a2 – a + 2d2 – 2d = 0 As a is real, 1 – 8(d2 – d) ≥ 0

⇒ d2 – d – 81 ≤ 0

⇒ 2

21d

− ≤

83

⇒ 21 –

223

≤ d ≤ 21 +

223

As d is an integer, d = 0, 1 But d ≠ 0, therefore, d = 1. Thus 3a2 – a = 0 ⇒ a = 0 or a = 1/3. As a is an integer, a = 0.

Hence, required number is 2.

15. [1] We have E = 2n + 1C1 + 2n + 1C2 + … + 2n + 1Cn – 2n + 1C0 – 2n + 1C1 – … – 2n + 1Cn [using nCr = nCn – r] = – 2n + 1C0 = – 1. ∴ |E| = 1

16. [3] The given system of equations will have a non-trivial solution if

∆ =t2bac

at2cbcbt2a

++

+= 0

Clearly ∆ is a cubic polynomial in t and has 3 roots.

17. [3] None that it is not given that f is a differentiable function We have

f ′(4) =h

)4(f)h4(flim0h

−+→

= ( )h

)2(f)h4(flim22

0h

−+→

=h

8)h4(lim2/3

0h

−+→

=h

]1)4/h1[(8lim2/3

0h

−+→

=h

1......4h

2318

lim0h

−++

→=

h

.....h838

lim0h

+

→

18. [2]

2x + 2ydxdy = 0 ⇒

dxdy = –

yx

⇒ )3,1(dx

dy = –3

1

Therefore, the equation of the tangent at (1, 3 ) is

y – 3 = (–1/ 3 ) (x – 1)

and the point of intersection of this tangent with the x-axis is (4, 0). The equation of the normal at

(1, 3 ) is y – 3 = 3 (x – 1), and the point of intersection of this normal with the x-axis is (0, 0). Hence the required area is

21 . 4 3 = 2 3

19. [4] The two curves represent parabolas with vertices at (0, 0) and (3, 0). They intersect at (1, 1) and (1, –1), so the required area is

area of OPMQO = 2 (area of OPMO).


O

(1, 1)P

M(3, 0)

Q

Fig.

= 2

−+∫∫

3

1

1

0dx

2x3dxx

= 2

−−

3

1

2/31

0

2/3 )x3(32.

21x

32

= 2 .434

3222

32.

210

32 2/3 =

+=

−−

PHYSICS

1.[D]

d/211′

2′3 3′

2

1 1′3 3′

Cl 2′ 2

Cl

= 2 × d

2)A( 0 ×∈ = 4 .

dA0∈

= 4C

2.[A] 2t = n2

1m2 λ

+

t n mgF2

tmin = n4

λ = 38.14

105.5 7

×× −

= 99.6 nm

3.[D] I = dtdq ; q = it + a; V =

Cq

V = C

ait +

∴ V is proportional to time

4.[A] Phase difference corresponding to y1 = –π/2 and that for y2 = + π/2

∴ Average intensity between y1 and y2

= π1 ∫

π

π−

φ

φ

2/

2/

2max d

2cosI = Imax

π+π

2)2(

Hence required ratio = 21

π+

21

M.C.Q. Type questions :

5.[A,C]

a

l

r

dr

Consider a cylindrical element of radius r, thickness dr. If dR is the resistance of this element then

dR =drr2)r(

πρ l

Total resistance of the cylinder is given by

totalR1 = ∫ dR

1 =lαπ2

∫a

0

3drr

⇒ totalR1 =

lαπ2

4a 4

⇒ totalR1 = 4a

2παl

⇒ ltotalR = 22 )a(

2π

πα

⇒ R = 2A2πα

Since E =l

V (in magnitude)

⇒ E = I

ltotalR (By Ohm's Law)

⇒ E = 2AI2πα

6.[A,D] Consider only one plate as shown in figure.


The field on both sides of plate is shown in figure. Applying Ampere's Circuital Law to the contour C, we get

B

L

C B

2BL = µ0 (jL)

B =2

j0µ

Superimposing, the field due to two plates we get at P both fields cancel each other and at Q, they added to give B0 = µ0j

7.[B,D] v < g(t) ⇒ v < 2g due to Lenz's Law

Also, s <21 gt2

⇒ s < 2g due to Lenz's Law

8.[B,C]

C S

R3

R2

R1

C

R0 R1

V0

V

⇒ R0 =32

32

RRRR

+ and V0 =

32

2

RRVR

+

⇒ τ = C(R1 + R0) = C

+

+32

321 RR

RRR

⇒ q = q0 (1 – e–t/τ) ⇒ q = CV0 (1 – e–t/τ)

⇒ q =32

2

RRCVR

+(1 – e–t/τ)

9.[C] If it was A Y → 2Z Reactant : R = 60 × 8.5 = 510 MeV Product : P = 2 × 30 × 5 = 300 MeV ∆E = – 210 MeV ENDOTHERMIC If it was B W → X + Z R = 120 × 7.5 = 900 MeV P = 90 × 8 + 30 × 5 = 870 MeV ∆E = – 30 MeV ENDOTHERMIC If it was C W → 2Y

R = 120 × 7.5 = 900 MeV P = 2 × 60 × 8.5 = 1020 MeV ∆E = 120 MeV EXOTHERMIC If it was D X → Y + Z R = 90 × 8.0 = 720 MeV P = 60 × 8.5 + 30 × 5.0 = 660 MeV ∆E = – 60 MeV ENDOTHERMIC Column Matching : 10. [A] → p ; [B] → q, r, s; [C] → p,s ; [D] → q,r When cohesive forces are greater then adhesive

forces shape of meniscus is concave from liquid side and pressure is greater in concave side due to surface tension.

11. [A] → r ; [B] → p, s; [C] → q ; [D] → s

(A) T

dTV

dV

P=

⇒

T1

VdV

dT1

P=

(B) λ = ρπ 2d2

kT

(C) Ideal gas law are valid at all temperatures. (D) conceptual Numerical Response type questions : 12. [1]

v1

30º

30ºα

O

α

–v2N

Mv12 v'12

In the figure 12v→

= velocity of ball w.r.t. wedge

before coolision, and 12'v→

= velocity of ball w.r.t. wedge after collision, which must be in vergically upward direction as shown.

In elastic collision 12v→

and 12'v→

will make equal angle (say α) with the normal to the plane. We can show that α = 30º ∴ ∠MON = 30º

Now 2

1

vv = tan 30º =

31

≈ 1

13. [5]

A

h B

Ch


For a cylinder R

T

KK = 2 in case of

pure rolling upto B :

KT =32 mgh

and KR =31 mgh

After B : rotational kinetic energy is constant however translational kinetic energy increases.

At C : KT =32 mgh + mgh =

35 mgh

while KR =31 mgh

∴ R

T

KK = 5

14. [2] According to the formula, the position of the particle is at the centre of the path. It goes by 7 units to the right and then by another 7 units back to the initial position and then it goes to the negative side by 7/2 units for the first time.

So the minimum time is 2 ×4T plus the additional

time (t) to cover 7/2 units on the negative side.

Here ω =2π or

T2π =

2π or T = 4s

∴ tmin = 2 ×44 + t = 2 + t

Now 27 = 7sin

2π t ⇒

2π t = sin 30º =sin

6π

⇒t =31 s

∴ tmin = 2 +31 = 2.33s ≈ 2

15. [6] Let ω be the actual angular velocity of the

satellite from east to west and ωe be the angular speed of the earth (west to east). Then ωrelative = ω – (–we) = ω + ωe ⇒ ω = ωrel – we By the dynamics of circular motion

2RGMm = mω2 R or ω2 = 3

2e

RgR (Q GM =gRe

2)

⇒ ω = 3

2e

RgR ∴ ωrel = 3

2e

RgR + ωe

⇒ ωrel = 213

122

102104.610

××× + 7.27 × 10–5

(Q ωe = 864002π = 7.27 × 10–5)

⇒ ωrel =22.6×10–5 +7.27×10–5 = 30×10–5 rad s–1

∴ τ =relw

2π = 510302

−×π = 2.09×104 s = 5hr 48 min.

16. [1]

V1 =1

4

l8Prη

π , V2 =2

4

l8Prη

π , V3 =3

4

l8Prη

π

and V =l8

Pr 4

ηπ

Now V = V1 + V2 + V3 Substituting the values, we get

l =323121

321

lllllllll

++

=323121

321×+×+×

×× =116 m

17. [1]

At x1 = k3π and x2 = k2

3π

Sin k x1 or sin k x2 is not zero. Therefore, neither of x1 or x2 is a node

∆x = x2 – x1 =

−

31

23

kπ =

k67π

Since k

2π > ∆x >kπ

λ > ∆x >2λ

π

=k

2k

Therefore, φ1 = π

and φ2 = k.∆x =6

7π

∴ 2

1

φφ =

76

≈ 1

18. [1] With an AC source, current in the circuit is maximum when Z = Zmin = R (the resistance of coil)

∴ R =624 = 4Ω

When connected with an dc source of emf 12V.

i =Rr

12+

(r internal resistance of source)

=44

12+

= 1.5 A

19. [1] For incident electron

λ1 = ph =

Km2h =

mVe2h

and shortest wavelength of X-rays is λ2 = Vehc

∴ 2

1

λλ =

c1

me

2V

Substituting the values, we get 2

1

λλ = 1


PHYSICS

1.[C]

2.[B] v ∝ n T∝ because λ = constant

1N4N

++ =

289324 =

1718

17N + 68 = 18N + 18 50 = N

3.[B]

4.[A]

5.[B] u = 0 a = constant v0 = u + at = at

n' =

+vvv 0 n = n +

vtna

6.[C] LC = 1 MSD – 1VSD

LC =

−

30291 MSD =

301 MSD

720 MSD = 360°

1 MSD = 720360° = 1/2°

So LC = 301 ×

21° =

601° =

60'60 = 1'

7.[C] 8.[D] Ticker timer is a better device than a stop watch.

9.[C] M = 2

MM 21 + = 2

62.1230.10 + = 11.46 g

10.[B]

11.[B] YY∆ =

DD2∆ +

l

l∆

YY∆ = 2

4.001.0 +

8.005.0

= 2× 0.025 + 0.0625

YY∆ = 0.05 + 0.0625 = 0.1125

∆Y = 2× 1011 × 0. 1125 = 0.225 × 1011 So (2 ± 0.2) × 1011 N/m2

12.[B]

13.[B]

d

f F

f

It is only possible if object and image coincide.

14.[D] I = I 0 cos2 2φ

2I0 = I 0 cos2

2φ

cos 2φ =

21

φ = 90° = λπ2 ∆x

2π =

λπ2 ∆x

∆x = 4λ ⇒

Ddy =

4λ

y = d4Dλ = 4

7

10141105

−

−

××

×× = 45 × 10–3

y = 1.25 mm

15.[C] →τ = →p × →E = q (2

→a ) ×

→E

Here 2→a = (2 – 1) i + (– 1–0) j + (5 – 4) k

= i – j + k

→E = 0.20 i V/cm = 20 i V/m

∴ →τ = (4 × 10–6) ( )[ ]i20kj–i ×+ = 8 × 10–5

( )jk +

Magnitude of torque τ = 8 2 × 10–5 N-m 16.[D]

– σ

+ σ

+ σ

bc

a

C

B

A


PAPER - II AIEEE


Potential of shell A is,

VA = o4

1∈π

σπ+σπ−σπcc4

bb4

aa4 222

= 0∈

σ (a – b + c)

Potential of shell C is,

VC = 04

1∈π

σπ+σπ−σπcc4

cb4

ca4 222

= 0∈

σ

+− c

cb

ca 22

As VA = VC

∴ 0∈

σ (a – b + c) = 0∈

σ

+

− ccb

ca 22

or a – b = ( ) ( )c

baba +− or a + b = c

17.[C]

5

4

3

2

1

B

A

A B

1 2

2 3 5 4

3 4

≡

The capacity of each capacitor is, C0 = d

A0∈

From fig. it is clear that Ceq = 35 C0 =

35

dAo∈

18.[B] Resistivity of conductors increases with increase in temperature because rate of collisions between free electrones and ions increases with increase of temperature. However, the resistivity of semiconductors decreases with increase in temperature because more and more covalent bonds are broken at higher temperatures.

19.[B] let LA and LB be length of parts A and B

Then B

ARR =

B

ALL [as cross-section is same]

Now Lc = 2 LA and (volume)c = (volume)P i.e.Lc × Ac = 2 LA × Ac = LA × AA where Ac = AA are cross-sectional area of part C

and A. ∴ Ac = AA/2

∴ C

ARR =

cALAAL

//

c

A

ρ

ρ =

C

ALL ×

A

CAA

= A

AL2

L × A

AA

2A / = 41

20.[A] Current flowing through potential wire is –

I = rr15

E+

= r16

E

Potential drop across potential wire is,

V = I × 15 r = 16

E15

Potential gradient, K = 600

E1615

∴ 2E = Kl or

2E =

60016E15

× × l

or l = 215

60016×

× = 320 cm.

21.[A]

D i C

O

A BPa

b φ2

φ1φ2 φ1

Q

BAB = BCD = ( )2/b4i0

πµ

(sin φ1 + sin φ1)

= π

µ4

0 . bi4

22 ba

a

+

BBC = BDA = π

µ4

0 .ai4 .

22 ba

b

+

∴ B = BAB + BBC + BCD + BDA

= π

µ4

0 .22 ba

i4

+

+++

ab

ba

ab

ba

= π

µ4

0 . ab

bai8 22 +

22.[A] In tan A position,

π

µ4

0 3dM2 = BH tan 30° =

3BH …….(1)

Magnetic moment of second magnet, M' = (3m)(2 × 2l) = 6M In tan B position,

π

µ4

0 3dM6 = BH tanθ …… (2)

dividing eq. (2) by (1) we get


26 =

3/1tan θ or tan θ = 3 or θ = 60°

23.[B] Magnetic field of solenoid, B1 = l

110 iNµ

Magnet flux of coil, φ2 = N2 B1 A2 = N2

µl

110 iNA2

As φ2 = M i1, so M = 1

2iφ

= l

2210 ANNµ

∴ induced emf, |e| = Mdtdi1

or |e| = l

2210 ANNµ ×

dtid 1

= 30.0

102.13002000104 37– −×××××π × 25.04

= 4.8 × 10–2 Volt

24.[A] Z = ( )2CL2 XXR −+

Here XL = 2πfL = 2 × 3.14 × 500 × (8.1 × 10–3) = 25.4 Ω

and XC = Cf2

1π

= 6105.1250014.321

−××××

= 25.4 Ω

∴ Z = ( ) ( )22 4.254.2510 −+ = 10 Ω

Now irms = Z

Erms = 10

100 = 10 A

∴ VR = irms × R = 10 × 10 = 100 V

25.[D]

26.[C] g

2sinu2 θ = gh2u

( )8.9

452sin)14( 2 =

8.9)h(214 ⇒ h = 10 m

27.[B] Loss in P.E. = gain in K.E.

mg r = 21 mv2 ⇒ v2 = 2g r

ac = r

v2 = 2g

T – mg cos θ = rvm 2

⇒ T = 3 mg

28.[D]

29.[C] Loss in P.E. = gain in K.E. + work done against friction

mg R = 21 m (1.4 gR) + Wf

Wf = 0.3 mgR

Now, W B → C = mg R + 0.3 mgR = 1.3 mgR

30.[C] dtdm × u = M

dtdv

0.5 × 400 = 2000 × 5v∆ ⇒ ∆v = 0.5 ms–1

CHEMISTRY

31.[C] HCO3– can donate a proton to CO3

2– and it can accept a proton to form H2CO3.

32.[C] O F2 ⇒ O = +2 O2 F2 ⇒ O = +1 O2 [PtF6] ⇒ O2

+ + [PtF6]–1

O2+ ⇒ 2x = +1

x = +1/2 33.[A] ∆G = ∆H –T ∆S if ∆H = ⊕ & ∆S = ⊕ & T ∆S > ∆H than ∆G = Θ & process is spontaneous. 34.[C] Calcined Gypsum is calcium sulphate 35.[A] Inter particles forces between CH3COCH3 &

CHCl3 are strong H-bonding. Thus solution shows negative deviation. ∆Vmixing = Negative.

36.[C] Addition of inert gas at constant volume does not cause any effect on the equilibrium.

37.[A] For I order Reaction t1/2 is constant

24tg625.0

24tg25.1

g5.2

24tg5

24tgm10

2/1

2/1

2/1

2/1

=

↓

=

↓

=

↓

=

↓

= 96 hours.

38.[D] 22 Oof.wt

Agof.wtOof.wt.EqAgof.wt.Eq

=

6.1

W8

108=

WAg = 21.6 gm,


39.[B] Solution is decinormal, that is N/10 & x factor is 1, so conc. = 0.1M

[H+] = c.α = 0.1 ×1.3/100 = 13×10–4 pH = –log [13×10–4] = 2.89 40.[A] The no. of atoms in fcc lattice = 4 a = 400 pm = 4×10–10 m = 4×10–8cm

d = 3aNoMn

×× = 38–23 )104(106

604×××

×

d = 6.23 g/cm3.

41.[B] For any Value of l possible values of m are m = –l to + l l = 2, m = –2, –1, 0 +1, +2 So option is (B)

42.[B] C : O : H 6gm : 3.01 × 1023 atoms : 2 mole Ratio ½ : ½ : 2 of mole 1 : 1 : 4 COH4 or CH4O

43.[B]

3 2

1

44.[B] NaNO2/HCl gives HNO2 which gives different products with Pri. and Sec. amines.

45. [A] CH3+ Cl2 ν→h

CH2Cl +HCl

46.[D] 2 CuSO4 + 2 KCN → Cu(CN)2 + K2SO4 2Cu(CN)2 → 2CuCN ↓ + (CN)2 ↑ 3 KCN + CuCN → K3[Cu(CN)4]

47. [B] Blue print process occurs with the help of Iron Compound.

48.[A] Effective nuclear charge increases therefore ionic radius follow the order.

49.[B] xy2 xy + y t = 0 : 600 0 0 teqm : 600-P P P Now : 600 – P + P + P = 800 P = 200 mm Hg

Hgmm100400

200200)Pxy(

)Py)(Pxy(Kp2

=×

==

50.[C] Cell reaction : Zn + Cu+2 → Zn+2 + Cu Cell emf:

Ecell = E°cell – n059.0 log

+

+

2

2

CuZn

So doubling the conc. of ions. Ecell remains same.

51.[A] pKa = –log Ka = –log (1.8×10–5) = 4.7447

[CH3COOH] = L

mol4.05.060

12=

×

[CH3COONa] = 5.0

4.16×82

= 0.4 L

mol

Now, pH = pKa + log

acidsalt

= 4.7447 + log

4.04.0

= 4.7447

52.[B] FeC2O4 → Fe+3 + CO2 +2 +3 +3 +4

Increase = 1

Increase = 1

Total Increase in O.N. = 3 So valence factor of FeC2O4 = 3 KMnO4 →Mn+2 +7 +2 v.f. (KMnO4) = 5 gm E KMnO4 = gm E of FeC2O4 Mole × v.f. = Mole × v.f. 1× 5 = x × 3 ⇒ x = 5/3 53.[A] ∆G = ∆H – T ∆S ∆G = Θ , ∆G < 0 (Spontaneous process) ∆G = ∆H – T ∆S = ∆E + P ∆V – T ∆S (∆G)E,V = 0 + 0 – T ∆S (∆S)E,V = ⊕ ⇒ ∆G = Θ (Spontaneous process)

54.[B]

nCarbocatio2

3CH–HC–3CH

|

3CH|C–3CHH

O2H–3CH–

OH|CH–

3CH|

3CH|C–C3H

°

++→

1, 2-Methyl shift

)stableMore(nCarbocatio3

CH–

3CH|CH–

3CH|C–CHCH–

3CH|C

3CH|C–CH 33

H–33

°

+←= +

55.[C] Factual Q.


56.[D]

OH Zndust→Cl–CH

AlCl

3

3

→ CH3

4KMnO.alk → COOH

57.[D] 58.[A] 4 HCl + O2 → 2 H2O + 2 Cl2 Chlorine is in the form of cloud.

59.[A] Coordination no. = 6 Oxidation no. = 3 no. of d electron = 6 no. of Unpaired d electron = 0

60.[C] Resonance structure should have same number of electron pairs.

MATHEMATICS

61.[A]

P

h

O 60º30º A Let the height of the tower is h 1000 = h cot 30º – h cot 60º

⇒ h = º60cotº30cot

1000−

h = 500 3 m

62.[D] since variance is independent of change in origin. Hence variance of observations 101, 102, ...... 200 is same as variance of 151, 152, .....250.

∴ VA = VB

⇒ B

A

VV = 1

63.[C] centroid of triangle is given by

−++3

3ba,3

3c2

if centroid lie on y axis ⇒ abscissae = 0 c2 + 3 = 0 ⇒ no real c exist if centroid lie on x axis ⇒ ordinate = 0 ⇒ a + b – 3 = 0 ⇒ a + b = 3

64.[C] f(x) = sin x – cos x – kx + b f '(x) = cos x + sin x – k f '(x) = 2 sin(π/4 + x) – k if f(x) is decreases for all x ∴ f '(x) is –ve i. e. k > max. of 2 sin(π/4 + x) i. e. k > 2

65.[A] 0xlim→

−2

x

xxcose

2

= 0xlim→

−+

−22

x

xxcos1

x1e

2

1 + 0xlim→ 2

2

x2/xsin2 = 1 +

21 =

23

66.[A] Second determinant has been obtained from the first by the operation C1 → C1 + 2C2 – 3C3. so its value remains

unchanged

67.[C] given 3sin x – 4sin3x – k = 0 ⇒ 3sin x – 4sin3x = k ⇒ sin3x = k

.........(i) angle A and B satisfy the equation (i) ∴ sin 3A = sin3B = k ⇒ sin3A = sin3B But A > B ⇒ A B Now, sin3A = sin(π – 3B) 3A = π – 3B

A + B = 3π ⇒ C =

32π

68.[B] given statement (p q) ∨ ~ r → (p ∧ r) (F ↔ F) ∨ F → (F ∧ T) T ∨ F → F T → F = F 69.[C]

TFTFFFTTTTTFFFFTFTFFFTTT

p~q)qp(p~qp~qpqp ∧↔∨∧∨

Hence neither tautology nor contradiction 70.[D] Given sequence can be written as

25 ,

1320 ,

910 ,

2320 , ..............

or 820 ,

1320 ,

1820 ,

2320 ..........


which is a H.P. nth term of corresponding A.P.

an = 208 + (n – 1)

205 =

203n5 +

nth term of H.P. = na

1 = 3n5

20+

also 3n5

20+

= 175

5n + 3 = 20 × 5

17 = 68

n = 565 = 13

71.[B] Let any point P divides line joining A(–2, 4, 7) & (3, –5, 8) in ratio λ : 1

then P

−λ+λ

+λλ−

+λλ+−

178,

154,

132

if P lies on plane 2x – k = 0

⇒ x = 2k

2k =

132

+λλ+− given λ = 9

k =

1025 × 2 ⇒ k = 5

72.[C] |a × b| = |a| |b| sin θ

sin θ = 2.5

8 = 54

⇒ cos θ = 53

|a – b|2 = |a|2 + |b|2 – 2a.b = 4 + 25 – 2|a| |b| cos θ

= 29 – 2.2.553 = 29 – 12

|a – b| = 17 73.[C]

1

P2

(0, b)A

B(a, 0)

given length AB = 6 or a2 + b2 = 36

if PBAP =

21

co-ordinates of point P given by

3b2,

3a

locus of point P, a2 + b2 = (3x)2 + 2

2y3

= 36

or 16y

4x 22

+ = 1

b > a then eccentricity e = 2

2

ba1− =

23

74.[D] parabola y = x2 + 2px + p2 + 13 – p2 (y – (13 – p2)) = (x + p)2 vertex is given by (– p, 13 – p2) is 4 units above x-axis ⇒ 13 – p2 = 4 ⇒ p = ± 3 also lies in Ist quadrant ⇒ p < 0 ⇒ p = – 3 75.[C] If lines x2 + 2λx + 2y2 = 0 & (1 + λ) x2 – 8xy + y2 = 0 are equally inclined ⇒ Their bisectors eqn must be same

⇒ eqn

21yx 22

−− =

λxy

& 1)1(

yx 22

−λ+− =

4xy−

are same

⇒ x2 – y2 = λ−

xy & x2 – y2 = λ− /4

xy are same

⇒ – λ = λ−4

⇒ λ2 = 4 ⇒ λ = ± 2 76.[D] given circle in standard form is

x2 + y2 + 2

+

t1t x – 2

−

t1t y + 1 = 0

centre is given by

−−−

t1t,

t1t

or h = –

+

t1t & k = t –

t1

h2 – k2 = 2

t1t

+ –

2

t1t

− = 4

locus x2 – y2 = 4 which is a hyperbola


77.[C] ∫−

4

1

)x(f dx = 4

⇒ ∫−

2

1

)x(f dx + ∫4

2

)x(f dx = 4

⇒ ∫−

2

1

)x(f dx = 4 – ∫4

2

)x(f dx ...........(i)

Q )f(x)– (34

2∫ dx = 7 (given)

⇒ ∫4

2

dx3 – ∫4

2

)x(f dx = 7

⇒ 3[x]42 = 7 + ∫

4

2

)x(f dx

⇒ 3 × 2 = 7 + ∫4

2

)x(f dx ⇒ ∫4

2

)x(f dx = – 1

.....(ii) put this value from (ii) in (i), we get

∫−

2

1

)x(f dx = 4 – (–1) = 5

⇒ ∫−1

2

)x(f dx = – 5

78.[A] lines – 2x – y + 6 = 0 & 4x – 2y + 7 = 0 (make c1 & c2 +ve) now a1a2 + b1b2 = – 8 + 2 = – 6 < 0 bisector by +ve is acute and contains origin eqn

given by 5

6yx2 +−− = + 52

)7y2x4( +−

– 4x – 2y + 12 = 4x – 2y + 7 8x = 5

79.[C] [ ]∫ −2

1

1)x(gf f 'g(x). g'(x) dx

put f g(x) = t ⇒ f 'g(x) g'(x) dx = dt ∴ required integral [ ]

1

2)x(gflog

log (f g(2)) – log (f g(1)) = 0 Q g(1) = g(2)

80.[C] f(x) = ∫ ++

2

22

x1xsinx sec2x dx

= ∫

+−+

2

22

x1xcos1x sec2x dx

= ∫

+− 2

2

x1xcos1 sec2x dx

= ∫

+− 2

2

x11xsec dx

= tan x – tan–1x + c ∴ f(0) = 0 ⇒ tan0 –tan–10 + c = 0 ⇒ c = 0 ∴ f(x) = tan x – tan–1x ∴ f(1) = tan1 – tan–11 = tan1 – π/4

81.[B] we know x –[x] = x ∴ Domain of x is R & x ∈ [0, 1) but in f(x), x is in denominator & it should

not be equal to zero ∴ x 0 ⇒ x I and domain of sec–1x is R –(–1, 1) ∴ domain of f(x) is R – (–1, 1) – I

82.[C] Q sin–1x is defined for |x| ≤ 1 and sec–1x is defined for |x| ≥ 1 therefore both defined for |x| = 1 ⇒ x = 1,–1 ∴ Df = –1, 1 further f(–1) = sin–1 (–1) + sec–1 (–1) = – π/2 + π = π/2 and f(1) = sin–1(1) + sec–1(1) = π/2 + 0 = π/2 Hence Rf = π/2

83.[B] Let A(z1), B(z2) and C(z3) be the vertices of the triangle then

|z1| = |z2| = |z3| ⇒ |OA| = |OB| = |OC| O being the origin ⇒ O is circumcentre of the triangle, Also, the

triangle is equilateral, therefore circumcentre coincide with the centroid

⇒ origin is centroid

⇒ 3

zzz 321 ++ = 0

⇒ z1 + z2 + z3 = 0 84.[B] Total students n = 100 Average marks x = 72 Total boys n1 = 70 average marks of boys 1x = 75 Total girls n2 = 30

now x = n

xnxn 2111 +

∴ 2x = 2

11

nxnxn −

2x = 30

757072100 ×−× = 65


85.[C] Letters of word 'STATISTICS' are 1A, 2I, 1C, 3S, 3T total = 10 Letters of word 'ASSISTANT' are 2A, 1I, 1N, 3S, 2T total =9 common letters are A, I, S & T

probability of choosing A is = 101 ×

92 =

902

probability of choosing I = 102 ×

91 =

902

probability of choosing S = 103 ×

93 =

909

probability of choosing T = 103 ×

92 =

906

total = 902 +

902 +

909 +

906 =

9019

86.[D] A = – 2, –1, 0, 1, 2 ⇒ n(A) = 5 B = 0, 1, 2, 3 ⇒ n(B) = 4 C = 1, 2 ⇒ n(C) = 2 D = (1, 7),(2, 6),(3, 5),(4, 4),(5, 3)(6, 2),(7, 1), ⇒ n(D) = 7 (A ∪ B ∪ C) = – 2, – 1, 0, 1, 2, 3 ⇒ n(A ∪ B ∪ C ) = 6 n(D) = 7 n(B ∪ C) = 4

87.[B] this is of the form

f '(y)dxdy + f(y). f(x) = Q(x)

put tan y = z

∴ sec2ydxdy =

dxdz

∴dxdz

+ 2xz = x3

∴ ∫pdxe = ∫ xdx2e =

2xe

∴ soln is z 2xe = ∫ 3x x.e

2dx

= 21

∫ 2x x.e2

2x dx

= 21

∫ tdt.et = 21 et (t – 1) + c

∴ tan y 2xe =

21 2xe (x2 – 1) + c

⇒ tan y = c2xe− +

21 (x2 – 1)

88.[D] The given curves are y2 = 8x ...(1) and xy = – 1 ...(2) any tangent to (1) is

y = mx + m2

...(3)

=⇒=∴

+=

=

2a8a4mamxyis

ax42ytogenttanQ

we shall find m so that it touches (2) ∴ from (2) & (3)

1m2mxx −=

+ ⇒ m2x2 + 2x + m = 0 ...(4)

(3) touches (2) if quadratic (4) has equal roots ∴ D = 0 ⇒ 4 – 4m3 = 0 ⇒ m3 = 1 ⇒ m = 1 ∴ required common tangent is y = x + 2

89.[B] h(x) = (f(x))2 + (g(x))2 ∴ h'(x) = 2f(x)f '(x) + 2g(x) g'(x) = 2f(x)g(x) + 2g(x)g'(x) Q f '(x) = g(x) Also f '(x) = g(x) ∴ f"(x) = g'(x) ⇒ – f(x) = g'(x) ∴ h'(x) = 2f(x) g(x) – 2g(x)f(x) h'(x) = 0 ∴ h(x) = constant for all x but h(5) = 11 ∴ h(x) = 1 for all x ∴ h(10) = 11

90.[A] for m1 : Let I & N are assumed single letter in IN order Now total no. of letter are IN T E G E R = 6 letter

these 6 letter can be arranged in row

= 26

× 2 = 720 ways

∴ Letter of words INTEGER can be arranged in

a row = 27

= 2520 ways

Now no. of ways in which IN are not together is m1 = 2520 – 720 = 1800

Now for m2 : I- - - - - R → rest five can arrange

= 25

= 60 ways

∴ 2

1

mm =

601800 = 30


PHYSICS

1.[B] In case of damped vibrations, amplitude decreases exponentially with time

∴ A = A0e–bt or 0A

A = e–bt

or 21 = eb×2 and

0A´A = e–b×b

or 0A´A = (e–2b)3 =

3

21

=

81

2.[C] Here ek

1 = k1 +

k21 +

k41 + ....... ∞

= k1

∞+++ ......

41

21

11

=

− 2/111

k1 =

k2 i.e. ke =

2k

3.[ B] KEmax = 21 kA2 =

21 Mω2A2

8 × 10–3 = 21 × 1 × ω2 × (0.1)2

ω = 4 rad/sec y = A sin(ωt + φ) or y = 0.1 sin (4t + π/4)

4. [B] Here →L = m )vr(

→→× = m v y(–

→k )

(Where y is the vertical distance of particle from x axis)

Here m,v and y all are fixed so →L . remains

constant. 5. [B] Using, weight of floating body = weight of

liquid displaced.

we get V ρ g =

2V (13.6) g –

2V (0.8g)

(buoyant forces of mercury and oil act in opposite direction)

Then, ρ = 2

8.06.13 − = 6.4

6. [C] v´ = s

0

vvvv

−+ v We get

νν´ =

s

0

vvvv

−−

i.e. 89 =

´v340´v340

−+ ∴ (v0 = vs = v´)

i.e. v´ = 20 ms–1

7. [A] Ilog = 1

2

II dB; i.e. 4 = 10 log

)101010(I

492

×× −

Then I2 = 2.5 × 10–4 Wm–2

8. [A] Here work done = pdv and area under the curve given work done

∴ 10 + WCA = 5 or WCA = –5 J

9. [C] For equilibrium F = qE = mg

or neE = 34

πr3ρg or r = 3/1

g4neE3

πρ

10.. [A] i = 111

5.15.15.1++−+ = 0.5 A

As the current has to from A to C to B, for kirchhoff's law, VA = 0.5 × 1 + 1.5 = 1V (Q v = E – ir) VB = 0.5 × 1 + 1.5 = 1V VC = 0.5 × 1 – (–1.5) = 2V

11. [D] R = R1 + (273 – T) α ...(i) or 2R = R0[1 + T´α] ...(ii)

Deciding (i) by (ii) 21 =

α+α

T´1T)-1(273

or 1 + T´α = 2 + (273 – T)2α

or T´ = α

+α 1T)2-(273

12. [A] Current, I = 21 rrR

E2++

P.O. across cell 1 = Ir1 = 21

1

rrREr2

++

For zero p.d. the fall of potential should be equal to in emf.

E = 21

1

rrREr2

++ i.e. R = r1 – r2


PAPER - II

BIT-SAT


13. [B] Point P lies on the arms CD and AF so inclusion at P due to them is zero.

Magnetic induction at p due to currents in AB and BC is given by

B1 = B2 = )a2(4

iµ0

πsin 45º

= a28

iµ0

π (Q distance of p from AB or BC is

2a) similarly due to DE and EF

B3 = B4 = a4iµ0

π sin 45º = ⊗

πa24iµ0

Net induction = 2(B1 – B3)

= 2a28

iµ0

π –

a24iµ0

π = –

a8iµ2 0

π

14. [C] v = rω = r × T2π

or T = v

r2π = qB

m2π = q

km

Now mα = 4mp and qα = 2qv

Tp = kαq

mp and Tα = kα

α

qm

= kP

P

q2m4 = 2k

αqmP

⇒ Tα = 2Tp or Tp = 21 Tα

15. [D] i = i0(1 – e–t/τ)

i =

−

−RL/t

e1Rv =

−

−×−

6104.8t

3

e16

12 = 1

(Q i = 1A given) ⇒ t = 0.97 × 10–3 s , i.e. t ≈ 1ms

16. [A] Optical distance between fish and the bird is Differentiating w.r.t.

dtds =

dt´dy +

dtµdy

i.e. 9 = 3 + 34

dtdy

or dtdx = 6 ×

43 = 4.5 ms–1

17. [A] f1 =

−1

µµ

m

g

−

21 R1

R1

f1 =

−1

75.15.1 ×

−

− R1

R1 =

R5.31

i.e. f = 3.5 R. In the medium it behaves as a convergent lens.

18. [D] Reflection of light from plane mirror gives additional path difference of λ/2 between two waves.

∴ Total path difference = 2

3λ + 2π = 2λ

which satisfy the condition of maxima. Resultant intensity ∝ (A2 + A2) [Q I ∝ Α2] 4Α2 = 4Ι

19. [B] Here 1

2

λλ

)()(

20

10

λ−λλ−λ =

12

or 7.34.5

)104.5()105.3(

70

70

−

−

×−λ

×−λ = 12

or λ0 = 11.8 × 10–7 m

But ω = 0

hcλ

=)106.1)(108.11(

)103)(106.6(197

834

−−

−

×××× = 1.05

eV

20. [D] Let whole the energy of electrons be converted in x-rays. eV = hv

or eV = λhc

or λ = eVhc =

)1040)(100.1()103)(106.6(

319

834

××××

−

−

i.e. λ = 3.1 × 10–11 m or λ = 0.31 Å

21. [A] Here 88Ra222 →α 86R218 →−β 87Fr218

→α 85Al214 →−β

80Rn214 →α 84PO210 →α 82Pb206

4α deceys and 2β decays.

22. [A] A1 = 4, A2 = 3 and θ = 2π = 90º

∴ Resultant amplitude,

Α = º90cosAA2AA 2122

21 ++ = 2

221 AA +

= 22 34 + = 25 = 5 unit 23. [C] Using d sin θ = nλ

sin θ = θ = Dλ

∴ Ddy = nλ or y =

dnD

λ

i.e. 3

7

101)106(21

−

−

×××× = 1.2 × 10–3 = 1.2 mm

Distance between first minima on either side of centrar maxima ∆y = 2y = 2.4 mm

24. [B] For constructive inteference

Imax = ( )221 II + = ( )2II2 +


For destuctive interference

Imin = ( )221 II − = ( )2II2 −

Then max

min

II =

2

2

)II2()II2(

+

− = 2

2112

+

− = 341

25. [B] Using decay equation A2 = A1e–λt

or e–λt = 1

2

AA or λt = log

2

1

AA

time t = λ

21 A/Alog = 1.44T loge

2

1

AA

26. [B] Vx = ka

)a4( 2 σπ – b

)b4(k 2 σπ + b

)c4(k 2 σπ

= 4π 04

1πε

σ(a – b + c)

πε

=04

1kQ

= 0ε

σ (a – b + c)

27. [D] Let the given quantity be x1 then,

x = 3π (a2 – b2)h =

3π (a2h – b2h)

= 3π a2h –

3π b2h

Each term has the dimension of x1 then [x] = [a2h] = [L2L] = [L3] and also [x] = [b2h] = [L2L] = [L3] quantity is volume.

28. [B] U = Bx

xA2 +

or A = x

)Bx(U 2 +

Here dimensions of x2 and B should be same. i.e. [B] = [x2] = [L2]

Also [A] =

−

2/1

22

LTML [L2] = [ML7/2T–2]

Then [AB] = [ML7/2T–2] [L2] = [ML11/2T–2

29. [A] v = u

−

´tt1 or

dtdx = u

−

´tt1

integrating, x = u

−

´t2tt

2 + C

at t = 0, n = 0 and c = 0

∴ x = u

−

´t2tt

2 = 10t

−

10´tt

Putting t = 10

x = 10 × 10

−

10101 = 0

30. [A] using h = 21 gt2, we get t1 =

gh2

let t1 be the time taken from instants of jumping to the opening of parachute, then

t1 = 8.9402× = 2.86 sec

His velocity at this point is given by v1

2 = 2gh1 = 2 × 9.8 × 40 = 784 or v1 = 28 ms–1 for the remaining journey, v = v1 + at2

or t2 = a

uv − = 2282

−− = 13sec

∴ total time = t1 + t2 = 2.86 + 13 = 15.86 ≅ 16 s

31. [B] Let u be the velocity of projectile w.r.t. tanks velocity v then

ux = u cos 30 + v; uy = u sin 30º

and T = g

º30sinu2

Range, R1 = uxT = g

º30sinu2 (u cos 30º + v)

for y axis ux

´ = u cos 30º – v and uy´ = u sin 30º

T = g

º30sinu2

Range e, R2 = Tu´x

= g

º30sinu2 (u cos 30º – v)

Then R1 + R2 = gu4 2

(sin 30º cos 30º)

R1 – R2 = gu4 v sin 30º

Eliminating u we get

v2 = 30tan4

g)RR(

)RR(

21

221

+−

= 30tan4

10)200250(

)200250( 2

+− = 24 m2s–2

⇒ 4.9 ms–1

32. [D] Let α be the angle between velocities of pair of particles then relative velocity is given by

vr = α××−+ cosvv2vv 22 = 2 v sin2α

average vr = ∫∫

π

π

α

α2

02

0

d

)2/(sinv2 dα = π4 v


33. [B] For quarter revolution

∆→V =

→

2V – →

1V

angle between →

1V and →

2V is 90º

E

S

N

W

v1

–v1

v2

a

∴ ∆v = 2

122 vv + = 22 vv + = 2 V

Also tan–1

vv = 45º

∴ ∆→v = 2 v south west

34. [D] For vertical motion

H = 21 gt2 or t = g/H2

For horizontal motion, distance covered is given by

2πrn = ut or 2πrn = u g/H2

or u = g/H2r2

4π

35. [A] On descending (Mg – f) – Ma = 0 (where f is the upthrust due to buoyancy) On ascending, f – (M – m)g – (M – m)a = 0

m =

+ gaa2 M

36. [A] The masses will be lifted if the tension of the string is more than the gravitational pull of mass.

50 N20 N

50 N

50 – 2T = 0 or T = 25N So, 5 kg weight cannot be lifted but 2 kg weight

will be lifted

25 – 20 = 2a or a = 25 = 2.5 ms–2

37. [A] On cutting of string QR, the resultant force on m1 remains zero because its weight mg is balanced by the tension is the spring but on block m2 a resultant upward Force (m1 – m2)g is den sped. The block m1 will have no resultant acceleration where as m2 does have an upward

acceleration given by 2

21

mg)mm( −

38. [A] Here v = 2/1

Mpt2

or dtds =

2/1

Mpt2

or ds = 2/1

Mpt2

dt

integrating s = 2/1

Mp2

32 t3/2 + C

at t = 0, S = 0, so c = 0

S = 2/1

M9p8

t3/2

39. [B] Let a small displacement be given to the system

in vertical plane of frame such that ST remains horizontal then let vertical displacement of centres of rods up and QR be y then vertical displacment of centres of VT and RS will be 3y and that of TS will be 4y. Equating total vertical work to zero we get

P Q

W R

T S

W

W

W W

W

W

yyyy

(w + w)δy + (w + w)3 δy + w(4δy) – T(4δy) = 0 or 2w + 6w + 4w = 4T or T = 3w 40.[A] Normal reaction R = mg = 2 × 9.8 N Frictional force, F = µR = 0.2 × 2 × 9.8 = 3.92 N Distance traveled 2 × 5 = 10 m ∴ Work done = f × s = 3.92 × 10 = 39.2 J


CHEMISTRY

1.[D] Fe2O3 + 3CO → 2Fe + 3CO2 160 g 112 g

Pure Fe2O3 in ore = 10

80100× = 80 kg

Iron produced from 80 kg of Fe2O3

= 160112 × 80 = 56 kg

2.[D] ∆v = )x(m4

h∆π

= )1.010)(1011.9)(14.3(4

10626.61031–

34–

×××

−

= 5.79 × 106 ms–1

3.[B] Given n = 0.5

Then

+ 2

2

VanP (V – nb) = nRT

⇒ [P + 2

2

Va)5.0( ] [V – 0.5b] = 0.5 RT

⇒

+ 2V4

aP (2V – b) = RT

4.[B] N2O4(g) 2NO2(g) n0 : 1 0 ne : 1 – 0.5 2 × 0.5 = 0.5 1

Ce : 55.0 M

51

= 0.1 M = 0.2 M

then KC = ]ON[

]NO[

42

22 =

)1.0()2.0( 2

= 1.0

04.0 = 0.4

5.[A] →<<<

strengthAcidicHClOHClOHClOHClO 432

Weakest acid has strongest conjugate base.

6.[C] HCl : NA × VA = (0.4 × 1) × 50 = 20 meq. NaOH : NB × VB = (0.2) × 50 = 10 meq. Q NAVA > NBVB

∴ [H+] = BA

BBAA

VVVNVN

+− =

1001020 −

= 0.1 M = 10–1 M, ∴ pH = 1 7.[D]

(+1) H – O – O – S(x) – O – H

(–1)

(–1) (+1)

(–2)

O

O

(–2)

(–2) 2(+1) + 2 (– 1) + x + 3(–2) = 0 ⇒ x = + 6

8.[C] (+2)(+3) FeC2O4 → Fe3+ + CO2

3+ (+4)

–2e

–1e

∴ Valence factor (FeC2O4) = 1 + 2 = 3

∴ E( FeC2O4) = 3M

9.[D] The fractional of total volume occupied in simple

cube = cubeofvolume

particlesofvolume = 3

3

a2a

34

π

= 6π

10.[A] Cu2+ + e– → Cu+ ; 0

1E = 0.15 V

∴ 01G∆ = –1 × 0.15 × F = – 0.15 F

Cu+ + e– → Cu ; 02E = 0.5 V

∴ 02G∆ = – 1 × 0.5 × F = – 0.5 F

Cu2+ + 2e– → Cu ; 03E = ?

∴ 03G∆ = – 2 × 0

3E × F = – 2F 03E

Also, 03G∆ = 0

1G∆ + 02G∆

= – 0.15 F + (– 0.5 F) = – 0.65 F Now – FE2 0

3 = – 0.65 F

or 03E =

265.0 = 0.325 F

11.[A] 4hrs = 4 half lives 1

→ 2/1t

21

→ 2/1t

41 → 2/1t

81

→ 2/1t

161

fraction left after 4 half lives = 161 or

4

21

fraction reacted in 4 half lives = 1 –161 =

1615

12.[B] A solution showing +ve deviation has higher vapour pressure and lower boiling point.

13.[C] In multi molecular solutions the different layers

hold each other through van der Waal's forces. 14.[A] CS2 + 3O2 → CO2 + 2SO2 ; ∆rH = ? ∆rH° = 0

)P(f HΣ∆ – 0)R(f HΣ∆

= [2(– 297) + (– 393)] – (117) = – 1104 kJ mol–1

15.[C] Aspirin chemically acetyl salicylic acid OCOCH3

COOH


16.[A]

NH2

+ NaNO2 + 2HCl → °− C50

N2Cl

+ 2H2O + NaCl

(CH3)2N H + Cl – N = N →−HCl

(CH3)2N N = N

17.[A] R – C – O – C2H5 + CH3MgBr

O →

:O–

CH3 R

OC2H5

→ −− OHC 52

R – C

O

CH3

→ MgXCH3

XMgO

CH3 R

CH3 →

+H/HOH HO

R

CH3

CH3

18.[C] It is cannizzaro reaction

Cl

CHO

∆ →KOH

Cl

COO–

+

CH2OH

Cl

19.[C] Phenol is less acidic than acetic acid and p-nitrophenol.

20.[A]

OH

+ C2H5I OHHC

.Anhy

52

→

O – C2H5

21.[C] CH ≡ CH 4

42HgSO%1

SOH%40 CH2 = CH – OH

→ − mtautomerisenolKeto CH3 – C – H

O Acetaldehyde

22.[C] The rate of nitration is greater in

hexadeuterobenzene 23.[D] Halogenation on alkene occurs by electrophilic

addition.

24.[A] Twisted boat is chiral as it does not have plane of symmetry.

25.[C]

O C – CH3

Acetophenone has highest dipole moment.

26.[A]

N CH3

CH3CO

N,N-dimethyl cyclopropane carboxamide.

27.[B] NaCl + H2SO4 → NaHSO4 + HCl K2Cr2O7 + 2H2SO4 → 2KHSO4 + 2CrO3 + H2O CrO3 + 2HCl → CrO2Cl2 orange red vapour 28.[C]

Co

Cl

Cl

en

en

+

Cis-d-isomer

Co

Cl

Cl

en

en

+

Cis-l-isomer 29.[B] −

4MnO2 + 16H+ + −42OC →2Mn2+ +2CO2 +

8H2O 30.[C] Lanthanoid contraction takes place. 31.[D] In nitrogen d orbital is absent. 32.[A] HNO3 is acidic in nature. 33.[A] 2KBr + H2SO4 → K2SO4 + 2HBr 34.[A] Due to formation of chelate compound it act as

strong acid and proceed in forward direction. CH – O

BO – CH

CH – O O – CH +H+

35.[B] Na2SO3 + S →NaOH

Sod. thiosulphate Na2S2O3

36.[D] Critical temperature of water is more than O2

due to its dipole moment (Dipole moment of water = 1.84 D, Dipole moment of O2 = 0D.)

37.[D] By the process of zone refining, semiconductors

like Si, Ge and Ga are purified. 38.[C] Half filled orbitals are more stable in

comparison of partial filled. 39.[A] The dipole moment of

CH4 = 0 NF3 = 0.2 D NH3 = 1.47 D H2O = 1.85 D

40.[A] Molecule existence is possible in such case

when no. of bonding electron is greater than antibonding.


MATHEMATICS

1.[D] Clearly all term can Neither be positive nor negative

T1001 = sin log101000 = sin 3 > 0

T10001 = sin log1010000 = sin 4 > 0 ∴ (0 < 2 < π & π < 4 < 2π)

2. [A] sin x < x < tan x in (0, 1) ⇒ tan–1x < x < sin–1 x Altiter : f(x) = sin–1x – x

f '(x) = 2x–1

1 – 1 > 0 ∀ x ∈ (0, 1)

∴ f(x) is increasing function ∴ x > 0 ⇒ f(x) > f(0) = 0 ⇒ sin–1 x > x Similarly g(x) = x – tan–1 x is increasing fun

and x > 0 ⇒ x – tan–1x > 0

3.[A] log2 cos x + log2(1 – tan x) + log2(1 + tan x) – log2 sinx = 1 ⇒ log2(1 – tan2x) – log2 tan x = 1 (cos x > 0, sin x > 0, – 1 < tan x < 1)

⇒ (1 – tan2x) × xtan

1 = 2

tan2x + 2 tan x – 1 = 0 tan x = –1 ± 2

tan x = 2 – 1 (Q 0 < tan x < 1) ⇒ x = π/8 4. [B]

P

A

r

α B

r

α

AB = 2r sin α/2

h = AB tan α = 2r sin 2α tan α

5.[C] Let vertex P be (h, k), then perpendicular distance of P from the base x = a is |h – a|

∴ Since length of the base is 2a, we have

21 × 2a|h – a| = a2

⇒ |h – a| = a (a ≠ 0) So h – a = – a or h – a = a

∴ h = 0 or h = 2a ∴ locus of P is x = 0 or x = 2a

6. [B] Equation of angle bisector of the pair of straight

lines is bayx 22

−− =

hxy−

⇒ hx2 + (a – b)xy – hy2 = 0

⇒ h + (a – b) xy – h

2

xy

= 0

Now, y = mx is one of the bisector ∴ hm2 – (a – b)m – h = 0 h(m2 – 1) = (a – b)m

⇒ m

1m2 − = h

ba −

7.[C] x2 + y2 + 2gx + 2fy + c = 0 passes through all

the four quadrants ⇒ origin in an interior point ⇒ c < 0 8. [A] two normal are x – 1 = 0 and y – 2 = 0, their

point of intersection (1, 2) is the centre & radius of circle perpendicular distance from centre (1, 2) on tangent 3x + 4y = 6

= 169

62.43+

−+ = 1

∴ equation of circle is (x – 1)2 + (y – 2)2 = 12 ⇒ x2 + y2 – 2x – 4y + 4 = 0

9.[B] x2 – 4x + 6y + 10 = 0 ⇒ (x – 2)2 = –6(y + 1) tangent to the vertex is y + 1 = 0 circle drawn on focal distance as diameter

always touch the tangent at the vertex i.e. the line y + 1 = 0.

10. [B] Given ellipse is 25x2

+ 9y2

= 1

a2 = 25, b2 = 9, e = a

ba 22 − = 54

⇒ ae = 4 ∴ Foci of ellipse are (± ae, 0) = (± 4, 0) For hyperbola e = 2 ⇒ 2a = 4 ⇒ a = 2 Also b2 = a2(e2 – 1) ⇒ b2 = 4 × 3 = 12 ∴ equation of hyperbola

4

x2 –

12y2

= 1

⇒ 3x2 – y2 – 12 = 0


11.[B] ∴ →α &

→β are two mutually perpendicular

unit vector.

∴ →α ×

→β is a unit vector perpendicular to

both →α &

→β . So we can consider

→α ,

→β ,

→α ×

→β as i , j & k .

Given vector are coplanar so

bcc101caa

= 0

⇒ a(– c) + a(c – b) + c2 = 0 ⇒ c2 = ab

12. [D] S1 = x2 + y2 + z2 + 2x – 4y – 4z – 7 = 0 centre C1 = (–1, 2, 2) and radius r1 = 4 S2 = x2 + y2 + z2 + 2x – 4y – 16z + 65 = 0 centre C2 = (–1, 2, 8) radius r2 = 2 C1C2 = 6; r1 + r2 = 6 ∴ sphere S1 & S2 touches externaly ∴ point of contact divides C1C2 in the ratio 4 : 2 ∴ point of contact = (–1, 2, 6)

13.[D] y = f(x) = 3 – 8x4x

42 +−

⇒ (3 – y)x2 – 4(3 – y)x + 20 – 8y = 0 x ∈ R ∴ D ≥ 0 ⇒ 16(3 – y)2 – 4(3 – y) (20 – 8y) ≥ 0, y ≠ 3 ⇒ –y2 + 5y – 6 ≥ 0; y ≠ 3 ⇒ (y – 2) (y – 3) ≤ 0 ⇒ 2 ≤ y < 3

14. [B] x > 0, g(x) is bounded

∴ nx

nx

n e1)x(ge)x(flim

++

∞→

nx

nx

n e11e/)x(g)x(f

lim+

+∞→

= f(x)

[g(x) is bounded ⇒ nxe)x(g ⇒

infintefinite = 0]

15.[B] a1 = 1, an = n(1 + an–1)

⇒ 1 + an–1 = n

a n

∴ ∞→n

lim

+

1a11

+

2a11 ....

+

na11

∞→n

lim

+

1

1

aa1

+

2

2

aa1 ....

+

n

n

aa1

=∞→n

lim2

a 2 .3

a3 ........ 1n1a n

++ .

n21 a...aa1

= 1n

alim 1nn +

+

∞→ =

1n)n1)(a1(lim n

n +++

∞→

=

+

∞→ na

n1lim n

n=

−+

−+ −

∞→ 1na

1n1

n1lim 1n

n

=

+++−

+∞→ 1

a11

21....

1n1

n1lim 1

n

=

+++∞→ n

1....2

11

11limn

Q a1 = 1

= e

16.[A] 2 sgn 2x =

>=<−

0x;20x;00x;2

f(x) = |2 sgn 2x| + 2 =

=≠

0x20x4

∴ By defining f(0) = 4, f(x) will become continuous function at x = 0 as then

f(0 – 0) = f(0 + 0) = 4 Hence at x = 0 f(x) has removable discontinuity 17.[A] Q 1 < x < 2 ⇒ [x] = 1

∴ f(x) = cos

−

π 3x2

= sin x3

f ´(x) = cos x3 . 3x2

⇒ f ´

π32

= 3 cos2π .

3/2

2

π = 0

18.[C] yex = cos x ...(1) yex + y1ex = – sin x ...(2) again differentiating yex + y1ex + y1ex + y2ex = – cos x yex + 2y1ex + y2ex = – yex (from (1)) ⇒ 2y + 2y1 + y2 = 0 ...(3) again differentiating 2y1 + 2y2 + y3 = 0 again differentiating 2y2 + 2y3 + y4 = 0 ...(4) from (1) (2) & (3) 4y + y4 = 0

∴ y

y4 = – 4

19.[A] fog = I ⇒ fog(x) = I(x) = x ∴ fg(x) = x ⇒ f´g(x) × g´(x) = 1 ⇒ f´g(a) × g´(a) = 1 ⇒ f´(b) × 2 = 1

⇒ f´(b) = 21


20.[C] Let P(h, k) be one of point of contact then k = sin h ...(1) equation of tangent is y – k = cos h (x – h) which parries through origin ∴ k = h cos h ...(2) from (1) & (2)

2

2

hk + k2 = sin2h + cos2h = 1

⇒ k2 + k2h2 = h2 ⇒ x2 – y2 = x2y2 ∴ locus of (h, k) is x2 – y2 = x2y2 21.[B] x + y = 16, x, y > 0 s = x3 + y3 = x3 + (16 – x)3

dxds = 3x2 – 3(16 – x)2

dxds = 0

⇒ x2 = 256 – 32x + x2 ⇒ x = 8

2

2

dxsd = 6x – 6(16 – x) = 6[2x – 16]

at x = 8

2

2

dxsd = 0 and 3

3

dxsd = 12 ≠ 0

Hence there in no minimum exist.

22.[C]

a1

b1

a1

b1

22

−

− =

2

3

c1c2

−

− =

c2

⇒ b1 +

a1 =

c2

⇒ c = ca

ab2+

∴ c is H.M. of a & b

23.[B] I = ∫ + xcos1 2 sin 2x cos 2x dx

put 1 + cos2x = t2 ⇒ –2sin x cos x dx = 2tdt ⇒ – sin 2x dx = 2tdt

∴ I = – ∫ − )1xcos2).(dtt2.(t 22

= – ∫ − dt)3t2(.t2.t 2

= –2

−

3t3

5t2 35

+ c

= – 54 (1 + cos2x)5/2 + 2(1 + cos2x)3/2 + c

= (1 + cos2x)3/2

++− 2)xcos1(

54 2 + c

= 52 (1 + cos2x)3/2 (3 – 2cos2x) + c

24.[A] 1 – 3

x2 +

4x4

– .......... = cos x

∴ I = ∫ dxxcos = sin x

25.[A]

∫

∫

π

π

−

π→

π

−

−

2

2

x

4/

x

2/

tcos

2x dt

2t

dt)12(lim

00

x2.

2x

12limxcos

2x

π

−

−−

π→

= π−

−

π→ x4

xsin.2n2limxcos

2x

l = π

2nl

26.[A] y = 222 n

6

2

2n

4

2

2n

2

2n n31

n21

n11lim

+

+

+

∞→

....2nn2

2

2

nn1

+

log y = ∑=

∞→

+

n

1r2

2

n nr1loglim . 2n

r2

= ∑=

∞→

+

n

1r2

2

n nr1loglim .

nr2 .

n1

= ∫ +1

0

2 dx)x1log(x2

= ∫2

1

dttlog = ( )21ttlogt −

= 2 log 2 – 1 = loge4

⇒ y = 4/e 27.[B] required area

–1

0

1 y = 2x2

1/√2


2

1 – ∫2/1

0

2dxx2

⇒ 2

1 –32 2/1

03 ]x[

⇒ 2

1 – 32 .

221 =

232 =

622

28.[B] ´´y = (y´ + 3)1/3 ⇒ (y´´)3 = (y´ + 3)2

⇒ 3

2

2

dxyd

–

2

dxdy

– 6

dxdy – 9 = 0

∴ order is 2 & degree = 3 29.[A] x18 = y21 = z28 ⇒ 18 log x = 21 log y = 28 log z ⇒ logy x = 7/6, logz y = 4/3, logxz = 9/14 Now, 3, 3 logyx, 3 logz7, 7 logx2

= 3, 27 , 3 ×

34 , 7 ×

149

= 3, 27 , 4,

29

which are in A.P. 30.[B] log2x + log2y ≥ 6 Here x > 0, y > 0 ∴ log2xy ≥ 6 ⇒ xy ≥ 26 ⇒ xy ≥ 64 Now, A.M ≥ G.M.

∴ 2

yx + ≥ (xy)1/2

⇒ x + y ≥ 2(64)1/2 x + y ≥ 2 × 8 ⇒ x + y ≥ 16 ∴ (x + y)min = 16 31.[A] f(x) = x3 + x2 + 10x + sinx f´(x) = 3x2 + 2x + 10 + cos x

= 3 2

31x

+ +

329 + cos x > 0 ∀ x

⇒ f(x) is strictly increasing Also x → ∞ ⇒ f(x) → ∞, x → – ∞ ⇒ f(x) → – ∞ ∴ f(x) has only one real root. 32.[B] Let roots be (2k – 1) & (2k + 1) k ∈ N

the Sum of roots : 4k = – ab

Q a ∈ R+, b < 0 as k ≥ 1

We have – b = 4ak ⇒ – b ≥ 4a ⇒ |b| ≥ 4a b < 0 ∴ |b| = –b 33.[B] Let G.P. be a + ar + ar2 ...... G.P is infinite so – 1 < r < 1 G.P. is decreasing ⇒ r > 0 so 0 < r < 1 and therefore a > 0 f´(x) = 3x2 + 3 > 0 ⇒ f(x) is strictly increasing function ∴ f(x)max on [–2, 3] is f(3) = 27 & f´(0) = 3

∴ r1

a−

= 27 & a – ar = 3

⇒ r = 32 or

34 Q r < 1

∴ r = 32 & if r =

32 ; a = 9

∴ Sum of first three terms = 9 + 6 +4 = 19

34.[B] If z1, z2 & z3 vertex of equilateral triangle then 2

1z + 22z + 2

3z = z1z2 + z2z3 + z3z1 ∴ (a + i)2 + (1 + ib)2 + 0 = (a + i) (1 + ib) + 0 +

0 ⇒ a2 – 1 + 2ia + 1 – b2 + 2ib = a + iba + i – b ⇒ a2 – b2 + i(2a + 2b) = a – b + i(ab + 1) on comparing a2 – b2 = a – b and 2a + 2b = ab + 1 ⇒ (a – b) (a + b – 1) = 0 & 2a + 2b = ab + 1 ⇒ a = b or a + b = 1 ....(1) ⇒ 2a + 2b = ab + 1 .... (2) Now from (1) take a = b put in (2) 2a + 2a = a2 + 1 ⇒ a2 – 4a + 1 = 0 ⇒ a = 2 ± 3

Q a < 1 ⇒ a = 2 – 3

Q a = 2 – 3 & a = b = 2 – 3 It we take a + b = 1 & put in (2) then it becomes

ab = 0 which not possible because a & b lies between 0 and 1

35.[A] S =

+

+

+−∑

=

....87

43

21)1(

rr

r

n

0r

r nCr

= rn

r

n

0r

r C.21)1(∑

=

− + rn

rn

0r

r C.43.)1(

−∑

=

+ rn

rn

0r

r C.87.)1(

−∑

=

+ ....

= n

211

− +

n

431

− +

n

871

− + .....

= n21 + n22

1 + n321 + ..... =

121

n −


36.[B] The general term in the expansion of (x1 + x2 +

...xn)n given ..... n21 p...pp

n. 1p

1x 2p2x .... mp

mx ,

p1 + p2 + p3 .... + pm = n Now in (1 + x + y – z)9, coefficient of x3y4z = coeft of u0x3.y4z1 in (u + x + y – z)9

= 1430

9 × (–1)1 = –2 . 9C2 . 7C3

37.[B] x1

ex

− = B0 + B1x + B2x2 + .....

⇒ ex = B0 – B0x + B1x – B1x2 + B2x2 – B2x2 + ....

⇒ ex = B0 + (B1 – B0)x + (B2 – B1)x2 + .....

⇒ 1 + x + 2

x2+

3x3

+ 4

x4+ ....

= B0 + (B1 – B0)x + (B2 – B1)x2 + ... ∴ Bn – Bn –1 is coeff. of xn

On comparing coeff. of xn = n1

38.[A] x + y + z + 12 = 0 x, y, z are negative integers Let x = – a, y = –b, z = – c, a, b, c are +ve integer then required number of

points (x, y, z) = Number of positive integral solution of a + b + c = 12 = 12–1C3–1 = 11C2 = 55

39.[A] p1 = 12

662 p2 =

11

65

2

1

pp = 11

6512

662× = 1

40.[C] 2f(x) = )x/1(f1

)x(f)x/1(f)x(f −

= f(x) .

x1 – f

x1 + f(x)

⇒ f(x) + f

x1 = f(x) . f

x1

⇒ f(x) = 1 ± xn f(2) = 17 ⇒ 1 ± 2n = 17 ⇒ ± 24 = 16 ∴ +ve sign will be take ⇒ 2n = 16 ⇒ n = 4 Now, ∴ f(x) = 1 + x4 ⇒ f(5) = 54 + 1 = 626

41.[C] A is idenpotent ⇒ A2 = A

A2 =

20x1

20x1

=

40x31

≠ A

∴ not possible for any x 42.[A] for any a ∈z ⇒ a = 20a ⇒ a R a ∀ a ∈ z ∴R is reflexive a R b ⇒ a = 2kb, k ∈ z ⇒ b = a.2–k, – k ∈ z ⇒ b R a ∴ R is symmetric Let a R b, b R c ⇒ a = b2 1k , b = c2 2k a = 1k2 c2 2k = c2 21 kk + , k1 + k2 ∈ z ⇒ a R c ∴ R is transitive Hence R is an equivalence Relation. 43.[D] Q A.m ≥ G.m

∴

+

ab

ba

21 ≥

ba.

ab = 1 ⇒

ba +

ab ≥ 2

Similarly cb +

bc ≥ 2 &

ca +

ac ≥ 2

Adding we get

ba +

ab +

cb +

bc +

ca +

ac ≥ 6

⇒ a

cb + + b

ac + + c

ba + ≥ 6

∴ minimum value is 6.

44.[A] f(x) = x1 ⇒ f ´(x) = – 2x

1

∴ ab

)a(f)b(f−− = f ´(x1)

⇒ b1 –

a1 = (b – a)

− 2

1x1 ;

a < x1 < b ⇒ x1

2 = ab ⇒ x1 = ab

45.[C] ∫ xcot4 dx = ∫ xcot2 (cosec2 x –1)dx

= ∫ xcot2 cosec2x dx – ∫ − )1xec(cos 2 dx

= – 31 cot3x + cotx + x + c

∴f(x) = – 31 cot3x + cotx + x + c +

31 cot3x – cotx

= x + c


∴ f

2x =

2π + c ⇒

2π =

2π + c

⇒ c = 0 ∴ f(x) = x

LOGICAL REASONING 1.[B] The given sequence is a combination of two

series: I series : 11, 17, 23, (?) II series : 12, 18, 24 Pattern in both is + 6 So, missing number = 23 + 6 = 29

2.[B] Dum Dum is an airport in calcutta and Palam is an airport in Delhi.

3. [D] All except mustard are food grains, while

mustarel is an oilseed. 4.[C] 5. [B]

6.[D]

7.[C] The 3rd figure in each row comprises of parts which are not common to the first two figures.

8. [C]

9.[D]

10.[B] The number of each type of figures decrease by 1 at each step from left to eight in each row.

ENGLISH 1.[B] Assure : (persuade that all is well) Hence, irrelevant meaning. Ensure : (guarantee) This is a relevant option as it properly suits to

the meaningful expression. Insure : (to cover against any loss) Hence, irrelevant meaning.

Accept : Irrelevant meaning in respect of the sentence. 2.[B] burst : (punctured) Irrelevant meaning. bust : (collapsed) Quite relevant meaning. Hence, correct option. bursted : irrelevant because this is an improper

form of the verb 'burst'. busted : Irrelevant because this is an incorrect form of

the verb 'bust' 3.[A] deduce : (to conclude) (to infer) This word really suits to the given sentence

making it a meaningful one. deduct : (to take away) Irrelevant, because it means something else. e.g. Tax is 'deducted' from his salary every year. reduce : (to decrease) Different meaning makes the sentence

meaningless. Hence, this is an incorrect option. Conduce : (to suit) Irrelevant word.

4.[C] Sky (Firmament) Irrelevant 'opposite' Firmament : irrelevant 'opposite'. nadir : (lowest point) 'Zenith' means highest point'. Hence this is the most suitable word in opposite, naive (Simple) Irrelevant word that doesn't serve any purpose.

5.[A] Hungry : (a voracious eater) (a voracious reader)

Wild : (Irrelevant) It's Synonym is 'Savage'. Quick : (Soon) therefore, irrelevant. Angry : (furious) Hence, irrelevant.

6.[C] Journey man : (Irrelevant) Because it is a person who journeys regularly on

a particular route. Tramp : (Irrelevant) because it means 'a vagabond' who gets about

purposelessly. Itinerant : (Relevant) It's a person who moves from one place to

another during his travel. Mendicant : (Irrelevant) It's a religious preacher who goes from place to

place in the form of a beggar.

7.[B] To dislocate : irrelevant meaning. To lose one's temper : Quite Relevant It's often used when a person is about to get

angry.


To take off : irrelevant To be indifferent : Irrelevant 8.[A] took to : (to be accustomed to/to be addicted to) Correct because it suits to the sentence when

Gandhi Ji was addicted to smoking. took for : (to be mistaken while recognising) irrelevant took in : (to deceive someone) irrelevant took up : (to adopt) irrelevant 9.[A] Get someone to break the box. Correct answer because : • The given sentence is in Passive Voice which requires its answer in Active Voice. • The given verb is Causatives in Imperative. • This option is Causative Active Voice in imperative form. They have broken the box. Incorrect Answer : Because • The given verb is not causative. • The given sentence is not imperative. Have the broken box. Incorrect : Because • The verb is not causatives. • 'Broken' has been used as an adjective in this

option. Break the box Incorrect because : • The subject 'you' (implied) wan't break it but

will get someone to break it.

10.[B] He asked how shabby I was looking. (Incorrect option) because : The required answer (type of sentence) is

wrong. He exclaimed with disgust that I was looking

very shabby. (Correct answer because) • This option is exclamatory. • Past Indefinite Tense has been used. • The mood of the speaker is correct. He exclaimed with sorrow that they were

looking much shabby. (Incorrect option because) : • Mood of exclamatory sentence is wrong. • 'much' will be replaced with 'very' • 'They' won't be used as a singular subject is

required. He told that I was looking much shabby. Incorrect answer because – • Type of sentence is assertive whereas the

required type is 'exclamatory'. • 'much' is to be replaced with 'very'.

11.[C] Neigh : Correct spelling as it means 'the cry of horse'.

Reign : Correct spelling as it means 'the controlling chord of an animal.'

Niece : Incorrect spelling as the correct one is 'niece'. (Opposite of 'Nephew').

Neither : Correct spelling. It is a conjunction to be used with 'nor' for one of two options.

12.[C] I wonder : No error in it. What he has done with the book. No error I lend him (Erroneous) because there is an error of

'Tenses'. The word 'lend' is to be 'lent'. No error : There is an error. 13.[A] Distraught, awry : Correct answer : 'Distraught' means to 'get

upset' and 'awry' means in 'disorder'. Frustrated, Magnificently : Both are opposite. One is positive and the other

one is negative. Therefore, no meaningful sentence.

Elated, Wild : No co-ordination, therefore incorrect answer. Dejected, splendidly : No co-ordination, therefore incorrect answer. 14.[D] Interesting : " . . . and only a few were . . . . " phrase shows

that something opposite is required here. The given option is not opposite to 'trivial'. Hence, Irrelevant option.

Practical : Like aforesaid logic, this option also is

irrelevant. Complex : 'Complex' can't be opposite to 'trivial'.

Therefore, can't be relevant. Significant : This is the relevant option making the sentence

quite meaningful with two contradictory words, i.e. trivial and 'Significant connected with the phrase ". . . . . and only a few were . . . . . "

15.[A] Paths, grave : It is a meaningful pair of words to make the

sentence idiomatically correct. Ways, happiness : Not a meaningful pair. Hence, irrelevant option. Acts, prosperity : Not a meaningful pair. Hence, irrelevant option. Achievements, Suffering : Not a meaningful pair. Hence, irrelevant option. Hence, inappropriate


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