Jan 2010

XtraEdge for IIT-JEE 1 JANUARY 2010

Dear Students, Are You an Optimist or a Pessimist ?

I have been giving some thought lately to optimism and pessimism. Basically, these are attitudes — attitudes that shape and formulate our entire existence. I mean, have you ever met a happy pessimist? Of course not. In short, our optimism or pessimism is this: The way we interpret the past. The way we experience and view the present. The way we imagine the future Have you given much thought about how your attitude, whether you are an optimist or a pessimist, affects you business, organization or school? Have you thought about how it affects you personally? And what about the team you are a part of? What is optimism? It is the belief that things in our past were good for us, even if that means they were hard and taught us lessons. It is also the belief that things will be better in the future. Here are some contrasts between optimism and pessimism and how they affect us: Optimism breathes life into you each day. Pessimism drains you. Optimism helps you to take needed risks. Pessimism plays it safe and never accomplishes much. Optimism improves those around you. Pessimism drags them down. Optimism inspires people to great heights. Pessimism deflates people to new lows. There is only one way that optimism and pessimism are the same, and that is that they are both self-fulfilling. If you are an optimist, you will generally find that good things happen to you. And if you are a pessimist, you will find yourself in the not-so-good situations more often than not. So can a person just become an optimist? Yes! We can choose to look at the world any way we want to. We can choose to look at the world and think the worst, or we can tell ourselves the good things about each situation. As you find yourself looking at your enterprise, begin to view it through the eyes of an optimist, and you will reap the rewards listed above, and so will the people around you. There are tremendous benefits to being an optimist, as stated above. But there are some pessimists out there who will say, “But that isn’t realistic.” I say, “Who cares?” If things go awry, at least I have spent my time beforehand enjoying life and not worrying about it. And, being an optimist, I would view the “negative” situation as an opportunity to grow and learn. So I can even look forward to my failures because they will be steppingstones and learning tools to be applied to my future success. Have you ever met a successful pessimist? Become an optimist, and see your world change before your eyes! Have a blessed day! Let's make the uncommon knowledge common Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Volume - 5 Issue - 7

January, 2010 (Monthly Magazine)

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Volume-5 Issue-7 January, 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

Mock Test CBSE Pattern Cllass XII

Success Tips for the Month • It is more important to know where you

are going than to get there quickly.

• The secret of success is constancy to

purpose.

• Action without planning is the cause of all

failure. Action with planning is the cause

of all success.

• We cannot discover new oceans unless we

have the courage to lose sight of the

shore.

• One person with a belief is equal to 99

who have only interests

• A thousand mile journey begins with one

step. Start today.

• Keep your eyes on the stars and your feet

on the ground.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 IIT boys draw power & water from sewage Alumni of IIT-Madras to come together on December 26

IITian ON THE PATH OF SUCCESS 8 Dr. Alok Aggarwal

KNOW IIT-JEE 10 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 59

Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper Mock Test CBSE Pattern Paper-2 [Class # XII] Mock Test CBSE Pattern Paper-1 (Solution)[Class # XII]

Regulars ..........

DYNAMIC PHYSICS 17

8-Challenging Problems [Set# 8] Students’ Forum Physics Fundamentals Refraction at plane & curved surfaces Properties of Matter CATALYST CHEMISTRY 33

Key Concept Carbohydrates Salt Analysis Understanding : Organic Chemistry

DICEY MATHS 48

Mathematical Challenges Students’ Forum Key Concept Differential Equations Trigonomatrical Rations

Study Time........

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


IIT boys draw power & water from sewage

When the campus placements

happen in 2011 at IIT Kharagpur,

these five students will give it a

miss. They would have by then

started their own company. They

have developed a bio cell'

(battery) that can not only treat

sewage water but also generate

electricity a that could offer a one-

stop solution to the water and

energy crises.

Their bio-product has won them

rave accolades from the ministry

of science and technology and a

cash award to carry their research

forward.

It was while working on microbes

that can be used as purifying

agents that Manoj Mandelia, a

fourth year student of biotech

engineering, stumbled upon the

idea that if a bio cell can be

developed to treat sewage water

for use, it would solve one of the

biggest problems of the present

times. Mandelia, who is pursuing

an integrated M Tech programme

at the institute, started looking for

like minded boys for his project.

He soon found Prateek Jain

(agriculture and food engineering

department), Shobhit Singhal

(electrical engineering), Pulkit

Anand (energy engineering) and

Mohan Yama (PhD student of

biotechnology). Renowned

biotech faculty member Debabrata

Das enthusiastically joined them in

the pursuit.

The idea is simple. The specially

designed bio cell (LOCUS) will be

set up in the form of a plant,

through which the sewage water

of a housing complex would be

flowed in. The genius of this

invention which is awaiting patent

lies in the design of the cell that

will automatically grow millions of

anaerobic bacteria that multiply

through respiration. The bacteria

clean up the sewage water and in

the process generate free

electrons. If harnessed, these

electrons can generated

electricity.

"We worked on this concept for

nearly a year before we readied

the cell and applied to the ministry

to enter its annual business plan

competition that focuses on bio

technology products for

sustainable development," Mandelia

said.

Twenty teams, mostly corporate

houses dealing with bio products,

participated in this premier

competition. The IIT-Kgp team

managed to come second and won

a cash prize of Rs 3 lakh. "LOCUS

is a green tech development,

which is sustainable both

economically and environmentally

and serves as an ideal integration

to address the key issues of

wastewater treatment and energy

gap," reads the award citation.

The cell, at this stage, can clean up

50,000 litres of sewage water,

about the amount generated by

100 flats in a day. The water

produced this way can be

compared with that supplied

provided by a civic body, the

students say. "The purified water

has been tested and has been

certified to be clean and fit for

household use. It is, however, not

fit for drinking," Mandelia

explained.

The IIT-Kgp team has even

produced electricity with the bio

cell. "A township of 100,000

people needs about 2.3 megawatts

of electricity a day. It will be years

before we reach that stage. But

we have already been able to

generate electricity. By next year,

we aim to generate 350 units,

enough to meet 50% of the

demand of a 100-flat complex.

When we say this we are not

taking airconditioners into

consideration," said Prateek.


Alumni of IIT-Madras to come together on December 26

The "first ever" gathering of the

alumni of IIT-Madras (IITM) will be

held here on December 26.

According to a release here, more

than 32,000 students have

graduated from the institute in the

past 50 years and most were

expected to attend the reunion,

co-organised by the Office of

Alumni Affairs (OAA) and IIT-M

Alumni Association.

The day-long event would see the

screening of two films, one

highlighting alumni contributions

to the campus and the other their

socially-relevant work, especially

in the rural sector.

IIT-M director Prof MS Ananth is

scheduled to deliver the inaugural

address, the release said.

Alumni from different parts of the

world would make presentations

on various issues, the release said

adding an "IIT Madras Heritage

Quiz" would also be held.

IIT-M launches seed fund of $ 0.8 mn to help budding entrepreneurs

Students, researchers and faculty

members from IIT-Madras now

have the chance to turn innovative

ideas into sustainable businesses

with the Micro, Small and Medium

Enterprises (MSME) incubation

initiative. Intended to provide

seed-stage funding of up to $ 0.8

million (Rs 3.5 crore) per venture

for the first year, the initiative

being executed by the department

of management studies is designed

to act as a catalyst to provide

initial momentum.

Speaking at the launch of the

initiative here on Friday, L S

Ganesh, professor-in-charge of

Cell for Technology, Innovation,

Development and

Entrepreneurship Support

(C-TIDES), said IIT-Madras is

perhaps the first institution to

start a formal programme in

entrepreneurship in the 1980s.

Emphasizing the importance of

developing small and medium

enterprises, IIT-Madras director M

S Ananth said they generated 10

times the employment per unit of

investment. "We need to

decentralise these aspects to get

innovations started with centres

such as this. Also, innovators need

to be aware of their privileges and

what they are entitled to," he said.

According to professor Thillai

Rajan, assistant professor,

department of management

studies, detailed workshops would

be conducted for prospective

incubates before issuing a call for

proposals.

IIT-R shifting to Jodhpur gradually taking shape

JODHPUR: In what can be

referred to as one step further

towards shifting the IIT-R from

Kanpur to Jodhpur, a team

comprising administrative

in-charge of IIT-R (Kanpur)

Niraj Gupta and two others,

visited the MBM Engineering

College here on Friday. Sources

said they discussed the make-shift

requirements and took stock of

the facilities available in the

college. The team will submit the

report to the Director (IIT-R),

who is scheduled to visit the city

in the first week of December to

consolidate the ground for

shifting.

MBM dean Arvind Roy said that

the team was here to oversee the

proposed shifting early next year.

The team had a thorough visit of

the entire college campus

buildings, classrooms, laboratories

and other rooms housing different

departments and collected a map

of the college.

Roy added that their main focus,

however, was on the hostel

facilities, for which they also

visited the AIIMS site, where the

students are initially proposed to

be shifted, in case the building

of the IIT-R is not ready by then.

It is evident that there has been an

understanding in-principle that

IIT-R will be shifted to the said

college here from next year. A

Central team, headed by Union

additional secretary (HRD) Ashok

Thakur, recently visited the

college and the hostels which are

complete and said to have

expressed satisfaction with the

arrangements there.

The visit of this team of these sites

again has further strengthened

that fact that the plans of shifting

are gradually taking shape. The

team will explore and identify the


necessary additions and alterations

that will have to be made into the

existing structure of the college to

accommodate IIT-R and prepare a

report to be submitted to the

director.

IITs worried as reserved seats remain unfilled

Even as the process for applying to

the Joint Entrance Examination

(JEE) 2010 of the Indian Institutes

of Technology (IITs) has begun,

figures from JEE 2009 reveal that

137 seats meant for candidates

with physical disabilities (PD), got

wasted. The reason is that not

enough disabled candidates qualify,

and these seats cannot be

converted into general seats.

"Unless the rules are changed and

the IITs are allowed to transfer

the vacant PD seats to general or

non-PD seats within the category,

we will not be able to stop this

loss," said Anil Kumar, JEE 2010

chairman, IIT Bombay.

As per a judgment passed by the

chief commissioner for persons

with disabilities, IITs were directed

to treat disabled candidates on a

par with scheduled caste

(SC)/scheduled tribe (ST)

candidates and give them similar

relaxation from 2009, including

admission for preparatory

courses.

Currently, 3% seats in each

category [general category, SC, ST

and other backward classes

(OBC)] are reserved for disabled

students in each IIT. While the

total number of PD seats in 2009

across all IITs was 251, only 138

candidates could qualify at the JEE,

despite 12% more disabled

candidates taking the exam as

compared to 2008.

"Despite the 50% relaxation for

PD candidates (from the last

general candidate), only 44 could

be given admission across all the

IITs. This left around 207 seats

unfilled," said Kumar.

To make up for the shortfall,

scores of PD candidates were

relaxed by another 50% to enable

candidates to qualify for the

preparatory course, which was

started for disabled candidates in

2009. In case of OBCs, 51 seats

across the IITs in 2009 were

converted into general seats due

to unavailability of OBC students

even after giving full 10%

relaxation. Again, over 1,000 seats

for SCs and STs which remained

vacant were transferred to the

preparatory course and were filled

after lowering the bar by 50%.

IIT Guwahati director Gautam

Barua said that a resolution was

adopted at the last joint admission

board meeting that the PD

commissioner should be

approached for possible

conversion of PD seats to general

category in future. "However, the

issue is complex since there are

certain PD seats within SC and ST

and these cannot be converted,"

said Barua.

Signal engineering is the backbone of Rlys: SCR

Technology changes have helped

Indian Railway enhance its

productivity to meet the increase

in demand for rail traffic, Sudesh

Kumar, member (Electrical)

Railway Board has said.

Speaking at the 52nd Annual Day

celerations of the Indian Railways

Institute of Signal Engineering and

Telecommunications (IRISET) here

today, he said signalling has

evolved from an ordinary means

of communication to start a train

and later emerged as a symbol of

trust and safety on railway tracks

over the years.

Participating in the event, the new

South Central Railway (SCR)

General Manager M S Jayanth said

IRISET over a period of time has

evolved as a quality institute.

``Signal engineering happens to be

the backbone of railways, since

this is vital for reliability and safety

of rails,’’ Jayanth said.

According to him, advancement

made in telecommunication has

palyed vital role in the information

management of Indian Railways,

especially in passenger reservation

system, fright and other services.

IRISET Director V Balaram said

signalling is very specific to

railways.

IRISET is the only institute in

railways, which trains both officers

and supervisory staff. ``During the


last 52 years, about 10,036 officers

and 44,906 signal and telecom

supervisiors have been trained,’’

Balaram said.

V G K Murti, former Dean of IIT

Chennai, said in the earlier days

railways used to be the most

preferred place of employment for

most IIT students. He delivered

his speech targeting young

trainees of IRISET, by passing on

knowledge inputs. The function

was attended by many senior

officials of the SCR.

Enterprises need mobile strategy

Bangalore: One of the most

discussed issues when it came to

developing applications for mobile

phones was the lack of any

reasonable returns on apps made

for popular platforms like the

iPhone. A lot of developers said

they hardly made any money from

developing apps for these high-end

phones. Turns out they have got

their strategy wrong. Endeavour, a

city-based company, sees good

business in making apps for

enterprises.

T

he founders, all IIT graduates,

started the company in 2002.

According to Avinash Misra, one

of the founders, their first app was

targeted at doctors. "We made it

in co-ordination with a doctor

based in the US using which he

could make a profile of a patient.

It helps him determine the amount

of anaesthesia, and can also be

tweaked for other medical

requirements. This app is now

used by around 15,000 doctors

internationally. In the future, we

have plans to make it possible for

doctors to exchange patient

profiles," Avinash says.

The company has also developed

mobile applications for an

Australian newspaper and a top

computer manufacturer, among

others.

Making apps for the enterprise

instead of selling it to the public

via an appstore is a different

ballgame altogether. The app store

was an instant success as it

allowed a single developer to roll

out an app and be able to sell it

without worrying about

marketing. All the developer has

to worry about, is the competition

--similar apps available. But while

making apps for enterprises, it's

the functionality and the

adaptability that counts.

"The cost of an app can range

from $10,000 to $3,50,000

depending on the depth and the

functionality required. We don't

just create an app, we see to it

that their back-end is made

operational for such an

environment as well. It's very easy

to make a news application by

using the RSS feed available on the

site. But with the Australian

newspaper for which we made an

app, we not only created an app

but also made the changes

required to their backend

operation so that data was

delivered faster to the user. When

we approach a client we don't

introduce ourselves as mobile app

developers, but instead, tell them

we can come up with a mobile

strategy for their company. These

enterprise apps will never make it

to the app store, but are

distributed to all the employees in

a company," he adds.

The company has also developed a

few fun apps like upcoming app

Ambience which allows users mix

and match sound effects like that

of a waterfall, the sea, birds, etc to

create a relaxing atmosphere. "As

of now, most of the enterprises

still use Blackberries.But we're

seeing a slow shift towards touch

screen devices like the iPhone," he

says.

Avinash says they have not made

any apps for Indian companies till

now, as very few Indians use high-

end smartphones or other mobile-

enabled devices.

"But in 18 months, I'm sure India

will have one of the largest

markets in terms of mobile usage.

Indians just might skip the desktop

internet revolution and switch

over to the mobile internet

altogether," Avinash adds.


Dr. Alok Aggarwal received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1980. He obtained his Ph.D. in Electrical Engineering and Computer Science from Johns Hopkins University in 1984.

Dr. Alok Aggarwal is the Founder and Chairman of Evalueserve - a company that was started in December 2000 and that provides various kinds of research and analytics services to clients in North America, Europe and Asia Pacific from its five research centers in Delhi-Gurgaon, India; Shanghai, China; Cluj, Romania; Santiago-Valparaiso, Chile; and New York, USA.

Dr. Alok Aggarwal joined IBM Research Division in Yorktown Heights New York in 1984. During the fall of 1987 and 1989, he was on sabbatical from IBM and taught two courses (in two terms) at the Massachusetts Institute of Technology (MIT) and also supervised two Ph.D. students. During 1991 and 1996, along with other colleagues from IBM, he created and sold a "Supply Chain Management Solution" for paper mills, steel mills and other related industries. In July 1997, Dr. Aggarwal "Founded" the IBM India Research Laboratory that he set-up inside the Indian Institute of Technology Delhi. Dr. Aggarwal started this Laboratory from "ground zero" and by July 2000, he had built it into a 60-member team (with 30 PhDs and 30 Masters in Electrical Engineering, Computer Science, and in Business Administration). In August 2000, Dr. Aggarwal became the Director of

Emerging Business Opportunities for IBM Research Division worldwide.

Dr. Alok Aggarwal has published 86 Research papers and he has also been granted 8 patents from the US Patents and Trademark Office. Along with his colleagues at Evalueserve, in 2003, he has pioneered the concept of “Knowledge Process Outsourcing (KPO)” and wrote the first article in this regard. Dr. Aggarwal has served as a Chairperson of the IEEE Computer Society's Technical Committee on Mathematical Foundations of Computing and has been on the editorial boards of SIAM Journal of Computing, Algorithmica, and Journal of Symbolic Computation. During 1998-2000, Dr. Aggarwal was a member of Executive Committee on Information Technology of the Confederation of the Indian Industry (CII) and also of the Telecom Committee of Federation of Indian Chamber of Commerce and Industry (FICCI). He is currently a Chartered Member of The Indus Entrepreneur (TiE) organization.

In honouring Dr. Alok Aggarwal, IIT Delhi recognizes the outstanding contributions made by him as an Entrepreneur and Researcher. Through his achievements, Dr. Alok Aggarwal has brought glory to the name of the Institute.

Dr. Alok Aggarwal Electrical Engineering from IIT Delhi in 1980 Ph.D. in Electrical Engineering and Computer Science, Hopkins University (1984)

Director : IBM Research Division worldwide

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

Anything you can hunt, I can hunt better.


PHYSICS

1. A charge +Q is fixed at the origin of the co-ordinate system while a small electric dipole of dipole moment →P pointing away from the charge along the x-axis is

set free from a point far away from the origin. (a) Calculate the K.E. of the dipole when it reaches to

a point (d, 0). (b) Calculate the force on the charge +Q at this

moment. [IIT-2003] Sol. (a) Potential energy of the dipole-charge system Ui = 0 (Since the charge is far away)

Uf = – Q × 20 d

p4

1πε

∴ K.E. = |Uf – Ui | = 20 d

pQ4

1πε

(b) Electric field at origin due to dipole

→E = i

dp2

41

30πε

Now, force on charge Q is given by

→F =

→EQ = i

d4pQ2

30πε

2. Three infinitely long thin wires, each carrying current

i in the same direction, are in the x-y plane of a gravity free space. The central wire is along the y-axis while the other two are along x = ± d.

(i) Find the locus of the points for which the magnetic field B is zero. [IIT-1997]

(ii) If the central wire is displaced along the Z-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear mass density of the wires is λ, find the frequency of oscillation.

Sol. (i) We know that magnetic field due to an infinitely long current carrying wire at distance r is given by

B =

π rI2

4µ0

The direction of B is given by Right hand palm rule no. 1.

B

OP

A

ZX

d d

C

Z

d d B

θ θ

B

X Ar

f f Z r

Hence in case of three identical wires resultant field

can be zero only if the point P is between the two wires otherwise field B due to all the wires will be in the same direction and so resultant B cannot be zero. Hence, if point P is at a distance x from the central wire as shown in fig. then,

pB = PAB→

+ PBB→

+ PCB→

where PAB→

= magnetic field at P due to A

PBB→

= magnetic field at P due to B

PCB→

= magnetic field at P due to C.

Bp =

−++

+π xd1

x1

xd1

4I2µ0 (– k )

For PB→

= 0

On solving we get x = ± d 3 . (ii) The force per unit length between two parallel

current carrying wires is given by

rII2

4µ 210

π = f(say)

and is attractive if currents are in the same direction. So when the wire B is displaced along z-axis by a

small distance z, the restoring force per unit length F/l on the wire B due to wires A and C will be

KNOW IIT-JEE By Previous Exam Questions


l

F = 2f cos θ = 2rII2

4µ 210

π×

rz

=θ

rzcosas

or l

F =π4

µ0

)zd(I4

22

2

+z [as I1 = I2 = I and r2 = d2 + z2]

or l

F = π4

µ02

dI2

z [as d>>z and F is opposite to z]

...(1) Since F ∞ –z the motion is simple harmonic. Comparing eq. (1) with the standard equation of

S.H.M. which is F = – mω2z

i.e., l

F = –l

mω2z = – λω2z, we get

λω2 = π4

µ02

2

dI4 ⇒ ω =

λπ 2

20

dIµ

⇒ 2πn = dI

πλ0µ ⇒ n =

πλπ0µ

d21

3. 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 resistivity of the material of cylindrical shell is ρ, find the induced current in the shell. [IIT- 2005]

a

d

L

R

Sol. The magnetic field in the solenoid is given by B = µ0 ni

a

d

L

R

⇒ B = µ0 n i0 sin ωt [Q i = i0 sin ωt given]

The magnetic flux linked with the solenoid

φ = →B .

→A

= B A 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 electromagnetic induction, the induced emf produced in the cylindrical shell is

× ×

I

× ×

TOP VIEWS

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 = 2 π R, A = L × d

∴ R = ρ Ld

R2π … (ii)

The induced current I will be

I =R

|e| = R2

Ld]tcosina[ 02

0

π×ρ×ωωπµ

I = ρπ

ωωπµR2

tcosiLdna 02

0

⇒ I = R2

tcosiLdna 02

0

ρωωµ

4. An object is moving with velocity 0.01 m/s towards a

convex lens of focal length 0.3 m. Find the magnitude of rate of separation of image from the lens when the object is at a distance of 0.4 m from the lens. Also calculate the magnitude of the rate of change of the lateral magnification. [IIT - 2004]

Sol. Using lens formula

v1 =

4.01

− =

3.01

⇒ v = 1.2 m


Now we have

v1 –

u1 =

f1 , differentiating w.r.t. t

we have – 2v

1dtdv +

2u1

dtdu = 0

given dtdu = 0.01 m/s

⇒

dtdv = 2

2

)4.0()120( × 0.01 = 0.09 m/s

So, rate of separation of the image (w.r.t. the lens) = 0.09 m/s

Now, m =uv ⇒

dtdm = 2u

dtvdu

dtudv

−

2)4.0()01.0)(2.1()09.0)(4.0( − = – 0.35

So magnitude of the rate of change of lateral magnification = 0.35.

5. Two metallic plates A and B, each of area 5 × 10–4 m2,

are placed parallel to each other at a separation of 1 cm. Plate B carries a positive charge of 33.7×10–12C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 1016 photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 106 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remain constant at the value 2eV. Determine [IIT-2002]

(a) the number of photoelectrons emitted to t = 10 s, (b) the magnitude of the electric field between the

plates A and B at t = 10 s, and (c) the kinetic energy of the most energetic

photoelectron emitted at t = 10 s when it reaches plate B.

Neglect the time taken by the photoelectron to reach plate B. Takes ε0 = × 10–12 C2/N-m2.

Sol. (a) Number of electrons falling on the metal plate A = 1016 × (5 × 10–4)

∴ Number of photoelectrons emitted from metal plate A upto 10 second is

ne = 6

164

1010)105( ×× −

× 10 = 5 × 107

A B

d = 1 cm

(b) Charge on plate B at t = 10 sec Qb = 33.7 × 10–12 – 5 × 107 × 1.6 × 10–19 = 25.7 × 10–12 C also Qa = 8 × 10–12C

E = 0

B

2εσ –

0

A

2εσ =

0A21ε

(QB – QA)

= 124

12

1085.8105107.17

−−

−

×××× = 2000 N/C

(c) K.E. of most energetic particles = (hν – φ) + e(Ed) = 23 eV [(hν – φ) is energy of photo electrons due to light.

e(Ed) is the energy of photoelectrons due to work done on photoelectrons between the plates].

CHEMISTRY

6. At room temperature, the following reactions proceed nearly to completion :

2NO + O2 → 2NO2 → N2O4 The dimer, N2O4, solidified at 262 K. A 250 ml flask

and a 100 ml flask are separated by a stopcock. At 300 K, the nitric oxide in the larger flask exerts a pressure of 1.053 atm and the smaller one contains oxygen at 0.789 atm. The gases are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to 200 K. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at 220 K. (Assume the gases to behave ideally) [IIT-1992]

Sol. According to the gas equation, PV = nRT

or n = RTPV

At room temperature, For NO, P = 1.053 atm, V = 250 ml = 0.250 L

∴ Number of moles of NO = 3000821.0250.0053.1

××

= 0.01069 mol For O2, P = 0.789 atm, V = 100 ml = 0.1L


∴ Number of moles of O2 = 3000821.0

1.0789.0××

= 0.00320 mol According to the given reaction, 2NO + O2 → 2NO2 → N2O4 Composition of gas after completion of reaction, Number of moles of O2 = 0 1 mol of O2 react with = 2 mol of NO ∴ 0.00320 mol of O2 react with = 2 × 0.00320 = 0.0064 mol of NO Number of moles of NO left = 0.01069 – 0.0064 = 0.00429 mol Also, 1 mol of O2 yields = 1 mol of N2O4 ∴ Number of moles of N2O4 formed = 0.00320 mol N2O4 condenses on cooling, ∴ 0.350 L (0.1 + 0.250) contains only 0.00429 mol

of NO At T = 220 K, Pressure of the gas,

P = V

nRT = 350.0

2200821.000429.0 ×× = 0.221 atm

7. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V, while its temperature falls to T/2.

(a) How many degrees of freedom do the gas molecules have ?

(b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V. [IIT-1990]

Sol. (a) According to adiabatic gas equation, TVγ–1 = constant or T1V1

γ–1 = T2V2γ–1

Here, T1 = T ; T2 = T/2 V1 = V and V2 = 5.66 V

Hence, TVγ–1 = 2T × (5.66V)γ–1

= 2T × (5.66)γ–1 × Vγ–1

or (5.66)γ–1 = 2 ...(1) Taking log, (γ – 1)log 5.66 = log 2

or γ – 1 = 66.5log2log =

7528.03010.0 = 0.4

or γ = 1.4 If f, be the number of degrees of freedom, then

γ = 1 + f2 or 1.4 = 1 +

f2

or f2 = 1.4 – 1 = 0.4

or f = 4.0

2 = 5

(b) According to adiabatic gas equation, P1V1

γ = P2V2γ

Here, P1 = P V1 = V V2 = 5.66 V Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ

or P2 = γ)66.5(

P = 4.1)66.5(P =

32.11P [using eq.(1)]

Hence, work done by the gas during adiabatic expansion

= 1–

VPVP 2211

γ− =

1–4.1

V66.532.11

PPV ×−

= 4.0

2PVPV −

= 4.02

PV×

= 1.25 PV

8. (a) A white solid is either Na2O or Na2O2. A piece of

red litmus paper turns white when it is dipped into a freshly made aqueous solution of the white solid.

(i) Identify the substance and explain with balanced equation.

(ii) Explain what would happen to the red litmus if the white solid were the other compound.

(b) A, B and C are three complexes of chromium (III) with the empirical formula H12O6Cl3Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H2SO4, whereas complexes B and C lose, 6.75% and 13.5% of their original mass, respectively, an treatment with conc. H2SO4. Identity A, B and C. [IIT-1999]

Sol. (a) The substance is Na2O2 (i) Na2O2 + 2H2O → 2NaOH + H2O2 (strong base) (Weak acid) H2O2 + red litmus → White H2O2 → H2O + [O]


Nascent oxygen bleaches the red litumus. (ii) Na2O + H2O → 2NaOH NaOH solution turns colour of red litmus paper into

blue due to stronger alkaline nature. (b) A = [Cr(H2O)6]Cl3. It has no reaction with conc.

H2SO4 as its all water molecular are present in coordination sphere.

B = [Cr(H2O)5Cl]Cl2.H2O Conc. H2SO4 removes its one mol of H2O as it is

outside the coordination sphere. Molecular Weight of complex = 266.5

% loss = 5.266

18 × 100 = 6.75%

C = [Cr(H2O)4Cl]Cl2.2H2O Conc. H2SO4 removes its 2H2O which are outside of

the coordination sphere.

% loss = 2 × 5.266

18 × 100 = 13.5 %

Hence complexes A = [Cr(H2O)6]Cl3 B = [Cr(H2O)5Cl]Cl2.H2O C = [Cr(H2O)4Cl2]Cl.2H2O 9. (a) Write the chemical reaction associated with the

"brown ring test". (b) Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2–

and [Ni(CO)4]. Write the hybridization of atomic orbital of the transition metal in each case.

(c) An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involved in the formation of A and B. [IIT-2000]

Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3 2HNO3 + 6FeSO4 + 3H2SO4 → 3Fe2(SO4)3 + 2NO + 4H2O [Fe(H2O)6]SO4.H2O + NO Ferrous Sulphate → [Fe(H2O)5NO] SO4 + 2H2O (Brown ring) (b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its

coordination number is six. Co27 = 1s1, 2s22p6, 3s23p63d7, 4s2 Co3+ion = 1s2, 2s22p6, 3s23p63d6

3d 4s 4p

3d 4s 4p

d2sp3 hybridization

Hence

Co3+ion in Complex ion

3+

Co

NH3

NH3

NH3H3N

H3N NH3

or

H3N

H3N

NH3 NH3

NH3NH3

Co3+

In [Ni(CN)4

2– nickel is present as Ni2+ ion and its coordination numbers is four

Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p6, 3s23p63d8

3d 4s 4p

3d 4s 4p

dsp2 hybridization

Ni2+ ion =

Ni2+ion in Complex ion

Hence structure of [Ni(CN)4]2– is

Ni2+

N ≡ C

N ≡ C

C ≡ N

C ≡ N

In [Ni(CO)4, nickel is present as Ni atom i.e. its oxidation number is zero and coordination number is four.

3d 4s 4p

sp3 hybridization

Ni in Complex

Its structure is as follows :

Ni

CO

CO

CO

OC


(c) The transition metal is Cu2+. The compound is CuSO4.5H2O

CuSO4 + H2S → mediumAcidic pptBlack

CuS ↓ + H2SO4

2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 (B) white

I2 + I– → I3– (yellow solution)

10. A basic volatile, nitrogen compound gave a foul

smelling gas when treated with CHCl3 and alcoholic KOH. A 0.295 g sample of the substance dissolved in aqueous HCl and treated with NaNO2 solution at 0ºC liberated a colourless; odourless gas whose volume corresponds to 112 ml at STP. After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave yellow precipitate. Identify the original substance. Assume it contains one N-atom per molecule. [IIT-1993]

Sol. Clue 1. Nitrogen compound gave foul smelling gas when treated with CHCl3 and alc. KOH (carbylamine reaction), thus it is a primary amine.

Clue 2. This compound when treated with HCl + NaNO2 solution (nitrous acid test) at 0ºC liberates colourless and odourless gas.

CnH2n+1NH2 → + 2NaNOHCl AlcoholROH +

Nitrogen2N ↑

At STP, 112 ml of N2 is evolved from = 0.295 g CnH2n+1NH2

∴ 22400 ml of N2 is evolved from

= 112

22400295.0 × = 59 g CnH2n+1NH2

∴ CnH2n+1NH2 = 59 or n × C + (2n + 1) × H + N + 2 × H = 59 or 12n + 2n + 1 + 14 + 2 × 1 = 59

or n = 1442 = 3

Thus the molecular formula of nitrogen compound is C3H7NH2.

Clue 3. Alcohol obtained gives iodoform test positive, thus it is a secondary alcohol and its structure should be

CH3CHCH3

OH2-propanol

and hence the structure of (A) should be

CH3CHCH3

NH2

Propan-2-amine

MATHEMATICS

11. Find the values of a and b so that the function

f(x) =

π≤<π−π≤≤π+

π≤≤+

x2/,xsinbx2cosa2/x4/,bxcotx2

4/x0,xsin2ax

is continuous for 0 ≤ x ≤ π [IIT-1989] Sol. As, f(x) is continuous for 0 ≤ x ≤ π

∴ R.H.L.

π

=4

xat = L.H.L.

π

=4

xat

⇒

+

ππ b4

cot4

.2 =

π

+π

4sin.2a

4

⇒ 2π + b =

4π + a

⇒ a – b = 4π ....(i)

also, R.H.L

π

=2

xat = L.H.L

π

=2

xat

⇒

π

−π

2sinb

22cosa =

+

ππ b2

cot.2

.2

⇒ – a – b = b ⇒ a + 2b = 0 ...(ii)

From (i) and (ii), a = 2

3π and b = 43π−

12. Find dxdy at x = –1, when

x

2sin

)y(sinπ

+ 23 sec–1(2x) + 2x tan ln (x + 2) = 0

[IIT-1991] Sol. Here,

x

2sin

)y(sinπ

+ 23 sec–1(2x) + 2x tan (log (x + 2)) = 0

Differentiating both sides, we get


x

2sin

)y(sinπ

. log(sin y) . cos2π x .

2π

+

π x

2sin

1x2

sin)y(sin

−

π

. cos y . dxdy

+ 2x4|)x|2(

2.23

2 −+

)2x())2x(log(sec.2 2x

++

+ 2x log 2 . tan (log(x + 2)) = 0

putting,

π−=−=

3y,1x , we get

π−

−

3,1dxdy =

2

2

31

3

π−

π−

= 3

32 −ππ

13. ABC is a triangle such that

sin(2A + B) = sin(C – A) = –sin(B + 2C) = 21

If A, B and C are in Arithmetic Progression, determine the values of A, B and C. [IIT-1990]

Sol. Given that in ∆ABC, A, B and C are in A.P. A + C = 2B also A + B + C = 180º

⇒ B = 60º Also given that, sin (2A + B) = sin (C – A) = – sin (B + 2C) = 1/2 ...(1)

⇒ sin (2A + 60º) = sin (C – A) = – sin (60º + 2C) = 21

⇒ 2A + 60º = 30º, 150º neglecting 30º, as not possible

⇒ 2A + 60º = 150º

⇒ A = 45º again from (1), sin (60º + 2c) = –1/2

⇒ 60º + 2C = 210º, 330º

⇒ C = 75º or 135º Also from (1) sin (C – A) = ½ C – A = 30º, 150º, 195º for A = 45º, C = 75º and C = 195º (not possible)

∴ C = 75º Hence, A = 45º, B = 60º, C = 75º

14. If exp (sin2x + sin4x + sin6x + ...... ∞). ln 2 satisfies the equation x2 – 9x + 8 = 0, find the value of

xsinxcosxcos

+, 0 < x <

2π . [IIT-1991]

Sol. exp (sin2x + sin4x + sin6x + ...... ∞) loge2

⇒ 2log.

xsin1xsin

e2

2

e −

⇒ x2tan

e 2loge

⇒ xtan22 satisfy x2 – 9x + 8 = 0

⇒ x = 1, 8

∴ xtan22 = 1 and xtan2

2 = 8 ⇒ tan2x = 0 and tan2x = 3

⇒ x = nπ and tan2x = 2

3tan

π

and x = nπ ± 3π

Neglecting x = nπ as 0 < x < 2π

⇒ x = 3π ∈

π

2,0

∴ xsinxcos

xcos+

=

23

21

21

+

= 31

1+

× 1313

−

−

= 2

13 −

∴ xsinxcos

xcos+

= 2

13 −

15. Find the value of :

cos (2 cos–1 x + sin–1x) at x = 51 , where 0 ≤ cos–1x ≤ π

and –π/2 ≤ sin–1x ≤ π/2 [IIT-1981] Sol. cos2cos–1x + sin–1x

= cos

π

+−

2xcos 1 , as cos–1x + sin–1x =

2π

= – sin(cos–1x )

= – sin(sin–1 2x1− )

= – sin

−−

21

511sin

= – sin

−

562sin 1 =

562


Q.1 An imaginary closed loop is shown with current

carrying conductors then

a

cd

b

4A

3A

1A Closed Loop

(A) Line integral of magnetic field over the closed

loop abcda is zero (B) Surface integral of the magnetic field over the

closed loop abcda is non zero (C) Line integral of magnetic field and surface

integral of magnetic field over the closed loop abcda both are non zero

(D) Surface integral of magnetic field over the closed loop is zero but the line integral of

magnetic field over the closed loop is0

2 .C1ε

Here C - speed of light in air / free space / vacuum ε0 - Absolute permittivity of air / free space / vacuum

Q.2 A charged particle is entering through the tiny hole, in the given magnetic field between the plates →

× × ×

BPlate-2 × × ×

× × ×

× × ×

× × ×

× × × Plate-1

P2 P P1

Charged particle

Plane Mirror

d

(A) If the particle is positively charged it's motion

as seen in plane mirror is anticlockwise and on circular path

(B) If d ≤qBmv

then the particle will hit the upper

plate at P2 if it is negatively charged

(C) If d ≥qBmv

then the particle will hit the upper

plate at P1 if it is negatively charged

(D) If d ≥qBmv then it will never hit upper plate and

it's motion will be clockwise for positively charged particle as seen in plane mirror

Q.3 The principal axis of the given concave mirror is

along X-axis. The details about the mirror are also shown. In the shaded portion there is the coexistence of uniform magnetic field and electric field the details are written below

→E = E0 i ,

→B = kB0

E0 and B0 are positive constants

00 B:E = 19.6 : 1

A charged particle is projected from the point

(30 cm, 0, 0) with the velocity of jvv 0=→

, v0 is a positive constant. If the particle moves undeviated then the particle is -

Y-axis

X-axis10cm

C

Z-axis

20cm 30cm 40cmF

Concavemirror

F = Focus C= centre of curvature

(A) Positively charged (B) Negatively charged (C) May be positive of negative (D) Particle can not pass undeviated through the

pair of transverse magnetic field and electric field

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 SharmaDirector Academics, Jodhpur Branch

Physics Challenging Problems

Solutio ns wi ll b e pub lish ed in n ext issue

Set # 9


Q.4 In previous ques.(Q.3) if charged particle is projected from point (30cm, 0, 0) with the velocity

of jvv 0=→

and it goes un deviated then (here v0 is positive constant)

(A) Particle should be negatively charged and v0 = 6.19 m/s

(B) Particle should be positively charged and v0 = 6.19 m/s

(C) Particle is negatively charged and it will reach up to the maximum height of 10m parallel to Y-axis if gravity is taken into account

(D) Particle is negatively charged and when viewed through the concave mirror it is going parallel to negative Y-axis maximum 20m below

Q.5 An R-L series circuit is shown in figure

The R-L circuit is in discharging mode and current

i = 18 amp. then

B

3RR2

6R

R1A

L

L↑i

R1 = R2 = R = 1Ω L = 1/3 Henry

A, B = Terminals of resistance R1

G = Ground terminal

In column II, quantities are in SI units Column-I Column-II (A) Total energy dissipated (P) 0.25 in resistances (B) Time constant for (Q) 12 discharging mode (C) Potential drop across R1 (R) 108 initially (D) Potential drop across R2 (S) 18 initially

Q.6 Two charged particles having same de-Broglie wavelengths enters in given transverse magnetic field as shown below

→

× × × × × ×

B

O

y-axis

(x, 0, 0) z-axis

x-axis

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

Particle-A Particle-B

(A) Proton Electron

(B) Deutron Electron

(C) Monoionized Electron helium atom

(D) Doubly ionized Electron

helium atom

Q.7 A circular current carrying coil is placed in uniform

magnetic field as shown in left column

Column-I Column-II

(A) →

Bori

(P) Magnetic force τm = 0

(B) →

Bor

i (Q) Magnetic torque τm = 0

(C) →×××

oBr

×××

×××

×××

i (R) Expansion of coil

(D)

i

r o (S) Compression of coil

Q.8 Match the followings Column-I Column-II

(A) ∫→→

sd.E (P) C2µ0 qnet Gauss law

(C = speed of light in Air/Free space/Vacuum)

(B) ∫→→

sd.B (Q) Zero

(C) ∫→→

ed.B (R) Magnetic monopole is

impossible

(D) ∫→→

ed.E (S) Induced emf Faraday's law of

electromagnetic induction

(T) net0

2 i..C1ε

Ampere's circuital law


1. As information given in: 1st Bright Fringe occurs in front of a slit so,

As 2d.Dx n = nλ for nth Bright Fringe

So, 2d.Dx1 = 1λb ⇒ 2d

Dd = λb ⇒ λb = 2.

Dd2

For the missing wavelength, destructive interference should occur in front of slit

So, 2d.Dd = (2n – 1).

2gsinmisλ

⇒ λmissing = 2.Dd2 2

. )1n2(

1−

=1n2

2−

.Dd2 2

=1n2

2−

. λb

x1 =

O

Screen

Slit s1 1st B/F

d

Slit = s2

D For n = 1 λmissing = 2λb ≡ = more than λb

n = 2 λmissing = 32 . λb ≡ less than λb

(λmissing) max. = 2λb , Option (C) is correct

2. As(lmissing) max. = 2λb Option (A) is correct

3. If Slit width are not equal then I1 ≠ I2 and a1 ≠ a2 So Imin. = Intensity of Dark fringe = (a1 ~ a2)2 ≠ 0 So, dark fringe will be of blue colour Option (B) is correct

4. The equivalent circuit is

b

R

R

R R

a

2RR2R

)b,a(eq += =

2R5

If 2R51R

)b,a(eq == So R =

52

Ω

Option (A) is correct

5. As, R =52

Ω , Ω= 1R)b,a(

eq

So, vab = i. 1R)b,a(

eq = (1) = 1 volt

Option (B) is correct

6. If vab= 0, then irrespective to the value of capacitance C energy stored will be zero.

vab =21

2211

R/1R/1)R/(R/

+ε−+ε = 0

So, 1

1

Rε –

1

1

Rε = 0 ⇒

1

1

Rε =

2

2

Rε

⇒ 2

1

RR =

2

1

εε

ε2

R

R1

R2

ε1

C

Option (A) is correct

7. To calculate time constant Replace voltage source by short circuit mean by zero

resistance and then find Req with C and time constant τ = Req.C

τ =

+

+R

RRRR

21

21 C

R

R1

R2

C⇒

R2

R1R2/R1+R2

C

Option (D) is correct 8. Maximum current through the resistance Imax

=R

vab =R

R/1R/1/)R/()R/( 212211 +ε−+ε =Reqε

Option (C) is correct.

Solution Physics Challenging Problems

Set # 8

8 Question s were Published in Decemb er I ssue


1. A sphere of mass 50 g is attached to one end of a steel wire, 0.315 mm diameter and one metre long. In order to form a conical pendulum, the other end is attached to a vertical shaft which is set rotating about its axis. Calculate the number of revolutions necessary to extend the wire by 1 mm. Young's modulus of elasticity of steel = 2 × 1012 dynes/cm2 and g = 980 cm/sec2.

Sol. Let T be the tension in the wire, when the extension is 1 mm. According to definition,

Tθ

T cos θ

T sin θ

θ

mg

θ L = 1m

Y = strainTensilestressTensile =

L/A/T

l =

AT ×

l

L

= 2r

Tπ

×l

L

∴ T = LrY 2lπ =

1001.0)01575.0(102 212 ××π××

= 1.559 × 106 dynes

When the sphere is revolving, it is acted upon by two forces namely the tension T along the wire and its weight mg acting vertically downwards. Resolving T into vertical and horizontal components, we get

T cos θ = mg or cos θ = T

mg = 610559.198050×

×

∴ sin θ = )cos1( 2 θ− = 0.9998 = 1.00 (nearly)

∴ Radius of the circle described = r = (L + l) sin θ

= 100.1 cm

Now, T sin θ = r

mv2

or, 1.559 × 106 × 1 = 1.100v50 2×

∴ v2 = 50

1.10010559.1 6 ××

or v = 1766 cm/sec

∴ Period of revolution = v

r2π = 1766

1.1002 ×π

= 0.3561 s

So, Frequency of revolution = period

1 = 3561.01

= 2.808/sec

2. A wave travels out in all directions from a point source. Justify the expression y = (a0/r) sin K(r – vt), at a distance r from the source. Find the speed, periodicity and intensity of the wave.

Sol. If P be the power of the source then intensity

I = 2r4Pπ

or, I ∝ 2r1

But I ∝ a2, so a ∝r1 or a =

ra0

Where a0 is constant. The equation in standard form is, y = a sin K (r – vt) Therefore, above equation is written as :

y =r

a0 sin K(r – vt)

Now, comparing this equation with y = a sin (Kr – ωt)

We have, ω = Kv or n =π2

Kv and

K =λπ2 or λ =

K2π

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

Students' ForumPHYSICS


Speed c = nλ =

π2Kv ×

π

K2 = v

Also, T =n1 =

Kv2π

Thus, intensity is given by

I =21

ρa2 ω2 c =21

ρ 2

20

ra

. K2v2 . v

or I =21

2

3220

rvKaρ

3. A conducting bar of mass m, length l is pushed with a speed v0 on a smooth horizontal conducting rail containing an inductance L. If the applied magnetic field has inward field of induction B, find the maximum distance covered by the bar before it stops.

L v0 l

m B ⊗

Sol. If the bar slides a distance dx, the flux linkage

– dφ = Bldx

The induced e.m.f. =dtdφ = – Bl

dtdx

Since the induced e.m.f. across the inductor = – LdtdI

∴ – Bldtdx = – L

dtdI

or, LdI = Bl dx

or, LI = Bl x

or, I = LBl x

This induced current interacts with the applied magnetic field of induction and imparts a restoring (magnetic) force

F = IlB = –

x

LBl l B

or, mv dxdv = –

LB 22l x

or, ∫0

v0

v dv = – MLB 22l

∫s

0x dx = –

mL2sB 222l

∴ s = )mL(lB

v0

4. Two coherent light sources emit light of wavelength 550 nm which produce an interference pattern on a screen. The sources are 2.2 mm apart and 2.2 m from the screen. Determine whether the interference at the point O is constructive or destructive. Calculate the fringe width.

S1 2d S2

D

O

l

Sol. The path difference at O is given by

∆ = S2O – S1O

From figure, S2O = [l 2 + (2d)2]1/2

∴ S2O = l2/12d21

+

l= l

+

2d2211

l

Now, ∆ = l

−

+ 1d2

211

2

l=

l2)d2( 2

=2.22

)102.2( 23

×× −

= 1.1 × 10–6 m

The difference will be constructive if path difference is an integral multiple of wavelength i.e., n = 1, 2, 3, …….

∴ n =λ∆ = 7

6

105.5101.1

−

−

×× = 2

Fringe width,

β = d2Dλ = 3

7

102.22.2105.5

−

−

×××

= 5.5 × 10–4 m = 0.55 mm

5. Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find

(a) the energy of the photons causing photoelectric emission

(b) the quantum numbers of the two levels involved in the emission of these photons


(c) the change in the angular momentum of the hydrogen atom in the above transition, and

(d) the recoil speed of the emitting atoms assuming it to be at rest before the transition.

(Ionization potential of hydrogen is 13.6 volt and the mass of the hydrogen atom is 1.67 × 10–27 Kg, 1 eV = 1.6 × 10–19 J)

Sol. (a) According to Einstein’s photo-electric equation, the maximum kinetic energy Ek of the emitted electrons is given by

EKmax

= hν – W,

Where hν is the energy of photons causing the photo-electric emission and W is the work-function of the emitting surface.

Given that, EKmax = 0.73 eV and W = 1.82 eV

∴ hν = EKmax + W = 0.73 eV + 1.82 eV = 2.55 eV

(b) These photons (where energy is 2.55 eV) are emitted by hydrogen atoms.

As (I.E.)H = 13.6 eV, hence

E1H = – (I.E.)H = – 13.6 eV

The energy of higher levels is given by

HaE = 2

H1

nE

Hence, H2E = –

46.13 = – 3.4 eV

H3E = –

96.13 = – 1.5 eV

and H4E = –

166.13 = – 0.85 eV

The energy of the emitted photon is 2.55 eV

Now H4E – H

2E = – 0.85 – (– 3.4) = 2.55 eV

Thus, the quantum numbers of two levels involved in the emission of photon of energy 2.55 eV are 4 and 2.

(c) The electron transition causing the emission of photon of energy 2.55 eV is from n = 4 level to n = 2 level. Now, according to Bohr’s 2nd postulate, the angular momentum of electron in the hydrogen atom is (n h/2π). Thus, the change in angular momentum in above transition is

∆L =π2h4 –

π2h2 =

πh

(d) The momentum of the photon emitted from the hydrogen atom

pPh =c

hν =s/m103

J106.155.28

19

××× −

= 1.36 × 10–27 Kg.m/s

According to the law of conservation of momentum, the recoil momentum of a hydrogen atom will be equal and opposite to the momentum of the emitted photon.

( Php→

+ Ap→

= 0 or Ap→

= – Php→

)

Hence, the recoil speed of the atoms is :

V =mass

|Momentum| =A

A

m|p|

→

=A

Ph

m|p|

→

=kg1067.1

s/mkg1036.127

27

−

−

×−× = 0.814 m/s

SCIENCE TIPS

• A porcelain funnel used for filtration by suction is known as Bucher Funnel

• What is diazomethane ?

]NCHorNNCH[ 22–

2 ==+

• A drying chamber, containing chemicals such as concentrated sulphuric acid or silica gel is known as Desiccator

• Reforming of a gasoline fraction to increase branching in presence of AlCl3 is known as

Isomerization• A condenser consisting of glass tube surrounded by

another glass tube through which cooling water flows is known as Liebig condenser

• For wattles current what should be the value of the power factor of the circuit ? Zero

• For which colour is the critical angle of light, pasing from glass to air, minimum ? Violet

• Give an example of application of mutual induction in any device. Transformer

• What is the correct sequence of the semiconductors silicon, tellunium and germanium in the increasing order of their energy gap ?

Tellurium, germanium, silicon• Which ammeter is used to measure alternating

current ? Hot wire ammeter• What quantity has the ampere-second as its unit ?

Quantity of electricity


Laws of Refraction :

The incident ray, the refracted ray and normal on incidence point are coplanar.

µ1 sin θ1 = µ2 sin θ2 = ... = constant.

µ1

µ2

θ2

θ1

Snell's law in vector form :

µ1

µ22e

1e

n

Let, 1e = unit vector along incident ray

2e = unit vector along refracted.

n = unit vector along normal on incidence point.

Then µ1( 1e × n ) = µ2( 2e × n )

Some important points :

(a) The value of absolute refractive index µ is always greater or equal to one.

(b) The value of refractive index depends upon material of medium, colour of light and temperature of medium.

(c) When temperature increases, refractive index decreases.

(d) Optical path is defined as product of geometrical path and refractive index.

i.e., optical path = µx

(e) For a given time, optical path remains constant.

i.e., µ1x1 = µ2x2 = ... constant

∴ µ1 dtdx1 = µ2 dt

dx2

∴ µ1c1 = µ2c2 (where c1 and c2 are speed of light in respective mediums)

∴ 1

2

µµ =

2

1

cc

i.e., µ ∝ c1

(f) The frequency of light does not depend upon medium.

∴ c1 = fλ1, c2 = fλ2

∴ 2

1

µµ =

1

2

cc =

1

2

λλ

∴ µ ∝ λ1

When observer is rarer medium and object is in denser medium :

Then µ = depthapparent

depth real

Air Observer

P Object

Denser medium(µ)

Apparentdepth PReal

depth

When object is in rarer and observer is in denser

medium :

µ = position real

positionapparent

Refraction at plane & curved surface

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


The shift of object due to slab is x = t

µ1–1

t

P P´ Q

µ

Object shiftness

(a) This formula is only applicable when observer is in rarer medium.

(b) The object shiftiness does not depend upon the position of object.

(c) Object shiftiness takes place in the direction of incidence ray.

The equivalent refractive index of a combination of a

number of slabs for normal incidence is µ =

i

i

i

µtt

Σ

Σ

t1

t2

µ1

µ2

Here, Σti = t1 + t2 + ...

Σi

i

µt =

1

1

µt +

2

2

µt + ...

The apparent depth due to a number of media is Σi

i

µt

The lateral shifting due to a slab is d = t sec r sin(i – r).

i

rµ

t

d

Critical angle : When a ray passes from denser

medium (µ2) to rarer medium (µ1), then for 90º angle of refraction, the corresponding angle of incidence is critical angle.

90º

Rarer µ1

Denserµ2 c

Mathematically, sin c = 2

1

µµ

r

i c i i

i<c i=c

Rarer medium(µ1)

Denser medium (µ2)

(i) When angle of incidence is lesser than critical angle, refraction takes place. The corresponding deviation is

δ = sin–1

isin

µµ

1

2 – i for i < c

(ii) When angle of incidence is greater than critical angle, total internal reflection takes place. The corresponding deviation is

δ = π – 2i when i < c

The δ – i graph is :

(i) Critical angle depends upon colour of light, material of medium, and temperature of medium.

(ii) Critical angle does not depend upon angle of incidence

π/2 ci

δ

Refractive surface formula,

v

µ2 – uµ1 =

rµµ 12 −

Here, v = image distance,

u = object distance,

r = radius of curvature of spherical surface.


(a) For plane surface , r = ∞

(b) Transverse magnification,

m = sizeobject size ageIm =

uµvµ

2

1

(c) Refractive surface formula is only applicable for paraxial ray.

Lens :

Lens formula :

v1 –

u1 =

f1

(a) Lens formula is only applicable for thin lens.

(b) r = 2f formula is not applicable for lens.

(c) m = sizeobject size image =

uv

(d) Magnification formula is only applicable when object is perpendicular to optical axis.

(e) lens formula and the magnification formula is only applicable when medium on both sides of lenses are same.

(f)

f(+ve)

(i)

f(–ve)

(ii)

f(–ve) f(+ve)

(iii) (iv)

f(–ve)

(v)

f(+ve)

(vi) (g) Thin lens formula is applicable for converging as

well diverging lens. Thin lens maker's formula :

f1 =

−

1

12

µµµ

−

21 r1

r1

µ1 µ1

µ2

(a) Thin lens formula is only applicable for paraxial ray.

(b) This formula is only applicable when medium on both sides of lens are same.

(c) Intensity is proportional to square of aperture.

(d) When lens is placed in a medium whose refractive index is greater than that of lens. i.e., µ1 > µ2. Then converging lens behaves as diverging lens and vice versa.

(e) When medium on both sides of lens are not same. Then both focal lengths are not same to each other.

(f) If a lens is cut along the diameter, focal length does not change.

(g) If lens is cut by a vertical, it converts into two lenses of different focal lengths.

i.e., f1 =

1f1 +

2f1

ff1

+

f1

(h) If a lens is made of a number of layers of different refractive index (shown in figure)

µ1

µ6

µ2µ3

µ4µ5

+ + ++ + +

Then number of images of an object by the lens is

equal to number of different media.

(i) The minimum distance between real object and real image in is 4f.

(j) The equivalent focal length of co-axial combination of two lenses is given by

F1 =

1f1 +

2f1 –

21ffd


f1

o1

d

o2

d<f1 d<f2

f2

(k) If a number of lenses are in contact, then

F1 =

1f1 +

2f1 + ......

(l) (i) Power of thin lens, P = F1

(ii) Power of mirror is P = –F1

(m) If a lens silvered at one surface, then the system behaves as an equivalent mirror, whose power

P = 2PL + Pm

Here, PL = Power of lens

=

−

1

12

µµµ

−

21 r1

r1

Pm = Power of silvered surface = –mF1

Here, Fm = r2/2, where r2 = radius of silvered surface.

P = – 1/F

Here, F = focal length of equivalent mirror.

1. A thin converging lens forms the image of a certain object magnified p times. The magnification becomes q when the lens is moved nearer to the object by a distance a. Calculate the focal length of the lens.

Sol. The magnification (m) produced by a lens in terms of u and f i : given by

m = fu

f−

or fu =

m1m +

In 1st case, fu =

p1p + (Q m = p)

In 2nd case, f

au − =q

1q + or, fu –

fa =

q1q +

or, p

1p + –fa =

q1q + or, =

fa –

p1p + –

q1q +

or, fa =

pqppqqpq −−+ =

pqpq −

∴ f =pq

apq−

2. A convex refracting surface of radius of curvature 30 cm separates two media of refractive indices n1 = 4/3 and n2 = 3/2 respectively. Find the position of image formed by refraction of an object placed at a distance of (i) 280 cm and (ii) 80 cm, from the surface.

Sol. (i) Given that n1 = 4/3,

n2 = 3/2, |u| = 280 cm, |R| = 30cm

For refraction through a spherical surface :

v

n2 –un1 =

R)nn( 12 −

Here, u = – 280 cm, R = + 30 cm. Hence

v2

3 –2803

4×−

=3

)]2/3()3/4[(+−

or v2

3 +2101 =

1801

∴ v2

3 =180

1 –2101 =

12601

or, v = (3/2) × 1260 = 1890 cm = 18.9 m

As v is positive, hence the image is real and is formed in second medium at a distance of 18.9 m from the refracting surface.

(ii) In this case, u = – 80 cm, R = + 30 cm Again from the formula for refraction through a

surface, (3/2v) – [4/–3 × 80)] = [(4/3) – (3/2) / + 30] or (3/2v) + (1/60) = (1/180) or 3/2v = [(1/180) – (1/60)]

or v = (3/2) × (– 90) = – 135 cm

Solved Examples


As v is negative, hence the image is virtual and is

formed in the first medium of refractive index 4/3 at

a distance of 135 cm from the pole.

3. There is a small air bubble in side a glass sphere (n =

1.5) of radius 10 cm. The bubble is 4 cm below the

surface and is viewed normally from the outside

(Fig.). Find the apparent depth of the air bubble.

n2 = 1A

C

P

n1 = 1.5

O

I

Sol. The observer sees the image formed due to refraction

at the spherical surface when the light from the

bubble goes from the glass to air.

Here u = – 4.0 m, R = – 10 cm, n1 = 1.5 and n2 = 1

We have [(n2/v) – (n1/u) = (n2 – n1)/R

or (1/v) – (1.5/ –4.0 cm) = (1 – 1.5)/ (– 10 cm)

or (1/v) = (0.5/10 cm) – (1.5/4.0 cm)

or v = – 3.0 cm

Thus, the bubble will appear 3.0 cm below the

surface.

4. A convex lens focuses a distance object on a screen

placed 10 cm away from it. A glass plate (n = 1.5) of

thickness 1.5 is inserted between the lens and the

screen. Where should the object be placed so that its

image is again focused on the screen ?

Sol. The situation when the glass plate is inserted between

the lens and the screen, is shown in fig. The lens

forms the image of object O at point I1 but the glass

plate intercepts the rays and forms the final image at I

on the screen. The shift in the position of image after

insertion of glass plate

II1 O

Scre

en

10 cm

I1I = t

−

n11 = (1.5 cm)

−

5.111 = 0.5 cm.

Thus, the lens forms the image at a distance of 9.5 cm

from itself. Using

v1 –

u1 =

f1 , we get

u1 =

v1 –

f1 =

5.91 –

101

or u = – 190 cm.

i.e. the object should be placed at a distance of

190 cm. from the lens.

5. A candle is placed at a distance of 3 ft from the wall.

Where must a convex lens of focal length 8 inches be

placed so that a real image is formed on the wall ?

Sol. According to formula for refraction though a lens

f = 8"

36 – v v

d = 3 ft = 36"

v1 –

u1 =

f1 or

v1 –

)v36(1

−−=

81

or v1 +

v361−

=81 or

)v36(vvv36

−+− =

81

or, v2 – 36 v + 8 × 36 = 0

or v = 12" or 24" = 1 ft or 2 ft.

∴ u = 24" or 12" = 2 ft or 1 ft

Hence, lens should be placed at either 1 ft or 2 ft

away from the wall.


Key Concepts : Stress : The restoring force setup inside the body per unit

area is known as stress. Restoring forces : If the magnitude of applied

deforming force at equilibrium = F

Then, Stress = AF

In SI system, unit of stress is N/m2. Difference between pressure and stress : (a) Pressure is scalar but stress is tensor quantity. (b) Pressure always acts normal to the surface, but

stress may be normal or tangential. (c) Pressure is compressive in nature but stress may

be compressive or tensile. Strain :

Strain = dimension originaldimensionin change

(a) Longitudinal strain = LL∆

L F F

Longitudinal strain is in the direction of deforming force but lateral strain is in perpendicular direction of deforming force.

Poisson ratio :

σ = strain allongitudin

strain lateral = L/Ld/D

∆∆

Here ∆d = change in diameter.

(b) Volumetric strain = VV∆

F

F F

F V (c) Shear strain = φ

Shear strain

φ

Stress-strain graph : From graph, it is obvious that in elastic limit, stress is

proportional to strain. This is known as Hooke's law. ∴ Stress ∝ Strain ∴ Stress = E .strain

∴ E = strainstress

where E is proportionality dimensional constant known as coefficient of elasticity.

O

A

B C

Plastic regionBreaking

strength Elasticlimit

Strain

Stre

ss

Types of coefficient of elasticity :

(a) Young's modulus = Y = strain allongitudinstress ogitudinall

∴ Y =

LLA

F∆

= LA

FL∆

L

∆L

F

(b) Bulk modulus = B = strain volumetricstress volumetric

Compressibility = 1/B

Properties of Matter

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


(c) Modulus of rigidity = η = φA

F = strainshear stress shear

(d) For isothermal process, B = P. F

φφ

F (e) For adiabatic process, B = γP

(f) modulusbulk Isothermalmodulusbulk Adiabatic = γ

(g) Esolid > Eliquid > Egas (h) Young's modulus Y and modulus of rigidity η

exist only for solids. (i) Bulk modulus B exist for solid, liquid and gas. (j) When temperature increases, coefficient of

elasticity (Y, B, η) decreases.

(k) B1 +

η3 =

Y9

(l) Y = 2(1 + σ)η (m) Poisson's ratio σ is unitless and dimensionless.

Theoretically, –1 < σ < 21

Practically, 0 < σ < 21

(n) Thermal stress = Yα∆θ (o) Elastic energy stored,

U = 21 × load × extension =

21 Fx =

21 kx2

= stress × strain × volume For twisting motion,

U = 21 × torque × angular twist

= 21

τ × θ = 21 cθ2

Elastic energy density,

u = 21 × stress × strain J/m3 =

21 Y × strain2J/m3

Thermal stress = Yα∆θ and Thermal strain = α∆θ Work done in stretching a wire :

(a) W = 21 F∆L

(b) Work done per unit volume = 21 × stress × strain

(c) Breaking weight = breaking stress × area

Surface tension :

T = LF

Here L = length of imaginary line drawn at the surface of liquid. and F = force acting on one side of line (shown in figure)

(a) Surface tension does not depend upon surface area.

(b) When temperature increases, surface tension decreases.

(c) At critical temperature surface tension is zero.

F

F L

Rise or fall of a liquid in a capillary tube :

h = gr

cosT2ρ

θ

Here θ = angle of contact. r = radius of capillary tube ρ = density of liquid For a given liquid and solid at a given place, hr = constant Surface energy : Surface energy density is defined as work done

against surface tension per unit area. It is numerically equal to surface tension.

W = work = surface tension × area (a) For a drop of radius R, W = 4πR2T (b) For a soap bubble, W = 8πR2T Excess pressure :

(a) For drop, P = RT2

(b) For soap bubble, P = RT4

Viscosity : (a) Newton's law of viscous force :

F = – ηAdydv

where dydv = velocity gradient

A = area of liquid layer η = coefficient of viscosity The unit of coefficient of viscosity in CGS is poise.


(b) SI unit of coefficient of viscosity = poiseuille = 10 poise. (c) In the case of liquid, viscosity increases with

density. (d) In the case of gas, viscosity decreases with

density. (e) In the case of liquid, when temperature increases,

viscosity decreases. (f) In the case of gas, when temperature increases,

viscosity increases. Poiseuille's equation :

V = L8rP 4

ηπ

where V = the volume of liquid flowing per second through a capillary tube of length L and radius r

η = coefficient of viscosity and P = pressure difference between ends of the tube Stoke's law : The viscous force acting on a spherical body moving

with constant velocity v in a viscous liquid is F = 6πηrv where r = radius of spherical body Determination of η :

η = v9

g)(r2 2 σ−ρ

where r = radius of spherical body moving with constant velocity v in a viscous liquid of coefficient of viscosity η and density ρ

and σ = density of spherical body Critical velocity (v0) :

v0 = r

kρη

where k = Reynold's number for narrow tube, k ≈ 1000. (a) For stream line motion, flow velocity v < v0. (b) For turbulant motion, flow velocity v > v0. 1. A mass of 5 kg is suspended from a copper wire of 5

mm diameter and 2 m in length. What is the extension produced in the wire ? What should be the minimum diameter of the wire so that its elastic limit is not exceed ? Elastic limit for copper = 1.5 × 109 dynes/cm2. Y for copper = 1.1 × 1012 dynes/cm2.

Sol. Given that Y = 1.1 × 1012 dynes/cm2,

L = 2m = 200 cm, d = 5 mm = 0.5 cm

or r = d/2 = 0.25 cm, F = 5.0 × 1000 × 980 dynes.

Y =l2r

FLπ

or l =Yr

FL2π

= 22 101.1)25.0(142.320098010000.5×××

×××

= 4.99 × 10–3 cm Also, elastic limit for copper = 1.5 × 109 dynes/cm2 If d' is the minimum diameter, then maximum stress

on the wire =4/'d

F2π

= 2'dF4

π

Hence, 2'dF4

π= 1.5 × 109

or d'2 =9105.1

F4××π

=9105.1142.3

98010000.54××

×××

= 41.58 × 10–4 d' = 0.0645 cm.

2. A uniform horizontal rigid bar of 100 kg in supported horizontally by three equal vertical wires A, B and C each of initial length one meter and cross-section I mm2. B is a copper wire passing through the centre of the bar; A and C are steel wires and are arranged symmetrically one on each side of B YCu = 1.5 × 1012 dynes / cm2, Ys = 2 × 1012 dynes/cm2. Calculate the tension in each wire and extension.

Sol. The situation is shown in figure. Because the rod is horizontally supported, hence extensions in all the wires must be equal i.e., strains in all the wires are equal as initial lengths are also equal.

As Y =StrainStress

AS

BCu

C S

100 Kg

Hence, YCu =Strain

A/FCu … (1)

and Ys = StrainA/Fs … (2)

∴ S

Cu

YY

=S

Cu

FF

=25.1 =

43 or 4FCu = 3FS ...(3)

According to figure, we can write 2FS + FCu = 100 g or 2 × (4/3) FCu + FCu = 100 g

or [(8/3) + 1] FCu = 100 g

Solved Examples


∴ FCu = (3/11) × 100g

= (3/11) × 100 Kgwt = 27.28 Kgwt

and FS = (4/3) FCu = (4/3) × (3/11) × 100g

= (400/11)g = 36.36 Kgwt

Extension in each wire,

l =Cu

Cu

AYLF =

122 105.11010098027280

××

××−

= 0.178 cm

3. A copper rod of length L and radius r is suspended from the ceiling by one of its ends. Find: (a) the elongation of the rod due to its own weight when ρ and Y are density and Young's modulus of the copper respectively, (b) the elastic potential energy stored in the rod due to its own weight.

Sol. (a) Consider any length x of the rod from the fixed end. Weight of lower portion of rod will exert stretching force on the upper portion.

∴ Stress over the portion at a distance x from fixed end = (L – x) mg/πr2

where m is the mass per unit length of the rod (m = πr2ρ)

∴ Strain at x = (L – x)mg/πr2Y

[Q Strain = (Stress/Y)] Hence, increase in length in elementary length dx at

x =Yrmg)xL(

2π

− dx

∴ Total increase in length = ∫ π

−L

0 2Yrmg)xL(

dx =Yr

mg2π 2

L2=

Yr2gLr2

22

π

ρπ =Y2

gL2ρ

(b) Energy density at x =21 × Strain × Stress

=21 × 2r

mg)xL(π

− ×Yrmg)xL(

2π

−

=21 .

Yg)xL( 222ρ−

∴ Energy stored in the volume of length element

dx =21

Yg)xL( 222ρ− × πr2dx

∴Total potential energy stored in the rod of length L,

U =21

Ygr 222ρπ

∫ −L

0

2)xL( dx =61

YgLr 2322ρπ

4. A ring is cut from a platinum tube 8.5 cm internal and 8.7 cm external diameter. It is supported horizontally from a pan of a balance so that it comes in contact with the water in a glass vessel. What is the surface tension of water if an extra 3.97 gm weight is required to pull it away from water (g = 980 cm/sec2).

Sol. The ring is in contact with water along is inner and outer circumference. So when pulled out the total force on it due to surface tension will be

T

F F

T

F = T (2πr1 + 2πr2)

∴ T =)rr(2

mg

21 +π=

)7.85.8(14.3298097.3

+×××

= 72.13 dyne/cm.

5. Two long capillary tubes of diameter 5.0 mm and 4.0 mm are held vertically inside water one by one. How much high the water will rise in each tube ? (g = 10 m/s2, surface tension of water = 7.0 × 10–2 N/m).

Sol. Height of water column in a capillary tube of radius r is given by

h =gr

cosT2ρ

θ … (1)

where T is surface tension, ρ is density and θ is angle of contact of water-glass which can be assumed zero.

For the first tube, r = 2.5 mm = 2.5 × 10–3 m

∴ h = )Kg/N10()m/Kg101()105.2(

)m/N100.7(2333

2

××××

××−

−

= 5.6 mm According to equation (1), for the same liquid, we

have

hr =g

cosT2ρ

θ = Constant

If a liquid rises to a height h1 in a capillary tube of radius r1 and to a height of h2 in a capillary tube of radius r2, then

h1r1 = h2r2 or h2 =2

11

rrh =

0.25.26.5 × = 7.0 mm


Definition and Classification : Carbohydrates are polyhydroxy aldehydes,

polyhydroxy ketones, or compounds that can be hydrolyzed to them. A carbohydrate that cannot be hydrolyzed to simpler compounds is called a monosaccharide. A carbohydrate that can be hydrolyzed to two monosaccharide molecules is called a disaccharide. A carbohydrate that can be hydrolyzed to many monosaccharide molecules is called a polysaccharide.

A monosaccharide may be further classified. If it contains an aldehyde group, it is known as an aldose; if it contains a keto group, it is known as a ketose. Depending upon the number of carbon atoms. It contains, a monosaccharide is known as a triose, tetrose, pentose, hexose, and so on. An aldohexose, for example, is a six-carbon monosaccharide containing an aldehyde group; a ketopentose is a five-carbon monosaccharide containing a keto group. Most naturally occurring monosaccharides are pentoses or hexoses.

Carbohydrates that reduce Fehling’s (or Benedict’s) or Tollens’ reagent are known as reducing sugars. All monosaccharides, whether aldose or ketose, are known as reducing sugars. Most disaccharides are reducing sugars; sucrose (common table sugar) is a notable exception, for it is a non-reducing sugar.

(+)-Glucose : an aldohexose : Because it is the unit of which starch, cellulose, and

glycogen are made up, and because of its special role in biological processes, (+)-glucose is by far the most abundant monosaccharide- there are probably more (+)-glucose units in nature than any other organic group–and by far the most important monosaccharide.

Cyclic structure of D-(+)-glucose. Formation of glucosides : D-(+)-glucose is a pentahydroxy aldehyde. D-(+)-

glucose had been definitely proved to have structure. CHO

OH H OH OH

H HO

H H

CH2OH D-(+)-Glucose

By 1895 it had become clear that the picture of D-(+)-glucose as a pentahydroxy aldehyde had to be modified.

Among the facts that had still to be accounted for were the following:

(a) D-(+)-Glucose fails to undergo certain reactions typical of aldehydes. Although it is readily oxidized, it gives a negative Schiff test and does not form a bisulfite addition product.

(b) D-(+)-Glucose exists in two isomeric forms which undergo mutarotation. When crystals of ordinary D-(+)-glucose of m.p. 146ºC are dissolved in water, the specific rotation gradually drops from an initial + 112º to + 52.7º. On the other hand, when crystals of D-(+)-glucose of m.p. 150ºC (obtained by crystallization at temperatures above 98ºC) are dissolved in water, the specific rotation gradually rises from an initial + 19º to + 52.7º. The form with the higher positive rotation is called α-D-(+)-glucose and that with lower rotation β-D-(+)-glucose. The change in rotation of each of these to the equilibrium value is called mutarotation.

(c) D-(+)-Glucose forms two isomeric methyl D-glucosides. Aldehydes react with alcohols in the presence of anhydrous HCl to form acetals. If the alcohol is, say methanol, the acetal contains two methyl groups :

–C=O

H

–C–OCH3

H

OH

–C–OCH3

H

OCH3

CH3OH,H+ CH3OH,H+

Aldehyde Hemiacetal Acetal When D-(+)-glucose is treated with methanol and

HCl, the product, methyl D-glucoside, contains only one –CH3 group; yet it has properties resembling those of a full acetal. It does not spontaneously revert to aldehyde and alcohol on contact with water, but requires hydrolysis by aqueous acids.

Furthermore, not just one but two of these monomethyl derivatives of D-(+)-glucose are known, one with m.p. 165ºC and specific rotation + 158º, and the other with m.p. 107 ºC and specific rotation –33º. The isomer of higher positive rotation is called methyl α-D-glucoside, and the other is called methyl β-D-glucoside. These glucosides do not undergo mutarotation, and do not reduce Tollens’ or Fehling’s reagent.

Organic Chemistry

Fundamentals

CARBOHYDRATES

KEY CONCEPT


D-(+)-Glucose has the cyclic structure represented crudely by IIa and IIIa, more accurately by IIb and IIIb.

H H

HO HH

OH OH HOH

1 2345

6 CH2OHH OH

OHOH H

HHH

HO

CH2OH O

12 3

45

6

IIa IIb

α-D-(+)-Glucose (m.p. 146 ºC, [α] = +112º)

O

HO H

HO HH

H OH HOH

1 2345

6 CH2OHH OH

HOH H

HOHH

HO

CH2OH O

12 3

45

6

IIIa IIIb

β-D-(+)-Glucose (m.p. 150 ºC, [α] = +19º)

O

D-(+)-Glucose is the hemiacetal corresponding to

reaction between the aldehyde group and the C-5 hydroxyl group of the open-chain structure. It has a cyclic structure simply because aldehyde and alcohol are part of the same molecule.

There are two isomeric forms of D-(+)-glucose because this cyclic structure has one more chiral centre than Fisher’s original open-chain structure. α-D-(+)-Glucose and β-D-(+)-glucose are diastereomers, differing in configuration about C-1. Such a pair of distereomers are called anomers.

As hemiacetals, α-and β-D-(+)- glucose are readily hydrolyzed by water. In aqueous solution either anomer is converted –via the open-chain form–into an equilibrium mixture containing both cyclic isomers. This mutarotation results from the ready opening and closing of the hemiacetal ring.

The typical aldehyde reactions of D-(+)-glucose –osazone formation, and perhaps reduction of Tollens’ and Fehling’s reagents– are presumably due to a small amount of open-chain compound, which is replenished as fast as it is consumed. The concentration of this open-chain structure, however, is too low (less than 0.5%) for certain easily reversible aldehyde reactions like bisulfite addition and the Schiff test.

Disaccharides : Disaccharides are carbohydrates that are made up of

two monosaccharide units. On hydrolysis a molecule of disaccharide yields two molecules of monosaccharide.

We shall study four disaccharides : (+)-maltose (malt sugar), (+)-cellobiose, (+)-lactose (milk sugar), and (+)-sucrose (cane or beet sugar).

(+)-Maltose : (+)-Maltose can be obtained, among other products,

by partial hydrolysis of starch in aqueous acid. (+)-Maltose is also formed in one stage of the fermentation of starch to ethyl alcohol; here hydrolysis is catalyzed by the enzyme diastase, which is present in malt (sprouted barley).

Let us look at some of the facts from which the structure of (+)-maltose has been deduced.

(+)-Maltose has the molecular formula C12H22O11. It reduces Tollens’ and Fehling’s reagents and hence is a reducing sugar. It reacts with phenylhydrazine to yield an osazone, C12H20O9(=NNHC6H5)2. It is oxidized by bromide water to a monocarboxylic acid, (C11H21O10)COOH, maltobionic acid. (+)-Maltose exists in alpha ([α] = + 168º) and beta ([α] = + 112º) forms which undergo mutarotation in solution (equilibrium [α] = + 136º).

(+)-Cellobiose : When cellulose (cotton fibers) is treated for several

days with sulfuric acid and acetic anhydride, a combination of acetylation and hydrolysis takes place; there is obtained the octaacetate of (+)-cellobiose. Alkaline hydrolysis of the octaacetate yields (+)-cellobiose itself.

Like (+)-maltose, (+)-cellobiose has the molecular formula C12H22O11, is a reducing sugar, forms an osazone, exists in alpha and beta forms that undergo mutarotation, and can be hydrolyzed to two molecules of D-(+)-glucose. The sequence of oxidation, methylation, and hydrolysis (as described for (+)-maltose) shows that (+)-cellobiose contains two pyranose rings and glucoside linkage to an –OH group on C–4.

(+)-Cellobiose differs from (+)-maltose in one respect : it is hydrolyzed by the enzyme emulsin (from bitter almonds), not by maltase. Since emulsin is known to hydrolyze only β-glucoside linkages.

(+)-Lactose : (+)-Lactose makes up about 5% of human milk and

of cow’s milk. It is obtained commercially as a by-product of cheese manufacture, being found in the whey, the aqueous solution that remains after the milk proteins have been coagulated. Milk sours when lactose is converted into lactic acid (sour, like all acids) by bacterial action (e.g., by Lactobacillus bulgaricus).


(+)-Lactose has the molecular formula C12H22O11, is a reducing sugar, forms an osazone, and exists in alpha and beta forms which undergo mutarotation. Acidic hydrolysis or treatment with emulsin (which splits β linkages only) converts (+)-lactose into equal amounts of D-(+)-glucose and D-(+)-galactose. (+)-Lactose is evidently a β-glycoside formed by the union of a molecule of D-(+)glucose and a molecule of D-(+)-galactose.

(+)-Sucrose : (+)-Sucrose is our common table sugar, obtained

from sugar cane and sugar beets. Of organic chemicals, it is the one produced in the largest amount in pure form.

(+)-Sucrose has the molecular formula C12H22O11. It does not reduce Tollen’s or Fehling’s reagent. It is a non-reducing sugar, and in this respect it differs from the other disaccharides we have studied. Moreover, (+)-sucrose does not form an osazone, does not exist in anomeric forms, and does not show mutarotation in solution. All these facts indicate that (+)-sucrose does not contain a “free”aldehyde or ketone group.

(+)-Sucrose is made up of a D-glucose unit and a D-fructose unit; since there is no “free” carbonyl group, if must be both a D-glucoside and a D-fructoside.

Polysaccharides : Polysaccharides are compounds made up of many-

hundreds or even thousands-monosaccharide units per molecule.

Polysaccharides are naturally occurring polymers, which can be considered as derived from aldoses or ketoses by polymerization with loss of water. A polysaccharide derived from hexoses, for example, has the general formula (C6H10O5)n.

The most important polysaccharides are cellulose and starch. Both are produced in plants from carbon dioxide and water by the process of photosynthesis.

Starch : Starch occurs as granules whose size and shape are

characteristic of the plant from which the starch is obtained. When intact, starch granules are insoluble in cold water; if the outer membrane has been broken by grinding, the granules swell in cold water and form a gel.

In general, starch contains about 20% of a water-soluble fraction called amylose, and 80% of a water-insoluble fraction called amylopectin. These two fractions appear to correspond to different carbohydrates of high molecular weight and formula (C6H10O5)n. Upon treatment with acid or under the

influence of enzymes, the components of starch are hydrolyzed progressively to dextrin (a mixture of low-molecular-weight polysaccharides), (+)-maltose, and finally D-(+)-glucose. (A mixture of all these is found in corn sirup, for example.) Both amylose and amylopectin are made up of D-(+)-glucose units, but differ in molecular size and shape.

Cellulose : Cellulose is the chief component of wood and plant

fibers; cotton, for instance, is nearly pure cellulose. It is insoluble in water and tasteless; it is a non-reducing carbohydrate. These properties, in part at least, are due to its extremely high molecular weight.

Cellulose has the formula (C6H10O5)n. Complete hydrolysis by acid yields D-(+)-glucose as the only monosaccharide. Hydrolysis of completely methylated cellulose gives a high yield of 2, 3, 6-tri-O-methyl-D-glucose. Like starch, therefore, cellulose is made up of chains of D-glucose units, each unit joined by a glycoside linkage to C–4 of the next.

Cellulose differs from starch, however, in the configuration of the glycoside linkage. Upon treatment with acetic anhydride and sulfuric acid, cellulose yields octa-O-acetylcellobiose.

Reactions of cellulose : Like any alcohol, cellulose form esters. Treatment

with a mixture of nitric and sulfuric acid converts cellulose into cellulose nitrate. The properties and uses of the product depend upon the extent of nitration.

In the presence of acetic anhydride, acetic acid, and a little sulfuric acid, cellulose is converted into the triacetate. Partial hydrolysis removes some of the acetate groups, degrades the chains to smaller fragments (of 200–300 units each), and yields the vastly important commercial cellulose acetate (roughly a diacetate). Cellulose acetate is less flammable than cellulose nitrate and has replaced the nitrate in many of its applications, in safety-type photographic film, for example. When a solution of cellulose acetate in acetone is forced through the fine holes of a spinnerette, the solvent evaporates and leaves solid filaments. Threads from these filaments make up the material known as acetate rayon.

Industrially, cellulose is alkylated to ethers by action of alkyl chlorides (cheaper than sulfates) in the presence of alkali. Considerable degradation of the long chains is unavoidable in these reactions. Methyl, ethyl, and benzyl ethers of cellulose are important in the production of textiles, films, and various plastic objects.


Identification of acidic radicals For the identification of the acidic radicals, the

following scheme is followed. Group I : The radicals which are analysed by dilute

H2SO4 or dilute HCl. These are (i) carbonate (ii) sulphite, (iii) sulphide, (iv) nitrite, and (v) acetate

Group II : The radicals which are analysed by concentrated H2SO4 . These are (i) chloride, (ii) bromide, (iii) iodide (iv) nitrate, and (v) oxalate

Group III : The radicals which are not analysed by dilute and concentrated H2SO4. These are (i) sulphate, (ii) Phosphate, (iii) borate, and (iv) fluoride.

Group I : Add dilute HCl or H2SO4 to a small amount of

substance and warm gently, observe. 1. Carbonate or CO3

2– : The carbonates are decomposed with the

effervescence of carbon dioxide gas. Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2 ↑ When this gas is passed through lime water, it

turns milky with the formation of calcium carbonate.

Ca(OH)2 + CO2 → CaCO3 + H2O Lime water White ppt.

If the CO2, gas is passed in excess, the milky solution becomes colourless due to the formation of soluble calcium bicarbonate.

CaCO3 + H2O + CO2 → Ca(HCO3)2 White ppt. Soluble Note : Carbonates of bismuth and barium are not easily

decomposed by dilute H2SO4. Dilute HCl should be used.

Sulphur dioxide evolved from sulphites also turns lime water milky.

Ca(OH)2 + SO2 → CaSO3 + H2O White ppt. However SO2 can be identified by its pungent

odour of burning sulphur. PbCO3 reacts with HCl or H2SO4 to give in the

initial stage some effervescence but the reaction slows down due to formation of a protective insoluble layer of PbCl2 or PbSO4 on the surface of remaining salt or mixture.

2. Sulphite : The sulphites give out sulphur dioxide gas having

suffocating smell of burning sulphur. CaSO3 + H2SO4 → CaSO4 + H2O + SO2 ↑ When acidified potassium dichromate paper is

exposed to the gas, it attains green colour due to the formation of chromic sulphate.

K2Cr2O7 + H2SO4 + 3SO2 → K2SO4 + Cr2 (SO4)3 + H2O The sulphite also gives white precipitate with

BaCl2, Soluble in dil. HCl Na2SO3 + BaCl2 → 2 NaCl + BaSO3 ↓ 3. Sulphide, S–2: The sulphide salts form H2S which smells like

rotten eggs. Na2S + H2SO4 → Na2SO4 + H2S ↑ On exposure to this gas, the lead acetate paper

turns black due to the formation of lead sulphide. Pb(CH3COO)2 + H2S → PbS ↓ + 2CH3COOH black ppt.

The sulphides also turn sodium nitroprusside solution violet (use sodium carbonate extract for this test).

Na2S + Na2[FeNO(CN)5] → Na4 [Fe(NOS) (CN)5] Sulphides of lead, calcium, nickel, cobalt,

antimony and stannic are not decomposed with dilute H2SO4. Conc. HCl should be used for their test.However brisk evolution of H2S takes place even by use of dilute H2SO4 if a pinch of zinc dust is added.

Zn + H2SO4 → ZnSO4 + 2H HgS + 2H →Hg + H2S ↑

4. Nitrite, NO2– :

The nitrites yield a colourless nitric oxide gas which in contact with oxygen of the air becomes brown due to the formation of nitrogen dioxide.

2KNO2 + H2SO4 →`K2SO4 + 2HNO2 Nitrous acid 3HNO2 → H2O + 2NO + HNO3 2 NO + O2 → 2NO2 ↑ brown coloured gas

Inorganic Chemistry

Fundamentals

SALT ANALYSIS

KEY CONCEPT


On passing the gas through dilute FeSO4 solution, brown coloured complex salt is formed.

FeSO4.7H2O + NO → [Fe(H2O)5NO].SO4 + 2H2O Brown coloured (panta aquo nitroso ferrous sulphate) When a mixture of iodide and nitrite is treated

with dilute H2SO4, the iodides are decomposed giving violet vapours of iodine, which turns starch iodide paper blue.

2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 2KI + H2SO4 → K2SO4 + 2HI 2HNO2+ 2HI → 2H2O + I2 + 2NO Violet vapours I2 + Starch → Blue colour

5. Acetate : Acetates decompose to give acetic acid vapours

having characteristic smell of vinegar. 2CH3COONa + H2SO4 → 2CH3COOH + Na2SO4 All acetates are soluble in water and their aqueous

solution on addition to neutral FeCl3 solution develops a blood red colour due to the formation of ferric acetate.

FeCl3 + 3CH3COONa → (CH3COO)3Fe + 3NaCl

Blood Red colour Acetates are also decomposed with oxalic acid

and give off acetic acid. 2CH3COONa + H2C2O4 → Na2C2O4 + 2CH3COOH Note : The ferric chloride solution supplied in the

laboratory is always acidic due to hydrolysis. It is made neutral by the addition of dilute solution of NH4OH drop by drop with constant stirring till the precipitate formed does not dissolve. The filtrate is called neutral ferric chloride solution.

Before testing acetate in the aqueous solution by FeCl3, it must be made sure that the solution does not contain

(i) CO3–2, (ii) SO3

–2 (iii) PO4

–3, (iv) I–

Since these also combine with Fe+3. Therefore , the test of acetate should be performed by neutral ferric chloride solution only after the removal of these ions by AgNO3 solution.

Group II: Add concentrated H2SO4 to a small amount of the salt

or mixture and warm gently, observe.

1. Chloride Cl–: Colourless pungent fumes of hydrogen chloride are

evolved. NaCl + H2SO4 → NaHSO4 + HCl ↑ The gas evolved forms white fumes of ammonium

chloride with NH4OH. NH4OH + HCl → NH4Cl + H2O White fumes The gas evolved or solution of chloride salt forms

a curdy precipitate of silver chloride with silver nitrate solution.

AgNO3 + HCl → AgCl ↓ + HNO3 Yellowish : green chlorine gas with suffocating

odour is evolved on addition of MnO2 to the above reaction mixture.

NaCl + H2SO4 –→ NaHSO4 + HCl MnO2 + 4HCl –→ MnCl2 + 2H2O + Cl2 Note : The curdy precipitate of AgCl dissolves in

ammonium hydroxide forming a complex salt. AgCl + 2NH4OH → Ag(NH3)2Cl + 2H2O The solution having the silver complex on

acidifying with dilute nitric acid gives again a white precipitate of silver chloride.

Ag(NH3)2Cl + 2HNO3 → AgCl + 2NH4NO3 Chromyl chloride Test : When solid chloride is

heated with conc. H2SO4 in presence of K2Cr2O7, deep red vapours of chromyl chloride are evolved.

NaCl + H2SO4 → NaHSO4 + HCl K2Cr2O7 + 2H2SO4 → 2KHSO4 + 2CrO3 + H2O CrO3 + 2HCl → CrO2Cl2 + H2O Chromyl chloride These vapours on passing through NaOH

solution, form the yellow solution due to the formation of sodium chromate.

CrO2Cl2 + 4NaOH → Na2CrO4 +2NaCl+ 2H2O Yellow colour The yellow solution neutralised with acetic acid

gives a yellow precipitate of lead chromate with lead acetate.

Na2CrO4 + Pb(CH3COO)2 –→ PbCrO4 + 2CH3COONa Yellow ppt. Note : This test is not given by the chloride of mercuric,

tin, silver, lead and antimony. The chromyl chloride test is always to be

performed in a dry test tube otherwise the


chromyl chloride vapours will be hydrolysed in the test tube.

CrO2Cl2 + 2H2O →H2CrO4 + 2HCl Bromides and iodides do not give this test. 2. Bromide, Br– : Reddish- brown fumes of bromine are formed. NaBr + H2SO4 → NaHSO4 + HBr 2HBr + H2SO4 → Br2 + 2H2O + SO2 More reddish brown fumes of bromine are

evolved when MnO2 is added. 2NaBr + MnO2 + 3H2SO4 → 2NaHSO4 + MnSO4 + 2H2O + Br2 The aqueous solution of bromide or sodium

carbonate extract gives pale yellow precipitate of silver bromide which partly dissolves in excess of NH4OH forming a soluble complex.

NaBr + AgNO3 → AgBr ↓ + NaNO3 Pale yellow ppt. AgBr +2NH4OH → Ag(NH3)2Br + 2H2O 3. Iodide, I– : Violet vapours of iodine are evolved. 2KI + H2SO4 → 2KHSO4 + 2HI 2 HI + H2SO4 → I2 + SO2 + 2H2O Violet vapours with starch produce blue colour. I2 + Starch → Blue colour More violet vapours are evolved when MnO2 is

added. 2KI + MnO2 + 3H2SO4 –→ 2KHSO4 + MnSO4 + 2H2O + I2

Aqueous solution of the iodide or sodium carbonate extract gives yellow precipitate of AgI with silver nitrate solution which does not dissolve in NH4OH.

NaI + AgNO3 → AgI + NaNO3 Yellow ppt.

Note : Sodium carbonate extract of bromide and iodide

on addition of CHCl3 and chlorine water gives brown or violet layer to CHCl3 respectively.

2NaBr + Cl2 → 2NaCl + Br2 ; Br2 + CHCl3 → Brown 2NaI + Cl2 → 2NaCl + I2 ; I2 + CHCl3 → Violet Excess of chlorine water should be avoided as the

layer may become colour less due to conversion of Br2 into HBrO and I2 into HIO3.

Br2 + 2H2O + Cl2 → 2HBrO + 2HCl

I2 + 5Cl2 + 6H2O → 2HIO3 + 10 HCl

4. Nitrate, NO3– :

Light brown fumes of nitrogen dioxide are evolved.

NaNO3 + H2SO4 → NaHSO4 + HNO3 4 HNO3 → 2H2O + 4 NO2 + O2 These fumes intensify when copper turnings are

added. Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O Ring Test : An aqueous solution of salt is mixed

with freshly prepared FeSO4 solution and conc. H2SO4 is poured in test tube from sides, a brown ring is formed on account of the formation of a complex at the junction of two liquids.

NaNO3 + H2SO4 → NaHSO4 + HNO3 6 FeSO4 + 2HNO3 + 3H2SO4 → 3Fe2 (SO4)3 + 4H2O + 2NO [Fe(H2O)6]SO4. H2O + NO → Ferrous sulphate [Fe(H2O)5 NO]SO4 + 2H2O Brown ring

The nitrates can also be tested by boiling nitrate with Zn or Al in presence of concentrated NaOH solution when ammonia is evolved which can be detected by the characteristics odour.

Zn + 2NaOH → Na2ZnO2 + 2H Al + NaOH + H2O → NaAlO2 + 3H NaNO3 + 8H → NaOH + 2H2O + NH3 Note : Ring test is not reliable in presence of

nitrite, bromide and iodide. 5. Oxalate, C2O4

–2 : A mixture of CO and CO2 is given off. The CO

burns with blue flame. Na2C2O4 + H2SO4 → Na2SO4 + H2C2O4 H2C2O4 + [H2SO4] → CO + CO2 + H2O + [H2SO4] 2CO + O2 → 2CO2

A solution of oxalates give the white precipitate with CaCl2 solution. This precipitate get dissolved in dil. H2SO4 and decolourises KMnO4 (acidified) solution.

Na2C2O4 + CaCl2 → CaC2O4 ↓ + 2NaCl CaC2O4 + H2SO4 → CaSO4 + H2C2O4 5H2C2O4 + 2KMnO4 + 3H2SO4 → 2 MnSO4 + K2SO4+ 8 H2O + 10CO2

Group III : 1. Sulphate ,SO4

–2 : Add conc. HNO3 to a small amount of substance or

take sodium carbonate extract and then add BaCl2 solution. A white precipitate of BaSO4 insoluble in conc. acid is obtained.


Na2SO4 + BaCl2 → 2NaCl + BaSO4 White ppt. Note : Silver and lead if present, may be precipitated

as silver chloride and lead chloride by the addition of barium chloride. To avoid it, barium nitrate may be used in place of barium chloride.

2. Borate : To a small quantity of the substance (salt or mixture),

add few ml. of ethyl alcohol and conc. H2SO4. Stir the contents with a glass rod. Heat the test tube and bring the mouth of the test tube near the flame. The formation of green edged flame indicates the presence of borate.

2Na3BO3 + 3H2SO4 → 3Na2SO4 + 2H3BO3 H3BO3 + 3C2H5OH → (C2H5)3BO3 + 3H2O Ethyl borate

3. Phosphate : Add conc. HNO3 to a small amount of substance or

take sodium carbonate extract, heat and then add ammonium molybdate. A canary yellow precipitate of ammonium phospho molybdate is formed.

Ca3(PO4)2 + 6HNO3 → 3Ca (NO3)2 + 2H3PO4 H3PO4 + 12 (NH4)2 MoO4 + 21 HNO3 → (NH4)3 PO4. 12 MoO3 + 21 NH4NO3 + 12 H2O Canary yellow ppt. Note : Arsenic also yields a yellow precipitate of

(NH4)3. AsO4.12 MoO3 (Ammonium arseno molybdate).Thus in presence of As, phosphate is tested in the filtrate of second group.

The precipitate of ammonium phosphomolybdate dissolves in excess of phosphate. Thus, the reagent (ammonium molybdate) should always be added in excess.

HCl interferes in this test. Hence, before the test of phosphate is to be performed, the solution should be boiled to remove HCl.

Reducing agents such as sulphites, sulphides, etc., interfere as they reduce Mo+6 to molybdenum blue (Mo3O8.xH2O). The solution, therefore, turns blue. In such cases, the solutions should be boiled with HNO3 so as to oxidise them before the addition of ammonium molybdate.

4. Fluoride : Take small amount of the substance in dry test tube

and add an equal amount of sand and conc. H2SO4.Heat the contents and place a glass rod moistened with water over the mouth of the test tube. A gelatinous waxy white deposit on the rod is formed.

2NaF + H2SO4 → Na2SO4 + H2F2 SiO2 + 2H2F2 → SiF4 + 2H2O

3SiF4 + 4H2O → H4SiO4 + 2H2SiF6 Silicic acid (white)

Note : The test should be performed in perfectly dry test

tube otherwise waxy white deposit will not be formed on the rod.

HgCl2 and NH4Cl also give white deposits under these conditions, but these are crystalline in nature.

Sodium carbonate extract : One part of the given substance is mixed with

about 3 parts of sodium carbonate and nearly 10 to 15 ml. of distilled water. The contents are then heated for 10-15 minutes and filtered. The filtrate is known as sodium carbonate extract or soda extract and this contains soluble sodium salts due to exchange of partners in between sodium carbonate and salts.

CaCl2 + Na2CO3 → CaCO3 + 2NaCl Insoluble Sodium chloride (soluble) PbSO4 + Na2CO3 → PbCO3 + Na2SO4 Insoluble Sodium sulphate (Soluble) BaCl2 + Na2CO3 → BaCO3 + 2 NaCl Insoluble Sodium chloride (Soluble)

The carbonates of the cations of the mixtures are mostly insoluble in water and are obtained in the residue. On the other hand, sodium salts of the anions (acidic radicals) of the mixture being soluble in water are obtained in the filtrate.

The sodium carbonate extract is basic in nature and before it is used for the analysis of a particular acidic radical, it is first neutralised by the addition of small quantity of an appropriate acid. The acid is added to the extract till the effervescence cease to evolve.

Advantages of preparing sodium carbonate extract-

The preparation of sodium carbonate extract affords a convenient method for bringing the anions of the mixture into solution which were otherwise insoluble with cation of salt.

It removes the basic radicals (usually coloured) which interferes in the usual tests of some of the acidic radicals.

The residue can be used for the tests of basic radicals of I to VI groups. Such a solution does not involve the problem of removing interfering radicals like oxalate, fluoride, borate and phosphate.


1. 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

55.744.9 = 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 = .wtEmpirical.wtMolecular =

53106 = 2

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

]O[6

oxidationVigrous

)A(108HC →

HOOC HOOC

C6H4

(B)

+ 2H2O

Meq. of dicarboxylic acid = Meq. of NaOH

E10001.0 × = 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.

COOH

COOH

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

6[O]COOH

COOH+ 2H2O

CH3

CH3m-xylene

6[O]COOH

COOH

+ 2H2O

CH3

CH3

6[O]

COOH

COOH

+ 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.

COOHCOOH

∆ CO–H2O CO O

One acid which on nitration gives a mono nitro

compounds is p-dicarboxylic acid.

COOHHNO3

COOH

∆; H2SO4

COOH

NO2

COOH One acid which on nitration gives three mono nitro

compounds will be the m-isomer.

COOHHNO3

COOH

H2SO4 COOHNO2

COOH

COOH

COOHNO2

COOH

COOHNO2

, ,

2. An organic compound (A) contains 69.42% C, 5.78%

H and 11.57% N. Its vapour density is 60.5. It evolves NH3 when boiled with KOH. On heating with P2O5, it gives a compound (B) having C = 81.55%, H = 4.85% and N= 13.59%. On reduction

UNDERSTANDINGOrganic Chemistry


with Na + C2H5OH (B) gives a base, which reacts with HNO2 giving off N2 and yielding an alcohol (C). The alcohol can be oxidised to benzoic acid. Explain the above reactions and assign structural formulae to (A), (B) and (C)

Sol. (i) Calculation of empirical formula of (A) : Element At.

wt. % Relative no. of

atoms Simplest

ratio C 12 69.42

1242.69 = 5.785

826.0785.5 = 7

H 1 5.78 178.5 = 5.78

826.078.5 = 7

N 14 11.57 14

57.11 = 0.826 826.0826.0 = 1

O 16 13.23 16

23.13 = 1.827 826.0827.0 = 1

Hence, empirical formula of (A) = C7H7NO Empirical formula wt. = 84 + 7 + 14 + 16 = 121 (ii) Calculation of molecular weight of (A) : Molecular weight = 2 × V.D. = 2 × 60.5 = 121 (iii) Determination of molecular formula of (A):

n = .wtEmpirical.wtMolecular =

121121 = 1

Hence, molecular formula = empirical formula i.e., C7H7NO (iv) Calculate of empirical formula of (B) : Element At.

wt. % Relative no. of

atoms Simplest

ratio C 12 81.55

1255.81 = 6.80

97.080.6 = 7

H 1 4.85 185.4 = 4.85

97.085.4 = 5

N 14 13.59 14

59.13 = 0.97 97.097.0 = 1

Hence, empirical formula of (B) = C7H5N (v) Determination of structural formulae : (a) Since compound (A) on heating with KOH gives

NH3, a characteristic test of amide, hence the compound (A) is an amide (–CONH2).

(b) Since compound (B) is obtained by heating (A) with P2O5, a dehydrating agent.

OHNHCONHC 2)B(57

OP

)A(77

52 + →

The above reaction confirms that (A) is an amide, and the remaining reaction are :

C7H5N

COOH

Alcohol

[H] HNO2 N2 + (C) [O]

(B)

The formula of benzoic acid indicates that the compound (A) is an aromatic amide. Hence, the reactions are :

CONH2

Benzamide

KOH + NH3 ↑Boil

(A)

COOK

CONH2

Benzonitrile

P2O5 + H2O∆

(A)

C≡N

(B)

CH2NH2

Benzyl amine

Na +C2H5OH+ 4[H]

(Base)

C≡N

(B)

CH2NH2

Benzyl alcohol

HNO2 + N2 + H2O

CH2OH

(C)

CH2OH

Benzoic acid

2[O]– H2O

(C)

COOH

3.

)ClHC(CB)HC(A

136

HCl126 + →

B → KOH.alc D isomer of A D →Ozonolysis E (it gives negative test with Fehling solution but responds to iodoform test) A →Ozonolysis F + G Both gives positive Tollen's test but do not give iodoform test. F + G → NaOH.Conc HCOONa + primary alcohol Identify to A to G Sol.

)ClHC(CB)HC(A

136

HCl126 + →

B →Ozonolysis E (it gives negative test with Fehling solution but responds to iodoform test) A →Ozonolysis F + G Both gives positive Tollen's test but do not give iodoform test. F + G → NaOH.Conc HCOONa + primary alcohol Both F and G are aldehydes because they give

positive Tollen's test and do not give iodoform test. These aldehydes give Cross Cannizzaro's reaction. So they do not have α-hydrogen atoms. In cross Cannizzaro's reaction HCOONa is formed along with p-alcohols. So in these an aldehyde is HCHO and


another is (CH3)3C.CHO. F and G are obtained by ozonolysis of A. Therefore compound 'A' is

CH2 = CH – C(CH3)3. Structure of compound 'A' is

CH3 CH3 – C – CH = CH2 CH3

Compound 'A' on reaction with HCl gives comp. B and C which have molecular formula C6H13Cl. Thus,

CH3 – C – CH = CH2 → CH3 – C — CH – CH3

CH3

CH3

HCl

CH3

CH3 Cl Comp. 'B'

CH3 – C – CH2 – CH2Cl

CH3

CH3 Comp. 'C'

+

Compound 'B' gives 'D' on dehydrohalogenation with

alc. KOH.

CH3 – C — CH – CH3

CH3

CH3

alc. KOH

CH3

CH3Cl

Sec. carbonium ion

CH3 – C – CH

CH3

Compound 'D'

CH3 – C — CH – CH3 Boil (–Cl–)

+

+

CH3

CH3

H+

C = C CH3

CH3

CH3

CH3

Compound 'D' on ozonolysis to give compound 'E'

Compound 'E'

C = CCH3

CH3

CH3

CH3

Ozonolysis 2CH3 – C – CH3

O

Compound 'E' has methyl ketonic groups (–COCH3)

so it gives positive iodoform test and does not give the test with Fehling solution due to absence of –CHO group.

Compound 'A' on ozonolysis to give compounds F and G as follows :

(CH3)3CCH = CH2 → Ozonolysis

Comp. 'G'

(CH3)3C – CHO + CH2O'F'

Compound G and F gives crossed Cannizzaro's

reaction with conc. NaOH solution.

HCOONa + CH3 – C – CH2OH

Ozonolysis

CH3

CH3 Comp. 'G'

CH3 – C — CHO + CH2O + conc. NaOH →'F'

CH3

CH3

Hence,

Compound 'A' =

CH3 – C – CH = CH2 (C6H12)

CH3

CH3

Compound 'B' =

CH3 – C – CH – CH3

CH3

CH3 Cl

Compound 'C' =

CH3 – C – CH2 – CH2Cl

CH3

CH3

Compound 'D' =

C = CCH3

CH3

CH3

CH3

Compound 'E' =

CH3 – C – CH3

O

Compound 'F' =

H – C – H

O

Compound 'G' =

CH3 – C – CH2OH

CH3

CH3

4. A hydrocarbon (A) of the formula C8H10, on

ozonolysis gives compound (B), C4H6O2, only. The compound (B) can also be obtained from the alkyl bromide, (C) (C3H4Br) upon treatment with Mg in dry ether, followed by treatment with CO2 and acidification. Identify (A), (B) and (C) and also equations for the reactions.

Sol. A(C8H10) OH)ii(

O)i(

2

3 →)B(

264 OHC

Since compound (A) adds one mol of O3, hence it should have either C = C or a – C ≡ C – bond. If it was alkene its formula should be C8H16 (CnH2n), and if it was alkyne it should have the formula C8H14; it means it is neither a simple alkenen or simple alkyne. However it is definite that the compound has an unsaturated group, it appears that it is a cyclosubstituted ethyne.

H – C ≡ C – H 106HC

H2

+

− → C3H5 – C ≡ C – C3H5

the C3H5 – corresponds to cyclopropyl (∆) radical, hence compund (A) is

CH – C≡C – CH

CH2

CH2

CH2

CH2 1,2-dicyclopropyl ethyne

The ozonolysis of above compound would give two moles of cyclopropane carboxylic acid (C4H6O2).


CH – C≡C – CH

CH2

CH2

CH2

CH2 (A)

(i) O3

CH – C — C – CH CH2

CH2

CH2

CH2(A)

H2O

O

O O

Warm

CH – C – C – CH

CH2

CH2

CH2

CH2O O

(B)

+ H2O2

CH2

CH2

CH – COOH 2

Compound (B) is prepared from cyclopropyl bromide

as follows :

CH – Br CH2

CH2

CH2

CH2

Cyclopropyl magnesium bromide

CH . MgBr Mg

etherC=O∆

O

CH .COOMgBr

CH2

CH2

CH2

CH2

Addition compound

CH–COOH HOH

dil. HCl; –MgBrOH

5. 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 decolourises bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to gives (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)

Alc. KOH∆; –KCl

C4H8 (C)

(Alkene) Aq.KOH ∆; –KCl ∆; –H2O

Cu C4H9OH (B)

(Alcohol)

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.

(A) =

3

33

CH|

CHCCH|Cl

−− and (B) =

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 +

++

→H

Hg

OH|

CHCCH 23 =−

→

)D(O||

CHCCH 33 −−

Hence (D) is acetone and (E) is formaldehyde. Therefore, alkene (C) is 2-methyl propene. (CH3)2–C=CH2

(D) reacts with hydroxyl amine (NH2OH) to form oxime (F).

CH3

CH3C = O + H2 NOH

–H2O CH3

CH3C = NOH

(D) (F)

(B) =

3

33

CH|

CHCCH|

OH

−− and (A) =

3

33

CH|

CHCCH|Cl

−−

Reactions :

)A(3

33

CH|

CHCCH|Cl

−−KCl;–

KOH.Aq

∆ →

)B(3

33

CH|

CHCCH|OH

−−

OH

Cº300/Cu

2− →

)C(3

223

CH|

OHCHCCH +=−

OH;KCl

/KOH.Alc

2−−

∆ →

)C(3

23

CH|

CHCCH =−

CH3 – C = CH2(I) O3

(II) H2O/ZnC = O + H – C – H

CH3

CH3

CH3

O

(C)(D)

(E)

C = O + H2NOH ∆ –H2O

C = NOHCH3

CH3 (D)(F)

CH3

CH3

)E(HCHO2 + NaOH →

)G(3OHCH +

)H(HCOONa

CH3 – C ≡ CH + H2O +

++

→H

Hg

CH3 – C – CH3

O

(D)


1. Given that φ (x) = )ca()bx()cx()bx(

−−−− f (a) +

)ab()cb()ax()cx(

−−−− f (b)+

)bc()ac()bx()ax(

−−−− f (c) - f (x) where

a < c < b and f ′′(x) exists at all points in (a, b) .

Prove that there exists a number µ,

a < µ < b , such that

)ca()ba(

)a(f−−

+ )ab()cb(

)b(f−−

+ )bc()ac(

)c(f−−

= 21 f ′′(µ).

2. An unbiased die is tossed until it lands the same way

up twice running. Find the probability that it requires

r tosses.

3. Given the base of a triangle and the sum of its sides

prove that the locus of the centre of its incircle is an

ellipse.

4. Let f (x) = ax2 + bx + c & g (x) = cx2 + bx + a, such

that | f (0) | ≤ 1, | f (1) | ≤ 1 and |f (-1) | ≤ 1 , prove that | f (x) | ≤ 5/4 and | g (x) | ≤ 2.

5. In order to find the dip of an oil bed below the

surface of the ground, vertical borings are made from the angular points, A, B, C of a triangle ABC which is in horizontal plane. The depth of the bed at these points are found to be x, x + y and x + z respectively.

Show that the dip θ (angle with horizontal) of the oil bed which is assumed to be a plane is given by tan θ .

sin A = Acosbcyz2

bz

cy

2

2

2

2−+ where b and c are the

lengths of the sides CA and AB respectively and A is

the angle between CA and AB.

6. Evaluate : ∫ +−

x5cos21x7cosx8cos

7. Let f (x) be an even function such that f ′ (x) is

continuous, find y for which 2

2

dxyd = ∫

−

x

x

dt)t(f

8. Prove the inequality (aα + bα)1/α < (aβ + bβ)1/β,

for a > 0, b > 0 & α > β > 0.

9. A circle of radius 1 rolls (without sliding) along

the x-axis so that its centre is of the form (t, 1)

with t increasing. A certain point P touches the

x-axis at the origin as the circle rolls. As the circle

rolls further, the point P passes through the point

(x, 1/2). Find x, when it passes through (x, 1/2)

first time.

10. Find all positive integers n for which

1n − + 1n + is rational.

`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 MaheshwariJoint Director Academics, Career Point, KotaSo lu tions will be p ublished in next issue

9Set


1.

A

B

C

d

a

db

c Plane through mid pt of AB, ⊥ to CD is

( rr

– 1/2 ( ar + b

r)).( c

r – dr

) = 0; Let centroid

4

dcbarrrr

+++ = 0 at origin

( rr

+ 1/2 ( cr + d

r)). ( c

r – dr

) = 0

| rr

+ cr |2 = | r

r + d

r|2

it is the locus of pt. equidistance from – cr

& – dr

similarly.

| rr

+ cr

|2 = | rr

+ dr

|2 = | rr

+ ar

|2 = | rr

+ br

|2

so the pt. is equidistant from – ar , – b

r, – c

r , – dr

(i.e. circumcentre of tetrahedron ar

, – br

, – cr

, – dr

)

2. As the function is symmetrical about x = a & x = b

lines

so f (a + x) = f (a − x) ................(1) &

f (b + x ) = f (b − x) ................(2)

As it is defined for x ∈ R.

Let x = b − a − t in (1)

f (b − t) = f (2a − b + t)

use (2) in it

f (b + t) = f (2a − 2b + b + t)

so the function is periodic & its possible period

may be |2a − 2b| = 2b - 2a (as b > a).

3. 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 ≤ s 3

3)cs()bs()as(

−+−+−

A2 ≤ s3

3s2s3

− =

3

4

3s ⇒ A ≤

33s2

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 = 312

p2 ; for a = b = c

Now again p ≥ A312

pmin = A312 , and

again equality holds if a = b = c.

4. ac4b2 − ≤ | b | 2bac41− ≤ | b | 2b

ac41+

≤ | b |

+ 2b

ac21

so ac4b2 − ≤ | b | + bac2

so that a2

ac4ba2b 2 −

±− ≤ a2

b + a2

b + bc

= ab +

bc

MATHEMATICAL CHALLENGES SOLUTION FOR NOVEMBER ISSUE (SET # 8)


Hence the solutions of az2 + bz + c = 0 satisfy

condition | z | ≤ + ab +

bc .

5. P (a cos θ, b sin θ)

Equation of AC ⇒ ax cos θ +

by sin θ = 1

A

B CD

P(θ)

Point A : (0 , b cosec θ)

Equation of BC ⇒ y = −b

Point C = ax cos θ − sin θ = 1

x = θθ+

cosa)sin1(

Point C

−

θθ+ b,

cos)sin1(a

Area A = 21 AD . BC = AD . DC

=

θ+

sinbb .

θθ+

cos)sin1(a

= θθ

θ+cossin

)sin1(ab 2

A = ab θθ

θ+cossin

)sin1( 2

θd

dA =

ab . θθ

θ−θθ+−θθθ+22

2222

cossin)sin(cos)sin1(sincos)sin1(2

= θθ

θ+22 cossin

)sin1(ab [2cos2θ sinθ − (1 + sinθ) cos2θ

+ (1 + sinθ) sin3θ)]

= θθ

θ+22 cossin

)sin1(ab

[– cos2θ + sin2θ + sin3θ + cos2θ sinθ]

for max / min . θd

dA = 0

sinθ (cos2θ + sin2θ) + sin2θ − cos2θ = 0

sinθ + sin2θ − (1 − sin2θ) = 0

⇒ 2sin2θ + sinθ – 1 = 0

(2 sinθ − 1) (sin θ + 1) = 0

as sin θ ≠ −1

sin θ = 1/2 ; θ = π/6

when θ > π/6 ; θd

dA > 0

when θ < π/6 ; θd

dA < 0

so θ = 6π ; is the pt of min.

min. area.

Amin = θθ

θ+cossin

)sin1(ab 2

= 2/3.2/1)2/11(.ab 2+ =

23.

21.4

9ab ×

= 3 3 ab sq. units. 6. ax2 + 2hxy + by2 = 0; (y – M1x) (y – M2x) = 0

where M1 + M2 = – bh2 & M1M2 =

ba

Now as given the second pair must be given by

(y – M1x)(M2y + x) = 0

M2y2 + (1 – M1M2)xy – M1x2 = 0

Compare it with a´x2 + 2h´xy + bý2 = 0

´b

M2 = ´h2MM1 21− =

áM1−

so ´b

M2 = –á

M1− = á´b

MM 21

−+ =

´)a´b(bh2−

−

= ´h2MM1 21− =

´h2b/a1−


M2 = –´)ab(b

´hb2−

& M1 = – b´h2

á)ab( −

Since M1M2 = ba so

´)a´b(b´hb2

−.

b´h2á)ab( − =

ba

Thus áb´b´ha

− =

abab´h−

7. LHS = coeff. of xn in [nC0(1 + x)m + nC1(1 + x)m+1 +

.... + nCn(1 + x)m + n]

= coeff. of xn in (1 + x)m[nC0 + nC1(1 + x) + ..... + nC0

(1 + x)n]

= coeff. of xn in (1 + x)m(2 + x)n

= coeff of xn in (1 + x)m ∑=

−n

0r

rrnr

n 2.xC

= nC0 . mC0 + nC1 .mC1 . 2 + nC2 . mC2 . 22 + .... + nCnmCn . 2n

8. In = ∫−

−1

1

n2 )x1( cos mx dx

= 1

1

n2

mmxsin)x1(

−

− + ∫

−

−−1

1

1n2 )x1(xmn2 sin mx dx

= 0 + mn2

( )

−+−−+

−− ∫

−

−−

−

−1

1

1n22n221

1

1n2 dxmxcos)x1()x1(x)1n(2(m1

nmxcos

x1(x

= 2mn2

∫−

−−1

1

2n2 )x1( [ ]22 x1x)2n2( −++− cosmx dx

= 2mn2

∫−

−−1

1

2n2 )x1( [ ]1x)1n2( 2 ++− cos mx dx

= 2mn2

−−−−− ∫∫

−

−

−

−1

1

2n21

1

1n2 dxmxcos)x1()2n2(dxmxcos)x1()1n2(

m2In = (2n (2n − 1) In–1 − 4n (n − 1) In–2.

Hence proved.

9. tn = 2+

+)1n(.n

1n22

= 2n1 − 2)1n(

1+

Sn = 1 − 221 + 22

1 – 231 + 23

1− 24

1 .................

Sn = 1 − 2)1n(1+

Required sum = ∞→n

Lim Sn = 1.

10. Let the given circle be x2 + y2 = r2 & parametric

angles of A, B, C are respectively θ1 , θ2 & θ3. Let the slopes of the given two lines are m1 & m2. Sides AB & BC are parallel to these lines.

A(θ1)

C(θ3)B(θ2)

Equation of AB;

x cos 2

21 θ+θ + y sin 2

21 θ+θ = r cos 2

21 θ−θ

so m1 = – cot 2

21 θ+θ = θ1 + θ2 = α

similarly : m2 = − cot 2

32 θ+θ =θ2 + θ3 = β

Here α, β are constants as m1 & m2 are constants.

Now equation of AC ;

x cos

θ+θ2

31 + y sin 2

31 θ+θ = r cos 2

31 θ−θ

x cos

θ+θ2

31 + y sin 2

31 θ+θ= rk

where k = cos 2

β−α (i . e. constant)

so foot of the perpendicular from centre of given

circle on AC

θ+θθ+θ2

kr,2

coskr 3131 is

which lies on x2 + y2 = (rk)2.


1. There are two die A and B both having six faces. Die A has 3 faces marked with 1, 2 faces marked with 2 and 1 face marked with 3. Die B has 1 face marked with 1, 2 faces marked with 2 and 3 faces marked with 3. Both dice are thrown randomly once. If E is the event of getting sum of the numbers appearing on top faces equal to x and let P(E) be the probability of event E, then

(i) find x, when P(E) is maxm.

(ii) find x, when P(E) is minm

Sol. X can be 2, 3, 4, 5, 6.

The no of ways in which sum 2, 3, 4, 5, 6.

can occur are the coefficients of x2, x3, x4, x5, x6 is (3x + 2x2 + x3) (x + 2x2 + 3x3)

= 3x2 + 8x3 + 14x4 + 8x5 + 3x6

This shows that sum that occurs most often is 4, and sum that occurs minimum times is 2 or 6.

2. Six points (xi, yi), i = 1, 2, 3, .... 6 are taken on the

circle x2 + y2 = 4 such that ∑=

6

1iix = 8 and ∑

=

6

1iiy = 4.

The line segment joining orthocentre of a ∆ made by any three points and the contrioid of the ∆ made by other three points passes through a fixed points (h, k), then find h + k.

Sol. Let ∑=

6

1iix = α and ∑

=

6

1iiy = β

let 0 be orthocentre of ∆ made by (x1, y1), (x2, y2) and (x3, y3)

⇒ 0 is (x1 + x2 + x3, y1 + y2 + y3) = (αi , β1)

similarly let G be the centroid of the ∆ made by other three points.

G is

++++3

yyy,

3xxx 654654

G is

β−βα−α

3,

311

The point dividing OG. in the ratio 3 : 1 is

βα

4,

4 ≡ (2,1) ⇒ h + k = 3

3. Suppose a function f(x) satisfies the following

conditions f(x + y) = )y(f).x(f1)y(f)x(f

++ ∀ x, y and

f´(0) = 1. Also – 1 < f(x) < 1 for all x ∈R, then find the set of values of x where f(x) is differentiable and also find the value of

∞→xlim [f(x)]x.

Sol. First put x = 0, y = 0 ⇒ f(0) = 0

Now, f´(x) = h

)x(f)hx(flim0x

−+→

= h

)x(f)h(f).x(f1)h(f)x(f

lim0x

−+

+

→

=

+−−

−−

→ )h(f).x(f1)x(f1

0h)0(f)h(flim

2

0x

= 1 – f2(x)

integrating we get 21 ln

−+

)x(f1)x(f1 = x + c

⇒ f(x) = xx

xx

eeee

−

−

+−

clearly f(x) is differentiable for all x ∈R.

x

x)]x(f[lim

∞→ =

x

xx

xx

x eeeelim

+

−−

−

∞→

=

+

−−

−

∞→ xx

xx

x eeee

lim

e x = 1

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Students' Forum

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4. Let z1, z2, z3 be the complex nos. represent the vertices of the ∆ABC which is circumscribed by the circle |z| = 1. Altitude from A meets the side BC at D and circum circle at E. Let P be the image of E about BC, then find

(i) the complex no. of point P.

(ii) the complex no. of point E.

A(z1)

B (z2)

C(z3)

E(z4)

D

OP

Sol. (i) we know that the image of orthocentre about any

side of the ∆ lies on the circum circle of ∆.

Point P = z1 + z2 + z3

(ii) Let O (origin) be the circum centre of ∆.

∠BOE = π – 2B and ∠AOC = 2B

3

1

zz = ei 2B ...(1)

2

4

zz = ei(π–2B) ...(ii)

3

1

zz .

2

4

zz = –1 ⇒ z4 = –

1

23

zzz

5. If A be the area bounded by y = f(x), y = f–1(x) and

the line 4x + 4y – 5 = 0 where f(x) is a polynomial of 2nd degree passing through the origin and having

maximum value of 41 at x = 1, then find A.

O

B C A

P Q x=1

y=f(x)

y = xy = f–1(x)

Sol. Let f(x) = ax2 + bx

given that 41 = a + b ...(1)

0 = 2a + b ...(2)

from (1) and (2)

a = –41 , b =

21

f(x) = 4

xx2 2−

since 4x + 4y – 5 = 0 passes through A

41,1 and

B

1,

41 so area bounded is OAB = 2OAC

= 2[ar(OCP) + ar(CAQP) – ar(OAQ)]

= 2

−−

++×× ∫

1

0

2dx

4xx2

83

41

85

21

85

85

21

= 9637 (unit)2

6. Let P(x) be a polynomial of degree n such that

P(i) = 1i

i+

for i = 0, 1, 2 ..... n. If n is odd than find

the value of P(n + 1).

Sol. Let Q(x) = (x + 1) P(x) – x

clearly Q(x) is polynomial of degree n + 1. Also

Q(i) = (i + 1) 1i

i+

– i = 0 for i = 1, 2, 3 .....n

Thus we can assume

Q(x) = kx(x – 1) (x – 2) ...... (x – n) where k is a constant.

Now Q(–1) = k(–1)(–2)(–3) ...... (–1 – n)

1 = (–1)n + 1 k(n + 1) !

⇒ k = !)1n(

1+

(Q n is odd)

Thus P(x) =

+

+−−−

+x

!)1x()nx)....(2x)(1x(x

1x1 ,

where n is odd

∴ P(n + 1) = 1


Differential Equation :

An equation involving independent variable x, dependent variable y and the differential coefficients

dxdy ,

2

2

dxyd , .... is called differential equation.

Examples :

(1) dxdy = 1 + x + y

(2) dxdy + xy = cot x

(3) 3

4

4

dxyd

– 4

dxdy + 4y = 5 cos 3x

(4) x22

2

dxyd +

2

dxdy1

+ = 0

Order of a Differential Equation :

The order of a differential equation is the order of the highest derivative occurring in the differential equation. For example, the order of above differential equations are 1, 1, 4 and 2 respectively.

Degree of a Differential Equation :

The degree of the differential equation is the degree of the highest derivative when differential coefficients are free from radical and fraction. For example, the degree of above differential equations are 1, 1, 3 and 2 respectively.

Linear and Non-linear Differential Equation :

A differential equation in which the dependent variable and its differential coefficients occurs only in the first degree and are not multiplied together is called a linear differential equation. The general and nth order differential equation is given below :

a0(x)n

n

dxyd + a1(x)

1n

1n

dxyd

−

− + .... + an – 1

dxdy

+ an(x)y + φ(x) = 0

Those equations which are not linear are called non-linear differential equations.

Formation of Differential Equation :

(1) Write down the given equation.

(2) Differentiate it successively with respect to x that number of times equal to the arbitrary constants.

(3) And hence on eliminating arbitrary constants results a differential equation which involves x, y,

dxdy , 2

2

dxyd .....

Solution of Differential Equation :

A solution of a differential equation is any function which when put into the equation changes it into an identity.

General and particular solution :

The solution which contains a number of arbitrary constant equal to the order of the equation is called general solution by giving particular values to the constants are called particular solutions.

Several Types of Differential Equations and their Solution :

(1) Solution of differential equation

dxdy = f(x) is y = ∫ + cdx)x(f

(2) Solution of differential equation

dxdy = f(x) g(y) is ∫ )y(g

dy = ∫ + cdx)x(f

(3) Solution of diff. equation dxdy = f(ax + by + c) by

putting ax + by + c = v and dxdy =

− a

dxdv

b1

)v(bfa

dv+

= dx

Thus solution is by integrating

∫ + )v(bfadv = ∫dx

DIFFERENTIAL EQUATIONS

Mathematics Fundamentals

MATH


(4) To solve the homogeneous differential equation

dxdy =

)y,x(g)y,x(f , substitute y = vx and so

dxdy = v + x

dxdv .

Thus v + xdxdv = f(v)

⇒ x

dx = v)v(f

dv−

Therefore solution is ∫ xdx = ∫ − v)v(f

dv + c

Equation reducible to homogeneous form :

A differential equation of the form

dxdy =

222

111

cybxacybxa

++++ ,

where 2

1

aa ≠

2

1

bb , can be reduced to homogeneous

form by adopting the following procedure :

Put x = X + h, y = Y + k,

so that dXdY =

dxdy

The equation then transformed to

dXdY =

)ckbha(YbXa)ckbha(YbXa

22222

11111

++++++++

Now choose h and k such that a1h + b1k + c1 = 0 and a2h + b2k + c2 = 0. Then for these values of h and k, the equation becomes

dXdY =

YbXaYbXa

22

11

++

This is a homogeneous equation which can be solved by putting Y = vX and then Y and X should be replaced by y – k and x – h.

Special case :

If dxdy =

cy´bxácbyax

++++ and

áa =

bb = m (say), i.e.

when coefficient of x and y in numerator and denominator are proportional, then the above equation cannot be solved by the discussed before because the values of h and k given by the equations will be indeterminate.

In order to solve such equations, we proceed as explained in the following example.

Solve dxdy =

4y3x7y6x2

+−+− =

4y3x7)y3x(2

+−+−

== 2

´bb

áaobviously

Put x – 3y = v

⇒ 1 – 3dxdy =

dxdv (Now proceed yourself)

Solution of the linear differential equation :

dxdy + Py = Q, where P and Q are either constants or

functions of x, is

∫ dxPye = ∫

∫ dxPQe dx + c

Where ∫ dxPe is called the integrating factor.

Equations reducible to linear form :

Bernoulli's equation : A differential equation of

the form dxdy + Py = Qyn, where P and Q are

functions of x alone is called Bernoulli's equation.

Dividing by yn, we get y–n

dxdy + y–(n – 1). P = Q

Putting y–(n – 1) = Y, so that ny)n1( −

dxdy =

dxdY ,

we get dxdY + (1 – n)P. Y = (1 – n)Q

which is a linear differential equation.

If the given equations is of the form

dxdy + P. f(y) = Q . g(y), where P and Q are

functions of x alone, we divide the equation by g(y) and get

dxdy

)y(g1 + P.

)y(g)y(f = Q

Now substitute )y(g)y(f = v and solve.

Solution of the differential equation :

2

2

dxyd = f(x) is obtained by integrating it with respect

to x twice.


Some Important Definitions and Formulae : Measurement of angles : The angles are measured

in degrees, grades or in radius which are defined as follows:

Degree : A right angle is divided into 90 equal parts and each part is called a degree. Thus a right angle is equal to 90 degrees. One degree is denoted by 1º.

A degree is divided into sixty equal parts is called a minute. One minute is denoted by 1 .

A minute is divided into sixty equal parts and each parts is called a second. One second is denoted by 1´´.

Thus, 1 right angle = 90º (Read as 90 degrees) 1º = 60 (Read as 60 minutes) 1 = 60´ (Read as 60 seconds). Grades : A right angle is divided into 100 equal

parts and each part is called a grade. Thus a right angle is equal to 100 grades. One grade is denoted by 1g.

A grade is divided into 100 equal parts and each part is called a minute and is denoted by 1´.

A minute is divided into 100 equal parts and each part is called a second and is denoted by 1"

Thus, 1 right angled = 100g (Read as 100 grades) 1g = 100 (Read as 100 minutes) 1 = 100´´ (Read as 100 seconds) Radians : A radian is the angle subtended at the

centre of a circle by an arc equal in length to the radius of the circle.

Domain and Range of a Trigono. Function : If f : X → Y is a function, defined on the set X,

then the domain of the function f, written as Domf is the set of all independent variables x, for which the image f(x) is well defined element of Y, called the co-domain of f.

Range of f : X → Y is the set of all images f(x) which belongs to Y, i.e.,

Range f = f(x) ∈ Y : x ∈ X ⊆ Y The domain and range of trigonmetrical functions

are tabulated as follows :

Trigo. Function

Domain Range

sin x R, the set of all the real number

–1 ≤ sin x ≤ 1

cos x R – 1 ≤ cos x ≤ 1

tan x R –

∈

π+ In,

2)1n2( R

cosec x R – n π, n ∈ I R – x : –1 < x < 1

sec x R –

∈

π+ In,

2)1n2( R – x : –1 < x < 1

cot x R – n π, n ∈ I R

Relation between Trigonometrically Ratios and identities:

tan θ = θθ

cossin ; cot θ =

θθ

sincos

sin A cosec A = tan A cot A = cos A sec A = 1 sin2θ + cos2θ = 1 or sin2θ = 1 – cos2θ or cos2θ = 1 – sin2θ 1 + tan2θ = sec2θ or sec2θ – tan2θ = 1 or sec2θ – 1 = tan2θ 1 + cot2θ = cosec2θ or cosec2θ – cot2θ = 1 or cosec2θ – 1 = cot2θ Since sin2A + cos2A = 1, hence each of sin A

and cos A is numerically less than or equal to unity. i.e.

| sin A| ≤ 1 and | cos A | ≤ 1 or –1 ≤ sin A ≤ 1 and – 1 ≤ cos A ≤ 1 Note : The modulus of real number x is defined as

|x| = x if x ≥ 0 and |x| = – x if x < 0. Since sec A and cosec A are respectively

reciprocals of cos A and sin A, therefore the values of sec A and cosec A are always numerically greater than or equal to unity i.e.

sec A ≥ 1 or sec A ≤ – 1 and cosec A ≥ 1 or cosec A ≤ – 1 In other words, we never have –1 < cosec A < 1 and –1 < sec A < 1.

TRIGONOMETRICAL RATIOS

Mathematics Fundamentals

MATH

XtraEdge for IIT-JEE JANUARY 2010 57

Trigonometrical Ratios for Various Angles :

θ 0 6π

4π

3π

2π π

23π 2π

sin θ 0 21

21

23 1 0 –1 0

cos θ 1 23

21

21 0 –1 0 1

tan θ 0 3

1 1 3 ∞ 0 ∞ 0

Trigonometrical Ratios for Related Angles :

θ – θ 2π ± θ

π ± θ 2

3π ± θ 2π ± θ

sin – sin θ cos θ m sin θ – cos θ ± sin θ

cos cos θ m sin θ – cos θ ± sin θ cos θ

tan – tan θ m cot θ ± tan θ m cot θ ± tan θ

cot – cot θ m tan θ ± cot θ m tan θ ± cot θ

Addition and Subtraction Formulae : sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B m sin A sin B

tan (A ± B) = BtanAtan1BtanAtan

m

±

cot (A ± B) = AcotBcot

1BcotAcot±

m

sin (A + B) sin (A – B) = sin2A – sin2B = cos2B – cos2A cos (A + B) cos (A – B) = cos2A – sin2B = cos2B – sin2A Formulae for Changing the Sum or Difference into Product :

sin C + sin D = 2 sin 2

DC + cos 2

DC −

sin C – sin D = 2 cos 2

DC + sin 2

DC −

cos C + cos D = 2 cos 2

DC + cos 2

DC −

cos C – cos D = 2 sin 2

DC + sin 2

CD −

Formulae for Changing the Product into Sum or Difference : 2 sin A cos B = sin (A + B) + sin (A – B) 2 cos A sin B = sin (A + B) – sin (A – B) 2 cos A cos B = cos (A + B) + cos (A – B) 2 sin A sin B = cos (A – B) – cos(A + B)

Formulae Involving Double, Triple and Half Angles :

sin 2θ = 2 sin θ cos θ = θ+

θ2tan1

tan2

cos 2θ = cos2 θ – sin2 θ = 2 cos2θ – 1

= 1 – 2 sin2θ = θ+θ−

2

2

tan1tan1

sin 2θ = ±

2cos1 θ− ; cos

2θ = ±

2cos1 θ+

tan 2θ = ±

θ+θ−

cos1cos1

tan 2θ = θ−

θ2tan1

tan2

sin 3θ = 3 sin θ – 4 sin3θ

or sin3θ = 41 (3 sin θ – sin 3θ)

cos 3θ = 4 cos3θ – 3 cos θ

or cos3θ = 41 (3 cos θ + cos 3θ)

tan 3θ = θ−

θ−θ2

3

tan31tantan3

π

+π≠θ6

n

Trigonometrical Ratios for Some Special Angles :

θ 21º7 15º 22

21º

sin θ 22

624 −− 2213 −

222 −

cos θ 22

624 ++ 2213 +

222 +

tan θ ( 3 – 2 )

( 2 –1) 2 – 3 2 – 1

θ 18º 36º

sin θ 4

15 − 4

5210 −

cos θ 4

5210 + 4

15 +

tan θ 5

51025 − 525 −

Important Points to Remember : Maximum and minimum values of

a sin x + b cos x are + 22 ba + , – 22 ba + respectively.


sin2x + cosec2x ≥ 2 for every real x. cos2x + sec2x ≥ 2 for every real x. tan2x + cot2x ≥ 2 for every real x

If x = sec θ + tan θ, then x1 = sec θ – tan θ

If x = cosec θ + cot θ, then x1 = cosec θ – cot θ

cos θ . cos 2θ . cos 4θ . cos 8θ

.... cos 2n–1θ = θθ

sin22sin

n

n

sin θ sin (60º – θ) sin (60º + θ) = 41 sin 3θ

cos θ cos (60º – θ) cos (60º + θ) = 41 cos 3θ

tan θ tan (60º – θ) tan (60º + θ) = tan 3θ

Conditional Identities : 1. If A + B + C = 180º, then sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C sin (B + C – A) + sin (C + A – B) + sin (A+B – C) = 4 sin A sin B sin C cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C cos 2A+ cos 2B – cos 2C = 1 – 4 sin A sinB cosC 2. If A + B + C = 180º, then

sin A + sin B + sin C = 4 cos2A cos

2B cos

2C

sin A + sin B – sin C = 4 sin2A sin

2B sin

2C

cos A+ cos B+ cos C = 1 + 4 sin2A sin

2B sin

2C

cos A+ cos B – cos C = –1 + 4cos2A cos

2B sin

2C

CsinBsin

Acos + AsinCsin

Bcos + BsinAsin

Ccos = 2

3. If A + B + C = π, then sin2A + sin2B – sin2C = 2 sin A sin B cos C cos2A + cos2B + cos2C = 1– 2 cos A cos B cos C sin2A + sin2B + sin2C = 2 + 2 cos A cos B cos C cos2A + cos2B – cos2C = 1 – 2 sin A sin B cos C 4. If A + B + C = π, then

sin2

2A +sin2

2B + sin2

2C = 1– 2 sin

2A sin

2B sin

2C

cos2

2A +cos2

2B +cos2

2C =2 + 2 sin

2A sin

2B sin

2C

sin2

2A +sin2

2B –sin2

2C = 1 – 2cos

2A cos

2B cos

2C

cos2

2A + cos2

2B – cos2

2C = 2cos

2A cos

2B sin

2C

5. If x + y + z = π/2, then sin2x + sin2y + sin2z = 1 – 2 sin x sin y sin z cos2x + cos2y + cos2z = 2 + 2 sin x sin y sin z sin 2x + sin 2y + sin 2z = 4 cos x cos y cos z 6. If A + B + C = π, then tan A + tan B + tan C = tan A tan B tan C cot B cot C + cot C cot A + cot A cot B = 1

tan2B tan

2C + tan

2C tan

2A + tan

2A tan

2B = 1

cot 2A + cot

2B + cot

2C = cot

2A cot

2B cot

2C

7. (a) For any angles A, B, C we have sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C tan(A + B + C)

= AtanCtanCtanBtanBtanAtan1CtanBtanAtan–CtanBtanAtan

−−−++

(b) If A,B, C are the angles of a triangle, then sin(A + B + C) = sin π = 0 and cos (A + B + C) = cos π = –1 then (a) gives sin A sin B sin C = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C and (a) gives 1 + cos A cos B cos C = cos A sin B sin C + sin A cos B sin C + sin A sin B cos C Method of Componendo and Dividendo :

If qp =

ba , then by componendo and dividendo we

can write

qpqp

+− =

baba

+− or

pqpq

+− =

abab

+−

or qpqp

−+ =

baba

−+ or

pqpq

−+ =

abab

−+


PHYSICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. If x = ba

cosbsina+

θ+θ , then –

(A) The dimension of a and x are same (B) The dimension of b and x are same (C) Both (A) and (B) (D) x is dimensionless

2. ABCD is a smooth horizontal fixed plane on which mass m1 = 0.1 kg is moving in a circular path of radius 1 m. It is connected by an ideal string which is passing through a smooth hole and

connects mass m2 =2

1 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 –

A B

C D

m2

m1

(A) 10 m/s (B) 10 m/s

(C) 101 m/s (D) None of these

3. If x grams of steam at 100ºC becomes water at 100ºC which converts y grams of ice at 0ºC into water at 100ºC, then the ratio x/y will be –

(A) 31 (B)

427

(C) 3 (D) 274

4. A gas is at pressure P and temperature T. Coefficient of volume expansion of one mole of gas at constant pressure is –

(A) T1 (B) T (C)

2T1 (D) T2

5. A source of light is placed at double focal length

from a convergent lens. The focal length of the lens is f = 30 cm. At what distance from the lens should a flat mirror be placed so that ray reflected from the mirror are parallel after passing through the lens for the second time ?

(A) Beyond 2 F (B) Between lens and F (C) Between F and 2F near 2F (D) Between F and 2F equidistant from F and 2F

XtraEdge Test Series # 9

IIT-JEE 2010

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 14 are multiple choice questions with multiple (one or more than one) correct answer. +4 marks and

-1 mark for wrong answer. • Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - II • Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, +1 marks will be given for a correctly marked answer in any row.


6. Consider a toroid of circular cross-section of radius b, major radius R much greater than minor radius b, (see diagram) find the total energy stored in magnetic field of toroid –

(A) 0

222

2RbB

µπ (B)

0

222

4RbB

µπ

(C) 0

222

8RbB

µπ (D)

0

222 RbBµπ

7. A small charged ball is hovering in the state of

equilibrium at a height h over a large horizontal uniformly charged dielectric plate. What would be the instantaneous acceleration of the ball if a disc of radius r = 0.001 h is removed from the plate directly underneath the ball –

(A) 2g 2

hr

(B)

2g 2

rh

(C) 4g 2

hr

(D)

4g 2

rh

8. For pair production i.e. for the production of

electron and positron incident photon must have minimum frequency of the order of –

(A) 1018/sec (B) 1021/sec

(C) 1025/sec (D) 1030/sec

9. For the vectors →a and

→b shown in figure,

→a = 3 i + j and |

→b | = 10 units while θ = 23º,

then the value of R = | →a +

→b | is nearly –

θ

y

O x

→b

→a

(A) 12 (B) 13 (C) 14 (D) 15

Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

10. Path of three projectiles are shown. If T1, T2 and T3 are time of flights and ignoring air resistances -

y

1 23

R1.5R

2R

x

(A) T1 > T3 (B) T1 < T3

(C) T2 =2

TT 31 + (D) T1 = T2 = T3

11. Two blocks of masses 2 kg each are moving in opposite direction with equal speed collides at t = 5 sec. The magnitude of relative velocity (v) is plotted against time 't'. The loss in kinetic energy is K and coefficient of restitution in e, then –

4 m/s

t = 5 sec t

v

(A) K = 8J (B) K = 16J (C) e = 0.5 (D) e = 0 12. A constant voltage is applied between two ends of

a uniform conducting wire. If both the length and radius of the wire are doubled –

(A) the heat produced in the wire will be doubled (B) the electric field across the wire will be

doubled (C) the heat produced will remain unchanged (D) the electric field across the wire will become

half 13. A solenoid is connected to a source of constant

emf for a long time. A soft iron piece is inserted into it. Then –

(A) self inductance of the solenoid gets increased (B) flux linked with the solenoid increases hence

steady state current gets decrease (C) energy stored in the solenoid gets increased (D) magnetic moment of the solenoid increased


14. In radioactivity decay according to law N = N0e–πt which of the following is/are true ?

(A) Probability that a nucleus will decay is 1 – e–λt (B) Probability that a nucleus will decay four half

lives is 15/16 (C) Fraction nuclei that will remain after two half

lives is zero (D) Fraction of nuclei that will remain after two

half-lives is 1/4 This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 15 to 17)

Al foil

V

10×106Ω

ε

106Ω

Voltmeter is ideal and two aluminum foil is given at

certain separation. In this set up upper aluminum foil jumps to the lower Al foil at the potential difference between the plates (or foils) of 500 V.

15. When emf ε of the battery is 400 volt and foil has

not jumped, approximate reading of the voltmeter is -

(A) 500 V (B) 400 V (C) 0 V (D) 250 V

16. What is the reading of voltmeter just after the foil has jumped and connected the two plates -

(A) 500 V (B) less than 500 V (C) more than 500 V (D) 600 V

17. What is the emf of the battery just after the foil has jumped and connected the two plates -

(A) 500 V (B) 600 V (C) 700 V (D) 800 V Passage : II (No. 18 to 20)

Figure (i), below depicts three hypothetical atoms. Energy levels are represented as horizontal segments. The distance between the segments is representative of the energy difference between the various levels. All possible transitions between energy levels are indicated by arrows.

Atom # 1 Atom # 2 Atom # 3

Fig (i)

Scientists can observe the spectral lines of atoms that are dominant in far-away galaxies. Due to the speed at which these galaxies are travelling, these lines are shifted, but their pattern remains the same. This allows researchers to use the spectral pattern to determine which atoms they are seeing. Table - 1 below shows spectroscopic measurements made by researchers trying to determine the atomic makeup of a particular far-away galaxy. Light energy is not measured directly, but rather is determined from measuring the frequency of light, which is proportional to the energy.

Table – 1

856390

Frequencies Measured868440880570879910

18. For each of three hypothetical atoms (Atom 1, Atom 2 and Atom 3), Figure 1 depicts the -

(A) number of electrons and the amount of energy the atom contains

(B) distance an electron travels from one part of the atom to another

(C) energy released by the atom as an electron as it moves from one energy state to another

(D) frequency with which the atom’s electrons move from one energy state to another

19. Based on the spectroscopic measurements shown

in Table - 1, which of the atoms in Figure (i) (Atom 1, Atom 2, or Atom 3) is most similar to the one the scientists were observing, and why ?

(A) Atom 2, because it contains four different energy levels

(B) Atom 3, because it contains four different energy levels

(C) Atom 1, because the frequencies listed in Table 1 indicate a high level of atomic activity

(D) Atom 3, because there is a comparatively small difference between exactly two of the four frequencies listed in Table 1

20. The laws of atomic physics prohibit electron movements between certain energy states. In atomic physics, these prohibitions are called “forbidden transitions.” Based on Figure (i), which of the following is most accurate ?

(A) Atom 2 has the same number of forbidden transitions as Atom 1


(B) Atom 2 has more forbidden transitions than Atom 3

(C) Atom 3 has the same number of forbidden transitions as Atom 1

(D) Atom 1 has fewer forbidden transitions than Atom 2

This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

D C B A

P P P P

Q Q Q

R R R R

S S S S Q P R S

Q

21. Column I Column II (A) Pair production (P) Few MeV (B) Inverse photoelectric (Q) 20 KeV effect (C) De-excitation of Be+3 (R) 54 eV atom from second excited state (D) Kα – X-ray photons (S) 0.1 eV of molybdenum Z = 42

22.

60°

30°

(A)

µ1

µ2

µ1

θc

(B)

µ2

α

µ1

µ2 β

(C) α > θc Column I Column II (Α) θc (P) 45°

(Β) sin–1

31 (Q) Critical angle

(C) Refractive index of 1 (R)

31

with respect to 2 (D) Total internal reflection (S) α = β

CHEMISTRY


1. Brown ppt. (A) dissolve in HNO3 gives (B) which gives white ppt. (C) with NH4OH. (C) on reaction with HCl gives solution (D) which gives white turbidity on addition of water. What is (D) ?

(A) BiCl3 (B) Bi(OH)3 (C) BiOCl (D) Bi(NO3)3 2. Histidine (A) has pKa values as indicated

N NH

CH2 CH C OH

O

NH3 + 6.04

9.171.82

(A) What will be its form at pH = 4 ?

(A)

HN NH

CH2–CHCOH

O

NH3+ ⊕

(B)

HN NH

CH2CHC O

O

NH3

–

+ +

(C) N NH

CH2CHC O

O

NH3

–

+

(D)

N NH

CH2CHC O

O

NH2

–

3. The correct order of M–C π bond and strength in given metal carbonyl is-

(A) [Fe(CO)4]2– > [Co(CO)4]– > [Ni(CO)4] (B) [Ni(CO)4] > [Co(CO)4]– > [Fe(CO)4]2– (C) [Fe(CO)4]2– > [Ni(CO)4] > [Co(CO)4]– (D) [Ni(CO)4] > [Co(CO)4]– = [Fe(CO)4]2–

4. Compound 'A' (molecular formula C3H8O) is treated with acidified potassium dichromate to form a product B (molecular formula C3H6O). 'B' form a shining silver mirror on warming with ammonical silver nitrate 'B' when treated with an aqueous solution of H2NCONHNH2.HCl and sodium acetate gives a product 'C'. Identify the structure of C.


CH3C NCONHNH2

CH3

(A)

(B) CH3CH2CH N NHCONH2

CH3 C NNHCONH2

CH3

(C)

(D) CH3CH2CH = NCONHNH2

5. Vander Waal's equation for a real gas is

+ 2

2

VanP (V – nb) = nRT

Plot of quantity Q = ba

ab+

with temperature is

(A)

T

Q (B)

Q

T

(C)

T

Q (D)

T

Q

6. A solution containing NaOH and Na2CO3 was titrated against HCl using phenolphthalein as an indicator. The tire value of HCl solution was found to be x ml. At the end point, methyl orange was added and the titration continued. A further y ml of HCl solution was required to get the end point with methyl orange. The volume of HCl solution used with Na2CO3 during the whole process is

(A) 2x (B) 2y (C) x (D) y – x 7. For crystallisation of a solid from the aqueous

solution, if the values of ∆H and ∆S are –x J mol–1 and – y J K–1 mol–1 respectively, which of the following relationships is correct

(A) x = T × y (B) x > T × y (C) x < T × y (D) None of these 8. Reduction of but-2-yne with Na and liquid NH3 gives

an alkene which upon catalytic hydrogenation with D2/Pt gives an alkane. The alkene and alkane formed respectively are

(A) cis but-2-ene and racemic-2, 3-dideuterobutane (B) trans but-2-ene and meso 2, 3-dideuterobutane (C) trans but-2-ene and racemic 2, 3-dideuterobutane (D) cis but-2-ene and meso 2, 3-dideuterobutane

9.

+

O1. AlCl3

2. H+/H2O A

The product 'A' is

(A) Me

PhOH

(B)

O

Ph

OH

(C)

OH

H

MePh (D)

PhH


10. The following complexes are given (1) trans – [Co(NH3)4Cl2]+ (2) cis – [Co(NH3)2 (en)2]3+ (3) trans – [Co(NH3)2(en)2]3+ (4) NiCl4

2– (5) TiF6

2– (6) CoF6

3– Choose the correct code : (A) (1), (2) are optically active, (3) is optically inactive (B) (2) is optically active, (1), (3) are optically inactive (C) (4), (6) are coloured and (5) is colourless (D) (4) is coloured and (5), (6) are colourless

11. Cellulose is made of glucose units joined together by β-1, 4-glycosidic linkages. These molecules are held by-

(A) ionic bonds (B) hydrogen bonds and van der Waal's forces (C) weak van der Waal's forces only (D) All of these

12. Which of the following statements is/are correct. ? (A) The monoatomic gas He has lower entropy than

the triatomic gas CO2, which has lower entropy than gaseous benzene

(B) For single atoms, the absolute entropy increases as the number of electrons and the protons increases

(C) Among the solid elements the absolute entropy generally increases as the atomic number increases

(D) All metallic solids have entropies below 85 J. mole–1K–1

13. In the Libermman test for phenols, the blue or green colour produced is due to the formation of -

(A) OH OH

(B) O = = NOH


(C) O = = N – – ONa

– +

(D) O = = N – – OH

14. Which of the following statements are correct for SN2

reaction. (A) Increasing the polarity of solvent causes a large

increase in the rate of SN2 attack by NH3 on alkyl halide

(B) Increasing the polarity of solvent causes a large decrease in the rate of SN2 attack by OH– ion on trimethyl sulphonium ions

(C) Increasing the polarity of solvent causes a small decrease in the rate of SN2 attack by trimethyl amine on trimethyl sulphonium ion

(D) Increasing the polarity of solvent causes a large increase in the rate of SN2 attack by OH– ion on trimethyl sulphonium ion

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 15 to 17)

R C H

O

+ R C H

OOH

R C O

O

+ R CH2OH All non-enolisable aldehydes undergo such type of

ractions in strongly basic medium. These type of reactions are disproportionation reaction in which one molecule gets oxidised and other reduced.

(i) R C H

O

+ OH C H

–O

R

OH

(ii) R C H

O

C H

–O

R

OH

+

R C OH +

O

R C H

–O

H

(iii)

R C O

O

R C OH +

O

R CH2O

+ R CH2OH

15. Aldehydes with H at α-carbon do not undergo this type of reaction

R CH2

OH

O

C H (no reaction)

because - (A) aldehyde is enolised in basic condition (B) bond energy of C–H increase (C) steric hinderance increase (D) all the above 16. Identify the products in the following reactions ?

2H

O

C D + OH → acid salt(i) + alcohol(ii)

2D

D

C O + OH → acid salt(iii) + alcohol(iv)

(i) (ii) (iii) (iv)

(A) D

O

C O– CH2D–OH D

O

C O– CH2DOH

(B) H

O

C O–

CH3OH H

O

C O– CH2DOH

(C) H

O

C O– CH2DOH

D

O

C O– CH2DOH

(D) D

O

C O– CH2DOH

D

O

C O–

CD3OH

17. Consider the following aldehydes :

H C H

O

(i)

CCl3 C H

O

(ii)

CH2 C H

O

(iii)

C H

O

(iv) Cl

The reaction is not given by- (A) (i), (ii) (B) (ii), (iii) (C) (iii), (i) (D) only (iii) Passage : II (No. 18 to 20)

Equilibrium constant are given (in atm) for the following reaction at 0ºC :

SrCl2. 6H2O(s) SrCl2.2H2O(s) + 4H2O(g) : Kp = 5 × 10–12 Na2HPO4.12H2O(s) NaSO4.7H2O(s) +5H2O(g) : Kp = 2.43 × 10–13


Na2SO4.10H2O(s) Na2SO4 (s) + 10H2O(g) : Kp = 1.024 × 10–27 The vapour pressure of water at 0ºC is 4.56 torr.

18. Which is the most effective drying agent at 0ºC ?

(A) SrCl2.2H2O (B) Na2HPO4.7H2O

(C) Na2SO4 (D) all equal

19. At what relative humidities will Na2SO4.10H2O lose water of hydration when exposed to air at 0ºC?

(A) above 33.33% (B) below 33.33% (C) above 66.66% (D) below 66.66%

20. At what humidities will Na2SO4 absorb moisture when exposed to air at 0ºC ?

(A) above 33.33% (B) below 33.33% (C) above 66.66% (D) below 66.66% This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

D C B A

P P P P

Q Q Q

R R R R

S S S S Q P R S

Q

21. Column –I Column-II (A) NH3 → NO3

– (P) M/20 (B) Fe2S3 → 2FeSO4 + SO2 (Q) M/5 (C) KMnO4 in acidic medium (R) M/8 (D) CuS → CuSO4 (S) M 22. Column-I Column-II (A) CuCl2.2H2O (P) Colourless and diamagnetic (B) Cu2Cl2 (Q) Green coloured and paramagnetic (C) CuO (R) Calamine (D) ZnCO3 (S) Black in colour, basic in nature

MATHEMATICS


1. z1 and z2 lie on a circle with centre at the origin. The point of intersection z3 of the tangents at z1 and z2 is given by -

(A) 21

)zz( 21 + (B) 21

21

zzzz2

+

(C) 21

+

21 z1

z1 (D)

21

21

zzzz +

2. The set of all x satisfying the equation

10)x(logxlog 23

23x −+ = 1/x2 is

(A) 1, 9 (B) 1, 9, 1/81 (C) 1, 4, 1/81 (D) 9, 1/81

3. Value of S = 1 × 2 × 3 × 4 + 2 × 3 × 4 × 5 + … + n (n + 1) (n + 2) (n + 3) is -

(A) 51 n (n + 1) (n + 2) (n + 3) (n + 4)

(B) !5

1 (n+3C5)

(C) 51 (n+4C4)

(D) none of these

4. If 1

1tantan1

1tantan1 −

θ−

θ

θ

θ−=

−abba

,

then - (A) a = b = 1 (B) a = cos 2θ, b = sin 2θ (C) a = sin 2θ, b = cos 2θ (D) a = 1, b = sin 2θ 5. A natural number x is chosen at random from the

first one hundred natural numbers. The probability

that 30x

)40x)(20x(−

−− < 0 is

(A) 1/50 (B) 3/50 (C) 3/25 (D) 7/25


6. Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaffs as seen from A are 30º and 60º and as seen from B are 60º and 45º. If AB is 30 m, then distance between the flagstaffs in metres is -

(A) 30 + 15 3 (B) 45 + 15 3

(C) 60 – 15 3 (D) 60 + 15 3 7. If g(x) is a polynomial satisfying g(x) g(y) = g(x) +

g(y) + g(xy) – 2 for all real x and y and g(2) = 5 then

3xlim→

g(x) is -

(A) 9 (B) 25 (C) 10 (D) none of these

8. If y = tan–1 x

1x1 2 −+ , then -

(A) y′ (1) = 1 (B) y′ (1) = 1/4 (C) y′ (1) = 0 (D) y′ (1) does not exist 9. The degree of the differential equation satisfying

2x1− + 2y1+ = a(x – y) is -

(A) 1 (B) 2 (C) 3 (D) none of these


10. The number of ways of choosing triplets (x, y, z) such that z ≥ max x, y and

x, y, z ∈ 1, 2, …, n, n + 1 is -

(A) n+1C3 + n+2C3 (B) 61 n (n + 1) (2n + 1)

(C) 12 + 22 + … + n2 (D) 2(n+2C3) – n+1C2

11. If f(x) = xcotxcos1xeccosxcosxcos

11xsec

22

222

2

then -

(A) ∫π

π−

4/

4/

dx)x(f = )83(161

+π

(B) f ′ (π/2) = 0 (C) Maximum value of f(x) is 1 (D) Minimum value of f(x) is 0

12. If I = ∫ 2sec x cosec4 x dx = K cot3 x + L tan x +

M cot x + C then - (A) K = –1/4 (B) L = 2 (C) M = – 2 (D) none of these

13. ∫π

+

2/

0 xtan1dx is -

(A) a multiple of π/4 (B) a multiple of π/2 (C) equal to π/4 (D) a multiple of π 14. The orthogonal trajectories of the system of curves

2

dxdy

= a/x are -

(A) 9a(y + C)2 = 4x3 (B) y + C =a32− x3/2

(C) y + C =a3

2 x3/2 (D) none of these

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 15 to 17) A polynomial p(x) = a0 xn + a1 xn–1 + … + an–1 x + an, a0 ≠ 0 is said

to be a reciprocal equation if ai = an – i for 0 ≤ i ≤ [n/2] where [x] denote the greatest integer ≤ x.

15. If p(x) is a reciprocal polynomial of odd degree, then

one of the roots of p(x) = 0 is - (A) – 1 (B) 1 (C) 0 (D) (n + 1)/2 16. Sum of the rational roots of

x5 =x78133

78x133−

−

is - (A) 2/9 (B) 9/2 (C) 13/6 (D) 6/13 17. Let m, n ∈ N and p(x) = 1 + x + … + xm. p(x) will divide p(xn) if (A) hcf (m, n) = 1 (B) hcf (m + 1, n) = 1 (C) hcf (m, n + 1) = 1 (D) hcf (m+1, n+1) = 1


Passage : II (No. 18 to 20)

A straight line is called an asymptote to the curve y = f(x)( if the distance from the variable point M of the curve to the straight line approaches zero as the point M recedes to infinity along some branch of the curve. We have three kinds of asymptotes; vertical, horizontal and inclined.

There are three types of asymptotes vertical, horizontal and inclined.

Vertical asymptotes If at least one of +→ax

lim f(x) or

−→axlim f(x) is equal to infinity then x = A is a vertical

asymptote. If ax

lim±→

f(x) = A then y = A is a

horizontal asymptote. If limits x

)x(flimax ±→

=k2, ax

lim±→

[f(x) – kx] = b2 then y = k2x + b2 is an inclined asymptote.

18. Let y =1x

x3−

+ 3x be curve 1 and y = xe1/x be curve 2

then -

(A) curve 1 has no horizontal asymptote and curve 2 has no vertical asymptote

(B) y = 3x + 3 and y = x + 2 inclined asymptotes to curve 1 curve 2

(C) y = 3x + 3 and y = x + 1 are inclined asymptotes to curve 1 and curve 2

(D) y = x + 1 and y = 3x + 3 are inclined asymptotes to curve 1 and curve 2

19. Let y =3x

x5−

then

(A) There are no vertical asymptotes

(B) There are no horizontal asymptotes

(C) There are no inclined asymptotes

(D) x = 3 and y = 5 are the only asymptotes

20. Let y = 1x2 + sin 1/x

(A) There are no horizontal asymptotes

(B) There is only one horizontal asymptote

(C) There are only two horizontal asymptotes

(D) There is one vertical asymptote


DCBA

PPPP

Q Q Q

R R R R

S S S S Q P R S

Q

21. Column-I Column-II (A) locus of point of intersection (P) x2 + y2 = 2a2

of the lines x = at2, y = 2at (B) locus of the point of (Q) y2 = 4ax intersection of the perpendicular tangents to the circle x2 + y2 = a2

(C) locus of the point of (R) x2 + y2 = ax intersection of the lines x cos θ = y cot θ = a (D) The locus of the mid (S) x2 – y2 = a2

points of the chords of the circle x2 + y2 – 2ax = 0 passing through the origin

22. Column-I Column-II

(A) f(x) = ,1x)3x(

2

2

++ (P) 0 ≤ f(x) ≤ 3

– ∞ < x < ∞

(B) R = (x, y): x, y ∈ R, (Q) 3 ≤ f(x) ≤ 9

x2 + y2 ≤ 25

R′ = (x, y): x, y ∈ R,

y ≥ 4x2/9

and let (x, f(x)) = R ∩ R′

(C) f(x) =x3cos2

9−

(R) 0 ≤ f(x) ≤ 10

(D) f(x) = (S) 0 ≤ f(x) ≤ 5

3 2 sin 22 x)16/( −π


PHYSICS


1. Two blocks A and B, have equal masses. They are placed next to each other on a horizontal frictionless fixed table. Compare the velocities of the blocks as each of them reaches the opposite end of the table –

A A

B B

4F

F

(A) vA = 2 vB (B) vA = 4 vB

(C) vA = 8 vB (D) vA = 16 vB

2. Centre of mass of two thin uniform rods of same length but made up of different materials and kept as shown, can be, if the meeting point is the origin of co-ordinates –

x L

L

y

(A)

2L,

2L (B)

2L,

3L2

(C)

3L,

3L (D)

6L,

3L

3. A body of mass 1 kg starts moving from rest at t = 0, in a circular path of radius 8 m. Its kinetic energy varies as a function of time as : KE = 2t2 Joules, where t is in seconds. Then - (A) Tangential acceleration = 4 m/s2

(B) Power of all forces at t = 2 sec is 8 watt (C) First round is completed in 2 sec (D) Tangential force at t = 2 sec is 4 newton

4. With what minimum velocity should block be projected from left end A towards end B such that it reaches the other end B of conveyer belt moving with constant velocity v. Friction coefficient between block and belt is µ.

IIT-JEE 2011

XtraEdge Test Series # 9

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 14 are multiple choice questions with multiple (one or more than one) correct answer. +4 marks and

-1 mark for wrong answer. • Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - II • Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, +1 marks will be given for a correctly marked answer in any row.


m

vB

L

µ v0A

(A) gLµ (B) gL2µ

(C) gL3µ (D) 2 gLµ 5. A block of mass m is attached to a pulley disc of

equal mass m, radius r by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of 15 m/s. Its velocity when the string becomes taut will be –

(A) 5 m/s (B) 6 m/s (C) 7.5 m/s (D) 10 m/s

6. A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. A stone of mass 0.5 kg and density 500 kg/m3 is completely immersed in water without touching the walls of beaker. Now the balance reading will be - (A) 2 kg (B) 1 kg (C) 2.5 kg (D) 3 kg

7. A uniform rod of mass M1 is hinged at its upper

end. A particle of mass M2 moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision, the value of

2

1MM is –

vM2

M1

(A) 43 (B)

34 (C)

32 (D)

23

8. Two identical spheres move in opposite directions

with speeds v1 and v2 and pass behind an opaque screen, where they may either cross without

touching (event 1) or make an elastic head-on collision (event 2) - (A) we can never make out which event has

occurred (B) we can not make out which event has

occurred only if v1 = v2 (C) we can always make out which event has

occurred (D) we can make out which event has occurred

only if v1 = v2

9. The escape velocity from the earth is about

11 km/s. The escape velocity from a planet having twice the radius and the same mean density as the earth is - (A) 22 km/s (B) 11 km/s (C) 5.5 km/s (D) 15.5 km/s


10. The potential energy of a particle of mass 0.1 kg, moving along x-axis is given by U = 5x (x – 4) J, where x is in meters. It can be concluded that - (A) The particle is acted upon by a constant force (B) The speed of the particle is maximum at

x = 2 m (C) The speed of the particle is maximum at

x = 0 m (D) The period of oscillation of the particle is

5π sec

11. A particle of mass m is at rest in a train moving with constant velocity with respect to ground. Now the particle is accelerated by a constant force F0 acting in the direction of motion of train for time t0. A girl in the train and a boy on the ground measure the work done by this force. Which of the following are incorrect ? (A) Both will measure the same work (B) Boy will measure higher value than the girl (C) Girl will measure higher value than the boy (D) Data are insufficient for the measurement of

work done by the force F0 12. In figure, two blocks M1 and m2 are tied together

with an inextensible and light string. The mass M1 is placed on a rough horizontal surface with coefficient of friction µ and the mass m2 is hanging vertically against a smooth vertical wall. The pulley is frictionless –


M1

m2Smooth

Rough (µ)

Choose the correct statement (s) related to the

tension T in the string - (A) When m2 < µM1, T = m2g (B) When m2 < µM1, T = M1g (C) When m2 > µM1, µM1g < T < m2g (D) When m2 > µM1 , m2g < T < µM1g

13. Overall changes in volume and radii of a uniform

cylindrical steel wire are 0.2% and 0.002 % respectively when subjected to some suitable force. If Young’s modulus of elasticity of steel is Y = 2.0 × 1011 N/m2, then - (A) Longitudinal tensile stress acting on the wire

is 4.08 × 108 N/m2 (B) Longitudinal tensile stress acting on the wire

is 3.92 × 108 N/m2 (C) Longitudinal strain is 0.204 % (D) Longitudinal strain is 0.196 %

14. A particle falls freely near the surface of the earth.

Consider a fixed point O (not vertically below the particle) on the ground - (A) Angular momentum of the particle about O is

increasing (B) Torque of the gravitational force on the

particle about O is decreasing (C) The moment of inertia of the particle about O

is decreasing (D) The angular velocity of the particle about O

is increasing This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 15 to 17) A siphon tube is discharging a liquid of specific

gravity 0.8 from a reservoir as shown in figure. (Take g = 10 m/s2).

1.5 m

B

AC 1 m

4 m

D

15. Velocity of the liquid coming out of siphon at D

is -

(A) 4 5 m/s (B) 130 m/s

(C) 5 2 m/s (D) 10 m/s

16. Gauge pressure at the highest point B is - (A) – 52 kPa (B) – 44 kPa (C) – 20 kPa (D) – 12 kPa

17. Gauge pressure at point C is - (A) – 32 kPa (B) 8 kPa (C) 20 kPa (D) 0


Two pulse are traveling in opposite direction with speed 1 m/s. Figure shows the shape of pulse at t = 0.

2 4 6 8 10 12 14 16– 5mm

5mm10 mm

Distance (in cm)

x

1 m/s

y

18. Speed of particle at x = 2 cm and t = 0 is - (A) 1 m/s (B) 0.75 m/s (C) 0.5 m/s (D) 0.25 m/s 19. Displacement of particle at x = 8 cm and t = 6 sec

is - (A) 10 mm (B) 5 mm (C) – 5 mm (D) zero

20. Speed of particle at x = 8 cm and t = 6 sec - (A) zero (B) 0.125 m/s (C) 0.25 m/s (D) None of these



D C B A

P P P P

Q Q Q

R R R R

S S S S Q P R S

Q

21. In the situation shown, all surfaces are frictionless and triangular wedge is free to move. In column II the direction of certain vectors are shown. Match the direction of quantities in column I with possible vector in column II.

θ

A

Column-I Column-II

(A) Acceleration of the block (P) θ

A relative to ground

(B) Acceleration of block (Q) > θ

A relative to wedge

(C) Normal force by (R) θ

block on wedge (D) Net force on the wedge (S) 22. A rigid cylinder is kept on a smooth horizontal

surface as shown. If column I indicates velocities of various points (3-centre of cylinder, 2-top point, 4-bottom point, 1-on the level of 3 at the rim) on it shown. Choose correct state of motion from column – II.

4

3

2

1

y

x

Column – I Column – II

(A) →

1v = i + j , (P) Pure rotation about

→

2v = 2 i centre

(B) →

1v = i + j , (Q) Rolling without

→

3v = – i slipping to left

(C) →

2v = i , →

3v = 0 (R) Rolling without slipping to right

(D) →

4v = 0, (S) Not possible

→

1v = – i – j

CHEMISTRY Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. A compression of an ideal gas is represented by curve AB, which of the following is wrong

B(vB)

A(vA) log V

log P

(A) number of collision increases B

A

VV times

(B) number of moles in this process is constant (C) it is isothermal process (D) it is possible for ideal gas

2. A compound containing only sodium, nitrogen and oxygen has 33.33% by weight of sodium. What is the possible oxidation number of nitrogen in the compound?

(A) –3 (B) + 3 (C) –2 (D) + 5

3. How many moles of nitrogen is produced by the oxidation of one mole of hydrazine by 2/3 mole bromate ion

(A) 31 (B) 1 (C) 1.5 (D)

32

4. For the reaction BaSO4 (s) BaSO4 (aq) The equilibrium moles of BaSO4(aq) were

0.2 moles. The equilibrium constant of the above reaction is -


(A) 0.2 (B) 0.2 mol L–1 (C) 2 × 10–4 mol L–1 (D) Data insufficient 5. Which of the following correctly explain the nature

of boric acid in aqueous medium -

(A) H3BO3 → OH2 H3O+ + H2BO3–

(B) H3BO3 → OH2 2 2H3O+ + HBO32–

(C) H3BO3 → OH3 2 3H3O+ + BO33–

(D) H3BO3 → OH2 B(OH)4– + H+

6. The shape of TeCl4 is - (A) Linear (B) Square planar (C) Tetrahedral (D) See-Saw 7. Arrange NH4

+, H2O, H3O+, HF & OH– in increasing order of acidic nature -

(A) OH– < H2O < NH4+ < HF < H3O+

(B) H3O+ > HF > H2O > NH4+ > OH–

(C) NH4+ < HF < H3O+ < H2O < OH–

(D) H3O+ < NH4+ < HF < OH– < H2O

8. The correct order of increasing boiling point is - (A) NH3 > HF > H2O (B) H2O > HF > NH3

(C) NH3 > H2O > HF (D) HF > H2O > NH3 9. Oxidation states of carbon and nitrogen in KCN are,

respectively - (A) – 3, + 2 (B) + 2, – 3 (C) + 1, – 2 (D) zero each


10. Which of the following samples of reducing agents is/are chemically equivalent to 25 mL of 0.2 N KMnO4, to be reduced to Mn2+ + H2O ?

(A) 25 mL of 0.2 M FeSO4 to be oxidized to Fe3+ (B) 50 mL of 0.1 MH3AsO3 to be oxidized to H3AsO4 (C) 25 mL of 0.2 M H2O2 to be oxidized to H+ and O2 (D) 25 mL of 0.1 M SnCl2 to be oxidized to Sn4+

11. A sample of water has a hardness expressed as 77.5 ppm Ca2+. This sample is passed through an ion exchange column and the Ca2+ is replaced by H+. Select correct statement(s)

(A) pH of the water after it has been so treated is 2.4 (B) Every Ca2+ ion is replaced by one H+ ion (C) Every Ca2+ ion is replaced by two H+ ions (D) pH of the solution remains unchanged

12. The order of Keq values for the following keto-enol equilibrium constants is

CH3 CHOk1 CH2 CH OH

CH3k2CH2C

O

C

O

CH3

CH3 CH C

OH

C

O

CH3

CH3k3CH3C

O

C

OH

CH3 CH2

(A) K1 > K2 > K3 (B) K2 > K3 > K1 (C) K2 > K1 > K3 (D) K1 > K3 > K2 13. In the purification Zr and B, which of the following

is/are true ?

(A) Zr + 2I2 → ZrI4(g) overpassedWhotwhitethe →

the pure Zr is deposited on W

(B) 2B + 3I2 → 2BI3(g) overpassedWhotwhitethe →

the pure B is deposited on W

(C) Zr + 2I2 → ZrI4 (s) Wwithmixedheatedthen&

→

ZrI4 is reduced to ZrI2

(D) none of these 14. Which of the following statements is correct ? (A) At 273ºC, the volume of a given mass of a gas at

0ºC and 1 atm. pressure will be twice its volume (B) At –136.5ºC, the volume of a given mass of a gas

at 0ºC and 1 atm. pressure will be half of its volume

(C) The mass ratio of equal volumes of NH3 and H2S under similar conditions of temperature and pressure is 1 : 2

(D) The molar ratio of equal masses of CH4 and SO2 is 4 : 1

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15 to 17)

Entropy is measure of degree of randomness. Entropy is directly proportional to temperature. Every system tries to acquire maximum state of randomness or disorder. Entropy is measure of unavailable energy. Unavailable energy = Entropy × Temperature

The ratio of entropy of vapourisation and boiling point of substance remains almost constant.


15. Which of the following process have ∆S = – ve ? (A) Adsorption (B) Dissolution of NH4Cl in water (C) H2 → 2H (D) 2NaHCO3(s) → Na2CO3 + CO2 + H2O

16. Observe the graph and identify the correct statement(s)

T1 T2Temperature

Entro

py

∆SfusionΑ

Β ∆Svap

(A) T1 is melting point, T2 is boiling point (B) T1 is boiling point, T2 is melting point (C) ∆Sfusion is more than ∆Svap (D) T2 is lower than T1

17. The law of Thermodynamics invented by Nernst, which helps to determine absolute entropy is

(A) Zeroth law (B) 1st law (C) 2nd law (D) 3rd law Passage : II (No. 18 to 20)

Effect of temperature on the equilibrium process is analysed by using the thermodynamics.

From the thermodynamics relation ∆G° = – 2.3 RT logK........(1) ∆G° = Standard free

energy change ∆G° =∆H° – T∆S°….(2) ∆H° = Standard heat of

the reaction From (1) & (2) – 2.3 RT log K = ∆H° – T∆S° ; ∆S° : Standard

Entropy change,

log K = RT3.2H°∆ +

R3.2S°∆ ........(3)

Clearly if a plot of log K vs 1/T is made then it is

a straight line having slope = R3.2H– °∆ &

y–intercept = R3.2

S°∆ .

If at a temperature T1 equilibrium constant be K1 and at temperature T2 equilibrium constant be K2 then, the above equation reduces to :

⇒ log K1 = 1RT3.2

H– °∆ + R3.2

S°∆ ........(4)

⇒ log K2 = 2RT3.2

H– °∆ + R3.2

S°∆ ........ (5)

Subtracting (4) from (5) we get.

⇒ log 1

2

KK =

R3.2H°∆

21 T1–

T1

18. If standard heat of dissociation of PCl5 is 230 Cal.

then the slope of the graph of log K vs T1 is -

(A) + 50 (B) – 50

(C) 10 (D) None of these

19. For exothermic reaction of ∆So < 0 then the

sketch of log K vs T1 may be -

(A)

1/T

log K (B)

1/T

log K

(C)

1/T

log K (D)

1/T

log K

20. If for a particular reversible reaction if Kc = 57 at

355°C and Kc = 69 at 450°C then - (A) ∆H < 0 (B) ∆H > 0 (C) ∆H = 0 (D) ∆H sign can't be determined This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q

R R R R

S S S S Q P R S

Q

21. Match the temperature (in column I) with its value (in column II)

Column –I Column II (A) Critical temperature (P) a/Rb (B) Boyle temperature (Q) 2a/Rb (C) Inversion temperature (R) T/Tc

(D) Reduced temperature (S) 8a/27Rb


22. Match the half-reaction (in column I) with equivalent mass (molar mass = M) (in column II)

Column –I Column II (A) Cr2O7

2– → Cr3+ (P) M (B) C2O4

2– → CO2 (Q) M/2 (C) MnO4

– → MnO2 (R) M/6 (D) HC2O4

– → C2O42– (S) M/3

MATHEMATICS


1. If iz3 + z2 – z + i = 0, then |z| equals - (A) 4 (B) 3 (C) 2 (D) 1 2. If log3 x + log3 y = 2 + log3 2 and log3 (x + y) = 2

then - (A) x = 1, y = 8 (B) x = 8, y = 1 (C) x = 3, y = 6 (D) x = 9, y = 3

3. The exponent of 7 in 100C50 is - (A) 0 (B) 2 (C) 4 (D) none of these

4. The equation cos 2x + a sin x = 2a – 7 possesses a solution if -

(A) a < 2 (B) 2 ≤ a ≤ 6 (C) a > 6 (D) a is any integer 5. In a triangle ABC, 2a2 + 4b2 + c2 = 4ab + 2ac, then

numerical value of cos B is equal to - (A) 0 (B) 1/8 (C) 3/8 (D) 7/8

6. The tangent at the point P(x1, y1) to the parabola y2 = 4ax meets the parabola y2 = 4a (x + b) at Q and R, the coordinates of the mid-point of QR are -

(A) (x1 – a, y1 + b) (B) (x1, y1) (C) (x1 + b, y1 + a) (D) (x1 – b, y1 – b) 7. Equation of the line of shortest distance between the

lines 1z

3y

2x

=−

= and 2

2z51y

32x +

=−−

=− is -

(A) 3(x – 21) = 3y + 92 = 3z – 32

(B) 3/1

)3/31(z3/131y

3/1)3/62(x +

=−

=−

(C) 3/1

)3/32(z3/1

)3/92(y3/121x +

=−

=−

(D) 3/11z

3/13y

3/12x −

=+

=−

8. A vector c, directed along the internal bisector of

the angle between the vectors a = 7i – 4j – 4k and

b = –2i – j + 2k, with |c| = 5 6 , is -

(A) 35 (i – 7j + 2k) (B)

35 (5i + 5j + 2k)

(C) 35 (i + 7j + 2k) (D)

35 (–5i + 5j + 2k)

9. If a and b are two unit vectors such that a + 2b and

5a – 4b are perpendicular to each other then the angle between a and b is -

(A) 45º (B) 60º

(C) cos–1 (1/3) (D) cos–1 (2/7)


10. Suppose a, b, c are positive integers and f(x) = ax2 – bx + c = 0 has two distinct roots in (0, 1), then -

(A) a ≥ 5 (B) b ≥ 5 (C) abc ≥ 25 (D) abc ≥ 250

11. The coefficient of xk (0 ≤ k ≤ n) in the expansion of E = 1 + (1 + x) + (1 + x)2 + … + (1 + x)n is - (A) n+1Ck+1 (B) nCk (C) n+1Cn–k (D) nCn–k–1 12. The equation of a tangent to the hyperbola

3x2 – y2 = 3, parallel to the line y = 2x + 4 is - (A) y = 2x + 3 (B) y = 2x + 1 (C) y = 2x – 1 (D) y = 2x + 2


13. The plane passing through the point (–2, –2, 2) and containing the line joining the points (1, 1, 1) and (1, –1, 2) makes intercepts of lengths a, b, c respectively on the axes of x, y and z respectively, then -

(A) a = 3b (B) b = 2c (C) a + b + c = 12 (D) a + 2b + 2c = 0 14. Let a = 4i + 3j and b be two vectors perpendicular

to each other in xy-plane. The vectors c in the same plane having projections 1 and 2 along a and c are -

(A) –32 i +

211 j (B) 2i – j

(C) –52 i +

511 j (D)

32 i +

211 j

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15 to 17) For k, n ∈ N, we define B(k, n) = 1.2.3 … k + 2.3.4 …(k + 1) + … + n(n + 1) … (n + k – 1), So (n) = n and Sk(n) = 1k + 2k + … + nk To obtain value of B(k, n), we rewrite B(k, n) as

follows: B(k, n) = k! ( ) ( ) ( ) ( )[ ]1kn

k2k

k1k

kkk ... −+++ +++

= k! ( )kn1k

++

=1k

)kn).....(1n(n+

++

15. S3(n) + S1(n) equals -

(A) B(2, n) (B) 21 B(2, n)

(C) 61 B(2, n) (D) none of these

16. S3(n) + 3S2(n) equals -

(A) B(3, n) (B) B(3, n) – 2B(2, n)

(C) B(3, n) – 2B(1, n) (D) B(3, n) + 2B(1, n)

17. ( )1k1+ Sk(n) + ( )1k

2+ Sk–1(n) + … +

( )1kk+ S1(n) + ( )1k

1k++ S0(n) equals -

(A) (n + 1)k (B) (n + 1)k – 1 (C) nk – (n – 1)k (D) (n + 1)k – (n – 1)k


f(x) = sin cot–1 (x + 1) – cos (tan–1 x) a = cos tan–1 sin cot–1 x b = cos (2 cos–1 x + sin–1 x) 18. The value of x for which f(x) = 0 is (A) –1/2 (B) 0 (C) 1/2 (D) 1

19. If f(x) = 0 then a2 is equal to - (A) 1/2 (B) 2/3 (C) 5/9 (D) 9/5

20. If a2 = 26/51, then b2 is equal to - (A) 1/25 (B) 24/25 (C) 25/26 (D) 50/51 This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q

R R R R

S S S S Q P R S

Q

21. Centre of circle Column-I Column-II (A) |z – 2|2 + |z – 4i|2 = 20 (P) 1 – i

(B) 1z1z

+− = 2 (Q) 5/3 + 0i

(C) z z – (1 + i)z (R) – 4 – i – (1 – i) z + 7 = 0

(D) arg

−+++

i25zi43z (S) 1 + 2i

22. cos α + cos β = a, sin α + sin β = b Column-I Column-II (A) cos (α + β) (P) 2ab/(a2 + b2) (B) sin (α + β) (Q) b/a (C) cos (α – β) (R) (a2 – b2)/(a2 + b2)

(D) tan 2

β+α (S) (a2 – b2 – 2)/2


PHYSICS 1. Write the name of the quantity whose SI Unit is Amp

m–1. 2. Name the material for which magnetic susceptibility

is negative.

3. Write the value of ∫→→ds.B

Here →B = magnetic field strength.

→ds = small area vector over a closed surface.

4. If an electron is having the energy of 10 eV then find

its De-Broglie wavelength.

5. If a thin foil of metal, parallel to capacitor plates get introduced between the two plates of air capacitor then write the effect on capacitance C of air capacitor.

A A

a b

d

C = ε0A/d

6. Why the 'CORE' of the transformer is laminated. 7. Draw the symbol of photodiode. 8. Write the name of maxwell's fourth equation for

electro-magnetic waves.

9. G/D is known as ground current detector. Find the reading of it for the given circuit diagram

+3v

+3v +3v

1Ω

G/D

General Instructions : Physics & Chemistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. • Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. • Use of calculators is not permitted.

General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. • All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and

2 question of six marks each. You have to attempt only one of the alternatives in all such questions. • Use of calculators is not permitted.

MOCK TEST PAPER-2

CBSE BOARD PATTERN

CLASS # XII

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS SOLUTIONS WILL BE PUBLISHED IN NEXT ISSUE


10. α particle, β particle and Deutron are placed on the vertices of an equilateral triangle of side 'a' find the electric potential energy of the given system. Also find the work done to place these particles from equilateral triangle of side 'a' to the equilateral triangle of side '2a'.

aDeutron

α particle

β particle

11. Draw the truth table of the following gates

(i)

Y B

A

(ii)

Y B

A

12. In which case the bulb will glow and why ? (i) If VA = 0 volt (ii) If VA = 3 volt Here VA is the potential of A and B is the bulb

+5V

B

RC

A

13. Calculate the no. of electric field lines emitted by the

stationary proton. 14. An equilateral triangle current carrying coil of side 'b'

is placed near the intinite current carrying conductor which is stationary then in which direction the triangular loop will move (towards the conductor or away from the conductor) and why ?

i1 i2

a

15. Find the value of potential difference across tertninals A and B.

C

C

C

B 5V

A

16. If potential difference across terminals X and B is

10V then find the potential difference across A and B.

R R R

RR R

R

R

A BX

10V 17. If the potential difference across the resistance R = 10 Ω is 100 volt then (i) find the current passing through R–L–C series

circuit. (ii) Write the power factor of circuit.

~

R L C

V= 100√2 sin 314t volt 18. Write Einstein's eqn for photoelectric effect. A photon of energy 5eV falls on the surface a metal

whose work function is 2eV. Calculate the value of stopping potential for metal.

19. Appratus of YDSE is shown in figure, the

interference fringes are observed on the screen. The relation between x and y is given below

y → f(t) : y = xt for 0 < t ≤ 4 sec.

S1

S2

Screen

x

y

Find the intensity ratio of maxima to minima on the screen at

(i) t = 1s (ii) t = 4s


20. A convex lens is made of two different materials which are having the absolute refractive indices as µ1 and µ2, this is placed in a media of ARI µ3 as shown below, then

Incident rays

Incident rays

µ3 µ3

µ1

µ2

Lens Medium - 1Medium - 2

Surrounding Medium - 3

(i) Which is the most dense media (ii) Arrange µ1, µ2 and µ3 in increasing order 21. A monochromatic light Ray is incident as shown in

figure then

Normal

incident ray ip

i

Reflected ray

Medium -1 airMedium -2 water

Refracted ray

ip, polarizing angle for air water (i) Calculate the value of ip if wµa = 3/4 (ii) Find the value of φ, φ is the angle between

reflected ray and refracted ray. (iii)Out of reflected and refracted ray which of the

following is plane polarized. (iv)Write the relation between polarizing angle and

critical angle for the two media interface. 22. Calculate the output voltage V0

Ge. Diode

2Ω

1Ω

7Ω

10.3V

V0

23. Draw the wave shape of the output signal Y. The

wave shapes of inputs A and B are shown.

YB

At

t 0

1 B

0

1 A

24. (i) Find the terminal voltages of Batteries BT-1 and BT-2.

(ii) Why terminal voltage of battery BT-1 is more than it's emf and in case of battery BT-2 the terminal voltage is less than it's emf.

Load Resistance RL = 2Ω

1Ω10V 2Ω 20V

BT-1 BT-2

25. Explain the followings (i) Why a metallic spring get shrinked when current

is flown through it (ii) Why N-P-N transistor is preferred over P-N-P

transistor in electronic industry 26. State inconsistency in Ampere's circuital law. What is

meant by displacement current? Prove that displacement current is equals to conduction current.

27. Find the electric flux passing through the cube for the

given arrangement of charges.

+1C –2C

–4C +3C

-6C +7C

+5C –8Ca

28. What type of feedback is used in transistor Amplifier.

Draw the circuit diagram of transistor oscillator and explain it's working. What type of feedback is used in transistor oscillator. Write the expression for frequency of signals generated by transistor oscillator.

29. Explain the principle, construction and working of

Vande Graff Generator. 30. Explain construction and working of Cyclotron.

Why cyclotron can not be used to accelerate the electrons.


CHEMISTRY

1. What are the co-ordination no. of Ca+2 & F– ions in CaF2 structure ?

2. Name of ionisation isomer of [Cr(H2O)5Br]SO4.

3. Write IUPAC name of following (a) O2N – C6H4 – OCH3(p)

(b)

CH2 = C — C — C — C — OH

OH CH3 Br CH3

CH3 CH3 H

4. Why aromatic ketones are not react with NaHSO3 ? 5. Why sulphanilic are amphoteric in nature ? 6. Give monomer of Glyptal. 7. Explain terms Antacids. 8. Why carbohydrates are optically active ? 9. In a compound AX, the radius of A+ ion is 95 pm and

that of X– is 181 pm. Predict the crystal structure of AX and write the co-ordination number of each of the ions.

10. MgO has structure of NaCl and TlCl has the structure

of CsCl. What are the co-ordination number of ions in MgO & TlCl ?

11. What is meant by mole fraction of solute & solvent? 12. A solution containing one mole per litre of each

Cu(NO3)2, AgNO3, Hg2(NO3)2 is being electrolyzed by using inert electrodes, the value of standard electrode potential in volts (reduction potential) are

Ag/Ag+ = 0.80 V, 2Hg/Hg2++ = + 0.79 V

Cu/Cu+2 = 0.34 V, Mg/Mg++ = – 2.37 V with increasing voltage, give the sequence of

deposition of metal on cathode ? 13. Give the structure of dichromate dianion. 14. Which compound is form when excess of KCN is

added to aqueous solution of CuSO4. 15. Give the number & structure of possible enantiomeric

pairs that can be produced during monochlorination of 2-methyl butane.

16. Write the chemical equations for all the steps involved in the rusting of iron.

17. Give the reactions : (a) When phenol is treated with excess of Br2 water. (b) Diethyl ether heated with conc. HI. 18. From Fehling's solution, Schiff's reagent, Tollen's

reagent & Grignard reagent, which reagent react with both aldehyde & ketone.

19. Why 2-pentanone give iodoform test but 3-pentanone not ?

20. Give the order of basicity in the following compound.

N

H(I)

N

(II)

N

H (III)

N

H (IV)

O

21. 0.2 molal acid HX is 20% ionised in solution,

Kf = 1.86 K molality–1. Calculate the freezing point of the solution.

22. How long a current of 3-ampere has to be passed

through a solution of AgNO3 to coat a metal surface of 80 cm2 with a 0.005 mm thick layer. Density of silver is 10.5 g/cm3.

23. Explain that the rate of physisorption increases with

decrease in temperature. 24. Explain that boric acid is monobasic & weak lewis

acid. 25. Give the reason : (a) VOCl2 & CuCl2 give same colour in aqueous

solution. (b) CuSO4 decolourise on addition of KCN. 26. Complete the reaction.

CHO

CHO

CHO

CHO

(i) NaOH/100ºC(ii) H+/H2O


27. Aspartame, an artificial sweetner, is a peptide having following structure.

H2N – CH – CONH – CH – COOCH3

CH2 – COOH

CH2 – C6H5

(i) Identify the four functional group. (ii) Write the zwitter ionic structure. (iii) Write the structure of the amino acids obtained

from the hydrochloride of aspartame.

28. The rate constant for the first order decomposition of a certain reaction is described by the equation

log (k) = 14.34 – T

1025.1 4× .

(i) What is the energy of activation for this reaction. (ii) At what temperature will its half-life period be

256 minutes. 29. The Haber process can be represented by following

CaCO3 → CaO + CO2

H2O

A

BNH3 + H2O

C + H2O + CO2 NaHCO3 + D

NH3 + H2O + E

NaCl

Identify A, B, C, D & E. 30. An alkene 'A' on ozonolysis yields acetone and an

aldehyde, the aldehyde is easily oxidised to an acid (B). When (B) is treated with Br2 in presence of phosphorus it yields a compound (C) which on hydrolysis give a hydroxy acid (D). This acid can also be obtained from acetone by the reaction with HCN followed by hydrolysis. Identify compound A, B, C, & D.

MATHEMATICS

Section A

1. Which one of the following graphs represent the

function of x ? Why ?

x

y

(a)

x

y

(b) 2. What is the principal value of

cos–1

π

32cos + sin–1

π

32sin ?

3. A matrix A of order 3 × 3 has determinant 5. What is

the value of |3A| ? 4. For what value of x, the following matrix is singular?

+−42

1xx5

5. Find the point on the curve y = x2 – 2x + 3, where the

tangent is parallel to x-axis.

6. What is the angle between vectors →a &

→b with

magnitude 3 and 2 respectively ? Given →a .

→b = 3.

7. Cartesian equations of a line AB are.

2

1x2 − = 7

y4 − = 2

1z +

Write the direction ratios of a line parallel to AB.

8. Write a value of ( )∫ 4xlog3 xe dx

9. Write the position vector of a point dividing the line

segment joining A and B with position vectors →a &

→b externally in the ratio 1 : 4,

where →a = k4j3i2 ++ and

→b = kji ++−

10. If A =

514412

and B =

−

312213

.

Write the order of AB and BA.


Section B

11. Show that the function f : R → R defined by

f(x) = 3

1x2 − , x ∈ R is one-one function. Also find

the inverse of the function f. OR Examine which of the following is a binary operation

(i) a * b = 2

ba + , a, b ∈ N

(ii) a * b = 2

ba + , a, b ∈ Q

for binary operation check the commutative and associative property.

12. Prove that

tan–1

1663 = sin–1

135 + cos–1

53

13. Using elementary transformations, find the inverse of

−−

2162

OR Using properties of determinants, prove that

ababbaba

accacacabccbcbbc

22

22

22

−+++−+++−

= (ab + bc + ca)3

14. Find all the points of discontinuity of the function f

defined by

f(x) = 2x,0

2x1,2x1x,2x

≥<<−

≤+

15. If xpyq = (x + y)p+q, prove that dxdy =

xy

OR

Find dxdy , if y = tan–1

−−+

−++22

22

x1x1

x1x1 , 0 < |x| < 1

16. Evaluate ∫ −+++

)5x)(3x()4x)(1x(

22

22 dx

17. A water tank has the shape of an inverted right

circular cone with its axis vertical and vertex lower

most. Its semi vertical angles is tan–1

21 . Water is

poured into it at a constant rate of 5 cubic meter per minute. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 10m.

18. Evaluate the following integral as limit of

sum ∫ −2

1

2 )1x3( dx

19. Evaluate ∫π 2/

0dxxsinlog

20. Find the vector equation of the line parallel to the line

51x − =

2y3 − =

41z + and passing through (3, 0, –4).

Also find the distance between these two lines.

21. In a regular hexagon ABCDEF, if →

AB = →a and

→BC =

→b , then express

→CD ,

→DE ,

→EF ,

→FA ,

→AC ,

→AD ,

→AE and

→CE in terms of

→a and

→b .

22. A football match may be either won, drawn or lost by

the host country's team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches.

OR A candidate has to reach the examination centre in

time. Probability of him going by bus or scooter or by

other means of transport are 103 ,

101 ,

53 respectively.

The probability that he will be late is 41 and

31

respectively, if he travels by bus or scooter. But he reaches in time if the uses any mode of transport. He reached late at the centre. Find the probability that he travelled by bus.

Section C

23. Find the matrix P satisfying the matrix equation

2312

P

−

−35

23 =

−1221

24. Find all the local maximum values and local minimum values of the function

f(x) = sin 2x – x, –2π < x <

2π


OR A given quantity of metal is to be cast into a solid

half circular cylinder (i.e., with rectangular base and semicircular ends). Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its circular ends is π : (π + 2).

25. Sketch the graph of

f(x) =

>−≤+−

2x,2x2x,2|2x|

2

Evaluate ∫4

0)x(f dx. What does the value of this

integral represent on the graph ? 26. Solve the following differential equation

(1 – x2)dxdy – xy = x2,

given y = 2 when x = 0 27. Find the foot of the perpendicular from P(1, 2, 3) on

the line

3

6x − = 2

7y − = 27z

−−

Also obtain the equation of the plane containing the line and the point (1, 2, 3)

28. Let X denote the number of colleges where you will

apply after your result and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that

P(X = x) =

===

4or 3 x if– x) k(52 x if2kx

1or 0 x ifkx, k is +ve constant

(a) Find the value of k. (b) What is the probability that you will get

admission in exactly two colleges ? (c) Find the mean and variance of the probability

distribution. OR The bags A and B contain 4 white 3 black balls and 2

white and 2 black balls respectively. From bag A two balls are transferred to bag B. Find the probability of drawing

(a) 2 white balls from bag B ? (b) 2 black balls from bag B ? (c) 1 white & 1 black ball from bag B ?

29. A catering agency has two kitchens to prepare food at two places A and B. From these places 'Mid-day Meal' is to be supplied to three different schools situated at P, Q, R. The monthly requirements of the schools are respectively 40, 40 and 50 food packets. A packet contains lunch for 1000 students. Preparing capacity of kitchens A and B are 60 and 70 packets per month respectively. The transportation cost per packet from the kitchens to schools is given below :

Transportation cost per packet (in rupees)

To From

A B

P 5 4

Q 4 2

R 3 5 How many packets from each kitchen should be

transported to school so that the cost of transportation is minimum ? Also find the minimum cost.

MEMORABLE POINTS

MECHANICS

1. Weight (force of gravity) decreases as you move away from the earth by distance squared.

2. Mass and inertia are the same thing. 3. Constant velocity and zero velocity means the

net force is zero and acceleration is zero. 4. Weight (in newtons) is mass x acceleration

(w=mg). Mass is not weight! 5. Velocity, displacement [s], momentum, force

and acceleration are vectors. 6. Speed, distance [d], time, and energy (joules) are

scalar quantities. 7. The slope of the velocity-time graph is acceleration. 8. At zero (0) degrees two vectors have a

resultant equal to their sum. At 180 degrees two vectors have a resultant equal to their difference. From the difference to the sum is the total range of possible resultants.

9. Centripetal force and centripetal acceleration vectors are toward the center of the circle- while the velocity vector is tangent to the circle.

10. An unbalanced force (object not in equilibrium) must produce acceleration.

11. The slope of the distance-tine graph is velocity. 12. The equilibrant force is equal in magnitude but

opposite in direction to the resultant vector. 13. Momentum is conserved in all collision systems. 14. Magnitude is a term use to state how large a

vector quantity is.


PHYSICS 1. Relation is not valid because charge is independent

of motion of particle.

2. No change because focal length of mirror is independent of refractive index.

3. Galium Aresnide

4. Yes, it get observed but the fringes are of diffraction type. These are not the interference fringes.

5. Zener diode is used as a voltage regulator to obtain constant voltage output.

6. Davison – Germer experiment. 7. n = 0 8. Because for alloys temperature coefficient of

resistivity is nearly constant for wide range of temperature.

9. (i) V-m = electric flux (ii) C-m = Dipole moment 10. (i) NPN transistor (ii) Yes, the transistor is properly biased because

emitter-base junction is in forward bias and the collector base junction is in reverse bais

11. irms = ∫T

0

2 dt)f(iT1

i = 1 + 23 sin (314 t + 30º) G = 1, i0 = 23 from the above formulae

irms

2/1202

2iG

+ = (1 + 9)1/2 = 10

12. At P1 drift speed of the electron is maximum for a

conductor vd ∝ A1 (I = neAVd)

At P3 current density is minimum.

13. From Einsten's equation Kmax = hv – hv0 eV0 = hv – hv0

V0 = e

hν – e

h 0ν = e

hν – eφ

φ = work function. From the graph. metal-2 has high value of φ ∴ threshold wavelength of metal-1 is high. (ii) θ1 = θ2 because slope of the graph is constant

for all metals.

14.

G

( )

( )R.BK2

K1 Rh Ep

15. (i) Angle of Dip: It is the angle which the direction

of resultant intensity of earth's magnetic field subtends with horizontal line in magnetic meridian at the given place.

(ii) BH = B cos φ ⇒ tan φ = H

v

BB

Bv = B sin φ Bv = BH ⇒ tan φ = 1 If φ = 45º 16. (i) Algebric sum of currents meeting at a point is

equal to zero. (ii) Current in 2Ω resistor is 3A ∴ Potential difference is = 2 × 3 = 6V

17. (i) Lenz's law : The direction of magnetic induction in a circuit is such that so as to oppose the cause of change in magnetic flux.

(ii) On increasing current i inwards magnetic field increases. Direction of induced current is such that to produce outwards magnetic field i.e. anticlockwise direction.

18. (i) The order of colours in secondary rainbow is opposite to that of primary rainbow

(ii) This is due to total internal reflection.

MOCK TEST PAPER SOLUTION FOR PAPER – 1 PUBLISHED IN DECEMBER ISSUE


19. (i) Potential difference on 1µF and 3µF is same

∴ from energy = 21 CV2

E ∝ C

1

2

EE =

31

(ii) Parallel combination of 1 µF and 3 µF is 4 µF

12 4

30V

3µF

30V Q = CV = 90µC

P.d. on 12µF ⇒ V = cd =

1290 =

430 = 7.5 Volt

(iii) Energy supplied by battery E = QV = 9 ×10–6 × 30 = 2.7 × 10–3 J 20. (i) Magnetic field in the solenoid is along the axis

∴ angle between →v and

→B is 0º

∴ F = q(→v ×

→B ) = 0

(ii) When current flows through the spring current in different coils of spring flows in same direction. Therefore due to magnetic force spring gets compressed.

21. (i)

5Ω 10Ω

x (100 – x)

let P be the null point 21 =

x–100x

2x = 100 – x, x = 33.33 cm

(ii)

Y B

A

111101010100YBA

22. (i) When unpolarised light is incident on a

transparent surface of refractive index µ at a certain angle ip such that the refracted light ray and the reflected light ray are perpendicular to each other, the reflected light is plane polarised as shown here.

The refractive index is related to the angle ip called the polarising angle by a relation known as

Brewster's law µ = tan ip.

Plane polarisedlight

µ

ip

(ii) sin θc = µ1 =

pitan1

θc = sin–1(cot ip) (iii) It is not correct.

23. Need for modulation (i) Frequency of signal : The audio frequency signal

(20 Hz to 20 KHz) cannot the transmitted without distortion over long distance due to less energy carried by low frequency audio waves.

(ii) Number of channels: Audio frequencies are concentrated in the range 20 Hz to 20 kHz. This range is so narrow that there will be overlapping of signals. In order to separated the various signals it is necessary to convert all of them to different portions of the electromagnetic spectrum

EC

EC

Mo dulating signal Carrier

A.M. wave

24. To convert a galvanometer into an ammeter of range I a small resistance S is connected in parallel with the galvanometer so that the current passing through the galvanometer G becomes equal to its null scale deflection value Ig.

I Ig I

S

(I – Ig)

G


IgG = (I – Ig) × S ⇒ S = IgI

IgG−

25. (i) Equivalent ckt is

R

R R R/2R/2

RR

4R/3 4R/3

R

Ans. 8R/7 (ii) Material will be constantan because for alloys

value of α varies very slightly with temperature as compared to metals.

26. (i) Angle between electric field line and

equipotential surface is 90º. (ii) Electric field directed in the direction of

decreasing potential. So electric potential is maximum at point a and proton will have maximum potential energy at point a.

(iii) Electric field is maximum at point c. Thus proton will have maximum force at c.

27. (i)

+q2a

2a2a

+q

–q –q

Pnet

Pt

–2q can be assumed as two –q charges placed at the point p = q(2a)

Pnet = º60cosP2PP 222 ++ = 3 P = 3 q(2a) (ii) Electric potential energy

+q2a

2a2a

+q

–2q

U = a2

)q2(kq − +a2

)q2(kq − +a2

kq2

= a2kq2 2− –

a2kq2 2− +

a2kq2

= a2kq3 2−

28. Working : During the positive half cycle of the input signal, the forward bias across the emitter-base junction will be increased while during the negative half cycle of the signal, the forward bias across emitter-base junction is decreased. Hence more electrons flow from the emitter to the collector via the base during positive half cycle. The increased collector current will produce a large voltage drop across the load resistance RL. However during the negative half cycle of the collector, current decreases resulting in the decreased output voltage. Hence an amplified output is obtained across the load.

Signal

Vi

t

IB

~ IE

VCE

IC

RL Out

put IC

ICtO

BE

29. Cyclotron

D2

D1path of accelerated proton

high frequencyAC source

positive chargedbeam

B

Principle : Cyclotron device is based on the fact

that heavy positive ions can be accelerated to the high energies with a comparatively smaller alternating potential difference by making them to cross the field again and again using strong magnetic field. Here the magnetic field used in cyclotron maintain the charged particles in circular paths while the electric field imparts them energy periodically.

Construction : Cyclotron consists of two D shaped hollow metallic enclosures D1 and D2 called dees. These dees have their diometric edge parallel to one other and are separated by a small gap. These dees are connected to the terminals of a high frequency alternating potential difference. This potential difference creates an electric field of high frequency in the gap between the dees. The whole apparatus is placed between N-S poles of a powerful electro magnet which produces strong magnetic field.


Cyclotron can not be used to accelerate electrons because it is used to accelerate heavy ions.

30. Ampere's circuital law : The circulation of →B along

a closed loop of any arbitary shap is µ0 times the algebric sum of current embraced by the loop.

Determination of magnetic field for a solenoid :

b c

l

aP

d

b c

a d

B

dl

Axis Turns of Solenoid

Let p be the point where B is to be determined. From Ampere's law

∫→→ld.B = µ0inet ∫

→→b

ad.B l + ∫

→→c

bd.B l

+ ∫→→d

cd.B l + ∫

→→a

dd.B l

[ ∫→→b

ad.B l = ∫

→→d

cd.B l = 0;

→B ⊥

→ld

and ∫→→d

cd.B l = 0, Q

→B = 0]

∫→→a

dd.B l = B ∫

a

d

0cosdl = BI ⇒ µ0(nil)

⇒ B = µ0ni

CHEMISTRY

1. CH3 – C = CH – C – CH3

|| O

| CH3

5 4 3 2 1

4-methyl-pent-3-en-2-one

2. CH3 – CH2 – CH = CH2 + HCl → CH3 – CH2 – CH – CH3

| Cl

3. Enzyme streptokinase can dissolve blood clots. So it is useful in medicines for checking heart attacks caused by blood clotting.

4. When two different molecules participate in the polymerization process it is called copolymeri-zation.

5. A metal which is more electropositive than iron

such as Al, Zn, Mg can be used in cathodic protection of iron against rusting.

6. Lower value of bond dissociation energy of F2 is

due to the strong repulsion between the non bonding electrons of F atoms in the small sized F2 molecule. Also there is no multiple bonding due to absence of d-orbitals.

7. Hydrolysis of sucrose produces change in optical

nature form dextro rotatory to laevorotatory, the process is called inversion of sugar.

C12H22O11 ((cane sugar) + H2O ) →+H

C6H12O6 + C6H12O6

Glucose Fructose

Invert sugar

8. Three types of lattice imperfections are possible

(a) Schottky defect (b) Frenkel defect (c) Interstitial defects 9. Mechanism of the formation of diethyl ether from

ethanol : The formation of ether is a nucleophilic bimolecular reaction (SN2) involving the attack of alcohol molecule on a protonated alcohol, as indicated below

(i) CH3 – CH2 – O – H + H+ → CH3 – CH2 – +O – H

H

Ethanol

(ii) CH3 CH2 – O + CH3 – CH2 – O

H

|H

+

H →

CH3 CH2 – O – CH2 CH3 + H2O |

H

+

(iii) CH3 CH2 – +O – CH2 CH3 →

|H

Diethyl ether (Ethoxyethane)

CH3 CH2 – O – CH2 CH3 + H+


10. The order of basicity in gaseous phase is (i) (CH3)3 N > (CH3)2 NH > CH3NH2 > NH3 due to

+I effect of alkyl group, there is more density at N at tertiary amine.

(ii) The order of basicity in aqueous state is (CH3)2 NH > CH3 NH2 > (CH3)3 N > NH3

The inductive effect, solvation effect, H-bonding and steric hinderance of the alkyl group decides the basic strength of alkyl amines in the aqueous state.

11. (i) Monomer of Teflon is tetrafluro ethylene.

Teflon is a addition polymer. (ii) Monomer of Bakelite is formaldehyde and

phenol. Bakelite is a condensation polymer. (iii)Monomer of natural rubber is isoprene (2-

methyl-1, 3-butadiene). Natural rubber is a addition polymer.

12. (i)

ONa

+ CO2

atm74K400

− →

Sod. phenoxide

OH

COONa

+ → HO2H

OH

COOH

sod. salicylate salicylic acid

(ii)

OH

Phenol

NaOHaq,3CHClK340

→

ONa

CHCl2

NaOH →

ONa

CHO +

→O3H

OH

CHO

Salicylaldehyde

13. (i) Treat the compound with Lucas reagent (conc. HCl + anhy. ZnCl2) 2-propanol gives turbidity in 5 min whereas ethanol gives no turbidity at room temperature

CH3CH2OH + HCl → 2ZnCl No reaction

CH3CHCH3 + HCl|

OH

→ 2ZnCl

CH3 – CH – CH3 + H2O |

Cl

Turbidity appears in 5 min (ii) Acetaldehyde reduces Tollen’s reagent to silver

mirror but acetone does not. CH3CHO + 2 [Ag (NH3)2]+ + OH– –→ CH3COO + 2H2O + 2Ag ↓ + 4NH3 Silver mirror

CH3COCH3 → reagents’Tollen No action 14.

Multimolecular Colloids

Macromolecular Colloids

The particles of this type of colloids are aggregates of atoms or molecules with diameter less than 1 nm. Examples : Sol of sulphur conists of colloidal particles which are aggregate of 58 molecules.

The particles of this type of colloids are themselves large molecules of colloidal dimension. Examples : Starch, proteins etc.

The atoms of molecules are held together with Van der Waal’s forces

Covalent bonds are present in one chain and different chains have the force like H-bonds, dipole-dipole interaction and salt bridge etc.

15. Pyrophosphoric acid is prepared by the removal of H2O from two molecules of orthophosphoric acid (having tetrahedral shape). Hence two tetrahedra are attached through an oxygen.

P

OH

O

O

O P

OH

O

O


OR

O = P – O – P = O

OH |

| OH

OH |

| OH

16. Cu2+ (aq) + 2e– → Cu (s)

]Cu[

]Cu[log2059.0EE

2

Cu/CuCu/Cu 22

++°= ++

= 0.34 + 11.0log

2059.0 = 0.34 +

101log

2059.0

= 0.34 + )1(2059.0

−× [Q log 10 = 1]

= 0.34 – 0.0295 = 0.3105 V When the concentration of Cu2+ ions is decreased,

the emf for copper electrode decreases. 17. Adsorption isobar for physical adsorption shows

that the extent of adsorption decreases with the increase in temperature. The adsorption isobar of chemical adsorption shows that the extent of adsorption first increases and then decreases with the increase in temperature. The initial unexpected increase in the extent of adsorption with temperature is due to the fact that the heat supplied acts as activation energy required for chemical adsorption which is much more than that of physical adsorption.

Physical adsorption isobar

Temperature

Ext

ent o

f ad

sorp

tion

Chemical adsorption

isobar

Temperature

Ext

ent o

f ad

sorp

tion

18. In the complex [Ni (CO)4], the oxidation state of Ni

is ‘0’. Its electronic configurations is [Ar] 3d8 4s2. A.O. of Ni (28)

↑↓↑↑↑↓↑↓↑↓p4s4d3

sp3 hybridised obritals of Ni

↑↓↑↓↑↓↑↓↑↓

sp3 hybridisation

Formation of Ni (CO)4

××××↑↓↑↓↑↓↑↓↑↓

Four electrons pairs from four CO molecules The resulting complex has tetrahedral shape and is

diamagnetic due to absence of unpaired electrons. 19. (i) Markownikoff’s rule : According to this rule,

when addition across an unsymmetrical double bond takes place, the positive part of the addendum goes to the carbon atom with the larger number of hydrogen atoms.

CH3 – CH = CH2 + HBrpropene Peroxide

ofAbsence →

CH3 – CH – CH3

Br|

2-Bromopropane

(ii) Hofmann Bromide Reaction : When amide is

treated with bromide in on alkaline solution, an amide yields an amine containing one carbon less than the starting amide.

R – C – NH2 + Br2 + 4 KOH

O||

Amide

RNH2 + K2 CO3 + 2KBr + 2H2OAmine

For example :

CH3 CH – C – NH2 + Br2 + 4 KOH

O||

Propanamide

CH3 CH2 NH2 + K2 CO3 + 2KBr + 2H2O

Ethylamine 20. (a) (i) CH2 = CH2 + H2SO4

Ethene conc. Sulphuric acid

–→ CH3CH2HSO4

Ethyl hydrogen

O2HBoil → CH3CH2OH + H2SO4

Ethanol

(ii) OH

+ CH3 COCl

Phenol

Acetyl chloride

Pyridine

OCOCH3

+ HCl

Phenyl ethanoate

(iii)

CH3

HC = O

EthanaletherDryMgI3CH

→+

CH3

HC

OMgICH3


O2H,H

I)OH(Mg

+

− →

CH3

HC

OHCH3

2-Propanol

21. (i) A is CH2 = CH2 B is CH2 – CH2

Br|

Br|

(ii) A is

N2Cl– +

(Benzene diazonium chloride)

B is

CN

(Benzonitrile)

(iii) A is

NH3+

HSO4–

(Anilium hydrogen

sulphate)

B is

NH2

SO3H

(Sulphanilic acid)

22. (a) Electronic configuration of Ti in [Ti (H2O)6]3+

is Ti3+ (d1) Two vacant d orbitals are available for

octahedral hybridization with 4s and 4p orbitals. (b) The colour of the complex is purple. The colour of complex is due to the jumping of

electron from lower level to higher level. When an electron from a lower energy of orbital is excited to higher energy of level, the energy of excitation corresponds to the frequency of light absorbed. This frequency lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed.

23. (a) According to electrochemical theory during the

formation of just the impure iron surface behaves like a small electrochemical cell in the presence of water containing dissolved oxygen or carbon dioxide. In such alls, pure iron acts as anode and impure surface acts as cathode. Moisture containing dissolved oxygen or CO2 in the electrolytic solution. Hence rusting is an electrochemical phenomenon.

(b) On dilution, the degree of ionization of the weak electrolyte increases. This increases the molar conductance of the solution sharply.

24. In [Co (NH3)6]3+, the oxidation state of Co is + 3. Co (Z = 27) atom is ground state

↑↓↑↑↑↑↓↑↓3d 4s 4p 4d

Co3+ ion

↑↑↑↑↑↓

↑↑↑↑↑↓

Hybridisation sp3 d2 hybridisation

Formation of [Co(NH3)6]3+ ion

↑↑↑↑↑↓ × × × × ×Six pairs of electrons fromsix NH3 molecules

Due to sp3

d2 hybridisation, the complex ion has octahedral shape. Due to the presence of four unpaired electrons, the complex ion is paramagnetic.

26. (i) A deep red sol of ferric hydroxide is obtained

by the hydrolysis of ferric chloride. The sol particles are propositively charged because of preferential adsorption of Fe3+ ions.

(ii) Adsorption is an endothermic process. So the rate of physical adsorption decreases with the rise in temperature in accordance with Le Chatelier’s principle.

(iii)River water contains charged colloidal particles of sand, clay, etc. As river water comes in contact with saline sea water, the electrolytes of sea water coagulate the suspended colloidal particles which settle down at the point of contact resulting in the rise of river bed. So water adopts a different course and a delta is formed in due course of time.

27. Solid catalysts are used in a number of gaseous reactions. Such catalytic reactions called heterogeneous reactions. Examples of heterogeneous catalysis are

(i) Manufacture of ammonia from N2 and H2 by Haber’s process in the presence of catalyst.

N2(g) + 3H2(g) → )s(Fe 2NH3(g) (ii) V2O5 catalyst is used in the manufacture of

H2SO4 by contact process


2SO2(g) + O2 →)s(5O2V

2SO3(g) Solid catalyst helps in the following ways : (a) Simultaneous adsorption of reactants increases

the concentration at the surface of the catalyst which increases the reaction rate.

(b) Adsorption of reactant molecules makes the attack of other molecules on it easier.

(c) Some adsorbed molecules dissociate into atoms which are very reactive.

(d) Heat of adsorption released provides activation energy for the reaction.

28. (a) A is aldehyde or Ketone. A gives Tollen’s test hence it is an aldehyde

CH3 – CH = CH – CH2 – CHO

(b) (i)

CH2 CH3

Ethylbenzene →−OH/4OKMn

COOH

Benzoic acid

→ + NaOHCaO

Benzene

(ii) 2CH3CHO

Acetaldehyde )oncondensatiAldol(NaOHDil →

CHOCHCHOHCH1234

23 −−−)duction(Re

4NaBH →

CH3 – CHOH – CH2 CH2OH

Butane-1, 3-diol

4 3 2 1

(iii)

C = O + H2 –→ CH3 CHOH CH3

CH3

CH3

Acetone

nDehydratio42SOH.Conc → CH3 CH = CH2

Propene

29. Preparation of K2 Cr2 O7 : Chromite ore is fused with molten NaOH in the presence of air to get sodium dichromate.

4 Fe Cr2 O4 + 16 NaOH + 7O2 –→ Chromite ore 8 Na2 Cr O4 + 2Fe2 O3 + 8H2 O Sod. chromate The fused mass is dissolved in water. The filtrate is

treated with dil H2 SO4

2Na2 CrO4 + H2 SO4 –→ Na2 Cr2 O7 + Na2 SO4 + H2O

Na2 SO4 is separated by fractional crystallisation. Sod. dichromate is converted into potassium

dichromate by heating with KCl. Na2 Cr2 O7 + 2KCl –→ K2 Cr2 O7 + 2NaCl Potassium dichromate being less soluble is obtained

by fractional crystallisation. (i) Cr2 O7

2– + 14 H+ + 6I– –→ 2Cr3+ + 7H2 O + 3I2 (ii) Cr2 O7

2– + 4Fe2+ + 14 H+ –→ 2Cr3+ + 6Fe3+

+ 7H2 O Uses the potassium dichromate : (i) In volumetric analysis for the estimation of Fe2+

and I– ions. (ii) In chrome tanning in leather industry. 30. (i) At elevated temperatures, sulphur vapours

exists, as S2 molecules which are paramagnetic like O2.

(ii) This is due to reluctance of silicon to form pπ – pπ multiple bonds because of large size of silicon atom. Hence, silicon exists only in diamond structure.

(iii) Xe has relatively lower ionization energy among inert gases and thus the outermost shell electrons of Xe are excited to d-subshell and thereby showing unpaired electronic structure.

(iv) Nitrogen shows a little tendency for catenation, due to weakness of N – N single bond whereas phosphorus shows a clear tendency for catenation due to its unexpectedly high bond energy.

MATHEMATICS

Section – A

1. Let (a, b) ∈ R [Q (1, 2) ∈ R ] ∴ (b, a) ∈ R [Q (2, 1) ∈ R ] Hence R is symmetric.

2. ∫ −− = xtan2

xdxxtan.x 12

1 – ∫ +dx

1x1.

2x

2

2

(Integrating by parts)

= ∫ +−+

−− dx1x

11x21xtanx

21

2

212

= ∫

+−

1−− dx

1x11

2xtanx

21

212

= .C)xtanx(21xtanx

21 112 +−− −−


3. The given curves are x2 + y2 = 2ax ….(1) Here, a is arbitrary constant Diff. (1) w.r.t. x

2x + 2y a2dxdy

=

Substituting for 2a in (1), we get

x2 + y2 = x

+

dxdyy2x2

⇒ 2xydxdy + x2 – y2 = 0

which is the reqd. diff. eq. 4. Let x = tan–1 (–1) tan x = –1

tan x = – tan 4π

tan x = tan

π

−π4

[Q tan(π – θ) = –

tanθ]

tan x = tan 4

3π

x = 4

3π

∴ the principle value of tan–1 (–1) is 4

3π

5. A =

2013

and B =

−3012

2A – B + C = 0 ⇒ C = –2A + B

⇒ C = –2

2013

+

−3012

⇒ C =

−+

−−−

3012

4026

C =

−−−

1018

6. Q | adj A | = | A | n – 1 64 = | A | n – 1 64 = | A |3 – 1 Order of matrix n=3 | A |2 = 64 | A | = 8

7. Let A =

−−

x110723534

Q the matrix A is singular

∴ | A | = O ⇒

−−

x110723534

⇒ 4 (–2x + 7) –3 (3x – 70) + 5 (–3 + 20) = 0 ⇒ x = 19

8. →→→

++ cba = ikkjji +++++ = 2 ( kji ++ )

→→→

++ cba = 1112 ++ = 32

∴ unit vector = |cba|

cba→→→

→→→

++

++ = 32

)kji(2 ++

= 3

kji ++

9. Here |a|→

= 3 and |b|→

= 2 ; →→b.a = 3

cos θ = |b||a|

b.a→→

→→

= 2.3

3 = 23

θ = 3π

10. Let the equal angle = x ∴ l = cos α ; m = cos α and n = cos α ∴ l2 + m2 + n2 = 1 cos2 α + cos2 α + cos2 α = 1 ⇒ 3cos2α = 1

⇒ cos α = 3

1± Hence direction cosines

3

1± ,

31

± , 3

1±

Section – B

11. We have f(x) = 2x, g (y) = 3y + 4 and h (z) = sin z L.H.S. = ho (gof) (x) = h ( g ( f ( x )) = h ( g (2x)) = h (3 (2x) +4) = h (6x + 4) = sin (6x + 4) R.H.S. = (hog) of (x) = (hog) (f (x)) = (hog) 2x = h (g (2x) = h (3 (2x) +4) = h (6x + 4) = sin (6x + 4) ∴ L.H.S. = R.H.S.

12. Let y = cot–1

+−

x1x1

Put : x1x1

+− = u ...(1)


∴ y = cot–1 u Diff. w.r.t. u

22

x1x11

1u11

dudy

+−

+

−=

+−

=

= )x1(2

)x1()x1()x1(

)x1(2

2

22

2

++

=−++

+− ...(2)

Diff. (1) w.r.t. x

22 )x1(2

)x1(1).x1()1).(x1(

dxdu

+=

+−−−+

= …(3)

From (2) and (3),

dxdu

dudy

dxdy

⋅=

= – .x1

1)x1(

2)x1(2

)x1(22

2

+=

+−

⋅+

+

13. The given diff. eq. is (1 + e2x) dy + ex (1 + y2) dx = 0 ...(1)

⇒ 0dxe1

ey1

dyx2

x

2=

++

+

Integrating, we get

tan–1 y + ∫ =+

ct1

dt2

(where t = ex)

⇒ tan–1y + tan –1 t = c ⇒ tan–1y + tan –1 ex = c when x = 0, y =1 ⇒ tan–1 + tan–1 e° =c

⇒ 244

c π=

π+

π=

tan–1y + tan–1 ex = 2π .

This is the reqd. sol of (1) OR Sol. The given diff. eq. is

x .0x2ydxdy 3 =−−

⇒ 2x2yx1

dxdy

=− ...(1)

This is a linear diff. eq.

On comparing by, QPydxdy

=+

Here, P = – x1 , Q = 2x2

I.F. = xlogdxx1

Pdxeee −−

== ∫∫

= 1xloge

− = 1x −

∴ The reqd. sol. of eq. (1) is

y . x–1 = ∫ +− cdxx.x2 12

= ∫ + cdxx2 = x2 + c

⇒ y = x3 + cx.

14. ∫π 4/

0

dxx3sinx2sin

= 21

∫π 4/

0

dx)x2sinx3sin2(

= 21

∫π

−4/

0

)x5cosx(cos

= 21 4/

05x5sinxsin

π

−

= 21

−

π−

π 04

5sin51

4sin

= 25

32

1511

21

4sin

51

4sin

21

=

+=

π

+π

OR

Sol. Let I = ∫π

+4/

0

dx)xtan1log( …..(1)

By the formula, ∫∫ −=a

0

a

0

dx)xa(fdx)x(f

= ∫π

−

π+

4/

0

dxx4

tan1log

= ∫π

+−

+4/

0

dxxtan1xtan11log

= ∫π

+

4/

0

dxxtan1

2log

= ∫π

+−4/

0

dx)]xtan1log(2[log

= ∫ ∫π π

+−4/

0

4/

0

dx)xtan1log(dx2log

= log2 [ ] Ix 4/0 −π [By (1)]

∴ 2I = log2

−

π 04

⇒ I = 2log8π


15 Let I = ∫ +−

+ dx3x2x2

1x32

Here : 3x + 1 = 43 .

dxd (2x2 – 2x + 3) +

25

∴ I = ∫ +−

+−dx

3x2x225)2x4(

43

2

= ∫ ∫ +−+

+−

−

3x2x2dx

25dx

3x2x22x4

43

22

= cI25I

43

21 ++ …(1)

where ∫ +−

−= dx

3x2x22x4I

21

Put : 2x2 – 2x + 3 = t ⇒ (4x – 2)dx = dt

∴ ∫ == |t|logt

dtI2

= log | 2x2 – 2x + 3 | ....(3)

and ∫+−

=

23xx2

dxI2

2 = ∫

+

−

22

25

21x

dx21

=

−⋅ −

2/521x

tan)2/5(

121 1

=

+∫ −

axtan

a1

axdx,formulatheBy 1

22

=

−−

51x2tan

51 1 …(3)

From (1), (2) and (3), we get

c5

1x2tan25|3x2x2|log

43I 12 +

−++−= −

16 Given : x = a (θ – sin θ); y = a (1 – cosθ)

⇒ θd

dx = a (1– cos θ); θd

dy = a sin θ

⇒ )cos1(a

sinad/dxd/dy

dxdy

θ−θ

=θθ

= = cot 2θ .

Now dxd.

2cot

dd

dxyd2

2 θ

θ

θ=

= –21 cosec2

)cos1(a1.

2 θ−θ

= 2

eccosa4

1 4 θ− .

At θ =2π ;

π

−=4

eccosa4

1dx

yd 42

2= –

a1 .

17 Q f(x) is continuous at x = π ∴ )(f)x(flim)x(flim

xxπ==

+π→−π→

xcoslim1Kxlimxx +π→−π→

=+

K · π + 1 = cos π π K + 1 = – 1 π K = –2

K = π−2

OR Sol. Since the function is defined at x = 0 33xlim)x(flim 3

0x0x=+=

→→

Also f(0) = 1 ∴ )0(f)x(flim

0x=

→

Hence f(x) continuous at x = 0.

18. tan–1

42x1xtan

2x1x 1 π

=

++

+

−− −

4

2x1x.

2x1x1

2x1x

2x1x

tan 1 π=

++

−−

−

++

+−−

−

4

tan

)2x)(2x()1x)(1x()2x)(2x(

)2x)(2x()2x)(1x()2x)(1x(

π=

+−+−−+−

+−−+++−

11x4x

2xx2xx22

22=

+−−−−+−+

=

π 14

Q

13

4x2 2=

−−

2x2 – 4 = –3 2x2 = –3 + 4 = 1 2x2 = 1

x2 = 21

⇒ x = ± 2

1

19. A =

−−

−−

7321311154

| A | = 4 (21 – 3) + 5 (–7 – 2) –11 ( 3 + 6) = 72 – 45 – 99 = – 72

Adj. A =

T

71538226689918

−−−−−−


∴ adj A =

−−−−−−

72291569386818

∴ A–1 =

−−−−−−

−=

72291569386819

721

AadjA

OR

Sol. ∆ =

++

+

zayxzyaxzyxa

operate R1 → R1 – R2

∆ =

++

−

zayxzyaxoaa

operate C2 → C2 + C1

∆ =

++++

zaxyxzxyaxooa

∆ = a [(a + y + x) (a + z) – z (y + x)] = a [a2 + az + ( y + x) a + (y + x ) z – z (y + x)] = a2 ( a + x + y + z) Proved. 20. The given plane is 3x + 2y + 2z + 5 = 0 ...(i) line through P (2, 3, 4) and parallel to the line :

2z

62y

33x

=−

=+ is

k2

4z6

3y3

2x=

−=

−=

− (say) ….(ii)

Any point on it is Q (3k +2, 6k + 3, 2k + 4) Let it lie on (i) ∴ 3(3k + 2) +2 (6k +3) +2 (2k + 4) +5 = 0 ⇒ 25 k + 25 = 0 ⇒ k = – 1 ∴ Q (–1, –3, 2) ∴ The required distance = PQ

= 749)24()33()12( 222 ==−++++

21. Q |a|→ = |b|

→ = 1 (Given)

|→a +

→b |2 = |

→a |2 + |b|

→2 +2

→→ba

= 1 + 1 + 2 |b||a|→→

cos θ

= 2 + 2 (1) (1) cos θ = 2 (1 + cosθ)

= 2 . 2 cos2 2θ

2|ba|→→

+ = 4 cos2 2θ

|ba|→→

+ = 2 cos 2θ

cos2θ =

21 |ba|

→→+ Proved.

22. p = P (correct forcasting) = 1/3 q = P (two incorrect forecasting) = 2/3 n = 4 Let r be the number of correct forecast P (at least three correct results) = P ( r = 3) + P (r = 4)

= 4C3 444

44

334

31

32C

31

32

+

−−

[Q P (r) = nCr qn– r Pr]

= 4C1

40

04

31

31

32C

31

32

+

= 4 · 32 ·

271 +1 · 1 ·

811

= 91

819

811

818

==+

OR Sol. Let A, B and C be the events of candidate going by

bus, scooter and other means of transport. Let E be the event of getting late.

P (A) = 103 , P (B) =

101 , P (C) =

53

P (E/A) = 41 , P(E/B) =

31 , P(E/C) = 0

P (that he traveled by bus) = P (A/E)

= )C/E(P)C(P)B/E(P)B(P)A/E(P)A(P

)A/E(P)A(P++

= 0·

53

31·

101

41·

103

41·

103

++ =

301

403

403

+

= 139

13120

403

12049

403

=×=+

.

Section – C

23. Let I = dx)xsin2)(xsin1(

xcos2/

0∫

π

++

Put : sin x = t ⇒ cos x dx = dt

Also x = 0 ⇒ t = 0 and x = 2π ⇒ t = 1


∴ ∫ ++=

1

0)t2)(t1(

dtI = dtt2

1t1

11

0∫

+−

+

(Resolving into partial fractions) = [log | 1 + t | – log | 2 + t |] 1

0 = (log 2 – log 3) – (log 1 – log 2) = 2 log 2 – log 3 = log 22 – log 3

= log 34 .

24. x2 +y2 = 16 …(1) y2 = 6x. ...(2) (1) and (2) intersect, where

Y

Y'

X X' O D C

x = 4 x = 2

B (2, –2 3 )

x = 0

A (2, 2 3 ) 1

2

x2 + 6x – 16 = 0 ⇒ (x + 8) (x – 2) = 0 ⇒ x = 2 (Q x ≠ –8) Q A (2, 2 3 ) and B (2, –2 3 ) Also C (4, 0). Area OBCAO = 2 (Area ODA + Area DCA)

=

+∫ ∫

2

0

4

212 dxydxy2

=

−+∫ ∫

2

0

4

2

2 dxx16dxx62

=

+−

+

−

4

2

122

0

2/3

4xsin

216

2x16xx

32.62

=

+−++ −−

21sin8

232.21sin8022.

3622 11

= )6

.1634(2

.163

316 π+−

π+

=

π+

316

334 sq. units.

∴ Reqd. Area = Area of circle –

π+

316

334

= 16π – π−3

163

34

= )38(34

334

332

−π=−π sq. units.

OR

Sol. 1by

ax

2

2

2

2=+ ….(1)

1by

ax

=+ …..(2)

XX'

Y

Y'

A (a, 0)

B (0, b)

1

2

x = a x = 0

O

We shall find the shaded area (Area of the smaller region)

= ∫ −a

021 dx)yy(

= ∫

−−−

a

02

2dx

ax1b

ax1b

= ∫ ∫

−−−

a

0

a

0

22 dxax1bdxxa

ab

= a

0

2a

0

1222

a2xxb

axsin

2a

2xax

ab

−−

+

− −

=

−−

−

a2aab1sin

2a

ab 2

12

= units.sq)2(ab41

2ab

2.

2ab

−π=−π

25. Let S, V, r and h be the surface area, volume, radius of the base and height of the given cylinder respectively. Then

S = 2πrh + πr2 (Given) [Q Cylinder is open at the top]

⇒ h = r2rS 2

ππ−

V = πr2h = r2rSr

22

ππ−

π = ]rSr[21 3π−


Diff. w.r.t r

drdV = r3

drVdand]r3S[

21

2

22 π−=π−

For max. or min.,

drdV = 0

⇒ S – 3πr2 = 0

⇒ r = π3S

For this value of r, 0dr

Vd2

2<

∴ V is maximum and

h = rr2

rr3 22=

ππ−π [Using (1) and (2) ]

26. Sol. Part-I

Given matrix A =

−

−

111312

111

| A | = 111312

111−

−

= 1 (1 + 3) + 1 ( 2 + 3) +1 ( 2 – 1 ) = 4 + 5 + 1 = 10 ∴ A–1 exists C11 = (–1)2 (1 + 3) = 4 C12 = (–1)3 (2 + 3) = –5 C13 = (–1)4 (2 – 1) = 1 C21 = (–1)3 (–1 –1) = 2 C22 = (–1)4 (1 – 1) = 0 C23 = (–1)5 (1 + 1) = –2 C31 = (–1)4 (3 – 1) = 2 C32 = (–1)5 (–3 –2) = 5 C33 = (–1)6 (1 + 2) = 3

A–1 =

−−=

321505224

101Aadj

A1

Part II The given equation can be written as AX = B ⇒ X = A–1 B

Where A =

=

=

−−

204

Bzyx

x131111121

∴ A–1 =

−

−

352202

154

101 [Using part I]

−

−=

204

352202

154

101

zyx

=

=

++−++−

144

18

101

6084082016

101

∴ ,59

1018x == ,

52

104y ==

57

1014z == .

27. Let A and B the events of getting letter from Tata

Nagar and Calcutta

∴ P(A) = 21 , P (B) =

21

Let E be the event of visibility of letter TA

P (E | A) = 2

72

= 1.2.3.4.5.6.7

2.2 = 1260

1

[Q Total no. of events in Tata Nagar = 2

7as

TA TA NAGAR has only 2 A's ]

P (E | B) = 2

71

= 1.2.3.4.5.6.7

2 = 2520

1

[Q Total no. of events in Calcutta = 2

7as

TACALCUT has only 2 C's ] (i) P (that letter has come from Tata Nagar)

= (A | E) = )B|E(P)B(P)A|E(P)A(P

)A|E(P).A(P+

=

25201·

21

12601·

21

12601·

21

+=

252012

12601

+

= 32

32520

12601

=×

(ii) P (that letters has come from Calcutta) = P (B | E)

= 1 – P (A | E)

= 1 – 32 =

31

OR Sol. White balls = 4 Red balls = 6 Total balls = 4 + 6 = 10 Let X be the number of drawing 3 white balls


∴ X = 0, 1, 2, 3

P (X = 0) = 3

103

60

4

CC·C = 1 .

305

9015

8.9.101.2.3.

1.2.34.5.6

==

P (x = 1) =3

102

61

4

CC·C = 4 ·

3015

9045

8.9.101.2.3.

1.25.6

==

P (x = 2) 3

101

62

4

CC·C

= 309

9027

8.9.101.2.3.6

1.23.4

==

P (X = 3) = 3

100

63

4

CC·C

=

301

903

8.9.101.2.3.1.

1.2.32.3.4

==

Therefore, required probability distribution is X 0 1 2 3

P (X) 305

3015

309

301

Calculation for mean and variance

X P(x) XP(x) X2P(x)

0 305 0 0

1 3015

3015

3015

2 309

3018

3036

3 301

303

309

Total 1 56

3036

= 23060

=

Mean µ = Σ X P(x) = 56 = 1.2

Variance = Σ X2 P (x) – [ Σ X P (X)]2

= 2 – (1.2)2 = 2 – 1.44 = 0.56 28. Equation of the given line is 2x = y = z

2z

2y

1x

== …..(1)

Equation of line passes through the point A (3, 4, 5) and parallel to line ( 1 ) is

)say(2

5z2

4y1

3xλ=

−=

−=

− …..(2)

∴ Any point on the line (2) is P (λ + 3, 2λ + 4, 2λ + 5) Since the point P lies on the plane x + y + z = 2 ….(3) ∴ (λ + 3) + (2λ + 4) + (2λ + 5) = 2 λ + 3 + 2λ + 4 + 2λ + 5 = 2

5λ = 2 – 12 5λ = –10 ⇒ λ = –2 Putting the value of λ, we get. The coordinate of P (–2 + 3, –4 + 4, –4 + 5) = P (1, 0 1) The required distance = |AP|

= 222 )15()04()13( −+−+−

= 16164 ++

= 36 = 6 unit. 29. Given problem can be tabulated as

Proteins Carbohydrates Cost Wheat 0.01 0.025 Rs. 4kg = 0.4 P/g Rice 0.05 0.5 Rs. 6 kg = 0.6 P/g

Min50g Min200 g Let the quantity of wheat = x gms and the quantity of rice = y gms Min cost z = 0.4x + 0.6y subject to constraints 0.1x + 0.05y ≥ 50 0.25x + 0.5y ≥ 200 x, y ≥ 0 Table for 0.1x + 0.05y = 50

x 0 1000 y 500 0

Table for 0.25x + 0.5y = 200

x 0 800 y 400 0

200 400 600 800 1000

X

1000

800

600

400

200

(1000, 0)

(0, 500)

The corner points of feasible region is (0, 500) and

(1000, 0) Now evaluate z at the corner points

Corner point Z = 0.4x + 0.6y (0, 500) (1000, 0)

Z = 0 + 300.0 = 300 ← Min Z = 400.0 + 0 = 400

Minimum cost = 300 paise = Rs. 3. when x = 0 gms, y = 500 gms.


XtraEdge Test Series ANSWER KEY

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans D B A A D D A B A C ,D Ques 11 12 13 14 15 16 17 18 19 20 Ans A,D A,D A ,C ,D A ,B ,D C B A C D B 21 A → P B → Q C → R D → Q 22 A → Q B → Q C → R D → Q,S

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 Ans A B A B A B B C A B ,D Ques 11 12 13 14 15 16 17 18 19 20 Ans B A,B ,C ,D C A,B ,C A D D A B A 21 A → R B → P C → Q D → R 22 A → Q B → P C → S D → R

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans B B A B D D C B A A,B ,C ,D Ques 11 12 13 14 15 16 17 18 19 20 Ans A,B ,C ,D A,C A,C A,B ,C A C B C D C 21 A → Q B → P C → S D → R 22 A → R B → S C → Q D → P

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans A D B B D C A A A B ,D Ques 11 12 13 14 15 16 17 18 19 20 Ans A,C A,C A,C A ,C ,D D A A D D B 21 A → Q B → P C → R D → S 22 A → S B → S C → P D → P,Q,R

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 Ans A B B D D D A B B A,D Ques 11 12 13 14 15 16 17 18 19 20 Ans A,C B A,B A,B ,C ,D A A D B B B 21 A → S B → P C → Q D → R 22 A → R B → P,S C → Q D → Q

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans D C A B D B A A B A,B ,C Ques 11 12 13 14 15 16 17 18 19 20 Ans A,C B ,C A ,B ,C B ,C A C B A C B 21 A → S B → Q C → P D → R 22 A → R B → P C → S D → Q

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