Pace Test Paper

CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA

ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI IIT JEE - 2013 FULL TEST 2 MARKS: 238 (ADVANCED PATTERN) PAPER - I

SECTION - I PHYSICS

PART I: Single Correct Answer Type This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (+3, 1)

1. Three charge q, Q and 4q are placed in a straight line of length l at points distance 0, 2l and l

respectively from one end. In order to make the net force on q zero, the charge Q must be equal to

(a) q (b) 2q (c) q2 (d) q

2. A point charge q is placed at the midpoint of a cube of side L. The electric flux emerging from the

cube is

(a) 0

q

(b) 20

q6L

(c) 2

0

6qL

(d) ZERO

3. The charge flowing through a resistor R varies as 2Q t t t . The total heat produced in R is

(a) 3R

(b) 3R

2

(c) 3R

3

(d) 3R

6

4. A 2.0 V potentiometer is used to determine the internal resistance of a 1.5 V cell. The balancing

point of the cell in open circuit is 75 cm. When a resistor of 10 is connected across the cell, the balance point shifts to 60 cm. the internal resistance of the cell is

(a) 1.5 (b) 2.5 (c) 3.5 (d) 4.5 5. A current carrying wire RS is placed near an another long current carrying

wire PQ. If free to move, wire RS will have (a) translational motion only (b) rotational motion only (c) translational as well as rotational motion (d) neither translational nor rotational motion 6. A long straight wire along the z-axis carriers a current I in the negative z

direction. The magnetic vector field B

at a point having coordinates (x, y) on the z = 0 plane is

(a) 0 2 2 I yi xj

2 x y

(b) 0 2 2 I xi yj

2 x y

(c) 0 2 2 I xj yi

2 x y

(d) 0 2 2 I xi yj

2 x y

1I2I

P

Q

R S


7. A proton, a deutron and an -particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If p dr , r and r denote respectively the radii of the trajectories of these particles, then

(a) P dr r r (b) d pr r r (c) d pr r r (d) p dr r r 8. In the arrangement shown in the figure AC and BD are straight lines and CED and AFB are

semicircular with radii r and 4r, respectively. The entire setup is lying in the same plane. If I is current entering at A what fraction of I will flow in the ACEDB such that resultant magnetic field at O is zero

(a) 1 (b) 35

(c) 45

(d) 15

PART II: Multiple Correct Answer(s) Type This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) or (D) out of which ONE or MORE are correct. (+4, 1) 9. A parallel plate capacitor having capacitance C has two plates X and Y each having a charge Q. The

plate X is now connected to the positive terminal of a battery and the plate Y to the negative terminal

of the same battery of emf QEC

. Then it is observed that

(a) an energy 2CE is supplied by the battery. (b) Q amount of charge flows, through the capacitor, from the positive terminal to the negative terminal of the battery. (c) the total charge on the plane X is 2Q. (d) the total charge on the plate Y is zero. 10. Consider a conducting sphere A having radius a, charge Q placed concentrically inside a conducting

shell B having radius b (> a) and earthed. Let C is the common centre of the sphere A and the shell B. If E and V be the respective electrostatic field and potential at a distance r a r b from C and

V be the electrostatic potential difference between the two shells, then

(a) E 0 (b) 20

1 QE4 r

(c)

0

Q 1 1V4 a b

(d)

0

Q 1 1V4 r b

A BC DE

O I

F

4r

r

AB

C

b

aQ


11. Three ammeters 1 2 3A ,A and A having resistance 1 2 3R ,R and Rrespectively are connected between two points X and Y, such that 1Aand 2A are in series which in turn are in parallel with 3A . When some potential difference is applied across the terminals X and Y, the respective readings of the ammeters are 1 2 3I , I and I . Which of the following mathematical relations support the argument given here?

(a) 1 2I I (b) 1 1 2 2 3 3I R I R I R (c) 323 1

RII R

(d) 313 1 2

RII R R

12. A current I is being driven through a battery of emf E and internal

resistance r, as shown. Then which of the following statement(s) best suit the above mentioned condition?

(a) The battery absorbs energy at the rate of EI. (b) The battery stores chemical energy at the rate of 2EI I r . (c) The potential difference across the battery is E Ir . (d) Some heat is produced in the battery.

PART III: PARAGRAPH TYPE This section contains 3 multiple choice questions relating to ONE paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (+3,0)

PARAGRAPH I

Two capacitors 1C 1 F and 2C 3 F are each charged to a potential 0V 100V but with opposite polarity, as shown. Switches

1S and 2S are now closed. Based on the above facts, answer the following questions.

13. What is the potential difference between points e and f? (a) 25 V (b) 50 V (c) 75 V (d) 100 V 14. What is the final charge on 1C ? (a) 50 C (b) 75 C (c) 150 C (d) 300 C 15. What is the final charge on 2C ? (a) 50 C (b) 75 C (c) 150 C (d) 300 C

2A1A

3A

X Y

I + E,r

a

b c

de

f

2C1C

1S

2S


PART IV: Integer Answer Type

This section contains 4 questions. The answer to each question is a single/two digit integer, ranging from 0 to 99 (both inclusive) (+5,1) 16. A capacitor of capacitance 1C 1 F withstands the maximum voltage 1V 0.6 kV while another

capacitor of capacitance 2C 2 F withstands the maximum voltage 2V 4 kV . What maximum kilovolt will the system of these two capacitors withstands if they are connected in series?

17. A parallel plate capacitor of plate area 3A 10 metre 2 and plate separation 2d 10 m is charged to

0V 100 V . Then after removing the charging battery, a slab of insulating material of thickness 2b 0.5 10 m and dielectric constant K = 7 is inserted between the plates. Calculate the free charge,

in pC, on the plates of the capacitor, electric field intensity, in 1kVm , in the dielectric, potential difference, in volt, between the plates and capacitance, in pF, with dielectric present.

18. A box contains a inductor of inductance L milli henry, a capacitor of capacitance C microfarad and a

resistor of resistance R ohm. When a 250 V dc is applied to the terminals of the box, a current of 1 A flows in the circuit. When an ac source of 250 V rms flows. It is observed that the current rises with frequency and becomes maximum at 4500 rads 1 . Find the values of C.

19. A coil of 15 turns and radius 10 cm surrounds a long solenoid of radius 2 cm and 1000 turns per

metre as shown. The current in the solenoid changes as I (in ampere) = 4 cos (250t). Find the induced peak emf, in millivolt, in the 15 turn coil as a function of time. (Take 2 10 )

~

R

I

15 turn coil


PART V: Matrix Match Type

This section contains 1 question. Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column - II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column- II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example: If the correct matches are A - p , s and t ; B - q and r ; C - p and q ; and D - s and t. (+8, 1) 20. Column 1 Column 2

J

K L

M

10 20 V

P

10

20

30

(a) JK process (P) W > 0 (b) KL process (Q) W < 0 (c) LM process (R) Q > 0 (d) MJ process (S) Q < 0

PAPER I SOLUTION 1. (a) netF 0

2

2 2Qq 4qk k 0

2

where 0

1k4

24Qq 4q 0 Q q 2. (a) According to Gauss Theorem

Flux 0

q

3. (d)

dQIdt

I 2 t

Since at 0t t 2

, I becomes zero. Hence we shall calculate the heat produced from

t 0 to 0t 2

t0

2

0H I R dt

2

2

0H R 2 t dt

2 1 2

0

2 tH R

6

3

H R 06

3RR

6

4. (b)

Since

r R'

l ll

75r 1 1060

15R 1060

5R 2.52

5. (c)

Since, the magnetic field due to wire PQ is non uniform. Neither the force nor the torque on wire RS is zero. Therefore, it will have both translational as well as rotational motion.

q Q 4ql/2 l/2

RP

Q

S

1F

2F

3F

1I

2I

6. (a) 7. (a)

Radius of the circular path is given by mc 2KmrBq Bq

Here, K is the kinetic energy to the particle.

Therefore, mrq

if K and B are same.

p d1 2 4r : r : r : : 1: 2 :1

1 1 2

Hence, p dr r r 8. (d)

If 1I is current in ACDB then 0 01 2I I

4 r 4 4r

11

2

I 1 III 4 5

1I 1I 5

9. (a, b, c, d)

10. (b, c, d)

11. (a, b, d)

12. (a, b, c, d) 13. (d) Let 1V and 2V be the voltage across 1 and 2 F capacitors respectively. Then 1 2V V 300V

In series the charge Q on each capacitor is same, therefore we have 1 2

Q Q 300C C

1 2

1 1 Q 300C C

Here 1 2C 1 F,C 2 F

6 6

1 26 61 2

300 1 10 2 103000C CQC C 1 10 2 10

6Q 200 10 C 200 C 11

Q 200 CV 200VC 1 F

22

Q 200 CV 100 VC 2 F

14. (c)

When the plates of same polarity are connected together, the capacitors are in usual parallel combination. In parallel the potential difference across each capacitor is the same The common potential difference

1 2

1 2 1 2

Q Q 2QVC C C C

2 200 C 400 C 400V V1 F 2 F 3 F 3

1Q'

2Q '

1C

2C

V '

+ -

If '1Q and '2Q are new charges on the respective capacitors, then

'1 1400 400Q C V 1 F V C3 3

'2 2400 800Q C V 2 F V C3 3

15. (a)

When the opposite plates of SITUATION(I) are connected together, the charge will flow from the capacitor of higher potential to that of lower potential until their potentials are same. The net charge is then 1 2Q Q If V '' is common potential difference across the capacitors, then

1 21 2

200 F 200 CQ QV" 0C C 1 F 2 F

The charge after sharing

"1 1 1Q C V" C 0 0 "2 2 2Q C V" C 0 0

16. (9) Given, 1 2V 6kV,V 4kV Charge on first capacitor

31 1 1q C V 1 F 6 10 V 1q 6000 C Charge on second capacitor

2 2 2q C V 32q 2 F 4 10 V 8000 C In series the charge on each capacitor remains same, but maximum charge on first capacitor will be 6000 C . Therefore charge on second capacitor must also be 6000 C . The potential across second capacitor.

2

2

q 6000 CV 3000V 3kVC 2 F

So, maximum voltage across system will be 1 2V V 6kV 3kV 9kV

17. (16) The capacitance 0C before the slab is introduced

12 2 120

0 2

8.9 10 10AV 8.9 10 faradd 10

free charge

200 V

1C

2C

100 V

++++

----

----

++++

Charge Flow

(a) (b)

2Q"

1Q"

2C

1C

++

--

++

--

100 V"

1C 2C

2V

1V

q

12 120 0q C V 8.9 10 100 890 10 C 890pC Now, Electric field intensity

4 1 100 2V 100E 1 10 Vm 10kVmd 10

Electric field intensity in dielectric

4

3 1 10E 1 0E 1.43 10 Vm 1430VmK 7

1

Potential difference between the plates with dielectric present is given by 0V E d b Eb

4 2 2 3 2V 1 10 10 0.6 10 1.43 10 0.5 10 V 57V The free charge on the plate is the same as before. The capacitance with dielectric present is

10

12q 8.9 10 CC 16 10 farad 16pFV 57C

18. 1

Since, the capacitor is a dc blocking element, but still the current in the circuit for a dc source is not zero, so all the elements must not be series. Further in case of ac, current rises with frequency and has a maximum value at 14500rads , L and C should be in series. The circuit diagram therefore, be as shown in figure. Case I : When dc is applied, we have

V 250R 250I 1

Case II: When AC is applied, then for 12250rads , let the applied voltage be

0V V sin t . Then 0RVI sin tR

R250 2I sin t 2 sin t

250 rmsas V 250 volt

and 0L CC L

VI I sin tX ~ X 2

{ ' ~ ' indicates positive difference between LX and CX }

0LC L

VI cos tX X

So, the total current through the circuit at any instant is

0 0R CC L

V VI I I sin t cos tR X X

0I I sin t

L C

250 V

I

R R

~

L C

RI

CASE I : When dc is applied CASE II : When ac is applied

RI

where

0 0 2 2C L

1 1I VR X ~ X

rms rms 2 2C L

1 1I VR X ~ X

Given that, rms rmsV 250 volt, I 1.25A

2

2 2C L

1 1 1.25250R X ~ X

2 2

2C L

1 1.25 1250 250X ~ X

C L1000X ~ X

3

1 1000~ LC 3

..(1)

Current in the circuit will be maximum at,

L CX X 00

1LC

01LC

2 20

1 1LC4500

.(2)

Solving equations (1) and (2) with 12250rads , we get

6C 10 F 1 F and L 0.049H So, C 1,L 49 and R 250

19. (24) Since 0 solenoidBA nI A

20 solenoidn r I , where n 1000 According to Faradays Laws, we have

20 solenoidNd dIN n rdt dt

, where N 15

Since I 4cos 250 t dI 1000 250tdt

2715 4 10 1000 0.02 1000 sin 250t 2 4 160 4 10 10 sin 250t 5600 4 10 sin 250t

52400 10 sin 250 t

324 10 sin 250t volt

peak 24 mV

20. a S ; b P, R ;c R ;d Q,S


SECTION - II CHEMISTRY

PART I: Single Correct Answer Type This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (+3, 1) 1. The correct order of bond angles of the XNX bond (where X is the surrounding atom or group of atoms) (a) 4NH >N(CH3)3> NH3> NF3 (b)

4NH < N(CH3)3< NH3< NF3

(c) NH3> NF3> 4NH >N(CH3)3 (d) NH3> NF3> N(CH3)3>4NH

2. Which of the following compound responds to iodoform reaction? (a)

O C2H5 C C2H5

(b) 3 2 2CH CH CH COOH

(c) O

CH3 C CH2 C OC2H5 O

(d) Br

CH3 CH CH3

3. Which one of the following is not correct systematic name of corresponding compound? (a) NaMn(CO)5: Sodium pentacarbonylmangnate(I) (b) SnCl4(Et2NH)2: Tetrachlorobis(diethylamine)tin(IV) (c) Ni(CO)2(PPh3)2: Dicarbonylbis(triphenylphosphine)nickel(0) (d) Na2[Fe(CN)5NO]: Sodium pentacyanonitrosylferrate(II) 4. If o is the threshold wavelength for photoelectric emission, the wavelength of light falling on the surface of

a metal and m is the mass of the electron, then the velocity of ejected electron is given by

(a) 2/1

o )(mh2

(b) 2/1

o )(mhc2

(c) 2/1

o

o

mhc2

(d) 2/1

o

11mh2

5. The vapour pressure of the distillate obtained on condensation of the vapour formed above a solution

containing 2 mole of a liquid A ( oAP = 100 mm of Hg) and 3 mole of a liquid B (oBP = 150 mm of Hg) at same

temperature would be (a) 130 mm of Hg (b) 135 mm of Hg (c) 140 mm of Hg (d) 145 mm of Hg 6. An inorganic salt (A) on reaction with H2O gives a solution (B) and a gas (C). Gas (C) on passing into

ammonical AgNO3 gives white precipitate of (D). CO2 gas turns solution (B) milky. The Compound (A) would be

(a) BaSO4 (b) Na2SO4 (c) CaC2 (d) CaCO3 7. In the estimation of nitrogen by Dumas method 1.18 g of an organic compound gave

224 ml of N2 at NTP. The percentage of nitrogen in the compound is about (a) 20 (b) 11.8 (c) 47.5 (d) 23.7 8. Aluminium is not present in the mineral (a) Feldspar (b) Fluorspar (c) Cryolite (d) Mica


PART II: Multiple Correct Answer(s) Type This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) or (D) out of which ONE or MORE are correct. (5, 2) 9. Which of the following statements regarding phenols is correct? (a) Phenols are stronger acid than water and alcohols. (b) Phenols are weaker acids than carboxylic acids. (c) Phenols are soluble in both aqueous NaOH and aqueous NaHCO3. (d) Phenoxide ions are better resonance stabilised than the corresponding phenols. 10. When H2S gas is passed separately through an acidic solution of CuCl2, HgCl2, BiCl3 and CoCl2, which of the

following will precipitate out (a) CuS (b) HgS (c) Bi2S3 (d) CoS 11. [Ni(CN)4]2 and [NiCl4]2 have dissimilarity in (a) magnetic moment (b) coordination number and oxidation states (c) structure (d) colour 12. Which of the following is correct under the same conditions of pressure and temperature? (a) H2 gas diffuses four times faster than oxygen gas (b) H2 gas diffuses 2.83 times faster than CH4 gas (c) He escapes at a rate 2 times as fast methane does (d) He escapes at a rate 4 times as much as sulphurdioxide

PART III: PARAGRAPH TYPE This section contains 3 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (+4, 1)

PARAGRAPH FOR QUESTIONS 13 TO 15 The aldehydes not having -hydrogens react in presence of strongly basic medium to give cannizzaro reaction. The mechanism followed is: Step I

OHHH

O

H

O-

OH

H

Step II

H

OH

O

H

O-

OH

H H

Step III

H HOH

O

H

O-

OH

H

Step IV

HOH

O

H

O-

OH

H Pr oton3transfer HCOO CH OH

Second step is rate determining step.


13. Which among the following can give cannizzaro reaction?

(a) CH3

O

(b)

O

(c) O

CH3

(d)

O

14. OHStrongly basic

O||

H C H Ph CHO P

Among the products alcohol is?

(a) CH3 OH (b) Ph

OH

(c)

OH

(d) HOH

O

15. Though CCl3CHO has no -hydrogen it doesnt give cannizzaro reaction. The reason is:

(a) There is no H attached to

O

(b) CCl3 is a weak base (c) CCl3 is a strong base (d) None of these

PART IV: Integer Answer Type

This section contains 4 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive) (+4, 0)

16. The potential energy electron in an exited state of Be3+ is27.2 eV. Find the maximum types of photons produced when it return to ground state.

17. A compound of Vanadium has a magnetic moment 1.73BM. The oxidation state of V in the compound is. 18. For a cell Cu+2 +In+2 = Cu+ +In+3

Given E0 (Cu2+/Cu+) = 0.15V E0(In+2/In+)= -0.40 V

And E0(In+3/In+)= -0.42V The equilibrium constant is x1010. Then the value of x is_________ 19. The kinetic datas for the reaction: 2A + B2 2AB are as given below: [A] mol L1 [B2] mol L1 Rate mol L1 min1 0.5 1.0 2.5 103 1.0 1.0 5.0 103 0.5 2.0 1 102 The sum of order of reaction with respect to A and B2 will be



This section contains 1 question. Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column - II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column- II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example: If the correct matches are A - p , s and t ; B - q and r ; C - p and q ; and D - s and t. (+8, 0) 20. Column I (Chemical Tests) Column II (Acid radicals)

(A) CO2 (p) Gives black ppt. with Pb(CH3COO)2 solution (B) SO2 (q) Turns lime water milky (C) H2S (r) Turns alkaline nitroprusside solution voilet (D) 24SO

(s) Gives a white ppt. with BaCl2

(t) Smell of rotten eggs.


PAPER I (SOLUTION) 1.

N

H

H H H

10928 N

H H H

107

N

H3C

more than 107 & less than 109 as methyl group is bulkier than hydrogen and one lone pair of electrons is present.

CH3 CH3

N

F F F

Less than 107

Therefore, 4NH >N(CH3)3> NH3> NF3 (a) 2. Due to presence of active methylene group in (c), halogenation will take place at methylene carbon

(CH2) rather than at methyl carbon. In the presence of alkali (NaOH), CH3CH(Br)CH3 will be converted to CH3CH(OH)CH3 which will

respond to iodoform test. (d) 3. Correct IUPAC name of NaMn(CO)5 is Sodium pentacarbonylmangnate(I) (a)

4. 2o

mv21hchc

o

2 11hcmv21

v = 2/1

o

o

mhc2

(c) 5. Mole fraction of A in the distillate or in the vapour above the liquid mixture

= 134

15053100

52

10052

Mole fraction of B in distillate = 139

Therefore, total vapour pressure above distillate = 150139100

134

= 134.61 mm of Hg

(b) 6. CaC2 + H2O Ca(OH)2 + C2H2 (A) (B) (C) C2H2 + 2AgNO3 +2NH4OH C2Ag2 + 2NH4NO3 + 2H2O (D) (c) 7. (d)


8. The composition of ores are Feldspar KAlSi3O8, Fluorspar CaF2 Cryolite Na3AlF6, Mica = Al2O3.2SiO2.2H2O (b) 9. (a),(b),(d) 10. Comparatively cobalt sulphide is more soluble so it can not be precipitated by such a small

concentration of sulphide ion which is produced by passing H2S gas in acidic solution. (a), (b), (c) 11. In [Ni(CN)4]2 , the hybridization of nickel is dsp2 (structuresquare planar) having no unpaired

electrons while in [NiCl4]2, the nickel is present in sp3 hybridization (structure tetrahedral) with 2 unpaired electrons. Hence these two complexes differ in magnetic moment, structure and colour but have similar coordination number and oxidation states.

(a), (c), (d) 12. (a, b, c, d) 13. (d) 14. (b) 15. (b) 16. 6 17. 4 18. 1 19. 3 2.5 103 = k[0.5][1.0] (1) 5 103 = k[1.0] [1.0] (2) 1 102 = k[0.5] [2.0] (3) Dividing equation (1) and (2)

hence = 1 Dividing equation (1) and (3)

= 2

20. (A q, s), (B q, s), (C p, r, t), (D s)

21

21

0.20.1

101105.2

2

3

21

41

CENTERS : MUMBAI / AKOLA / NASHIK / PUNE / DELHI / KOLKATA / LUCKNOW / GOA

PART I : Single Correct Answer Type

This section contains 6 multiple choice questions. Each question has four choices (A),(B), (C) and (D) out of which ONLY ONE is correct. (+3, 1)

1 If [.] denotes the integral part of x and

sin sin x 1x 1

f x x1 X

, then

(A) f x is continuous in R(B) f x is continuous but not differentiable in R(C) f ' x exists x R (D) f x is discontinuous at all integral point in R.

2 If [X] represents the greatest integer less than or equal to x, then the solution of the

equation 22 4x 4 4x 6 0 is

(A) 1 3x2 4 (B)

1 3x2 4 (C)

31 x4

(D) 2, 3

3 If

2

sin x x , x 52

f x 5 b 1 , x 5x 11x 24

ab , x 5x 3

, is continuous at x 5, a, b R

then ([.] denotes greatest integer function)

(A) 25 6a , b

108 5 (B)

6 17a , b13 29

(C) 1 25a , b2 36

(D) 23 6a , b

100 5

4 If AB, BE and CF are the altitudes of a triangle ABC whose vertex A is the point 4, 5 . The coordinates of the points E and F are 4,1 and 1, 4 respectively,,then equation of BC is(A) 3x 4y 28 0 (B) 4x 3y 28 0 (C) 3x 4y 28 0 (D) x 2y 7 0

5 2 2ax 2hxy by 2gx 2fy c 0 represent two parallel lines if

(A) 2 2h ab, f bc (B) 2 2h ab, hf gb, f bc

(C) 2 2h ab, hf gb, f bc (D) none of these

6 2x a 1 x 1 x has exactly three real roots then one value of a(A) 1 (B) 2 (C) 3 (D) 4

SECTION III : MATHEMATICS


PART II : Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A),(B), (C) and (D) out of which ONE or MORE are correct. (+5, 2)

71

x 0

sin xlim

x

(where [.] denotes greatest integer function) is

(A) left hand limit is 2 (B) left hand limit is 1(C) right hand limit is 1 (D) limit exists and both are equal to 1

8 Let 2 n2 20 1 2 0 1 21 x 1 x a a x a x ..... if a , a , a are in A.P., then the value of n is(A) 3 (B) 5 (C) 4 (D) 2

9 Which of the following pair(s) of function have same graphs?

(A) secx tan x cos x sin xf x , g xcos x cot x secx cosec x

(B) 2 2 2f x sgn x 4x 5 , g x sgn cos x sin x 3

where sgn denotes signum function.

(C) 2ln x 3x 3 2f x e , g x x 3x 3

(D) 2sin x cos x 2cos xf x , g x

sec x cosec x cot x

10 Which of the following statement(s) is True?(A) 2P n : n n 41 is prime for every n N .

(B) n

n

n 1

1 2 ln2n!

is rational.

(C) 1 1 1ln .........

13 1235 123457

simplifies to an integer..

(D) Coefficient of 5x in 61 x is 10 6C

PART III : Paragraph Type

This section contains 4 multiple choice questions relating to three paragraphs with twoquestions on each paragraph. Each question has four choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct. (+4, 1)

PASSAGE 1Consider a parabola P which touches y 0 at (1, 0) and x 0 at (0, 2).

11 The vertex of P is

(A) 16 2,25 25

(B) 4 2,5 5

(C) 12 6,25 25

(D) none of these


12 Latus rectum of P is

(A) 9

5 5 (B) 16 5

25

(C) 1625 (D)

825

PASSAGE 2

2x ax

bx cx 2

f where a,b,c R, 1 0 f and y 1 is an asymptotes of

1y x , y x f f is inverse function of xf

13 f x is equal to

(A) x 1xx 2

f (B) 2x 1x

x x 2

f

(C) x 1xx 2

f (D) none of these

14 1 : R 1 R f then 1 xf is(A) one-one onto (B) one-one into(C) many-one into (D) many-one onto

PART IV : Integer Answer Type

This section contains 4 questions. The answer to each question is a single digit integer,ranging from 0 to 9 (both inclusive). (+4, 0)

15 Let two parallel lines 1L and 2L with positive slope are tangent to the circle2 2

1C : x y 2x 16y 64 0. If 1L is also tangent to the circle 2 22C : x y 2x 2y 2 0 and

equation of 2L is a a x by c a a 0 where a, b, c N , then find the value of a b c .

16 If n n

n mm p

p 1 m pC . C 19

then find value of n.

17 Let C be the set of 6 consonants {b, c, d, f, g, h} and V be a set of 5 vowels {a, e, i, o, u}and W be the set of seven letter words that can be formed with these 11 letters using boththe following rules.(a) The vowels and consonants in the word must alternate.(b) No letter can be used more than once in a single word.

If the number of words in the set W are K then K12000

is.

18 A cricket player played n n 1 matches during his career and made a total of2 n n(n 12n 39)(46 53 1)

5 runs. If Tr represent the runs made by the player in rth match such

that 1T 6 and r

r r 1T 3T 6 , 2 r n then find n.

19 Find the number of solutions of 2log x log x x



This section contains 1 question. Each question contains statements given in two columns, which have tobe matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column -II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE ORMORE statement(s) in Column- II. The appropriate bubbles corresponding to the answers to thesequestions have to be darkened as illustrated in the following example: If the correct matches are A - p , sand t ; B - q and r ; C - p and q ; and D - s and t. (+8, 0)

20 Column I Column II

(A) If 2 1 2 1a cosec cot 3 sec tan 2 then (P) x 2

(B) If 1 1 11tan x tan tan 3y

and 2y y 56 0 then (Q) x 5

(C) If 1 1cos x tan y and 2y 3 then (R)1x2

(D) If 1 1 3sin tan sin 04 y 6

and 2x y then (S)

1x2

PAPER - I (SOLUTION)

1

Sol At nx n,f n sin f nn 1 n 1

n 1f n sinn n

f(x) is discontinuous at all n I

2Sol 22y 7y 6 0 where y 4x

2y 3 y 2 0 y 2 or 32 , but y is an integer

4x 2 ; Hence 2 4x 3 or 3x

2 4

3

Sol x 5 x 5

f 5 lim sin x x lim sin x 4 12 2

2 2 2

2

x 5 x 5

ab x 11x 24 ab x 3 x 8f 5 lim lim 3ab .

x 3 x 3

25 b 1 1 3ab

6 25b ,a

5 108

4

Sol Slope of 5 4 9

AF 34 1 3

Slope of 1 1

CF ,3 3

Also F lies on CF

Equation of CF : x 3y 11 0 ...(1)

A 4,5

E 4,1 1, 4 F

B D C

slope of 5 1 4 1

AE4 4 8 2

O

Slope of 1

BE 2,12

Also E lies on BE

Equation of BE : 2x y 7 0 ...(2)

solving (1) and (2), we get point O 2, 3 Since CB is passing through intersection of AC & FC

Equation of CB : x 2y 6 x 3y 11 0

or x 1 y 2 6 11 0 ...(3)

slope of 1

CB3 2

; also slope of 5 3 4

AO4 2 3

since

1 4CB AO ; 1

3 2 3 or 2;

Equation of BC : 3x 4y 28 0.

5

Sol 2 2 2hx f h ab x gx hf gb f bc

yb

6Sol 21, 1 1a x a x x

0, 0, 4D a

21, 1 1A x a x x 0, 0, 4.D a

7

Sol1 1

h 0 h 0

sin h sin hLHL lim lim 2

h h

,1

h 0

sin hRHL lim 1.

h

8

Sol 4 2 n n n 2 20 1 2 0 1 21 x 2x C C x C x ... a a x a x ... By comparing, n0 1 2 2a 1,a n,a 2 C

n n 12n 1 2

2

2n 5n 6 0

n 2,3.

9

[Sol.(A) We have f(x) = xcotxtan

xcosxsec , g(x) =

xeccosxsin

xsecxcos

Clearly both f(x) and g(x) are identical functions as x 2

k k I.

(B) As x2 4x + 5 = (x 2)2 + 1 > 0Hence f(x) = 1 x R.

Also cos2 x + sin2

2

x > 0

Hence g(x) = 1 x R. f(x) and g(x) are identical.

(C) f(x) = )3x3xln(2

e

As x2 + 3x + 3 = 43

23x

2

> 0 x R.

Hence f(x) = x2 + 3x + 3 x R. f(x) is identical to g(x).

(D) We have f(x) = xeccosxcos

xsecxsin , g(x) =

xcotxcos2 2

Clearly both f(x) and g(x) are identical functions as x 2

k k I. ]

10[Sol.(A) for n = 41, P(n) is not prime

(B) 0!38n

!24n

!14n.......

!32n

!22n

!12n 3232

llllll

= (eln 2 1) + (eln 4 1) = 1 + 3 = 4 + Q (B) is correct

(C) ........!76

!54

!32

Tn = )!1n2(1

)!n2(1

)!1n2(11n2

= .....!51

!41

!31

!21

= .....!31

!21

!111 = e

1

(D) Coefficient of x5 in (1 x)6 = 10C5 (D) is not correct ]

11Sol. A

12Sol. B

Sol 13 to 14

2 2x af x

bx cx

1 0 1f a

21

2xf x

bx cx

f x

lim 1x

f x

0 & 1b c

12

xf x yx

1f x

1 2 for any f x x

1 2x yx y

2 1 12 11 1

yyxy y

121 y

15 [Ans. 14 ]

16 [Ans. n = 3]

Sol. nCm . mCp = !m)!mn(

!n !p!)pm(

!m

= !p!)pn(!n

. !)pm(!)mn(!)pn(

= nCp .

n pCm p

Now,

n

1p

n

pmpm

pnp

n C.C

=

n

pmp

nC (n pC0 + n pC1 +

n pC2 + ........ + n pCn p)

=

n

1p

pnp

n 2.C =

n

0p

pnp

n 2.C 2n

= 3n 2n

further 3n 2n = 19 n = 3

17 [Ans. 3][Sol. Consonant b, c, d, f, g, h (6)

Vowels a, e, i, o, u (5)w : Case I: If word begins with consonantsthen ( 6C4 4!) (

5C3 3!) = 360 60= 21600Case II : If word begins with vowels

( 5C4 4!) ( 6C3 3!) = 120 120 = 14400

Total = 36000

36000K12000

K 3 Ans. ]

18 [Ans. 6]

[Sol. We have Tr 3Tr1 = 6r

Dividing both sides by 3r, we get

rr

3T

1r1r

3T

= 2r

n

2r1r1r

rr

3T

3T

=

.P.G

n

2r

r2

3T

3T 1

22

+ 22

33

3T

3T

+ 33

44

3T

3T

+ 1n1n

nn

3T

3T

3T

3T 1

nn = 4(2n1 1)

nn

3T

36 = 4(2n1 1) ( T1 = 6)

Tn = 2 3n + 4 3n 2n 1 4 3n = 2n + 1 3n 2 3n

Tn = 2(6n 3n)

So, Sn = 2

n

1

n

1

nn 36

=

)13(23)16(

562 nn

Sn = 53

13564 nn Now on comparing, we get

5)13564()39n12n( nn2 = 5

3 13564 nn

n2 12n + 39 = 3 (n 6)2 = 0Hence n = 6 Ans. ]

19 [Ans. 0]Sol. log x is defined for x > 0 only, |log x log x2| = x x 0

No solution.

20Sol (A) (q),(B) (p,q),(C) (r),(D) (p)

(A) 2 1 2 11 cot cot 3 1 tan tan 2x 5.x

.P.G

n

2r

r2

(B) 1 11

tan tan 31

xy

xy

if 1 0xy

5 & 2.x

(C)2

1 1cos21

xy

12

x

(D) 1 1 3sin 1 sin 06y

4y and 2.x


ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI FULL TEST 2 (ADVANCED PATTERN)

PAPER I (ANSWER KEY)

PHYSICS 1. A 2. A 3. C OR D 4. B 5. C 6. A 7. A 8. D 9. A, B, C, D 10. B, C, D 11. A, B, D 12. A, B, C, D 13. B 14. A 15. C 16. (9) 17. (16) 18. (1) 19. (24) 20. a S ; b P, R ;c R ;d Q,S

CHEMISTRY

1. (A) 2. (D) 3. (A) 4. (C) 5. (B) 6. (C) 7. (D) 8. (B) 9. (A,B,D) 10. (A,B,C) 11. (A,C,D) 12. (A, B, C, D) 13. (D) 14. (B) 15. (B) 16. (6) 17. (4) 18. (1) 19. (3) 20. (A Q, S), (B Q, S), (C P, R, T), (D S)

MATHS

1. D 2. A 3. A 4. A 5. C 6. D 7. A,C 8. A,D 9. A,B,C,D 10. B,C 11. A 12. B 13. A 14. B 15. 14 16. 3 17. 3 18. 6 19. 0 20. A Q;B P,Q;C R; D P


PAPER II

PHYSICS 1. D 2. A 3. C 4. C 5. D 6. C 7. B 8. D 9. A, C 10. B, C 11. A, B, C, D 12. B, C 13. A 14. A 15. A 16. (1) 17. (2) 18. (8) 19. (2) 20. a P ;b P,Q,S ;c Q ;d Q,R

CHEMISTRY

1. (B) 2. (C) 3. (B) 4. (A) 5. (C) 6. (B) 7. (B) 8. (B) 9. (A,B,C,D) 10. (B, C, D) 11. (A, C, D) 12. (A, D) 13. (C) 14. (D) 15. (A) 16. (1) 17. (1) 18. (1) 19. (4) 20. (A) (R), (S), (B) (R), (C) (P), (Q), (D) (R), (S)

MATHS

1. C 2. D 3. C 4. C 5. A 6. C 7. B 8. C 9. A,B,C 10. A,B,C,D 11. A,B,C 12. B,C 13. C 14. C 15. D 16. 1210 17. 1 18. 6 19. 4 20. A R;B S;C P,S;D Q, R

Pace Test Paper

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