Joint Entrance Exam/IITJEE-2017 Paper Code -...

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VMC/IIT JEE-2017 1 JEE Entrance Examination

Joint Entrance Exam/IITJEE-2017

Paper Code - C 2nd April 2017 | 9.30 AM – 12.30 PM

CHMISTRY, MATHEMATICS & PHYSICS

Important Instructions: 1. Immediately fill in the particulars on this page of Test Booklet with only Blue / Black Ball Point Pen

provided in the examination hall.

2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 90 questions. The maximum marks are 360.

5. There are three parts in the question paper A, B, C, consisting of Chemistry, Mathematics and Physics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

6. Candidates will be awarded, marks as stated above in instruction No. 5 for correct response of each question. ¼ (one fourth) marks of the total marks allotted to the question (i.e. 1 mark) will be deducted for indicating incorrect response for each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

7. There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8. For writing particulars/marking responses on Side -1 and Side-2 of the Answer Sheet use only Black Ball Point Pen provided in the examination hall.

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12. The CODE for this booklet is C. Make sure that the CODE printed on Side – 2 of the Answer Sheet and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet.

13. Do not fold or make any stray mark on the Answer Sheet.

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Joint Entrance Exam/IITJEE-2017 CHMISTRY, MATHEMATICS & Physics

PART-A CHEMISTRY

1. The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (Kf for benzene = 5.12 K kg mol–1)

(1) 94.6 % (2) 64.6 % (3) 80.4% (4) 74.6% 1.(1) f fT i K m

0.2 10000.45 5.1260 20

i

i = 0.527

2/ 21

2AcOH AcOH

12

i

0.527 12

0.946 or 94.6 %

2. On treatment of 100 mL of 0.1 M solution of 3 2CoCl .6H O with excess 3AgNO ; 221.2 10 ions are precipitated. The complex is:

(1) 2 5 2 2[Co(H O) Cl]Cl .H O (2) 2 4 2 2[Co(H O) Cl ]Cl.2H O

(3) 2 3 3 2[Co(H O) Cl ].3H O (4) 2 6 3[Co(H O) ]Cl

2.(1) 3 2 30.01

.6CoCl H O AgNO nAgCl

22

223

1.2 10 2 10 0.026.0 10

moles

0.01 0.02n 2n

Complex is 2 2 25 .Co H O Cl Cl H O

3. Which of the following compounds will form significant amount of meta product during mono-nitration

reaction?

(1) (2) (3) (4)

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3.(4)

4. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are:

(1) Cl and 2ClO (2) ClO and 3ClO

(3) 2ClO and 3ClO (4) Cl and ClO

4.(4) 2 2

&cold dilCl NaOH Na Cl Na Cl O H O

5. Both lithium and magnesium display several similar properties due to the diagonal relationship; however,

the one which is incorrect, is: (1) nitrates of both Li and Mg yield 2NO and 2O on heating

(2) both form basic carbonates (3) both form soluble bicarbonates (4) both form nitrides 5.(2) 1, 3 & 4 are correct statements. 6. A water sample has ppm level concentration of following anions

24 3F 10; SO 100; NO 50

The anion/anions that make/makes the water sample unsuitable for drinking is/are:

(1) Only 24SO (2) Only 3NO

(3) Both 24SO and 3NO (4) Only F

6.(4) Permissible limit of F– is 1 ppm. 7. The formation of which of the following polymers involves hydrolysis reaction ? (1) Terylene (2) Nylon 6 (3) Bakelite (4) Nylon 6, 6 7.(2)

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8. The Tyndall effect is observed only when following conditions are satisfied : (a) The diameter of the dispersed particles is much smaller than the wavelength of the light used. (b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.

(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.

(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

(1) (b) and (c) (2) (a) and (d) (3) (b) and (d) (4) (a) and (c) 8.(3) Refer NCERT Class XII Page no. 139

Tyndall effect is observed only when the following two conditions are satisfied.

(i) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (ii) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. 9. apK of weak acid (HA) and bpK of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their

salt (AB) solution is: (1) 1.0 (2) 7.2 (3) 6.9 (4) 7.0 9.(3) pH of a salt made up of weak acid and weak base is calculated by using expression

1 1 3.2 3.47 7 6.92 2 2 2a bpH pK pK

10. The major product obtained in the following reaction is :

(1) (2) (3) (4) 10.(2) DIBAL-H reduces only ester group to an aldehyde and alcohol.

11. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution.

(1) (2) (3) (4)

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11.(2)

12. The correct sequence of reagents for the following conversion will be

(1) 3 3 32Ag NH OH ,CH MgBr, H / CH OH

(2) 3 3 32Ag NH OH , H / CH OH, CH MgBr

(3) 3 3 3 2CH MgBr, H / CH OH, Ag NH OH

(4) 3 3 32CH MgBr, Ag NH OH , H / CH OH

12.(2)

13. Which of the following species is not paramagnetic? (1) B2 (2) NO (3) CO (4) O2

13.(3) (1) x y

2 2 2 2 1 1**2 1s 1s 2s 2s 2p 2pB , , , , ,

B2 contains two unpaired electrons hence paramagnetic.

(2) N O odd electron species

hence paramagnetic

(3) x y z

2 2 2 2 2 2 2**1s 1s 2s 2s 2p 2p 2pCO , , , , ,

CO contains no unpaired electron hence diamagnetic. (4)

z x y x y2 *2 2 *2 2 2 2 *1 *1

2 1s 1s 2s 2s 2p 2p 2p 2p 2pO , , , , , ,

O2 contains two unpaired electron hence paramagnetic.

14. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to

decolourize the colour of bromine?

(1) (2) (3) (4)

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14.(2)

15. Which of the following reactions is an example of a redox reaction?

(1) 6 2 2 2XeF 2H O XeO F 4HF (2) 4 2 2 6 2XeF O F XeF O

(3) 2 5 6XeF PF XeF PF (4) 6 2 4XeF H O XeOF 2HF

15.(2) Any reaction having a substance in its elemental form is a redox reaction.

4 1 6 0

2 24 2 6Xe F O F Xe F O

16. U is equal to: (1) Isothermal work (2) Isochoric work (3) Isobaric work (4) Adiabatic work

16.(4) For adiabatic process : q = 0

So from Ist law U q w

We can write U w 17. Which of the following molecules is least resonance stabilized?

(1) (2) (3) (4)

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17.(1)

18. The increasing order of the reactivity of the following halides for the 1NS reaction is:

I. 3 2 3CH C HCH CH

|Cl

II. 3 2 2CH CH CH Cl III. 3 6 4 2p H CO C H CH Cl

(1) (II) < (III) < (I) (2) (III) < (II) < (I) (3) (II) < (I) < (III) (4) (I) < (III) < (II)

18.(3) Order of reactivity for NS 1 mechanism is in accordance with order of stability of carbocation involved :

So, order of reactivity

19. 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar

mass of M2CO3 in g mol–1 is: (1) 11.86 (2) 1186 (3) 84.3 (4) 118.6

19.(3) 2 3 2 2M CO 2HCl 2MCl H O CO

No. of moles of 2 3M CO No. of moles of 2CO evolved

1

0.01186M

(M = molar mass of 2 3M CO )

51 10

M 84.30.01186 1186

20. Sodium salt of an organic acid ‘X’ produces effervescence with conc. H2SO4. ‘X’ reacts with the

acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. ‘X’ is :

(1) Na2C2O4 (2) C6H5COONa (3) HCOONa (4) CH3COONa

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20.(1) 2 2 4 2 4 2 4 2 2(X)

Na C O H SO (conc) Na SO H O CO CO

2 2 2 4 2 4white ppt

CaCl Na C O CaC O 2NaCl

H2 22 4 4 2C O MnO CO Mn

21. The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon

(22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is:

(1) 10 kg (2) 15 kg (3) 37.5 kg (4) 7.5 kg

21.(4) Mass of 1H in body 3 31075 10 g 7.5 10 g

100

No. of moles of 1H replaced by 2 3H 7.5 10

So mass increased 37.5 10 g 7.5kg

22. The major product obtained in the following reaction is :

(1) 6 5 2 6 5

tC H CH O Bu CH C H (2) 6 5 2 6 5tC H CH O Bu CH C H

(3) 6 5 6 5C H CH CHC H (4) 6 5 2 6 5tC H CH O Bu CH C H

22.(3)

23. Given : 1(graphite) 2 2 rC O (g) CO (g); H 393.5 kJ mol

12 2 2 r

1H (g) O (g) H O(l); H 285.8 kJ mol

2

12 2 4 2 rCO (g) 2H O(l) CH (g) 2O (g); H 890.3 kJ mol

Based on the above thermochemical equations, the value of rH at 298 K for the reaction

(graphite) 2 4C 2H (g) CH (g) will be:

(1) – 144.0 kJ 1mol (2) +74.8 kJ 1mol

(3) + 144.0 kJ 1mol (4) – 74.8 kJ 1mol

23.(4) 1(graphite) 2 2 rC O (g) CO (g); H 393.5 kJ mol ……….(i)

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12 2 2 r

1H (g) O (g) H O(l); H 285.8 kJ mol

2 ……….(ii)

12 2 4 2 rCO (g) 2H O(l) CH (g) 2O (g); H 890.3 kJ mol …….(iii)

2(g) 2 ( ) 4(g) 2(g)CO 2H O CH 2Ol …………(iv)

4 1 2 32r r r rH H H H

= –393.5 + (–285.8×2) + 890.3 = –74.8 KJ / mol 24. In the following reactions, ZnO is respectively acting as a/an: (a) 2 2 2ZnO Na O Na ZnO (b) 2 3ZnO CO ZnCO

(1) acid and base (2) base and acid (3) base and base (4) acid and acid

24.(1) 2 2 2BaseAcid

Zn O Na O Na Zn O

2 3AcidBase

Zn O CO Zn CO

Non metal oxides are generally acidic while alkali metal oxides are basic.

25. The radius of the second Bohr orbit for hydrogen atom is: (Planck's Const. 34h 6.6262 10 Js; mass of

electron = 319.1091 10 kg; charge of electron 19e 1.60210 10 C; permittivity of vacuum 12 1 3 2

0 8.854185 10 kg m A )

(1) 2.12 Å (2) 1.65 Å (3) 4.76 Å (4) 0.529 Å

25.(1) 2

nn

r 0.529Z

2r 0.529 4 2.116Å

26. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of

R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then

ln(k2/k1) is equal to : 1 18.314R J mol K

(1) 4 (2) 8 (3) 12 (4) 6

26.(1) From Arrhenius equation aE / RTk Ae

Since 2a a1 2

1

k 1ln E E

k RT

31 10 10

48.314 300

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27. A metal crystallizes in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be:

(1) 2

a (2) 2a (3) 2 2a (4) 2 a

27.(1) In FCC distance of closest approach between two atoms 2r

In FCC, atoms are in close contact along face diagonal of FCC unit cell.

4r 2a

2a a

2r2 2

28. The group having isoelectronic species is: (1) 2O , F , Na ,Mg (2) 2 2O , F , Na , Mg

(3) O , F , Na, Mg (4) 2 2O , F , Na, Mg

28.(2) Isoelectronic species have same no. of electrons. All these species contains 10 electrons.

29. Given 32 / /

1.36 , 0.74Cl Cl Cr Cr

E V E V

2 3 22 7 4/ /

1.33 , 1.51Cr O Cr MnO Mn

E V E V

Among the following, the strongest reducing agent is:

(1) Cl (2) Cr (3) Mn2+ (4) Cr3+

29.(2) Re d1

EReducing strength

30. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The

number of possible stereoisomers for the product is: (1) Four (2) Six (3) Zero (4) Two

30.(1)

No of stereoisomers n 22 2 4 [n = No. of chiral carbon atoms]

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PART-B MATHEMATICS

31. The integral

34

4

1dxcos x

is equal to :

(1) 4 (2) 1 (3) 2 (4) 2

31.(4)

34

4

1dxIcos x

34

4

using property1

dxI a b xcos x

34

2

4

221

I dxcos x

34

2

4

I cos ec x dx

43 4 1 1 2/

/cot x

32. Let nI 1ntan x dx, n . If 5 5

4 6I I a tan x bx C, where C is a constant of integration, then the

ordered pair (a, b) is equal to :

(1) 1 15

,

(2) 1 05

,

(3) 1 15

,

(4) 1 05

,

32.(4) 4 6 5 5tan x dx tan xdx a tan x bx c

Differentiating both sides 4 6 4 2 45 5tan x tan x atan x sec x bx 4 2 45 1 5atan x tan x bx 4 6 45 5 5atan x a tan x bx

1 05

a b

33. The area (in sq. units) of the region 2: 0 3 4 and 1x, y x , x y , x y y x is :

(1) 73

(2) 52

(3) 5912

(4) 32

33.(2) Solving 2 4 and 3x y x y

We get, 2

34x x

2 4 12 0x x 6 2 0x x 2x 1y

Solving 1 and 3y x y x

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We get, 1 3x x 1 2x y

Area = 1 2 2 4

0 1 0

1 34xx dx x dx dx

2 21 2 33 2

0 1 0

2 533 2 12 2

/ x xx x x

34. A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with

replacement, then the variance of the number of green balls drawn is :

(1) 4 (2) 625

(3) 125

(4) 6

34.(3) n = 10

1525

p

1025

q

var( )X npq

15 101025 25

3 2105 5

125

35. If (2 sin ) ( 1) cos 0dyx y xdx

and (0) 1,y then 2

y

is equal to :

(1) 13

(2) 43

(3) 13

(4) 23

35.(3) ( 1) cos2 sin

dy y xdx x

cos1 2 sin

dy x dxy x

( 1) (2 sin )n y n x nc ( 1) (2 sin )y x c (0) 1y (2) (2) = c 4

( / 2) ?y (y + 1) (2 + 1) = 4

13

y

36. Let be a complex number such that 2 1 z where 3z .

If 2 2

2 7

1 1 1

1 1 3 ,

1

k

then k is equal to :

(1) –1 (2) 1 (3) –z (4) z 36.(3) 2 1 z

3 12i

2 2

2

1 1 1

1 1 3

1

k

1 1 2 3R R R R

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2 2

2

3 0 0

1 1

1

2 4 33 1 3 3 1 2

3 2 1 3 3z k k z

37. Let 2 2a i j k and ˆ ˆb i j

. Let c be a vector such that | | 3c a

, 3a b c and the

angle between andc a b be 30°. Then a c is equal to :

(1) 5 (2) 18

(3) 258

(4) 2

37.(4) ˆˆ ˆ2 2a i j k

ˆ ˆb i j

3c a

3a b c

sin 3a b c

6a b c …(i)

ˆˆ ˆ

ˆ ˆˆ ˆ ˆ ˆ2 1 2 2 2 1 2 21 1 0

i j ka b i j k i j k

3a b

From (i) 3 6c

2c

Now 3c a

2 2 2 9c a a c

4 9 2 9a c

2a c

38. The radius of a circle, having minimum area , which touches the curve 24y x and the lines

| |y x is :

(1) 4 2 1 (2) 4 2 1 (3) 2 2 1 (4) 2 2 1

38.(1) 2 4 0x x

1 1 162

x

1 17

2

(4 ) 02

r r

4 2r r

4 4( 2 1)12 1

r

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39. If for 10,4

x

, the derivative of 13

6tan1 9

x xx

is ( )x g x , then ( )g x equals :

(1) 33

1 9x

x (2) 3

31 9x

(3) 39

1 9x (4) 3

31 9

x xx

39.(3) 10,4

x

3/ 2 33 0,8

x

1 30, tan8

13

6tan1 9

x xyx

Let 1 3 / 2tan 3x 12

2 tantan1 tan

1tan (tan 2 ) 2 1 3/ 22 tan (3 )x

1/ 23

2 3321 9

y xx

3

91 9

xx

40. If two different numbers are taken from the set {0, 1, 2, 3, ……., 10} ; then the probability that their sum

as well as absolute difference are both multiple of 4, is :

(1) 1445

(2) 755

(3) 655

(4) 1255

40.(3) 0 4 8 0A , , remainder when divided by 4.

2 6 10 2B , ,

1 5 9 1C , ,

3 7 1D , orx, y A x, y B

3 3

2 211

2

6 610 55112

C CC .

41. 3

2

cot coslim2x

x xx

equals :

(1) 18

(2) 14

(3) 124

(4) 116

41.(4) 3/ 2

cot cos( 2 )x

x xLtx

30

cot ( / 2 ) cos[ / 2 ]8h

h hLth

30

1 sin [1 cos ]8 h

h hLth

116

42. The value of 21 10 21 10 21 10 21 10 21 101 1 2 2 3 3 4 4 10 10.....C C C C C C C C C C , is :

(1) 20 92 2 (2) 20 102 2 (3) 21 112 2 (4) 21 102 2

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42.(2) 21 10 21 10 21 10 21 10 21 101 1 2 2 3 3 4 4 10 10....C C C C C C C C C C

21

10 20 10 20 102 2 2 1 2 1 2 1 2 22

43. For three events A, B and C, P(Exactly one of A or B occurs) = P (Exactly one of B or C occurs) = P

(Exactly one of C or A occurs) = 14

and P (All the three events occur simultaneously) = 116

.

Then the probability that at least one of the events occurs, is :

(1) 764

(2) 316

(3) 732

(4) 716

43.(4) 124

P A P B P A B

124

P B P C P B C

124

P C P A P C A

116

P A B C

38

P A P B P C P A B P B C P C A

3 1 78 16 16

P A B C

44. Let a vertical tower AB have its end A on the level ground. Let C be the mid point of AB and P be a point on the ground such that 2 .AP AB If ,BPC then tan is equal to :

(1) 29

(2) 49

(3) 67

(4) 14

44.(1) tan

1tan2

; 1tan4

tan tan

1 1 122 4 4

1 9 918 8

45. The eccentricity of an ellipse whose centre is at the origin is 1 .2

If one of its directrices is 4,x then

the equation of the normal to it at 31,2

is :

(1) 4 2 7x y (2) 2 4x y (3) 2 2y x (4) 4 2 1x y

45.(4) 2 2

2 2 1x ya b

412

a 2a

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2

221 be

a 2 3b

2 2

14 3x y

Equation of normal

2 2

2 2

1 1

a x b y a bx y

4 3 1312

x y

4 2 1x y

46. If, for a positive integer n, the quadratic equation,

1 1 2 ...... 1 10x x x x x n x n n

has two consecutive integral solutions , then n is equal to : (1) 10 (2) 11 (3) 12 (4) 9

46.(2) 1

1 10n

rx r x r n

2 2 1 1 10x r x r r n

2

2 231

03

n nnx x.n

2 23 3 31 0x nx n

2 29 12 372 1

3D n n

| |

2372 3 9n 23 372 9 363n 2 121n 11n 47. The following statement p q p q q is :

(1) equivalent to p q (2) a fallacy

(3) a tautology (4) equivalent to p q

47.(3) p q p q q

p q p q p q p q q p q p q q

T T T T T T

T F F T F T

F T T T T T

F F T F T T

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48. The normal to the curve 2 3 6y x x x at the point where the curve intersects the y-axis, passes

through the point :

(1) 1 1,2 3

(2) 1 1,2 3

(3) 1 1,2 2

(4) 1 1,2 2

48.(4) 6 6y 0, 1

1y

2 3 3 2 1x x y x x y

6 3 2 1y

1y

'0 1xy Slope of normal 1

1y x

1y x

49. For any three positive real numbers , and ,a b c 2 2 29 25 25 3 15 3 .a b c ac b a c Then :

(1) , anda b c are in A.P. (2) , anda b c are in G.P.

(3) , andb c a are in G.P. (4) , andb c a are in A.P.

49.(4) 2 2 2225 9 25 75 15 (3 )a b c ac b a c

2 2 2225 9 25 75 45 15a b c ac ab bc

2 2 2(15 ) (3 ) (5 ) 45 75 15a b c ab ac bc

15a = 3b = 5c = k

15 15k ka

53 15k kb

35 15k kc

b, c, a are in A.P. 50. If the image of the point (1, 2,3)P in the plane, 2 3 4 22 0x y z measured parallel to the line,

1 4 5x y z is Q, then PQ is equal to :

(1) 42 (2) 6 5 (3) 3 5 (4) 2 42

50.(4) 1 2 3 31 4 5

x y

( 1, 4 2,5 3) 2 2 12 6 20 12 22 0 6 6 0 1 (2, 2, 8)

2 2 21 4 5 42

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51. If 2 25(tan cos ) 2 cos 2 9x x x , then the value of cos 4x is :

(1) 29

(2) 79

(3) 35

(4) 13

51.(2) 2 25(tan cos ) 2cos 2 9x x x

2 4

22

sin cos5 2cos 9cosx x x

x

2cos x t

215 2(2 1) 9t t t

t

2 25 5 5 4 7t t t t

29 15 3 5 0t t t 3 (3 5) (3 5) 0t t t

13

t

1 1cos 2 2 13 3

x

21 2 7cos 4 2 1 1

3 9 9x

52. Let , ,a b c R . If 2( )f x ax bx c is such that a + b + c = 3 and ( ) ( ) ( ) ,f x y f x f y xy

10

1

, , then ( )n

x y R f n

is equal to :

(1) 190 (2) 255 (3) 330 (4) 165

52.(3) 2 21( )2

f x x x bx

1 1 5(1) 3 32 3 2

f b b

21 5( )2 2

f x x x

21 52 2

f n n n

1 10.11.1 5 10.11.2 6 2 2

385 275 660 330

2 2 2

53. The distance of the point (1, 3, 7) from the plane passing through the point (1, 1, 1) , having normal

perpendicular to both the lines 1 2 4 2 1 7and1 2 3 2 1 1

x y z x y z

is :

(1) 583

(2) 1074

(3) 2074

(4) 1083

53.(4) 1 2 32 1 1

2 3 : 6 1 : 1 4 5 : 7 : 3 5 7 3 5 7 3 5x y z 5 2 3 5 0x y z

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5 21 21 5 1025 49 9 83

54. If S is the set of distinct values of 'b' for which the following system of linear equations 1x y z 1x ay z 0ax by z has no solution, then S is : (1) a finite set containing two or more elements (2) a singleton

(3) an empty set (4) an infinite set

54.(2) 21 1 11 1 ( ) (1 ) ( )

1a a b a b a

a b

a b 1 a b 2a 2( 2 )a a t 2( 1) 0a 1a 1x y z 0x by z Two plane should be parallel b = 1

55. If 2 34 1

A ,

then adj 23 12A A is equal to :

(1) 51 8463 72

(2) 72 6384 51

(3) 72 8463 51

(4) 51 6384 72

55.(4) 2 34 1

A

2 2 3 2 3 4 12 6 3 16 9

4 1 4 1 8 4 12 1 12 13A

2 48 27 24 363 : 12

36 39 48 12A A

2 72 633 12

84 51A A

Adj 2 51 633 12

84 72A A

56. A hyperbola passes through the point 2 3P , and has foci at 2 0, . Then the tangent to this

hyperbola at P also passes through the point : (1) 3 2, (2) 2 3, (3) 3 2 2 3, (4) 2 2 3 3,

56.(4) Clearly 2 2

2 22 ; for 1x yaea b

2 2 2

2 3 11a a e

2 2 2 22 3 1

a a e a

2 2

2 3 14a a

Solve to get 2 1 8a ,

2 8a Rejected as e can’t be less than 1

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2 21 3a , b

2 2

11 3x y

32 1

3x y

57. Let k be an integer such that the triangle with vertices 3 5k , k , , k and 2k , has area 28 sq.

units. Then the orthocentre of this triangle is at the point :

(1) 314

,

(2) 122

,

(3) 122

,

(4) 314

,

57.(2) 3 1

1 5 11 282

2 1

k kk

k

3 1

1 5 11 562 1

k kk

k

21 2 3 5 1 10 1 56k k k k k

21 5 3 10 1 56k k

25 3 10 56k k or 2 25 3 66 0 or 5 3 46 0k k k k D < 0 Solving we get k = 2 Hence, the vertices are 2 6 5 2 2 2, , , , ,

Solving the equation of two altitudes we get orthocente as 122

,

58. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the

maximum area (in sq. m) of the flower-bed, is : (1) 25 (2) 30 (3) 12.5 (4) 10 58.(1) Given 2 20r r …(i)

Area 212

r A

From (i)

20 2rr

22 220 21 20 2 10

2 2r r rA r r r

r

210A r r

10 2 0dA rdr

5r

2

2 2 0d Adr

5r will given maximum area

20 2 5

2 rad5

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21 5 2 252

A .

59. The function 1 1: ,2 2

f R defined as 2 ,

1xf xx

is :

(1) Surjective but not injective (2) Neither injective not surjective (3) Invertible (4) Injective but not surjective

59.(1) 21xf xx

1 12 2

f : R ,

2

22

1 1 2

1

x . x xf ' x

x

Non-monotonic not injective

21xyx

2 0x y x y 0D 21 4 0y

1 12 2

y ,

60. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are

ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in the party, is :

(1) 469 (2) 484 (3) 485 (4) 468 60.(3) Let mM denotes male relative of man 3X wM denotes female relative of man 4X

mW denotes male relative of woman 4Y

wW denotes female relative of woman 3X

Case I 3 3m wM W 1

Case II 3 3w mM W 4 43 3 16C C

Case III 2 1 2 1m w w mM M W W 3 4 3 43 1 2 1 144C C C C

Case IV 1 2 2 1m w m wM M W W 3 4 4 31 2 2 1 324C C C C

Total 324 144 16 1 485

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PART-C PHYSICS

61. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer?

(speed of light = 3 × 108 ms–1 ) (1) 12.1 GHz (2) 17.3 GHz (3) 15.3 GHz (4) 10.1 GHz 61.(2) This question involves the use of relativistic Doppler’s effect. The usual non-relativistic Doppler formula will

NOT be applicable here as the velocity of observer is not small as compared to light.

The relativistic Doppler's formula is 1( ) ( )1

observed actual

Where VC

V is relative velocity of observer w.r.t. the source and is taken to be positive if observer and source are moving

towards each other. So, here 1 1/ 2( ) (10 .)1 1/ 2

observed GHz

17.3 .GHz

62. The following observations waver taken for determining surface tension T of water by capillary method:

Diameter of capillary, 2D 1.25 10 m rise of water, 2h 1.45 10 m.

Using 2g 9.80 m /s and the simplified relation 3rhgT 10 N / m,2

the possible error in surface tension

is closest to: (1) 1.5% (2) 2.4% (3) 10% (4) 0.15%

62.(1) 3 310 / 10 /2 4

rhg dhgT N m N m

% error

T d hT d h

2 2

2 20.01 10 0.01 10100 1001.25 10 1.45 10

TT

0.8% 0.7% = 1.5 %

(There is no error in g because its value is not calculated through experiments)

63. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths 1 1 2r / is given by:

(1) 2r3

(2) 3r4

(3) 13

r (4) 4r3

63.(3) 2

43 3

hc E EE

1

2 hc E E E 1

2

13

64. A body of mass 2m 10 kg is moving in a medium and experiences a frictional force 2F –k .v Its initial

speed is 0 10v ms–1. If, after 10 s, its energy is 20

1 ,8

mv the value of k will be:

(1) 3 110 kg s (2) 4 110 kg m (3) 1 1 110 kg m s (4) 3 110 kg m

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64.(2) 2F kv 2ma kv

2 2k dv ka v vm dt m

210 0

v tdv k dtmv

10

1 v k tv m

1 0.1 ktv m 1 1

0.1 10000.1v kt k

m

2 20

1 12 8

m v v

0 52vv

1 5 1 0.5 50000.1 1000

kk

0.55000

k 410 /k kg m

65. andp vC C are specific heats at constant pressure and constant volume respectively. It is observed that

p vC C a for hydrogen gas

p vC C b for nitrogen gas

The correct relation between a and b is:

(1) a b (2) 14a b (3) 28a b (4) 114

a b

65.(2) If &P vC C are specific heat capacity per gram

2p vRa C C (for hydrogen) (As R is per mole of H2 i.e. 2 gm of H2)

28p vRb C C (for Nitrogen) (As R is per mole of N2 i.e. 28 gm of N2)

14a b

66. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum?

(1) 32

(2) 1 (3) 32

(4) 32

66.(4) 2dm R dx

22

4dmRdI dm x (Parallel axis theorem)

2

2

l

lI dI

2 2

2 22 2

2 24

l l

l lR R dx R x dx

2 2 3 222

22

4 3

ll

ll

R R xx R

2 2 2 3 2 2

4 3 4 4 12R R R l mR mll

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2 2

4 12mR mlI

2 22

4 12

m ml r m

l

For I to be max.

2

21 0

4 6dI m mldl l

2 3

4 6

m ml

1/31/3

3 3 32 2

m ml l

2

mR l

2 mRl

2 /31/3 1/3 1/32 2 2

3 3m nR

m

1/3 1/62

3mR

1/31/3

1/3 1/6

1/3 1/6

32 3 3

2 223

mlR m

32

lR

67. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At

sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by:

(1) log1.3log 2

t T (2) log 1.3t T (3) log 1.3Tt (4) log 2

2 log1.3Tt

67.(1) ln 2

T

0 0.3N N

N

01.3NN

0tN N e

11.3

te

ln(1.3) ln(1.3)ln(2)

t T

68. Which of the following statements is false?

(1) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.

(2) A rheostat can be used as a potential divider (3) Kirchhoff’s second law represents energy conservation (4) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of

magnitude 68.(1) From the given figure it is clear that if galvanometer is connected between AD and cell between BC , then

31

2 4

RRR R

….(i)

If cell and galvanometer are interchanged, then for balance

1 2

3 4

R RR R

….(ii)

Since equations (i) and (ii) are same, null point is undisturbed if cell and galvanometer are interchanged

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69. A capacitance of 2 µF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 µF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

(1) 16 (2) 24 (3) 32 (4) 2

69.(3) 8 24eqFC F

70. In the given circuit diagram when the current reaches steady state in the

circuit, the charge on the capacitor of capacitance C will be :

(1) 1

2( )rCE

r r (2) 2

2( )rCE

r r

(3) 1

1( )rCE

r r (4) CE

70. (2) At steady state current through the capacitor = 0

2

EI

r r

P.d. across AB = QC

QE IrC

2

Er QEr r C

2

2

rQ CEr r

71.

In the above circuit the current in each resistance is: (1) 0.25 A (2) 0.5A (3) 0 A (4) 1 A

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71.(3) If 2 A BV V V V and 2CV V

Potential difference across each resistor = 0 current across each resistor = 0

72. In amplitude modulation, sinusoidal carrier frequency used is denoted by c and the signal frequency is

denoted by m . The bandwidth ( )m of the signal is such that m c . Which of the following frequencies is not contained in the modulated wave? (1) c (2) m c (3) c m (4) m

72.(4) The frequencies in amplitude modulated wave is between c m and c m .

73. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input

and the output voltages will be: (1) 90º (2) 135º (3) 180º (4) 45º

73.(3) 0 ( )cc c cv V i R c ci R

)(i BE B Bv v i R B Bi R

0 c cv

i B B

v i RAv i R

c

B

RR

“–ve” sign indicates output is exactly out of phase i.e. phase difference = 180 74. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100

gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75ºC. T is given by: (Given : room temperature = 30ºC, specific heat of copper = 0.1 cal/gmºC) (1) 885ºC (2) 1250ºC (3) 825ºC (4) 800ºC

74. (1) Heat gained by (water + calorimeter) = Heat lost by copper ball 100 0.1 75 30 170 1 75 30 100 0.1 75 T

450 7650 10 75T 75 810T 885T C

75. In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm

away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is: (1) 7.8 mm (2) 9.75 mm (3) 15.6 mm (4) 1.56 mm

75.(1) 1 11

n Dyd

, 2 22

n Dyd

Given 1 2y y

1 1 2 2 n n

1 2

2 1

45

nn

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Therefore 9

1 34 650 10 1.5

0.5 10y

1 7.8y mm

76. An electric dipole has a fixed dipole moment P

, which makes angle with respect to x-axis. When

subjected to an electric field 1ˆE Ei

, it experiences a torque 1

ˆT k

. When subjected to another electric

field 2 1 ˆ3E E j

it experiences a torque 2 1T T

. The angle is:

(1) 45º (2) 60º (3) 90º (4) 30º 76.(2) T p E

ˆ ˆcos sinp p i p j

sinpE T and 3 cosp E T

tan 3 60º

77. A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle with the vertical is:

(1) 2 sin3gl

(2) 3 cos2gl

(3) 2 cos3gl

(4) 3 sin2gl

77.(4) 2

sin2 3l mlmg

3 sin2

gl

78. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the

bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

(1) PK

(2) 3PK

(3) 3PK (4) 3

PK

78.(4) /

P V P PVK VV V V K K

(3 )3

PV PV T TK K

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79. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is:

(1) Virtual and at a distance of 40 cm from convergent lens. (2) Real and at a distance of 40 cm from the divergent lens. (3) Real and at a distance of 6 cm from the convergent lens. (4) Real and at a distance of 40 cm from convergent lens. 79.(4) 1 25f cm , 2 20f cm

For diverging lens 25V cm For converging lens, (15 25) 40u cm

1 1 1 1 1 1 1 4040 20 20 40 40

v cmv v

Image is real 80. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It

produces continuous as well as characteristic X-rays. If min is the smallest possible wavelength of X-

ray in the spectrum, the variation of minlog with log V is correctly represented in:

(1) (2)

(3) (4)

80.(4) min hceV

minln ln ln hc Ve

y k x

81. The temperature of an open room of volume 330m increases from 17°C to 27°C due to the sunshine. The

atmospheric pressure in the room remains 51 10 Pa . If in and fn are the number of molecules in the

room before and after heating, then f in n will be :

(1) 231.38 10 (2) 252.5 10 (3) 252.5 10 (4) 231.61 10

81.(3) 0 0f if i

PV PVn n N NRT RT

, (N0 = Avogadro’s number )

5

23 250

10 30 1 16.02 10 2.5 108.314 300 290

N

82. In a coil of resistance 100 , a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is:

(1) 225Wb (2) 250Wb

(3) 275Wb (4) 200Wb

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82.(2) qR

area of I t graph Δ=

R

1 1

10 0.5 10 0.5 100 250 Wb2 100 2

83. When a current of 5mA is passed through a galvanometer having a coil of resistance 15 , it shows full

scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 10V is :

(1) 32.045 10 (2) 32.535 10 (3) 34.005 10 (4) 31.985 10 83.(4) 5gI mA

g g sV I R R

310 5 10 15 sR 315 2 10 2000sR

31985 1.985 10sR 84. A time dependent force 6F t acts on a particle of mass 1 kg. If the particle starts from rest, the work

done by the force during the first 1 sec. will be: (1) 22 J (2) 9 J (3) 18 J (4) 4.5 J 84.(4) 6F t

Fdt P 1

0

66 32

P t d t

2 23 4.5

2 2 1PKm

4.5W K J

85. A magnetic needle of magnetic moment 2 26.7 10 Am and moment of inertia 6 27.5 10 kg m is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:

(1) 8.89 s (2) 6.98s (3) 8.76 s (4) 6.65s 85.(4) sinT MB MB (if is small)

6

27.5 102 2 0.665

6.7 10 0.01

ITMB

Time taken for 10 oscillation 6.65 86. The variation of acceleration due to gravity g with distance d from centre of the earth is best

represented by (R = Earth’s radius) :

86.(3) 3 , 0 GMg r r RR

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2 ,GMg r R

r

87. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity

vs time?

(1) (2)

(3) (4)

87.(2) During the whole journey acceleration remains constant a g 0V V gt

88. A particle A of mass m and initial velocity v collides with a particle B of mass 2m which is at rest.

The collision is head on, and elastic. The ratio of the de-Broglie wavelengths toA B after the collision

is:

(1) 2A

B

(2) 2

3A

B

(3) 1

2A

B

(4) 1

3A

B

88.(1)

2A Bmmv mV V (Conservation of momentum)

Also B AV V V (e = 1)

/ 12/ 2

BA A B B

B B A A A

m Vh p p Vh p p mV V

Solving (i) and (ii) 4,3 3A BV VV V

1 4 / 3 22 / 3

A

A

VV

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89. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy – time graph of the particle will look like:

(1) (2)

(3) (4)

89.(3) K is maximum at mean position and minimum at extreme position and extreme position is reached at T/4. 90. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his

density remains same, the stress in the leg will change by a factor of :

(1) 19

(2) 81 (3) 181

(4) 9

90.(4) Stress = weight

area

30

20

9 W

9 A

0

0

W9

A

As volume increases by (9)3 times and area increases by (9)2 times.

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