JEE MAIN – 2017 QUESTION PAPER & SOLUTION CHEMISTRY · JEE MAIN 2017 CODE (C) CENTERS : MUMBAI...

JEE MAIN 2017 CODE (C)

CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 1

JEE MAIN – 2017 QUESTION PAPER & SOLUTION

CHEMISTRY

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

(1) 94.6% (2) 64.6% (3) 80.4% (4) 74.6% Sol. (1)

0.2 / 600.45 5.12 i20 /1000

0.45 20 60i 0.5275.12 0.2 1000

1 0.527 1 0.523 22

0.946 94.6% 2. On treatment of 100 mL of 0.1 M solution of CoCl3.6H2O with excess AgNO3; 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 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 Sol. (1)

22

2ppt 23

1.2 10n 2 106 10

3 2compoundn 10 10 10

One mole compound gives two mores ppt. 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)

NHCOCH 3

(2)

OH

(3)

OCOCH 3

(4)

NH2

Sol. (4)

NH2

3 2 4HNO ,H SO

NH3

NH3

NO2

2N Oheat

-I Meta directing group (47%)



4. The product obtained when chlorine gas reactions with cold and dilute aqueous NaOH are : (1) Cl and 2ClO (2) ClO and 3ClO

(3) 2ClO and 3ClO (4) Cl and ClO Sol. (4) 2NaOH(Cold and Dilute) Cl NaCl NaClO 5. Both lithium and magnesium display serval similar properties due to the diagonal relationship;

however, the one which is incorrect, si : (1) nitrates of both Li and Mg yield NO2 and O2 on heating (2) both form basic carbonates (3) both form soluble bicarbonates (4) both form nitrides Sol. (2) or (3) 'Mg 'forms:- 3 22

4MgCO .Mg OH .4H O which is basic carbonate Lithium doesn’t form basic carbonate. 6. A water sample has ppm level concentration of following anions 2

4 3F 10; SO 100; NO 50 The anion / anions that makes / makes the water sample unsuitable for drinking is / are :

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

(3) 24 3both SO and NO (4) only F

Sol. (4) Maximum permissible value for drinking: ppm for F 1ppm ppm for 2

4SO 5000ppm ppm ,, 3NO 50 ppm As given ppm for F is 10 ppm That is why answer is (4) 7. The formation of which of the following polymers involves hydrolysis reaction? (1) Terylene (2) Nylon 6 (3) Bakelite (4) Nylon 6, 6 Sol. (2) Nylon – 6 Formation involves hydrolysis Caprolactum

NH

O

n

533K,

O

CH3NH

CH3 n

Caprolactum Nylon - 6 Ans is (2)



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 particles 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) Sol. (3) Condition for Tyndall effect are :

(1) The diameter of the dispersed particle is not much smaller than the wavelength of light used (2) The refractive indices of the disperse phase and dispersion medium differ greatly in magnitude

9. apK of a 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 Sol. (3)

1 1pH 7 3.2 3.42 2

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

DIBAL H

O

CH3

O

(1) CHO

CHO

(2) COOH

CHO

OH

(3) CHO

CHO

OH

(4) COOH

CHO

Sol. (2) DIBAL–H reduces esters to aldehydes.

O

O

COOH

DIBAL H

OH

COOH

CHO



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

(1)

OHOH2C

OHOH

CH2OCH3

OH

(2)

OCH2OH

OCOCH 3

OH

OH

HOH2C

(3)

OCH2OH

OH

OH

HOH2C

(4)

OCH2OH

OCH3

OH

OH

HOH2C

Sol. (2)

O CH2OH

O CH3

O

CH2OH

OH

KOH

O CH2OH

OH

CH2

OH

HO

hemiacetal group is present

Ag

Tollen's test positive

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

O

CHO

CH3OH

OH

CH3

CH3

(1) 3 3 32

Ag NH OH , CH MgBr, H CH OH

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

(3) 3 3 3 2CH NgBr, H CH OH, Ag NH OH

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

Sol. (2)



O

CO

H

3 2Aq CH OH

O

COO-

3H CH OH

O

CO

OCH3

2

3

H O HCH MgBr

OH

C

CH3

OHCH3 CH3

13. Which of the following species is not paramagnetic? (1) 2B (2) NO (3) CO (4) 2O Sol. (3) For CO

2

2Px2 2 2 2 2z2

2Py

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

All electrons are paired. 14. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water,

fails to decolourize the colour of bromine?

(1) Br

O

CH3

(2) Br

O

(3) Br

C6H5

(4) Br

O CH3



Sol. (2)

O

Br

CH3

CH3

CH3 OK OCH3

CH3CH3

2Br / Water

No deccolourlization 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 2 2 6 2XeF O F XeF O (4) 6 2 4XeF H O XeOF 2HF Sol. (2)

4 1 6 0

4 2 2 6 2XeF O F XeF O

oxidation

Reduction 16. U is equal to: (1) Isothermal work (2) Isochoric work (3) Isobaric work (4) Adiabatic work Sol. (4) By 1st law of thermodynamics U q For adiabatic q 0 Hence U 17. Which of the following molecules is least resonance stabilized?

(1)

CH3

CH3

O (2)

(3) (4) N

Sol. (1) All other compounds are aromatic except

O

C H

3

CH

3



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

3 2 3CH CHCH CH

C

3 2 2CH CH CH Cl

(I) (II)

(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 Sol. (3)

3 3 3CH CH CH CH

C

3 2 2CH CH CH Cl

(I) (II)

CH2Cl

OCH3(III)

Carbocation would be most stable in (III) then (I) & (II). Rate of SN1 reactivity is II I III 19. 1 gram of a carbonate 2 3M CO on treatment with excess HCl produces 0.01186 mole of 2CO . The molar mass of 2 3M CO in g mol-1 is (1) 11.86 (2) 1186 (3) 84.3 (4) 118.6 Sol. (3) 2 3 2 2M CO 2HCl MCl H O CO

2 3

2

N CO

CO

nn 1

12x 60 10.01186

12x 600.01186

x 12.15 Molar mass of 2 3N CO 2 12.15 60



20. Sodium salt of an organic acid ‘X’ produces effervescence with conc. 2 4H SO . ‘X’ reacts with the acidified aqueous 2CaCl solution to give a white precipitate which decolourises acidic solution of 4KMnO ’X’ is: (1) 2 2 4Na C O (2) 6 5C H COONa (3) HCOONa (4) 3CH COONa Sol. (1) 2 4conc.H SO 2 2

2 2 4 2 2 4 2 4white

Na C O CO CO , C O Ca CaC O

2 22 4 4 2 2C O MnO H CO Mn H O

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 2 H atoms is: (1) 10 kg (2) 15 kg (3) 37.5 kg (4) 7.5 kg Sol. (4) Weight of person 75Kg

Hydrogen 10 75 7.5Kg100

Atoms if 3

1 231

7.5 10H 6.022 101

New mass of 21 H mass of 1 atom No. of atoms

3 23

23

0 7.5 10 6.022 10 26.023 10

315 10 gm Weight gain due to 2 3 3H 15 10 7.5 10 37.5 10 gm 7.5Kg 22. The major product obtained in the following reaction is:

Br

C6H5C6H5

H

t BuOK

(1) t

6 5 2 6 5C H CH O Bu CH C H (2) t6 5 2 6 5C H CH O Bu CH C H

(3) 6 5 6 5C H CH CHC H (4) t6 5 2 6 5C H CH O Bu CH C H

Sol. (3)

HBr

C6H5 C6H5

carbon carbon

BrOK

H

C6H5 C6H5

H

C6H5

HH

C6H5



This is a typical example of E – 2 elimination Ansis 3 23. Given 2 2graphiteC O g CO g ;

0 1rH 393.kJ mol

2 2 21H g O g H O l ;2

0 1rH 285.8 kJ mol

2 2 4 2CO g 2H O l CH g 2O g ; 0 1

rH 890.3kJ mol Based on the above thermochemical equations, the value of 0

rH at 298 K for the reaction 2 4graphiteC 2H g CH g will be:

(1) 1144.0kJ mol (2) 174.8kJ mol (3) 1144.0kJ mol (4) 174.8kJ mol Sol. (4) 2 2graphiteC O g CO g ……… (1)

2 2 21H g O g H O2

………. (2)

2 2 4 2CO g 2H O CH g 2O g …….. (3) 2 4graphiteC 2H g CH g ………. (4)

1 2 2 3 gives (4) 2 2 2 2 2 4graphiteC O g 2H g O g CO g 2H O CH g

2 2CO g 2H O

rH 393.5 285.8 2 890.3 74.8KJ 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 Sol. (1) 2 2 2Acidic Basic

ZnO Na O Na ZNO 2 3Basic AcidicZNO CO ZnCO

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) 0

2.12A (2) 0

1.65A (3) 0

4.76A (4) 0

0.529 A



Sol. (1)

o

24r 0.529 A1

o

2.12 A 26. Two reactions 1R and 2R have identical pre – exponential factors. Activation energy of 1R exceeds that of 2R by 10 kJ mol-1. If 1k and 2k are rate constants for reactions 1 2R and R respectively at 300

K, then 2 1ln k k is equal to: 1 1R 8.314 J mol K (1) 4 (2) 8 (3) 12 (4) 6 Sol. (1)

10000

42 8.314 300

1

R e eR

2

1

Rln 4R

27. A metal crystallises 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) a2

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

Sol. (1)

2a

a

The closest distance between two atoms in metallic crystal is 2a2

a2

Ans. is (1) 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 Sol. (2) 10 electrons are present in each species 29. Given 3

2

0 0Cl Cl Cr Cr

E 1.36 V,E 0.74 V

2 3 22 7 4

0 0Cr O Cr MnO Mn 1.51V

E 1.33V, E .

Among the following, the strongest reducing agent is: (1) Cl (2) Cr (3) 2Mn (4) 3Cr



Sol. (2) 3

oCr/Cr

E 0.74

Oxidation potential is maximum hence stronger reducing agent. 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 Sol. (1)

CH3

CH3 CH3HBr

RooR

Antimarkovinikov's CH3

CH3 CH3

BrH

* *

In products two chiral carbon are generated & molecule is unsymmetric Total isomers are 22 4 Ans is (1)



MATHEMATICS

31. The integral

34

4

dx1 cos x

is equals

(1) 4 (2) 1 (3) 2 (4) 2 Sol. (4)

34

4

dxI1 cos x

34

4

dxI1 cos x x

34

4

dxI1 cos x

Add

34

2

4

2dx2I1 cos x

34

2

4

I cosec xdx

I 2 32. Let n

nI tan x dx, n 1 .If 5 54 6I I a tan c 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

Sol. (4) 4

nI tan x dx n 1

4 64 6I I tan x tan x dx

4 2tan x sec x dx

2tan x 1,sec x dx 4t dt

5t c5

41 tan x C5

1a , b 05



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

(1) 73

(2) 52

(3) 5912

(4) 32

Sol. (2) 2x 0, x y 3, x 4y & y 1 x

0,3

1,2

2y x / 4y 1 x

2,1

3,0x y 3

1 22 2

0 1

x xA 1 x dx 3 x dx4 4

1 23 2 3

3/2

0 1

2 x x xx x 3x3 12 2 12

2 1 8 1 11 6 2 23 12 12 2 12

2 8 623 12 12

2 223 12

52

34. A bod 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

Sol. (3)

3P G5

2P y5

Variance n pq

Variance 3105 5

125



35. If dy2 sin x y 1 cos x 0dx

and y 0 1 , then y2

is equal to

(1) 13

(2) 43

(3) 13

(4) 23

Sol. (3)

dy2 sin x y 1 cos x 0dx

dy cos x dxy 1 1 sin x

n y 1 n 2 sin x C

Now, y 0 1 C 2 n2

n y 1 n 2 sin x 2 n 2

4y 12 sin x

Now, 4 1y 12 3 3

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

2 2

2

1 1 11 1 3k1

Then k is equal to: (1) 1 (2) 1 (3) z (4) z

Sol. (3)

On solving 1 3i2 2

2 2

2

1 1 11 1 3 11

2k 1 z 37. Let ˆ ˆ â 2i j 2k

and ˆ ˆb i j

. Let c

be a vector such that c a 3

, a b c 3

and the angle

between c

and a b

be 30 . Then a.c

is equal to

(1) 5 (2) 18

(3) 258

(4) 2

Sol. (4)

ˆ ˆ â 2i j 2k

, ˆ ˆb i j

,

ˆ ˆ î j kˆ ˆ â b 2 1 2 2i 2 j k

1 1 0

Now, 2

c a 3 c 2c.a 0

, Now, a b c 3



a b c sin 3

, 3 c sin 3 c 26

2

c 2c.a 0 c.a 1

38. The radius of a circle, having minimum area, which touches the curve 2y 4 x and the lines

y x is:

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

Sol. (1) Let radius r Centre 0,4 r y x 0

4 r r2

4 r 2r

4r 4 2 12 1

39. If for 1x 0,4

, the derivative of 13

6x xtan1 9x

is x.g x ,then g x equals:

(1) 3

3x1 9x

(2) 3

31 9x

(3) 3

91 9x

(4) 3

3x x1 9x

Sol. (3)

13

6x xtan1 9x

12

2 3x xtan

1 3x x

Put 3x x tan

Then 1 13 2

6x x 2 tantan tan1 9x 1 tan

1tan tan 2

12 2 tan 3x x 1y 2 tan 3x x Differentiating

3

dy 9 xdx 1 9x

3

9g x1 9x



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

Sol. (3) Case I Both number are of form 3

24K C Case II Both number are of form 3

24k 2 C

Probability 655

41. 3

x2

cot x cos xlim2x

equals:

(1) 18

(2) 14

(3) 124

(4) 116

Sol. (4)

3

x2

cos x cos x sin xt

x sin x2

3x

2

1 cos xt 1 sin x8

42

2x

2

sin x 1 cos x1 2 2t8 x x2 2

1 118 2 1

16

42. The value of 21 10 21 10

1 1 2 2C C C C 21 10 21 103 3 4 4C C C C ...... 21 10

10 10C C is:

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

Sol. (2) 21 21 21

1 2 10C C ....... C

10 10 10 101 2 3 10C C C ...... C

21

102 1 2 12

20 102 2



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

Sol. (4)

1P A P B 2P A B4

_______(1)

1P B P C 2P B C4

________(2)

1P A P C 2P A C4

_________(3)

1P A B C16

Adding (1) +(2)+ (3) 3P A P B P C P A B P B C P A C8

3 1 7P A B C8 16 16

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 AP 2AB . If BPC , then tan is equal to:

(1) 29

(2) 49

(3) 67

(4) 14

Sol. (1)

2x P

Cx2

x2

B

A C

x

x / 2 1tan

2x 4

1 tan1 4tan 12 1 tan

4

1 1 1tan tan2 8 4

2tan9



45. The eccentricity of an ellipse whose centre is at the origin is 12

. If one of its directrices is x 4 ,

then the equation of the normal to its at 31,2

is:

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

1e2

a 4 a 2e

2 2 1b a 1 e 4 1 34

3a cos , bsin 1,2 3

Normal: 2 2ax sec bycos ec a b 4x 2y 1 46. If, for a positive integer n, the quadratic equation, x x 1 x 1 x 2 ..... x n 1 x n 10n

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

Sol. (2)

2

2 2n n 1

nx n x 10n3

2

2n 1

x nx 10 03

1

2

2 n 31n 4 13

2 23n 4n 124 3 n 11

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

Sol. (3) p Q p q ~ p ~ p q ~ p q q T T T F T T T F F F T F F T T T T T F F T T F T



p q ~ p q q T T T T 48. The normal to the curve y x 2 x 3 x 6 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

Sol. (4) Put x 0, y 1 In y ' x 2 x 3 y x 3 y x 2 1 y ' 1 Hence normal is x y 1 49. For any three positive real numbers a,b and c, 2 2 29 25a b 25 c 3ac 15b 3a c Then:

(1) a,b and c are in A.P. (2) a,b and c are in G.P. (2) b,c and a are in G.P. (4) b,c and a are in A.P.

Sol. (4) 2 2 2225a 9b 25c 45ab 15bc 75ac 0 2 2 215a 3b 3b 5c 5c 15a 0 15a 3b 5c only

k k ka , b ,c15 3 5

50. If the image of the point P 1, 2, 3 in the plane, 2 3 4 22 0 x 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 Sol. (4)

x 1 y 2 z 31 4 5

Image Q 1, 4 2,5 3

Mid point by 2 4 4 5 6PQ , ,2 2 2

Lie on 2x 3y 4z 22 0 2 3 2 2 2 5 6 22 3 22 2 6 12 2 P 1, 2,3 Q 3,6,13

PQ 4 64 100 2 42



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

(1) 29

(2) 79

(3) 35

(4) 13

Sol. (2) 2cos x t 2 2 25 tan x 5cos x 2 2 cos x 1 9

5 1 t5t 2 2t 1 9

t

25 5t 5t 4t 7 t

29t 12t 5 0 29t 15t 3t 5 0 3t 3t 5 1 3t 5 0

1 5t ,3 3

2 1cos x3

2cos 2x 2cos x 1

2 113 3

2cos 4x 2cos 2x 1

212 1

3

2 719 9

52. Let , , ,a b c R . If 2f x ax bx c is such that 3a b c and

, ,f x y f x f y xy x y R , then 10

1nf n

is equal to :

(1) 190 (2) 255 (3) 330 (4) 165 Sol. (3) f ' x y f ' x y

Put x 0, f ' y f ' 0 y [Clearly f 0 0 ]

f ' y y k

2yf y ky c

2

f 0 0 c 0

2yf y ky

2

1f 1 k 32

5k2



2x 5f x x

2 2

10

n 1

f n 330

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 41 2 3

x y z

and 2 1 7

2 1 1x y z

, is :

(1) 583

(2) 1074

(3) 2074

(4) 1083

Sol. (4) Equation of plane

x 1 y 1 z 1

1 2 3 02 1 1

Distance from 1,3, 7 is 1083

54. If S is the set of distinct values of ‘b’ for which is 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 Sol. (2)

1 1 11 a 1 0a b 1

By solving a 1and b R Then the planes are x y z 1 x y z 1 x by z 0 If b 1 , infinite solution If b 1 , no solution



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

Sol. (4)

2 2 3 2 3 16 9A

4 1 4 1 12 13

2 72 633A 12A

84 51

2 51 63Adj 3A 12A

84 72

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

Sol. (4)

2 2

2 2

x y 1a b

2 2

2 3 1a b

2 2a b 4 2 2a 4 b 2 2 2 22b 3 4 b b 4 b

2b 3 and 2a 1

So equation is 2 2x y 1

1 3

Hence point is 2 2,3 3

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

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

(1) 31,4

(2) 12,2

(3) 12,2

(4) 31,4

Sol. (2)

Area 21 5k 13k 10 282

K 2 So vertices are 2, 6 5,2 2, 2

Orthocenter is 12,2



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 Sol. (1) Length of arc 20 2r r

10 r2

r

2Clearly A 10r r

Or 2A 25 r 5 So max value is 25 when r 5

59. The function 1 1f : ,2 2

R defined as 2

xf x1 x

, is :

(1) surjective but not injective (2) neither injective nor surjective (3) invertible (4) injective but not surjective Sol. (1)

1 1f.R ,2 2

Consider 2

xf x1 x

To find range let x tan / 2

f x is same as sin2

range 1/ 2,1/ 2

For injection

2

22

1 xf ' x1 x

clearly non-monotonic

Hence not injective 60. A man X has 7 friends, 4 of them are ladies and 3 are men. His wire 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 this party, is :

(1) 469 (2) 484 (3) 485 (4) 468 Sol. (3) HUSBAND WIFE 3M 3F 24 2

3C 4

2M1F 2F1M 24 3 22 1C . C 18

1 M2 F 1F 2m 24 3 21 2C . C 12

3F 3M 233C 1

485

20 2r

r r



PHYSICS 61. An observers 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 8 13 10 ms ) (1) 12.1 GHz (2) 17.3 GHz (3) 15.3 GHz (4) 10.1 GHz Sol. (2)

0 s1f f1

, where 0vc

0cv2

( moving towards source)

0 z

112f 10GH 17.3GHz112

62. The following observations were 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 g = 9.80 m/s2 and the simplified relation 3rhgT 10 N m2

, the possible error in surface

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

0.01 0.01 0.01% errors 100 1001.25 1.45 9.8

1.5% 63. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths 1 2r , is

given by:

(1) 2r

3 (2) 3r

4 (3) 1r

3 (4) 4r

3

Sol. (3)

1

hcE

2

1

3

2

E hc3

or 1

2

1 3



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

initial speed is 10 10ms . If, after 10 s, its energy is 2

01 m8

, the value of k will be:

(1) 3 110 kgs (2) 4 110 kg m (3) 1 1 110 kg m s (4) 3 110 kg m Sol. (2)

2dv k vdt m

5 10

2 210 0

1 k dtv 10

2

1 1 k 105 10 10

2

410k 10 kg / m100

65. CP and CV are specific heats at constant pressure and constant volume respectively. It is observed that

C C a p v for hydrogen gas C C b p v for nitrogen gas The correct relation between a and b is:

(1) a = b (2) a = 14 b (3) a = 28 b (4) 1a b14

Sol. (2) For Hydrogen

1

p vJ KC C a2gm

1

P vJ KC C b2gm

1

p vJ kC c b28gm

a b2 28

Or 14a b

ba14

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 Rl such that the moment of inertia is minimum?

(1) 32

(2) 1 (3) 32

(4) 32

Sol. (4)

2 2mr mlI

4 12

2

2m lr4 3

, as 2r l m



2m m l

4 l 3

For minimum I , dI 0dl

2

m 2l 0l 3

l 3r 2

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

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

(1) log1.3t Tlog 2

(2) t T log 1.3 (3) Tt

log 1.3 (4) T log 2t

2 log1.3

Sol. (1) t

0N N e

t00

N N e1.3

t ln 1.3

ln 1.3t T

ln 2

hn2

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) Krichhoff’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 Sol. (1) Theoretical 69. A capacitance of 2 F is required in an electrical circuit across a potential difference of 1.0 kV. A

large number of 1 F capacitance 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 Sol. (4) Let there be P branches in parallel and every branch consists of ‘S’ capacitors in series.

eq1 FC 2 f , P 2 F

S

P = 2S … (i)

a capacitor cannot withstand more than 300 V, So 1000 300S



10S S 43

… (ii)

No of capacitors = P S 32 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

rCEr r

(2)

2

2

rCEr r

(3)

1

1

rCEr r

(4) CE

Sol. (2)

At steady state

r 22

i irr r

ri , 0

2 2C r r 2V V i r

2

2

rr r

2C

2

C rQ CVr r

71.

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



72. In amplitude modulation, sinusoidal carrier frequency used in 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 Sol. (4) Refer Theory 73. In a common emitter amplifier circuit using an n–p–n transistor, the phase difference between the

input and the output voltage will be (1) 900 (2) 1350 (3) 1800 (4) 450 Sol. (3) Refer Theory 74. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass

100gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 750C. T is given by:

(Given: room temperature = 300C, specific heat of copper = 0.1 cal/gm0C) (1) 8850C (2) 12500C (3) 8250C (4) 8000C Sol. (1) Heat released by Cu ball = Heat absorbed by water + Calorimeter Cu Cu w w cal wM S T m S T m S T 100 0.1 T 75 170 1 75 30 100 0.1 75 30

0T 885 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 Sol. (1)

1 1 2 2n D n Dd d

1 2

2 1

n 4n 5

8

1 14

n D 4 1.5 65 10y 7.8mmd 5 10



76. An electric dipole has a fixed dipole moment P

, which makes an angle with respect to x – axis. When subjected to an electric field 1E Ei

, it experiences a torque 1T k

. When subjected to

another electric field 2 1E 3E j it experiences a torque 2 1T T

. The angle is

(1) 450 (2) 600 (3) 900 (4) 300 Sol. (2) P E

1 2E & E

don’t have k component and 1 2&

only have k component so P

will be i j plane.

Let x yP P i P j

1 x yk P i P j Ei

yP E … (1)

2 x y 1k P i P j 3E j

x3P E 060 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) 2g sin

3

l (2) 3g cos

3

l (3) 2g cos

3

l (4) 3g sin

2

l

Sol. (4) I

3g sin2

78. An external pressure P is applied on a cube at 00C 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) P

3 K

Sol. (4)

P PVB or VV BV



Also, V V. T V.3 T

pVV.3 TB

Or PT3B

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 Sol. (1)

For 21 1 1L :f v u

2

1 1 120 v 40

2v 40cm 80. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X –

rays. It produces 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 Y is correctly represented in

(1)

(2)

(3)

(4)

Sol. (4)

chcev

chcV constante

cln ln V ln cln ln V cosntant



81. The temperature of an open room of volume 30 m3 increases from 170C to 270C due to sunshine. The atmospheric pressure in the room remains 51 10 Pa.

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

f 0f 0

pv 1 1n nR T T

5 3

1 1

1 10 pa 30m 118.314J mol x 300

290

41.47 mol 2341.87 6.02J 10 252.5 10 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 :

Time

Current(amp.)

10

0.5 sec

(1) 225 Wb (2) 250 Wb (3) 275 Wb (4) 200 Wb Sol. (2)

d2.5 2.5 100R

83. When a current of 5 mA 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 – 10 V is :

(1) 32.045 10 (2) 32.535 10 (3) 34.005 10 (4) 31.985 10 Sol. (4) V ig R G

3

10 G R5 10

1 10 0.5 2.52

2000 – 15 = R 1985 84. A time dependent force F = 6t acts on a particle of mass 1 kg. if the particle starts from set, 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 Sol. (4) F 6t

dvm 6tdt



v t

0 0dv 6tdt

2v 3t Work done k

2 2f i

1 1m v mv2 2

21 1 3 02

4.5J 85. A magnetic needle of magnetic moment 2 26.7 10 Am and moment of inertial 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) 6.89 s (3) 8.76 s (4) 6.65 s Sol. (4)

6 2

2

I 7.5 10 102 . 2 10MB 6.7 10 0.01

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) :

(1)

g

OR

d

(2)

g

OR

d

(3)

g

OR

d

(4)

g

O d

Sol. (3) Theoretical 87. A body is thrown vertically upwards. Which one of the following graphs correctly represent the

velocity vs time?

(1)

t

(2)

t



(3)

t

(4) t

Sol. (2) u Initial velocity is +ve

dV gdt

V

t

0U

dV g dt

V U gt V U gt (straight line with negative slope)

88. A particle A of mass m and initial velocity collides with a particle B of mass m2

which is at rest.

The collision is head on, and elastic. The ratio of the de-Broglie wavelength A to B after the collision is :

(1) A

B

2

(2) A

B

23

(3) A

B

12

(4) A

B

13

Sol. (1)

1 2mmv mv v2

21

vv v2

… (1)

2 1v 0 v v … (2) On solving 1 2v v 3, v 4 v 3

1 1 1

2 2

h m v 22 h m v

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)

KE

0 T t (2)

KE

0 T t T2

(3)

KE

0 t T2

T4

T (4)

KE

0 t TT2

T

Sol. (3)



90. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that this density remains same, the stress in the leg will change by a factor of :

(1) 19

(2) 81 (3) 181

(4) 9

Sol. (4)

Stress MgA

v gA

3

2

q vgq A

q

JEE MAIN – 2017 QUESTION PAPER & SOLUTION CHEMISTRY · JEE MAIN 2017 CODE (C) CENTERS : MUMBAI...

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Transcript of JEE MAIN – 2017 QUESTION PAPER & SOLUTION CHEMISTRY · JEE MAIN 2017 CODE (C) CENTERS : MUMBAI...