solutionsvidyamandir classes vmc/jee mains-2017 1 solutions solutions joint entrance...

Vidyamandir Classes

VMC/JEE Mains-2017 1 Solutions

SOLUTIONS Joint Entrance Exam/IITJEE-2017

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

Vidyamandir Classes


Joint Entrance Exam | JEE Mains 2017

PART-A CHEMISTRY

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.(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.(4)

4.(4) 2 2

&cold dilCl NaOH Na Cl Na Cl O H O

5.(2) 1, 3 & 4 are correct statements. 6.(4) Permissible limit of F– is 1 ppm.

Vidyamandir Classes


7.(2)

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

11.(2)

12.(2)

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.(2)

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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.(4) For adiabatic process : q = 0

So from Ist law U q w

We can write U w

17.(1)

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

So, order of reactivity

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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.(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.(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.(3)

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

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

Vidyamandir Classes


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.(1) 2

nn

r 0.529Z

2r 0.529 4 2.116Å

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

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

29.(2) Re d1

EReducing strength

30.(1)

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

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PART-B MATHEMATICS

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.(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.(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

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.(3) n = 10

1525

p

1025

q

var( )X npq

15 101025 25

3 2105 5

125

35.(3) ( 1) cos2 sin

dy y xdx x

cos1 2 sin

dy x dxy x

( 1) (2 sin )n y n x nc

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

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

13

y

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

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.(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

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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

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.(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.(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.(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.(4) 124

P A P B P A B

124

P B P C P B C

Vidyamandir Classes


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.(1) tan

1tan2

; 1tan4

tan tan

1 1 122 4 4

1 9 918 8

45.(4) 2 2

2 2 1x ya b

412

a 2a

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.(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

Vidyamandir Classes


11n 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 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.(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.(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 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

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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.(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.(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

5 21 21 5 1025 49 9 83

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.(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

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56.(4) Clearly 2 2

2 22 ; for 1x yaea b

2 2 2

2 3 11a a e

2 2 2 2

2 3 1a 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 2 21 3a , b

2 2

11 3x y

32 1

3x y

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.(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

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20 2 52 rad

5

21 5 2 252

A .

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.(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.(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.(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.(3) 2

43 3

hc E EE

1

2 hc E E E 1

2

13

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.(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

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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

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.(1) ln 2

T

0 0.3N N

N

01.3N

N

0tN N e

11.3

te

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

t T

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

Vidyamandir Classes


69.(3) 8 24eqFC F

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

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

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

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 RA

v i R

c

B

RR

“–ve” sign indicates output is exactly out of phase i.e. phase difference = 180

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

Vidyamandir Classes


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

Therefore 9

1 34 650 10 1.5

0.5 10y

1 7.8y mm

76.(2) T p E

ˆ ˆcos sinp p i p j

sinpE T and 3 cosp E T

tan 3 60º

77.(4) 2

sin2 3l mlmg

3 sin2

gl

78.(4) /

P V P PVK VV V V K K

(3 )3

PV PV T TK K

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.(4) min hceV

minln ln ln hc Ve

y k x

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

Vidyamandir Classes


82.(2) qR

area of I t graph Δ=

R

1 1

10 0.5 10 0.5 100 250 Wb2 100 2

83.(4) 5gI mA

g g sV I R R

310 5 10 15 sR 315 2 10 2000sR

31985 1.985 10sR 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.(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.(3) 3 , 0 GMg r r RR

2 ,GMg r R

r

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

89.(3) K is maximum at mean position and minimum at extreme position and extreme position is reached at T/4.

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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