TARGET IIT-JEEerp.motioniitjee.com/images/041d65de-15e8-4f23-b77... · class xii [main] answer key with solution target iit-jee physics mathematics chemistry date : 24-07-2016 [ all

CLASS XII [MAIN]

ANSWER KEY WITH SOLUTION

TARGET IIT-JEE

PHYSICS

MATHEMATICS

CHEMISTRY

DATE : 24-07-2016

[ All Batch ]

SECTION – A1. A 2. C 3. B 4. D 5. D

6. A 7. D 8. A 9. D 10. C

11. C 12. B 13. D 14. B 15. C

16. B 17. B 18. C 19. B 20. C

21. C 22. C 23. D 24. A 25. A

26. A 27. B 28. B 29. C 30. D

SECTION – A

1. B 2. A 3. A 4. A 5. C

6. B 7. B 8. D 9. B 10. C

11. C 12. D 13. D 14. A 15. D

16. A 17. B 18. B 19. A 20. D

21. C 22. B 23. D 24. C 25. D

26. B 27. A 28. B 29. C 30. A

394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , [email protected] 1

SECTION – A

1. A 2. C 3. A 4. D 5. B

6. A 7. B 8. C 9. D 10. A

11. C 12. A 13. B 14. D 15. A

16. D 17. D 18. C 19. B 20. D

21. C 22. A 23. D 24. D 25. D

26. C 27. A 28. B 29. B 30. B


MATHEMATICS

SOLUTIONS

SECTION – ASingle Correct

1. A

dtdx

= 2at and dtdy

= 2a

= dxdy

= t1

at2a2

[Slope of tangent at the point (at2, 2at)]Since tangent line is perpendicular to x-axis,then

dxdy

= t1= t = 0

Therefore required point = (0, 0)

2. C

R(b + c) = bca

R(b + c) = 2RsinA bc

sin A = bc2cb

1

(using AM GM equality holds if b = c) A = 90° and b = c (C)

3. B

– 3 < < 3y = x3 + x2 + x + 5

dxdy

= 3x2 + 2x + 1

dxdy

> 0 3x2 + 2x + 1 > 0

42 – 12 < 02 < 3

– 3 < < 3

4. Dy2 = 8x … (1)Let slope of tangent is m tangent to the parabola is

y = mx + a/m here a = 2 Equation of tangent is

y = mx + 2/m … (2)Put value of y from (2) in xy = –1

x (mx + 2/m) = –1m2x2 + 2x = –m

m2x2 + 2x + m = 0 (2) touches the xy = –1 D = 0

4 – 4 × m2 × m = 0 m3 = 1 m = 1 equation of tangent is

y = x + 2

5. D

0 < x < 4

0 < x < 414.3

0 < x < 0.8[x] = 0

4/

0

n2n )000x(d)xtanx(tan

=

4/

0

22n dx)xtan1(xtan

=

4/

0

22n dxxsec.xntan ; t = tanx

= 1

0

2n dtt = 1

0

1n

1nt

= 1n1

6. AA = x2

dtdA

= .2x. dtdx

= × 2 × 7.5 × 3.5= 52.5 cm2/sec

7. DIf Statement-I is false but Statement-II istrue

n

n

'SS

= 17n41n7

thenn

n

'TT

= 17)1–n2(41)1–n2(7

is not 47

S-I is not true.S-II is true.

8. ASlope of C1 is sin x and for x > 0 slope of C2

is 23 . Thus for point of contact


sin x = 23 x =

3

or 32

Hence point of contact is

21,

3 or

23,

32

For

21,

3 , we get a = 21

– 32

For

23,

32

, we get a = 23

– 3

9. D

g(2) = 2

20 , g ‘(x) = 4x1

x

.1 g ‘(2) = 172

f ‘(x) = eg(x).g ‘(x)

f ‘(2) = eg(2).g ‘(2) = e0. 172

= 172

10. Cy2 = x3 ......(i)equation of tangent at P(4m2, 8m3) isy = 3mx – 4m3 ......(ii)on solving (i) & (ii)x3 = (3mx – 4m3)2 x = 4m2, m2

Put x = m2 in equation (ii)y = – m3 so let Q(m2, –m3)

so m23

dxdy

)m,m( 32

& slope of normal at (m2, –m3) = 32

m

since tangent at P & normal at Q are same

m32

= 3m 9m2 = 2 27m2 = 6

11. COA = HAR = 2R cosA (distance of orthocentre fromthe vertex A is 2R cosA)

cosA = 21

A = 3

(C)

12. BThe equation of the curve is y = e2x + x2,when x = 0, y = 1.

dydx = 2e2x + 2x = 2 at the point (0, 1)

slope of the normal = (1/2) and thenormal passes through the point (0, 1)the normal has the equation y1 = (1/2)x x + 2y 2 = 0,

required distance = 2/ 5 .

13. Dindependent of a|sin x| + |cos x| period = /2

14. B

x y = a

1

2 x +

12 y

dydx = 0

O Px

Q

y

dydx = –

yx

Equation of tangent at (x, y) is

Y – y = – yx

(X – x)

Yy – y = –

Xx

+ x

Xx

+Yy

= x + y = a [from Eq. (i)]

Xa x +

Ya y = 1

OP = a xand OQ = a ythen OP + OQ = a x + a y

= a( x y)

= a · a = a.

15. C

b = caac2

H.P..

log (a + c) + log

caac4–ca

log (a + c) + log (a – b)2 – log (a + c)= 2 log (a – c)= 2 log (c – a) c > a


16. BEquation of the joining the points (0, 3) and(5, –2) is y = 3 – x. If this line is tangent to

y= )1x(ax , then (3 – x) (x + 1) = ax should

have equal roots. Thus (a – 2)2 = 12

a = 2 ± 2 3

17. B

1xx

12

h1zz

dz

lim

hx

x2

0h

1

0)10(1)hx()hx(

1

lim2

0h

= 1xx

12

18. C

dydx = 3ax2 + 2bx + c ...(i)

Since, the curve y = ax3 + bx2 + cx + 5touches x-axis at P(–2, 0), the curve meetsy-axis in (0, 5).

(0,5)

dydx = 0 + 0 + c = 3 (given) ...(ii)

and(–2,0)

dydx = 0 12a – 4b + c = 0

12a – 4b + 3 = 0 [from Eq. (ii)] ...(iii)and (– 2, 0) lies on the curve, then

0 = – 8a + 4b – 2c + 5 – 8a + 4b – 1 = 0 [ c = 3] 8a – 4b + 1 = 0 ...(iv)

From Eqs. (iii) and (iv) we get a = – 12

, b = – 34

–10a – 100b + 1000c = – 10 × –12

–

100 × –34

+ 1000 × 3

= 5 + 75 + 3000= 3080.

Now d = 3080

So d

1000

= 3.

19. Barea of hexagon

= – 91

Bsinca

21Asinbc

21Csinab

21

= – 91

[3] = 32

(where is the area of the triangle ABC)

triangle of areaheaxgon of area

= 32

20. C

1

03x

2

2)x2(e

dxxI 3 ; t = x3

=

1

0t )t2(e3dt

=

1

0t1 ))t1(2(e

dt31

(by 3rd property)

=

1

0

t

)t1(ee

31

e31I2 I1

21. C

= 2h·6

= 3h ; (where h is the altitude from A);

Also = 2r·21 (using = r·s)

2r·21 = 3h h

r = 7

2

now APQ and ABC are similiar

hrh = 6

PQ h

r1 = 6PQ

721 = 6

PQ 7

5= 6

PQ PQ = 7

30

Ans.


22. C

Put 1 + sin 2x sin = t2 cos d= x2sindtt2

f(x) =

x2sin1

x2sin1

2

x2sin)1t(2 . t

1. x2sin

dtt2

f(x) =

xsinxcos

)xsinx(cos2

2

x2sin)1t(4

dt

f(x) = – 34

cot x cosec x

dx)x(f = 34

cosec x + c.

23. DS1 = T1 + T3 + T5 …… + T2n–1& S2 = T2 + T4 + T6 ……. + T2n S2 = rT1 + rT3 + rT5 …….. + rT2n–1 S2 = rS1

1

2SS

= r

24. A1 – x2014 = t –2014 x2013 dx = dt

=

2013

1

0

2013/1

x2014dtt

= 20141

1

0 20142013

20131

t1

t dt

= 20141

dtt1

t

120132014

20141t1t

201411

0 20132012

2014201312013

1

= 20131

11

2014

20120 2013

1 tdt

t

t = x2013 1 1

2013 2014

0

1 x dx

25. AHere e2 = dfNow dx2 + 2ex + f = 0 given

dx2 + 2 df x + f = 0 x = – df

Putting in ax2 + 2bx + c = 0 we get

a df

+ c = 2b df

da

+fc

= eb2

da

, eb

,fc

are in A.P..

ad

, be

, cf

are in H.P..

26. A

Here, f’(x) = (f(x))2 > 0, axat

))x(g(fdxd

= f’(g(x)) axLim

ax)a(g)x(g

As f’(g(x)) 0g(x) must be differentiable at x = a

27. B

Clearly x = 2cos11

= 2sin1

sin2 = x1

& y = 2sin11

= 2cos1

cos2 = y1

& z = 22 cossin11

= y1.

x11

1

z = 1xyxy

xyz = xy + z

28. BF(a) = F(b) (Apply RHVT)

29. C

f’(x) = 0 x = n

a.....aa n21 = 9 Also

(a1 ........ an)1/n = 9

a1 = a2 = ....... = an Sum = 81

9/119/1

.

30. D

a

0

dxxaxa1

=

dsec.tan2tan2 2

8/

0

(sin2 = 2tan1

tan2)

= a

02/32/3 x)ax(

32.

a1

=8/

0tan

= 12aa)a2(a32 2/32/32/3

= 1212a2.a32 2/3 1a

34

169a


Section–A1. B

E = kq/r2 outside a uniformly charged sphereand also for a point charge.

2. A

R

M R m

dx

x

F = 2R

2R

mGM dxR

x

= 2

GMm2R

3. AThe given circuit is equivalent to

As 10 × 4 = 5 × 8 this is balancedWheatstone network

Therefore R = 189)810()45(

= 6 ohm

4. A

Use C = dA0

A = 0

dC

5. CThe gravitational field at any point on thering due to the sphere is equal to the fielddue to single particle of mass M Placed atthe centre of the sphere. Thus, the force onthe ring due to the sphere is also equal tothe force on it by particle of mass M placedat this point. By Newton's third law it is equalto the force on the particle by the ring. Nowthe gravitational field due to the ring at adistance d = 3 a on its axis is given as

g = 23/222 8aGm3

)d(aGmd

a a

M md = a3

The force on sphere of mass M placed here is

F = Mg = 28aGMm3

6. B

Orbital velocity = hRRg

20 where R is radius

of earth.

If h = 0, RgRRgv 0

20

0

If h = 2R

, 32

2

02

0 RgRR

Rgv

= 03

2v .

(B)

7. B

PVR

2

Resistance of first bulb is 1

2

1 PVR ,

and resistance of the second bulb is

2

2

2 PVR

In series same current will pass througheach bulb

Power developed across first is 1

22'

1 PVIP

and that across second is 2

22'

2 PVIP

1

2'2

'1

PP

PP

as 12 PP 1PP

1

2

1PP

'2

'1 '

2'

1 PP

The bulb rated 220 V & 40 W will glow more.

8. DGMm

r

= 2GMm 1 mvR 2

r = 20

GMv

2e

2GMvR

2 2 20 e2v v v

PHYSICS


9. B

a b c

d f g

i 2 1

3 7

1 V 2 V

4 V

8 V 1 V

Along agbfcdVa + 1 - 2 - 4 + 8 - 1 = Vd

Va + 2 = Vd Vd - Va = 2 V (B)

10. CFringe width . Therefore, and hence decreases 1.5 times when immersed in liquid.The distance between central maxima and10th maxima is 3 cm in vacuum. Whenimmersed in liquid it will reduce to 2 cm.Position of central maxima will not changewhile 10th maxima will be obtained at y= 4cm.

11. C

i = 50

20 RPotential drop across R = Potential dropacross AB

50 .R

20 R = 30 R = 30

12. DIf d sin = ( - 1)t, central fringe is ob-tained at O.If d sin > ( - 1)t, central fringe is obtainedabove O andif d sin < ( - 1)t, central fringe is obtainedbelow O. (D)

12. DAt earth's equator effective value of gravityis

geq = gs – 2 Re

If geff at equator to be zero, we havegs – 2 Re = 0

or = e

e

Rg

14. A

22 r16kq

)r4(kqE

q

y

x

2r

(0, 0)

+ + +++++

+++++

+++++

+ + + q

– ––––

––––

(4r, 0)

15. DAfter charging, isolatedSo Q = constant d increasing C

V C1

V increasing

16. AWe can observer that 10, 10 and 20are shorted and bypassed, so no current will

flow in then and I = 90V45

= 0.5 A.

+

–A

20

9010 1045V

Alternative :

+–

A90

10 10

20

–

–

I

–

A5.090

V45I

17. B

Shift 1 7 1 1 4 1

5t . D t . D DS

d d d

t = 8 × 10–3 mm = 8 m

18. BFrom symmetry flux through each point ofthe sphere is same.

Flux through whole sphere = o

q

q R


Total surface area = 4144 2

R m2

flux through 0.2 m2 is

= o

o qq

202.0

4/

(B)

19. A

So, eg. circuits diagrams

(A) Req = 3 I = eqR

9 = 3A

Heat produced in cell=I2r=9 × (2/3) = 6 W

(B)

I2 = 21

1

RRIR

= 5

835

)8/35(3

= 5

7 = 1.4 A

So, (A) is correct

20. D

0 0 0

56 24 24

q q q

21. C

there is zero potential difference across 4and 6 resistance.

i = 202 = 10 A

power by batteryPb = i = 20 × 10 = 200 W Ans.

22. B

Given C = 0 Ad

If separation is halved d' = d/2

C' = 0A/d' = 0 A 2d

= 2C

23. D

Efficiency = = output powerinput power

= 2i R

i i = R r

= R

R r 3R + 3r = 5R or 2R = 3r

0.6 = R

R r

= 6R

6R r = 0.9 = 90% Ans.

24. CThe potential on the surface of the sphere 1is given by

v1 = bq

41

aq

41 2

0

1

0

. .. .. .. . (a)

The potential on the surface of the sphere 2is given by,

V2 = bq

41

bq

41 2

0

1

0

v = v1 – v2

v = bq

41

aq

41 11

v=

b

1a1

4q1

(C)


25. D

P = I2R = I22 = 18I = 3A

Pnet =

R

2I 2

2 + [I2R]

= q4RI2

+ I2R

= 218

+ 18 = 27

26. B

ER

GMm

2For escaping to infinity

Energy required = ER

GMm 2

(B)

27. Athe ring can be treated as electric dipole.

28. BEnergy stored in capacitor is in the form ofelectric field or electric energy which is apotential energy

29. C E = j

30. A

v 20 v v 52 2

= 0 ;

v – 20 + v + 2 (v–5) = 0 = 4v – 20–10=0

v = 304 =

152 v – 5 =

15 52

= 15 10

2

= 52 i =

5 / 21 amp. Ans.


SECTION - A1. A

2. CCharge Transfer

3. A

4. D

5. B

6. A(SnCl2)n & (BeCl2)n

7. B

8. C

9. D

10. A

(A)Cocl2

(B)Nicl2

Co(CN)Buff. ppt

2 Ni(CN)Green ppt

2(C) (F)

K4[Co(CN) ]Brown

6

K [Ni(CN) ]Yellow

2 4

(D) (G)

K [Co(CN) ] Yellow

3 6(E)

KCN KCN

KCN(ex) KCN

Exposure

&

[o]

11. C

12. A

13. B

14. D

15. A

16. D

17. D

18. C

19. B

20. D

21. Cproduct A is obtained by OMDM (i.e. additionof HOH acc. to morkovnikov’s without rear-rangement.)product B is obtained by HBO (Hydroborationoxidation) (i.e. final product is addition ofHOH acc. to antimorkovnikov’s without rear-rangement.)

22. A23. D

OH

O MgBr

+ PhMgBr

Acid base reaction

Benzene

H

+

24. D

CHEMISTRY


25. D

26. CRate of dehydration stability of carbocation

27. A

A B Kc = 2

1KK

= xaxb

a - x b + x x = 21

21KK

bKaK

Therefore, (A) option is correct.

28. B20 × M × 2 = 20 × 0.02 × 5M = 0.05

29. B

CH2=CH—CCH

30. B

Cis-2-butenePd/BaSO4

H2

Br /CCl (Anti addition)2 4

(Racemic Mixture)

Documents

TARGET IIT-JEEerp.motioniitjee.com/images/041d65de-15e8-4f23-b77... · class xii [main] answer key with solution target iit-jee physics mathematics chemistry date : 24-07-2016 [ all