Concept Recap Test Mains 1 Sol

7/28/2019 Concept Recap Test Mains 1 Sol

1/18

AITS-CRT(Set-I)-PCM(S)-JEE(Mains)/13

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1

ANSWERS, HINTS & SOLUTIONS

CRT(SetI)

PHYSICS CHEMISTRY MATHEMATICS

Q. No. ANSWER ANSWER ANSWER

1. C C A

2. B C A

3. C B C

4. B B D

5. A B A

6. B B A

7. A D B

8. B D A

9. D A B

10. A A C

11. C B A

12. C A D

13. B B C

14. A D A

15. B B A

16. C C C

17. C C C

18. C B B

19. B B C

20. C C C

21. D B A

22. A C C

23. D D A

24. A C B

25.C A D

26. A A C

27. A B A

28. C A D

29. C A C

30. D C B

PPhhyyssiiccss

ALL

INDIA

TEST

SERIES

FIITJEE JEE (Main)-2013

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ongTermC

la

ssroomP

rogramsandMediu

m/

ShortClassroomP

rogram4

inTop10,

10inTop20,

43inTop100,

75in

Top200,

159inTop500Ra

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totalselecti

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in

IIT-JEE

2012


2/18



2

PART I

1. At the highest point of the curve, B,2BV gr

= ; mechanical energy is also conserved.

3. The total time from A to C

4AC AB BC BC

Tt t t t = + = +

Where T = Time period of oscillation of spring-mass system tBC can be given by

2sin

BCBC AB t

T

=

Putting1

,2

BC

AB= we get

12

BC

Tt =

AC2 m

t3 k

=

4. The force acts along the tangential direction.

5. Using impulse momentum theorem

cm cm

JJ mV V

m= = ;

Using angular momentum theorem about centre of mass2m

J2 12

=

6J

m =

speed of point A is2J

m

6. After long timemv0 = 2mv

0vv2

= .

7. Potential difference = 0 Charge = 0

8. dI neAv=

I

constantne

=

Avd = constant

9. Use Kinematics : Acceleration =2

2

d x

dt


3/18



3

11. 2 1 2 1 2

0 2 0 1

q q q q1 1 1mv

2 4 r 4 r + =

By solving we get

21 2

0 1 2

1 1 1 1mv q q

2 4 r r

=

; v = 90 m/s

12. at t = 0, resistance of inductor =

14. From the F.B.D.S of the two blocks,V g

10g

kx

kx

V g

Vg + kx = 10g

And Vg = kxx = 10 cm

15. Volume equality gives

2 3 =1

2 h 3

h = 4m

tan =4 a

3 g=

2m

3m

h

3mA

a

16. s E' = +

or, s E' =

2 2 10

6 24 24

= + =

2 2 6

6 24 24

= =

2

T 4.8'

= =

hour

2

T 8'

= =

hour

17. W = P0V = P0V0

and 0 0 0 0 0U 2P (2V V ) 2P V = = 0 0Q W U 3P V= + =

18. AB CDR RKA

= =

( )eq BCR 2KA

=


4/18



4

we have,100 0

2

KA 2KA

+

= CT 0

KA

TC = 28

19. 0ms A(T T )

dT A 1

dt m d

20. 1KA(200 )Q

t 2L

=

1 2A( )(2K)2L

=

2( 18)(1.5K)A2L

=

or,1 1 2 2200 2 2 1.5 27 = =

1 2116 , 74C C = =

21. Area under acceleration-time graph gives change in velocity

Hence total4 4

A 4 12

=

= 8 4Vf Vi = 4Vf 3 = 4Vf= 7 m/s

22. 0 = u cos 30 g sin 30 tucos30

t

gsin30

=

..(A)

21Hcos30 usin30 t gcos30 t2

=

By equation (A) and (B), we get

{ }2 2u cot 2gH

H 1 v 30g 2 5

= + = =

23. Maximum restoring force develops at the end where force is applied. This force decreaseslinearly such that it becomes zero at the other end so stress also decreases linearly.

24.1

2

KA2

=1

2

mv2

+1

2

2mv2

mv2 = 13

KA2

Work done by friction on Q =1

22mv

2

=1

3KA

2


5/18



5

25. Using formulae B = 0I(sin sin )

4 r

We get 0I

(1 cos )4 d

.

26. Use F q(v B)=

and write the appropriate equations.

27. According to Mosleys law

2

1

(z 1)

2z 1

21

(z 1)4

(z 1)

= =

Solving (z1 1) = 2 (z 1)z1 = (2z 1)

Similarly2

z 2

2

2

(z 1) 1

4(z 1)

= =

Solving

22(z 1) (z 1) =

2

z 1z

2

+ =

.

28. maxK h=

here maxK (e)Ed=

h eEd =

12404eV

200=

6.2eV 4eV= = 2.2 eV.

29. [ ]1 1 1 1

1 ( 1)f R R

= =

when one surface is silvered

eq

1 2( 1)

f R

=

eq

ff

2=

concave mirror of focal length 10 cm.

30. For [ ] 11 1 R

P 1 ff R 2( 1)

= = (since one side is silvered)

for Q [ ] 21 1 1 R

1 ff r 1

= =

For R [ ] 31 1 1 R

1 ff R r 2[ 1]

= =


6/18



6

CChheemmiissttrryy PART II

1. Period of the revolution T =2 r

V

2 2 1 2 1

1 2 1 1 2

T 2 r V r V

T V 2 r r V

= =

Also Vn = V1 / n

Vn = r1 n2

22 2

11 1 1 2

T r 2 8V

T r v / v 1

= =

Hence, (C) is the correct option.

2. Magnetic moment = ( )n n 2+ B. M. where n is the number of unpaired electrn

V (Z = 23) (Ar) 3d34s2 n = 3, 15 BM = x

Cr (Z = 24) (Ar) 3d54s1 n = 6, 48 BM = y

Mn (Z = 25) (Ar) 3d5

4s2

n = 5, 35 BM = z

3.[ ]1

H AHA H A ; K

HA

+ +

+ =

[ ]2H B

HB H B ; KHB

+ +

+ =

By mass balance

[ ] [ ] 1initial eqHA HA A C = + =

[ ] [ ] 2initial eqHB HB B C = + =

By change balance[ ] [ ]1 2k HA k HB

HH H

++ +

= +

[ ] [ ]2

1 2H k HA k HB+ = +

{ } { }1 1 2 2k C A k C B = +

1 1 2 2 n nH k c k c .................k c+ = + +

n

i ii 1

k c=

= for n number of weak acids.

4.

O O

CH CH3NO2

Cl

O

OCH3

O2N


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7

5.

OH

CH3

H

D

[ ]( )0X +

TsCl

(i) Tosylation OTs

CH3

H

D

( )3

2

(2)CH COO

(i) SN invers ion

C OH3C

O

CH3

D

H

HO

OH

( )3( )4

H+H

CH3

OH

D

18

3CH COOH+

[] = (X0)

6.3 2 3 2

NaHCO NaOH Na CO H O+ +

10 0.15 -12.6 0.1 = 0.24Meq. Of NaHCO3 = 0.24 50 = 12

And mass of NaHCO3 = 1.008 g.

Also meq. of (NaHCO3 + Na2CO3) = 12.4 0.1 50 = 62Meq. of Na2CO3 = 50 and mass of Na2CO3 = 2.65 gNa2CO3 = 53%NaHCO3 = 20.16%

7. Initial solution is aqueous NaCl so it is neutral solution pH = 7But during electrolysis following reaction takes place.

2NaCl + 2H2O H2 + Cl2 + 2NaOH

and due to NaOH solution becomes basic.So new pH = 7 + 4 = 11

pOH = 3 [OH] = 103 [NaOH] = 103 M NaOH = 103 0.15

= 1.5 104

= eq of NaOH

By Faradays Law, No of eq =it

96500

1.5 104 =1 t

96500

t = 14.5 sec

8. Rate = N (N = No. of atoms of element yet not decayed)R1 = N1, R2 = N2Number of atoms decayed in time (t2 t1)

( ) ( )2 1 2 12 1 R R t R RR R0.693

= = =

Number of atom decayed in time (t2 t1) t (R2 R1)Hence (D) is the correct option.

9. The balanced chemical reactions are

( ) 42 5 y 2 5xC H TiCl xC H Ti yCl+ + +

Cl Ag AgCl ++


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8

Moles of ( )2 5 yx0.612

C H TiCl48 29x 35.5y

=+ +

Moles of C2H5 = ( )0.612x 0.1

.......... i48 29x 35.5Y 30

=+ +

Moles of Cl

=0.612y

48 29x 35.5y+ +

= mole of AgCl

= ( )1.435

.......... ii143.5

from eqation (i) and (ii)

x 1

y 3= or y = 3x

Substituting in Eq (i) gives:

0.612x 1

48 29x 106.5x 300=

+ +

48 + 135.5x 183.6 xx = 1, y = 3

10. Equivalence point is reached when 16.24 ml of 0.02 M NaOH is added, there is 50%neutralisation of p-hydroxy benzoic acid. Which is converted into sodium salt. Thus at 50%neutralisation

OH

COOH

OH

COO

=

and is a buffter

[ ]

6 4

16 4

HOC H COO

pH PKa log HOC H COOH

= +

PKa1 = 4.5%The HOC6H4COO

formed is amphiprotic

OH

COO

2H O+

O

COO

3H O++

OH

COO

2H O+

OH

COOH

OH+

Basic

pH = 1 2PKa PKa

2

+


9/18



9

7.02 = 24.57 PKa

2

+

PKa2 = 9.47

14.

NH2

COOH

H

OHH

COOH

( )2NaNO , HCl/

Diazotisation

N

COOH

H

OHH

COOH

Nhydride

shift

CH2

COOH

C OH

COOH

CH2

COOH

C O

COOH

keto acid

2CO

C

CH2

COOH

OHC

CH3

COOH

O

15.

Cl

Cl

( )2Me Et O1 eq

MgCl

Cl

Cyclooctatetraene (Tubshaped not planer) Nonaromatic

16. CdMe2 (dimethyl cadmium) is a poor nucleophile so it only reacts with highly electrophilic site of

acid halide. It doesnot react with carbonyle compound.

CClO

CO H

( ) 22

i CdMe

(ii) H O

CCH3O

CO H

17. O

Dil KOH

OH

O

Aldol condensation producing chiral centre


10/18



10

19.OH

CH3I

H

D

I

* ( )OH , H +

O

CH3I

H

D

I

* ( )18

NGP OH

I

CH3I

H

D

*

OH18

O

CH3I

H

D

OH18

3H O+

OH

CH3I

H

D

OH

* *

18

20.

I( )

2Me CuLi

Corey house synthesis CH3 2

Br / h / Mg

CH2MgBr1,4 Addition

O

Me

CH2

BrMgO

Me

Clemmensen reduction CH2 Me

3H O+

CH2 Me

O

27.2 1 1 1

1 1

Fe 2e Fe, E ; G nFE

G 2FE

+ + = =

3 22 2 2Fe e Fe ; E ; G 1 F E

+ ++

G is an additive property but E does not, thats why it is essential to convert E into G andfinally G into E)For the reaction

3Fe 3e Fe+ +

1 2G G nFE + =

-2FE1 + (FE2) = 3FEE = (2E1 + E2)/3

28. Dehydration of IV is most facile since, it gives an aromatic compound dehydration of III gives aconjugate compound dehydration of III gives a conjugated diene which is stablised by resonance

OH 2H o

(iv)


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11

OH 2H o

(iii) Dehydration of II gives only cyclohexane which is not stabilised by resonance

OH 2H o

(ii) In contrast phenol (1) does not undergo dehydration. Thus the ease of dehydration isIV > III > II > I

29. (a) ( ) ( )

( )

3 3X UnstableY

NaOH AgNO AgOH NaNO

Delig. white crystal

+ +

2 2Brown

2AgOH Ag O H O +

(X) is a powerful cautery and breaks down the proteins of skin flesh to a pasty mass. i.e. X iscastic soda.


12/18



12

MMaatthheemmaattiiccss PART III

1.2

cot x

2

[lncosec x 2sinxcosx ]e dx

sin x

+ (put cot x = t)

cosec2

x dx = dt dx =

2

1dt

1 t

+=

t 2

2

2te ln(1 t )

1 t

+ + + dt = e

tln (1 + t

2) + c = 2ecotx ln (cosec x) + c.

2. Let v and s be the volume and total surface area of closed right circularcylinder of radius r and height h.

v = r2hs = 2rh + 2r2given v = 2156

r2h = 2156 .. (1)

s = 2rh + 2r2 = 2r 22

21562 r

r

+

h

s = 24312 44 rr 7

+

2

ds 4312 88r

dr 7r= + ..... (2)

ds0

dr=

2

4312 88r 0

7r + =

2

4312 88r

7r=

3 37 4312r 788

= = r = 7

2

2 3

d s 8624 88

7dr r= +

2

2 3

r 7

d s 8624 88

7dr 7=

= + > 0

s is minimum when r = 7 units.

3. Consider g(x) = f(x) x 1

2

for any x = a

g(a) = f(a) a 1

2

g(f(a)) = f(f(a)) f(a) 1

2

= 1 + a f(a) 1

2

=1

2+ a f(a)

g(f(a)) = g(a) between any point a and f(a) there lies a root of g(x) .... (1)g(f(x)) = g(x) g(f(x)) + g(x) = 0 g(f(f(x)) + g(f(x)) = 0 g(f(f(x)) = g(x) g(1 + x) = g(x) g is periodic function with period 1 ..... (2)Considering any n, n + 1 interval, we can see from (1) and (2) that g(x) is identically zero.

g(x) = 0 f(x) = x +1

2.


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13

4. { }100

0

x dx = { }100

0

x x dx =100 100

0 0

x dx x dx

=

2

2

1 4 a 101003 / 2

0 0 1 4 19

2x 0 dx 1 dx 2 dx ..... 9 dx

3 + + + +

= ( ) ( ) ( ) ( ) ( )3 2 2 2 2 2 2 2 2 2 22 10 0 1 2 1 2 3 2 2 3 2 3 4 3 ..... 9 10 93

+ + + + + +

= ( )2 2 2 2 2210 9 10 1 2 .... 93

= + + + + =( ) ( )3 9 9 1 18 12 10 900

3 6

+ + + =

2000 9 10 19900

3 6

+

=2000 1710

9003 6

=

2000 3690 310

3 6 6 = .

5. 1 2 3 4

1 2 3 4 1 2 3 4

1000x 100x 10x xN

x x x x x x x x

+ + +=

+ + + + + += 1000

( )

( )2 3 4

1 2 3 4

900x 990x 999x

x x x x

+ +

+ + +

maximum value of1 2 3 4

N

x x x x+ + +

= 1000

6. In triangle ABD, we have BP =6 4 8

6 32

+ + = .

In triangle BCD, we have

DQ =10 6 8

10 22

+ + = .

PQ = 8 (3 + 2) = 3

area of trapezium C1PC2Q = 1 21

(r r ) PQ2

+ ,

where r1 =9 3 5 1 15

9 3

= A B

C

D

P

Q

C1

C2

and r2 =12 2 6 4

212

= .

area of quadrilateral C1PC2Q =1 15 15

2 3 32 3 2

+ = +

sq. units.

7. Put x = y 1

y

1 =2

1 1f y 1 dy

y y

+

=

0 0

2

1 1 dyf y dy f y

y y y

+

Putting z = 1y

=

0

0

1 1f y dy f z dzy z

+ = 1.

8. In given ABC bothA

2and

B

2lie strictly between 0,

2

and sin x is always increasing in

0,2

where as cos x is always decreasing in 0,2

.


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14

So ifA

2>

B

2

sinA

2> sin

B

2or x1 > x2

and

3 4

1 1

x x

> as x3 < x4

x12007

.x42006

= x22007

.x32006

is not valid

similarly forA

2 x4

3 4

1 1

x x<

so again equality for x12007

.x42006

= x22007

.x32006

is not possible.

So only one possible case exist for x1 = x2 and x3 = x4 or3 4

1 1

x x= or

A B

2 2=

BAC is isosceles with ABC = CAB so BC = AC = 1 unit.

9.2 2

2 2 2

(r 1) f(r) (r 1)

r r r

+

2r

f(r)lim

r= .

10. Let y = 1 1

x

f(y) +1

1fy

=

1 xf f1 x

x x 1

+

=11 1

f f 21xx 1 x

+ =

..... (1)

put z =1

1 x

f(z) +1

f 1z

= ( )

11f f 1x

1 x1 x

+ = + ..... (2)

subtract

f(x) 1 11

f 111 x xx

= +

f(x) =1 1 1

x2 1 x x

+ +

.

Alternate:

Putting1

x 2,2

= and 1 successively

f(2) + f(1/2) = 3 ..... (1)f(1/2) + f(1) = 3/2 ..... (2)and f(1) + f(2) = 0 ..... (3)

Solving, we get f(2) = 3/4.


15/18



15

11.1 1 1 2

a2 2

+ += = 2

h 1 >2

2

h > 1 + 1

h > 2S(1,1)

12. Let P (, ), 2 + 2 = 5 and image of P(, ) with respect line x y + 1 = 0 be Q lies on7x + y + 3 = 0

( )2 1x y

1 1 2

+ = =

x = 1, y = + 1 7( 1) + + 1 + 3 = 0 = 3 7

2

+ 9 + 49 2

42 = 5 50 2

42 + 4 = 0( )

242 42 800

2, 2, 1100

=

13. 2ae = 48 e = 4e =

|Z 4| [1, 9]

14. The required condition is ( )( )a 2a a a 2a ae e e 1 4e 2e e 1 0 + + <

( )( )2a a 2a ae 2e 1 2e 5e 1 0 + + < Let x = ea

( ) ( )2 2x 1 2x 5x 1 0 + < = ( )25 17 5 17

x y x x 02 2

+


16/18



16

17. ( ) ( ) ( ) ( ) ( )( )8

P A B P A P B P A 1 P B25

= = =

P(A) =12

31.

18. f(2x + 3) + f(2x + 7) = 3 ..... (1)

Replace x by x + 1, f(2x + 5) + f(2x + 9) = 2 ..... (2)Now replace x by x + 2, f(2x + 7) + f(2x + 11) = 2 ..... (3)from (1) (3) we get f(2x + 3) f(2x + 11) = 0

f(2x + 3) = f(2x + 11) T = 4.

19.( ) ( )

22 4 2 2

2 4 2 4

cos x cos x .cosxdx 1 t 1 tI dt

sin x sin x t t

+ + = =

+ +

=( )( )

( )( )2 2 22

2 4 22 2

1 t 2 t 2 t2 tdt 2 dt dt

t t tt 1 t

=

++

= ( ) 22 23dt dt

2. 4 dt dttt 1 t ++ =2

2 2

1 dt

6 t dt 4 dt1 t t

+ + =2 2

dt dt

2 6 dtt 1 t ++

= ( )12

6 tan t t ct

+ + = ( ) ( )1 1sinx 2 sinx 6 tan sinx c

+

20.

sin

2

1

tdtA

1 t

=+

Let2

1 1t , dt dx

x x= = =

cossc

2 2

12

1 1 dx. .

x x x 1

x

+ = ( )cosec

21

dxB

x 1 x

= +

A + B = 0 2 2

A B 2 2

2 2 2

A A B A A A

e B 1 1 A 1 0

1 A B 1 1 2A 1

+

= = =

+

21. y = mx + 29m 4+ Equation of circle with TT as diameter

(x2

9) + (y 29m 4+ )2 9m2 = 0

x2

+ y2

2 29m 4.y+ 5 = 0

T

(3, 3m + 29m 4+ )

(3, 3m + 29m 4+ )

T

A A

22. 2 2 2 2 2 2 2 2 2y mx a m b y m x 2xy m a m b= + + =

m2 (x2 a2) 2xym + y2 + b2 = 0

( )2 2

2 2 2 2 2 21 2 2 2

y bm m c y b c x a

x a

+ = = + =


17/18



17

23. ( ) 1 1 1f x sin x cos x tan x = + + Domain = [1, 1]

fmax = 3/4, fmin = /4

24. Let a, b, c

be the position vectors of the points w.r.t. origin (O). We now that the position vector

of Incentre (I), where the internal bisectors of angle ofABC meet it

OI

=a b c + + + +

IA

= OA OI

=( )a b (a c) +

+ +

IA

=(a b) (a c) +

+ +

similarly I

=(b c) (b a) +

+ +

and IC

=(c a) (c b) +

+ +

Hence IA

+ IB

+ IC

= 0

25. f(x + y) = f(xy)Put y = 0

f(x) = f(0) x R f(x) is a constant function.Since, f(2000) = 1999 f(2001) = 1999.

26. y = ax2+ bx + c, vertex is (4, 2)

4 = b

2a, b = 8a,

24ac b2

4a

= c = 2 +

2b

4a= 2 +16a

Now = abc = 8a2(2 + 16a) = 16(a2 + 8a3)

d

da

= 16(2a +24a

2) < 0 a [1, 3].

|max = 144, |min = 3600 Difference = 3456.

27. I(n) =

/ 2

n

0

. sin d

I(n) = ( )/ 2

n 2 2

0

. sin 1 cos d

= I(n 2) ( )/ 2

n 2

0

. cos .cos . sin d

= I(n 2) .cos . ( )/ 2n 1 n 1

0

/ 2sin sin. sin cos . d

0n 1 n 1

+ +

= I(n 2) ( )

/ 2 / 2n n 1

0 0

1 1. sin d cos .sin d

n 1 n 1

+

= I(n 2) ( )

( )( )( )

n / 21 1.I n .sin0n 1 n 1 n

+

( ) ( )( ) ( )

n 1I n I n 2

n 1 n 1 n= +

I(n) I(n 2).

2

n 1 1

n n

= .


18/18


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18

28. |z|2

|z| 2 < 0

( ) ( )| z | 2 | z | 1 0 + < |z| < 2

Now |z2

+ z sin| |z|2 + |z sin | |z|2 + |z| < 4 + 2 = 6.

29. x2

+ r2 2xr and y2 + s2 2ys

2(xr + ys) x2 + y2 + r2 + s2xr + ys 1 maximum value of xr + ys = 1

30. Let A (a cos , a sin )M (h, k)

h = a cos + a,k = a sin (h a)

2+ k

2= a

2

h2 + k2 2a h + a2 = a2 x2 + y2 = 2ax

AM

Concept Recap Test Mains 1 Sol

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