concept recap test mains 6 sol

7/28/2019 Concept Recap Test Mains 6 Sol

1/13

AITS-CRT(Set-VI)-PCM(S)-JEE(Main)/13

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com

1

ANSWERS, HINTS & SOLUTIONS

CRT(SetVI)

ANSWERS KEY

PHYSICS CHEMISTRY MATHEMATICS

Q. No. ANSWER ANSWER ANSWER

1. B B B

2. A D A

3. D D B

4. B D A

5. A D C

6. A B B

7. B A B

8. C A C

9. C D A

10. B C B

11. C B A

12. B A C

13. A C A

14.D B B

15. A B A

16. D B B

17. D A B

18. B D A

19. C C B

20. C B C

21. A D B

22. C B D

23. B C A

24. D D D25. A D A

26. D B C

27. A B C

28. D C C

29. C A A

30. A A B

ALL

INDIA

TEST

SERIES

FIITJEE JEE (Main)-2013

FromL

ongTermC

la

ssroomP

rogramsandMediu

m/

ShortClassroomP

rogram4

inTop10,

10inTop20,

43inTop100,

75in

Top200,

159inTop500Ra

nks&3542

totalselecti

ons

in

IIT-JEE

2012


2/13



2

PPhhyyssiiccss PART I

1.th

m

hc hc = =

th 5525 =

2. ( )V VS 1 A= ; ( )R rS 1 A=

( ) ( )V rD A 1.54 1.52 20= = = .02 20 = 0.4

3.d D

dd

= =

;

dy y xx 2

D

= = =

( )Y 0.1 10 20 4

2 2 202.5 25 25 5

= = = = =

= 0.25 y = 0.1

2

max

2

I I cos 5

=

4. Using the relation, stress =2

F mr

A A

=

we get2m

SA

l=

5. For tendency of forward slipping friction acts backward.

6. There will be no effect of magnetic force on time periodbecause the magnetic force will be perpendicular to theinclined plane.

q,m

7. Wavelength of first overtone in open pipe is = l

1

1

Vf

l=

wavelength of the closed pipe for fundamental is

44

l = l

= ;2

2

V Vf V

4 4 0.25l= = =

;

2 1f f 5 =

1 1

340 340340 5 340 5

l l = =

8. From work energy in the frame which is attached to point A

2 2 2xm m

1 1 1max k x m(0) m(0)

2 2 2+ =

max2ma

xk

= .


3/13



3

9.av rms rmP V I cos= ;

2

= avP 0= .

11. P = i 2 R 2 21i 6 i 24= 1i

i2

= R 24 =

12.2

.5 1.255 r

= +

13.2

0

qF qE

2A= =

14. at t = 0,Inductor will acts as an open circuit, no current will flow.

15.2T

T = eff

g

g geff=

2

3 g

T2 =3

2T

16. dp =p p

dx dyx y

+

17. In all the options given, the term force is used.But gravitational potential at a point is defined in terms of the work done in displacing a unit massfrom infinity to that point without change in kinetic energy.

18. PV = RTPM

RT =

P =

constant or

B B

A A

P1

P

= r > q.

3.2Cl

h

CH2 Cl

+ Cl +

Cl

+

Cl4 stereo product 2 geometrical isomers

5. In polar protic solvent less hydrogen bonding nucleophile is more nucleophilic.

6. Reaction is an eg. of nucleophilic addition.Rate of NA. R : acid halide > aldehyde > ketone > acid.

7.14

3

||OH aldol

23 3CH CHO

O

CH CH CH CHO CH CH CH CHO CH CH CH CH CH C H

= = = =

8.

OH

( )d

CH3 CH

Et

O C

O

CH3 CH3 CH

Et

OH + CH3 C O

O

Saponification

reaction9. Product (i) and (ii) will be obtained by normal Backmann rearrangement mechanism of product

(iii) is:

O

O

CH3

O

2NHOH

O

O

CH3

N H

5PCl

O

O

N C CH3O

ON

CH3

10. Keto group increases the acidic character.

So,

CH2OH

O will react fastest with sodium.


6/13



6

12.2 2 2 21 2

1 1 1 1 1R R

n n 1 n

= =

1/2R

nR 1

=

13. Binding energy of w = 7.5 120 = 900 MeVBinding energy of 2y = 2(8.5 60) = 1020 MeV.Binding energy (2y) > Binding energy (w)Thus energy will be released.

14. P & Q contain n1 & n2 moleculesPV nKT

1

2

n 5

n 3=

Internal energy is conserved, hence

( )1 1 1 2 23 3 3

n KT n n KT KT2 2 2

= = + =

2700T K8

= =

( )1 2

1 1 1 1

n nPV T

P V n T

+ =

On solving, we get 1.8 105 N/m2

15. When two Fe3+ ions are dopped, three Fe2+ ions are replaced to neutralize the charge hence onecationic valency creates.Let the moles of Fe3+ are x and moles of Fe2+ ions are (0.93 x) in one mole of Fe0.93O.3x + 2(0.93 x) = 23x + 1.86 2x = 2x = 2 1.86 = 0.14

In 0.1 mole of compound 0.014 moles of Fe

3+

ions are present. Therefore 0.007 moles of cationicvacancies i.e. 0.007 6.023 1023 total no. of cationic vacancies are present.

16. V nRT1000

=

B

B

W 1000V RT

m

=

B

B B

W 1000RT C 1000RT

V m m

= =

When is plotted against C, the slope will be equal to

B

1000RT

m

3

B

1000 0.082 3004 10

m =

6Bm 6 10=


7/13



7

17. 2

2

2

Hcathode anodecell 2

Hanode cathode

H P0.59E log

2 H P

+

+

=

5 3bOH CK 0.1 10 10 M

= = = 14

11

3anode

10

H 10 M10

+

= = 7 4

acathodeH CK 0.1 10 10 M+ = = =

( )

( )

24

cell 211

10 0.10.059E log

2 10 1

=

= 0.40 V

18. Molecules with 2 b.p. and 3 l.p. have linear shape.

19. ( ) ( ) ( ) spAgBr s Ag aq. Br aq. K+ +

( ) ( ) ( ) ( )

32

2 3 2 3 f 2Ag s 2S O aq. Ag S O aq. K

+

+

( ) ( ) ( ) ( ) ( )32

2 3 2 3 sp f 2AgBr s 2S O aq. Ag S O aq. Br aq K K K 25

+ + = =

( )

2

2

xK

0.1 2x=

( )

2

2

x25 ,x 0.045 M

0.1 2x= =

20. Benzoic acid C6H5COOH, M.W. = 122.Heat liberated by the combustion of 1.89 g of benzoic acid - ms t = 18.94 0.998 0.632 = 11.946 cal

Heat of combustion per mole of benzoic acid = 11.946 1221.89

= - 771.1 kcal

22.N

B

B

N

NH

HB

X

H

X

H

(i)

N

B

B

N

N

B

X

H

H

X

H

H

(ii)

N

B

B

N

N

B

X

X

H

H

H

H

(iii)

N

B

B

N

N

B

H

H

X

H

X

H

(iv)

23. 50 ml 2 N hydro solution 50 ml of 2 N I2 solution.

50 ml 2 N Cl2 solution 50 ml 2 N CaOCl2 solution. 6.35 g CaOCl26.35

% 100 63.510

= =

24. Fluorine does not form stable oxoacid.


8/13



8

26. f fT i K m i 1.86 0.1 = = = 0.372 = 0.186 ii = 2This shows that complex gives 2 ions in solution. Thus the formula of the complex is[Co(NH3)4Cl2]Cl.

( ) ( )3 2 3 24 4Co NH Cl Cl Co NH Cl Cl+ +

27.( ) ( ) ( ) ( )

2 4 2x zA y

CaC O CO CO CaO + +

( )( )

2 2B

CaO H O Ca OH+

( ) 2 3 22Milkiness

Ca OH CO CaCO H O+ +

28. WolffKishner reduction involves (NH2NH2/KOH) a very strong base so, along with the reduction ofO

to

group, it also undergo E2 elimination in presence of strong base.

Hence, (C).

29. Lanthanide contraction is observed in these ions, i.e., ionic radius decreases as atomic numberincreases.


9/13



9

MMaatthheemmaattiiccss PART III

2. Domain of the function x {1, 1}

3. Two lines are

x 0 y 0 z n

0 n

+

= = and

x 0 y m z 0

0 m n

= =

applying the formulae of shortest distance we get2 2 2 2

1 1 1 1

m n c+ + =

5. Curves are x + y = 252x + 2y = 8 x + y = 4x y = 3(1, ) lies in the shaded region then

( )4 6, 1 (, 2) lies in side so (1, 2)

6. Let n(B) = x n(A) = x + 2n(F) = y n(C) = y + 3n(D) = z n(E) = z + 5x + y + z + x + 2 + y + 3 + z + 5 = 40x + y + z = 15 but x 1 y 1 and z 1So total ways of distribution is 14C2 = 91

8. For the lines to be coplanes

AB

, a

and b

should be coplanes

2 0 k

1 k 2 0

4 1 1 k

=

2(k k2 + 2) + k(1 4k) = 02k 2k2 + 4 + k 4k2 = 0

(1, 2, 1)

a 2i 0j kk= +

B(3, +1, k) b i kj 2k= +

6k2 k + 4 = 06k2 + k 4 = 0This equation has two distinct rootsSo two such planes exist.

9. y = sin2x + cosec2 x + 2 + cos2 x + sec2 x + 2 + tan2 x + cot2 x + 2= 1 + 6 + 1 + tan2 x + 1 cot2 x + tan2 x + cot2 x= 9 + 2(tan2 x + cot2 x)= 9 + 2 [(tan x cot x)2 + 2]= 13 + 2 (tan x cot x)2ymin = 13 = p

p4

3

=

10. C1 : z z 2 z 1+ = Put z = x + iy2x = 2|x 1 + iy|x2 = (x 1)2 + y2


10/13



10

y2 = 21

x2

C2 : arg(z (1 i) = its a ray emanating from (1, 1) and making angle with the positive real axis.C1 and C2 have exactly one point common.C2 must be tangent to C1

y + 1 = m(x + 1)Solving C1 and C2m = 1y = x = 1P(z0) = 1 + i

0z 2=

11. For second equation to passes real rootsD 0D = p2 4qas p, q [1, 10] and p, q Nsuch 32 quadratics are possibleand x2 + 5x + 3 = 0 difference of roots = |4 cos 2|x2 + px + q = 0 difference of roots = |4 cos 2|So p2 4q = 25 12p2 4q = 13So (p, q) an be (5, 3), (7, 9) two possible cases

So required probability =2 1

32 16=

12. a x a b =

( )a x b 0 =

x b a =

x b a= + = ( ) ( ) ( ) 2 i 1 2 j 1 3 k+ + + +

Now a x 0 = 2 + + 2 + 4 + 3 + 9 = 014 = 7

=1

2

3 1 x i 0 j k2 2

= + +

9 1 5x

4 4 2= + =

13. Let cos1 x + cot1 x = f(x)Domain x [1, 1] and f(x) is decreasing function in its domain.

at x = 1max

3 7f(x)

4 4

= + =

at x = 1minf(x) 0

4 4

= + =

( )7

f x4 4

k can be 1, 2, 3, 4, 5


11/13



11

14.12

sinA13

= cos A can be5

13or

5

13 A is in 3rd or 4th quadrant

15sinB

17= cos B can be

7

17 or

7

17B is in 1st or 2nd quadrant

cos (A B) = cos A cos B + sin A sin Btotal 2 values are possible.

15.A 6

tan2 P

=

p = 6 cotA

2

q = 6 cotB

2

r = 6 cotC

2

A

B C

6

P

A/2

p + q + r =A B C

6 cot cot cot2 2 2

+ +

pqr = 3 A B C6 cot cot cot2 2 2

and we know in ABC A B C A B Ccot cot cot cot cot cot2 2 2 2 2 2

+ + =

pqr36

p q r=

+ +

16. g(f(x)) = x if x = 0 so f(x) = 6

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

1

f ' 0

( )( )( )

1g' f x

f ' x= ( )

( )

1g' 6 1

f ' 0= =

17. ( )( ) ( )f 9 f 4

f ' c9 4

=

492 3 / 2 2 3 / 2

0 0

t 2t t 2t2 3 2 3

c c5

+ =

=

81 1618 8

2 35

=

45 81192 3

5 30

=

18. 2x + 3y z = 9 on solving x = 1x + 2y z = 5 y = 32x + y 4z = 3 z = 2

19. Sn = C0C1 + C1C2 + C2C3 + .. + Cn 1C0

= C0Cn 1 + C1Cn 2 + C2Cn 3 + .. + Cn 1 C0 = 2n n 1C

n 1

n

S 15

S 4+ =

2n 2n

2nn 1

C 15

4C

+

= on solving n = 2 or 4

20. For ellipse 2 4 5 < 5 < 2 4 < 5 1 < ( 2)2 < 9 ( 2)2 < 9 3 < ( 2) < 3


12/13



12

1 < < 5 R+ 0 < < 5

21. Clearly ABC is equilateral

arABC =23

z wz4

=23 3

z 48 34

=

= |z|2 = 64|z| = 8

2 /3

2/3

z(wz)

w

22.n

n x2 16

+

rn n

r 1 rx

T 2 C6+

=

8

n n n n7 87

1 12 C 2 C

66

=

n n7 86 C C= n = 55

23. Given z3 + iwz2 = (1 + iw)z1z3 z1 = iw(z1 z2)

3 1

2 1

z ziw

z z

=

i / 63 1

2 1

z ze

z z =

3 1 2 1z z z z = with A 6

=

24.n

n nr r

r 0

2C C

r 1=

+

+

= ( )n n n n

n n n n 0 1 2 n0 1 2 n

C C C CC C C .......... C 2 .....

1 2 3 n 1

+ + + + +

+

= ( )n n n n 2 n n

0 1 2 n1 x C C x C x ..... C x+ = + + +

= ( ) ( )1 1

n 2 n0 2 2 n

0 0

1 x dx C C x C x ..........C x dx+ = + +

=n 1

0 1 nC C C2 1 .....n 1 1 2 n 1

+ = + + +

+ +

nn

rr 0

r 3C

r 1=

+ + =

( )n 1n 2 1 22

n 1

+ +

+=

( ) n nn 1 2 4.2 2

n 1

+ + +

=( ) n 10n 5 2 2 7.2 2

n 1 10

+ =

+=

914 2 2

10

n = 9

25. tan C =4

3 cos C =

3

5

2 2 2a b ccosC

2ab

+ =

( ) ( )

( )( )

2 2 23 n 1 n 2 n

5 2 n 1 n 2

+ + + =

+ +


13/13



13

on solving n = 13sides a = 14, b = 15, c = 13

1ar ABC absinC

2 = =

1 414 15

2 5 = 84 unit2

26.2 3 1 3

1 sin x sinx 12 4

+ + +

2 3 1 3sin x sinx sinx 02 2 4

+ =

3 1 3sin x sin x sin x 0

2 2 2

=

1 3sin x sin x 0

2 2

=

1sinx

2= x =

6

and

5

6

3sinx

2= x =

3

and

2

3

27. Centroid of the given triangle = (4, 2)So centroid of the image triangle is itself the image of the original centroidG1 (4, 2)G2 (4, 2)G3 (2, 4)

Area ofG1, G2, G3 =

4 2 11

4 2 12

2 4 1

= 20 unit2

29. Let ( ) ( ) ( )

3 2

f x x bx cx 1. f 0 1 0, f 1 b c 0= + + + = > = < so, ( )1,0 . So, ( ) ( )1 1 22tan cosec tan 2sin sec +

( )1 1 1 121 2sin 1

2 tan tan 2 tan tan sinsin sin1 sin

= + = +

22

= =

( )as sin 0 <

30. By applying Pythagoreans theorem on various right triangles as BD and AC are diameters.DEB, DAB, AEC and ABC are right trianglesIn DEB, DE2 + BE2 = BD2In DAB, AD2 + AB2 = BD2

DE2

+ BE2

= AD2

+ AB2

(1)In AEC, AE2 + CE2 = AC2In ABC, AB2 + BC2 = AC2 AE2 + CE2 = AB2 + BC2 (2)

Adding (1) and (2) AE2 + CE2 + BE2 + DE2 = AD2 + AB2 + AB2 + BC2 = 256.

concept recap test mains 6 sol

Documents