AITS-2014-CRT-I-JEEM+ADV_Advanced_PAPER-1_Solutions_ANSWER_SOLUTION.pdf

AITS-CRT-I(Paper-1)-PCM (Sol)-JEE (Adv)/14

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1

ANSWERS, HINTS & SOLUTIONS

CRT–I (Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. C C D

2. A D A

3. B D B

4. B C A

5. A B B

6. D D A

7. B C C

8. A. C A

9. B B C

10. C B A

11. A, B, C A, C A, B, C

12. A, B, C, D B, C A, B, C, D

13. B, C, D A, C A, D

14. A, B, C A, B, C B, C

15. A, B, C A, C B, C

1. 2 2 4

2. 1 1 2

3. 5 4 6

4. 1 3 2

5. 2 6 3

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PPhhyyssiiccss PART – I

SECTION – A

1. relU g t 10 1 = 10 m/s

1

1S g(1) 5m2

2rel 1 re re

1S S u a t2

= 5 + 10 3 + 0 = 35 m

2. sin = h

mgh = mg cos (/2)

h cos2

tan = 2

/2

/2

3. Centre of mass does not move in the absence of external force so m1x1 = m2x2

1 2

2 1

x mx m

4. 2 2

1dA r d rAdt 2 dt 2

L = 2Mr L = 2MA

5. F 2

Gm(M m) 0r

2

dF G (M 2m) 0dm r

M = 2m m = (M/2)

6. A g sA

sg

7. h

2

0

gh r (2 rdh) gh

2

2 hgh r 2 r g2

hr 22

r = h



3

8. Avg. speed 2 2

r mr(v t) (v t)3

t

2rv 5 9

rv 2m/ s

9. avg

1 t 1v v 3v2 2 2vt 4

10. m mg T a2 2

…(i)

T cos 60 = macos 60

…(ii)

Solving (i) and (ii) acceleration of ring = 2g9

11. 2 2

2

2kq mvmg TR R

…(i)

mgR + 2 21 1mu mgR mv2 2

…(ii)

if T = 0 2

min

2Kqu 5gRmR

12. dUF 5(2x 4)dx

At mean position F = 0 x = 2m minU 20 J a = 50 2(x 2) = 10 rad/sec T = /5 sec

13. If the volume immersed initially is (V/3). Then V g mg3 …(i)

If the volume immersed when the system accelerates is V then

g mgV ' g mg2 2

VV '3

14. The potential of the two surface will be equal when the whole charge Q flows from inner to other

shell.

SECTION –C

1. Consideration refraction at glass-water interface

Ruv

122

r

)2/3()3/4(r2

3v3

4

r

(4/5)r

(27r/20)



4

r6

1r2

3v3

4

v = r5

Now refraction at water air surface

u = - 5r9r

54r

1212

uv

)3/4(1

r9354

v1

r27

20v1

v = -20

r27

so height above centre = 2r - 20

r27

= 20

r27r40 =

2013

r = 13 cm.

2. Mg T = Ma …(i) T mg = ma …(ii)

T = 2Mmg(M m)

9T 2Mmg 2mg 2 10mA A(M m) A 1M

M = 1.86 Kg

m

M

3. Relative motion between block and table

will start when 2 2m r sin (mg m r cos ) …(i) t …(ii) solving (i) and (ii) t 500 22.4 sec .

4 cm

3 cm

m2r

4. = gR = 1 rad/sec

5. N1 + N2 = mg . . . (1) N1 = N2 . . . (2)

N2 = 2

mg1

N1 = 2

mg1

N1 N2

N1

N2

mg Torque about centre of mass



5

(N1 + N2)R = 21 mR2

= 1 22 N NmR

=

2

2 1 g1 R

2 = 20 2

= 20

2

=

2 20 1 R

2 2 1 g

Hence number of turns

N =2

=

2 20R 1

4 1 g.2

N =

2 20R 1

8 g 1

CChheemmiissttrryy PART – II

SECTION – A

1. In ‘A’ bond is at bridge head which is not stabilized in the given structure;

In ‘B’ the product is CN

ACOH ;

In ‘D’ the product is

2.

O

H2N OHO NH2 OH

IMPT

OH NH

OH

2H O

N OH

N OH2

H

2H ON

2H OH

N

OH

5. CH3Cl is formed by SN2 while CH3CHClC2H5 is formed by SNi. 6.

5 3 2PCl g PCl g Cl g

Let initially no. of moles xat equilibrium volume V x 1 x x



6

2 2

cxK

x 1 V

, also Kp = Kc(RT) gn

on reducing volume to V2

, initial concentration of PCl5 doubles and its degree of dissociation is

increased but Kp does not change, as it is characteristic constant for a reaction at constant temperature.

7. 2 6 2 3 6

Al Cl 12H O 2H Al OH 6HCl ; 10. In blast furnace, In combustion zone, CO is produced ultimately which reduce Fe2O3 to Fe3O4 in reduction zone,

which is uppermost zone. 11. O

O

3 2 5CH COOH C H OH

O

O

P =

Q =

R =

S =

Oxidation by XO2 5 3C H OH CH CHO

;

12. Due to hydrogen bonding Gauche conformation of H2C CH2 NH2

NH2

and

H2C CH2F

OH

are more stable than their anti-conformations.

In case of ClCH2CH2Cl, on increasing temperature, % of Gauche conformation increases. Hence

dipole moment increases. In case of option (D) boat conformation is most stable. 13. Mn appears colourless in reducing flame in Borax Bead Test. 3

4 46 6 3Prussian blue

Fe K Fe CN Fe Fe CN K

2 2 3 2 2 5 2 4 2Na S O .5H O Na S Na SO H O 14. At pH = pI, amino acid exists as Zwitter Ion, they are highly soluble in polar solvents.



7

15. pH of 10–8 M HCl (aq) is 6.96 and pH of 10–8 M NaOH (aq) is 7.04

SECTION – C 1. = CRT × i hence i = 1.1; so = 0.1 hence [H+] = c = 0.1 0.1 = 10–2 M pH = 2 2. It has only one CH2OH group. 3. Three equivalents to remove three hydrogens from CH3 group and one for attack on C of CHO

group. 4. BaCrO4 – Yellow, soluble in dil HNO3 Hg2CrO4 – Red, soluble in conc. HNO3 ZnS – White, soluble in Conc. HNO3 BaSO4 – White, insoluble in all mineral acids BaS2O3, CH3COOAg and AgNO2 all are white solid and are soluble in dilute HNO3 solution.

5. Except

Cl

and Cl

MMaatthheemmaattiiccss PART – III

SECTION – A

1. [y + [y]] = 2 cosx [y] = cosx

y = 13

[sinx + [sinx + [sinx]]] = [sinx]

[sinx] = cosx Number of solution in [0, 2] is 0 Hence total solution is 0. both are periodic with period 2 2. (x 0)2 + (y k)2 = k2 x2 + (y k)2 = k2

2x + 2 (y k) dy 0dx

dy xdx y k

y k = xdxdy

k = y + xdxdy



8

x2 + 2 2

xdx xdxy y ydy dy

x2 + 2 2

2 2 2dx dx 2xydxx y xdy dy dy

x2 = y2 + 2xydxdy

(x2 + y2) dy 2xy 0dx

3. Given < < < also sin = sin = sin = sin = k and , , , are smallest positive angles = , = 2 + , = 3 as sin = sin and > sin = sin and > sin = sin and > Putting these values in the given expansion, we have given expression

= 2 sin cos 2 1 sin 2 1 k2 2

4. x3 – x < – a2 + a – 23 3

f(x) = x3 – x f(x) = 3x2 – 1 = 0

x = 13

In positive region minimum value of f(x) = 1 1 23 3 3 3 3

So, – a2 + a – 23 3

> – 23 3

a2 – a < 0 a (0, 1) 5. 3x 2 4x 4 5x 1 4x 5 …(1) v u p q u2 – v2 = x + 6 = p2 – q2 u – v = p – q …(2) solving (1) and (2) we get, 2 4x 4 2 5x 1 x = 3 6. Limit be equal to y

logy = n

1 n 1 n 2lim log log ...n n n

= n

nr 1

1 n rlim logn n

= 1n

nr 1 0

1 rlim log log dx1 1 xn n



9

= . 2 1log22

= [log4 – loge] = log 4

e

logy = log 4e

7. 0 argz 4 , represent the region of complex plane lying in the first quadrant and bounded by x-

axis and the line y = x |2z – 4i| = 2|z – 2i| least value of |z – 2i| is length of perpendicular from (0, 2) to y = x, which is 2 So the least value of 2 |2z – 4i| is 4 8. Let f(x) = 2x3+ 3x2 – 12x + 3 f(x) = 0 has three real roots (, , )

+ + = – 32

Centroid = , ,3 3 3

= 1 1 1, ,2 2 2

which lies on x = y = z.

9. If a > b > c and a2 = b = c then b < 1 cot–1x < 1 x > cot1

10. a 4sinA a 4sinA

R 2

so for any point (x, y) inside the circumcircle, x2 +y2 < 4 |xy| < 2

11. a, b and A are given in ABC,

cos A = 2 2 2b c a

2bc

c2 2bc cos A + (b2 a2) = 0 c2 (2b cos A)c + (b2 a2) = 0 which is quadratic in c and gives two values of c and let these are c1 and c2 c1 + c2 = 2b cos A and c1 c2 = (b2 a2) 2 2

1 2 1 2c c 2c c cos2A = (c1 + c2)2 2c1c2 (1 + cos 2A) = 4b2 cos2 A 2 (b2 a2) (2 cos2 A) = 4b2 cos2 A 4b2 cos2 A + 4a2 cos2 A = 4a2 cos2 A

12. 1 1 12 5 5 8 8 11

n terms

= 5 (n 1)3 2 (n 1)35 2 8 5 11 8

3 3 3 3

=

3n 2 2 3n 2 2 n3 3n 2 23 3n 2 2



10

= n n n2 3n 2 3n

13. 2sin24 R 2

R 2 1 cosec24

(A) is true

2

sin cos2 2 1 cosecsinA

(D) is true 14. x2 + x + 1 = |[x]| Hence only 2 solutions exist i.e. x = – 1 and another lie – 2 < x < – 1 15. sin x + cos x = 1 – a sin x cos x a2 sin2x cos2x – 2(a + 1) sin x cos x = 0

sin2x 2a sin2x 2 a 12

= 0

Hence sin2x = 0 a R

x = n2 , n I

And sin2x = 2

4 a 1a a (–, 2 – 2 2 ] [2 + 2 2 , )

SECTION – C

1. a = 2 2r 1 r 1

1 1, br (2r 1)

a = 2 2 2 21 1 1 1 ...1 2 3 4

and b = 2 2 2 21 1 1 1 ...1 3 5 7

b = 2 2 2 2 2 2 21 1 1 1 1 1 1... ...1 2 3 4 2 4 6

b = 2 2 2 2 2 2 21 1 1 1 1 1 1... 1 ...1 2 3 4 2 2 3

b = a 14

a

b = 3a4

a 4b 3

2. 1

0 0

f (x)dx t f (t)dt



11

= 00

t f t t sin t dt

= 2

2 1 12

= 2

22

k = 2 3. Case I: When we use 6, 7 or 8 at ten thousand place then number of numbers = 3 4P4 = 72 Case II: When we use 5 at ten thousand place and 6, 7 or 8 at thousand place then number of

numbers is = 1 3 3P3 = 18 hence the required numbers of numbers is 72 + 18 = 90 4. f(x) is an odd function g(x) is f–1(x)

Area = 2 2

1

0 0

f y dy x f x dx

= 2

20

0

x f x x sin x dx

= 22

5. 2 2 2a b b c c a

= 2 2 22 a b c 2 a b b c c a

= 2(4 + 9 + 16) – 2 a b b c c a

= 58 – 2 a b b c c a

.

Now 2a b c 0

a2 + b2 + c2 – 2 a b b c c a

– 2 a b b c c a

29

2 2 2a b b c c a

87

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AITS-2014-CRT-I-JEEM+ADV_Advanced_PAPER-1_Solutions_ANSWER_SOLUTION.pdf