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Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 7 8

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Date : 10 - 04 - 2014TARGET : JEE 2014

SCORE-II : TEST # 01

SECTION-I

1. Ans. (C)

–99 2

–98

cot x cosec xdx

1+cot x

Let 1+ cotx–98= t

98 cotx–99

cosec2xdx = dt

981 dt 1n 1 cot x c

98 t 98

2. Ans. (C)

1r

2r 3T tan

1 r r 1 r 2 r 3

1r 1 r 3 r r 2

tan1 r 1 r 3 r r 2

Tr = tan–1((r+1) (r+3)) – tan–1(r(r+2))

T1 = tan–1(8) – tan–1(3)

T2 = tan–1(15) – tan–1(8)

|

|

Tn = tan–1

((r + 1) (r + 3)) – tan–1

(r(r + 2))

Sn = cot–1

3

3. Ans. (A)

A2 + B4 = (AT)2

(AT)2 + (BT)4 = A2

B4 = –(B

T)4

|B|4 = |–B4|

|B|4 = –|B|

4

|B| = 0

SOLUTION

PART-1 : MATHEMATICS ANSWER KEYPAPER-1

Q. 1 2 3 4 5 6 7 8 9 10

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

Q. 11 12 13

A. B D B

Q. 1 2 3 4 5 6 7 8

A. 3 4 3 2 4 4 1 3SECTION-IV

SECTION-I

4. Ans. (A)

|3r cos + (3r sin)i + cos i sin

r r

| = 2

2 2cos sin

3r cos 3r sin 4r r

2 2 2

2

19r 6 cos sin 4

r

2

2

16cos2 4 9r

r

2

2

14 9r

rcos2

6

2

22

2

19r 1r 3 9r 6

2 r

maximum value of 2 1

cos26 3

5. Ans. (A,B,C)

(A) ƒ(x) is continuous everywhere because

denominator never become zero.

(B) 1 6

ƒ(x) tan 1(3 | x 2 |)

ƒ(x) not possible & it is an into function.

(C) graph is symmetric about the line x = 2,

because of |x – 2|

6. Ans. (B,C)

a2= (36 – ) & b

2 = (25 – )

(a) 25 – = (36 – ) (1 – e2)

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PAPER – 1

2 25 11e 1 e

36 36

Equation of directrices are

36ax

e 17 / 36

(b) a2 – b2 = constant

(c) focii are (± a e,0) = 11,0

(d) Eccentricity is depending on .

7. Ans. (A,B,D)

(A) a + b + c = = 2s= perimeter

s2

(B) x3 – x

2 + x – = (x – a) (x – b) (x – c)

put x = s = 2

3 3

s a s b s c8 4 2

3

2 8 2

= s(s – a) (s – b) (s – c)

3

2 2 8

(c)

3

2 8r

s

2

(d) 3

abcR

44

2 2 8

8. Ans. (A)

Total (n (A B))

n(A) = Two particular cricket players are

together n(A) = 8!2!

n(B) = Two particular football players are

together = 8!2!

n(A B) = 7!2!2!

n(A B) = 8!2!2 – 7!2!2!

number of ways = 9! – 24(7!) = 7!(44)

9. Ans. (A)

C1(ƒ1ƒ2) C2(ƒ3ƒ4) C3(ƒ5ƒ6)

(3! × 6!)3

10. Ans. (B)

1 2

0 1

dx 2 x 1 dx 2

2

1

x1 2 x

2

11 2 2

2

11. Ans. (B)

0 1 2 3

(3,4)

12. Ans. (D)

Domain of ƒ(x) is 1 < x < 2

1 1ƒ '(x) 0

x 1 2 x

3x

2

b 3b 3

2 2

13. Ans. (B)

1 2

area = 2

1

ƒ x dx 1

SECTION-IV

1. Ans. 3

Let [x] = 2

log23 = |x – 4| x = 4 + log23

x = 4 – log23

one solution

Let [x] = 3 log32 = |x – 4|

x = 4 + log32 & x = 4 – log32

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PAPER – 1

only one solution

Let [x] = 4

0 = |x – 4| x = 4

Three solutions are possible

2. Ans. 4

Sample space = 10 × 10 = 100

favourables cases when a = 1, b = 1 to 10

10 cases.

when a = 2, b = 2,4,6,8,10

5 cases

a = 3, b = 3,6,9 3 cases

a = 4, b = 4,8 2 cases

a = 5, b = 5,10 2 cases

cases = 22 & when a = b = 6,7,8,9,10

Probability 27

100

3. Ans. 3

(a)A

O

B(b)

C(c)

| a | | b | | c |

2

a.b b.c c.a2

Distance between any of two centroids is

a b a c

3 3

2 2 2 22 b c 2b.c 2

9 9

3

3

4. Ans. 2

ƒ(–1) = ƒ(1) a + b = 0 b = –a

1 1 3aƒ ' 0 a 0 a 2

2 2 4

5. Ans. 4

Let x = a – d,y= a, z = a + d

(2a – 2d) (2a + 2d) (a) = k(a – d)a(a + d)

k = 4

6. Ans. 4

Equation of tangent to y2 = 4ax is

a ay mx k mh

m m ...(1)

for x2 = 4ay

yx am

m

replace mby 1

m

ax ym

m

ah km

m ...(2)

(1) + (2) h + k = m(h – k)

h k

mh k

put value of min equation (i)

h k h a h kk

h k h k

k(h2 – k

2) = h(h + k)

2 + a(h – k)

y(x2 – y

2) = x(x + y)

2 + a(x – y)

2

7. Ans. 1

Let p = cos & q = sin

(1 – 2 cos2)

2 + (2 sin cos)

2

cos22 + sin22

= 1

8. Ans. 3

(x – 2)2 + (y – 1)2 < 1

(1,0)Let x – 1 = X & y – 1 = Y

|X| + |Y| > 1

& (X – 1)2 + Y

2 < 1

3

4

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PAPER – 1

SECTION-I

1. Ans. (C)

Sol. A

B C

450

0sin 45 = 1sin d

cosdt

=

0.4

2

d

dt

=

0.4

2sec

Here

2

2 1.4cos 1 sin 1

2

0.2

2

d 0.4 2

dt 0.22

= 2 rad/second

2. Ans. (D)

Sol. Iwire

= 4V

0.4 50 = 0.2 A

Potential difference across voltmeter, V = Ir – 2

2 sint = 0.2 × 50 x –2

10 10

4V

V4V x0V (let)

2V

2 cost = 10 V

V = 20 (cost) cm/s

3. Ans. (C)

4. Ans. (B)

Sol.

R

2R 2RA cos

2 3

R

R

;

2 2

3t t

5. Ans. (A,B,D)

Sol. At t= 0, capacitor act a short circuit & current in resistor branch = 0.

At t= , iR =2 mA; i

C = 0.

6. Ans. (A, D)

Sol. Superposition of waves (i) & (iii) will give travelling wave having amplitude of 2a

{waves are along x-axis but particle displacements are along y & z-axis respectively}

z1 + z

2 = a sin sin

2

x xt tv v

SOLUTION

PART-2 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

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

Q. 11 12 13

A. D B C

Q. 1 2 3 4 5 6 7 8

A. 3 6 2 5 9 6 8 4

SECTION-I

SECTION-IV

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PAPER – 1

7. Ans. (B, C)

Sol.

x

BAC

at any distance x from edge let there be a cross section for which FBD is given:

for complete rod : ma = F a = F

m...(1) C B

Ta

x

for portion BC : mBC

a = T T = mx

L

a; T =

mx

L

F

m = F

Lx stress =

Fx

AL

(C) d

dx

=

Fxstress FxLAY Y LAY

0

2

Fx Fdx

LAY L AY

(D) Energy stored = 0

1

2

L

2

2 2

F

L A Y x2 dx A =

2

2 2

1

2F A

L A Y

3

3

L

=

2

6

F L

AY

8. Ans. (A)

9. Ans. (D)

Sol. (1) The system can be considered as being a new solenoid whose length is twice as large as that of onesolenoid, with a density of turns the same as that in one solenoid and the with same cross-sectional area.Since the inductance of one solenoid is L = µ

0N

0V, where N

0 is the number of turns per unit length, and

V is the solenoid volume, which in this case is twice the volume of one solenoid, we have L1 = 2L

0. The

same result can be achieved by considering the self-induction emf that is generated in the two solenoidsconnected in series. The total emf is equal to the sum of the emf’s generated in each solenoid ; hence,

1 0

dI2L

dt ,

which yields L1 = 2L

0.

(2) When the solenoids are connected in parallel, the self-induction emf in each solenoid is :-

1 0

d I / 2L

dt = 0

1 dIL

2 dt .

Because the solenoids are connected in parallel, the total emf has the same value. Thus, with a current Iin a circuit that is external with respect to the solenoid , the induced emf is one-half the value for theinductance L

0. Hence,

2 0

1L L

2

(3) In this case, the number of turns per unit length is twice as large as that of one solenoid, and since theinductance is proportional to N

02, we have (provided that the current remains unchanged)

L3 = 4L

0

(4) If one solenoid is fitted onto the other and the senses of the turns coincide and the solenoids are connectedin parallel, the current through each solenoid is I/2 if the current in the circuit is I, while the flux associatedwith current I/2 is /2. The total flux is and the flux coupled with each solenoid is = N

0. In each

solenoid there is generated a self-induction emf equal to the one induced in a separate solenoid whencurrent I varies. Since the solenoids are connected in parallel, this emf is the common emf of bothsolenoids Hence,

L4 = L

0.

(5,6) In both cases the induction flux in the solenoids is zero, so that L5 = L

6 = 0

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PAPER – 1

10. Ans. (A) ; 11. Ans. (D) ; 12. Ans. (B) ; 13. Ans. (C)SECTION-IV

1. Ans. 3

Sol.6 10

310 2ABV

2 2 and AB C L C LR X X Z X X R

2. Ans. 6

Sol. Image end nearest to the lens is formed by part having maximum u

max 2 1 1 4u ;

min 2 1 0 2u

max

1 1 21u

v R R

{ for one end nearest to lens}

min

1 1 21u

v R R

{for other end furthest to lens}

3. Ans. 2

4. Ans. 5

Sol. observer for source 'A' 0 0

33

34sound medium

sound medium source

v vf f f

v v v

observer for source 'B' 0 0

35

34sound medium

sound medium source

v vf f f

v v v

Beat frequency = 1 2 0

35 335

34f f f

5. Ans. 9

Sol. We have taken x =0 as antinode so that

Amplitude = (Amax

sinkx)

2K

where 2

max

max

32 4sin sin

3 3 2

2

AA A A

6. Ans. 6

Sol.2 2

2kq kq 14 mv

a 3a 2

7. Ans. 8

Sol. Conserving angular momentum m (v1 cos60°) 4R = mv

2R 2

1

v

v = 2.

Conserving energy of the system. 2 21 2

1 1

4 2 2

GMm GMmmv mv

R R

2 22 1

1 1 3

2 2 4

GMv v

R

21

1

2

GMv

R ;

61

1 800064 10 /

2 2v m s X = 8

8. Ans. 4

Sol. Capacitance of spherical capacitor C =

04 ab

b a

; C = 2F

All capacitors are in parallel so that equivalent capacitance = 4C = 8FTotal charge drawn through battery : Q = Cv Q = 0.5v × 8 F = 4C

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PAPER – 1

SOLUTION

PART-3 : CHEMISTRY ANSWER KEY

SECTION-I1. Ans. (C)

Kc = 3

0R

2

0Q

Temperature is constant hence Kc is alsoconstant

At new eq.

Kc = [2R]3 [Q]' 2

Q' = 1

2 2 Q

0

2. Ans. (B)

3. Ans. (B)

4. Ans. (A)

5. Ans. (B, C, D)

6. Ans. (A, C)

7. Ans. (D)

8. Ans. (B)

Sol. e s ; G0 = –2.76 kcal/L

G0 = –2.3 RT log keq

–2.76 × 103 = – 2.3 × 2 × 300 log keq

log keq

= 2

keq

= 102

s 100

e 1 %s =

100

101 × 100 = 99%

9. Ans. (C)

Sol. – 2.76 = G0f(s) – G0

f(e)

G°f(e) = –7 + 2.76 = –4.24 kcal

G0 = 99

100 × G°

f(s) +

1

100 × G0

f(e)

= 99

100 × (–7) +

1

100 × –4.24

693 4.240 697.24

kcal100 100 100

0 ( 27.9724 6.9724)S

300

= 21kcal

70 cal / k300

10. Ans (C)

11. Ans (D)

12. Ans. (C)

13. Ans. (A)

SECTION-IV

1. Ans. 0

For zero order reaction

xt

k

7 /8

1/ 4

7at 78

1t 2a4

2. Ans. 3

3. Ans. 2

4. Ans. 3

5. Ans. 6

6. Ans. 3

7. Ans. 5

8. Ans. 6

Q. 1 2 3 4 5 6 7 8 9 10

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

Q. 11 12 13

A. D C A

Q. 1 2 3 4 5 6 7 8

A. 0 3 2 3 6 3 5 6

SECTION-I

SECTION-IV

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SECTION-I

1. Ans. (C)

Corresponding to a root xi

6 – xi is also the root

sum of roots = 6 × 4 = 24

2. Ans. (C)

z = 3, x + y = 1

Any point P on this line is (x, 1 – x, 3)

d2 = x2 + (x – 1)2 + 9 = 2x2 – 2x + 10

min

19d

2

3. Ans. (A)

RBW can be arranged in 3! = 6 ways

for one such way

3C1.

3C1.

3C1

2C1.

2C1.

2C1

1C1.

1C1.

1C1

— — — — — — — — —

1 2 3 4 5 6 7 8 9

Total ways = (3! × 3! × 3!) × 3!= 36(3!)2

4. Ans. (D)

Equation of pair of image is (Replace y by – y)

y2 + 2xy + 4y + 8x = 0

SOLUTION

PART-1 : MATHEMATICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

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

Q. 11 12 13 14

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

A B C D

Q P Q R,S

Q. 1 2 3 4 5

A. 2 5 6 3 2

SECTION-II

SECTION-IV

SECTION-I

Q.1

5. Ans. (B)

from figure it is clear

y = –x

|z + 4 + 3i| = 3

6. Ans. (C)

a = 6; 2

2b 32b 64 b 8

a 3

2 2x y

136 64

7. Ans. (D)

ƒ(1) < 0 3 – 3sin – 2cos2 < 0

2sin2 – 3sin + 1 < 0

(2 sin – 1) (sin – 1) < 0

1sin 1

2

52n ,2n 2n

6 6 2

8. Ans. (D)

Solving the curves

x2 – 2x3 = –1

2x3 – x2 – 1 =0

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Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 7 9

PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE

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SCORE-II : TEST # 01

PAPER-2

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PAPER – 2

(x – 1) (2x2 + x + 1) = 0

x = 1, y = 1

m1 = 1 &

2

1,1

dy 1m

dx 3

9. Ans. (B)

xdy – dx = –ex+y dx

d(e–y.x) = exdx

x = ex+y

+ cey

x = ey(ex + c)

10. Ans. (A,B)

AA' = I, BB' = I

(AB) (AB)' = ABB'A' = I

[(AB) (AB)] [(AB) (AB)]' = AB AB(AB)'(AB)'

= ABABB'A'B'A' = I

||AB| A| = |±A| = (±1)n|A| = 1 or –1

also ||AB|B| = 1 or –1

11. Ans. (A,B,D)

2acb

a c

and b2 = –2ac

2b

b a b c 0a c

12. Ans. (B,C)

1A 2 1

2 2

/2

–1 0 1A

2

1 2y sin x x sin y

/ 2

y 0

A 2 1 sin y dy

Paragraph for Question 13 & 14

13. Ans. (D)

3 / 2 1x 1 log 4sin log x x 1 cos

x 1

x 1lim e

1 / 24sin logx 1x 1 log x 1 x 1 log x 1 cos

x 1 x 1

x 1lim e

x 1 log x 1 0

x 1lim e

2

1

x 11

x 1 0e e 1

14. Ans. (C)

g(1) = a = 1

n

sin a 1lim 1

sin a 1

SECTION – II

1. Ans. (A) (Q); (B) (P); (C) (Q);(D) (R,S)

B = [aij]

4×4, a

ij = 1

|B| = 0

B2 = 4B

B3 = 4B

2 = 16B

|B2 – 4B + I| = 0

|B3 + B

2 – 8B| = 0

from B2 – 4B + 3I = 3I

B2 – 4BI + 3I = 3I

B2 – 3BI – BI + 3I = 3I

B(B – 3I) – (B – 3I) = 3I

BA I I

3

1 BA I

3

A–1

+ I = B

3

|A–1 + I| = 0

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PAPER – 2

SECTION-IV

1. Ans. 2

c = ae and 22b

4ca

c2 = a2e2 = a2 – b2 & b2 = 2ae

c2 = a2 – 2ae

2a a

2 1 0c c

a 2 4 4 2 2 21 2

c 2 2

a a1 2 2

c c

2. Ans. 5

The relevent terms are

5C3x2(x2)3 and 4C1(x3)3(–x2)

5C33= – 4C13

103 = –43

53 + 23 = 0

3

3

25

3. Ans. 6

Let AB = x

Area of base = 23 3

. 14 4

volume is maximum if the face ACD is r

BCD

height 3

2

A

x

B

D

C

1 3 3 1

V3 4 2 8

4. Ans. 3

22 ˆxa yb c 2i

x2 + y2 + 1 – 2xy a.b 2y b.c 2x a.c

= 4

x2 + y2 = 3

5. Ans. 2

Let ƒ–1(x) = t, ƒ–1(2a) = & ƒ–1(8a) =

3 2

2 4

4t t 2tI dt

t t

3

2 4 2

4t 2t 1dt dt

t t 1 t

4 21 1

4 2log tan tan

34 22a ƒ( )

3

34 28a ƒ( )

3

3

1 1

3

8a3I log tan tan

2a3

= log2 (a , )

n = 2

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PAPER – 2

SECTION–I

1. Ans. (A)

Sol.F / A W W YA

Y slope/ A

9 2

6 3

1 40 20Y (slope) 2 10 Nm

A 10 (2 1) 10 10

2. Ans. (D)

Sol.1 2

1 2

r i

v vA A

v v

; 12 2

vv energy reflected =

1

9 × energy incident

3. Ans. (C)

Sol. Optical path difference between (OPD) P & Q

(O.P.D.) = 2.25 0 ×1 + (3.5

0) × 2 + 3

0 × 3 = 18.25

0 and

0

2

2x

4. Ans. (C)

Sol. Net resistance is constant so current is constant But j = 2 2

i E ER

4 r kr

E = constant.

5. Ans. (A)

Sol. Current = 1

V VS

Sd d

6. Ans. (A)

Sol. In the standard PO experiment, reading are taken such that the galvanometer deflects in opposite direction.

The key connected to galvanometer is switched on after the key of the battery so as to reduce the

inductive effect in the circuit while taking the reading. The same is true while switching off the set-up.

Note : Refer the PO Box experiment.

7. Ans. (C)

Sol.

aCM

= r

2

qE mg mgrr

m mr / 2

qEqE mg 2 mg 3 g

m

3 gm 3 0.2 10 2E 12

q 1

SOLUTION

PART-2 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

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

Q. 11 12 13 14

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

A B C D

Q,R,S S P,S,T S

Q. 1 2 3 4 5

A. 4 4 1 1 3

Q.1

SECTION-I

SECTION-IV

SECTION-II

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PAPER – 2

SECTION–II

1. Ans. (A) (Q, R, S); (B) (S); (C) (P,S,T); (D) (S)

SECTION-IV

1. Ans. 4

2. Ans. 4

Sol. f = 2( 2 )

v

e where e = end correction = 0.6 r

f = 2( 2 0.6 )

v

r =

2( 1.2 )

v

r

f

f

=

v

v

–

( 1.2 )

1.2

r

r

=

v

v

–

1.2

1.2

r

r

here v

v

= 0 (given)

f

f

× 100 = –

1.2100

1.2

r

r

for maximum % error : = 0.1, r = 0.05

max

100f

f

=

0.1 1.2 0.05

94 1.2 5

× 100 = 0.16%

3. Ans. 1

Sol. The change in internal energy of the gas is U = 1.5 nR (T)= 1.5 (1 mol) ( 25

3 J/mol-K) (2K)

= 25 J.

The heat given to the gas = 42 J

The work done by the gas is W = Q – U= 42 J – 25 J = 17 J.

If the distance moved by the piston is x, the work done is W = (100 kPa) (8.5 cm2) x.

Thus, (105 N/m2) (8.5 × 10–4 m2) x = 17 J or, x = 0.2m = 20 cm.

4. Ans. 1

Sol. = t (–1) = 1 + d() = t(d) d = add() = td

When temperature of the plate is increased, its refractive index increases and, as a result, the fringes

cross at a particular point. Here the number of fringes crossing through a point is due to change in

refractive index of the plate. Shift in the location of a particular fringe due to changes in refractive index

is, 2 1( )µ µ tD

yd

So, the number of crossed fringes, 2 1( )µ µ t

n

5. Ans. 3

Sol. Atom will exist if size of nucleus is less than the orbit of -meson

15 1/3 0.531.6 10 37.5

265Z Z

Z

KOTA / HS - 13/1401CT313079

LEADER & ENTHUSIAST COURSE 10-04-2014

TM

Path to success KOTA (RAJASTHAN)

PAPER – 2

8. Ans. (D)

Sol.

N

mg

µ NS

mR2

N = mg cos + m2R

& mg sin = µsN

9. Ans. (A)

Sol. From the graph U = 3 sinx + 3 3 cosU

F xdx

max 3F

10. Ans. (B,C)

Sol. T 2g

g T–2 = g 2 T

g T

Therefore 2 T

P 100T

1

0.1 2(1)P

100 400

× 100 = 0.6%, 2

0.1 2(1)P

400 400

× 100 = 0.42%

3

0.1 2(1)P

100 200

× 100 = 1.1%, 4

0.1 2(1)P

400 800

× 100 = 0.28%

11. Ans. (A,C,D)

Sol. At steady state I(3) + I(2) = 15 I = 3

KVL C D E a b C

VC – I (3) +

11

q – 7 +

5

q=V

C

C1 C27V

3

15V

a b

A B

CD

E

2

2–q +q –q +q

I

I11

q +

5

q= 7 + 3 × 3 = 16 q = 55 C.

Now KVL for a b

Va – 7 +

5

q = V

b V

a – V

b = 7 –

5

q= 7 –

55

5=

55

5= –4V

Potential difference across C1 :

11

q=

55

11= 5V P.d. across C

2 :

5

q = 11V

Potential difference across terminal = 15 – I(2) = 15 – 3 × 2 = 9V

12. Ans. (A, B)

13. Ans. (D)

14. Ans. (C)

01CT313079KOTA / HS - 14/14

10-04-2014TARGET : JEE 2014

TM

Path to success KOTA (RAJASTHAN)

PAPER – 2

SECTION-I

1. Ans (D)

2. Ans. (C)

3. Ans. (B)

2O gas

gas

r M3

r 2 32 M

gas = 48

Vm, real =

1000

800 × 48 = 60

Vm, ideal =

RT 0.08 50040

P 1

Z = 60 3

40 2

4. Ans. (D)

5. Ans. (C)

6. Ans. (C)

7. Ans. (B)

8. Ans. (D)

9. Ans. (B)

10. Ans. (B)

11. Ans. (B, C)

12. Ans. (B,D)

13. Ans (B)

[HX] =

8

80100

× 1000 = 1 M

pH = 1

2 (pK

a + logC)

(since ak

C 10–3)

pH = 2

SOLUTION

PART-3 : CHEMISTRY ANSWER KEY

14. Ans.(B)

At equivalence point, 1 1 2 2

Acid Base

M V M V

, V2 = 400

ml

HX + NaOH NaX + H2O

0.1mol 0.1 mol 0 0

[NaX] = T

n

V=

1110 1000

400 100 5

pH = 1

2(pK

w + pK

a + log c)

= 8.65

SECTION-II

1. Ans. (A) (S, T); (B) (P, Q, R);(C) (P, R, T); (D) (P, T)

SECTION-IV

1. Ans (0.4) [OMR Ans. 4]

2. Ans. 2

3. Ans. 1

4. Ans. 2

5. Ans. 3

Q. 1 2 3 4 5 6 7 8 9 10

A. D C B D C C B D B B

Q. 11 12 13 14

A. B,C B,D B B

A B C D

S,T P,Q, R P,R,T P,T

Q. 1 2 3 4 5

A. 4 2 1 2 3

Q.1

SECTION-I

SECTION-IV

SECTION-II