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[1]TAITS / JM / PT-3 / 2016IITians TAPASYA...IITians creating IITians

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IITians TAPASYA

SOLUTION OF JEE MAIN PART TEST - 3 PHYSICSQ.1 Ans. (C)

Sol. PQ nC T

P v2 f RC C 1f 2

f 2 R2

Slope is more if f is more.

Q.2 Ans.(B)

Sol. Apply rPV const.

1 1 2 2P V P V

Q.3 Ans.(C)

Sol. Q U W

f i

U Q QQ mL 1 2256 2256JW P V P V V

5 610 1671 1 10 167JU 2256 167 2089J

Q.4 Ans.(C)

Sol.R/2=2

3M/4 com M/4

x

centre of mass will fall at a distance of 4 km.

3M M2 x4 4

x = 6 kmQ.5 Ans. (B)

Sol. P0 P020cm40cm

P0 = 76 cm of Hg

P1

P276 40x

40+x

P2 = P1 + 76 ---- (i)

Temp. of air remains constant in both sections.

P1 + (40 + x) = P0 (40) ----- (ii)

P2 + (40 x) = P0 (40) ----- (iii)

Find x using (i), (ii) and (iii).



IITians TAPASYA

Q.6 Ans.(C)

Sol. At constant volume work done is O.

So, Heat given is used to increas temperature only.

Q.7 Ans.(D)

Sol. Use average form of newtons cooling law.

avg s

s

s

T k T Tt ms

30 25 k 27.5 T ....(i)2 50 30 1

30 25 K 27.5 T ....(ii)2 1005 30 1

divide (i) by (ii) ans solve.

Q.8 Ans.(B)

Sol. Work done = Area under PV graph.

Q.9 Ans.(D)

Sol. v

dx

Air

Rotating Blades of turbine

reaching blade

Kinetic energy carried by wind.

21K mv2

rate at which kinetic energy obtained by wind

2dk V dmdt 2 dt

dm dxdm Adx A AVdt dt

3dk 1 AVdt 2

If efficiency of turbine is then power output 31P AV2



IITians TAPASYA

Q.10 Ans.(C)

Sol. lW nRT n2

l l3RT 3W 3WVM nM n2 m n2

Q.11 Ans.(A)

Q.12 Ans. (B)

Sol.Vx

Force applied on wall

l l

2x x

x

2mV mV2V

2F V and V T

Q.13 Ans.(C)

Sol. x dxA B

cm

dm xx

dm

bxdm dx a dxL

Q.14 Ans.(A)

Sol. Use dimensional analysis.

Q.15 Ans.(A)

Sol. mT constant Q.16 Ans.(C)

Sol.

2R

l l l2 2 2 2

1 2T T TK RT K 4 R R K 4 R



IITians TAPASYA

Q.17 Ans.(C)

Sol.

50 40 K 40 20 ....(i)5

40 30 K 35 20 ....(ii)t

Q.18 Ans. (B)

Sol. 1 2v v

is valid always in elastic collision.

Q.19 Ans.(A)

Sol. Maximum potential energy is stored when both the blocks move together during the collision.

2 2rel

m mn1 1 EMax.PE V v2 2 m m 1

Q.20 Ans.(A)

Sol.

u

u

time interval between two successive collisions 2ug

change in momentum in each collision = 2mu

2muavg. force g mg2u

Q.21 Ans.(B)

Sol. vv (u+v )1

v1Initial Final

Apply conservation of linear momentum.

(m + M) V = m (u + v1) + MV1Q.22 Ans.(C)

Sol. h Mu v

v

m

2 2mu m m v

1 1mgh mu m m v2 2

Q.23 Ans.(A)

Sol. If P VC C R, then it is Non ideal situation for the gas. For gas to behave as ideal. Pressure

is low and temp is high. Gas may deviate from ideal behaviour if pressure is increased and

temp. is decreased.



IITians TAPASYA

Q.24 Ans.(C)

Sol. Observe the distribution of mass along x-axis.

Q.25 Ans.(C)

Sol.

vv 4v2v

Before After2hTg

separation between wall and ball = 2h4v vg

Q.26 Ans.(A)

Sol. Since length of each segment is same then mass is also same.

cm

2R2m 2m oRy

4m

Q.27 Ans.(B)

Sol.

cm 1 2M x m x m' xxm m' mx 05

Q.28 Ans.(A)

Sol.x

0 Co t lCtT x

Q.29 Ans.(A)

Sol. heat current is same in all layers.

lTi KA

Q.30 Ans.(B)

Sol.l

l

C

S

Q TK A ...(i)30Q TK A ...(ii)60

from (i) and (ii) KC = 2Ks

C S C S Seq

K A K A K K 3KKA A 2 2

l

s3KQ T2A ...(iii)t 2

from (ii) and (iii) t = 20 min.



IITians TAPASYA

SOLUTION OF JEE MAIN PART TEST - 3 MATHEMATICSQ.61 Ans.(B)

Sol. a b c, a 0 b 0 c 0

2 2 2

2 2 2

a b c (a b c)(a b c ab bc ca)b c a 1(a b c) [(a b) (b c) (c a) ] 0c a b 2

Q.62 Ans. (A)

Sol. 1 1sin(cot (1 x)) cos(tan x)

2 22 2

1 1 1x 2x 2 x 1 x21 (1 x) 1 x

Q.63 Ans. (D)

Sol. xaxis(y=0)

P(1,3)P(1,3)

P(1,3)(Image of P)

Q(K,0)

R(6,7)

Eqn QR : y 7 = 7 3 (x 6)6 1

, put y = 0 5 57 2x 12 x K2 2

Q.64 Ans. (A)

Sol.

1 1

2 2

3 3

1L : x 3y p m3

aL : ax 2y q m2

L : ax y r m a

1 2

1 2

1 3

1 32

If L Lm m 1 a 6If L Lm m 1 a 3a 9q 18 0 a 6,3

Q.65 Ans. (A)

Sol. 21 2 1 2 9 4

A A A4 3 4 3 8 17

2 9 4 5 4 8 0 8 4 2 1A 4A 5I 48 16 0 17 12 5 8 0 2 0



IITians TAPASYA

Q.66 Ans. (D)

Sol. Consistency (infinite no. of Solutions)

x y z 0

1 2 3a 8 0

1 3 5 0a 8

2 5 a

Z1 2 6

b 15 01 3 9 0

b 152 5 b

a = 8, b = 15

Q.67 Ans. (D)

Sol. x y3x 4y 12 0 14 3

B 30,2

3b2

A(8,0)a = 8

Slope 3 0 320 8 16

Q.68 Ans. (D)

Sol. 1 1sin x 2tan x Domain x [ 1, 1]

12tan x

1sin x1sin x

1sin x

O1

2

2

1 3 solutions, {1, 0, 1}

Q.69 Ans. (A)

Sol.3 4 1

[3p 3q 2r 4p 2q p 3q 2r][p q r] 3 2 3

[3 0 1 ]2 0 2

3p 3q 2r 14p 2q 0p 3q 2r 0p 1, q 2, r 32p q r 3



IITians TAPASYA

Q.70 Ans. (B)

Sol. [0, 2 ] 22cos 3sin 0 22sin cos2 0 22sin 3sin 2 0

24sin 1 0 (2sin 1)(sin 2) 0

1sin2

1sin sin 22 (not possible)

5 7 11, , ,6 6 6 6

5, ; [0, ] or [0, 2 ]6 6

Q.71 Ans. (B)

Sol. 1 sin cos1 cos sin 01 sin cos

2 2

2

sin costan 1tan 1

0, ,2 4

cos sin cossin cos sin 0cos sin cos

2 22cos (sin cos ) 0 2 2cos 0 or sin cos tan 1

0, ,2 4

Q.72 Ans. (B)

Sol.90 2

90 2 90

A

B

C

D

x = 2a(2a,0)

(a,0)

2

a2a

BC AC , A B BD = 2a

a 1tan2a 2



IITians TAPASYA

BD 2atan2 CDCD tan2

22tan 4tan2

1 tan 3

3aCD2

5aAC (Base) AD CD2

BD (height) = 2a

Area = 21 5a 5a2a

2 2 2

Q.73 Ans. (A)

Sol. sin2x 2cosx 4sinx 4 0 2sinxcosx 2cosx 4sinx 4 0 (2cos x 4)(sinx 1) 0

cosx 2 (not possible)sinx 1

If 5 9x [0, 5 ], x , ,2 2 2

Q.74 Ans. (A)

Sol. 1 122x 3y C 0 m3

2 2

3 3

2 3

1x 5y C 0 m51x 5y C m5

m m

angle between lines 1 31 21 2 1 3

m mm mtan1 m m 1 m m

1 2 (angle depends an slopes only)

Q.75 Ans. (B)

Sol. A I or I

Let a bAc d

22

2

1 0a bc ab bdA I0 1ca cd cb d

2

2

a bc 1 b(a d) 0d bc 1 c(a d) 0



IITians TAPASYA

b and c cant be equal to zero as A will be I or Ia d 0 a d

Tr(A) a d 0

det 2(A) ad bc d bc 1

Q.76 Ans. (D)

Sol.

P(2, 3)

Q(4, 5)

L=0(3, 4)

D

image

Slope 5 3PQ 14 2

PQ L, slope L 1

eqn of line L = y 4 = 1 (x 3) x + y 7 = 0

R(0, 0)

I

(3, 4)x y 7 0

( )

image of (0, 0) is I ( )

2 20 0 7( 2)

1 1 1 1

7, 7

I = (7, 7)

Q.77 Ans. (B)

Sol. x ay 0.z 0 0.x y az 0

ax 0.y z 0

For unique solution 0

31 a 0 1 a 00 1 a 0a 1a o 1



IITians TAPASYA

Q.78 Ans. (B)

Sol. x12

2 1sin x

1cos x

O 1

1 1sin x cos x1x , 12

Q.79 Ans. (B)

Sol. C (h,k) A ( 3, 2) B ( 2,1)

Centroid h 5 k 3,3 3

lies on 3x 4y 2 0

h 5 k 33 4 2 03 3

3h 4K 3 0

3x 4y 3 0

Q.80 Ans. (C)

Sol. Least value of xyz occurs when x = y = z = 2

Q.81 Ans. (C)

Sol. 2 0 1 0 1 1 0A AA I1 0 1 0 0 1

2A I 0 _(i)4 2A A I 0 4 2 2A I 0 ( A I) 4A I 0 _(ii)

2 2A(A I) 0 A I 3 2 4A AI A I A I 0 3A I A(A I) _(iii)

Q.82 Ans. (D)

Sol. 8A 0, B(1, 3) C(82,30)3

Slope (AB) = 13 = Slope (BC)

A, B, C are collinear (lie on striaght line)



IITians TAPASYA

Q.83 Ans. (A)

Sol.

L 0

L 0Y

Q(2, 1)

X

P(1, 2)

B(0, 4)

F

A(2,0) 2 2

BP AP A(2, 0), B(0, 4)x yL 12 4

2x y 4 0Foot of Q( 2,1) on line 2x y 4 0x 2 y 1 4 1 4( )

2 1 2 14 12x y5 5

Q.84 Ans. (B)

Sol. 1 122xsin 2tan x if x 1

1 x

given 1 1x 5 f(x) 2tan x 2tan x

Q.85 Ans. (A)

Sol.3 1 2

2

2

adjA order 3 matrix| adj A | | A | | A |

| 5adj A | 125 | adj A | 125 | A | 51| A |

251| A |5

Q.86 Ans. (B)

Sol. Put x = 1

2 2 1a 12 4 3 0 12

6 1 1a 24

Q.87 Ans. (B)

Sol. 3 x y 1 inclination = 120o

Line L must have inclination o o o120 60 60 to intersect x-axis. i.e slope 3

equation of L is : y 2 3 (x 3)

Q.88 Ans. (C)

Sol. 2



IITians TAPASYA

3cos cos 2cos .cos2 2 2

__(i)

1sin sin 2sin .cos2 2 2

__(ii)

(ii) (i)

1 1tan tan2 3 3

2

2 22tan 1 tan 7sin2 cos2

1 tan 1 tan 5

Q.89 Ans. (B)

Sol. BD and BE are intresect at B

Coordinates of B are (3, 2)mAB = 1/5

tanq = 103151

23

= 2m31

m23

1 = m32m23

or + 1 = m32m23

m = 1/5 (rejected) or 5equation of BC = y + 2 = 5 (x + 3) 5x + y + 17 = 0Alternatively : Take image of (2, 1) in the line BD to get a point on BC

Q.90 Ans. (D)

Sol. Q x2 + ax + sin1 (x2 4x + 5) + cos1 (x2 4x + 5) = 0 ..........(1)for equation (1) to be defined.

1 x2 4x + 5 1 Q x2 4x + 5 > 0 x R

x2 4x + 4 0

(x 2)2 0

x = 2

equation (1) will be defined if x = 2, we get.

4 + 2a + 2 = 0.

2a =

24 a =

42 Ans.

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