3MCQS FOR JEE MAIN/ AIPMT/ BITSAT (SOLUTION)

1. (d)2. (a) Number of significant figures in 23.023= 5

Number of significant figures in 0.0003 = 1Number of significant figures in 2.1 103 = 2

3. (b) [momentum] = [M][L][T1] = [MLT1]

Plancks constant = En

= 121

21TML

T]LT][M[ -

-

-=

4. (a) 2rmp

=rl

llD+D+D=r

rDr

r2mm

Putting the values cm06.0=Dlcm6=l ; cm005.0r =D ; r = 0.5 cm

m = 0.3 gm; Dm = 0.003 gm we get 100

4=rrD

%4100 =rrD\

5. (c)0 0

1c=

m e 2

0 0

1 c=m e

20 0

1 [c] = m e

[c] = LT1 or [c]2 = L2T2

6. (b) As the ball moves down from height h toground the P.E. at height h is converted to K.E.at the ground (Applying Law of conservationof Energy).

Hence, 21

mAvA2 = mAghA

or vA = 2gh ;

Similarly, vB = gh2 or vA = vB7. (a)

8. (c) The distance covered in nth second is

a)1n2(21uSn -+=

where u is initial velocity & a is acceleration

then 2a19u26 += ....(1)

2a21u28 += ....(2)

2a23u30 += ....(3)

2a25u32 += ....(4)

From eqs. (1) and (2) we get u = 7m/sec,a = 2m/sec2\ The body starts with initial velocity u =7m/secand moves with uniform acceleration a = 2m/sec2

9. (d) The only force acting on the ball is the force ofgravity. The ball will ascend until gravity reducesits velocity to zero and then it will descend. Findthe time it takes for the ball to reach its maximumheight and then double the time to cover theround trip.Using vat maximum height = v0 + at = v0 gt,we get: 0 m/s = 50 m/s (9.8 m/s2) tTherefore,t = (50 m/s)/(9.8 m/s2) ~ (50 m/s)/ (10 m/s2) ~ 5sThis is the time ball takes to reach its maximumheight.Therefore, the total round trip time is 2t ~ 10s.

10. (b) Let after a time t, the cyclist overtake the bus.

Then t20t22196 2 =+

or t2 20 t + 96 = 0

1296440020t

-=\

.sec82

420 == .sec12and