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Question

Blocks A and B of mass m are placed inside an elevator which is going upward with an acceleration of 2 m/s2 and a convex mirror having focal length 12 cm is placed below block B. All the surfaces are smooth and the pulley is light. The mass pulley system is released from rest with respect to the elevator at time t=0 s, when the distance of the block B from the mirror is 42 cm. Find the distance between the image of the block B and mirror at t=0.2 s.


A
5.6 cm
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B
6.6 cm
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C
7.6 cm
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D
8.6 cm
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Solution

The correct option is D 8.6 cm
FBD of blocks A and B in elevator frame.
FP Pseudo Force.
For block A,
T=ma ............(1)
For block B,
mg+FPT=ma
mg+maeT=ma ............(2)
[ae acceleration of elevator]
On adding (1) and (2) and solving, we get,
a=6 m/s2
Distance travelled by block B in t=0.2 s,
S=12at2=12×6×(0.2)2=0.12 m=12 cm
Distance of block B from mirror =4212=30 cm
Here, u=30 cm
f=+12 cm
So, 1v+1u=1f
1v130=112
1v=760
v=607 cm=8.6 cm

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