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Q.

An elevator ascends with an upward acceleration of 2.0 ms–2. At the instant its upward speed is 2.5 ms–1, loose bolt is dropped from the ceiling of the elevator 3.0 m from the floor. If g = 10 ms–2, then

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a

the time of flight of the bolt from the ceiling to floor of the elevator is 0.11 s

b

the displacement of the bolt during the free fall relative to the elevator shaft is 0.75 m

c

the distance covered by the bolt during the free fall relative to the elevator shaft is 1.38 m

d

the distance covered by the bolt during the free fall relative to the elevator shaft is 2.52 m

answer is B, C.

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Detailed Solution

Velocity of bolt relative to elevator = 2.5 –2.5 = 0

Acceleration of bolt relative to elevator,

a=10--2=12ms-2g=10m/s2

Using the relation, s=ut+12at2

We have, 3.0=0×+12×12×t2

t=12s=0.707s=0.7s

Displacement =2.5×0.71+12×10×0.712

=-1.775+2.521=0.746=0.75m

Distance covered =2×u22g+displacement

=2×2.522×10+0.75

=0.63+0.75=1.38m

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