An elevator car, whose floor to ceiling distance is equal to 2.7 … - Vidyadip

MCQ

An elevator car, whose floor to ceiling distance is equal to $2.7 m$, starts ascending with constant acceleration of $1.2 ms$. $2 sec$ after the start, a bolt begins fallings from the ceiling of the car. The free fall time of the bolt is

A
$\sqrt{0.54} s$
B
$\sqrt{6} s$
✓
$0.7 s$
D
$1 s$

✓

Answer

Correct option: C.

$0.7 s$

(c) $t=\sqrt{\frac{2 h}{(g+a)}}=\sqrt{\frac{2 \times 2.7}{(9.8+1.2)}}$ $=\sqrt{\frac{5.4}{11}}=\sqrt{0.49}=0.7 \mathrm{sec}$
As $u=0$ and lift is moving upward with acceleration

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