An elevator car whose floor to ceiling distance is 27 ,m starts a… - Vidyadip

MCQ

An elevator car whose floor to ceiling distance is $27\,m$ starts ascending with a constant acceleration of $1.2\,m / s ^2$. After $2\,s$ of the start, a bolt falls from the ceiling of the car. The free fall time of the bolt is $\left(g=9.8\,m / s ^2\right)$

A
$\sqrt{\frac{2.7}{9.8}}\,s$
B
$\sqrt{\frac{5.4}{9.8}}\,s$
C
$\sqrt{\frac{5.4}{8.6}}\,s$
✓
$\sqrt{\frac{5.4}{11}}\,s$

✓

Answer

Correct option: D.

$\sqrt{\frac{5.4}{11}}\,s$

d
(d)

Relative to lift

$u_r =0$

$a_r =(9.8+1.2)=11\,m / s ^2$

$s_r =\frac{1}{2} a_r t^2$

$2.7 =\frac{1}{2} \times 11 \times t^2$

or $t=\sqrt{\frac{5.4}{11}}\,s$

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