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Q.

An elevator car whose floor to ceiling distance is 2.7m starts ascending with a constant acceleration of $1.2 {ms}^{- 2}$ . 2 s after the start, a bolt falls from the ceiling of the car. The free fall time of the bolt is (Take, $g = 9.8 {ms}^{- 2}$ )

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a

$\sqrt{\frac{2.7}{9.8}} s$

b

$\sqrt{\frac{5.4}{9.8}} s$

c

$\sqrt{\frac{5.4}{8.6}} s$

d

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

answer is D.

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Detailed Solution

Relative to lift,

$u_{r} = 0, a_{r} = (9.8 + 1.2) = 11 {ms}^{- 2}$

$s_{r} = \frac{1}{2} a_{r} t^{2} \Rightarrow 2.7 = \frac{1}{2} \times 11 \times t^{2} \Rightarrow t = \sqrt{\frac{5.4}{11}} s$

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An elevator car whose floor to ceiling distance is 2.7m starts ascending with a constant acceleration of 1.2 ms-2. 2 s after the start, a bolt falls from the ceiling of the car. The free fall time of the bolt is (Take, g = 9.8 ms-2)