An elevator car whose floor to ceiling distance is equal to 2 m starts ascending with constant acceleration 2 $m / s^{2}$ . $5 s e c$ after the start, a bolt begins falling from ceiling of car. Time after which bolt hits floor of elevator starting from fall. $(g = 10 m / s^{2})$

0.7

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\frac{1}{\sqrt{2}}

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1.73

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\frac{1}{\sqrt{3}}

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Solution

The correct option is D $\frac{1}{\sqrt{3}}$ s

By placing the frame of reference on the lift, the kinematic parameters of the bolt becomes,
$u = 0 m / s, a = 12 m / s^{2}, s = 2 m$
$s = u t + \frac{1}{2} a t^{2}$
$\Rightarrow 2 = \frac{1}{2} \times 12 \times t^{2} \Rightarrow t = \frac{1}{\sqrt{3}} s e c$

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An elevator car whose floor to ceiling distance is equal to 2 m starts ascending with constant acceleration 2 m/s2. 5 sec after the start, a bolt begins falling from ceiling of car. Time after which bolt hits floor of elevator starting from fall. (g=10 m/s2)

The correct option is D 1√3 s By placing the frame of reference on the lift, the kinematic parameters of the bolt becomes, u=0 m/s,a=12 m/s2,s=2 m s=ut+12at2 ⇒2=12×12×t2⇒t=1√3 sec

An elevator car whose floor to ceiling distance is equal to 2 m starts ascending with constant acceleration 2 $m / s^{2}$ . $5 s e c$ after the start, a bolt begins falling from ceiling of car. Time after which bolt hits floor of elevator starting from fall. $(g = 10 m / s^{2})$