A coin is dropped in a lift. It takes time t1 to reach the floor when lift is stationary. It takes time t2 when lift is moving up with constant acceleration. Then:

Question

A coin is dropped in a lift. It takes time t1 to reach the floor when lift is stationary. It takes time t2 when lift is moving up with constant acceleration. Then: (A) t1 > t2 (B) t2 > t1 (C) t1 =t2 (D) t1>>t2

Tardigrade Admin · Accepted Answer

Time taken by coin to reach the floor is given by

h = \frac{1}{2} g t^{2} (∵ u = 0)

\Rightarrow t = \frac{2 h}{g}

In stationary lift,

t_{1} = \frac{2 h}{g}

In upward moving lift with constant acceleration

a

,

g^{'} = g + a

∴ t_{2} = \frac{2 h}{g + a}

Clearly,

g^{'} > g

Thus,

t_{2} < t_{1}

Q. A coin is dropped in a lift. It takes time $t_{1}$ to reach the floor when lift is stationary. It takes time $t_{2}$ when lift is moving up with constant acceleration. Then:

$t_{1} > t_{2}$

$t_{2} > t_{1}$

$t_{1} = t_{2}$

$t_{1} >> t_{2}$

Q. A coin is dropped in a lift. It takes time t1​ to reach the floor when lift is stationary. It takes time t2​ when lift is moving up with constant acceleration. Then:

t1​>t2​

t2​>t1​

t1​=t2​

t1​>>t2​

Time taken by coin to reach the floor is given by h=21​gt2(∵u=0) ⇒t=g2h​​ In stationary lift, t1​=g2h​​ In upward moving lift with constant acceleration a, g′=g+a ∴t2​=g+a2h​​ Clearly, g′>g Thus, t2​<t1​

Q. A coin is dropped in a lift. It takes time $t_{1}$ to reach the floor when lift is stationary. It takes time $t_{2}$ when lift is moving up with constant acceleration. Then:

$t_{1} > t_{2}$

$t_{2} > t_{1}$

$t_{1} = t_{2}$

$t_{1} >> t_{2}$