Question

If a coin is dropped in a lift it takes ${{t}_{1}}$ time to reach the floor and takes ${{t}_{2}}$ time when lift is moving up with constant acceleration. Then which one of the following relations is correct?

& \text{A) }{{t}_{1}}={{t}_{2}} \\\ & \text{B) }{{t}_{1}}>{{t}_{2}} \\\ & \text{C) }{{t}_{1}}<{{t}_{2}} \\\ & \text{D) }{{t}_{1}}>>{{t}_{2}} \\\ \end{aligned}$$

Answer 1

We need to understand the relation between the downward motion of an object when it is in a system which is at rest and a system when it is under a constant acceleration. These relations can help us get the required solution for this problem.

Complete answer:
We are given two instances in which a coin is dropped. Initially, the coin is dropped in a lift, when the lift is at the state of rest. Then, the lift is set to a constant accelerated motion and the coin is dropped. We need to find the forces acting on the coin in the two situations to find the required time differences between the two incidents.
Let us consider the coin falling when the lift is at rest. The only force acting on the coin is the force of gravity. The time taken for this motion can be given by the equations of motion as –

& a=g \\\ & \text{Applying the equations of motion,} \\\ & S=ut+\dfrac{1}{2}g{{t}^{2}} \\\ & \therefore {{t}_{1}}=\sqrt{\dfrac{2S}{g}} \\\ \end{aligned}$$ Now, we can consider an acceleration in the upward direction which is against the gravitational acceleration, but is greater because the lift is moving upwards. We can find the time taken for this as – $$\begin{aligned} & a'=g+a \\\ & \text{Applying the equations of motion,} \\\ & S=ut+\dfrac{1}{2}g{{t}^{2}} \\\ & \Rightarrow S=\dfrac{1}{2}(a+g){{t}_{2}}^{2} \\\ & \therefore {{t}_{2}}=\sqrt{\dfrac{2S}{g+a}} \\\ \end{aligned}$$ From the above equations we can understand that the time taken for the coin to reach the floor when the lift is at rest is greater than the time taken when the lift is under acceleration. $$\therefore {{t}_{1}}>{{t}_{2}}$$ This is the required solution. **The correct answer is option B.** **Note:** We are considering the Newtonian mechanics in which the change in the distance to be travelled by the coin remains a constant in the two systems. The acceleration is considered to be not much close to ‘g’ so that the height of fall remains the same.