Question

A coin is thrown vertically upward in a lift which is ascending with acceleration \(a\). The coin takes time \(t_{1}\) to reach the maximum height \(h\). In the later instant, the coin is dropped from the same height in a lift which is descending with acceleration \(a\). The time taken by the coin to reach the floor of the lift is \(t_{2}\), then

• A. \(t_{2}\lt t_{1}\)
• B. \(t_{2}>t_{1}\)
• C. \(t_{2}=t_{1}\)
• D. \(t_{2}/t_{1}=0\)

Hint:

Use the effective gravity inside the lift: it is larger when the lift accelerates up and smaller when it accelerates down.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Inside a lift accelerating with \(a\), the coin feels an effective gravity \(g_{eff}\).
Upward acceleration gives \(g_{eff}=g+a\); downward acceleration gives \(g_{eff}=g-a\).

Step 2: Case 1: lift ascending
The coin is thrown up and reaches the height \(h\) in time \(t_1\). Equivalent to a free fall of \(h\) under \(g+a\):
\(h=\dfrac{1}{2}(g+a)t_1^{2}\).

Step 3: Case 2: lift descending
The coin is dropped from rest relative to the lift through the same height \(h\) under \(g-a\):
\(h=\dfrac{1}{2}(g-a)t_2^{2}\).

Step 4: Compare
\(\dfrac{t_2^{2}}{t_1^{2}}=\dfrac{g+a}{g-a}>1\), so \(t_2>t_1\).

Final Answer:
The time in the descending lift is longer, so \(t_2>t_1\). \[ \boxed{\text{(B) }t_{2}>t_{1}} \]