Question:

An elevator car whose floor to ceiling distance is \(2.7\ \text{m}\) starts ascending with a constant acceleration of \(1.2\ \text{m/s}^2\). \(2\ \text{s}\) after a bolt begins to fall from the ceiling of the car. The free fall time of the bolt is \((g = 9.8\ \text{m/s}^2)\)

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In non-inertial frames, use pseudo forces: relative acceleration = \(g + a\) when lift accelerates upward.
Updated On: Apr 23, 2026
  • \(\frac{2.7}{9.8}\ \text{s}\)
  • \(\frac{5.4}{9.8}\ \text{s}\)
  • \(\frac{5.4}{8.6}\ \text{s}\)
  • \(\frac{5.4}{11}\ \text{s}\)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
Relative acceleration of bolt w.r.t. elevator = \(g + a\) (since elevator accelerates upward).
Step 2: Detailed Explanation:
Elevator acceleration \(a = 1.2\ \text{m/s}^2\) upward. Bolt accelerates downward with \(g\).
Relative acceleration = \(g + a = 9.8 + 1.2 = 11\ \text{m/s}^2\).
Initial relative velocity = 0. Distance = \(2.7\ \text{m}\).
Using \(s = \frac{1}{2}at^2\): \(2.7 = \frac{1}{2} \times 11 \times t^2\)
\(\Rightarrow t^2 = \frac{5.4}{11} \Rightarrow t = \sqrt{\frac{5.4}{11}}\ \text{s}\).
Step 3: Final Answer:
Thus, time = \(\sqrt{\frac{5.4}{11}}\ \text{s}\).
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