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CGP EDU Academic Team
Published on: September 12, 2026
An elevator car, whose floor to ceiling distance is equal to 2.7 m, starts ascending with constant acceleration of 1.2 ms –2 . 2 sec after the start, a bolt begins fallings from the ceiling of the car. The free fall time of the bolt is
Text Solution
Verified by ExpertsThe correct answer is:
C
\(l = \sqrt{\frac{2h}{g+o}}\) \(= \sqrt{\frac{2 \times 2.7}{(9.8 + 1.2)}} = \sqrt{\frac{5.4}{11}} = \sqrt{0.49} = 0.7 \text{ sec}\)
As \nu = 0 and lift is moving upward with acceleration
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