MHT CET · Physics · Oscillations
The time period of a simple pendulum inside a stationary lift is \(\sqrt{3}\) second. When the lift moves upwards with an acceleration \(\mathrm{g} / 3\), the time period will be ( \(\mathrm{g}=\) acceleration due to gravity)
- A 1.5 s
- B 2 s
- C \(\sqrt{3} \mathrm{~s}\)
- D 3 s
Answer & Solution
Correct Answer
(A) 1.5 s
Step-by-step Solution
Detailed explanation
\( T_1 = 2\pi\sqrt{\frac{L}{g}} = \sqrt{3} \) \( g' = g + \frac{g}{3} = \frac{4g}{3} \)
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