Time period of a simple pendulum in a stationary lift is ‘T’. If the lift accelerates with \(\frac{g}{6}\) vertically upwards then the time period will be (Where g = acceleration due to gravity)
\(\sqrt{\frac{6}{5}}T\)
\(\sqrt{\frac{5}{6}}T\)
\(\sqrt{\frac{6}{7}}T\)
\(\sqrt{\frac{7}{6}}T\)
The Correct Option is C
Solution and Explanation
\(T' = 2\pi \sqrt{\frac{I}{g_{\text{eff}}}}\)
\(T' = 2\pi \sqrt{\frac{I}{g + \frac{g}{6}}} = 2\pi \sqrt{\frac{6I}{7g}}\)
\(T' = \sqrt{\frac{6}{7}}T\)
So,the correct option is C
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