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Physics · PYQ NEET UG 2022
हिन्दीTwo pendulums of length \( 121\text{ cm} \) and \( 100\text{ cm} \) start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:
Correct answer · Option 3
11
Explanation
Time periods are proportional to \( \sqrt{l} \). \( T_1 \propto \sqrt{121} = 11 \), \( T_2 \propto \sqrt{100} = 10 \). Let \( n_2 \) be the vibrations of the shorter pendulum: \( n_2 T_2 = (n_2 - 1) T_1 \implies n_2 \times 10 = (n_2 - 1) \times 11 \implies 10n_2 = 11n_2 - 11 \implies n_2 = 11 \).
