A pendulum suspended from the ceiling of the train beats second when the train is at rest. What will be the time period of the pendulum if the train accelerates at $10 m s^{- 2}$ ? Take $g = 10 m s^{- 2}$ .

(2 / \sqrt{2}) s

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2 s

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2 \sqrt{2} s

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none of these

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Solution

The correct option is D none of these
When train moves with acceleration a, net effective acceleration of pendulum = $\sqrt{a^{2} + g^{2}} = 2 \sqrt{10}$
Thus, $T = 2 π \sqrt{l / a_{e f f}} = 2 π \sqrt{L / 2 \sqrt{1} 0} = 1 / \sqrt{2} \times 1 = 1 / \sqrt{2}$ , which is none of the options

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A pendulum suspended from the ceiling of the train beats second when the train is at rest. What will be the time period of the pendulum if the train accelerates at 10ms−2? Take g=10ms−2.

The correct option is D none of theseWhen train moves with acceleration a, net effective acceleration of pendulum = √a2+g2=2√10Thus, T=2π√l/aeff=2π√L/2√10=1/√2×1=1/√2, which is none of the options

A pendulum suspended from the ceiling of the train beats second when the train is at rest. What will be the time period of the pendulum if the train accelerates at $10 m s^{- 2}$ ? Take $g = 10 m s^{- 2}$ .

The correct option is D none of these
When train moves with acceleration a, net effective acceleration of pendulum = $\sqrt{a^{2} + g^{2}} = 2 \sqrt{10}$
Thus, $T = 2 π \sqrt{l / a_{e f f}} = 2 π \sqrt{L / 2 \sqrt{1} 0} = 1 / \sqrt{2} \times 1 = 1 / \sqrt{2}$ , which is none of the options