A simple pendulum has time period ' $t$ '. Its time period in a lift which is moving upward with acceleration $3 m s^{- 2}$ is :

t \sqrt{\frac{9.8}{12.8}}

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t \sqrt{\frac{12.8}{9.8}}

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t \sqrt{\frac{9.8}{6.8}}

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t \sqrt{\frac{6.8}{9.8}}

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Solution

The correct option is A $t \sqrt{\frac{9.8}{12.8}}$
$t \propto \frac{1}{\sqrt{9.8}}, t^{'} \propto \frac{1}{\sqrt{12.8}} (∵ g^{'} = 9.8 + 3 = 12.8)$
$∴ \frac{t^{'}}{t} = \sqrt{\frac{9.8}{12.8}} \Rightarrow t^{'} = \sqrt{\frac{9.8}{12.8}} t$

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A simple pendulum has time period 't'. Its time period in a lift which is moving upward with acceleration 3 ms−2 is :

The correct option is A t√9.812.8t∝1√9.8,t′∝1√12.8(∵g′=9.8+3=12.8)∴t′t=√9.812.8⇒t′=√9.812.8t

A simple pendulum has time period ' $t$ '. Its time period in a lift which is moving upward with acceleration $3 m s^{- 2}$ is :

The correct option is A $t \sqrt{\frac{9.8}{12.8}}$
$t \propto \frac{1}{\sqrt{9.8}}, t^{'} \propto \frac{1}{\sqrt{12.8}} (∵ g^{'} = 9.8 + 3 = 12.8)$
$∴ \frac{t^{'}}{t} = \sqrt{\frac{9.8}{12.8}} \Rightarrow t^{'} = \sqrt{\frac{9.8}{12.8}} t$