If a simple pendulum is taken to a place where g decreases by 2%, then the time period

Increases by 1 %

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Decreases by 1 %

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Increases by 2 %

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Decreases by 2 %

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Solution

The correct option is A
Increases by 1 %

$T = 2 π \sqrt{\frac{1}{g}}$

$∴ \frac{d T}{T} = - \frac{1}{2} \frac{d g}{g} = - \frac{1}{2} \times - 2$

$\Rightarrow$ increases by 1%

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If a simple pendulum is taken to a place where g decreases by 2%, then the time period

The correct option is A Increases by 1 % T=2π√1g ∴dTT=−12dgg=−12×−2 ⇒ increases by 1%

The correct option is A
Increases by 1 %

$T = 2 π \sqrt{\frac{1}{g}}$

$∴ \frac{d T}{T} = - \frac{1}{2} \frac{d g}{g} = - \frac{1}{2} \times - 2$

$\Rightarrow$ increases by 1%