The length of a simple pendulum is made one-fourth. Its time period becomes:

four times

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one-fourth

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double

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half

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Solution

The correct option is D half
$T = 2 π \sqrt{\frac{L}{g}}$
If length is reduced to a fourth $T^{'} = 2 π \sqrt{\frac{L}{4 g}} = \frac{2 π}{2} \sqrt{\frac{L}{g}} = T / 2$

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The length of a simple pendulum is made one-fourth. Its time period becomes:

The correct option is D halfT=2π√LgIf length is reduced to a fourth T′=2π√L4g=2π2√Lg=T/2

The correct option is D half
$T = 2 π \sqrt{\frac{L}{g}}$
If length is reduced to a fourth $T^{'} = 2 π \sqrt{\frac{L}{4 g}} = \frac{2 π}{2} \sqrt{\frac{L}{g}} = T / 2$