The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is

Greater than T

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Less than T

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Equal to T

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Cannot be compared

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Solution

The correct option is A
Greater than T

Inside the mine g decreases

Hence from T = 2 $π \sqrt{\frac{l}{g}}$ : T increases

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The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is

The correct option is A Greater than T Inside the mine g decreases Hence from T = 2 π√lg : T increases

The correct option is A
Greater than T

Inside the mine g decreases

Hence from T = 2 $π \sqrt{\frac{l}{g}}$ : T increases