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Question

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


A
Greater than T
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B
Less than T
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C
Equal to T
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D
Cannot be compared
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Solution

The correct option is A Greater than $$T$$
Inside the mine $$g$$ decrease
hence form $$T=2\pi \sqrt {\dfrac lg}: T$$ increase

Physics

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