If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

T = 2 π \sqrt{\frac{R_{e}}{g}}

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T = 2 π \sqrt{\frac{2 R_{e}}{g}}

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T = 2 π \sqrt{\frac{R_{e}}{2 g}}

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T = 2

seconds

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Solution

The correct option is A $T = 2 π \sqrt{\frac{R_{e}}{g}}$
The time period of body performing oscillation in the tunnel dug through earth is given by:
$T = 2 π \sqrt{\frac{R_{e}}{g}}$

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If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

The correct option is A T=2π√RegThe time period of body performing oscillation in the tunnel dug through earth is given by:T=2π√Reg

The correct option is A $T = 2 π \sqrt{\frac{R_{e}}{g}}$
The time period of body performing oscillation in the tunnel dug through earth is given by:
$T = 2 π \sqrt{\frac{R_{e}}{g}}$