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Simple pendulum of large length is made equal to the radius of the earth. Its period of oscillation will be :

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$84.6 \min$

$59.8 \min$

$42.3 \min$

$21.15 \min$

answer is B.

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Detailed Solution

$T = 2 π \sqrt{\frac{1}{g (\frac{1}{R} + \frac{1}{R})}} \Rightarrow T = 2 π \sqrt{\frac{R}{2 g}} = \frac{84.6}{\sqrt{2}} \min$

or $T = 59.8 \min$

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