In the following four periods
(i) Time of revolution of a satellite just above the earth’s surface $(T_{s t})$
(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth $(T_{m a})$
(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/kg $(T_{s p})$
(iv) Period of an infinite length simple pendulum in the earth’s real gravitational field $(T_{i s})$

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$T_{s p} < T_{i s}$

$T_{s t} > T_{m a}$

$T_{s t} = T_{m a} = T_{s p} = T_{i s}$

$T_{m a} > T_{s t}$

answer is C.

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Detailed Solution

$(i) T_{st} = 2 π \sqrt{\frac{(R + h)^{3}}{GM}} = 2 π \sqrt{\frac{R}{g}}$

$[As h << R and GM = {gR}^{2}]$

$(ii) T_{ma} = 2 π \sqrt{\frac{R}{g}}$

$(iii) T_{sp} = 2 π \sqrt{\frac{1}{g (\frac{1}{l} + \frac{1}{R})}} = 2 π \sqrt{\frac{R}{2 g}}$

$[As l = R]$

$(iv) T_{is} = 2 π \sqrt{\frac{l}{g}} [As l = \infty;$

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