If $T_{1}$ = time period of a simple pendulum of infinite length, $T_{2}$ = time period of simple harmonic motion of a body dropped in a tunnel dug along diameter of earth and $T_{3}$ = time period of circular motion of a satellite revolving near the surface of the earth, then:

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$T_{1} < T_{2} < T_{3}$

$T_{1} = T_{2} = T_{3}$

$T_{1} > T_{2} = T_{3}$

$T_{1} = T_{2} > T_{3}$

answer is B.

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Detailed Solution

$T_{1} = 2 π \frac{1}{g (\frac{1}{l} + \frac{1}{R})} = 2 π \sqrt{\frac{R}{g}}, ∴ l \to \infty$

$T_{2} = \frac{2 π}{ω_{2}}, F_{2} = \frac{- G m M x}{R^{3}} \Rightarrow F_{2} = - m w_{2}^{2} x \Rightarrow T_{2} = 2 π \sqrt{\frac{R}{g}}$

$∴ T_{3} = 2 π \sqrt{\frac{R}{g}}$

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