A simple pendulum has a periodic time ' T₁ ' when it is on the surface of earth of radius ' R '. Its periodic time is ' T₂ ' when it is taken to a height ' R ' above the earth's surface. The value of (T₂)/(T₁) is (a) √(2) (b) 1 (c) 2 (d) 1/2

Answer: (c). At h = R: gh = g/4. (T₂)/(T₁) = √(g/gh) = √(4) = 2 So the answer is (c) 2.

A simple pendulum has a periodic time ' T₁ ' when it is on the… | MHT CET 2024 (16 May, Shift 1)

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Acceleration Due to Gravity

A simple pendulum has a periodic time ' $T_1$ ' when it is on the surface of earth of radius ' $R$ '. Its periodic time is ' $T_2$ ' when it is taken to a height ' $R$ ' above the earth's surface. The value of $\frac{T_2}{T_1}$ is

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Answer: (c)

At $h = R$ : $g_h = \frac{g}{4}$ .
$\frac{T_2}{T_1} = \sqrt{\frac{g}{g_h}} = \sqrt{4} = 2$
So the answer is (c) 2.

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