Time period of a simple pendulum is T₁ when on the earth's surface and T₂ when taken to a height '2R' above the earth's surface, where 'R' is the radius of the earth. The ratio T₁: T₂ is (a) 1: 2 (b) 1: 3 (c) 1: 4 (d) 1: 5

Answer: (b). gh = (g)/((1 + h/R)²); for h = 2R, gh = g/9. T = 2π√(L/g) ⇒ T ∝ 1/√(g). (T₂)/(T₁) = √(g/g/9) = 3 So the answer is (b) T₁: T₂ = 1: 3.

Time period of a simple pendulum is T₁ when on the earth's surface… | MHT CET 2026 (18 Apr, Shift 1)

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Acceleration Due to Gravity

Time period of a simple pendulum is $T_1$ when on the earth's surface and $T_2$ when taken to a height '2R' above the earth's surface, where 'R' is the radius of the earth. The ratio $T_1 : T_2$ is

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Answer: (b)

$g_h = \frac{g}{(1 + h/R)^2}$ ; for $h = 2R$ , $g_h = \frac{g}{9}$ .
$T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}}$ .
$\frac{T_2}{T_1} = \sqrt{\frac{g}{g/9}} = 3$
So the answer is (b) $T_1 : T_2 = 1 : 3$ .

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