A simple pendulum has a time period $T_{1}$ when on the earth's surface and $T_{2}$ when taken to a height $R$ above the earth's surface, where $R$ is the radius of the earth. The value of $T_{2} / T_{1}$ is :

1

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\sqrt{2}

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4

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2

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Solution

The correct option is D $2$
If acceleration due to gravity is $g$ at the surface of earth, then at height $R$ , it's value becomes $g^{'} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + R})}^{2} = \frac{g}{4}$
$T_{1} = 2 π \sqrt{\frac{l}{g}}$ and $T_{2} = 2 π \sqrt{\frac{l}{g / 4}}$
$∴ \frac{T_{2}}{T_{1}} = 2$

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Q. A simple pendulum has a time period

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A simple pendulum has a time period $T_{1}$ when on earth's surface and $T_{2}$ when taken to a height R above the earth's surface, where R is the radius of the earth. The value of $\frac{T_{2}}{T_{1}}$ is

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A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T2/T1 is :

The correct option is D 2If acceleration due to gravity is g at the surface of earth, then at height R, it's value becomes g′=g(RR+h)2=g(RR+R)2=g4 T1=2π√lg and T2=2π√lg/4 ∴T2T1=2

A simple pendulum has a time period $T_{1}$ when on the earth's surface and $T_{2}$ when taken to a height $R$ above the earth's surface, where $R$ is the radius of the earth. The value of $T_{2} / T_{1}$ is :