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

2

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1

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\sqrt{2}

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4

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Solution

The correct option is D $2$
The value of g at a height h above earth's surface
$= \frac{g}{(1 + \frac{h}{R})^{2}}$

So, at a height R,
$g^{'} = \frac{g}{4}$

We know that time period of simple pendulum $\propto \frac{1}{\sqrt{g}}$
So,
$T_{2} = 2 T_{1}$
$⟹ \frac{T_{2}}{T_{1}} = 2$

Hence, (A) is the correct answer.

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Similar questions

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 $\frac{T_{2}}{T_{1}}$ is :

Q. 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

\frac{T_{2}}{T_{1}}

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

The correct option is D 2The value of g at a height h above earth's surface=g(1+hR)2So, at a height R,g′=g4We know that time period of simple pendulum ∝1√gSo, T2=2T1⟹T2T1=2Hence, (A) is the correct answer.

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