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

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Solution

The correct option is C $1 / 2$
We know that time period of pendulum , T= $2 π \sqrt{\frac{l}{g}}$

$T \propto \sqrt{\frac{1}{g}}$

$g \propto (\frac{1}{1 + \frac{h}{R}})^{2}$

So, $T \propto 1 + \frac{h}{R}$

$\frac{T_{1}}{T_{2}} = \frac{1 + \frac{h_{1}}{R}}{1 + \frac{h_{2}}{R}}$
Putting $h_{1}$ = 0, $h_{2}$ = R
We get,
$\frac{T_{1}}{T_{2}} = \frac{1}{2}$
$S o, o p t i o n A i s c o r r e c t$

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A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to height R above the earth's surface, where R is the radius of the earth. The value of T1T2 is ______________.

The correct option is C 1/2We know that time period of pendulum , T= 2π√lg T∝√1g g∝(11+hR)2So, T∝1+hR T1T2=1+h1R1+h2R Putting h1 = 0, h2 = RWe get, T1T2=12So,optionAiscorrect

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