Q36 / 214 · NTA Abhyas (NEET) 2020 · Physics · Oscillations

A simple pendulum is taken to $64 km$ above the earth's surface. It's time period will be

Correct answer
A
Increase by $1 %$
B
Decrease by $1 %$
C
Increase by $2 %$
D
Decrease by $2 %$

Solution

g_{e f f}

at height of h

g_{e f f} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}} = \frac{g R_{e}^{2}}{{(R_{e} + h)}^{2}}

\Rightarrow T = 2 T_{1} \sqrt{\frac{l}{g}} = 2 T_{1} \sqrt{\frac{l {(R_{e} + h)}^{2}}{R_{e}^{2}}}

\Rightarrow \frac{T_{2} α}{T_{1}} \frac{R_{e} + h}{R_{e}} = \frac{6400 + 64}{6400}

\Rightarrow \frac{T_{2}}{T_{1}} = (1 + \frac{1}{100})

∴

Time period will increase by 1%.

Hence, the correct answer is option: 'A'.

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