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  • The time Period of a simple Pendulum on the surface of the Earth is <img alt="10 \mathrm{sec}" src="https://entrancecorner.oncodecogs.com/gif.latex?10%20%5Cmathrm%7Bsec

The time Period of a simple Pendulum on the surface of the Earth is 10 \mathrm{sec}. its time period on the surface of the Moon

Option: 1

23.5 Sec


Option: 2

25 Sec


Option: 3

24.4 Sec


Option: 4

23 Sec


Answers (1)

best_answer

Earth

ge=g
T=2 \pi \sqrt{\frac{l}{g}} \\
T_e=2 \pi \sqrt{\frac{l}{g e}} \\
T_e=10 \mathrm{sec}
Then 10=2 \pi \sqrt{\frac{l}{g e}}               --------eq(1)

Moon

T_m=2 \pi \sqrt{\frac{l}{g m}} \\                         {\left[\because g_m=\frac{ge}{6}\right]} \\
T_m=2 \pi \sqrt{\frac{l}{g e / 6}} \\
T_m=2 \pi \sqrt{\frac{6 l}{g e}} \\
T_m=2 \pi \sqrt{\frac{l}{g e}} \times \sqrt{6} -----------eq(2)
2 \pi \sqrt{\frac{l}{g e}}=10 Put the value of eq(2)

T_m=10 \times \sqrt{6} \\

 T_m=10 \times 2.44 \\

T_m=24.4 \mathrm{sec} \\

Posted by

Suraj Bhandari

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