The time period of a simple pendulum on the surface of the earth is $4$ s. Its time period on the surface of the moon is:

4

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8

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10

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12

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Solution

The correct option is C $10$ s
Acceleration due to gravity on the surface of the moon is $\frac{1}{6}$ that of surface of the earth.

$∴ g_{m} = \frac{1}{6} g_{e}$

On earth, $T_{e} = 2 π \sqrt{\frac{L}{g_{e}}}$

On Moon, $T_{m} = 2 π \sqrt{\frac{L}{g_{m}}}$

$∴ \frac{T_{m}}{T_{e}} = \sqrt{\frac{g_{e}}{g_{m}}} = \sqrt{6}$

$T_{m} = \sqrt{6} T_{e} = \sqrt{6} \times 4 s = 10 s$

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The time period of a simple pendulum on the surface of the earth is 4s. Its time period on the surface of the moon is:

The correct option is C 10sAcceleration due to gravity on the surface of the moon is 16 that of surface of the earth.∴gm=16geOn earth, Te=2π√LgeOn Moon, Tm=2π√Lgm∴TmTe=√gegm=√6Tm=√6Te=√6×4s=10s

The time period of a simple pendulum on the surface of the earth is $4$ s. Its time period on the surface of the moon is: