The acceleration due to gravity on the surface of the moon is $1.7 m s^{- 2}$ . The time period of a simple pendulum on the moon if its time period on the earth is $3.5$ s is (Given, $g = 9.8 m s^{- 2})$

2.2

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4.4

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8.4

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16.8

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Solution

The correct option is B $8.4$ s
For moon, $g_{m} = 1.7 m s^{- 2}$
For earth, $g_{e} = 9.8 m s^{- 2}, T_{e} = 3.5 s$

But, $T_{m} = 2 \sqrt{\frac{l}{g_{m}}}$ and $T_{e} = π \sqrt{\frac{l}{g_{e}}}$

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

or $T_{m} = \sqrt{\frac{g_{e}}{g_{m}}} \times T_{e} = \sqrt{\frac{9.8}{1.7}} \times 3.5 = 8.4 s$

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The acceleration due to gravity on the surface of the moon is 1.7ms−2. The time period of a simple pendulum on the moon if its time period on the earth is 3.5s is (Given, g=9.8ms−2)

The correct option is B 8.4sFor moon, gm=1.7ms−2For earth, ge=9.8ms−2,Te=3.5sBut, Tm=2√lgm and Te=π√lge∴TmTe=√gegmor Tm=√gegm×Te=√9.81.7×3.5=8.4s

The acceleration due to gravity on the surface of the moon is $1.7 m s^{- 2}$ . The time period of a simple pendulum on the moon if its time period on the earth is $3.5$ s is (Given, $g = 9.8 m s^{- 2})$