Ncert Exercises & Solutions

14.15 The acceleration due to gravity on the surface of the moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of the moon if its time period on the surface of the earth is 3.5 s ? (g on the surface of the earth is 9.8 m s–2)

Given:

Acceleration due to gravity on the surface of moon, g' = 1.7 {ms}^{- 2}

Acceleration due to gravity on the surface of earth, g = 9.8 {ms}^{- 2}

Time period of a simple pendulum on earth, T = 3.5 s

T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \sqrt{\frac{1}{g}}

\frac{T_{1}}{T_{2}} = \sqrt{\frac{g_{2}}{g_{1}}} = \sqrt{\frac{g'}{g}}

T_{2} = T_{1} \sqrt{\frac{g}{g'}} = 3.5 \sqrt{\frac{9.8}{1.7}} = 8.4 \sec

Hence, the time period of the simple pendulum on the surface of the moon is 8.4 sec.

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