Question

# A pendulum clock which keeps correct time at the bottom of mountain losses $$30\ sec/day$$ when it taken to the top of mountain. If the height of the mountain is $$1.1\mu\ km$$ find the value of $$\mu (R_{e} = 6400\ km)$$.

Solution

## $${ T }_{ e }=2\pi \sqrt { \dfrac { l }{ g } }$$As we know$$g={ g }_{ 0 }\left( 1-\dfrac { 2h }{ R } \right)$$$${ T }_{ mountain }=2\pi \sqrt { \dfrac { l }{ { g }_{ 0 }\left( 1-\dfrac { 2h }{ 6400 } \right) } }$$Consider a simple pendulum which is having time period of $$2$$ secNow,    $$2=2\pi \sqrt { \dfrac { l }{ g } } \quad \rightarrow \left( I \right)$$            $$2+\dfrac { 23 }{ 86400 } =2\pi \sqrt { \dfrac { l }{ g\left( 1-\dfrac { 2h }{ 6400 } \right) } } \quad \rightarrow \left( II \right)$$On dividing both equations, we get$$\dfrac { 2 }{ 2+\dfrac { 30 }{ 86400 } } =\sqrt { 1-\dfrac { 2h }{ 6400 } }$$$$0.9=\sqrt { 1-\dfrac { 2h }{ 6400 } }$$$$0.81=1-\dfrac { 2h }{ 6400 }$$$$2h=\left( 6400-5184 \right)$$$$h=608$$ Km.Physics

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