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Question
A pendulum clock giving correct time at a place where g = 9.800 m/s2 is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.
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Solution
Let T1 be the time period of pendulum clock at a place where acceleration due to gravity \[\left( g_1 \right)\] is 9.8 ms−2.
Let T1 = 2 s
`g_1 = 9.8"ms"^(-2)`
Let T2 be the time period at the place where the pendulum clock loses 24 seconds during 24 hours.
Acceleration due to gravity at this place is \[\left( g_2 \right)\]
\[T_2 = \frac{24 \times 3600}{\frac{\left( 24 \times 3600 - 24 \right)}{2}}\]
\[ = 2 \times \frac{3600}{3599}\]
As
\[T \propto \frac{1}{\sqrt{g}}\]
\[\therefore \frac{T_1}{T_2} = \sqrt{\left( \frac{g_2}{g_1} \right)}\]
\[\Rightarrow \frac{g_2}{g_1} = \left( \frac{T_1}{T_2} \right)^2 \]
\[ \Rightarrow g_2 = \left( 9 . 8 \right) \left( \frac{3599}{3600} \right)^2 \]
\[ = 9 . 795 \text { m/ s}^2\]
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