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.

Answer 1

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 T_{1 =} 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 

\[T \propto \frac{1}{\sqrt{g}}\]

\[\therefore \frac{T_1}{T_2} = \sqrt{\left( \frac{g_2}{g_1} \right)}\]

\[\Rightarrow   \frac{g_2}{g_1} =  \left( \frac{T_1}{T_2} \right)^2 \] 

\[ \Rightarrow  g_2  = \left( 9 . 8 \right)   \left( \frac{3599}{3600} \right)^2 \] 

\[               = 9 . 795  \text { m/ s}^2\]