A pendulum clock giving correct time at a place where $g = 9.800 m s^{- 2}$ 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

For the pendulum,

$\frac{T_{1}}{T_{2}} = \sqrt{(\frac{g_{2}}{g_{1}})}$

Give that, $T_{2} = 2 s e c, g_{1} = 9.8 m / s^{2}$

$T_{1} = \frac{24 \times 3600}{\frac{(24 \times 3600 - 24)}{2}}$

$= 2 \times \frac{3600}{3599}$

Now, $\frac{g_{2}}{g_{1}} = {(\frac{T_{1}}{T_{2}})}^{2}$

$∴ g_{2} = (9.8) {(\frac{3599}{3600})}^{2}$

$= 0.795 m / s^{2}$

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A pendulum clock giving correct time at a place where g=9.800 m s−2 is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.

For the pendulum, T1T2=√(g2g1) Give that, T2=2 sec, g1=9.8 m/s2 T1=24×3600(24×3600−24)2 =2×36003599 Now, g2g1=(T1T2)2 ∴ g2=(9.8) (35993600)2 =0.795 m/s2

A pendulum clock giving correct time at a place where $g = 9.800 m s^{- 2}$ is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.