A pendulum clock that gives correct time at a location where g =...

The time period of a simple pendulum is given by $T = 2\pi \sqrt{\dfrac{l}{g}}$.

Taking the natural logarithm and differentiating, we get the fractional change in the time period:

$\dfrac{\Delta T}{T} = -\dfrac{1}{2} \dfrac{\Delta g}{g}$

The time lost by the clock in a total time $t$ is given by $\Delta t = t \dfrac{\Delta T}{T}$.

Given that the clock loses $24$ seconds in $24$ hours ($86400$ seconds), the fractional change in the time period is:

$\dfrac{\Delta T}{T} = \dfrac{24}{86400} = \dfrac{1}{3600}$

Substituting this into the fractional change equation:

$\dfrac{1}{3600} = -\dfrac{1}{2} \dfrac{\Delta g}{g}$

$\Delta g = -g \times \dfrac{2}{3600} = -9.800 \times \dfrac{1}{1800} \approx -0.00544 \text{ m/s}^2$

The acceleration due to gravity at the new location is:

$g' = g + \Delta g = 9.800 - 0.00544 = 9.79456 \text{ m/s}^2$

Rounding to three decimal places, we get $g' \approx 9.795 \text{ m/s}^2$.