 Solution

For uniform Earth density, the shell theorem says that only mass within radius $r$ contributes to the gravitational field there. Since enclosed mass is proportional to $r^3$,At depth $h$, $g_h/g_0=1-h/R$. A simple pendulum has small-amplitude period $P=2\pi \sqrt{l/g}$, so the pendulum clock below a uniform-density surface runs at the fraction $P_0/P_h=\sqrt{1-h/R}$ of its surface-calibrated rate. After a surface-clock hour, its lag isThe mine clock is slower because gravity is weaker under the stated uniform-density assumption. The expansion $\Delta t\simeq 3600h/(2R)$ gives the same result. This is a change in the mechanical pendulum period, not a relativistic clock effect; a realistic Earth density profile would require a different interior-gravity calculation.