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Taking the earth’s radius as $$6400 Km$$ and assuming that the value of $$g$$ inside the earth is proportional to the distance from the earth’s centre, at what depth below the earth’s surface would a pendulum which beats seconds at the earth’s surface lose $$5$$ min in a day?

Discussion:

Let us assume $$g$$ is the acceleration due to gravity at depth $$d$$ below the surface of the earth and $$g$$ is the acceleration due to gravity on the surface.

Let the corresponding time periods be $$T_0$$ and $$T$$.
$$g=g_0(1-\frac{d}{R})$$
where $$g$$ and $$g_0$$ are the acceleration due to gravity at depth respectively, and $$R$$ is the radius of the earth
$$T=T_0\sqrt{\frac{g_0}{g}}=T_0(1-d/R)^{-1/2}$$ $$=T_0(1+\frac{d}{2R})$$
Time registered for the whole body will be proportional to the time period. Thus,
$$\frac{T}{T_0}=\frac{t}{t_0}=1+d/2R$$
$$= \frac{86400}{86400-300}=1+\frac{d}{2R}$$
Substituting
$$R=6400Km$$, we find $$d=44.6Km$$