Question

A pendulum clock which keeps correct time at the surface of the earth is taken into a mine, then
A. It keeps correct time.
B. it gains time.
C. it loses time.
D. None of these.

Answer 1

Hint: Taking the pendulum into a mine means that we are going inside the surface of the earth. The acceleration due to gravity depends directly on the mass of Earth and depends inversely on the separation from the centre.

Formula used:
When we reach a depth d inside Earth, the acceleration due to gravity will be defined as:
$g' = g \left(1 - \dfrac{d}{R} \right)$;

Complete answer:
If we assume the radius of the earth to be R, then we know that acceleration due to gravity (for an object at the surface) is given as:
$g = \dfrac{GM}{R^2}$
In this case, the time period of a pendulum (on the surface of Earth) is given as:
$T = 2 \pi \sqrt{\dfrac{l}{g}}$
When we go deep inside the surface of the Earth, the acceleration due to gravity is defined by the formula:
$g' = g(1 - \dfrac{d}{R} )$ .
This clearly helps us see that going in by a depth of d, the factor of d/R (small positive) is getting subtracted from 1 leading to g' being smaller than original g (at the surface).
The new time period of the pendulum will be simply given by:
$T' = 2 \pi \sqrt{\dfrac{l}{g'}}$
As g' is smaller than g, T' will be bigger than T. This means that the pendulum will complete 1 oscillation by taking more time. Therefore, in a day, it will be able to perform a lesser number of oscillations and hence, it will lose time.

Therefore, the correct answer is option (C). it loses time.

Note:
In the formula acceleration due to gravity we see an inverse dependence on the separation distance. It might appear that as the separation distance is decreasing g must increase but that is not the case. As we go beneath the surface of the Earth, the effective mass is going to reduce too (and will not be M). Some integral calculations result in the formula that has been used here.