Question

A pendulum is taken 1 km inside the earth from mean sea level. Then the pendulum
A. Loses 13.5 s per day
B. Gains 13.5 s per day
C. Loses 7 s per day
D. Gains 7 s per day

Answer 1

Use the formula for acceleration due to gravity at depth d below the surface of earth. Subtract the expression for period of pendulum inside 1 km inside the earth and period at the mean sea level to obtain the gain in the period.

Formula used:
${g_d} = g\left( {1 - \dfrac{d}{R}} \right)$
Here, g is the acceleration due to gravity at the surface, d is the depth from the mean sea level and R is the radius of earth.
$T = 2\pi \sqrt {\dfrac{l}{g}}$
Here, l is the length of a simple pendulum.

Complete step by step answer:
We know that the acceleration due to gravity decreases as we move towards the centre of earth. The variation of acceleration due to gravity with depth is given as,
${g_d} = g\left( {1 - \dfrac{d}{R}} \right)$ …… (1)
Here, g is the acceleration due to gravity at the surface, d is the depth from the mean sea level and R is the radius of earth.
Substitute $9.8\,m/{s^2}$ for g, 1 km for d and $6.4 \times {10^6}\,m$ for R in the above equation.
${g_d} = \left( {9.8\,m/{s^2}} \right)\left( {1 - \dfrac{{1\,km}}{{6400\,km}}} \right)$
$\Rightarrow {g_d} = \left( {9.8\,m/{s^2}} \right)\left( {0.9998} \right)$
$\Rightarrow {g_d} = 9.798\,m/{s^2}$
This is the acceleration due to gravity inside 1 km.
The period of simple pendulum at mean sea level is given by the expression,
$T = 2\pi \sqrt {\dfrac{l}{g}}$ …… (1)
Here, l is the length of a simple pendulum.
The period of simple pendulum at mean sea level is 1 second. Therefore, $T = 1\,s$.
Suppose the period of earth 1 km inside the earth from mean sea level is $T'$. Therefore,
$T' = 2\pi \sqrt {\dfrac{l}{{{g_d}}}}$
Use equation (1) to rewrite the above equation as follows,
$T' = 2\pi \sqrt {\dfrac{l}{{g\left( {1 - \dfrac{d}{R}} \right)}}}$
$\Rightarrow T' = 2\pi \sqrt {\dfrac{l}{g}} \left( {\sqrt {\dfrac{R}{{R - d}}} } \right)$ …… (2)
Subtract equation (1) from equation (2).
$T' - T = 2\pi \sqrt {\dfrac{l}{g}} \left( {\sqrt {\dfrac{R}{{R - d}}} } \right) - 2\pi \sqrt {\dfrac{l}{g}}$
$\Rightarrow T' - T = 2\pi \sqrt {\dfrac{l}{g}} \left( {\sqrt {\dfrac{R}{{R - d}}} - 1} \right)$
Substitute 6400 km for R and 1 km for d in the above equation.
$T' - T = \left( {1\,s} \right)\left( {\sqrt {\dfrac{{6400\,km}}{{6400\,km - 1\,km}}} - 1} \right)$
Since the above quantity is positive, the pendulum gains period as we move inside the earth.
This is the gain in period in 1 second. The gain in period of the pendulum in 24 hours is,
$T' - T = \left( {7.8134 \times {{10}^{ - 5}}\,s} \right) \times 24\,h \times 60\,\min \times 60\,\operatorname{s}$
$\Rightarrow T' - T = 6.75\,s$

So, the correct answer is “Option D”.

Note:
As we move inside the earth, the acceleration due to gravity decreases, and at the centre, it is zero. Therefore, the pendulum will feel weightless and the period of the pendulum will be infinite.