A pendulum clock is taken \[1\,{\text{km}}\] inside the Earth from mean sea level. Then the pendulum clock
A. Loses \[13.5\,{\text{s}}\] per day
B. Gains \[13.5\,{\text{s}}\] per day
C. Loses \[7\,{\text{s}}\] per day
D. Gains \[7\,{\text{s}}\] per day

Hint: Use the formula for time period of the simple pendulum. Also use the formula for variation of acceleration due to gravity with the depth below the surface of the Earth. First determine the value of acceleration due to gravity and time period pendulum clock 1 km below the sea level. Then take the difference of the time period of the pendulum clock at the sea level and below the sea level.

Formulae used:
The time period \[T\] of a simple pendulum is
\[T = 2\pi \sqrt {\dfrac{L}{g}} \] …… (1)
Here, \[L\] is the length of the simple pendulum and \[g\] is acceleration due to gravity.
The expression for variation of acceleration due to gravity at depth is
\[g' = g\left( {1 - \dfrac{d}{R}} \right)\] …… (2)
Here, \[g'\] is acceleration due to gravity at depth \[d\], \[g\] is acceleration due to gravity at sea level and \[R\] is radius of the Earth.

Complete step by step answer:
We have given that the pendulum clock is taken \[1\,{\text{km}}\] inside the Earth from the mean sea level.
\[d = 1\,{\text{km}}\]
We have asked to calculate the change in time period of the pendulum clock when it is moved under the surface of the Earth.We know that the time period of the pendulum clock on the surface of the Earth is 24 hours.
\[T = 24\,{\text{h}}\]

Let us first calculate the value of acceleration due to gravity \[g'\] under the surface of the Earth.Substitute \[1\,{\text{km}}\] for \[d\] and \[6400\,{\text{km}}\] for \[R\] in equation (2).
\[g' = g\left( {1 - \dfrac{{1\,{\text{km}}}}{{6400\,{\text{km}}}}} \right)\]
\[ \Rightarrow g' = g\left( {\dfrac{{\left( {6400\,{\text{km}}} \right) - \left( {1\,{\text{km}}} \right)}}{{6400\,{\text{km}}}}} \right)\]
\[ \Rightarrow g' = \left( {\dfrac{{6399}}{{6400}}} \right)g\]
This is the expression for acceleration due to gravity under the sea level.

Let us now calculate the time period of the pendulum clock under the surface of the Earth.Substitute \[T'\] for \[T\] and \[g'\] for \[g\] in equation (1).
Substitute \[\left( {\dfrac{{6399}}{{6400}}} \right)g\] for \[g'\] in the above equation.
\[T' = 2\pi \sqrt {\dfrac{L}{{\left( {\dfrac{{6399}}{{6400}}} \right)g}}} \]
\[ \Rightarrow T' = 2\pi \sqrt {\dfrac{{6400L}}{{6399g}}} \]
Substitute \[2\pi \sqrt {\dfrac{L}{g}} \] for \[T\] in the above equation.
\[ \Rightarrow T' = T\sqrt {\dfrac{{6400}}{{6399}}} \]

Let us now calculate the difference between the time periods of the pendulum clock at the sea level and below the sea level.
\[ \Rightarrow T' - T = T\sqrt {\dfrac{{6400}}{{6399}}} - T\]
\[ \Rightarrow T' - T = T\left( {\sqrt {\dfrac{{6400}}{{6399}}} - 1} \right)\]
Substitute \[24\,{\text{h}}\] for \[T\] in the above equation.
\[ \Rightarrow T' - T = \left( {24\,{\text{h}}} \right)\left( {\sqrt {\dfrac{{6400}}{{6399}}} - 1} \right)\]
\[ \Rightarrow T' - T = \left[ {\left( {24\,{\text{h}}} \right)\left( {\dfrac{{60\,{\text{min}}}}{{1\,{\text{h}}}}} \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right)} \right]\left( {\sqrt {\dfrac{{6400}}{{6399}}} - 1} \right)\]
\[ \Rightarrow T' - T = - 6.75\,{\text{s}}\]
\[ \therefore T' - T \approx - 7\,{\text{s}}\]
Therefore, the pendulum clock when shifted under the sea level, the time period of the pendulum clock loses \[7\,{\text{s}}\] per day.

Hence, the correct option is C.

Note:The students may think that how we can determine whether the pendulum clock loses the time or gains the time. From the final answer, we can see that the difference between the time periods of the pendulum clock below the sea level and at the sea level is negative which shows that the time period of the pendulum clock below the sea level is less than at the sea level. Hence, the pendulum clock loses time.