A pendulum clock keeps the correct time at 20 degrees Celsius. Correction to be made during summer per day where the average temperature is 40 degrees Celsius, ( \[\alpha = {10^{ - 5}}{/ ^\circ}C\] ) will be:
A) 5.64 sec
B) 6.64 sec
C) 7.64 sec
D) 8.64 sec

Hint: It is important to understand how the temperature change will change the time. Temperature affects the length of the pendulum. And the time-period of a pendulum is directly proportional to the length of the pendulum. First, calculate the change in the time period of the pendulum. We know that the time taken by the clock pendulum from one extreme to another is one second, and we also know that there are $24 \times 60 \times 60\sec $ in a day. Calculate the change in time after calculating the change in the time period.

Complete step by step solution:
Step 1: Suppose the length of the pendulum is ${l_{20}}$ at 20 degrees Celsius. Express the formula for the linear expansion of the pendulum in terms of linear expansion coefficient.
$\therefore {l_{40}} = {l_{20}}(1 + \alpha \Delta \theta )$ , where ${l_{40}}$ is the length of the pendulum at 40 degrees Celsius, $\alpha $ is the thermal linear expansion coefficient, and $\Delta \theta $ is change in temperature.
Step 2: Write the expressions for the time period for both the lengths of the pendulum.
The time period when the temperature was 20 degrees Celsius: ${T_{20}} = 2\pi \sqrt {\dfrac{{{l_{20}}}}{g}} $
The time period when the temperature was 40 degrees Celsius: ${T_{40}} = 2\pi \sqrt {\dfrac{{{l_{20}}(1 + \alpha \Delta \theta )}}{g}} $
Step 3: Now divide ${T_{40}}$ by ${T_{20}}$
$\therefore \dfrac{{{T_{40}}}}{{{T_{20}}}} = \dfrac{{2\pi \sqrt {\dfrac{{{l_{20}}(1 + \alpha \Delta \theta )}}{g}} }}{{2\pi \sqrt {\dfrac{{{l_{20}}}}{g}} }}$
$ \Rightarrow \dfrac{{{T_{40}}}}{{{T_{20}}}} = \sqrt {(1 + \alpha \Delta \theta )} $
\[ \Rightarrow \dfrac{{{T_{40}}}}{{{T_{20}}}} = {(1 + \alpha \Delta \theta)^{1/2}}\]……equation (1).
Step 4: use binomial theorem on the right-hand side
\[\therefore {(1 + \alpha \Delta \theta )^{1/2}} = 1 + \dfrac{1}{2}\alpha \Delta \theta + ....\]
We will leave a higher power term because $\alpha $ is a very small quantity then the higher power of $\alpha $ will be negligible. Therefore,
\[\therefore {(1 + \alpha \Delta \theta )^{1/2}} = 1 + \dfrac{1}{2}\alpha \Delta \theta \]
Step 5: Now put the above expression in equation 1.
\[\therefore \dfrac{{{T_{40}}}}{{{T_{20}}}} = 1 + \dfrac{1}{2}\alpha \Delta \theta \]
\[ \Rightarrow {T_{40}} = {T_{20}}\left( {1 + \dfrac{1}{2}\alpha \Delta \theta } \right)\]
\[ \Rightarrow {T_{40}} = {T_{20}} + {T_{20}}\dfrac{1}{2}\alpha \Delta \theta \]
$ \Rightarrow \dfrac{{{T_{40}} - {T_{20}}}}{{{T_{20}}}} = \dfrac{1}{2}\alpha \Delta \theta $
The above expression shows the change in time in one second. We are asked for a change in time in a day therefore we multiply the right-hand side with $24 \times 60 \times 60\sec $ . Let us say the change in time in a day is $\Delta T$ then,
$\therefore \dfrac{{{T_{40}} - {T_{20}}}}{{{T_{20}}}} = \Delta T = \dfrac{1}{2}\alpha \Delta \theta .T$
Step 6: substitute the values in the above expression
\[\therefore \Delta T = \dfrac{1}{2} \times {10^{ - 5}} \times 20 \times 24 \times 60 \times 60\]
\[ \Rightarrow \Delta T = 86400 \times {10^{ - 4}}\]
\[ \Rightarrow \Delta T = 8.64sec\]

Hence the correct option is option D.

Note: Don’t substitute the time of a day in hours or minutes because we are using the SI unit system here. Always divide complex terms with the shorter term to keep calculation easier. In the binomial theorem, we neglected the higher power terms because on adding them it will change negligibly.