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**Hint:**To solve these types of questions we need to have a knowledge about the relationship between increase in time period of a pendulum clock with a change in temperature. If the temperature decreases, the length of the pendulum decreases which affects its time period.

**Complete step-by-step solution:**

To solve his question, we need to know about the formula that relates fractional increment in time period of a pendulum with the change in temperature. The formula is as follows:

$\dfrac{\Delta t}{t}=\dfrac{1}{2}\alpha \Delta T$

Here $\dfrac{\Delta t}{t}$ is the fractional increase in time period of the pendulum clock where $\Delta t$ is the difference between initial and final time period and $t$ is the total time period,

$\alpha $ is the coefficient of linear expansion of copper rod from which the pendulum clock is made,

$\Delta T$ is the change in temperature which led to the change in time period of the pendulum clock.

We are given that the pendulum clock having a copper rod keeps correct time at \[20{}^\circ C\] and gains $15$ seconds per day if cooled to $0{}^\circ C$.

On substituting the values in the equation, we get the following:

$\begin{align}

& \dfrac{15}{24\times 60\times 60}=\dfrac{1}{2}\alpha \left( 20-0 \right) \\

& \Rightarrow \dfrac{15}{86400}=10\alpha \\

& \Rightarrow \alpha =\dfrac{15}{86400\times 10} \\

& \therefore \alpha =1.7\times {{10}^{-5}}\text{ }{}^\circ {{\text{C}}^{-1}} \\

\end{align}$

Thus, the coefficient of linear expansion of copper is $1.7\times {{10}^{-5}}\text{ }{}^\circ {{\text{C}}^{-1}}$.

**Note:**If the temperature is increased, the length of a pendulum clock would increase which will increase the time period of the pendulum clock because time period of a pendulum clock is directly proportional to the square root of the length of the pendulum. While if the temperature is decreased, the length of the pendulum will also decrease which will result in a decrease in time period.

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