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Question: A pendulum clock having copper rod keeps correct time at \[20{}^\circ C\]. It gains \(15\) seconds p...

A pendulum clock having copper rod keeps correct time at 20C20{}^\circ C. It gains 1515 seconds per day if cooled to 0C0{}^\circ C. Calculate the coefficient of linear expansion of copper.

Explanation

Solution

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:
Δtt=12αΔT\dfrac{\Delta t}{t}=\dfrac{1}{2}\alpha \Delta T
Here Δtt\dfrac{\Delta t}{t} is the fractional increase in time period of the pendulum clock where Δt\Delta t is the difference between initial and final time period and tt is the total time period,
α\alpha is the coefficient of linear expansion of copper rod from which the pendulum clock is made,
ΔT\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 20C20{}^\circ C and gains 1515 seconds per day if cooled to 0C0{}^\circ C.
On substituting the values in the equation, we get the following:
1524×60×60=12α(200) 1586400=10α α=1586400×10 α=1.7×105 C1 \begin{aligned} & \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{aligned}
Thus, the coefficient of linear expansion of copper is 1.7×105 C11.7\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.