Question

Question 13: Will a clock using a pendulum supported on a long thin brass rod that is accurate at 20°C run fast or slow on a hot day (30°C)? Explain.

Short answer

The pendulum clock will run slow due to the linear expansion of the brass rod on a hot day.

Step-by-step solution

Linear expansion

Most of the solid substances linearly expand on heating. When the temperature of a solid object of length\({l_0}\)is increased by\(\Delta T\), the length of the object increases by \(\Delta l\) amount.

The change in the length of the object is given as:

\(\Delta l = \alpha {l_0}\Delta T\)

Here, \(\alpha \) is known as the coefficient of linear expansion. The coefficient of linear expansion of brass is \(19 \times {10^{ - 6}}\;{\left( {{\rm{C}}^\circ } \right)^{ - 1}}\).

Explanation for the change in time period of the pendulum clock

If the length of the brass rod at \(20^\circ {\rm{C}}\) is\({l_0}\),then the time period of the pendulum clock at this temperature is given as:

\(T = 2\pi \sqrt {\frac{{{l_0}}}{g}} \)

Here, g is the acceleration due to gravity whose value is fixed. Thus, the time period of the pendulum clock is directly proportional to the square root of the length of the brass rod.

On a hot day, when the temperature is \(30^\circ {\rm{C}}\), due to a rise in temperature, the brass rod will undergo linear expansion, and the length of the rod will increase by some amount in accordance with relation (i). Thus, according to relation (ii), the time period of the pendulum clock will also get increased, i.e., it will take much longer to complete one oscillation. This means that the clock will run slow on a hot day.