Question

A clock, with a brass pendulum, keeps correct time at $20^{\circ}C$, but loses $8.212\;s$ per day, when the temperature is $30^{\circ}C$. The coefficient of linear expansion of brass is:
A. $25\times 10^{-6}\;^{\circ}C^{-1}$
B. $19\times 10^{-6}\;^{\circ}C^{-1}$
C. $20\times 10^{-6}\;^{\circ}C^{-1}$
D. $11\times 10^{-6}\;^{\circ}C^{-1}$

Answer 1

We know that the coefficient of linear expansion is the change in length of a material with respect to the original material with a change in temperature. Using this definition, determine an expression for the length of the pendulum at $30^{\circ}C$. Then substitute this expression for length, in the expression for the time period of oscillation of a simple pendulum. Using a certain binomial expression simplifies the relation to ease your calculation. Finally, substitute the given values and this should lead you to the appropriate result.

Formula used:
Time period of a simple pendulum: $T = 2\pi\sqrt{\dfrac{l}{g}}$
Linear coefficient of expansion: $\alpha = \dfrac{l^{\prime}}{l\Delta\theta}$

Complete step-by-step answer:
Let us begin by understanding what the coefficient of linear expansion is.
The linear expansion coefficient is the rate at which a material expands, and is an intrinsic property of every material.
Now, let the time period of oscillation of the simple pendulum at $\theta_1=20^{\circ}C$ be given as:
$T_{20} = 2\pi\sqrt{\dfrac{l}{g}}$, where l is the length of the pendulum and g is the acceleration due to gravity.
Now, the clock begins to lose time because of the expansion of the pendulum in length owing to the increase in temperature to $\theta_2 = 30^{\circ}C$. The new length of the pendulum can be given as:
$l_{30} = l(1+\alpha\Delta \theta)$, where $\alpha$ is the linear coefficient of expansion and $\Delta \theta = \theta_2-\theta_1$
This means that the time period of oscillation for the simple pendulum now becomes:
$T_{30} = 2\pi\sqrt{\dfrac{l(1+\alpha\Delta \theta)}{g}} = \left(2\pi\sqrt{\dfrac{l}{g}}\right).\sqrt{ 1+\alpha\Delta \theta} = T_{20} (1+\alpha\Delta \theta)^{1/2}$
Now, truncating the binomial expansion at: $(1+\alpha\Delta \theta)^{1/2} = 1 +\dfrac{\alpha\Delta\theta}{2}$
$T_{30} = T_{20}.\left(1+\dfrac{\alpha\Delta\theta}{2}\right) \Rightarrow \dfrac{T_{30}-T_{20}}{T_{20}} = \dfrac{1}{2}\alpha\Delta\theta \Rightarrow \Delta T = \dfrac{1}{2}\alpha\Delta\theta T_{20}$
From this we can get the expression for the coefficient of linear expansion:
$\Rightarrow \alpha = \dfrac{2\Delta T}{T_{20}\Delta \theta}$
From the question we have:
$T_{20} = 24\;h = 24 \times 60 \times 60 = 86400\;s$
$\Delta T = 8.212\;s$
$\Delta \theta = 30 -20 = 10^{\circ}C$
$\Rightarrow \alpha = \dfrac{2 \times 8.212}{86400 \times 10} = 1.9 \times 10^{-5} = 19 \times 10^{-6}\;^{\circ}C^{-1}$
Therefore, the correct choice would be: B. $19 \times 10^{-6}\;^{\circ}C^{-1}$

So, the correct answer is “Option B”.

Note: Note that sometimes linear expansion is referred to as the compressive strain though the material ends up expanding. This is because the linear strain is caused by the thermal stress whereas the compressive strain is caused by the rod in response to the linear strain, and both are equivalent in magnitude. This is obvious because the material exerts forces to prevent its expansion, by means of which it is in fact trying to compress its structure to hold off expansion caused by the thermal stress incident on it.