Question

A metal rod having a linear coefficient of expansion $2\times {{10}^{-5}}{}^\circ {{C}^{-1}}$ has a length $1\;m$ at $25{}^\circ C$ , the temperature at which it is shortened by $1\;mm$ is:
(A) $50{}^\circ C$
(B) $-50{}^\circ C$
(C) $-25{}^\circ C$
(D) $-12.5{}^\circ C$

Answer 1

Hint : The change in length due to thermal expansion is proportional to the change in temperature and the linear coefficient of expansion of the material. By the formula for linear thermal expansion, we can find the change in the temperature. As we are already given the initial temperature, we can find the final temperature.

Complete Step By Step Answer:
Let us note down the given data;
Length of the metal rod $L=1m$
Linear coefficient of expansion $\alpha =2\times {{10}^{-5}}{}^\circ {{C}^{-1}}$
The initial temperature of the rod $T=25{}^\circ C$
Change in length of the rod $\Delta L=1mm=1\times {{10}^{-3}}m$
Whenever an object is subjected to heat, it tends to change its dimensions with respect to length, area, or volume. If the length of an object changes due to the addition of heat, then the thermal expansion is known as linear thermal expansion.
The change in the length due to linear thermal expansion is mathematically calculated as,
$\Delta L=L\alpha \Delta T$
Where, $\Delta L$ = ${{L}_{2}}-{{L}_{1}}$ = Change in the length of the object
$L$ = Original length of the object
$\alpha$ = Linear coefficient of expansion
$\Delta T$ = Change in the temperature of the object due to heat
Here, we need to find the value of temperature change.
Hence, rearranging the equation to make temperature difference subject of the equation,
$\Delta T=\dfrac{\Delta L}{L\alpha }$
Substituting the given values in the equation,
$\therefore \Delta T=\dfrac{1\times {{10}^{-3}}m}{\left( 1m \right)\left( 2\times {{10}^{-5}}{}^\circ {{C}^{-1}} \right)}$
$\therefore \Delta T=0.5\times {{10}^{2}}{}^\circ C$
Shifting the decimal point,
$\therefore \Delta T=50{}^\circ C$
Now, there is a decrease in the length of the rod, which implies there will be a decrease in the temperature of the rod.
$\therefore \Delta T=-\left( {{T}_{f}}-{{T}_{i}} \right)$
$\therefore \Delta T={{T}_{i}}-{{T}_{f}}$
Substituting the value of initial temperature and the change in temperature,
$\therefore 50{}^\circ C=25{}^\circ C-{{T}_{f}}$
$\therefore {{T}_{f}}=-25{}^\circ C$
Hence, the correct answer is Option $(C)$ .

Note :
From the linear thermal expansion, we can understand that if the length of the object is to be decreased, the temperature has to be decreased and vice-versa. Hence, if the length is decreased, we must include a negative sign in the change of temperature.