Question

Coefficient of cubical expansion of a solid is $0.000027{}^{\circ }{{C}^{-1}}$. If the temperature is measured on Fahrenheit scale, numerical value of coefficient of linear expansion of solid is :
A. $0.000009{}^{\circ }{{F}^{-1}}$
B. $0.000005{}^{\circ }{{F}^{-1}}$
C. $0.000015{}^{\circ }{{F}^{-1}}$
D. $0.000018{}^{\circ }{{F}^{-1}}$

Answer 1

Hint: Coefficient of volume/linear expansion is defined as the fraction by which the volume/length of solid changes for a unit change in its temperature. The coefficient of volume expansion is three times the coefficient of linear expansion for a given solid. Use the relation between the Celsius and Fahrenheit scale.

Formula used:
$\dfrac{\Delta V}{V}=\gamma \Delta T$
$\alpha =\dfrac{\gamma }{3}$
Here, $\gamma $ is a proportionality constant called coefficient of volume expansion and $\alpha $ is coefficient of linear expansion.
$F=\dfrac{9C}{5}+32$
where $F$ represents temperature in ${}^{\circ }F$ and $C$ represents temperature in ${}^{\circ }C$.

Complete step by step solution:
It is found that ratio of the volume expansion (contraction) of a solid to its original volume is directly proportional to the change in its temperature.
Therefore, we can write that $\dfrac{\Delta V}{V}\propto \Delta T$.
$\dfrac{\Delta V}{V}=\gamma \Delta T$ …. (i)
Here, $\gamma $ is a proportionality constant called coefficient of volume expansion.
From (i), we can write that
$\gamma =\dfrac{\Delta V}{V}.\dfrac{1}{\Delta T}$.
With this, the coefficient of volume expansion is defined as the fraction by which the volume of solid changes for a unit change in its temperature. Similar, we have some known coefficient of linear expansion ($\alpha $). Coefficient of linear is the fraction by which the length of the solid changes for a change in temperature of one unit.And it is found that, $\alpha =\dfrac{\gamma }{3}$.
In the question it is given that $\lambda =0.000027{}^{\circ }{{C}^{-1}}$.
$\alpha =\dfrac{0.000027{}^{\circ }{{C}^{-1}}}{3}\\
\Rightarrow \alpha = 0.000009{}^{\circ }{{C}^{-1}}$.
This means that if we increase the temperature by $1{}^{\circ }C$, then the length of the solid will increase by a fraction of $0.000009$. The relation between the Fahrenheit scale and the Celsius scale is given as $F=\dfrac{9C}{5}+32$ ….. (ii).
where F represents temperature in ${}^{\circ }F$ and C represents temperature in ${}^{\circ }C$.
Therefore, if we change the temperature by $1{}^{\circ }C$, then in Fahrenheit scale it is equal to a change in temperature of $\dfrac{9}{5}{}^{\circ }F$. Hence, the linear expansion of the solid can be written as
$\alpha =\dfrac{0.000009}{\dfrac{9}{5}{}^{\circ }F}\\
\therefore \alpha =0.000005{}^{\circ }{{F}^{-1}}$.

This means that the correct option is B.

Note: By looking at the equation (ii), students may think that the change in
temperature by $1{}^{\circ }C$ is equal to $\left( \dfrac{9}{5}+32 \right){}^{\circ }F$.However, the equation (2) gives us the absolute value of the temperature in the degree Fahrenheit for a given value of the temperature in degree Celsius. While calculating the change in temperature, the constant term (i.e. 32) will not be counted.