Question

Mercury having volume 0.1 m${}^3$ at 10 ${}^{\circ }$C is heated to 30 ${}^{\circ }$C. Calculate the increase in volume of the mercury. Given, coefficient of cubic expansion $=1.8\times {10}^{-4}$ ${}^{\circ }$C${}^{–1}$.

• A. $0.36\times {10}^4$ m${}^3$
• B. $0.36\times {10}^3$ m${}^3$
• C. $0.36\times {10}^{-4}$ m${}^3$
• D. $0.36\times {10}^{-3}$ m${}^3$

Answer 1

The correct option is D $0.36\times {10}^{-3}$ m${}^3$

Initial volume of mercury $V_0=0.1m^3$​
Rise in temperature, $\Delta T$ = final temperature - initial temperature = 30 - 10 = 20 degree celsius
Coefficient of cubic expansion, $\gamma =1.8\times {10}^{-4}$ per degree celsius. We need to calculate the change in volume of mercury, $\Delta V$.
Using,
$\gamma =\frac{\Delta V}{V_0\times \Delta T}$ $\Delta V=\gamma \times V_0\times \Delta T=1.8\times {10}^{-4}\times 0.1\times 20$
$=0.36\times {10}^{-3}$ m${}^3$
Therefore increase in volume of the mercury $=0.36\times {10}^{-3}$ m${}^3$.