Question

A solid metal cube having total surface area $24m^2$ is uniformly heated. If its temperature is increased by ${10}^{\circ }C$, calculate the increase in volume of the cube.
(Given $\alpha =5.0\times {10}^{-4}$${/}^{\circ }C)$

• A. $6.0\times {10}^{4}c{m}^3$
• B. $4.8\times {10}^{5}c{m}^3$
• C. $2.4\times {10}^{6}c{m}^3$
• D. $1.2\times {10}^{5}c{m}^3$

Answer 1

The correct option is D $1.2\times {10}^{5}c{m}^3$

Surface area of cube of side $l$, $6\times l^2=24$

$\Rightarrow l=2m$

Volume of the cube, $l^3=(2{)}^3=8m^3$

Volumetric strain $=3\times$linear strain

$\therefore \frac{\Delta V}{V}=3\times \frac{\Delta l}{l}$

$\Rightarrow \Delta V=3\times \left(\alpha \Delta T\right)\times V$

$\Delta V=3\times 5\times {10}^{-4}\times 10\times \left(8\right)$

$=120\times {10}^{-3}{m}^3$

$=120\times {10}^{-3}\times {10}^{6}c{m}^3$

$=1.2\times {10}^{5}c{m}^3$