Question

The pressure required to stop the increase in volume of a copper block when it is heated from ${50}^{o}C$ to ${70}^{o}C$. Coefficient of linear expansion of copper is $8\times {10}^{-6}/{}^{o}C$ and bulk modules of elasticity $=3.6\times {10}^{11}N/m^2$, is

• A. $2.8\times {10}^{5}N/m^2$
• B. $1.72\times {10}^{8}N/m^2$
• C. $6.3\times {10}^{3}N/m^2$
• D. $8\times {10}^{-6}N/m^2$

Answer 1

Answer: (B)

$1.72\times {10}^{8}N/m^2$

If $V$ be the initial volume of the copper block and $\Delta V$ be the increase in volume when heated from temperature $t_1$ to $t_2$, then $\Delta V=V\gamma (t_2-t_1)$ where $\gamma =$ coefficient of volume exapansion

Thus, volume strain $=\frac{\Delta V}{V}=\gamma (t_2-t_1)$

The relation between coefficient of volume expansion and coefficient of linear expansion is given by, $\gamma =3\alpha$

So, volume strain $=\frac{\Delta V}{V}=3\alpha (t_2-t_1)$

Bulk modulus, $K=\frac{P}{\frac{\Delta V}{V}}=\frac{P}{3\alpha (t_2-t_1)}$

or $P=3K\alpha (t_2-t_1)=3(3.6\times {10}^{11})(8\times {10}^{-6})(70-50)=1.72\times {10}^{8}N/m^2$