Question

At a temperature $T^{\circ }C$, a liquid is completely filled in a spherical shell of copper. If $\Delta T$ is the increase in temperature of the liquid and the shell, then the outward pressure dP on the shell due to the increase in temperature is given by: (Given, $K=$ bulk modulus of the liquids, $\gamma =$ coefficient of volume expansion, $\alpha =$ coefficient of linear expansion of the material of the shell )

• A. $\frac{K}{2}(\gamma -3\alpha )\Delta T$
• B. $K(3\alpha -\gamma )\Delta T$
• C. $3\alpha (K-\gamma )\Delta T$
• D. $\gamma (3\alpha -K)\Delta T$

Answer 1

The correct option is A $\frac{K}{2}(\gamma -3\alpha )\Delta T$

We have,
$\Delta V_1=V\times 3\alpha \Delta t$
Also,
$\Delta V_2=V\times \gamma \Delta t$
The developed stress increases the volume of vessel and decreases the volume of liquid.
Let the change in volume due to stress = $\Delta V$
The final volume of both vessel and liquid is the same
$\therefore$ The final volume of vessel $\Rightarrow V_1=V+\Delta V_1+\Delta V$
And the final volume of liquid $\Rightarrow V_2=V+\Delta V_2-\Delta V$
But $V_1=V_2$
$V+\Delta V_1+\Delta V=V+\Delta V_2-\Delta V$
$\therefore 2\Delta V=\Delta V_2-\Delta V_1$
According to Hooke's law,
$K=\frac{dp}{\frac{\Delta V}{V}}=\frac{Vdp}{\Delta V}$
$\therefore dp=K\frac{V\Delta V}{V}=\frac{K\frac{V\Delta T}{2}(\gamma -3\alpha )}{V}$
$\therefore dp=\frac{K}{2}(\gamma -3\alpha )\Delta T$