Question

The density of a liquid of coefficient of cubical expansion $\gamma$ is $\rho$ at $0^{\circ} C$. When the liquid is heated to a temperature $T$, the change in density will be :

• A. $-\frac{\rho \gamma T}{(1+\gamma T)}$
• B. $\frac{\rho \gamma T}{(1+\gamma T)}$
• C. $-\frac{\rho(1+\gamma T)}{\gamma T}$
• D. $\frac{\rho(1+\gamma T)}{\gamma T}$

Answer 1

Answer: (A)

On expansion, volume of a given mass of a substance increases. So, density should decrease
$\rho=\frac{m}{V}$
$\Rightarrow \rho \propto \frac{1}{V}$
$\frac{\rho^{\prime}}{\rho} =\frac{V}{V^{\prime}}=\frac{V}{V+\Delta V}=\frac{V}{V+\gamma V \Delta T}$
$=\frac{1}{1+\gamma \Delta T}$
$\frac{\rho^{\prime}}{\rho} =\frac{1}{1+\gamma \Delta T}$
$\Rightarrow \rho^{\prime} =\frac{\rho}{1+\gamma \Delta T}$
Change in density,
$\rho^{\prime}-\rho=\frac{\rho}{1+\gamma \Delta T}-\rho$
$=\frac{\rho-\rho-\rho \gamma \Delta T}{1+\gamma \Delta T}$
$=\frac{-\rho \gamma \Delta T}{1+\gamma \Delta T}$
As Initial temp $=0^{\circ} C$
$\Rightarrow \Delta T =T$
$\Rightarrow \Delta \rho=\frac{-\rho \gamma T}{(1+\gamma T)}$