Question

The density of a liquid of coefficient of cubical expansion $\gamma $ is $\rho $ at $0{}^\circ C$. When the liquid is heated to temperature $T$, the change in density will be
A.$-\dfrac{\rho \gamma T}{\left( 1+\gamma T \right)}$
B.$\dfrac{\rho \gamma T}{\left( 1+\gamma T \right)}$
C.$-\dfrac{\rho \left( 1+\gamma T \right)}{\gamma T}$
D.$\dfrac{\rho \left( 1+\gamma T \right)}{\gamma T}$

Answer 1

Hint: We can solve this question with the help of the formula for volume expansion. Now we know the relation between volume and density hence we can find out the final density of the liquid and eventually calculate the change in density using the formula of volume expansion.
Formula used:
$V={{V}_{0}}\left( 1+\gamma \Delta T \right)$

Complete answer:
Let us first write down the formula for volume expansion for the liquid:
$V={{V}_{0}}\left( 1+\gamma \Delta T \right)$
Here, $V$ is the final volume of the liquid after expansion,
${{V}_{0}}$ is the initial volume of the liquid before expansion,
$\gamma $ is the coefficient of volume expansion of the liquid,
And $\Delta T$ is the change in temperature when it is heated which will be:
$\Delta T=T-0{}^\circ $
So substituting the value in the equation we get,
$\begin{align}
  & V={{V}_{0}}\left( 1+\gamma \left( T-0{}^\circ \right) \right) \\
 & \Rightarrow V={{V}_{0}}\left( 1+\gamma T \right) \\
\end{align}$
Now, as we got the final volume of the liquid, the next step would be to find a relation between volume and density to proceed further. We know that density is inversely proportional to volume, hence we can write it down as:
$\rho \propto \dfrac{1}{V}$
Which means that:
$\rho V=\text{constant}$
Therefore, we can write down the following relation,
\[\begin{align}
  & {{\rho }_{f}}V=\rho {{V}_{0}} \\
 & \Rightarrow {{\rho }_{f}}=\dfrac{\rho {{V}_{0}}}{V} \\
 & \Rightarrow {{\rho }_{f}}=\dfrac{\rho {{V}_{0}}}{{{V}_{0}}\left( 1+\gamma T \right)} \\
 & \Rightarrow {{\rho }_{f}}=\dfrac{\rho }{\left( 1+\gamma T \right)} \\
\end{align}\]
The change in density will be the difference between final and initial density and can be written as follows:
\[\begin{align}
  & {{\rho }_{f}}-\rho =\dfrac{\rho }{{{V}_{0}}\left( 1+\gamma T \right)}-\rho \\
 & \Rightarrow {{\rho }_{f}}-\rho =\dfrac{\rho -\rho \left( 1+\gamma T \right)}{\left( 1+\gamma T \right)} \\
 & \Rightarrow {{\rho }_{f}}-\rho =\dfrac{\rho -\rho -\rho \gamma T}{\left( 1+\gamma T \right)} \\
 & \therefore {{\rho }_{f}}-\rho =\dfrac{-\rho \gamma T}{\left( 1+\gamma T \right)} \\
\end{align}\]
Therefore, the change in density is \[\dfrac{-\rho \gamma T}{\left( 1+\gamma T \right)}\],

Hence the correct option is $A$.

Note:
When a liquid is heated, it expands or contracts differently based on the volume expansion coefficient which varies from one substance to the another. And hence, the change in density of the liquid would also eventually depend on the volume expansion coefficient as we have seen above.