Question

A thin walled cylindrical metal vessel of linear coefficient of expansion \[{10^{ - 3}}\,^\circ {{\text{C}}^{ - 1}}\] contains benzene of volume expansion coefficient \[{10^{ - 3}}\,^\circ {{\text{C}}^{ - 1}}\]. If the vessel and its contents are now heated by \[10\,^\circ {\text{C}}\], the pressure due to the liquid at the bottom:
A. Increases by \[2\% \]
B. Decreases by \[1\% \]
C. Decreases by \[2\% \]
D. Remains unchanged

Answer 1

Hint: First of all, we will find the coefficient of volume expansion of the vessel which is three times the linear expansion. After that we will find the change in volume of the liquid. By manipulating the expression, we will find the difference in pressure.

Formula used:
The formula which gives the change in volume is as follows:
\[\Delta V = V \times \left( {{\gamma _{{\text{vessel}}}} - {\gamma _{{\text{liquid}}}}} \right) \times \Delta T\] …… (1)
Where,
\[\Delta V\] indicates the change in volume.
\[V\] indicates the initial volume.
\[{\gamma _{{\text{vessel}}}}\] indicates the volume expansion of the vessel.
\[{\gamma _{{\text{liquid}}}}\] indicates the volume expansion of the liquid.
\[\Delta T\] indicates the change in temperature.

Complete step by step answer:
In the given question, we are supplied with the following data:
There is a thin walled cylindrical metal vessel, whose linear coefficient of expansion is \[{10^{ - 3}}\,^\circ {{\text{C}}^{ - 1}}\].
It contains a liquid called benzene whose volume expansion is \[{10^{ - 3}}\,^\circ {{\text{C}}^{ - 1}}\].
After that both the vessel and its contents are heated by \[10\,^\circ {\text{C}}\].
We are asked to find the difference in pressure due to the liquid at the bottom.

To begin with, we will first find the volume expansion of the vessel.
${\gamma _{{\text{vessel}}}} = 3 \times {\alpha _{{\text{vessel}}}} \\
\Rightarrow {\gamma _{{\text{vessel}}}} = 3 \times {10^{ - 3}} \\$
We know that when the liquid is heated, the level of the liquid contained inside falls. So, it will result in the decrease of pressure.

Now, we will use equation (1) to find the change in the volume of the liquid.
$\Delta V = V \times \left( {{\gamma _{{\text{vessel}}}} - {\gamma _{{\text{liquid}}}}} \right) \times \Delta T \\
\Rightarrow \Delta V = V \times \left( {3 \times {{10}^{ - 3}} - {{10}^{ - 3}}} \right) \times \left( {10 - 0} \right) \\
\Rightarrow \Delta V = V \times 3 \times {10^{ - 3}} \times 10 \\
\Rightarrow \dfrac{{\Delta V}}{V} = 0.02 \\$

Now, we can modify the following fraction as follows:
$\dfrac{{\Delta V}}{V} = 0.02 \\
\Rightarrow \dfrac{{A \times \Delta h}}{{A \times h}} = 0.02 \\
\Rightarrow \dfrac{{\Delta h}}{h} = 0.02 \\
\Rightarrow \dfrac{{\rho g\Delta h}}{{\rho gh}} = 0.02 \\$
\[ \Rightarrow \dfrac{{\Delta P}}{P} = 0.02\]
Where,
\[\Delta P\] is the change in pressure.
\[P\] is the original pressure.

Now, we will convert the fractional value as obtained to its corresponding percentage.
Percentage of decrease in the pressure is:
$= 0.02 \times 100\% \\
= 2\% \\$
Hence, the decrease in pressure due to the liquid at the bottom is \[2\% \].

So, the correct answer is “Option C”.

Note:
It is important to remember that the coefficient of volume expansion is three times the coefficient of linear expansion in case of a solid.
Most of the students seem to have confusion regarding this in this part. Most of the students think that it is one-third, however this is not correct. The correct relation is three times.