Question

A copper bar of length L and area of cross section A is placed in a chamber at atmospheric pressure. If the chamber is evacuated, the percentage change in its volume will be (compressibility of copper is $8 \times {10^{ - 12}}{m^2}/N$ and 1 atm$ = {10^5}N/m$)
(a)\[8 \times {10^{ - 7}}\]
(b)\[8 \times {10^{ - 5}}\]
(c)\[1.25 \times {10^{ - 4}}\]
(d)\[1.25 \times {10^{ - 5}}\]

Answer 1

Hint: If a material is kept under some pressure(from all sides) it undergoes compression and if the pressure is released then the material has a tendency to come back to its original state/structure. For longitudinal stress and strain we use the Young’s Modulus to evaluate the situation, whereas in case of Volume stress we use Bulk Modulus to evaluate the situation.

Formula Used:
1. Bulk’s Modulus of elasticity: $B = \dfrac{{ - \Delta P}}{{(\dfrac{{\Delta V}}{V})}}$ ……(1)
Where,
$\Delta P$ is the change in pressure
$\Delta V$ is the change in volume
V is the initial volume
2. Compressibility is defined as inverse of B: $C = \dfrac{{ - (\dfrac{{\Delta V}}{V})}}{{\Delta P}}$ ……(2)
3. Percentage of any quantity A: \[\% (A) = \dfrac{{\Delta A}}{A} \times 100\] ……(3)

Complete step by step answer:
Given:
1. Length of copper bar: L
2. Area of cross section of copper bar: A
3. Initial pressure inside the chamber: ${P_1} = 1atm$
4. Final pressure inside the chamber: ${P_2} = 0atm$
5. Compressibility of copper: $C = 8 \times {10^{ - 12}}{m^2}/N$

To find: The percentage change in the volume of the copper bar.

Step 1:
Convert given pressure from atm units to N/m units using the conversion 1 atm$ = {10^5}N/m$:
$
  {P_1} = 1atm \times {10^5}N/m \\
  {P_1} = {10^5}N/m \\
 $
$
  {P_2} = 0atm \times {10^5}N/m \\
  {P_2} = 0N/m \\
 $
Find the change in pressure ($\Delta P$):
\[
  \Delta P = {P_2} - {P_1} \\
  \Delta P = - {10^5}N/m \\
 \]

Step 2:
Consider case 2.
Put the values in expression for compressibility eq (2):
$8 \times {10^{ - 12}}{m^2}/N = \dfrac{{ - (\dfrac{{\Delta V}}{V})}}{{ - {{10}^5}N/m}}$
Rearrange to find $\dfrac{{\Delta V}}{V}$:
$8 \times {10^{ - 7}}m \times 100 = - \dfrac{{\Delta V}}{V} \times 100$

Step 3:
Find the percentage change in volume using eq (3):
$\% (\dfrac{{\Delta V}}{V}) = - 8 \times {10^{ - 5}}m$

Final Answer
The percentage change in the volume of the copper bar: (b)\[8 \times {10^{ - 5}}\]

Note: Higher is the compressibility of material, the easier it is to compress it on the other hand higher the bulk modulus of a material, higher is the resistance of the material to undergo compression.