Question

The volume of oil contained in a hydraulic press is $0.2{m^3}$ . The compressibility of oil is $20 \times {10^{ - 6}}$ per atmosphere. The decrease in volume of the oil when subjected to 200 atmospheres is (1 atmosphere $ = 1.02 \times {10^5}N/{m^2}$ ).
(A) $4 \times {10^{ - 4}}{m^3}$
(B) $8 \times {10^{ - 4}}{m^3}$
(C) $16 \times {10^{ - 4}}{m^3}$
(D) $2 \times {10^{ - 4}}{m^3}$

Answer 1

Hint
We know what volume strain and volume stress are. Then we will write that the ratio of within elastic limit the ratio of volume stress and volume strain is known as bulk modulus (K). We know the compressibility is the reciprocal of bulk modulus. Compressibility = 1/Bulk modulus. Putting all the values in the equation we will get the required solution.

Complete step by step answer
When a body is subjected to uniform pressure acting normally at every point on its surface, then the body undergoes a change in volume without changing its shape. So, if an external deforming force is applied on a body and as a result the volume of the body changes but the shape does not change. Then the fractional change in volume with respect to its initial volume is called volume strain.
Volume strain$ = \Delta V/V$
When an external deforming force acts on a body producing some deformation then the resistive elastic force generated in a body per unit area of cross section and normal to that area of cross section.
If ‘F’ be the elastic force and ‘A’ be the cross section then stress $ = \dfrac{F}{A} = P$ = Pressure
Within elastic limit the ratio of volume stress and volume strain is known as bulk modulus (K)
 So, we can write here $K = \dfrac{{F/A}}{{\Delta V/V}} = \dfrac{P}{{\Delta V/V}}$
Compressibility is defined as the change in volume due to a unit change in pressure, if a unit volume of a substance is taken initially. Like bulk modulus, compressibility is also a characteristic property of all substances- solids, liquids and gases.
Given that,
The volume of oil contained in a hydraulic press is (V) $ = 0.2{m^3}$.
Compressibility of oil (K) $ = 20 \times {10^{ - 6}}$ per Atm
Pressure (P) = 200 Atm
Compressibility = 1/Bulk modulus $ = \dfrac{{\Delta V/V}}{P}$
$\Delta V = compressibility \times P \times V$
$ \Rightarrow \Delta V = 20 \times {10^{ - 6}} \times 200 \times 0.2$
$ \Rightarrow \Delta V = 8 \times {10^{ - 4}}{m^3}$
Hence, the solution is $8 \times {10^{ - 4}}{m^3}$ (Option-B).

Additional Information
The unit of compressibility is ${m^2}/N$ .

Note
Bulk modulus and compressibility are not independent properties. So, one can easily calculate compressibility from bulk modulus. Here it is given to calculate the decrease in volume so we don’t need to subtract the final value of volume from initial volume. Because in this formula $\Delta V/V$ , $\Delta V$ means decrease in volume.