JEE MAIN - Physics (2025 - 28th January Evening Shift - No. 22)
The volume contraction of a solid copper cube of edge length 10 cm , when subjected to a hydraulic pressure of $7 \times 10^{6} ~\mathrm{Pa}$, would be __________ $\mathrm{mm}{ }^3$.
(Given bulk modulus of copper $=1.4 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ )
Explanation
To find the volume contraction of the copper cube, we need to use the formula for volumetric strain under pressure, which involves the bulk modulus (K). The volumetric strain is given by the formula:
$ \text{Volumetric strain} = \frac{\Delta V}{V} = -\frac{P}{K} $
Where:
$ \Delta V $ is the change in volume or volume contraction,
$ V $ is the original volume of the cube,
$ P $ is the hydraulic pressure applied,
$ K $ is the bulk modulus of the material.
Given:
Edge length of the cube, $ a = 10 \, \text{cm} = 0.1 \, \text{m} $
Hydraulic pressure, $ P = 7 \times 10^6 \, \text{Pa} $
Bulk modulus of copper, $ K = 1.4 \times 10^{11} \, \text{Pa} $
Calculate the original volume ($ V $) of the cube:
$ V = a^3 = (0.1 \, \text{m})^3 = 0.001 \, \text{m}^3 $
Calculate the volumetric strain:
$ \text{Volumetric strain} = -\frac{P}{K} = -\frac{7 \times 10^6}{1.4 \times 10^{11}} = -5 \times 10^{-5} $
Calculate the volume contraction ($ \Delta V $):
$ \Delta V = V \times \text{Volumetric strain} = 0.001 \, \text{m}^3 \times (-5 \times 10^{-5}) = -5 \times 10^{-8} \, \text{m}^3 $
Convert $\Delta V$ from cubic meters to cubic millimeters:
1 cubic meter = $10^9$ cubic millimeters. Therefore,
$ \Delta V = -5 \times 10^{-8} \, \text{m}^3 \times 10^9 \, \text{mm}^3/\text{m}^3 = -50 \, \text{mm}^3 $
Thus, the volume contraction of the copper cube, when subjected to the given hydraulic pressure, is $50 \, \text{mm}^3$.
Comments (0)
