Question

A wire of area of cross-section $3.0m{m}^2$ and natural length $50cm$ is fixed at one end and a mass of $2.1kg$ is hung from the other end. Find the elastic potential energy stored in the wire in steady state. Young’s modulus of the material of the wire $=1.9\times {10}^{11}N/m^2$. Take $g=10m/s^2$

• A. $1.96\times {10}^{-5}J$
• B. $1.93\times {10}^{-4}J$
• C. $3.89\times {10}^{-4}J$
• D. $2.05\times {10}^{-5}J$

Answer 1

The correct option is B $1.93\times {10}^{-4}J$

Draw a diagram of given problem.
Diagram

Calculate tension of wire.
Formula used: $T=mg$
Given, mass, $m=2.1kg$

As
$T=mg$
$T=2.1\times 10=21N$

Calculate stress of wire.
Formula used: $\text{Stress}=\frac{T}{A}$

Given,
Cross sectional area, $A=3\times {10}^{-6}{m}^2$

$\text{Stress}=\frac{T}{A}$

$=\frac{21}{3\times {10}^{-6}}$

$=7.0\times {10}^{6}N/m^2$

Calculate strain of wire.
Formula used: $\text{strain}=\frac{\text{stress}}{\text{young's modulus}}$

Given, Young modulus,

$Y=1.9\times {10}^{11}N/m^2$

$\text{strain}=\frac{\text{stress}}{Y}$

$=\frac{7.0\times {10}^6}{1.9\times {10}^{11}}$

$=3.7\times {10}^{-5}$

Calculate potential energy of wire.
Formula used:
$U=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}$

Given, length of wire, $L=50cm=0.5m$

Energy stored in the wire,

$U=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}$

$U=\frac{1}{2}(7.0\times {10}^6)(3.7\times {10}^{-5})(1.5\times {10}^{-6}$

$U=1.9\times {10}^{-4}J$