Question

The length of a rod is $20\text{cm}$ and area of cross-section $2{\text{cm}}^2$. The Young's modulus of the material of wire is $1.4\times {10}^{11}{\text{N/m}}^2$. If the rod is compressed by a load of $5\text{kg-wt}$ along its length, then find the energy stored in the rod (in Joules).

• A. $8.57\times {10}^{-6}$
• B. $22.5\times {10}^{-4}$
• C. $9.8\times {10}^{-5}$
• D. $45.0\times {10}^{-5}$

Answer 1

The correct option is A $8.57\times {10}^{-6}$

Given,
Length of a rod $\left(l\right)$= $20\text{cm}$
Area of cross - section $\left(A\right)$=$2{\text{cm}}^2$
Young's modulus of the material of wire$\left(Y\right)$= $1.4\times {10}^{11}{\text{N/m}}^2$
Applied load ($F)$=$5\text{kg-wt}$ or $5\times 9.81\text{N}$= $49.05\text{N}$
We know that,
Energy stored in the wire $=\frac{1}{2}\text{load}\times \text{elongation}=\frac{1}{2}\times F\times \left(\frac{FL}{AY}\right)=\frac{1}{2}\times \frac{F^{2}L}{AY}$
$=\frac{1}{2}\times \frac{(49.05{)}^2\times 20\times {10}^{-2}}{2\times {10}^{-4}\times 1.4\times {10}^{11}}\approx 8.57\times {10}^{-6}\text{J}$