Question

A thin rod of negligible mass and cross-sectional area $4\times 10$${}^{-6}{m}^2$ , suspended vertically from one end, has a length of 0.5 m at 100${}^o$C . The rod is cooled to 0${}^o$C .Young's modulus $=$ 10${}^{11}$Nm${}^{-2}$, coefficient of linear expansion $=$ 10${}^{-5}$K${}^{-1}$ and g $=10m{s}^{-2}$. The total energy stored in the rod is :

• A. $0.1J$
• B. $0.2J$
• C. $0.3J$
• D. $0.4J$

Answer 1

The correct option is A $0.1J$

Given:

$A=4\times {10}^{-6}L=.5mY={10}^{11}\alpha ={10}^{-5}g=10m{s}^{-2}\Delta T=100$

Total energy $=\frac{1}{2}k(\Delta l{)}^2$

$\Delta l$ is due to temperature change.

$l^{\prime }=l(1+\alpha \Delta T)$ where $l^{\prime }$ is new length

$l^{\prime }=l+l\alpha \Delta T$

$l^{\prime }-l=l\alpha \Delta T$

$\Delta l=l\alpha \Delta T$

$=\frac{1}{2}\times {10}^{-5}\times 100=5\times {10}^{-4}$

$k_{rod}=\frac{YA}{l}$

Total Energy $=\frac{1}{2}\frac{YA}{l}(\Delta l{)}^2=\frac{1}{2}\times \frac{{10}^{11}\times 4\times {10}^{-6}}{0.5}\times (5\times {10}^{-4}{)}^2=0.1J$