Question

Young modulus of elasticity of brass is ${10}^{11}N/m^2$. The increase in its energy on pressing a rod of length $0.1m$ and cross-sectional area $1c{m}^2$ made of brass with a force of $10kg$ along its length, will be $x\times {10}^{-7}$. Find $x$.

Answer 1

Increase in energy

$\Delta E=\frac{1}{2}\times$ stress $\times$ strain $\times$ volume

$\Delta E=\frac{1}{2}\times \frac{F}{A}\times \frac{F}{A\times Y}\times (A\times L)$

$\Delta E=\frac{1}{2}\times \frac{F}{A}\times \frac{FL}{Y}$

$=\frac{1}{2}\times \frac{10}{{10}^{-4}}\times \frac{10\times 0.1}{{10}^{11}}$

$\Delta E=5\times {10}^{-7}$J.