Question

What is the percentage increase in length of a wire having diameter $1.5\text{mm}$, stretched by a force of $50\text{kg-wt}$?
(Young's modulus of elasticity of wire $\left(Y\right)=15\times {10}^{11}{\text{dyne/cm}}^2\text{and}g=10{\text{m/s}}^2$)

• A. $0.19$
• B. $0.019$
• C. $1.9$
• D. $19$

Answer 1

The correct option is A $0.19$

Young modulus of elasticity $\left(Y\right)$ is given by $Y=\frac{FL}{A△L}$
$\Rightarrow$ percentage increase in length, $\frac{△L}{L}\times 100=\frac{F}{AY}\times 100$ ...$\left(i\right)$
Diameter $\left(d\right)=1.5\text{mm}$.
Area $\left(A\right)$ $=\frac{\pi }{4}{d}^2=\frac{\pi }{4}(1.5\times {10}^{-3}{)}^2=1.76\times {10}^{-6}{\text{m}}^2$

Given:
$Y=15\times {10}^{11}{\text{dyne/cm}}^2$
$[\because {\text{1 dyne/cm}}^2=0.1{\text{N/m}}^2]$
$\Rightarrow Y=1.5\times {10}^{11}{\text{N/m}}^2$

$F=50\text{kg-wt}=50\times 10\text{N}=500\text{N}$

From Equation $\left(i\right)$
Percentage increase in length $(\frac{△L}{L}\times 100)=\frac{F}{AY}\times 100$
$=\frac{500\times 100}{1.76\times {10}^{-6}\times 1.5\times {10}^{11}}$
$=0.19$ %