Question

A stress of $2kg/m{m}^2$ is applied on a wire. If $Y={10}^{12}dyne/c{m}^2$ then the percentage increase in its length will be

• A. $0.196\%$
• B. $19.6\%$
• C. $1.96\%$
• D. $0.0196\%$

Answer 1

Answer: (D)

$0.0196\%$
Given$:Stress=2kg/m{m}^2=2\times \frac{9.8}{{10}^{-6}}Pa=19.6\times {10}^{6}Pa$
Young's Modulus of elasticity $={10}^{12}dyne/c{m}^2$
Since$,1dyne={10}^{-5}N$
$={10}^{11}N/m^2$

Now, from Hooke's law, we know that
$Stress=YStrain$
Percentage change in length is strain $\%$
Strain$=\frac{\Delta L}{L}$
Strain$\%=\frac{Stress}{Y}\times 100$
So, percentage change in length $=\frac{19.6\times {10}^6\times 100}{{10}^{11}}=0.0196\%$