Question

Cross section area of a steel wire $(Y=2.0\times {10}^{11}{N/m}^2)$ is 0.1 ${cm}^2$. The required force, to stretch it to double will be -

• A. ${2\times 10}^{12}N$
• B. ${2\times 10}^{11}N$
• C. ${2\times 10}^{10}N$
• D. ${2\times 10}^{6}N$

Answer 1

The correct option is D ${2\times 10}^{6}N$

$\begin{array}{c}\text{For double the length}\\ \Delta l=l\text{.}\\ \text{where}l=\text{initial length}\end{array}$

$\begin{array}{c}Y=\frac{Fl}{A\Delta l}=\frac{F\times N}{0.1\times {10}^{-4}\times K}=2\times {10}^{11}\\ F=2\times {10}^{6}N\end{array}$