Question

A steel wire of cross-sectional area $3\times {10}^{-6}{m}^2$ can bear a maximum strain of ${10}^{-3}$. Young's modulus of steel is $2\times {10}^{11}N/m^2$. The maximum mass this wire can hold is

• A. $40$ kg
• B. $60$ kg
• C. $80$ kg
• D. $100$ kg

Answer 1

Answer: (B)

$60$ kg

$\begin{array}{c}A=3\times {10}^{-6}{m}^2\\ \frac{\Delta l}{l}={10}^{-3}\\ Y=2\times {10}^{11}N/m^3\\ \text{To find :}\text{maximun mass.}\\ \therefore \frac{F}{A}=Y\frac{\Delta l}{l}\end{array}$

$\begin{array}{c}\Rightarrow \frac{mg}{A}=Y\frac{\Delta l}{l}\\ \Rightarrow m=Y\left(\frac{\Delta l}{l}\right)\frac{A}{g}\\ \Rightarrow m=\frac{2\times {10}^{11}\times {10}^{-3}\times 3\times {10}^{-6}}{10}=6\times 10=60kg\\ \text{Ans}\left(B\right)\end{array}$