Question

The are of cross-section of steel wire is $0.1{cm}^2$ and Young's modulus of steel is $2\times {10}^{11}N{m}^{-2}$. The force required to stretch by $0.1$% of tis length is then

• A. $1000N$
• B. $2000N$
• C. $4000N$
• D. $5000N$

Answer 1

The correct option is B $2000N$

Here $A=0.1{cm}^2=0.1\times {10}^{-4}{m}^2,Y=2\times {10}^{11}N{m}^{-2}$
$\frac{\Delta L}{L}=0.1$%$=\frac{0.1}{100}=0.1\times {10}^{-2}$
As $Y=\frac{F/A}{\Delta L/L}$
$\therefore F=Y\frac{\Delta L}{L}A=2\times {10}^{11}(\times 0.1\times {10}^{-2})\times (0.1\times {10}^{-4}{m}^2)=2\times {10}^{3}N=2000N$