Question

The area of a cross-section of a steel wire is $0.1\, cm^{2}$ and Young’s modulus of steel is $2\times10^{11} N$ $m^{-2}$. The force required to stretch it by $0.1 %$ of its length is

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

Answer 1

Answer: (B)

Here, $A=0.1 \,cm^{2}$ $=0.1\times10^{-4}$ $m^{2}$,
$Y=2\times10^{11} N m^{-2}$
$\frac{\Delta L}{L}=0.1\%$ $=\frac{0.1}{100}$ $=0.1\times10^{-2}$
As $Y=\frac{F /A}{\Delta L/ L}$ $\therefore \quad$ $ F=Y$ $\frac{\Delta L}{L}A$
$=2\times10^{11} N m^{-2}\times0.1\times10^{-2}\times0.1\times10^{-4} m^{2}$
$=2\times10^{3} N$ $=2000 \, N$