Question

The area of cross section of a steel wire ($y=2\times {10}^{11}N/m^2$) is $0.1{cm}^2$. The force required to double its length 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 $2\times {10}^{6}N$

Given, $Y=2\times {10}^{11}N/m^2$

$A=0.1{cm}^2=0.1\times {10}^{-4}{m}^2$

Let, length of the wire is l. After applying force, length will be 2l.

$\therefore$ increase in length = $(2l-l)=l$

we know, $Y=\frac{\frac{F}{A}}{\frac{△l}{l}}$

$2\times {10}^{11}=\frac{\frac{F}{0.1\times {10}^{-4}}}{\frac{l}{l}}$

$\therefore F=2\times {10}^{11}\times 0.1\times {10}^{-4}$

$\therefore F=2\times {10}^{6}N$