Question

wire of length L and radius r fixed at one end and a force F applied to the other end produces an extension $l.$ The extension produced in another wire of the same material of length 2L and radius 2r by a force 2F,is

• A. $l$
• B. $2l$
• C. $4l$
• D. $\frac{l}{2}$

Answer 1

Answer: (A)

When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called the Youngs modulus $(Y)$ of the material of the body.
$Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{l/A}$
where, F is force, A the area, $l$ the change in length and L the original length.
$\therefore$ $Y=\frac{FL}{\pi r^{2}l}$ r being radius of the wire.
Given, $r_2=2r_1,L_2=2L_1,F_2=2F_1$
Since, Youngs modulus is a property of material, we have
$Y_1=Y_2$
$\therefore$ $\frac{F_1{L}_1}{{\pi }_1^2{l}_1}=\frac{2F_1\times 2L_1}{\pi {(2r_1)}^2{l}_2}$
$\Rightarrow$ $l_2=l_1=l$
Hence, extension produced is-same as that in the other wire.