Question

A wire of length $L$ and radius $r$ is rigidly fixed at one end. On stretching the other end of the wire with a force $F$, the increase in length is $l$. If another wire of the same material but double the length and radius is stretched with a force $2F$, then increase in length is

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

Answer 1

Answer: (D)

Let the length of wire $=L$ Radius $=r$
When a force $F$ is applied, then increase in length $=(l)$ Since, Young's modulus
$Y=\frac{\frac{F}{\pi r^2}}{\frac{l}{L}}$...(i)
$\Rightarrow Y=\frac{FL}{\pi r^{2}l}$
For another wire,
length of wire, $L^{\prime }=2L$
Radius of wire, $r^{\prime }=2r$
Applied force, $F^{\prime }=2F$
$\therefore Y=\frac{F^{\prime }{L}^{\prime }}{\pi r^{\prime 2}{l}^{\prime }}$ where, $l^{\prime }=$ elongation in length
$=\frac{2F\cdot 2L}{\pi (2r{)}^2\cdot l^{\prime }}=\frac{4FL}{4\pi r^2{l}^{\prime }}$
$Y=\frac{FL}{\pi r^2{l}^{\prime }}$...(ii)
From Eqs. (i) and (ii), we have $l^{\prime }=l$