Question

If work done in stretching a wire by 1mm is 2J. Then the work necessary for stretching another wire of same material but with double the radius and half the length by 1mm in joule is
(A) $\dfrac{1}{4}$
(B) 4
(C) 8
(D) 16

Answer 1

The work done in changing the length of wire due to stretching is related to the force applied and change in length of the wire. But force is related to stress and stress in turn is related to strain (change in length of wire) through the Young’s modulus.

Complete step by step solution:
The Young’s modulus is given by,
We know that the Young’s modulus is the same for a given material.
Let the length, radius and force for wire be ${l_1}$, ${r_1}$ and ${F_1}$ respectively and the length, radius and force for another wire be ${l_2}$, ${r_2}$and${F_2}$respectively.
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{r_1^2\Delta {l_1}}}{{{l_1}}}\dfrac{{{l_2}}}{{r_2^2\Delta {l_2}}}$
Given, $\Delta {l_1} = \Delta {l_2} = 1mm$
It is given that the radius of the stretched wire is doubled and its length is reduced by half. Thus we have, ${r_2} = 2{r_1}$ and ${l_2} = \dfrac{1}{2}{l_1}$
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{r_1^2 \times 1}}{{{l_1}}}\dfrac{{1{l_1}}}{{2 \times 4r_1^2 \times 1}}$
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{1}{8}$
The work done in stretching a wire by certain amount is given by, $W = \dfrac{1}{2}F\Delta l$
From the above relationship, we see that W is directly proportional to F. So,
$\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{{{F_1}}}{{{F_2}}}$
We are given that the work done in stretching a wire by 1mm is 2J.
${W_2} = 16J$

Option D is correct.

Note: Here we deal with the longitudinal strain related to the changes in the length of the wire and hence we use Young’s Modulus. Young’s modulus depends only on the material. For the same material, the ratio of work done depends mainly on the strain and physical dimensions of the wire.