Question

A $2\,m{m^2}$ cross-sectional area wire is stretched by $4\,mm$ by a certain wire. If the same material of wire of cross-sectional area $8\,m{m^2}$ is stretched by the same weight, the stretched length is:
(A) $2\,mm$
(B) $0.5\,mm$
(C) $1\,mm$
(D) $1.5\,mm$

Answer 1

HintWhen the two wires are stretched, and the two wires are made up of the same material then by using Young's modulus formula, the change in length of the wire can be determined by using the information given in the question.

Formulae Used:
Young’s modulus,
$Y = \dfrac{\sigma }{\varepsilon }$
Where, $Y$ is the young’s modulus, $\sigma $ is the stress and $\varepsilon $ is the strain.

 Complete step-by-step solution:
The cross-sectional area first wire is, ${A_1} = 2\,m{m^2}$
First wire is stretched by, $\Delta {L_1} = 4\,mm$
The cross-sectional area second wire is, ${A_2} = 8\,m{m^2}$
Young’s modulus,
$Y = \dfrac{\sigma }{\varepsilon }\,...............\left( 1 \right)$
Where, $Y$ is the young’s modulus, $\sigma $ is the stress and $\varepsilon $ is the strain.
Now,
$\sigma = \dfrac{F}{A}\,................\left( 2 \right)$
Where, $F$ is the force and $A$ is the area.
And,
$\varepsilon = \dfrac{{\Delta L}}{L}\,....................\left( 3 \right)$
Where, $\Delta L$ is the change in length and $L$ is the original length.
By substituting the equation (2) and equation (3) in equation (1), then
$Y = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta L}}{L}} \right)}}$
By rearranging the above equation, then
$Y = \dfrac{{F \times L}}{{A \times \Delta L}}\,...................\left( 4 \right)$
The above equation (4) is written for first wire and second wire, then
For first wire,
$Y = \dfrac{{F \times L}}{{{A_1} \times \Delta {L_1}}}\,..............\left( 5 \right)$
For second wire,
$Y = \dfrac{{F \times L}}{{{A_2} \times \Delta {L_2}}}\,.............\left( 6 \right)$
By equating the equation (5) and equation (6), then
$\dfrac{{F \times L}}{{{A_1} \times \Delta {L_1}}} = \dfrac{{F \times L}}{{{A_2} \times \Delta {L_2}}}$ (Here the wires are made of same material and stretched by same force, so $F$ and $L$ are same).
By cancelling the same terms, then
$\dfrac{1}{{{A_1} \times \Delta {L_1}}} = \dfrac{1}{{{A_2} \times \Delta {L_2}}}$
By rearranging the terms, then
$\dfrac{{\Delta {L_2}}}{{\Delta {L_1}}} = \dfrac{{{A_1}}}{{{A_2}}}\,.................\left( 7 \right)$
By substituting the area of first wire, area of second wire and the change in length of the first wire in the above equation (7), then
$\dfrac{{\Delta {L_2}}}{4} = \dfrac{2}{8}$
By simplifying the above equation, then
$\Delta {L_2} = 1\,mm$
Thus, the above equation shows the stretched length of the second wire.
Hence, the option (C) is the correct answer.

Note:- From equation (7) it is very clear that the length and area are inversely proportional. If the cross-sectional area of the object increases then the length of the object decreases. If the cross-sectional area of the object decreases then the length of the object is increased.