Question

Two wires are made of the same material and have the same volume. However wire 1 has a cross-sectional area $A$ and wire 2 has a cross sectional area $3A$. If the length of wire 1 increases by $\Delta x$ on applying force $F$, how much force is needed to stretch the wire 2 by the same amount?
(A) $F$
(B) $4F$
(C) $6F$
(D) $9F$

Answer 1

Hint
The length of wire 2 will be $\dfrac{1}{3}$ times the length of wire 1, as the area of the cross section of wire 2 is 3 times that of wire 1. Therefore by equating the values in the formula for the young’s modulus, we get the force in the case of the second wire.
Formula Used: In this solution we will be using the following formula,
$\Rightarrow Y = \dfrac{{FL}}{{A\Delta L}}$
where $Y$ is the young’s modulus, $F$ is the force on the wire, $L$ is the length of the wire, $\Delta L$ is the increase in length and $A$ is the area of cross-section of the wire.

Complete step by step answer
In the question we are given that the volumes of both the wires are the same. Now the area of cross-section of the second wire is 3 times the area of the cross-section of the first wire. Since the amount of material is the same, so we can write,
$\Rightarrow {A_1}{L_1} = {A_2}{L_2}$
Substituting ${A_1}$ as $A$ and ${A_2}$ as $3A$ we get,
$\Rightarrow A{L_1} = 3A{L_2}$
Cancelling $A$ we get the length of the second wire as, ${L_2} = \dfrac{{{L_1}}}{3}$
So the length of the second wire is $\dfrac{1}{3}$ times the length of the first wire.
Now since both the wires are made of the same material, so the young’s modulus of both the wires will be the same. Hence the young’s modulus of the first wire will be,
$\Rightarrow Y = \dfrac{{FL}}{{A\Delta x}}$
And the young’s modulus of the second wire will be,
$\Rightarrow Y = \dfrac{{F'\dfrac{L}{3}}}{{3A\Delta x}}$
On equating both the values we get,
$\Rightarrow \dfrac{{FL}}{{A\Delta x}} = \dfrac{{F'\dfrac{L}{3}}}{{3A\Delta x}}$
On cancelling the like terms on both the sides of the equation, we get,
$\Rightarrow F = \dfrac{{F'\dfrac{1}{3}}}{3}$
Therefore we get on simplifying,
$\Rightarrow F = \dfrac{{F'}}{{3 \times 3}}$
This gives us the force on the second wire as,
$\Rightarrow F' = 9F$
Hence the correct answer is option D.

Note
The young’s modulus or the modulus of elasticity of a material is the mechanical property that measures the stiffness of a material. It is given by the ratio of the stress to the strain of the material and has a unit of $N/{m^2}$.