Question

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

Answer 1

Here the wires are made of the same material and have the same volume. Since, the materials are the same, so the Young’s modulus for both the wires will be the same. Given some parameters, we can solve this problem by finding a relation between the two cases.

Complete step by step answer:
Using the formula of Young’s modulus, $Y=\dfrac{Fl}{A\Delta l}$
Volumes are same, volume can be written as, $V=Al$, so length from here will be $l=\dfrac{V}{A}$
Thus, $F=\dfrac{YA\Delta l}{l}$
Multiplying and divide by A on RHS we get,
$F=\dfrac{Y{{A}^{2}}\Delta l}{Al}=\dfrac{Y{{A}^{2}}\Delta l}{V}$
Now writing for first and second wire and dividing them we get,
$\dfrac{{{F}_{1}}}{{{F}_{2}}}=\dfrac{A_{1}^{2}}{A_{2}^{2}}$
But,

& {{A}_{1}}=A \\\ & {{A}_{2}}=3A \\\ \end{aligned}$$ Thus, $$\begin{aligned} & \dfrac{{{F}_{1}}}{{{F}_{2}}}=\dfrac{A_{{}}^{2}}{9{{A}^{2}}} \\\ & {{F}_{2}}=9{{F}_{1}} \\\ \end{aligned}$$ Given that $${{F}_{1}}=F$$, thus $${{F}_{2}}=9F$$ Thus, the force needed to stretch the second wire by the same amount is 9F. **So, the correct answer is “Option 1”.** **Additional Information:** Stress is the force applied to a material, divided by the material's cross-sectional area. The strain is the displacement of material that results from applied stress. **Note:** Young’s modulus is dependent on the nature of the material and it does not depend on other parameters such as length or area of the cross-section or the temperature. Here the area was uniform and thus we did not have to take into account deformation of shape. If we are given a problem in which two wires of different materials, then we have to use two sets of young’s moduli for each of them.