First slide

Stress and strain

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 $Δl$ on applying a force F, how much force is needed to stretch the second wire by the same amount?

Moderate

A

4F
B

6F
C

9F
D

F

Solution

According to the question,
For wire 1 Area of cross-section = A₁
Force applied = F₁
and increase in length $= Δl$
From the relation of Young's modulus of elasticity,
$Y = \frac{Fl}{AΔl}$
Substituting the values for wire 1 in the above relation, we get
$\Rightarrow Y_{1} = \frac{{FI}_{1}}{A_{1} ΔI} \dots . (i)$
For wire 2, Area of cross-section = A₂
Force applied = F₂
and increase in length $= Δl$
$∵ Volume, V = Al or I = \frac{V}{A}$
Substituting the value of I in Eqs. (i) and (ii), we get
$Y_{1} = \frac{F_{1} V}{A_{1}^{2} Δl} and Y_{2} = \frac{F_{2} V}{A_{2}^{2} Δl}$
It is given that the wires are made up of same material,
i.e. Y₁ = Y₂
$\begin{array}{l} \Rightarrow \frac{F_{1} V}{A_{1}^{2} Δ 1} = \frac{F_{2} V}{A_{2}^{2} Δ 1} \\ \Rightarrow \frac{F_{1}}{F_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}} = \frac{A^{2}}{9 A^{2}} = \frac{1}{9} (∵ A_{1} = A and A_{2} = 3 A) \\ or, F_{2} = 9 F_{1} = 9 F (Given, F_{1} = F) \end{array}$

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