First slide

Elastic Modulie

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 $∆ l$ 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 = $Δ l$
From the relation of Young's modulus of elasticity,
$Y = \frac{F l}{A Δ l}$
Substituting the values for wire 1 in the above relation, we get
$\Rightarrow Y_{1} = \frac{F l_{1}}{A_{1} Δ l}$ …(i)
For wire 2 Area of cross-section = A₂
Force applied = F₂
and increase in length = $Δ l$
Similarly, $Y_{2} = \frac{F_{2} l_{2}}{A_{2} Δ l}$ …(ii)
$∵$ Volume, $V = A l or l = \frac{V}{A}$
Substituting the value of I in Eqs. (i) and (ii), we get
$Y_{1} = \frac{F_{1} V}{A_{1}^{2} Δ l} and Y_{2} = \frac{F_{2} V}{A_{2}^{2} Δ l}$
is given that the wires are made up of same material,
i.e. Y₁ = Y₂
$\Rightarrow \frac{F_{1} V}{A_{1}^{2} Δ l} = \frac{F_{2} V}{A_{2}^{2} Δ l}$
$\Rightarrow \frac{F_{1}}{F_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}} = \frac{A^{2}}{9 A^{2}} = \frac{1}{9} (∵ A_{1} = A and A_{2} = 3 A)$
or, $F_{2} = 9 F_{1} = 9 F$ (Given, F₁ = F)

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