First slide

Elastic Modulie

Question

A steel wire of length I and cross-section area A is stretched by 1 cm under a given load. When the same load is applied to another steel wire of double its length and half of its cross-section area, the amount of stretching is

Easy

A

0.5 cm
B

2 cm
C

4 cm
D

1.5 cm

Solution

We know that, $Δ l = \frac{F L}{A Y}$
Here, $Δ l \propto \frac{L}{A}$
According to question, $\frac{Δ l_{1}}{Δ l_{2}} = \frac{L_{1}}{L_{2}} \times \frac{A_{2}}{A_{1}}$
$\Rightarrow \frac{Δ l_{1}}{Δ l_{2}} = \frac{1}{2} \times \frac{1}{2} \Rightarrow \frac{Δ l_{1}}{Δ l_{2}} = \frac{1}{4}$
$\Rightarrow Δ l_{2} = 4 Δ l_{1} \Rightarrow Δ l_{2} = 4 \times 1 = 4 cm$

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