Question

Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is
(a) 1/8
(b) 1/4
(c) 4
(d) 8

Answer 1

(a) 1/8

Explanation:

Given:

length of wire A, $l$

length of wire B, $2l$

radius of wire A, $r$

radius of wire B, $\frac{r}{2}$

Now the elongation in the wires on application of same force:

$\delta l_a=\frac{P.l}{\pi.r^2 .E}$ ...................(1)

where:

P = force applied on the wire

E = Young's modulus of elasticity

and

$\delta l_b=\frac{P.2l}{\pi.\frac{r^2}{4}.E  }$ ..........(2)

Now divide eq. (1) by (2)

$\frac{\delta l_a}{\delta l_b} =\frac{1}{8}$

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TOPIC: stress-strain

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Answer 2

By the same force, if the two wires are stretched, then the elongation in A divided by elongation in B is $\frac{1}{8}$.

Explanation:

Let the Young's modulus of the material of wire be Y.

Given: Force (F)

$A 1=\pi r^{2}$

$L 1=l$

$A 2=\pi(r / 2) 2=\frac{\pi r^{2}}{4}$

$L 2=2 l$

Let the elongation in A be x and that in B be y.

For both the wires, since the Young's modulus is the same:

$Y=\frac{F}{A 1} / \frac{x}{l}=\frac{F}{A 2} / \frac{y}{2 l}$

$x / y=\frac{A 2}{2 A 1}$

$\frac{x}{y}=\frac{1}{8}$.