One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

Properties of Matter PYQ — JEE Advanced 2013 · Shift 1 · Q47 with Solution

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Correct answer: C

For a wire under tension $F$ ,

\Delta l = \frac{F l}{A Y}

where:

Since both wires are made of copper, $Y$ is the same for both. Also, the two wires are connected in series, so the same force $F$ acts through both.

For the thin wire:

So its elongation is

\Delta l_\text{thin} = \frac{F L}{\pi R^2 Y}

For the thick wire:

A_2 = \pi (2R)^2 = 4\pi R^2

Thus,

\Delta l_\text{thick} = \frac{F(2L)}{4\pi R^2 Y} = \frac{F L}{2\pi R^2 Y}

\frac{\Delta l_\text{thin}}{\Delta l_\text{thick}} = \frac{\frac{F L}{\pi R^2 Y}}{\frac{F L}{2\pi R^2 Y}} = 2

The ratio of elongation in thin wire to thick wire is

2.00

So the correct option is C.