Question

The dimensions of four wires of the same material are given below. In which wire the increase in length will be maximum when the same tension is applied?

• A. Length 100 cm, diameter 1 mm
• B. Length 200 cm, diameter 2 mm
• C. Length 300 cm, diameter 3 mm
• D. Length 50 cm, diameter 0.5 mm

Answer 1

The Correct Option is D

As , $Y =\frac{\frac{F}{A}}{\frac{l}{L}} = \frac{F \times L}{\pi r^{2}\times l}$
$ l = \frac{F\times L}{\pi r^{2}\times Y} $
For given $F$ and $Y, l \propto \frac{L}{r^2} \propto \frac{L}{D^2}$
$\therefore \:\:\:\: l_{1}:l_{2}:l_{3}:l_{4} = \frac{L_{1}}{D_{1}^{2}} : \frac{L_{2}}{D_{2}^{2}} : \frac{L_{3}}{D_{3}^{2}} : \frac{L_{4}}{D_{4}^{2}} $
$\frac{L_{1}}{D_{1}^{2}}= {\frac{100 \, cm}{(1 mmm)^2}}, \frac{L_{2}}{D_{2}^{2}} = {\frac{2000 \, cm}{(2 mmm)^2}}, \frac{L_{3}}{D_{3}^{2}} = {\frac{300 \, cm}{(3 mmm)^2}} , \frac{L_{4}}{D_{4}^{2}} = {\frac{50 \, cm}{(0.5 mmm)^2}}$
So, $ l_{1}:l_{2}:l_{3}:l_{4} = 1 : \frac{1}{2} : \frac{1}{3} : 2$
Clearly $l_4$ is maximum. So correct option is (d) ..