Question

A 5 m long aluminum wire $(Y = 7 \times 10^{10}N/m^{2})$ of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10}N/m^{2})$ of the same length under the same weight, the diameter should now be, in mm

• A. 1.75
• B. 2.0
• C. 2.3
• D. 5.0

Answer 1

Answer: (C)

2.3

Answer 2

$l = \frac{FL}{\pi r^{2}Y} = \frac{4FL}{\pi d^{2}Y}$ [As $r = d/2$]

If the elongation in both wires (of same length) are same under the same weight then $d^{2}Y =$ constant

$\left( \frac{d_{Cu}}{d_{Al}} \right)^{2} = \frac{Y_{Al}}{Y_{Cu}} \Rightarrow d_{Cu} = d_{Al} \times \sqrt{\frac{Y_{Al}}{Y_{Cu}}} = 3 \times \sqrt{\frac{7 \times 10^{10}}{12 \times 10^{10}}} = 2.29mm$