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

Question

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

Dr. Abubacker · Accepted Answer

(c) $l = \frac{FL}{{\pi r}^{2}Y}$
$r^{2} \propto \frac{1}{Y}\ \ (F,\ L\ and\ l\ \ are\ constants)$
$\frac{r_{2}}{r_{1}} = \left\lbrack \frac{Y_{1}}{Y_{2}} ightbrack^{1/2} = \left\lbrack \frac{{7 	imes 10}^{10}}{{12 	imes 10}^{10}} ightbrack^{1/2}$
$r_{2} = 1.5 	imes \left( \frac{7}{12} ight)^{\frac{1}{2}} = 1.145\ mm$
$	herefore$ Diameter = 2.29 mm.

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

Explanation

Reviewed by Dr. Abubacker

Want more practice?