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

Options

A. 1.75
B. 1.5
C. 2.3
D. 5

Answer

c

Explanation

`I = (FL)/(pi r^(2) Y) implies r^(2) prop (1)/(Y)` (F,L and I are constants)
`(r_(2))/(r_(1)) = [(Y_(1))/(Y_(2))]^(1//2) = [(7 xx 10^(10))/(12 xx 10^(10))]^(1//2)`
`implies r_(2) = 1.5 xx ((7)/(12))^(1//2) = 1.145 mm`
`:.` Diameter = 2.29 mm.