A 5 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. 2.0
C. 2.3
D. 5.0

Answer

c

Explanation

Young's modulus, `Y = ("stress")/("strain") = (F//A)/(I//L) = (F L)/(pi r^(2) I)`
Given `Y_(1) = 7 xx 10^(10) Nm^(-2)` and `Y_(2) = 12 xx 10^(10) Nm^(-2)`
`r_(1) = (D_(1))/(2) = (3)/(2) mm` and `r_(2) = (D_(2))/(2)`
`:.` Taking ratio of Young's modulus and putting `r = (D)/(2)`,
we get `(y_(2))/(y_(1)) = ((D_(1))/(D_(2)))^(2) implies (12 xx 10^(10))/(7 xx 10^(10)) = ((3)/(D_(2)))^(2)`
`implies (3)/(D_(2)) = sqrt((12)/(7))`
`implies D_(2) = 3 sqrt((7)/(12)) = 2.3 mm`