A 5 m long aluminium wire `(Y=7xx10^10(N)/(m^2))` of diameter 3 mm supprts a 40 kg mass. In order to have the same elongation in a copper wire `(Y=12xx10^10(N)/(m^2))` of the same length under the same weight, the diameter should now, in mm

Options

A. 1.75
B. 1.5
C. 2.5
D. `5.0`

Answer

C

Explanation

`l=(FL)/(pir^(2)Y)`
`implies r^(2) prop (1)/(Y)` (F,L and l are constant )
`(r_(2))/(r_(1))=[(Y_(1))/(Y_(2))]^(1//2)=[( 7xx10^(10))/(12xx10^(10))]^(1//2)`
`implies r^(2)=1.5xx((7)/(12))^(1//2)=1.145` mm
`:.` diameter =2.29 mm .