Two wires of the same material and length but diameter in the ratic 1: 2 are stretched by the same load. The ratio of elastic potential energy per unit volume for the two wires is

Options

A. `1:4`
B. `1:2`
C. `2:1`
D. `1:1`

Answer

D

Explanation

By definition, Young.s modulus, `Y = (F//A)/(Delta L//L) = (FL)/(A Delta L) = (FL)/(pi r^(2) Delta L)`
where the symbols have their usual meaning. As both the wires are of the same material, therefore their Young.s moduli are same
i.e., `Y_(1) = Y_(2) or (F_(1)L_(1))/(pi r_(1)^(2) Delta L_(1)) = (F_(2) L_(2))/(pi r_(2)^(2) Delta L_(2))`
or `(F_(1))/(F_(2)) = ((L_(2))/(L_(1))) ((r_(1))/(r_(2)))^(2) ((Delta L_(1))/(Delta L_(2)))`
Substituting the given values, we get
`(F_(1))/(F_(2)) = ((2L)/(L)) ((r)/(2r))^(2) ((l)/(2l)) = (1)/(4)`
Elastic energy stored per unit volume in the wire is `u= (1)/(2) xx` stress `xx` strain `= (1)/(2) xx (F)/(A) xx (L)/(Delta L)`
`:. (u_(1))/(u_(2)) = ((F_(1))/(F_(2))) ((A_(2))/(A_(1))) ((L_(1))/(L_(2))) ((Delta L_(2))/(Delta L_(1)))`
`= ((1)/(4)) ((pi (2r)^(2))/(pi r^(2))) ((L)/(2L)) (2l)/(l)) = 1`