Two wires of the same material and length but diameters in the ratio 1:2 are stretched by the same force. The elastic potential energy per unit volume for the wires, when stretched by the same force will be in the ratio

Options

A. `4:01`
B. `1:01`
C. `2:01`
D. `4:01`

Answer

A

Explanation

Elastic potential energy in a streatched wire,
`U=1/2 Yxx"strain"`
`=1/2 xx F/A xx l/(Delta l) xx [(Deltal)/l]`
`=1/2 xx F/A xx l/(Delta l) xx (Deltal)^(2)/l^(2)`
`=1/2 xx F/A . L/(Deltal) xx (Deltal)/l xx (Deltal)/l`
`=1/2 F/A . l/(Deltal) xx F/(AY) xx F/(AY)=1/2 F^(3)/(A^(3)Y^(2)) xx l/(Deltal)`
`1/2 F^(3)/(A^(3)Y^(2)) xx 1/((Fl)/(AY)) = 1/2 F^(2)/(A^(2)Y)`
`therefore U+1/2 F^(2)/A^(2) . 1/Y =1/2 . F^(2)/[pi/4 d^(2)]^(2) .1/Y`
`=1/2 . F^(2)/[pi/4 4r^(2)]^(2) .1/Y`
`therefore U propto 1/r^(4)`
`therefore U_(1)/U_(2)= (1/r_(1)^(4))/(1/r_(2)^(4))=[r_(2)/r_(1)]^(4) =[2/1]^(4)`
`=16:1 ("given" d_(1)/d_(2) =1/2 or 1:2)`