Two steel wires of lengths 1 m and 2 m have diameters 1 mm and 2mm respectively . If they are stretched by force of 40 N and 80 N respectively, find the ratio of their elongations.

Explanation

We know that , `Y = (Fl)/(pi r^2 Delta l) implies Delta l = (Fl)/(pi r^2 Y)`
Therefore, for the two wires,
`(Deltal_(1)) = (F_1l_1)/(pi r_1^2Y) , (Delta l)_(2) = (F_2l_2)/(pi r_2^2Y)`
The ratio of their elongations
`((Delta l)_1)/((Delta l)_(2)) = (F_1l_1r_2^2)/(F_2l_2r_1^2) = 40/80 xx 1/2 xx (2/1)^2`
`((Deltal)_(1))/((Deltal)_(2)) = 1/1`
`(Deltal)_(1) , (Delta l)_(2) = 1 : 1`.