Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has a length l and diameter r while the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force the elongationn in A divided by the elongation in B is

Options

A. `1//8`
B. `1//4`
C. 4
D. 8

Answer

A

Explanation

`Y= (F//A)/(x//L)`
So, the elongation x is given as,
`x= ((F.L)/(A.Y)) = (F.L)/(pi R^(2).Y)`
For wires P and Q are made of same material or in other words same Y, ratio of elongation by the same force,
`(x_(P))/(x_(Q))= ((L_(P))/(R_(P)^(2))) ((R_(Q)^(2))/(L_(Q)))= ((L_(P))/(L_(Q))) .((R_(Q))/(R_(P)))^(2)`
Here `L_(P)= l, R_(P) = (r )/(2), L_(Q)= 2l and R_(Q) = (r )/(4)`
`rArr (x_(P))/(x_(Q)) = ((l)/(2l)) .((1)/(2))^(2)= (1)/(8)`
`rArr` (a) is correct