Wires `A` and `B` are made from the same material. `A` has twice the diameter and three times the length of `B`. If the elastic limits are not reached, when each is stretched by the same tension, the ratio of energy stored in `A` to that in `B` is

Options

A. `2:3`
B. `3:4`
C. `3:2`
D. `6:1`

Answer

B

Explanation

Potential energy of the wire
or `U=1/2xx"stress"xx"strain"xx"volume"`
or `U=1/2xxF/Axx(DeltaL)/LxxAL`
or `U=1/2FDeltaL`
Young's Modulus `=Y=(FL)/(ADeltaL)`
`=F/(pi(D//2)^(2))L/(DeltaL)`
`DeltaL=(4FL)/(piD^(2)Y)`
Both are made of same material.
So, tension is same on two wires.
`DeltaL_(A)=(4Fxx3L)/(pi(2D)^(2)Y)`
Where `L` is the length of wie B and D is the diameter of wire B.
`DeltaL_(B)=(4FL)/(piDF^(2)Y)`
Now ratio of energy stored for same tension.
`(U_(A))/(U_(B))=(DeltaL_(A))/(DeltaL_(B))=3/4=3:4`