If the ratio of diameters,lengths and Young's moduli of steel and brass wires shown in the figure are p, q and r respectively . Then the corresponding ratio of increase in their lenghts would be

Options

A. `(3q)/(5p^(2)r)`
B. `(5q)/(3p^(2)r)`
C. `(3q)/(5pr)`
D. `(5q)/(3pr)`

Answer

B

Explanation

As Young's modulus , `Y = (FL)/(ADeltaL) =(4EL)/(piD^(2)DeltaL)`
where the symbols have their usual meanings.

`therefore DeltaL =(4FL)/(piD^(2)Y)`
`therefore (DeltaL_(s))/(DeltaL_(B)) =(F_(S))/(F_(B)) (L_(S))/(L_(B))(D_(B)^(2))/(D_(S)^(2)) (Y_(B))/(Y_(S))`
where subscripts S and B refers to steel brass respectively.
Here, `F_ (s)=(2m + 3m)g = 5mg, F_(B)= 3mg`
`(L_(S))/(L_(B)) =q, (D_(S))/(D_(B)) = p,(Y_(S))/(Y_(B)) = r`
`therefore (DeltaL_(S))/(DeltaL_(B)) =((5mg)/(3mg)) (q) ((1)/(q))^(2) ((1)/(r)) =(5q)/(3p^(2)r)`