The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of

Options

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

Answer

D

Explanation

Let L and A be length and area of cross section of each wire, In order to have the lower ends of the wire to be at the same level (i.e., same elongation is produced in both wires), let weights `W_(s) and W_(b)` area added to steel and brass wires respectively. Then

By definition of Young.s modulus, the elongation produced in the steel wire is
`Delta L_(s) = (W_(s)L)/(Y_(s)A) ( "as" Y= (W//A)/(Delta L//L))`
and that in the brass wire is `Delta L_(b) = (W_(b) L)/(Y_(b)A)`
But `Delta L_(s) = Delta L_(b)` (given)
`:. (W_(s)L)/(Y_(s)A) = (W_(b)L)/(Y_(b)A) or (W_(s))/(W_(b)) = (Y_(s))/(Y_(b))`
As `(Y_(s))/(Y_(b)) = 2` (given)
`:. (W_(s))/(W_(b)) = (2)/(1)`