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

Answer

C

Explanation

Here `Y_(s) =2 Y_(B) , (Y_(S))/(Y_(B))=(2)/(1)`
Let `W_(S)` and `W_(B)` the weights hanged to streel and brass wires
`l_(S)=l_(B)=l, A_(S)=A_(B)=A, Delta l_(S)=Delta l_(B)=Delta l`
`Y=(Wl)/(A Delta l)` or `Delta l=(Wl)/(AY)`
As `Delta l_(S)=Delta l_(B)` `:. (W_(S)l)/(AY_(S))=(W_(B)l)/(AY_(B))`
`(W_(S))/(W_(B))=(Y_(S))/(Y_(B)) =(2)/(1)`