The Young's Modulus of steel is twice that of brass. Two wires of same length and of same area of cross-section, one of steel and another of brass are suspended from the same roof. If then the weights 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)=2Y_(B), (Y_(s))/(Y_(B))=(2)/(1)`
Let `W_(S)` & `W_(B)` the weights hanged to steel & brass wires
`l_(s)=l_(B)=l,A_(B)=A_(s)=A,Deltal_(s)=Deltal_(B)=Deltal`
`Y=(Wl)/(ADeltal) orDeltal=(Wl)/(AY)`
as`Deltal_(s)=Deltal_(B)`
`:. (W_(s)l)/(AY_(s))=(W_(B)l)/(AY_(B))`
`(W_(s))/(W_(B))=(Y_(B))/(Y_(B))=(2)/(1)`