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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 |
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Answer» `1:1` 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)` |
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