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A rigid bar of mass `M` is supported symmetrically by three wires each of length `l`. Those at each end are of copper and the middle one is of ion. The ratio of their diameters, if each is to have the same tension, is equal toA. `(Y_(copper))/(Y_(iron))`B. `sqrt((Y_(iron))/(Y_(copper)))`C. `(Y_(iron)^(2))/(Y_(copper)^(2))`D. `(Y_(iron))/(Y_(copper))` |
Answer» Correct Answer - B `Y=(F//pi(D//2)^(2))/(Delta l//l) = (4 F l)/(pi D^(2) Delta l)` `:. D=sqrt((4F l)/(pi Delta l Y))`. or `D prop sqrt((1)/(Y)` Hence, `(D_(copper))/(D_(iron)) = sqrt((Y_(iron))/(Y_(copper))` |
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