1.

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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