A rigid bar of mass M is supported symmetrically by three wires each of length I. Those at each end are of copper and the middle one is of iron. What is the ratio of their diameters `(D_("copper"))/(D_("iron"))` if each wire is to have ratio same tension?

Options

A. `(Y_("copper"))/(Y_("iron"))`
B. `sqrt((Y_("iron"))/(Y_("copper")))`
C. `(Y_("iron")^(2))/(Y_("copper")^(2))`
D. `(Y_("iron"))/(Y_("copper"))`

Answer

b

Explanation

`Y = (MgL)/(pi r^(2) (Delta l)) = (MgL)/(pi ((D^(2))/(4))(Delta l)) = (4 MgL)/(pi D^(2) (Delta l))`
`:. D^(2) = (4 MgL)/(pi Y(Delta l))`
`:. D = 2 sqrt((MgL)/(Delta Y (Delta l).pi))`
For all wire, `2 sqrt((MgL)/(pi (Delta l))` is constant
`:. D prop (1)/(sqrt(Y))`
`:. (D_("copper"))/(D_("iron") = sqrt((Y_("iron"))/(Y_("copper"))`