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 to

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

As the bar is supported symmetrically by the three wires, therefore extension in each wire is same
`Y = ("Stress")/("strain") = (F//A)/(Delta L//L) = (F)/(A) xx (L)/(Delta L)`
`= (F)/(pi(D//2)^(2)) xx (L)/(Delta L) = (4FL)/(pi D^(2) Delta L)`
`rArr D^(2) = (4FL)/(pi Delta LY) rArr D = sqrt((4FL)/(pi Delta LY))`
`D = (K)/(sqrt(Y))` (`K` is the proportionally constant)
`(D_("copper"))/(D_("iron")) = sqrt((Y_("iron"))/(Y_("copper")))`