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A rigid bar of mass M is supported symme...

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 same tension?

A

`(Y_("copper"))/(Y_("iron"))`

B

`sqrt((Y_("copper"))/(Y_("iron")))`

C

`(Y_("iron")^(2))/(Y_("copper")^(2))`

D

`(Y_("iron"))/(Y_("copper"))`

Text Solution

Verified by Experts

The correct Answer is:
B
`Y = (MgL)/(pi r^(2) (Delta l)) = (MgL)/(pi((D^(2))/(4))(Delta l)) = (4MgL)/(pi D^(2)(Delta l))`
`:. D^(2) = (4MgL)/(pi Y(Delta l)) :. D = 2sqrt((MgL)/(pi Y(Delta l)))`
Fol all wires, `2sqrt((MgL)/(pi(Deltal)))` is constant.
`:. D prop (1)/(sqrt(Y))` [`because` Tension = Mg].
`:. (D_("copper"))/(D_("iron")) = sqrt((Y_("iron"))/(Y_("copper")))`
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