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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 l. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

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
The situation is as shown in the figure.
Let T be tesnsion in each wire is same .
As `Y = (F//A)/(Delta L//L)`
If D is the diameter of the wire.
then `Y =(F//pi (D//2)^(2))/(DeltaL//L) = (4F//L)/(pi D^(2)DeltaL)`
As per the conditions of the problem, F (tension), lenght L, and extension `DeltaL` is same for each each wire.
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