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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
B
We know that Young’s modulus `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))`
As F and `(L)/(Delta L)` are constants
Hence, `D prop sqrt((1)/(Y))`
So the required ratio is, `(D_("copper"))/(D_("iron"))= sqrt((Y_("Iron"))/(Y_("copper")))`
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