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

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

We know that Young's modulus
`Y=("Stress")/("Strain")=(F//A)/(DeltaL//L)=F//Axx(L)/(DeltaL)`
`=(F)/(pi(D//2)^(2))xx(L)/(DeltaL)=(4FL)/(piD^(2)DeltaL)`
`implies " "D^(2)=(4FL)/(pi DeltaLY)implies D = sqrt((4FL)/(piDeltaLY))`
As F and `(L)/(DeltaL)` are constants.
Hence, `" "D prop sqrt((1)/(Y))`
Now, we can find ratio as `(D_("copper"))/(D_("iron"))=sqrt((Y_("iron"))/(Y_("copper")))`