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Question
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
`Y_(copper)/Y_(iron)`
`sqrt((Y_(iron))/(Y_(copper)`
`Y_(iron)^2/Y_(copper)^2`
`Y_(iron)/Y_(copper)^2`
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Solution
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 `underline(sqrt((Y_(iron))/(Y_(copper))`.
Explanation:
As the bar is supported symmetrically by the three wires, therefore extension in each wire is the same.
Let T be the tension in each wire and the diameter of the wire is D, then Young’s modulus is `Y = "Stress"/"Strain"`
= `(F/A)/((ΔL)/L)`
= `F/A xx L/(ΔL)`
= `F/(pi(D/2)^2) xx L/(ΔL)`
= `(4FL)/(piD^2ΔL)`
⇒ `D^2 = (4FL)/(piΔLY)`
⇒ `D = sqrt((4FL)/(piΔLY)`
As F and `L/(ΔL)` are constants.
Hence, `D ∝ sqrt(1/Y)`
or `D = K/sqrt(Y)` ......(K is the proportionality constant)
Now, we can find ratio as `D_(copper)/D_(iron) = sqrt(Y_(iron)/Y_(copper)`
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