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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 ...

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 of iron. The ratio of their diameters, if each is to have the same tension, is equal to

A

YcopperYiron\frac{Y_{copper}}{Y_{iron}}

B

YironYcopper\sqrt{\frac{Y_{iron}}{Y_{copper}}}

C

Yiron2Ycopper2\frac{Y_{iron}^{2}}{Y_{copper}^{2}}

D

YironYcopper\frac{Y_{iron}}{Y_{copper}}

Answer

YironYcopper\sqrt{\frac{Y_{iron}}{Y_{copper}}}

Explanation

Solution

: The situation is as shown in the figure.

Let T be tension in each wire.

As the bar is supported symmetrically by the three wires, therefore extension in each wire is same.

As Y=F/AΔL/LY = \frac{F/A}{\Delta L/L}

If D is the diameter of the wire, then

Y=F/π(D/2)2ΔL/L=4FLπD2ΔLY = \frac{F/\pi(D/2)^{2}}{\Delta L/L} = \frac{4FL}{\pi D^{2}\Delta L}

As per the conditions of the problem, F (tension), length L, and extension ΔL\Delta Lis same for each wire.

Y1D2\therefore Y \propto \frac{1}{D^{2}} or D1YD \propto \sqrt{\frac{1}{Y}}

DCopperDiron=YironYCopper\therefore\frac{D_{Copper}}{D_{iron}} = \sqrt{\frac{Y_{iron}}{Y_{Copper}}}