A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

Asked by Abhisek | 1 year ago |  74

##### Solution :-

As the tension on the wires is the same, the extension of each wire will also be the same. Now, as the length of the wires is the same, the strain on them will also be equal.

Now, we know :

Y = $$\dfrac{Stress}{Strain}$$

$$\dfrac{F}{A},\dfrac{1}{Strain}$$

$$\dfrac{4F}{πd^2}.\dfrac{1}{Strain}$$ . . . . . . . . . . (1)

Where,

A = Area of cross-section

F = Tension force

d = Diameter of the wire

We can conclude from equation ( 1 ) that Y ∝ ($$\dfrac{1}{d^2}$$)

We know that Young’s modulus for iron, Y1 = 190 × 109 Pa

Let the diameter of the iron wire = d1

Also, Young’s modulus for copper, Y= 120 × 109 Pa

let the diameter of the copper wire = d2

Thus, the ratio of their diameters can be given as :

$$\dfrac{d_{1}}{d_{2}} = \sqrt{\dfrac{Y_{1}}{Y_{2}}}$$

$$\sqrt{\dfrac{190×10^{9}}{120×10^{9}}} =1 : 25 : 1$$

Answered by Pragya Singh | 1 year ago

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