Ncert Exercises & Solutions

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:

1.	\(\dfrac{Y_{\text{copper}}}{ Y_{\text{iron}}}\)	2.	\({\sqrt{\dfrac{Y_{\text{iron}}}{Y_{\text{copper}}}}}\)
3.	\({\dfrac{Y^{2}_{\text{iron}}}{Y^{2}_{\text{copper}}}}\)	4.	\({\dfrac{Y_{\text{iron}}}{Y_{\text{copper}}}}\)

(2) Hint: The strain produced in each wire will also be the same.

Step 1: Find the diameter of a wire using Young's modulus.

We know that Young's modulus

Y = \frac{Stress}{Strain} = \frac{F / A}{∆ L / L} = F / A \times \frac{L}{∆ L}

= \frac{F}{π {(D / 2)}^{2}} \times \frac{L}{∆ L} = \frac{4 FL}{{πD}^{2} ∆ L}

D^{2} = \frac{4 FL}{π ∆ LY} \Rightarrow D = \sqrt{\frac{4 FL}{π ∆ LY}}

As F and

\frac{L}{∆ L}

are constants,

Hence,

D \propto \sqrt{\frac{1}{Y}}

Step 2: Find the ratio of the diameters.

Now, we can find ratio as

\frac{D_{copper}}{D_{iron}} = \sqrt{\frac{Y_{i r o n}}{Y_{c o p p e r}}}

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