Question

A rigid bar of mass $15kg$ is supported symmetrically by three wires each $2.0m$ 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.

Answer 1

Draw a rough sketch of the given situation
Find ratio of diameters}

Let
$d_1=\text{diameter of iron wire}$
$d_2=\text{diameter of copper wire}$

Tension $\left(T\right)$ is same. The extension in each case is the same. The strain will also be the same because wires are of the same length.

Young's modulus for iron,

$Y_1=\frac{T}{A_1}\times \frac{L}{\Delta L}$ ...$\left(i\right)$

Young's modulus for copper,

$Y_2=\frac{T}{A_2}\times \frac{L}{\Delta L}$ ...$\left(ii\right)$


Dividing equation $\left(i\right)$ by equation $\left(ii\right)$, we get

$\frac{Y_1}{Y_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}$

$\because$ diameter $r=\frac{d}{2}\Rightarrow d=2r$

$\frac{Y_1}{Y_2}=\frac{(d_2{)}^2}{(d_1{)}^2}\Rightarrow \frac{Y_1}{Y_2}=\frac{d_2^2}{d_1^2}\Rightarrow \frac{d_2}{d_1}=\sqrt{\frac{Y_1}{Y_2}}$ ...$\left(iii\right)$

$\Rightarrow \frac{d_2}{d_1}=\sqrt{\frac{Y_1}{Y_2}}$ ...$\left(iii\right)$

Young's modulus for iron $\left(Y_1\right)=190\times {10}^{9}Pa$
Young's modulus for copper $\left(Y_2\right)=110\times {10}^{9}Pa$

$\therefore$ From equation $\left(iii\right)$

$\frac{d_2}{d_1}=\sqrt{\frac{190\times {10}^{9}Pa}{110\times {10}^{9}Pa}}=\frac{d_2}{d_1}=\sqrt{\frac{19}{11}}=\frac{d_2}{d_1}=\frac{1.31}{1}$