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

The correct option is √(Y_copper/Y_iron).

Step-by-Step Explanation:

To find the ratio of the diameters of the wires, we start by analyzing the relationship between the physical quantities of the wire under tension. Young's modulus ($Y$) is defined as the ratio of tensile stress to tensile strain:

$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{T/A}{\Delta l/l}$

Where:
- $T$ is the tension in the wire,
- $A$ is the cross-sectional area of the wire,
- $l$ is the original length of the wire, and
- $\Delta l$ is the extension produced in the wire.

We can rewrite the equation to solve for the extension ($\Delta l$):

$\Delta l=\frac{T·l}{A·Y}$

Since the area of the wire of diameter $d$ is given by $A=\frac{\pi d^2}{4}$, substituting this value of $A$ into the equation gives:

$\Delta l=\frac{4Tl}{\pi d^{2}Y}$

Since the rigid bar is supported symmetrically and must remain horizontal, the extension in both the copper and iron wires must be the same:

$\Delta l_{\text{copper}}=\Delta l_{\text{iron}}$

It is given that each wire has the same initial length ($l$) and is subjected to the same tension ($T$). Equating the extensions of the copper and iron wires, we get:

$\frac{4Tl}{\pi d_{\text{copper}}^2{Y}_{\text{copper}}}=\frac{4Tl}{\pi d_{\text{iron}}^2{Y}_{\text{iron}}}$

Simplifying the equation by canceling the common terms ($4$, $T$, $l$, and $\pi$) from both sides:

$d_{\text{copper}}^2{Y}_{\text{copper}}=d_{\text{iron}}^2$

Rearranging the equation to find the ratio of the diameter of the iron wire to that of the copper wire ($d_{\text{iron}}/d_{\text{copper}}$):

$\frac{d_{\text{iron}}^2}{d_{\text{copper}}^2}=\frac{Y_{\text{copper}}}{Y_{\text{iron}}}$

Taking the square root on both sides, we obtain:

$\frac{d_{\text{iron}}}{d_{\text{copper}}}=\sqrt{\frac{Y_{\text{copper}}}{Y_{\text{iron}}}}$

Therefore, the ratio of their diameters is equal to √(Y_copper/Y_iron).