Q.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.

The key idea is Young's Modulus, which quantifies the stiffness of an elastic material.

1. For the rigid bar to remain horizontal and each wire to experience the same tension ($F$) and have the same length ($L$), the extension ($\Delta L$) in all wires must be identical.
2. Young's Modulus is given by $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$. Rearranging for extension, we get $\Delta L = \frac{FL}{AY}$.
3. Since $F$, $L$, and $\Delta L$ are the same for both copper (Cu) and iron (Fe) wires, the product $AY$ must be constant for all wires: $A_{Cu} Y_{Cu} = A_{Fe} Y_{Fe}$
4. Substituting the cross-sectional area $A = \frac{\pi d^2}{4}$, where $d$ is the diameter: $\frac{\pi d_{Cu}^2}{4} Y_{Cu} = \frac{\pi d_{Fe}^2}{4} Y_{Fe}$ This simplifies to $d_{Cu}^2 Y_{Cu} = d_{Fe}^2 Y_{Fe}$. The ratio of diameters is then $\frac{d_{Cu}}{d_{Fe}} = \sqrt{\frac{Y_{Fe}}{Y_{Cu}}}$. …

The core of this problem lies in understanding how a rigid bar supported symmetrically behaves under load, and how Young's Modulus relates stress and strain.

When a rigid bar is supported symmetrically by multiple wires and remains horizontal, it means that all supporting wires must undergo the same extension. If one wire extended more or less than another, the rigid bar would tilt. This is a crucial insight.

Next, we are told that each wire is to have the same tension. This simplifies the force aspect of the problem significantly. We don't need to calculate the actual tension or the total mass of the bar; we only need to know that the force ($F$) in each wire is identical.

Young's Modulus ($Y$) is a material property that quantifies its stiffness or resistance to elastic deformation under tensile or compressive stress. It is defined as the ratio of stress to strain:

We need to find the ratio of the diameters of the copper and iron wires. Since the cross-sectional area of a wire is $A = \pi (d/2)^2 = \frac{\pi d^2}{4}$, we can relate Young's Modulus, tension, length, and diameter.

Let's break down the solution:

1. Identify knowns and implied conditions:

   • The bar is rigid and supported symmetrically, implying that the extension ($\Delta L$) for all wires is the same.
   • All wires have the same original length ($L = 2.0 \text{ m}$).
   • Each wire is to have the same tension ($F$).
   • We need the Young's Moduli for copper and iron. Standard approximate values are:
     • Young's Modulus for Copper ($Y_{\text{Cu}}$) $\approx 110 \times 10^9 \text{ N/m}^2$
     • Young's Modulus for Iron ($Y_{\text{Fe}}$) $\approx 190 \times 10^9 \text{ N/m}^2$
   • We need to find the ratio of their diameters ($d_{\text{Cu}} : d_{\text{Fe}}$).

2. Express tension in terms of Young's Modulus for each wire:

   From the Young's Modulus formula, $Y = \frac{F/A}{\Delta L/L}$, we can rearrange to solve for the force (tension) $F$:

Since the tension $F$, original length $L$, and extension $\Delta L$ are the same for both copper and iron wires, we can write:
For a copper wire:

For the iron wire:

3. Equate the expressions for tension: Since the tension $F$ is the same for both, we can set equation (1) equal to equation (2):

We can cancel the common terms $\Delta L$ and $L$ from both sides:

This equation shows that for equal tension and extension, the product of Young's Modulus and cross-sectional area must be constant for each wire.

4. Substitute area in terms of diameter and solve for the ratio: …

Step 1: rigid symmetric bar => equal deltaL, equal L=2.0m in all 3 wires. Step 2: Y_CuA_Cu=Y_FeA_Fe (same F,L,deltaL). Step 3: using A=pi*d^ …