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

Options

A. `Y_("copper")//Y_("iron")`
B. `sqrt(Y_("iron")/Y_("copper"))`
C. `(Y_("iron")^(2))/(Y_("copper")^(2))`
D. `(Y_"iron")/(Y_"copper")`

Explanation

To solve the problem of finding the ratio of the diameters of the wires made of copper and iron, we can follow these steps:

### Step 1: Understand the relationship between stress, strain, and Young's modulus
Young's modulus (Y) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress = \(\frac{F}{A}\) (Force per unit area)
- Strain = \(\frac{\Delta L}{L}\) (Change in length per original length)

### Step 2: Express the area in terms of diameter
The area \(A\) of a wire with diameter \(D\) is given by:
\[ A = \frac{\pi D^2}{4} \]
Thus, the stress can be expressed as:
\[ \text{Stress} = \frac{F}{\frac{\pi D^2}{4}} = \frac{4F}{\pi D^2} \]

### Step 3: Substitute stress and strain into the Young's modulus formula
Using the definitions of stress and strain, we can rewrite Young's modulus as:
\[ Y = \frac{\frac{4F}{\pi D^2}}{\frac{\Delta L}{L}} = \frac{4FL}{\pi D^2 \Delta L} \]

### Step 4: Rearrange to find diameter in terms of Young's modulus
Rearranging the equation to solve for \(D^2\):
\[ D^2 = \frac{4FL}{\pi Y \Delta L} \]
Taking the square root gives:
\[ D = \sqrt{\frac{4FL}{\pi Y \Delta L}} \]

### Step 5: Establish the relationship between diameters of copper and iron
Since the tension \(F\) and the length \(L\) are the same for both wires, we can express the diameters of copper (\(D_c\)) and iron (\(D_i\)) as:
\[ D_c = k \sqrt{\frac{1}{Y_c}} \quad \text{and} \quad D_i = k \sqrt{\frac{1}{Y_i}} \]
where \(k\) is a constant that includes \(4F\) and \(\pi\Delta L\).

### Step 6: Find the ratio of diameters
The ratio of the diameters of copper to iron can be expressed as:
\[ \frac{D_c}{D_i} = \frac{\sqrt{\frac{1}{Y_c}}}{\sqrt{\frac{1}{Y_i}}} = \sqrt{\frac{Y_i}{Y_c}} \]

### Step 7: Conclusion
Thus, the ratio of the diameters of the wires made of copper and iron, if each is to have the same tension, is given by:
\[ \frac{D_c}{D_i} = \sqrt{\frac{Y_i}{Y_c}} \]

### Final Answer
The correct option is:
\[ \text{B) } \sqrt{\frac{Y_i}{Y_c}} \]

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