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_("copper"))/(Y_("iron"))`
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 copper and iron wires, we can follow these steps:

### Step 1: Understand the Setup
We have a rigid bar of mass \( M \) supported symmetrically by three wires: two copper wires at the ends and one iron wire in the middle. Each wire has the same length \( l \) and is subjected to the same tension \( T \).

### Step 2: Apply Hooke's Law
According to Hooke's Law, the stress in a wire is proportional to the strain experienced by the wire. The stress \( \sigma \) in a wire is given by:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force (tension) and \( A \) is the cross-sectional area of the wire.

### Step 3: Relate Stress and Young's Modulus
The relationship between stress, strain, and Young's modulus \( Y \) is given by:
\[ \sigma = Y \cdot \epsilon \]
where \( \epsilon \) is the strain, defined as:
\[ \epsilon = \frac{\Delta L}{L} \]
where \( \Delta L \) is the change in length and \( L \) is the original length of the wire.

### Step 4: Set Up Equations for Each Wire
For the copper wire (with diameter \( d_C \)):
\[ \frac{T}{A_C} = Y_C \cdot \frac{\Delta L}{L} \]
For the iron wire (with diameter \( d_I \)):
\[ \frac{T}{A_I} = Y_I \cdot \frac{\Delta L}{L} \]

### Step 5: Express Areas in Terms of Diameters
The cross-sectional area \( A \) of a wire can be expressed in terms of its diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]
Thus, we can write:
\[ A_C = \frac{\pi d_C^2}{4} \quad \text{and} \quad A_I = \frac{\pi d_I^2}{4} \]

### Step 6: Substitute Areas into the Stress Equations
Substituting the areas into the stress equations gives:
\[ \frac{T}{\frac{\pi d_C^2}{4}} = Y_C \cdot \frac{\Delta L}{L} \]
\[ \frac{T}{\frac{\pi d_I^2}{4}} = Y_I \cdot \frac{\Delta L}{L} \]

### Step 7: Equate the Two Equations
Since the tension \( T \) and the strain \( \frac{\Delta L}{L} \) are the same for both wires, we can equate the two equations:
\[ \frac{T}{\frac{\pi d_C^2}{4}} = \frac{T}{\frac{\pi d_I^2}{4}} \]
This simplifies to:
\[ \frac{d_I^2}{d_C^2} = \frac{Y_C}{Y_I} \]

### Step 8: Find the Ratio of Diameters
Taking the square root of both sides gives:
\[ \frac{d_I}{d_C} = \sqrt{\frac{Y_C}{Y_I}} \]
Thus, the ratio of the diameters is:
\[ \frac{d_C}{d_I} = \sqrt{\frac{Y_I}{Y_C}} \]

### Conclusion
The correct answer for the ratio of the diameters of the copper and iron wires, if each is to have the same tension, is:
\[ \frac{d_C}{d_I} = \sqrt{\frac{Y_I}{Y_C}} \]