A rigid bar of mass 15 kg is supported symmetrically by three wires, each of length 2 m. The wires at the endpoints are made of copper and the middle one is made of steel. If the tension in each wire is the same, then the diameter of copper wire to the diameter of steel wire is `["Given, "Y_("copper")=1.1xx10^(11)"N m"^(-2) and Y_("steel")=1.9xx10^(11)"N m"^(-2)]`

Options

A. `sqrt((19)/(11))`
B. `sqrt((11)/(19))`
C. `(11)/(19)`
D. `(19)/(11)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Forces Acting on the System
The rigid bar of mass \( M = 15 \, \text{kg} \) is supported symmetrically by three wires. The total weight acting downwards is given by:
\[ W = Mg = 15 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 147 \, \text{N} \]
Since the bar is in equilibrium, the total upward force provided by the three wires must equal the weight:
\[ 3T = W \quad \Rightarrow \quad T = \frac{W}{3} = \frac{147 \, \text{N}}{3} = 49 \, \text{N} \]

### Step 2: Relate Tension to Young's Modulus
The tension \( T \) in each wire creates a stress, which is defined as:
\[ \text{Stress} = \frac{F}{A} \]
where \( F \) is the force (tension \( T \)) and \( A \) is the cross-sectional area of the wire. The area \( A \) for a circular wire is given by:
\[ A = \frac{\pi d^2}{4} \]
Thus, the stress in each wire can be expressed as:
\[ \text{Stress} = \frac{T}{\frac{\pi d^2}{4}} = \frac{4T}{\pi d^2} \]

### Step 3: Use Young's Modulus Definition
Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} \]
For the wires, since the tension is the same, we can write:
\[ Y_{\text{copper}} = \frac{4T}{\pi d_c^2} \cdot \frac{L}{\Delta L_c} \]
\[ Y_{\text{steel}} = \frac{4T}{\pi d_s^2} \cdot \frac{L}{\Delta L_s} \]

### Step 4: Set Up the Ratio of Diameters
Since the tension \( T \) and length \( L \) are the same for both wires, we can set up the ratio:
\[ \frac{Y_{\text{copper}}}{Y_{\text{steel}}} = \frac{\frac{4T}{\pi d_c^2}}{\frac{4T}{\pi d_s^2}} = \frac{d_s^2}{d_c^2} \]
Rearranging gives:
\[ \frac{d_c^2}{d_s^2} = \frac{Y_{\text{copper}}}{Y_{\text{steel}}} \]
Taking the square root:
\[ \frac{d_c}{d_s} = \sqrt{\frac{Y_{\text{copper}}}{Y_{\text{steel}}}} \]

### Step 5: Substitute the Values of Young's Modulus
Given:
\[ Y_{\text{copper}} = 1.1 \times 10^{11} \, \text{N/m}^2, \quad Y_{\text{steel}} = 1.9 \times 10^{11} \, \text{N/m}^2 \]
Substituting these values into the equation:
\[ \frac{d_c}{d_s} = \sqrt{\frac{1.1 \times 10^{11}}{1.9 \times 10^{11}}} = \sqrt{\frac{1.1}{1.9}} = \sqrt{\frac{11}{19}} \]

### Final Result
Thus, the ratio of the diameter of the copper wire to the diameter of the steel wire is:
\[ \frac{d_c}{d_s} = \sqrt{\frac{11}{19}} \]