The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. `4:1`

Explanation

To solve the problem, we need to find the ratio of the weights added to the steel and brass wires so that their lower ends are at the same level. Here’s a step-by-step solution:

### Step 1: Understand the relationship between Young's Modulus, stress, and strain.
Young's Modulus (Y) is defined as the ratio of stress to strain. Mathematically, it can be expressed as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

### Step 2: Define stress and strain.
Stress (σ) is given by the formula:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force (weight in this case) applied to the wire, and \( A \) is the cross-sectional area of the wire.

Strain (ε) is defined as:
\[ \epsilon = \frac{\Delta L}{L} \]
where \( \Delta L \) is the change in length (stretch) and \( L \) is the original length of the wire.

### Step 3: Set up the equations for both wires.
Since both wires (steel and brass) have the same length (L) and cross-sectional area (A), and we want the stretches (ΔL) to be equal for both wires, we can write:
\[ \frac{W_1}{A} = Y_1 \cdot \frac{\Delta L}{L} \quad \text{(for steel)} \]
\[ \frac{W_2}{A} = Y_2 \cdot \frac{\Delta L}{L} \quad \text{(for brass)} \]

### Step 4: Relate the weights to Young's Modulus.
From the equations above, we can express the weights in terms of Young's Modulus:
\[ W_1 = Y_1 \cdot \frac{\Delta L}{L} \cdot A \]
\[ W_2 = Y_2 \cdot \frac{\Delta L}{L} \cdot A \]

### Step 5: Find the ratio of weights.
To find the ratio of the weights, we divide the two equations:
\[ \frac{W_1}{W_2} = \frac{Y_1}{Y_2} \]

### Step 6: Substitute the values of Young's Modulus.
We know from the problem statement that the Young's Modulus of steel (Y1) is twice that of brass (Y2):
\[ Y_1 = 2Y_2 \]

Substituting this into the weight ratio gives:
\[ \frac{W_1}{W_2} = \frac{2Y_2}{Y_2} = 2 \]

### Conclusion:
Thus, the ratio of the weights added to the steel and brass wires must be:
\[ W_1 : W_2 = 2 : 1 \]