Two light aluminium rods of same length are suspended from a ceiling. First rod has load W hanging from it, while second rod has load 2W hanging from it. If the ratio of energy stored per unit volume of first rod to second rod is 1:64, then the ratio of diameter of first rod to second rod will be :

Options

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

Answer

To solve the problem step by step, we will analyze the given situation involving two aluminum rods and the loads they carry.

### Step 1: Understand the Problem
We have two aluminum rods of the same length. The first rod has a load \( W \) and the second rod has a load \( 2W \). We need to find the ratio of the diameters of the two rods given that the ratio of energy stored per unit volume in the first rod to the second rod is \( 1:64 \).

### Step 2: Define the Energy Stored per Unit Volume
The energy stored per unit volume \( U \) can be expressed as:
\[ U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]

### Step 3: Relate Stress and Strain
Stress (\( \sigma \)) is defined as the force per unit area:
\[ \sigma = \frac{F}{A} \]
For a circular cross-section, the area \( A \) is given by:
\[ A = \frac{\pi D^2}{4} \]
Thus, the stress for each rod can be written as:
\[ \sigma_1 = \frac{W}{\frac{\pi D_1^2}{4}} = \frac{4W}{\pi D_1^2} \]
\[ \sigma_2 = \frac{2W}{\frac{\pi D_2^2}{4}} = \frac{8W}{\pi D_2^2} \]

### Step 4: Substitute Stress into Energy Equation
Now substituting stress into the energy stored per unit volume equation:
\[ U_1 = \frac{1}{2} \sigma_1 \epsilon_1 = \frac{1}{2} \left(\frac{4W}{\pi D_1^2}\right) \epsilon_1 \]
\[ U_2 = \frac{1}{2} \sigma_2 \epsilon_2 = \frac{1}{2} \left(\frac{8W}{\pi D_2^2}\right) \epsilon_2 \]

### Step 5: Use the Given Ratio of Energy Stored
We know:
\[ \frac{U_1}{U_2} = \frac{1}{64} \]
Substituting the expressions for \( U_1 \) and \( U_2 \):
\[ \frac{\frac{1}{2} \left(\frac{4W}{\pi D_1^2}\right) \epsilon_1}{\frac{1}{2} \left(\frac{8W}{\pi D_2^2}\right) \epsilon_2} = \frac{1}{64} \]
This simplifies to:
\[ \frac{4W \epsilon_1 D_2^2}{8W \epsilon_2 D_1^2} = \frac{1}{64} \]
Canceling \( W \) and simplifying gives:
\[ \frac{D_2^2 \epsilon_1}{2 D_1^2 \epsilon_2} = \frac{1}{64} \]

### Step 6: Relate Strain to Stress
Using the relationship \( \gamma = \frac{\sigma}{\epsilon} \), we can express \( \epsilon \) in terms of stress:
\[ \epsilon_1 = \frac{\sigma_1}{\gamma_1}, \quad \epsilon_2 = \frac{\sigma_2}{\gamma_2} \]
Assuming the material properties are the same for both rods, we can say \( \gamma_1 = \gamma_2 \). Thus:
\[ \epsilon_1 = \frac{4W}{\pi D_1^2 \gamma}, \quad \epsilon_2 = \frac{8W}{\pi D_2^2 \gamma} \]

### Step 7: Substitute Back into the Ratio
Substituting these into the ratio gives:
\[ \frac{D_2^2 \left(\frac{4W}{\pi D_1^2 \gamma}\right)}{2 D_1^2 \left(\frac{8W}{\pi D_2^2 \gamma}\right)} = \frac{1}{64} \]
This simplifies to:
\[ \frac{D_2^4}{D_1^4} = \frac{1}{32} \]

### Step 8: Solve for the Diameter Ratio
Taking the fourth root gives:
\[ \frac{D_2}{D_1} = \frac{1}{2} \implies \frac{D_1}{D_2} = 2 \]

### Final Answer
The ratio of the diameter of the first rod to the second rod is:
\[ \frac{D_1}{D_2} = 2:1 \]