Two unifrom brass rod A and B of length l and 2l and radii 2r respectively are heated to the same temperature. The ratio of the increase in the volume of A to that of B is

Options

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

Explanation

To solve the problem of finding the ratio of the increase in the volume of two brass rods A and B when heated to the same temperature, we will follow these steps:

### Step 1: Determine the original volumes of the rods
The volume \( V \) of a cylindrical rod is given by the formula:
\[ V = \text{Area} \times \text{Length} \]
The area \( A \) of a circle is given by:
\[ A = \pi r^2 \]
For rod A:
- Length \( L_A = l \)
- Radius \( r_A = r \)

Thus, the volume \( V_A \) of rod A is:
\[ V_A = \pi r^2 l \]

For rod B:
- Length \( L_B = 2l \)
- Radius \( r_B = 2r \)

Thus, the volume \( V_B \) of rod B is:
\[ V_B = \pi (2r)^2 (2l) = \pi (4r^2)(2l) = 8\pi r^2 l \]

### Step 2: Calculate the change in volume due to heating
The change in volume \( \Delta V \) due to thermal expansion is given by:
\[ \Delta V = V \cdot \gamma \cdot \Delta T \]
where \( \gamma \) is the coefficient of volumetric expansion and \( \Delta T \) is the change in temperature.

For rod A:
\[ \Delta V_A = V_A \cdot \gamma \cdot \Delta T = (\pi r^2 l) \cdot \gamma \cdot \Delta T \]

For rod B:
\[ \Delta V_B = V_B \cdot \gamma \cdot \Delta T = (8\pi r^2 l) \cdot \gamma \cdot \Delta T \]

### Step 3: Find the ratio of the increases in volume
Now, we will find the ratio of the increase in volume of rod A to that of rod B:
\[ \frac{\Delta V_A}{\Delta V_B} = \frac{\pi r^2 l \cdot \gamma \cdot \Delta T}{8\pi r^2 l \cdot \gamma \cdot \Delta T} \]

The \( \pi \), \( r^2 \), \( l \), \( \gamma \), and \( \Delta T \) terms cancel out:
\[ \frac{\Delta V_A}{\Delta V_B} = \frac{1}{8} \]

### Step 4: Final result
Thus, the ratio of the increase in the volume of rod A to that of rod B is:
\[ \frac{\Delta V_A}{\Delta V_B} = \frac{1}{8} \]

### Conclusion
The ratio of the increase in the volume of rod A to that of rod B is \( 1:8 \).

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