$$) of a rod due to a change in temperature can be given by:

$$ \Delta V = \beta V \Delta T $$

where:

$$\beta$$ = coefficient of volumetric expansion (which is the same for both rods since they are made of brass),

$$V$$ = initial volume of the rod,

$$\Delta T$$ = change in temperature (which is the same for both rods).

First, let's calculate the volume of each rod. The volume $$V$$ of a cylindrical rod is:

$$ V = \pi r^2 l $$

For rod $$A$$:

Length = $$l$$,

Radius = $$2r$$.

Therefore, the initial volume of rod $$A$$ ($$V_A$$) is:

$$ V_A = \pi (2r)^2 l = \pi \cdot 4r^2 \cdot l = 4 \pi r^2 l $$

For rod $$B$$:

Length = $$2l$$,

Radius = $$r$$.

Therefore, the initial volume of rod $$B$$ ($$V_B$$) is:

$$ V_B = \pi r^2 (2l) = 2 \pi r^2 l $$

Now, the increase in volume ($$\Delta V$$) of each rod due to heating is given by:

For rod $$A$$:

$$ \Delta V_A = \beta V_A \Delta T = \beta (4 \pi r^2 l) \Delta T = 4 \beta \pi r^2 l \Delta T $$

For rod $$B$$:

$$ \Delta V_B = \beta V_B \Delta T = \beta (2 \pi r^2 l) \Delta T = 2 \beta \pi r^2 l \Delta T $$

To find the ratio of the increase in the volume of rod $$A$$ to that of rod $$B$$, we take:

$$ \frac{\Delta V_A}{\Delta V_B} = \frac{4 \beta \pi r^2 l \Delta T}{2 \beta \pi r^2 l \Delta T} = \frac{4}{2} = 2 $$

Thus, the ratio of the increase in volume of rod $$A$$ to that of rod $$B$$ is: 2:1.

Therefore, the correct option is Option C: 2:1.

Heat And Thermodynamics

Continue this paper