thermal conductivity works in a cylindrical shape.

The heat flow rate through a cylindrical rod is given by Fourier's law of heat conduction:

$$ Q = \frac{KA(T_1 - T_2)}{L} $$

where:

$$Q$$ is the rate of heat transfer
$$K$$ is the thermal conductivity of the material
$$A$$ is the cross-sectional area of the rod
$$T_1$$ and $$T_2$$ are the temperatures at the ends of the rod
$$L$$ is the length of the rod

The cross-sectional area $$A$$ of a cylindrical rod with radius $$r$$ is given by:

$$ A = \pi r^2 $$

Now, suppose the initial length and radius of the rod are $$L$$ and $$r$$, and this gives us the initial heat flow rate $$Q_1$$:

$$ Q_1 = \frac{K \pi r^2 (T_1 - T_2)}{L} $$

If both the length and radius of the rod are doubled, the new length becomes $$2L$$ and the new radius becomes $$2r$$. Hence, the new cross-sectional area $$A'$$ of the rod becomes:

$$ A' = \pi (2r)^2 = 4\pi r^2 $$

Substituting these into the heat flow equation, we get the new heat flow rate $$Q_2$$:

$$ Q_2 = \frac{K A' (T_1 - T_2)}{2L} $$

$$ Q_2 = \frac{K (4\pi r^2) (T_1 - T_2)}{2L} $$

$$ Q_2 = 2 \frac{K \pi r^2 (T_1 - T_2)}{L} $$

$$ Q_2 = 2 Q_1 $$

Thus, the rate of flow of heat $$Q_2$$ when the length and radius of the rod are doubled is:

$$\mathrm{Q}_2 = 2 \mathrm{Q}_1$$

Therefore, the correct answer is:

Option D: $$\mathrm{Q}_2 = 2 \mathrm{Q}_1$$

Heat And Thermodynamics

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