Question

A cylindrical rod has temperatures '$T_1$' and '$T_2$' at its ends. The rate of flow of heat is '$Q_1$' $\text{cal s}^{-1}$. If length and radius of the rod are doubled keeping temperature constant, then the rate of flow of heat '$Q_2$' will be

• A. $Q_2 = \frac{Q_1}{2}$
• B. $Q_2 = \frac{Q_1}{4}$
• C. $Q_2 = 4Q_1$
• D. $Q_2 = 2Q_1$

Hint:

Since thermal resistance is defined as $R_{\text{th}} = \frac{L}{KA}$, doubling the length increases resistance by $2\times$, but doubling the radius increases the area by $4\times$, which cuts resistance by $4\times$. The net effect is that the total resistance is halved ($\frac{2}{4} = \frac{1}{2}$), meaning the heat flow current must double ($2\times$).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks how the rate of heat conduction through a cylindrical conductor changes when both its total length and its circular radius are simultaneously scaled up by a factor of 2, while maintaining the same temperature difference across its ends.

Step 2: Key Formula or Approach:
The steady-state rate of heat flow conduction ($Q$) through a material is given by Fourier's Law:
$$Q = \frac{KA(T_1 - T_2)}{L}$$ where $K$ is the thermal conductivity, $A$ is the cross-sectional area, $L$ is the length, and $(T_1 - T_2)$ is the temperature difference. For a cylinder, the cross-sectional area is $A = \pi r^2$. Substituting this into the main equation gives:
$$Q = \frac{K(\pi r^2)\Delta T}{L} \implies Q \propto \frac{r^2}{L}$$

Step 3: Detailed Explanation:
Let the initial heat flow rate be:
$$Q_1 \propto \frac{r^2}{L}$$ For the modified rod, the new parameters are $r' = 2r$ and $L' = 2L$. Let's set up the expression for the new heat flow rate $Q_2$:
$$Q_2 \propto \frac{(2r)^2}{2L}$$ $$Q_2 \propto \frac{4r^2}{2L} = 2 \cdot \left(\frac{r^2}{L}\right)$$ Taking the ratio of $Q_2$ to $Q_1$:
$$\frac{Q_2}{Q_1} = \frac{2 \cdot \left(\frac{r^2}{L}\right)}{\frac{r^2}{L}} = 2 \implies Q_2 = 2Q_1$$ This shows that the rate of heat flow doubles.

Step 4: Final Answer:
The new rate of heat flow is $Q_2 = 2Q_1$, which corresponds to option (D).