Q197 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
The thermal conduction rate '$P_{cond}$' of a slab is given by $P_{cond} = \dfrac{kA(T_1 - T_2)}{x}$, where ' $x$ ' is the thickness of the slab and 'A' is the area . The term that represents thermal resistance ' $R_T$'
- A$\dfrac{kA}{x}$
- Correct answerB$\dfrac{x}{kA}$
- C$\dfrac{k(T_1-T_2)}{x}$
- D$\dfrac{x}{k(T_1-T_2)}$
Solution
The rate of heat flow (thermal conduction rate) is given by $P_{cond} = \dfrac{kA(T_1 - T_2)}{x}$.
This can be rewritten as $P_{cond} = \dfrac{T_1 - T_2}{\dfrac{x}{kA}}$.
Comparing this with the electrical analogy $I = \dfrac{\Delta V}{R}$, where $I$ is the current, $\Delta V$ is the potential difference, and $R$ is the electrical resistance.
Here, $P_{cond}$ is analogous to current $I$, and the temperature difference $(T_1 - T_2)$ is analogous to potential difference $\Delta V$.
Thus, the thermal resistance $R_T$ is analogous to electrical resistance $R$.
Therefore, $R_T = \dfrac{x}{kA}$.
Answer: $\dfrac{x}{kA}$
This can be rewritten as $P_{cond} = \dfrac{T_1 - T_2}{\dfrac{x}{kA}}$.
Comparing this with the electrical analogy $I = \dfrac{\Delta V}{R}$, where $I$ is the current, $\Delta V$ is the potential difference, and $R$ is the electrical resistance.
Here, $P_{cond}$ is analogous to current $I$, and the temperature difference $(T_1 - T_2)$ is analogous to potential difference $\Delta V$.
Thus, the thermal resistance $R_T$ is analogous to electrical resistance $R$.
Therefore, $R_T = \dfrac{x}{kA}$.
Answer: $\dfrac{x}{kA}$
Hence, the correct answer is option: 'B'.