Q128 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
A block of metal A is connected in series with another block of metal B such that the two metal blocks have the same area of cross sections. The thermal conductivity of metal A is K and the free end of metal A is at $80^{\circ} \mathrm{C}$. The temperature of the interface is $60^{\circ} \mathrm{C}$ and the free end of metal B is at $20^{\circ} \mathrm{C}$. Assuming the two metals have the same thickness, the conductivity of metal B is:
- A$2 K$
- Correct answerB$\dfrac{K}{2}$
- C$\dfrac{K}{4}$
- D$4 K$
Solution
In a steady state, the rate of heat flow through two blocks connected in series is the same. Let $A$ be the cross-sectional area and $L$ be the thickness of each block. The rate of heat flow $H$ is given by $H = \dfrac{KA(T_1 - T_2)}{L}$.
For metal A, the temperature difference is $80^{\circ} \mathrm{C} - 60^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C}$. Thus, $H_A = \dfrac{KA(80 - 60)}{L} = \dfrac{20KA}{L}$.
For metal B, let $K_B$ be its thermal conductivity. The temperature difference is $60^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 40^{\circ} \mathrm{C}$. Thus, $H_B = \dfrac{K_B A(60 - 20)}{L} = \dfrac{40K_B A}{L}$.
Since the blocks are in series, $H_A = H_B$. Therefore, $\dfrac{20KA}{L} = \dfrac{40K_B A}{L}$.
Simplifying the equation, $20K = 40K_B$, which gives $K_B = \dfrac{20K}{40} = \dfrac{K}{2}$.
Answer: $\dfrac{K}{2}$
For metal A, the temperature difference is $80^{\circ} \mathrm{C} - 60^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C}$. Thus, $H_A = \dfrac{KA(80 - 60)}{L} = \dfrac{20KA}{L}$.
For metal B, let $K_B$ be its thermal conductivity. The temperature difference is $60^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 40^{\circ} \mathrm{C}$. Thus, $H_B = \dfrac{K_B A(60 - 20)}{L} = \dfrac{40K_B A}{L}$.
Since the blocks are in series, $H_A = H_B$. Therefore, $\dfrac{20KA}{L} = \dfrac{40K_B A}{L}$.
Simplifying the equation, $20K = 40K_B$, which gives $K_B = \dfrac{20K}{40} = \dfrac{K}{2}$.
Answer: $\dfrac{K}{2}$
Hence, the correct answer is option: 'B'.