Q136 / 160 · TS EAMCET 2024 (10 May Shift 1) · Physics
The thickness of a uniform rectangular metal plate is 5 mm and the area of each surface is $5 \mathrm{~cm}^2$. In steady state, the temperature difference between the two surfaces of the plate is $14^{\circ} \mathrm{C}$. If the heat flowing through the plate in one second from one surface to the other surface is 42 J , then the thermal conductivity of the metal is
- A$90 \mathrm{Wm}^{-1} \mathrm{~K}^{-1}$
- Correct answerB$30 \mathrm{Wm}^{-1} \mathrm{~K}^{-1}$
- C$45 \mathrm{Wm}^{-1} \mathrm{~K}^{-1}$
- D$60 \mathrm{Wm}^{-1} \mathrm{~K}^{-1}$
Solution
$\Delta \mathrm{x}=5 \mathrm{~mm}, \mathrm{~A}=5 \mathrm{~cm}^2, \Delta \mathrm{~T}=14^{\circ} \mathrm{C}, \mathrm{Q}=42 \mathrm{~W}$
$\therefore$ At steady state,
$Q=\frac{\mathrm{KA} \Delta \mathrm{~T}}{\Delta \mathrm{x}} \Rightarrow \mathrm{~K}=\frac{\mathrm{Q} \Delta \mathrm{x}}{\mathrm{~A} \Delta \mathrm{~T}}$
$\therefore$ Thermal conductivity, $\mathrm{K}=\frac{42 \times 5 \times 10^{-3}}{5 \times 10^{-4} \times 14}$ $=30 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$
Hence, the correct answer is option: 'B'.