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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Heat Transfer

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer

Three rods, each of length 20 cm and cross-sectional area 1 cm^2 ,...

Three rods, each of length 20 cm20 \text{ cm} and cross-sectional area 1 cm21 \text{ cm}^2, are joined together to form a triangle ABCABC. The thermal conductivities of these rods are K_{AB} = 50 \text{ J s}^{-1} \text{ m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}, K_{BC} = 200 \text{ J s}^{-1} \text{ m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}, and K_{AC} = 400 \text{ J s}^{-1} \text{ m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}. The junctions AA, BB, and CC are maintained at temperatures of 40^{}^{{}^{\circ}}\text{C}, 80^{}^{{}^{\circ}}\text{C}, and 80^{}^{{}^{\circ}}\text{C} respectively. Determine the rate of heat flowing through the rods ABAB, ACAC, and BCBC.

Options

  1. A0.1 W0.1 \text{ W}, 0.8 W0.8 \text{ W}, 00
  2. B100 W100 \text{ W}, 800 W800 \text{ W}, 00
  3. C1 W1 \text{ W}, 8 W8 \text{ W}, 00
  4. D2 W2 \text{ W}, 4 W4 \text{ W}, 00

Correct answer

C. 1 W1 \text{ W}, 8 W8 \text{ W}, 00

Step-by-step solution

The rate of heat flow through a rod is given by H=KAΔTLH = \dfrac{KA\Delta T}{L}.

Given:
Length of each rod, L=20 cm=0.2 mL = 20 \text{ cm} = 0.2 \text{ m}
Cross-sectional area, A=1 cm2=10−4 m2A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2
Temperatures at junctions: T_A = 40^{}^{{}^{\circ}}\text{C}, T_B = 80^{}^{{}^{\circ}}\text{C}, T_C = 80^{}^{{}^{\circ}}\text{C}

For rod ABAB:
\Delta T_{AB} = T_B - T_A = 80^{}^{{}^{\circ}}\text{C} - 40^{}^{{}^{\circ}}\text{C} = 40^{}^{{}^{\circ}}\text{C}
HAB=KABAΔTABL=50×10−4×400.2=2000×10−40.2=1 WH_{AB} = \dfrac{K_{AB} A \Delta T_{AB}}{L} = \dfrac{50 \times 10^{-4} \times 40}{0.2} = \dfrac{2000 \times 10^{-4}}{0.2} = 1 \text{ W}

For rod ACAC:
\Delta T_{AC} = T_C - T_A = 80^{}^{{}^{\circ}}\text{C} - 40^{}^{{}^{\circ}}\text{C} = 40^{}^{{}^{\circ}}\text{C}
HAC=KACAΔTACL=400×10−4×400.2=16000×10−40.2=8 WH_{AC} = \dfrac{K_{AC} A \Delta T_{AC}}{L} = \dfrac{400 \times 10^{-4} \times 40}{0.2} = \dfrac{16000 \times 10^{-4}}{0.2} = 8 \text{ W}

For rod BCBC:
\Delta T_{BC} = T_B - T_C = 80^{}^{{}^{\circ}}\text{C} - 80^{}^{{}^{\circ}}\text{C} = 0^{}^{{}^{\circ}}\text{C}
HBC=KBCAΔTBCL=0 WH_{BC} = \dfrac{K_{BC} A \Delta T_{BC}}{L} = 0 \text{ W}

The rates of heat flow through the rods ABAB, ACAC, and BCBC are 1 W1 \text{ W}, 8 W8 \text{ W}, and 00 respectively.
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