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NEET
Exam
NEET
Subject
Physics
Chapter
Thermal Properties of Matter

NEET · Physics · Thermal Properties of Matter

Three heat-conducting rods of identical length and uniform...

Three heat-conducting rods of identical length and uniform cross-sectional area are joined at a common point to form a Y-shaped junction. The thermal conductivities of the three rods are KK, 2K2K, and 3K3K. The free ends of the rods with conductivities KK, 2K2K, and 3K3K are maintained at steady temperatures of 100∘C100^{\circ}\text{C}, 40∘C40^{\circ}\text{C}, and 0∘C0^{\circ}\text{C} respectively. Assuming no heat loss from the lateral surfaces, the steady-state temperature of the central junction is:

Options

  1. A36∘C36^{\circ}\text{C}
  2. B30∘C30^{\circ}\text{C}
  3. C47∘C47^{\circ}\text{C}
  4. D60∘C60^{\circ}\text{C}

Correct answer

B. 30∘C30^{\circ}\text{C}

Step-by-step solution

Let the steady-state temperature of the central junction be TJT_J.

According to Kirchhoff's junction rule for heat (conservation of energy), the net rate of heat flow into the junction must be zero in steady state.
ΣH=0\Sigma H = 0
H1+H2+H3=0H_1 + H_2 + H_3 = 0

Using the formula for heat current H=KA(Tend−TJ)LH = \dfrac{KA(T_{\text{end}} - T_J)}{L}, we can write:
KA(100−TJ)L+2KA(40−TJ)L+3KA(0−TJ)L=0\dfrac{K A (100 - T_J)}{L} + \dfrac{2K A (40 - T_J)}{L} + \dfrac{3K A (0 - T_J)}{L} = 0

Since AA and LL are identical for all rods, we can divide the entire equation by KAL\dfrac{KA}{L}:
(100−TJ)+2(40−TJ)+3(0−TJ)=0(100 - T_J) + 2(40 - T_J) + 3(0 - T_J) = 0

Expanding the terms:
100 - TJT_J + 80 - 2TJ2T_J - 3TJ3T_J = 0
180 - 6TJ6T_J = 0
6TJ6T_J = 180
TJ=30∘CT_J = 30^{\circ}\text{C}

Answer: 30∘C30^{\circ}\text{C}
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