A circular ring (centre O) of radius a, and of uniform crosssection is made up of three different metallic rods AB, BC and CA (joined together at the points A, B and C in pairs) of thermal conductivities $α_{1}, α_{2} and α_{3}$ , respectively (see diagram). The junctions A, B and C are maintained at the temperatures 100°C, 50°C and 0°C, respectively. All the rods are of equal lengths and cross-sections. Under steady state conditions, assume that no heat is lost from the sides of the rods. Let Q₁, Q₂ and Q₃ be the rates of transmission of heat along the three rods AB, BC and CA. hen

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$Q_{1} = Q_{2} = Q_{3}$ and all are transmitted in the clockwise sense

$\frac{Q_{1}}{α_{1}} + \frac{Q_{2}}{α_{2}} = \frac{Q_{3}}{α_{3}}$

$Q_{1} and Q_{2}$ flow in clockwise sense and $Q_{3}$ in the anticlockwise sense

$Q_{1} : Q_{2} : Q_{3} :: α_{1} : α_{2} : 2 α_{3}$

answer is B, C, D.

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Detailed Solution

The flow of heat will always be in the direction of the temperature gradient from higher to lower temperature. Hence Q₁, in rod AB, Q₂ in rod BC will both be in clockwise sense while Q₃, in CA will be in anti-clockwise sense. Also, we have if L is the length of each rod and A its area of cross-section,

$\begin{array}{l} Q_{1} = \frac{α_{1} A (100 - 50)}{L} = (50 α_{1}) \frac{A}{L} \\ Q_{2} = \frac{α_{2} A (50 - 0)}{L} = (50 α_{2}) \frac{A}{L} \\ Q_{3} = \frac{α_{3} A (100 - 0)}{L} = (100 α_{3}) \frac{A}{L} \end{array}$