A composite block is made of slabs A, B, C, D and E of different thermal conductivity (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat 'Q' flows only from left to right through the blocks. Then in steady state

Engineering

Physics

Conduction of Heat

Question

heat flow through slab E is maximum

heat flow through C = heat flow through B + heat flow through D.

heat flow through A and E slabs are same

temperature difference across slab E is smallest.

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All the three system shown are in series hence rate of heat flow will be same through both A & E.

$R_{A} = \frac{L}{8 (KA)}; R_{B} = \frac{4 L}{3 KA}; R_{C} = \frac{4 L}{8 KA}; R_{D} = \frac{4 L}{5 KA}; R_{E} = \frac{L}{24 KA}$

Using parallel combination rate of heat flow across C = rate of heat flow through B+ rate of heat flow through D.

${Δθ}_{A} = {HR}_{A} = \frac{HL}{8 KA}$

${Δθ}_{B} = \frac{3 H}{16} R_{B} = \frac{3 H}{16} (\frac{4 L}{3 KA}) = \frac{HL}{4 KA}$

${Δθ}_{C} = \frac{H}{2} (R_{C}) = \frac{H}{2} (\frac{4 L}{8 KA}) = \frac{HL}{4 KA}$

${Δθ}_{D} = \frac{5 H}{16} (\frac{4 L}{5 KA}) = \frac{HL}{4 KA}$

${Δθ}_{E} = H (\frac{L}{24 KA}) = \frac{HL}{24 KA}$