Temperature difference of 120oC is maintained between ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length 3L/2, is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to: (a) 45oC (b) 75oC (c) 60oC (d) 35oC

Answer: (a). ABL/2L/2L3L/2PQ Same material and area, so resistance ∝ length. AP = QB = L/2; between P and Q: straight part L in parallel with bent rod 3L/2. RPQ = (L·3L/2)/(L+3L/2) = 3L/5 (in units of 1/KA) Total = L/2 + 3L/5 + L/2 = 8L/5 Δ TPQ = 120×(3/5)/(8/5) = 45° C So the answer is (a) 45° C.

Temperature difference of 120oC is maintained between ends of a… | JEE Main 2019 (9 Jan, Shift 1)

Q47JEE Main 2019 · 9 JAN · SHIFT 1Difficult

Heat Conduction

Temperature difference of 120oC is maintained between ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length $\frac{3L}{2},$ is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to:

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Answer: (a)

Same material and area, so resistance $\propto$ length. AP $= QB = L/2$ ; between P and Q: straight part $L$ in parallel with bent rod $3L/2$ .
$R_{PQ} = \dfrac{L\cdot\frac{3L}{2}}{L+\frac{3L}{2}} = \dfrac{3L}{5}$ (in units of $\frac{1}{KA}$ )
Total $= \dfrac{L}{2} + \dfrac{3L}{5} + \dfrac{L}{2} = \dfrac{8L}{5}$
$\Delta T_{PQ} = 120\times\dfrac{3/5}{8/5} = 45^\circ$ C
So the answer is (a) $45^\circ$ C.

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Heat Conduction question from JEE Main 2019 (9 Jan, Shift 1)

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