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A temperature difference of 120 °C is maintained between the two ends of a uniform straight rod AB of length 2L. Another rod PQ of the same material and the same cross-section, bent into a shape and of total length (3L)/2, is soldered across the rod AB at the two points P and Q, where AP=L/4 and PQ=L measured along AB. In the steady state the temperature difference between P and Q will be close to

Asked in JEE Main 9th Jan 1st Shift 2019 · Series and parallel resistance

• (1) 35 °C
• (2) 45 °C
• (3) 60 °C
• (4) 75 °C

Step-by-step solution

Why the other options are wrong

• (1) comes from no clean slip lands here; it sits just below the key as a near miss. The two obvious wrong routes are checkable: ignoring the bent rod entirely gives 120×L/(2L)=60 °C, and treating the bent rod as a short circuit gives 0.
• (3) comes from forgetting the bent rod and treating AB as a plain uniform rod: the stretch PQ is then half of 2L, so it takes half of 120 °C. Soldering a second path across PQ lowers the resistance of that stretch and so lowers its share of the drop.
• (4) comes from adding the two parallel lengths as if they were in series, L+(3L)/2=(5L)/2, and giving PQ that share of the total; parallel paths add conductance, not length.

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