The temperature difference of 120oC is maintained between two ends of a uniform rod AB of length 2L- Another bent rod PQ- of same cross-section as AB and length 3L2- is connected across AB see figure- In steady state- the temperature difference between P and Q will be close to-

The correct option is D 45^oC Δ TR_eq. =I = (120)58R =120× 58R Δ T_PQ = 120× 58R×35 R = 3608 = 45^oC ...

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Standard XII

Mathematics

Section Formula

The temperatu...

Question

The temperature difference of $120^{o} C$ is maintained between two ends of a uniform rod $A B$ of length $2 L$ . Another bent rod $P Q$ , of same cross-section as $A B$ and length $\frac{3 L}{2}$ , is connected across $A B$ (see figure). In steady state, the temperature difference between $P$ and $Q$ will be close to:

60^{o} C

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75^{o} C

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35^{o} C

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45^{o} C

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Solution

The correct option is D $45^{o} C$
$\frac{Δ T}{R_{e q .}} = I = \frac{(120) 5}{8 R} = \frac{120 \times 5}{8 R}$
$Δ T_{P Q} = \frac{120 \times 5}{8 R} \times \frac{3}{5} R = \frac{360}{8} = 45^{o} C$

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The temperature difference of 120oC is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length 3L2, is connected across AB (see figure). In steady state, the temperature difference between P and Q will be close to:

The correct option is D 45oCΔTReq.=I=(120)58R=120×58RΔTPQ=120×58R×35R=3608=45oC

The correct option is D $45^{o} C$
$\frac{Δ T}{R_{e q .}} = I = \frac{(120) 5}{8 R} = \frac{120 \times 5}{8 R}$
$Δ T_{P Q} = \frac{120 \times 5}{8 R} \times \frac{3}{5} R = \frac{360}{8} = 45^{o} C$