Temperature difference of 120°C 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 (3L/2) , is connected across AB (See figure). In steady state, temperature difference between P and Q will be close to :

Question

Temperature difference of 120°C 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 (3L/2) , is connected across AB (See figure). In steady state, temperature difference between P and Q will be close to : (A) 60° C (B) 75° C (C) 35° C (D) 45° C

Tardigrade Admin · Accepted Answer

(Δ T/Req) = I = ((120)5/8R) = (120×5/8R) Δ TPQ = (120 ×5/8R) ×(3/5)R = (360/8) = 45°C

$6 0^{\circ} C$

$7 5^{\circ} C$

$3 5^{\circ} C$

$4 5^{\circ} C$

$\frac{Δ T}{R _{e q}} = I = \frac{( 120 ) 5}{8 R} = \frac{120 \times 5}{8 R}$
$Δ T_{PQ} = \frac{120 \times 5}{8 R} \times \frac{3}{5} R = \frac{360}{8} = 4 5^{\circ} C$

Q. Temperature difference of 120∘C 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 23L​ , is connected across AB (See figure). In steady state, temperature difference between P and Q will be close to :

60∘C

75∘C

35∘C

45∘C

Req​ΔT​=I=8R(120)5​=8R120×5​ ΔTPQ​=8R120×5​×53​R=8360​=45∘C

$6 0^{\circ} C$

$7 5^{\circ} C$

$3 5^{\circ} C$

$4 5^{\circ} C$

$\frac{Δ T}{R _{e q}} = I = \frac{( 120 ) 5}{8 R} = \frac{120 \times 5}{8 R}$
$Δ T_{PQ} = \frac{120 \times 5}{8 R} \times \frac{3}{5} R = \frac{360}{8} = 4 5^{\circ} C$