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JEE Main 2021 (1 Sep, Shift 2) · Electrical Power & Heating Effect

Electrical Power & Heating Effect question from JEE Main 2021 (1 Sep, Shift 2)

Q37JEE Main 2021 · 1 SEP · SHIFT 2Difficult

Electrical Power & Heating Effect

Due to cold weather a 1 m water pipe of cross-sectional area 1 cm2 is filled with ice at −10∘-10^\circC. Resistive heating is used to melt the ice. Current of 0.5A is passed through 4 kΩ\Omega resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice = 3.33 ×\times 105 J kg−1{}^{-1}, specific heat of ice = 2 ×\times 103 J kg−1{}^{-1} and density of ice = 103 kg/m3)

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Answer: (b)

  1. Mass of ice m=ρAl=103×10−4×1=0.1m = \rho Al = 10^3 \times 10^{-4} \times 1 = 0.1 kg.
  2. Heat =mcΔT+mL=0.1(2×103)(10)+0.1(3.33×105)=2000+33300=35300= mc\Delta T + mL = 0.1(2 \times 10^3)(10) + 0.1(3.33 \times 10^5) = 2000 + 33300 = 35300 J.
  3. Power P=I2R=(0.5)2×4000=1000P = I^2R = (0.5)^2 \times 4000 = 1000 W.
  4. t=353001000=35.3t = \frac{35300}{1000} = 35.3 s. So the answer is (b).
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