A coil of wire of resistance 50 $Ω$ is embedded in a block of ice. If a potential difference of 210 V is applied across the coil, the amount of ice melted per second will be (Latent heat of fusion of ice = 80 cal/gm)

4.12 gm

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4.12 kg

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3.68 kg

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2.625 gm

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Solution

The correct option is D 2.625 gm
$\frac{Q}{t} = \frac{V^{2}}{4.2 R} = \frac{m}{t} L; \frac{m}{t} = \frac{V^{2}}{4.2 R L} \frac{(210)^{2}}{4.2 \times 50 \times 80} = 2.625 g m$

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A coil of wire of resistance 50 Ω is embedded in a block of ice. If a potential difference of 210 V is applied across the coil, the amount of ice melted per second will be (Latent heat of fusion of ice = 80 cal/gm)

The correct option is D 2.625 gmQt=V24.2R=mtL; mt=V24.2RL (210)24.2×50×80=2.625gm

A coil of wire of resistance 50 $Ω$ is embedded in a block of ice. If a potential difference of 210 V is applied across the coil, the amount of ice melted per second will be (Latent heat of fusion of ice = 80 cal/gm)

The correct option is D 2.625 gm
$\frac{Q}{t} = \frac{V^{2}}{4.2 R} = \frac{m}{t} L; \frac{m}{t} = \frac{V^{2}}{4.2 R L} \frac{(210)^{2}}{4.2 \times 50 \times 80} = 2.625 g m$