A coil of wire of resistance $5 Ω$ is embedded in a block of ice. If a potential difference of $210 V$ is applied across the coil, then the amount of ice melted per second will be

2.625 g

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3.68 k g

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4.12 g

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4.12 k g

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Solution

The correct option is A $2.625 g$
Using the relation,
$\frac{θ}{y} = \frac{V^{2}}{4.2 R} = \frac{m}{t} . L$
$∴ \frac{m}{t} = \frac{V^{2}}{4.2 R L} = \frac{(210)^{2}}{4.2 \times 50 \times 80} = 2.625 g$ .

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A coil of wire of resistance 5Ω is embedded in a block of ice. If a potential difference of 210 V is applied across the coil, then the amount of ice melted per second will be

The correct option is A 2.625 gUsing the relation,θy=V24.2R=mt.L∴mt=V24.2RL=(210)24.2×50×80=2.625 g.

A coil of wire of resistance $5 Ω$ is embedded in a block of ice. If a potential difference of $210 V$ is applied across the coil, then the amount of ice melted per second will be

The correct option is A $2.625 g$
Using the relation,
$\frac{θ}{y} = \frac{V^{2}}{4.2 R} = \frac{m}{t} . L$
$∴ \frac{m}{t} = \frac{V^{2}}{4.2 R L} = \frac{(210)^{2}}{4.2 \times 50 \times 80} = 2.625 g$ .