A wire of resistance 20Ω is covered with ice and a voltage of 210V is applied across the wire, then rate of melting the ice is
A
0.85 g/s
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B
1.92 g/s
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C
6.56 g/s
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D
All of these
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Solution
The correct option is D 6.56 g/s Here, H=(V2/R)t Also, H=ml where l=80cal=(80×4.2)J, latent heat of ice. So, ml=(V2/R)t Thus, the rate of melting the ice is, m/t=V2Rl=(210)220×(80×4.2)=6.56gm/s