A coil of wire of resistance `50 Omega` 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

A coil of water of resistance 50 Omega is embedded in a block of ice and a potential difference of 210 V is applied across it. The amount of ice which melts in 1 second is [ latent heat of fusion of ice = 80 cal g^(-1) ]

A coil of water of resistance 50 Omega is embedded in a block of ice and a potential difference of 210 V is applied across it. The amount of ice which melts in 1 second is [ latent heat of fusion of ice = 80 cal g^(-1) ]

A coil of enamelled copper wire of resistance 50Omega is embedded in a block of ice and a potential difference of 210V applied across it. Calculate the rate of which ice melts. Latent heat of ice is 80 cal per gram.

A coil of enamelled copper wire of resistance 50Omega is embedded in a block of ice and a potential difference of 210V applied across it. Calculate the rate of which ice melts. Latent heat of ice is 80 cal per gram.

A coil of enamelled copper wire of resistance 50 Omega is embedded in a block of ice and a potential difference of 210 V applied across it. Calculate, how much ice will melt in half minute. Latent heat of ice is 180 cal per gram.

If a wire of resistance 20 Omega is covered with ice and a voltage of 210 V is applied across the wire , then the rate of melting of ice is

If a wire of resistance 20 Omega is covered with ice and a voltage of 210 V is applied across the wire , then the rate of melting of ice is