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

Options

A. `8.85 gs^(-1)`
B. `1.92 gs^(-1)`
C. `6.56 gs^(-1)`
D. none of these

Answer

C

Explanation

The rate of heat developed is given by
`(V^(2))/( R) = ((210)^(2))/(20) = 2205 Js^(-1) = ( 2205 )/( 4.2) cals^(-1)`
Let `m` be the amount of ice melting per sec and , then
` m xx 80 = (2205) /(4.2) or m = 6.56 gs^(-1)`