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

  • Option 1)

    0.85g/s

  • Option 2)

    1.92g/s

  • Option 3)

    6.56g/s

  • Option 4)

    none of these

 

Answers (1)

As we learnt in 

Power dissipiated in external resistance -

P=(\frac{E}{R+r})^{2}R

-

 

 Rate of energy produces=\frac{V^{2}}{R}= \frac{210 \times 210}{20}= 2205 W

Energy required to melt one gm of ice = mL = 1 \times  80 \times 4.2 J = 336J/g

\therefore     Rate of ice melting = \frac{2205}{336}g/sec=6.56\ g/s


Option 1)

0.85g/s

Incorrect

Option 2)

1.92g/s

Incorrect

Option 3)

6.56g/s

Correct

Option 4)

none of these

Incorrect

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