An electric iron of resistance 20 Q. takes a current f 5 A. The heat developed in joules in 30s is

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Q: 75 (NDA-II/2009)
An electric iron of resistance 20 Q. takes a current f 5 A. The heat developed in joules in 30s is

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,3,9,3,4,3,2

keywords: 

{'joules': [0, 0, 1, 1], 'electric iron': [0, 0, 1, 1], 'resistance': [0, 0, 1, 2], 'heat': [10, 3, 13, 46], '30s': [0, 0, 1, 0], 'kj': [0, 1, 1, 1]}

Given that the electric iron has a resistance of 20 Ω and takes a current of 5 A.

To calculate the heat developed, we can use the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance.

Plugging in the values, we get P = (5 A)^2 * 20 Ω = 100 J/s = 100 W.

Since power is the rate at which energy is transferred or converted, we can find the total heat developed by multiplying the power by the time.

Heat developed = Power * Time = 100 W * 30 s = 3000 J = 3 kJ.

Therefore, the correct answer is option 3, which is 15 kJ.

The heat developed in 30 seconds is 15 kJ.