JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A motor operating on 100 V draws a current of 1 A. If the efficiency of the motor is \(91.6 \%\), then the loss of power in units of \(\mathrm{cal} / \mathrm{s}\) is
- A 43.8 kJ
- B 48.7 kJ
- C 87.7 kJ
- D 24.4 kJ
Answer & Solution
Correct Answer
(C) 87.7 kJ
Step-by-step Solution
Detailed explanation
Given \(\mathrm{V}=4.2\) volt \(\therefore\) Energy supplied by battery \(=\mathrm{vq}=4.2 \times 5800 \times 3600 \times 10^{-3} \mathrm{~J}=87.696 \mathrm{~kJ}\) \(\therefore\) Energy stored in the battery when fully charged \(=87.696 \mathrm{~kJ} \approx 87.7 \mathrm{~kJ}\)
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