JEE Main 2025 — Current Electricity Question with Solution

The motor operates at 100 V and draws 1 A, so the total electrical power input is:

$$P_{\text{in}}=IV=100\text{V}\times 1\text{A}=100\text{W}.$$

With an efficiency of 91.6%, only 91.6% of the input power is used for the motor's work, while the rest is lost as heat. The output power is:

$$P_{\text{out}}=0.916\times 100\text{W}=91.6\text{W}.$$

The power lost is:

$$P_{\text{loss}}=P_{\text{in}}-P_{\text{out}}=100\text{W}-91.6\text{W}=8.4\text{W}.$$

Since $1\text{W}=1\text{J/s}$ and $1\text{cal}\approx 4.2\text{J}$, we convert the lost power into calories per second:

$$P_{\text{loss}}(\text{in cal/s})=\frac{8.4\text{J/s}}{4.2\text{J/cal}}=2\text{cal/s}.$$

Thus, the correct answer is:

Option B: 2