1. Input electrical power

The motor operates at $V = 100\ \text{V}, \qquad I = 1\ \text{A}$

So the input power is $P_{\text{in}} = VI = 100 \times 1 = 100\ \text{W}$

2. Efficiency relation

Efficiency is $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = 91.6\% = 0.916$

Hence output power is $P_{\text{out}} = 0.916 \times 100 = 91.6\ \text{W}$

3. Power loss

Loss of power is $P_{\text{loss}} = P_{\text{in}} - P_{\text{out}} = 100 - 91.6 = 8.4\ \text{W}$

Since $1\ \text{W} = 1\ \text{J/s}$ we have $P_{\text{loss}} = 8.4\ \text{J/s}$

4. Convert into cal/s

Using $1\ \text{cal} \approx 4.2\ \text{J}$

Therefore, $P_{\text{loss}} = \frac{8.4}{4.2} = 2\ \text{cal/s}$

5. Check options

• A: $6.2$ ❌
• B: $2$ ✅
• C: $8.4$ ❌ (this is in W or J/s, not cal/s)
• D: $4$ ❌

Therefore, the correct answer is Option B.