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 cal/s is

The motor is supplied with a voltage $$V = 100\,\text{V}$$ and draws a current $$I = 1\,\text{A}$$, so the electrical power input is

$$P_{\text{in}} = V I = 100 \times 1 = 100\,\text{W}$$

The efficiency of the motor is $$\eta = 91.6\% = 0.916$$. Efficiency is defined as

$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$$

Hence the useful (output) power is

$$P_{\text{out}} = \eta P_{\text{in}} = 0.916 \times 100 = 91.6\,\text{W}$$

The power lost as heat (and other losses) is therefore

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

To express this loss in calories per second, use the relation $$1\,\text{cal} = 4.186\,\text{J}$$, so $$1\,\text{cal/s} = 4.186\,\text{W}$$. Converting:

$$P_{\text{loss}} = \frac{8.4\,\text{W}}{4.186\,\text{W per cal/s}} \approx 2.0\,\text{cal/s}$$

Therefore, the power loss is approximately $$2\,\text{cal/s}$$. This corresponds to Option C.