Physics Q39 – JEE-MAIN 2025

Home Physics Current ElectricityQuestion #56

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

Clue 1 of 3: Electrical Power and Efficiency

Understand the relationship between input power, output power, and efficiency in an electrical motor.

🥷

Ninja Strategy•Unit Conversion Check

Always pay close attention to the required units in the final answer, especially when a conversion factor (like J to cal) is involved.

🧠 Full Solution Path

Step 1: Calculate Input Power✦ Active

The input power to the motor is calculated using the formula $P_{i n} = V \times I$ , where $V$ is the voltage and $I$ is the current.

P_{i n} = 100 V \times 1 A = 100 W

Step 2: Calculate Power Loss○ Expand

Efficiency ( $η$ ) is defined as the ratio of output power to input power, or $1 - \frac{Power Loss}{Input Power}$ . Therefore, the power loss can be found using the formula $P_{l o s s} = P_{i n} (1 - η)$ .

P_{l o s s} = 100 W (1 - 0.916) = 100 W \times 0.084 = 8.4 W

Step 3: Convert Units to cal/s○ Expand

To convert power loss from Watts to calories per second, we use the conversion factor $1 W = 1 J/s$ and $1 cal \approx 4.2 J$ .

P_{l o s s (cal/s)} = 8.4 J/s \times \frac{1 cal}{4.2 J} = \frac{8.4}{4.2} cal/s = 2 cal/s

💡 Teacher's Secret Hint

Remember to use the appropriate conversion factor for Joules to calories, typically 4.18 J/cal or 4.2 J/cal for approximations.

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Physics Question 39 – JEE-MAIN 2025

Clue 1 of 3: Electrical Power and Efficiency

🧠 Full Solution Path