Q56 / 75 · JEE Main 2025 (3 Apr Shift 2) · Physics
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$4$
- B$8.4$
- Correct answerC$2$
- D$6.2$
Solution
$\begin{aligned} & P_{\text {input }}=V i=100 \mathrm{~W} \\ & \eta=\frac{P_{\text {out }}}{P_{\text {input }}}=0.916\end{aligned}$
$\begin{aligned} & P_{\text {out }}=91.6 \mathrm{~W} \\ & \text { Loss }=100-91.6=8.4 \mathrm{~J} / \mathrm{s}=2 \mathrm{cal} / \mathrm{s}\end{aligned}$
$\begin{aligned} & P_{\text {out }}=91.6 \mathrm{~W} \\ & \text { Loss }=100-91.6=8.4 \mathrm{~J} / \mathrm{s}=2 \mathrm{cal} / \mathrm{s}\end{aligned}$
Hence, the correct answer is option: 'C'.