Q175 / 180 · KCET 2022 · Physics
If voltage across a bulb rated $220 \mathrm{~V}, 100 \mathrm{~W}$ drops by $2.5 \%$ of its rated value, then the percentage of the rated value by which the power would decrease is
- A$2.5 \%$
- Correct answerB$5 \%$
- C$10 \%$
- D$20 \%$
Solution
Given, $P=100 \mathrm{~W}$
$$
V=220 \mathrm{~V}
$$
We know that, $P=\frac{V^2}{R}$
where, $R$ is resistance of bulb.
$$
\Rightarrow \quad R=\frac{V^2}{P}=\frac{220 \times 220}{100}=484 \Omega
$$
Now, according to question, voltage drops by $2.5 \%$ of its rated value.
New voltage, $V^{\prime}=220-2.5 \%$ of 220
$$
=220-\frac{2.5}{100} \times 220=214.5 \mathrm{~V}
$$
$$
\begin{aligned}
& \therefore \text { New power, } P^{\prime}=\frac{\left(V^{\prime}\right)^2}{R}=\frac{214.5 \times 214.5}{484}=95.06 \mathrm{~W} \\
& \begin{aligned}
\% \text { decrease in power } & =\frac{P-P^{\prime}}{P} \times 100 \\
& =\frac{100-95.06}{100} \times 100 \sim 5 \%
\end{aligned}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.