Q 11-01-048NEETJEE MainMedium
The potential difference across a resistor is $(50 \pm 1)\ \text{V}$ and its resistance is $(25 \pm 0.5)\ \Omega$. The power dissipated, $P = \dfrac{V^2}{R}$, is
Answer: (C) $(100 \pm 6)\ \text{W}$
$P = \dfrac{50^2}{25} = 100\ \text{W}$
$$\frac{\Delta P}{P}\times 100 = 2\,\frac{\Delta V}{V}\times 100 + \frac{\Delta R}{R}\times 100 = 2(2\%) + 2\% = 6\%$$
So $\Delta P = 6\%$ of $100\ \text{W} = 6\ \text{W}$, and $P = (100 \pm 6)\ \text{W}$.
Solution by Sreeraj P, M.Sc Physics