JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Current Electricity

The colour code green-black-red gives significant figures $5$ and $0$ with a multiplier of $10^2$, so the resistance is $50\times 10^2=5000\Omega$, the brown band being only the tolerance. The $2$ W rating caps the power, and with $P=I^{2}R$ the largest safe current is $I=\sqrt{P/R}=\sqrt{2/5000}=\sqrt{4\times 10^{-4}}=0.02$ A, that is $20$ mA. Draw more and the dissipation runs past $2$ W and cooks the resistor.