JEE Main · 2013 · 07 Apr 2013 · Offline · Current Electricity

Both appliances are rated at the supply voltage, so their resistances come straight from $R=V^2/P$: the bulb is $R_b=\frac{120^2}{60}=240\Omega$ and the heater $R_h=\frac{120^2}{240}=60\Omega$. With only the bulb glowing it sits in series with the $6\Omega$ lead wires, so the voltage across it is $V_1=120\times \frac{240}{246}\approx 117.1\text{V}$. Switch the heater in parallel with the bulb and that pair combines to $\frac{240\times 60}{300}=48\Omega$; the total becomes $6+48=54\Omega$ and the bulb voltage falls to $V_2=120\times \frac{48}{54}\approx 107.0\text{V}$. The decrease across the bulb is therefore close to $10\text{V}$.