1. Find the resistance of each appliance at rated voltage

Given supply voltage $V=120\text{ V}$.

For the $60\text{ W}$ bulb: $R_b=\frac{V^2}{P}=\frac{120^2}{60}=240\,\Omega$

For the $240\text{ W}$ heater: $R_h=\frac{V^2}{P}=\frac{120^2}{240}=60\,\Omega$

The resistance of lead wires is: $R_w=6\,\Omega$

2. Voltage across bulb before heater is switched on

Initially, only the bulb is on, so total resistance is: $R_{\text{total,1}}=R_w+R_b=6+240=246\,\Omega$

Current in the circuit: $I_1=\frac{120}{246}=\frac{20}{41}\approx 0.4878\text{ A}$

Voltage drop in the lead wires: $V_{w1}=I_1R_w=\frac{20}{41}\times 6=\frac{120}{41}\approx 2.93\text{ V}$

So voltage across the bulb initially is: $V_{b1}=120-2.93\approx 117.07\text{ V}$

3. Equivalent resistance when heater is connected in parallel with bulb

Parallel combination of bulb and heater: $R_p=\frac{R_bR_h}{R_b+R_h}=\frac{240\times 60}{240+60}=\frac{14400}{300}=48\,\Omega$

Now total resistance becomes: $R_{\text{total,2}}=R_w+R_p=6+48=54\,\Omega$

Total current: $I_2=\frac{120}{54}=\frac{20}{9}\approx 2.222\text{ A}$

Voltage drop in the lead wires now: $V_{w2}=I_2R_w=\frac{20}{9}\times 6=\frac{120}{9}=13.33\text{ V}$

Hence voltage across the parallel combination, and therefore across the bulb, is: $V_{b2}=120-13.33=106.67\text{ V}$

4. Decrease in voltage across bulb

$\Delta V=V_{b1}-V_{b2}=117.07-106.67=10.40\text{ V}$

So the decrease is approximately: $\boxed{10.4\text{ V}}$

Among the given options, this corresponds to Option D (listed as $10.04\text{ V}$, likely a typographical/rounding issue).

5. Option check

• A: zero $\to$ incorrect, voltage does decrease due to extra current through lead wires.
• B: $2.9$ V $\to$ this is the initial wire drop, not the decrease.
• C: $13.3$ V $\to$ this is the new wire drop, not the decrease.
• D: $10.04$ V $\to$ intended correct option; computed decrease is about $10.4$ V.