a
Power of bulb \(=60\, \mathrm{W}\) (given)

Resistance of bulb \(=\frac{120 \times 120}{60}=240\, \Omega\left[\because \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\right]\)

Power of heater \(=240\, \mathrm{W}\) (given)

Resistance of heater \(=\frac{120 \times 120}{240}=60\, \Omega\)

Voltage across bulb before heater is switched on,

\(\mathrm{V}_{1}=\frac{240}{246} \times 120=117.73\, \text { volt }\)

Voltage across bulb after heater is switched on,

\(\mathrm{V}_{2}=\frac{48}{54} \times 120=106.66\, \mathrm{volt}\)

Hence decrease in voltage

\(\mathrm{V}_{1}-\mathrm{V}_{2}=117.073-106.66=10.04\, \mathrm{Volt}\) (approximately)