The power developed across the resistor R is

\(P=\frac{V^2}{R}\), where V is the potential difference applied across R

If R1 and R2 are the resistances of P1 = 40 W and P2 = 60W bulbs respectively, then

\(P_1=\frac{V^2}{R_1}\), and

\(P_2=\frac{V^2}{R_2}\)

\(\Rightarrow\frac{P_1}{P_2}=\frac{V^2/R_1}{V^2/R_2}=\frac{R_2}{R_1}=\frac{\rho l/A_2}{\rho l/A_1}\)     \(\left[\because R=\rho\frac lA\right]\)

\(\therefore\frac{P_1}{P_2}=\frac{A_1}{A_2}=\frac{\pi d_1^2/4}{\pi d_2^2/4}=\frac{d_1^2}{d_2^2}\)

Since P1 < P2, then d1 < d2,

So, the 60-watt bulb has a thicker filament than the 40-watt bulb.