Two electric bulbs A and B rated 200 V – 100 W and 200 V – 60 W are connected in series to a 200 V line. Then the potential drop across:

The power $P$ of a bulb is related to its rated voltage $V$ and resistance $R$ by the formula $P = \frac{V^2}{R}$. Therefore, $R = \frac{V^2}{P}$.

For the 100 W bulb (Bulb A):

$$R_A = \frac{(200\, \text{V})^2}{100\, \text{W}} = \frac{40000\, \text{V}^2}{100\, \text{W}} = 400\, \Omega$$

For the 60 W bulb (Bulb B):

$$R_B = \frac{(200\, \text{V})^2}{60\, \text{W}} = \frac{40000\, \text{V}^2}{60\, \text{W}} = \frac{2000}{3}\, \Omega \approx 666.67\, \Omega$$

When bulbs are connected in series, the total resistance is the sum of individual resistances.

$$R_{total} = R_A + R_B = 400\, \Omega + \frac{2000}{3}\, \Omega = \frac{1200}{3}\, \Omega + \frac{2000}{3}\, \Omega = \frac{3200}{3}\, \Omega$$

The total current $I$ in the series circuit is given by Ohm's law, $I = \frac{V_{line}}{R_{total}}$, where $V_{line}$ is the supply voltage (200 V).

$$I = \frac{200\, \text{V}}{\frac{3200}{3}\, \Omega} = \frac{200 \times 3}{3200}\, \text{A} = \frac{600}{3200}\, \text{A} = \frac{3}{16}\, \text{A}$$

The potential drop across each bulb is given by $V = I \cdot R$.

Potential drop across the 100 W bulb ($V_A$):

$$V_A = I \cdot R_A = \frac{3}{16}\, \text{A} \times 400\, \Omega = \frac{1200}{16}\, \text{V} = 75\, \text{V}$$

Potential drop across the 60 W bulb ($V_B$):

$$V_B = I \cdot R_B = \frac{3}{16}\, \text{A} \times \frac{2000}{3}\, \Omega = \frac{2000}{16}\, \text{V} = 125\, \text{V}$$

We found that $V_A = 75\, \text{V}$ and $V_B = 125\, \text{V}$.
This means the potential drop across the 100 W bulb ($75\, \text{V}$) is smaller than the potential drop across the 60 W bulb ($125\, \text{V}$).

Comparing this with the given options:
(A) Each bulb is 200 V - Incorrect.
(B) 100 W bulb is greater than that across 60 W bulb - Incorrect.
(C) 100 W bulb is smaller than that across 60 W bulb - Correct.
(D) Each bulb is 100 V - Incorrect.

The final answer is $\boxed{\text{C}}$