Filament Resistance Relation of Bulbs at Room Temperature

To determine the relationship between the filament resistances at room temperature for the given bulbs, we analyze their rated powers and temperature dependence of resistance.

1. Power Rating and Operating Resistance:
   The rated power $P$ of an incandescent bulb designed to operate at a standard mains voltage $V$ is given by:
   $P = \frac{V^2}{R_{\text{hot}}}$
   where $R_{\text{hot}}$ is the resistance of the bulb's filament at its operating (hot) temperature.

Rearranging this expression for $R_{\text{hot}}$ gives:
$R_{\text{hot}} = \frac{V^2}{P}$

For the three bulbs rated at $100\text{ W}$, $60\text{ W}$, and $40\text{ W}$ operating at the same rated voltage $V$:
$R_{\text{hot}, 100} = \frac{V^2}{100}$
$R_{\text{hot}, 60} = \frac{V^2}{60}$
$R_{\text{hot}, 40} = \frac{V^2}{40}$

Since $100\text{ W} > 60\text{ W} > 40\text{ W}$, we have:
$R_{\text{hot}, 100} < R_{\text{hot}, 60} < R_{\text{hot}, 40}$

1. Resistance at Room Temperature:
   The resistance of metallic filaments increases with temperature according to:
   $R_{\text{hot}} = R_{\text{room}} (1 + \alpha \Delta T)$
   where $\alpha$ is the temperature coefficient of resistance and $\Delta T$ is the temperature difference between operating temperature and room temperature.

Since the operational design factors are directly proportional, a bulb with a smaller operating resistance $R_{\text{hot}}$ also possesses a smaller resistance $R$ at room temperature. Thus, the room temperature resistances $R_{100}$, $R_{60}$, and $R_{40}$ satisfy:
$R_{100} < R_{60} < R_{40}$

1. Reciprocal Relation:
   Taking the reciprocal of each resistance reverses the order of the inequalities:
   $\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$

Hence, the correct option is D.