Step 1: Understanding the Concept:
The brightness of the bulb depends on the power dissipated by it, which in turn depends on the root-mean-square (RMS) current flowing through the circuit ($P \propto I_{\text{rms}}^2$).
To decrease the brightness, we need to decrease the current in the circuit.
In an AC circuit with an inductor and a resistor (bulb), the current is inversely proportional to the total impedance $Z$.
Step 2: Key Formula or Approach:
The impedance $Z$ of an LR series circuit is $Z = \sqrt{R^2 + X_L^2}$, where $X_L = \omega L = 2\pi f L$ is the inductive reactance.
The RMS current is $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$.
To decrease current (and thus brightness), we must increase the impedance $Z$, which can be done by increasing the inductive reactance $X_L$.
Step 3: Detailed Explanation:
Let's analyze each option to see its effect on $X_L$ or $Z$:
(A) Inserting an iron rod: Iron is a ferromagnetic material. Inserting it increases the magnetic permeability ($\mu$) of the core. Since inductance $L \propto \mu$, $L$ increases. An increase in $L$ causes an increase in $X_L$ ($\omega L$), which increases the total impedance $Z$. A higher $Z$ leads to a lower current, decreasing the bulb's brightness. This is the correct option.
(B) Decreasing frequency ($f$): Since $X_L = 2\pi f L$, decreasing $f$ decreases $X_L$. This decreases $Z$, increases current, and increases brightness.
(C) Reducing number of turns ($N$): Since $L \propto N^2$, reducing $N$ decreases $L$. This decreases $X_L$ and $Z$, increasing current and brightness.
(D) Including capacitance with $X_C = X_L$: This creates a series resonant circuit. At resonance, the reactive components cancel out ($X_C - X_L = 0$), and the impedance is minimum ($Z = R$). This results in maximum current and maximum brightness.
Step 4: Final Answer:
Brightness decreases when an iron rod is inserted into the coil.