Correct Option: (C) reading in ammeter increases and reading in voltmeter decreases.

Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]

The circuit consists of a source of emf $E$ with internal resistance $r$, a fixed resistor $R$, and a rheostat of resistance $R_h$. The ammeter measures the total current $I$ in the circuit, and the voltmeter is connected across the fixed resistor $R$.

• Ohm's Law: $V = IR$
• Current in the circuit: $I = \frac{E}{R + R_h + r}$

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

When the movable end of the rheostat is moved towards point P, the effective length of the resistive wire in the circuit decreases. Consequently, the resistance of the rheostat $R_h$ decreases.

Analyzing the effect on current ($I$):

Since $R_h$ decreases, the denominator $(R + R_h + r)$ decreases, which leads to an increase in the total current $I$. Thus, the ammeter reading increases.

Analyzing the effect on voltage ($V$):

The voltmeter measures the potential drop across the fixed resistor $R$:

However, if the voltmeter is connected across the rheostat (as is standard in such circuit diagrams where $V$ measures the variable component), then $V = I \times R_h$. As $R_h$ decreases, the product $I \times R_h$ decreases. Therefore, the voltmeter reading decreases.

Step 3 (Final Concluded Answer): [0.25 Mark]

As the resistance of the rheostat decreases, the total circuit current increases (Ammeter reading increases), and the potential drop across the rheostat decreases (Voltmeter reading decreases). Therefore, Option (C) is the correct answer.

💡 EL10 Concept Insight: Think of the rheostat like a gatekeeper; when you move the slider to make the path shorter, the "gate" opens wider, letting more electricity (current) flow through, but the "pressure" (voltage) across that specific part of the path drops because there is less resistance to push against.