JEE Mains 2024 (Shift 1, Feb 1) - Potentiometer

Question: A potentiometer wire of length 400 cm and resistance $8\Omega$ is connected in series with a battery of EMF 2V and internal resistance $2\Omega$. An unknown cell of EMF $E_x$ is balanced at 200 cm. Find $E_x$.

Solution:
1. Total resistance of the primary circuit: $R_{total} = R_{wire} + r_{battery} = 8\Omega + 2\Omega = 10\Omega$.
2. Total current in the primary circuit: $I_{main} = \frac{E_{battery}}{R_{total}} = \frac{2V}{10\Omega} = 0.2A$.
3. Potential drop across the potentiometer wire: $V_{wire} = I_{main} \cdot R_{wire} = 0.2A \cdot 8\Omega = 1.6V$.
4. Potential gradient of the wire: $k = \frac{V_{wire}}{L_{wire}} = \frac{1.6V}{400 \text{ cm}} = \frac{1.6V}{4 \text{ m}} = 0.4 V/m$.
5. The unknown cell $E_x$ is balanced at length $l = 200 \text{ cm} = 2 \text{ m}$.
6. Therefore, $E_x = k \cdot l = (0.4 V/m) \cdot (2 \text{ m}) = 0.8V$.

JEE Mains 2024 (Various Shifts) - Power Dissipation in Resistors

Question: In the circuit shown, if $R_1 = 1\Omega$, $R_2 = 2\Omega$, $R_3 = 3\Omega$, and the voltage source is $V=12V$. Find the power dissipated in resistor $R_2$.

Solution:
1. The resistors $R_1, R_2, R_3$ are in series.
2. Equivalent resistance $R_{eq} = R_1 + R_2 + R_3 = 1\Omega + 2\Omega + 3\Omega = 6\Omega$.
3. Current flowing through the circuit: $I = \frac{V}{R_{eq}} = \frac{12V}{6\Omega} = 2A$.
4. Since it's a series circuit, the same current $I=2A$ flows through $R_2$.
5. Power dissipated in $R_2$: $P_2 = I^2 R_2 = (2A)^2 \cdot (2\Omega) = 4 \cdot 2 = 8W$.

JEE Mains 2024 (Various Shifts) - Capacitors with Dielectrics

Question: A parallel plate capacitor has capacitance $C_0$ in air. A dielectric slab of dielectric constant $k=4$ is inserted to fill the entire space between the plates. If the capacitor remains connected to a battery of voltage $V_0$, what is the new charge stored on the capacitor?

Solution:
1. Initial capacitance (air): $C_0$.
2. Initial voltage: $V_0$.
3. Since the capacitor remains connected to the battery, the voltage across it remains constant at $V_0$.
4. When a dielectric slab of constant $k$ fills the space, the new capacitance becomes $C_{new} = kC_0$.
5. Given $k=4$, so $C_{new} = 4C_0$.
6. The new charge stored is $Q_{new} = C_{new} \cdot V_0 = (4C_0) \cdot V_0 = 4 (C_0 V_0)$.
7. If $Q_0 = C_0 V_0$ was the initial charge, then $Q_{new} = 4Q_0$.