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The capacitance \( C \) of a parallel-plate capacitor is calculated using the formula: \(C = \epsilon_0 \frac{A}{d}\) where: \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m},\) \(A = \pi r^2 \quad \text{(area of one plate)},\) and \( d \) is the separation between plates. \(r = 6 \, \text{cm} = 0.06 \, \text{m},\) \(d = 0.12 \, \text{cm} = 0.0012 \, \text{m}\) Calculate the area: \(A = \pi r^2 = 3.142 \times (0.06)^2 = 3.142 \times 0.0036 = 0.0113112 \, \text{m}^2\) Calculate the capacitance: \(C = 8.85 \times 10^{-12} \times \frac{0.0113112}{0.0012} = 8.85 \times 10^{-12} \times 9426 \approx 8.34 \times 10^{-11} \, \text{F}\)

When capacitors are connected in series across a potential difference, there is a loss in their stored energy primarily due to the decrease in overall capacitance, which limits the amount of energy they can store, rather than due to charge flowing between the capacitors.

The force exerted on a charge in an electric field is given by the formula F = qE, where F is the force, q is the charge, and E is the electric field intensity. Given: the charge q is 1.6 \times 10\(^{-10}\) C, and the electric field intensity E is 2.0 \times 10\(^5\) NC\(^{-1}\). Substituting these values into the formula gives F = (1.6 \times 10\(^{-10}\) C) \times (2.0 \times 10\(^5\) NC\(^{-1}\)) = 3.2 \times 10\(^{-5}\) N.

Given: r = 20cm ⇒ 0.20m, K = \(\frac{1}{4\pi ε₀}\) = 9.0 x 10\(^9\)Nm\(^2\)C\(^{-2}\), q = 0.015C The electrical potential V = \(\frac{\text{kq}}{\text{r}}\) V = \(\frac{ 9.0 x 10^9 \times 0.015}{0.20}\) = \(\frac{1.35 \times 10^8}{0.20}\) = 6.75 x 10\(^8\)V

Effective capacitance across the parallel = C + C = 2C Effective capacitance across the circuit = \(\frac{1}{2C} + \frac{1}{2}\) = \(\frac{1 + C}{2C}\) ∴ Effective capacitance across the circuit = \(\frac{1 + C }{2C}\) = \(\frac{1}{1}\) 1 + C = 2C; 2C – C = 1 ∴ C = 1.0 \mu

A Wheatstone bridge is specifically designed for the accurate measurement of small and unknown resistances. It operates on the principle of balancing two legs of a bridge circuit, enabling precise determination of resistance values. Potentiometer: Used to measure potential difference or electromotive force (emf), not to measure resistance directly. Rheostat: A variable resistor utilized to control current, rather than for measuring resistance. Wheatstone Bridge: The most effective method for measuring very small resistances.

C. Ammonium chloride. It is the electrolyte in the wet Leclanché cell (aqueous solution).Others: Nickel hydroxide: In Ni-Cd/Ni-MH batteries. Lead dioxide: Cathode in lead-acid batteries. Carbon: Conductivity enhancer, not electrolyte.

Parallel connection does not add the voltages; it only increases the current capacity. Thus, total e.m.f. E = E\(_1\) = E\(_2\) = E\(_3\).

E = \(\frac{1}{2}\)QV; But, Q = CV ∴ E = \(\frac{1}{2}\)CV\(^2\); 40 = \(\frac{1}{2}\) \times 5 \times 10\(^{-6}\) \times V\(^2\) V\(^2\) = \(\frac{(2 \times 40)}{5 \times 10^{-6}}\) V\(^2\) = 16 \times 10\(^6\) ∴ V = \(\sqrt{(16 \times 10^6)}\) = 4000 V

The quantity of charge flowing through a conductor is given by the formula Q = It. Substituting the given values, we get Q = 10 \times 10 = 100 C.

\(\frac{1}{R_{XY}}\) = \(\frac{1}{3} + \frac{1}{2} + \frac{1}{6}\) = \(\frac{5}{6}\) \(\therefore\) R\(_{XY}\) = \(\frac{5}{6}\)\(\Omega\) \(\therefore\) R\(_T\) = \(\frac{5}{6}\) + 0.8 = \(\frac{5}{6}\) + \(\frac{4}{5}\) = \(\frac{10}{5}\) = 2\(\Omega\) But, V = IR\(_T\) where V = 5v and R\(_T\) = 2\(\Omega\) I = \(\frac{\text{V}}{R_T}\) = \(\frac{5}{2}\)A \(\therefore\) V\(_{across XY}\) = I R\(_{XY}\) = \(\frac{5}{2}\) x \(\frac{6}{5}\) = 3.0V.

V = IR Voltage across XY = 3V V = 3I = 3V I = \(\frac{3}{3}\) = 1.0 A

The resistance \(R\) of a wire is related to its resistivity \(\rho\), length \(L\), and cross-sectional area \(A\) by the formula: \(R = \frac{\rho L}{A}\) The conductivity \(\sigma\) is the reciprocal of the resistivity: \(\sigma = \frac{1}{\rho}\) \(\rho = \frac{R A}{L}\) \(\sigma = \frac{L}{R A}\) Substitute the given values (\(L = 5\) m, \(R = 10\) \Omega, \(A = 4.0 \times 10^{-3}\) m²): \(\sigma = \frac{5}{10 \times 4.0 \times 10^{-3}} = \frac{5}{0.04} = 125\) The conductivity of the wire is 125 \Omega\(^{-1}\)m\(^{-1}\) = 1.25 x 10\(^2\) \Omega\(^{-1}\)m\(^{-1}\)

3a. The e.m.f of the accumulator = 1.5V Tables of values R(\(\Omega\)) I(A) I\(^{-1}\)((A\(^{-1}\)) 0 0.78 1.28 1 0.50 2.00 2 0.38 2.63 3 0.30 3.33 4 0.25 4.00 5 0.22 4.55 S = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{4 - 2}{4 - 1}\) = \(\frac{2}{3}\) = 0.67 The intercept on the vertical axis = 1.32\(\Omega\) Precautions: (i) I ensured all electrical connections were securely tightened to prevent loose contacts, which could lead to fluctuating or inaccurate readings. (ii) I avoided parallax error by positioning my eyes directly in line with the scale when reading the ammeter or voltmeter for precise measurements. (iii) I checked and corrected for any zero error on the ammeter and voltmeter before taking readings to improve accuracy. (iv) I removed the key from the circuit when not taking measurements to avoid unnecessary heating and potential damage to components. (v) I took multiple readings at each data point and averaged the values to minimize random errors and ensure more reliable results. (vi) I ensured the apparatus was properly calibrated and regularly checked for consistency throughout the experiment. bi. 1. Rechargeability: A lead-acid accumulator is rechargeable, making it more suitable for long-term use, whereas a Leclanché cell is primarily a primary (non-rechargeable) cell. 2. Higher Current Supply: Lead-acid accumulators can deliver higher currents, making them ideal for applications requiring significant power, like in vehicles, while Leclanché cells are suitable for low-current devices. 3. Longer Lifespan: Due to its rechargeable nature, the lead-acid accumulator has a longer operational life compared to a Leclanché cell, which has a limited lifespan and must be replaced after depletion. 4. Stable Voltage: Lead-acid accumulators provide a relatively stable output voltage during discharge, while the voltage of a Leclanché cell drops significantly over time. bii. Since 3 \Omega and 4 \Omega are connected in parallel, \(\frac{1}{\text{R}}\) = \(\frac{1}{3}\) + \(\frac{1}{4}\) \(\frac{1}{\text{R}}\) = \(\frac{4 + 3}{12}\) \(\frac{1}{\text{R}}\) = \(\frac{7}{12}\) R = \(\frac{12}{7}\) = 1.71 \Omega Now, 1.71 \Omega is connected in series with 4 \Omega. ∴ The effective resistance in the circuit = 1.71 \Omega + 4 \Omega = 5.71 \Omega

Answer: B. 2.27A (I = \(\frac{\text{P}}{\text{V}}\) = \(\frac{500}{220}\) ≈ 2.27 A.)