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The gravitational force \( F \) between two masses is given by Newton's law of universal gravitation: \(F = G \frac{m_1 m_2}{r^2}\) where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses (P and Q), and \( r \) is the distance between their centers. Originally, \( F = 10 \, \text{N} \). If both masses are doubled (\( m_1' = 2P \), \( m_2' = 2Q \)) and \( r \) remains constant, the new force is: \(F' = G \frac{(2P)(2Q)}{r^2} = G \frac{4PQ}{r^2} = 4 \left( G \frac{PQ}{r^2} \right) = 4 \times 10 = 40 \, \text{N}\)

v\(_e\) = \(\sqrt{(2gR)}\) Since gR = GM \(\therefore\), v\(_e\) = \(\sqrt{(2GM)}\)

1a. Acceleration of a free fall due to gravity is the force of attraction of the Earth on a unit mass. b. g varies on the surface of the earth because of the shape of the earth and the rotation of the earth about its polar axis. ci. \theta = 30^\circ; H = 100 m; u = 100 m/s; h = ? Total height = H + h => h = \(\frac{u^2 sin^2 \theta }{2g}\) = \(\frac{100^2 sin^2 30^\circ}{ 2 \times 10}\) = \(\frac{10000 \times 0.25}{ 20}\) = \(\frac{2500}{ 20}\) = 125 m . Height from the ground = height of tower + the maximum height the stone reached = 100 + 125 = 225 m ii. t = \(\frac{u sin \theta }{\text{g}}\) = \(\frac{100 \times sin 30^\circ}{10}\) = \(\frac{100 \times 0.5 }{10}\) = 10 \times 0.5 = 5 s iii. Time of flight = time taken for a particle which is projected to return to its original level T = \(\frac{2u sin \theta}{\text{g}}\) = \(\frac{2 \times 100 \times sin 30^\circ}{10}\) = \(\frac{200 \times 0.5}{10}\) = 10 s Time of flight = 10 s But the total time taken to land on the ground = T + time used to return from the top of the tower to the ground level. Since it's falling S\(_y\) = 100 m S\(_y\) = ut sin \theta - \(\frac{1}{2}\) g t\(^2\) => -100 = 100t sin 30^\circ - \(\frac{1}{2}\) (10) t\(^2\) => -100 = 100t \times 0.5 - 5 t\(^2\) => -100 = 50t - 5 t\(^2\) => 5 t\(^2\) - 50t - 100 = 0 => t\(^2\) - 10t - 20 = 0 t = \(\frac{10 \pm \sqrt{10^2 - 4 \times -20 \times 1}}{2}\) = \(\frac{10 \pm \sqrt{100 + 80}}{2}\) t = \(\frac{10 + 13.416}{2}\) = \(\frac{23.416}{2}\) ≈ 11.7s (ignore the -ve part) t = 11.7 s => Total time = 10 + 11.7 = 21.7 seconds iv. Horizontal distance = the range = \(\frac{u^2 sin 2\theta}{\text{g}}\) = \(\frac{100^2 sin 60^\circ}{10}\) = \(\frac{10000 \times 0.866}{10}\) = 1000 \times 0.866 = 866 m

B. Increases When a positively charged rod is brought near the cap of a positively charged gold-leaf electroscope (as indicated by the diverging leaves with + signs in the diagram), electrostatic induction occurs. The rod attracts negative charges toward the cap and repels additional positive charges toward the leaves. This increases the positive charge on the leaves, enhancing their mutual repulsion and thus increasing the divergence.

The neutral point around a magnet is where the magnetic field strength is zero, meaning no magnetic effects (like deflection of a compass needle) are observed, often due to cancellation by Earth's magnetic field.

Soft iron has high magnetic permeability, so it easily attracts and redirects magnetic field lines, shielding areas from external magnetic fields. Aluminum, brass, and copper do not have this property and are ineffective for magnetic shielding.

This process disrupts the alignment of magnetic domains, effectively demagnetizing the bar.

Answer: C An electric bell works using an electromagnet.

a. Faraday's law of electromagnetic induction states that the induced e.m.f. in a circuit is directly proportional to the rate of change of the magnetic flux, or field lines, linking the circuit i.e e ∝ \(\frac{dφ}{\text{dt}}\). b. See diagram above. How it works: it has a primary coil of a few turns and a secondary coil of many turns. When the primary circuit is made, a current flows in P. A is then attracted, so breaking the circuit at the contact point. This makes the large magnetic flux in the secondary fall very quickly. A high voltage is then produced at its terminals X and Y, and a spark appears across the air gap as the circuit is made and broken repeatedly. c. The eddy current can be minimized by laminating the core to reduce energy losses. d. The spark which produce across the gap of the make and break device is reduced by the capacitor. Thus, it should be included. e. (i). It is used in the operation of X-ray tubes. (ii). It is used in the generation of high voltages. (iii). It is used in morse radio transmitter.

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.