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D = Lead: Lead is highly effective at shielding against ionizing radiation, particularly gamma rays, due to its high density and atomic number.

The sun produces energy primarily through the nuclear fusion of hydrogen nuclei, a reaction that occurs at extremely high temperatures, around 10\(^7\) K. H\(_1\) + H\(_1\) → \(_2\)H

At resonance in an a.c. circuit, during which the capacitive reactance X\(_c\) is equal to the inductive reactance X\(_l\), the phase angle Ø = 0^\circ

The force on a current carrying conductor in a magnetic field is greatest when the conductor lies perpendicular in the magnetic field

Transistors are versatile components used in electronic circuits mainly for two purposes: they can act as electronic switches, controlling the flow of current in a circuit, and they can amplify electrical signals, making them essential in audio devices, radios, and various electronic applications. Thus, option A correctly describes the main functions of a transistor.

Eddy currents are loops of electric current induced in conductors by a changing magnetic field, leading to energy losses as heat. Using insulated soft iron wires helps minimize these losses by reducing electrical conductivity and preventing the flow of these currents through insulation.

Resonance frequency in an a.c. circuit is given by F = \(\frac{1}{2\pi \sqrt{LC}}\) therefore, \(200\times 10^3\) = \(\frac{1}{2\pi \sqrt{L\times 20 \times 10^{-12}}}\) therefore (\(200\times 10^3)^2\) =\(\frac{1}{4 \times \pi^2 \times L \times 20 \times 10^{-12}}\) THEREFORE, L = \(\frac{1}{4\times10\times (200\times 10^3)^2 \times 20\times 10^{-12}}\), (\pi\(^2\) ≈10) Therefore L =\(\frac{1}{32}\)H

The magnetic force on a charge is given by F = qVB Sinθ, thus the force is proportional to both the magnitude by the charge (q) and the velocity (V) (i/e F α qV)

The pendulum that will resonate with P is T. being of the same length and therefore of the same frequency.

Since the number of pulleys present is 6, it implies that the velocity ratio V.R = 6 Efficiency = \(\frac{\text{M.A}}{\text{V.R}}\times \frac{100}{1}\) = \(\frac{Load}{Effort}\times\frac{1}{V.R}\times\frac{100}{1}\) = \(\frac{240}{80} \times\frac{1}{6} \times\frac{100}{1}\) = 50% ∴ Efficiency = 50%

\(n = \frac{sin i}{sin r}\) \(i = 90 - \alpha\) \(\therefore n = \frac{sin 90 - \alpha}{sin \beta}\) ⇒ \(n = \frac{cos \alpha}{sin \beta}\)

I = \(\frac{E}{R + r}\) For parallel of 5\Omega and 10\Omega \(\frac{1}{R}\) = \(\frac{1}{10}\) + \(\frac{1}{5}\) = \(\frac{1+2}{10}\) = \(\frac{3}{10}\) ∴R = \(\frac{10}{3}\) Total R = 3\(\frac{1}{3}\) + 3 x 1 = 7\(\frac{1}{3}\) ∴I = 6\(\div\) 7\(\frac{1}{3}\) = \(\frac{6}{1}\) x \(\frac{3}{22}\) = \(\frac{18}{22}\) = \(\frac{9}{11}\) ∴I = \(\frac{9}{11}\)

Energy = 1/2CV2 ∴Energy of P = 1/2 x 2 x V2 = V2 Energy of Q = 1/2 x 4 x V2 = 2V2 ∴ P/Q = V2/2V2 = 1/2 => P:Q = 1:2

C = \(\frac{\text{Q}}{\text{V}}\) => Q = CV : ∴ V = \(\frac{\text{Q}}{\text{C}}\) ∴V\(_1\) = \(\frac{\text{Q}}{C_1}\); V\(_2\) = \(\frac{\text{Q}}{C_2}\); V\(_3\) = \(\frac{\text{Q}}{C_3}\) for series \(\frac{1}{\text{C}}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\) + \(\frac{1}{C_3}\) = \(\frac{1}{2} + \frac{1}{6} + \frac{1}{3}\) = \(\frac{3 +1 + 2}{6}\) = \(\frac{6}{6}\) = 1 ∴ C = 1\muF = 1.0 x 10\(^{-6}\)F ∴ Q = CV = 1.0 x 10\(^{-6}\) x 12 = 12.0 x 10\(^{-6}\)C ∴V\(_1\) = \(\frac{\text{Q}}{C_1}\) = \(\frac{12 \times 10^{-6}}{2 \times 10^{-6}}\) = 6V V\(_2\) = \(\frac{\text{Q}}{C_2}\) = \(\frac{12 \times 10^{-6}}{6 \times 10^{-6}}\) = 2V V\(_3\) = \(\frac{\text{Q}}{C_3}\) = \(\frac{12 \times 10^{-6}}{3 \times 10^{-6}}\) = 4V ∴ 6V, 2V and 4V

0.0 E since one of the electric fields points in the opposite direction (left)