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No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option B.

F = \(\frac{Q^2}{4 \pi \varepsilon_o r^2}\) I = 9 x 109 x \(\frac{Q^2}{0.2^2}\) Q = \(\sqrt{\frac{0.04}{9 \times 10^9}}\) = 2.11 x 10-6

\(\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2}\) = \(\frac{1}{5} + \frac{1}{20}\) = \(\frac{4 + 1}{20} = \frac{5}{20}\) r = \(\frac{20}{5}\) = 4\(\Omega\) R = R1 + R2 = 5 + 4 = 9\(\Omega\) but V = IR I = \(\frac{V}{R} = \frac{9}{9}\) = 1A

e = \(\frac{RA}{L} = \frac{10 \times 0.005}{100} = 0.0005\)

\(\frac{ES}{EP} = \frac{Ns}{Np} = \frac{Es}{150} = \frac{40}{400}\) Es = \(\frac{150 \times 40}{400}\) = 15V

No expanded working has been attached for this question yet. The verified answer is Option B.

Electricity I = 0.36A \(\frac{x}{cm}\) \(\frac{I}{A}\) \(\frac{X^{-1}}{cm^{-1}}\) 20.0 1.14 0.050 35.0 0.90 0.029 45.0 0.84 0.022 60.0 0.78 0.017 80.0 0.66 0.013 Slope (s) = \(\frac{\bigtriangleup {X}^{-1}}{\bigtriangleup {I}}\) = \(\frac{0.08-0.01}{1.68-0.644} = \frac{0.07}{1.036}\) From the graph, When X\(^{-1}\) = 0 \(I_0\) of \(I_1\) = 0.448A x. Evaluate \(\frac{I_{0}}{I}\) = \(\frac{0.448}{0.36}\) = 1.244 Precaution; The key was opened when readings were not taken. I ensured tight connections of the connecting wire I avoided parallax errors when reading the potentiometer. (bi) Define emf of a battery: The e.m.f of a cell is the potential difference across its terminals when it is in an open circuit. i.e. not Supplying current to an external circuit. ii. A cell x emf is 1.018V. Lengt\(h_1\) on the potentiometer x = 50.0cm Lengt\(h_2\) on the potentiometer, y,= 75.0cm A cell Y emf is? Solution By formula; \(\frac{E_{1}}{E_{2}}\) = \(\frac{L_{2}}{L_{1}}\) \(\frac{1.018}{E_{2}} = \frac{75.0}{50.0}\) 75.\(E_2\) = 1.018 x 50 \(E_2\) = \(\frac{1.018\times 50}{75}\) \(E_2\) = 0.612V. see graph above

R = U\(\sqrt{\frac{2H}{g}} = 10\sqrt{\frac{2 \times 20}{10}}\) = 20

R = \(\frac{U^2 sin 2\theta}{g}\), 78 = \(\frac{30^2 sin 2\theta}{10}\) \(sin2\theta = \frac{780}{900}\) = 0.8667, \(2\theta = sin^{-1}0.8667 = 60^o\) \(\theta \frac{60^o}{2} = 30^o\)

\(\frac{Mv^2}{r}\) = qvB r = \(\frac{MV^2}{qVB} = \frac{9.11 \times 10^{31} \times (2.00 \times 10^7)^2}{1.61 \times 10^{-19} \times 2.00 \times 10^7 \times 1.2 \times 10^{-3}}\) = 9.43 x 10\(^{-2}\)

E \(\frac{Fl}{Ae}\) = \(\frac{35 \times 10^3 \times 50 \times 10^{-2}}{4 \times 10^{-4} \times 0.25 \times 10 ^{-3}}\) = 1.75 x 10\(^{11}\)Nm\(^{-2}\)

(a) Under very low pressure and very high voltage in at enclosed discharge tube ionized gas molecules or atom can conduct electricity. (b) Electric charge carriers in gases are ions and electrons.

(a) Graph slope represents Planck's constant. (b) Intercept OC represents work function of the metal. (c) Intercept OK represents the threshold frequency F\(_o\) of the incident radiation

Conditions for photo-electrons emission are: (i) the energy of the incident photons radiation mus', be greater than the work function of the metal. (ii) the wavelength of incident radiation must be less than the threshold wavelength for metal surface. (iii) the frequency of incident radiation must be greater than the threshold, frequency. (b) Particle characteristics of electroniagnetic waves are: (i) Thermionic emission (ii) Photoelectric effect (iii) Compton effect (iv) Emission and absorption of incandescence light. (v) Radiations from heated black bodies.