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

A photon of energy E = hf = 6.63 x 10-34 x 3 x 105 = 19.89 x 10-29 = 2.0 x 10-28J

After 5 half-lives, the fraction of any radio-active substance left = (1/2)5 = 1/32 Thus the percentage of the original nuclei = 1/32/1 x 100/1 assuming the original amount is unity percentage left = 1/32 x 100/1 = 3.12 = 3%

Power is transmitted at very high voltages primarily to reduce heating in the coils(cables). High Voltage: Lowers current, minimizing resistive (I\(^2\)R) losses, which cause heating. Eddy Currents: While they can be a concern in certain applications, such as transformers, they are not the primary issue in high-voltage transmission.

In a nuclear reaction, neutron is used as a bullet fired into 23592U, to trigger off nuclear chain reaction, which of course is controlled

n = \(\frac{\frac{A+D}{2}}{sin\frac{A}{2}}\) where A = the refracting angle of the prism(60º) D = angle of deviation of the ray (30^\circ) n = \(\frac{\frac{60+30}{2}}{sin\frac{60}{2}}\) = \(\frac{sin45}{sin30}\) v = \(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\) \(\frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}}\) = \(\frac{\sqrt{2}}{2}\) x \(\frac{2}{1}\) = \(\sqrt{2}\) n = \(\sqrt{2}\)

Let the combined parallel resistance of 3 Ω and x = Q Ω ∴ Total resistance in the circuit = 2 + Q + 1.5 = 3.5 + Q Ω Total current in the circuit = 2A P.D a cross the circuit 12 volts ∴ From Ohm’s law, V = IR 12 = 2(3.5 + Q) = 7.0 + 2Q ∴ 2Q = 12 – 7 = 5 ∴ Q = 5/2 = 2.5 Ω Thus from parallel formula: I/Q = 1/3 + 1/x 1/2.5 = 1/3 + 1/x ∴ 1/x = 1/2.5 - 1/3 = 3-2.5/7.5 = 0.5/7.5 X = 7.5/0.5 = 75/5 = 15 Ω ∴ X = 15 Ω

The energy stored in a given capacitor is given by E = \(\frac{1}{2}\) CV\(^2\) = \(\frac{1}{2}\) QV = \(\frac{1}{2}\)\(\frac{Q^2}{\text{C}}\) Where C = the capacitance of the capacitor Q = charge carried by the capacitor V = p.d across the plates of the capacitor, thus from the data given, C = 10\muF = 10 x 10\(^{-6}\)F Q = 100\muc = 100 x 10\(^{-6}\)C ∴ Energy stored = \(\frac{1}{2}\)\(\frac{Q^2}{\text{C}}\) = \(\frac{1}{2}\)\(\frac{(100 \times 10^{-6})^2}{10 \times 10^{-6}}\) = 0.0005 = 5 x 10\(^{-4}\)J

1 KW = 1000 watts, power P = IV, thus for a mains of 250 V, the maximum permissible Current I = \(\frac{\text{P}}{\text{V}}\) = \(\frac{1000}{250}\) = 4 A Thus, the fuse rating in the plug should be 4 A Putting safety into consideration, the appropriate rating of the fuse should be 5A That is = 4 x 1.25 = 5A ( 1.25 = safety factor)

The iris controls the amount of light by limiting the volume of light admitted. The lens then focuses the light onto the retina, forming an inverted image. The lens changes its shape to adjust the focus on both distant and near objects, a process called accommodation. This adjustment is done by two sets of muscles in the iris

Electric Power = IV = I2R = V2/R thus for a device rated 1000 W, 230V, from the relation P = V2/R, 1000 = (230)2/R => R = (230)2/1000 = 52900/1000 = 52.9Ω

In the provided diagram, the arrows on the field lines surrounding both point X and point Y are pointing outward. This indicates that both X and Y are negative charges. turn the positive to positive

Infrared rays produce a high heating effect. All red-hot bodies such as a heated iron ball, flame, and fire are sources of infrared rays.

IR = 0.6 X I ( R + r ) I X 3 = 0.6 X I ( 3 + r ) 3I / 0.6I = 3 + r 5 = 3 + r r = 2 ohms

The particle in the reaction is a neutron \(^{23}_{11}\)Na + \(^1_0\)n → \(^{20}_9\)F + \(^4_2\)He