Question Library

In longitudinal waves, such as sound waves in a liquid or gas, the displacement of the particles in the oscillation is always in the direction of propagation, so these waves do not exhibit polarization.

a)(i) Nuclear fission is a nuclear reaction in which a heavy nucleus is split into different lighter nuclei emitting neutrons with the release of a large amount of energy. (ii) Nuclear fusion is a nuclear reaction between two light nuclei at high temperatures formic a heavier nucleus with the release of a large amount of energy. (b)(i) Applications of atomic energy are: - for the production of nuclear weapons for warfare - in the generation of electric power - for health diagnosis and radiotherapy - to detect leakages. In underground pipes carrying oils or gases. (ii) Half life is the time required for radioactive substance to decay to half of its original quantity. (iii) T = \(\frac{0.693}{\lambda}\) (c)(i) \(^{226}_{88} X \) ---> \(^{222}_{88} Y + ^4_2He\) By the emission of \(\alpha\) the mass number of X decreases by 4 the atomic number by 2 to give daughter nucleus Y (ii) \(^{222}_{86} Y \) ---> \(^{222}_{87} Z + ^0_{-1} e\) By the emission of \(\beta\) particle the mass number of Y is unchanged while the atomic number of Y is unchanged while the atomic number is increased by 1. (iii) \(^{226}_{87} Z \) ---> \(^{222}_{87} Z + ^0_0\gamma\) By emission of \(\gamma\) radiation, the mass and atomic number of Z remain unchanged.

The energy store is equal to the work done in stretching the wire from E to F given by the \(\frac{1}{2}\)Fe where e = 0.10 - 0.50 = 0.05m. F = 0.2 + 0.1 = 0.3N = \(\frac{1}{2}\) x (0.2 + 0.1) x (0.10 - 0.50) = 7.5 x 10\(^{-3}\)J

The work done on this object is given by the area of the triangle, which is equal to 1/2 fx

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