Question Library

(a) Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f. is induced, the strength of which is proportional to the rate of the change of the flux linked with the circuit. The direction of the induced current is always such as to oppose the change producing it. As coil is rotating in the magnetic field, an e.m.f which is due to the changing flux induces between its terminals. The e.m.f. varies from maximum when the coil is horizontal to minimum when the coil is vertical at which position connections between brushes and terminals of coil are changed by commutator. This ensures that current continues to flow in the same direction. (c) – By increasing the number of turns – Winding the coil on the soft iron core – Increasing the strength of the magnet – turning the coil faster.

(a)(i) Fusion: Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of energy. Example: \(^{2} _{1} H + ^{2} _{1} H \to ^{4} _{2} He + Energy\) (ii) Fission: Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with fast neutrons or energetic particles releasing energy in the process. Example: \(^{235} _{92} U + ^{1} _{0} N \to ^{90} _{39} Kr + ^{144} _{50} Ba + ^{1} _{0} N + ^{1} _{0} N\) (iii) Radiation Hazards. Radioactive substances can be dangerous if one is exposed to their radiations for a long time. Some of the radiations are highly penetrating and if the body is over-exposed to them, they can destroy the cells in tissues and upset the natural chemistry of the body. Examples include: – Leukaemia – Skin burns (b) Advantages of fusion over fission include: (1) Easily achieved with lightest element so that nuclear repulsion is easily overcome as nuclei approach each other (2) Raw materials are cheaply available. (3) Produces less dangerous by products. (4) There is no upper limit to the mass of hydrogen that can be obtained. Very high temperature is required to overcome the coulomb repulsive forces between the two light nuclei. This poses severe technological problem due to the fact that materials to withstand this high temperature are difficult to come by. (c) \(I = I_{m} \sin 2 \pi ft ... (1)\) I - Instantaneous value of current \(I_{m}\) - Peak value of current F - frequency of current \(I = 30 \sin 100 \pi t ... (2)\) Comparing (1) and (2), (i) \(2 \pi f = 100\pi \therefore f = 50 Hz\) (ii) \(I_{m} = 30A\) (iii) \(I_{r.m.s} = \frac{I_{m}}{\sqrt{2}}\) = \(\frac{30}{\sqrt{2}} = 21.2A\)

S = Ut + \(\frac{1}{2}\) at\(^2\) but U = 0; from rest S = \(\frac{1}{2}\) at\(^2\) S = \(\frac{3 \times 8^2}{2}\) S = 96m

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

Upthrust = weight fluid displacement 0.040 - 0.024 = PVg 0.016 = 800 x V x 10 V = 2.0 x 106

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

F = m(v - u)/t; (v - u) = Ft/m = 8.0

μ= F/mg = 20/40 = 0.5

v2 = u2 + 2gh = 0 + 2 x 10 x 10 v2 = 200; K.E = 1/2mv2 = 1/2 x 5 x 200 = 500J

Under standard atmospheric pressure, pure water freezes at 0ºC. Water is most dense at around 4ºC. As it cools below this temperature, it begins to expand.

Using the genaral gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) T\(_1\) = 30º = 273 + 30 = 303k T\(_2\) = ? let V\(_1\) = V, V\(_2\) = \(\frac{2}{3}\)V, P\(_1\) = P, P\(_2\) = 2P From, \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) T\(_2\) = \(\frac{P_2 V_2 T_1}{P_1 V_1}\) = \(\frac{2P \times \frac{2}{3}V \times 303}{P \times V}\) = 404k(the Ps and Vs cancel out) 404k → ºC = 404 - 273 = 131ºC

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

Pressure at a depth for fluid with constant density such as mercury is given as \(p = p_{0} + \rho hg\) where \(p_{0}\) = atmospheric pressure. \(p_{0} = \rho hg =13600 \times \frac{76}{100} \times 10\) = \(103,360 Nm^{-2}\) \(p = 103,360 + (13600 \times \frac{10}{100} \times 10)\) \(p = 103,360 + 13,600\) \(p = 116,960 Nm^{-2} \approxeq 1.17 \times 10^{5} Nm^{-2}\)

L = λ/4; λ= 4 = 4m v = fλ = 85 x 4 = 340ms-1

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