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Speed of light in a medium is given by: v = \(\frac{\text{c}}{\text{n}}\) \(\frac{4}{3}\) = \(\frac{3 \times 10^8}{v}\) v = 2.25 x 10\(^8\)ms\(^{-1}\) \(\lambda\) = \(\frac{V}{F}\) = \(\frac{2.25 \times 10^8}{5 \times 10^{14}}\) \(\lambda\) = 0.45 x 10\(^{-6}\)m = 4.5\times10\(^{−7}\) m

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

\(\lambda\) = 4(L + C) ; where C = o \(\lambda\) = 4 x 20 = 80cm V = \(\lambda\)F = 80 x 10-2 x 480 = 384ms-1

The correct answer is C. i, ii, iii only, as the insulated charged body carries excess positive charges, creates an electric field, and possesses potential energy, but does not carry electric current(An insulated charged body carrying a positive charge does not carry electric current because it is not part of a circuit that allows the flow of charge.)

Electromotive Force (EMF) refers to the total energy supplied by a source (like a battery) to move a unit positive charge around a complete circuit. It is measured in volts and indicates how much energy per unit charge is available to drive the current through the circuit.

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.

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

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

Average velocity = \(\frac{total speed}{total time}\) Total distance = Area under the graph i.e area of trapezium = \(\frac{1}{2}\)(a + b) x h = 0.5(6 + 10) x 10 = 80m Total time = 10secs Average velocity = \(\frac{total speed}{total time}\) = \(\frac{80}{10}\) = 8m/s so closest answer is 8.5m/s

X is in the line of action of the force, Therefore moment is equal to zero. Moment = force x perpendicular distance. ( x is horizontal distance, not perpendicular distance)

(a) Velocity- time graph: (i) Acceleration : Gradient of AB in the graph above; (ii) retardation: Gradient of CD in the graph above; (iii) total distance covered is the area under the graph. (b) S = average velocity x time \(S = (\frac{u +v}{2}) \times t ... (1)\) but \(v = u + at ... (2)\), substitute (2) in (1) \(S = (\frac{u + u + at}{2}) \times t\) \(S = (\frac{2u + at}{2}) \times t \) \(S = ut + \frac{1}{2} at^{2}\) (c) P ----------- Q ----50m------- R 40 ms\(^{-1}\) 20 ms\(^{-1}\) Since the acceleration is uniform and using equation \(v^{2} = u^{2} + 2as\) for the motion along QR. \(0 = 20^{2} - 2a(50)\) (retardation) \(0 = 400 - 100a \implies 100a = 400\) \(a = 4 ms^{-2}\) (retardation, hence it is negative) \(20^{2} = 40^{2} + 2(-4) s\) \(400 = 1600 - 8s \implies 8s = 1200\) \(s = 150m\) (ii) Time taken to cover PQ Apply the equation v = u + at from P to Q: 20 = 40 + (-4)t 4t = 20 t = 5s Time to cover PQ = 5s (iii) Time taken to cover PR PR = PQ + QR = 150 + 50 = 200s Apply the equation v = u + at from P to R (initial velocity \(40\,\text{m s}^{-1}\), final velocity \(0\)): 0 = 40 + (-4)t 4t = 40 t = 10s.

(a) The boiling point of a liquid is the temperature at which its s.v.p is equal to the external atmospheric pressure. (b) Experiment to determine the boiling point of a small quantity of a liquid. Some liquid is introduced into the shorter length of the J - tube above the mercury. The beaker and its contents are heated and stirred continuously. At a stage when the levels of mercury in the two limbs of the J - tube are the same the temperature is read. At this temperature the s.v.p. of the liquid is equal to the atmospheric pressure. This temperature is the boiling point of the liquid. (c) Heat lost by copper = Heat gained by water \(mc \theta = m_{1} c_{1} \theta_{1} + ml\) \(0.3 \times 400 \times (950^\circ - 100^\circ) = 0.25 \times 4200 \times (100^\circ - 25^\circ) + 2.26 \times 10^{6} M\) \(M = \frac{102000 - 78750}{2.26 \times 10^{6}}\) \(M = 10.3g\) Therefore, the mass of water that will boil away is 10.3g. (d) Four effects of heat other than expansion are: (1) Change of state (2) Change in temperature (3) Chemical change (4) Change in electrical resistance of conductivity. (5) Melting sublimation.

(a) Capacitance of a capacitor is the ratio of the charge Q on either plate of the capacitor to the p.d in volts between the plates. Mathematically, \(Capacitance = \frac{Q}{V}\) Q - Charge ; V - voltage. (b) Three factors on which the capacitance of a parallel - plate capacitor depends on are: (1) Common area of the plates (2) Dielectric between the plates (3) Separation between the plates. (c) Work done W = QV Average work done = \(\frac{1}{2} QV\) But \(V = \frac{Q}{C}\) \(\therefore W = \frac{Q}{2} \times \frac{Q}{C} = \frac{Q^{2}}{2C}\) \(\therefore \text{Energy stored W = } \frac{Q^{2}}{2C}\) (d) \(\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}\) = \(\frac{1}{6} + \frac{1}{4} = \frac{10}{24}\) \(C = 2.4 \mu F\) \(Q = \frac{24}{10} \times 100 \times 10^{-6} = 240 \mu C\) = \(240 \times 10^{-6} C\) (ii) p.d across 4\(\mu F, V_{4} = \frac{Q}{C}\) \(V_{4} = \frac{240 \times 10^{-6}}{4 \times 10^{-6}} = 60V\) p.d across 6\(\mu F, V_{6} = \frac{Q}{C}\) = \(\frac{240 \times 10^{-6}}{6 \times 10^{-6}} = 40V\) (iii) Energy stored = \(\frac{Q^{2}}{2C}\) = \(\frac{(240 \times 10^{-6})^{2}}{2 \times 2.4 \times 10^{-6}}\) = \(1.2 \times 10^{-2} J\)

(a) Fission: Nuclear reaction in which a heavy nucleus is bombarded with neutrons and split into approximate equal parts emitting neutrons and releasing large amounts of energy. Fusion: Nuclear reaction between two light nuclei forming a heavier nucleus with the release of a large amount of energy. (b) Advantages of Fusion over Fission: (1)By - products are non-radioactive. (2) Easily achieved with lightest elements so that nuclear repulsion is easily overcome as nuclei approach each other. (3) Raw materials are cheaply available. Why fusion is not normally used : Very high temperature is required to overcome coulomb repulsive forces between the two light nuclei. It is difficult to get materials that will withstand such high temperature. Mathematically, \(T^{\frac{1}{2}} = \frac{0.693}{\lambda}\), where \(\lambda\) is the decay constant. Decay constant = \(\frac{\text{no of atoms disintegrating per sec}}{\text{no of atoms in the source at that instant}}\) Isotopes: Isotopes of an element are atoms which have the same proton numbers/atoms number but different nucleon numbers/mass number. Decay scheme: \(^{238} _{92} U \to ^{234} _{90} Th + ^{4} _{2} He + Energy\) \(^{234} _{90} Th \to ^{234} _{91} Pa + ^{0} _{-1} e + Energy\) \(^{234} _{91} \to ^{0} _{-1} e + ^{234} _{92} + Energy\) (d) Uses of radioisotopes: (1) Radio therapy (2) Radio dating (3) Used as tracers and preservatives in agriculture.