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Heat energy = Elect. energy 0.5 x 2.3 x 106 = 100t t = 1.15 x 104s

A simple form of interference pattern is obtained if two plane waves of the same frequency intersect at an angle. Interference is essentially an energy redistribution process.

n = \(\frac{3 \times\ 10^8}{2.4 \times\ 10^8}\) n = 1.25

V = \(\frac{W}{Q}\) V = \(\frac{8 \times\ 10^{-2}}{100 \times\ 10^{-6}}\) V = 8.0 x 10\(^2\) v

Energy = power x time = 1000 x 5 x 3600 = 1.8 X 107J

P = IVcos\(\theta\) = 5 x 10 x cos60 = 25W

So the diagram in the question is compressed to the one above. From the diagram: Cos\(\theta\) = \(\frac{6}{12}\) = \(\frac{1}{2}\) \(\theta\) = Cos\(^{-1}\)(0.5) = 60º

The weight mg acts vertically downward. Along the incline (parallel to QP): The component of weight down the plane: mg sin\theta Since the plane is rough and the block is sliding down, kinetic friction acts up the plane, opposing the motion. The normal reaction (N) acts perpendicular to the plane (not along QP). So, forces along QP (parallel to the incline) are: 1. mg sin\theta (down the plane) 2. Force of friction (up the plane)

Resolve T vertically and take a moment about x T sin30º x 20 = 50 x F T x 0.5 x 20 = 50F T x 10 = 50F T = \(\frac{50F}{10}\) = 5F

(a) Work is done whenever a force moves a body through a certain distance in the direction of the force. It is measured in joule and it is a scalar quantity. (i) Minimum effort (E) applied in pulling up the box = F = mgsin\(\theta\) = 100 + 700 sin30^\circ = 100 + 700 x 0.5 = 100 + 350=450N. (ii) Velocity ratio (V.R) of the plane = 1/sin\(\theta\) - 5/2.5 = 2 (iii) Mechanical advantage of the plane (M.A) = L/E = 1.56 . (iv) Efficiency of the plane = \(\frac{M.A}{V.R}\) x 100% = \(\frac{1.56}{2}\) x 100% = 78% (v) Energy lost in the system = work input - work output = (450 x 5) - (700/2 x 5) = 2250 - 1750 = 550J. (vi) Work output of the man = mgh = 70 x 10 x 2.5 = 1750J (vii) Total power developed by the man = \(\frac{\text{work done by effort}}{time}\) = \(\frac{450 \times 5}{50}\) = 45 watts

(a) Resonance occurs whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other body or system which is vibrating with the same frequency. At resonance, the amplitude is maximum. Two examples of resonance are tuning a radio and divers board. Procedure: In the diagram above the string is kept taut by using a constant weight (w) between the two bridges A and B. The length, L between the bridges is adjusted until the wire resonates with the vibrating tuning fork. This happens when a paper rider flies off or when a loud sound is heard. Then, measure length, L of the resonating wire and frequency f. The experiment is repeated for at least four other tuning forks of known frequencies. In each case the values of the frequency, f and the corresponding length, L are recorded. A graph of f is plotted against 1/L or L against 1/f. A straight-line graph passing through origin is obtained showing that frequency is proportional to the reciprocal of the length. (iii) Precautions: (1) A smooth pulley should be used (2) the tuning fork should be struck on a rubber bung (c) For a fundamental note f\(_o\) = \(\frac{1}{20}\) = \(\sqrt{\frac{T}{m}}\) where f\(_o\) = fundamental frequency L = Length of string T = Tension m = linear density or mass per unit length of the string (i) \(f_1 = \frac{1}{20} \sqrt{\frac{f}{m}}\) = 3(\(\frac{1}{20} \sqrt{\frac{1}{20}} \sqrt{\frac{1}{m}})\) f\(_1 = 3f_o\) i.e = the fundamental frequency, \(f_o\) is tripled (ii) \(f_2 = \frac{1}{20} (\frac{1}{20} \sqrt{\frac{T}{m}}) = \frac{1}{2} \times \frac{1}{20} \sqrt{\frac{1}{m}} = f_2 = \frac{1}{2} f_o\) i.e the fundamental frequency, \(f_o\) is halved

Explanation: When the key is closed, current flows in the circuit thereby magnetising the electromagnet. The soft iron armature is thus attached to the electromagnet, causing the hammer to strike the gong. When the armature is attracted, a gap is created at the contact and the circuit breaks. The electromagnet loses its magnetism and the armature falls back to its original position by the pull of this spring. This process of make and break is repeated thereby making the bell to ring continuously (i) E = IR - 0.2(2 + R) R = Ressistence of bell 3 = 0.2(2 + R) 15 = 2 + R R\(_\theta\) = 15 - 2 = 13e (ii) E = 1(0.5 _ R) 1.5 = I(0.5 _ 13) = I(13.5) I = \(\frac{15}{135}\) = 0.11A

(a) If a stable element is bombarded by \(\alpha\) - particles the nucleus will become unstable and decay into other elements with the release of energy. Example; \(^{14}_{7}N + ^4_2He \to ^{17}_8 O + ^1_1H\) + Energy (b) When the nucleus of Uranium is bombarded with a neutron, it splits into two nearly equal parts with a release of energy and neutrons. These excess neutron, if not controlled, further bombard the remaining nuclei, causing a chain reaction and consequently an explosion, provided the critical mass of the Uranium is, exceeded \(^{235}_{92}U + ^1_0U \to ^{236}_{92}U \to ^{141}_{56}Ba + ^{92}_{36}Kr + 3^1_0n\) + Energy (c)(i) Radioactive substances give off identifiable radiation i.e \(\alpha, \beta\) and 8 radiations (ii) They are chemically identical to their non-radioactive counterparts and later can be made radioactive by bombardment with nuclear particles (iii) Radioactive materials decay with specific half-life. (d) Three applications of atomic energy (i) It is used for the production of nuclear weapons for warfare (ii) It is used in the generation of electric power (iii) It is used to detect leakages in underground pipes carrying oil or gases. (e) Two Postulates of Bohr's Model of the Atom: (i) Electrons occupy certain fixed orbits around the nucleus of an atom. When an electron is in one of these orbits it does not radiate any energy (ii) The angular momentum is quantized in the allowed orbit. Two Limitations To Bohr's Model: (i) It could not interpret the details of the spectra of complex atoms (ii) There is no theoretical basis for selecting the allowed orbits postulated.

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

M1U = (M1 + M2)V = 4.2 x 10 = (4.2 = 2.8)V ∴ V = 4.2 x 10/7.0 = 6m/s