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(a)(i) Nuclear Fission is the splitting up of a heavy nucleus into lighter nuclei with release of a huge amount of energy and neutrons. (ii) (\(\alpha\)) graphite; (\(\beta\)) Boron rods; (\(\gamma\) )Liquid sodium Material Function Graphite Slows down high energetic neutrons in the reaction chamber. Boron rods To absorb excess neutrons so as to control fission rate Liquid sodium Extraction / conduction of excess heat from the reaction chamber. (b)(i) E\(_n\), n = \(\alpha\) ---------------------------------- 0.eV E\(_4\), n = 5 ---------------------------------------------------- - 0.54eV E\(_3\), n = 4 ---------------------------------------------------- - 0.85eV E\(_2\), n = 3 ---------------------------------------------------- 1.51eV E\(_1\), n = 2 ------------------------------------------------------ 3.39eV (ii) Using \(\lambda\) = \(\frac{hc}{\Delta E}\) = \(\frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{(-1.51 - ( - 13.60)) \times 1.6 \times 10^{-19}}\) = 1.02 x 10\(^{-7}\)m (c) \(\lambda\) = \(\frac{0.693}{ T^{\frac{1}{2}}}\) = \(\frac{0.693}{5572}\) = 0.0001244 years\(^{-1}\) N = Noe\(^{- \lambda t}\) e\(^{0.0001244t}\) = 1.2632 0.0001244t = 0.234 t = 1878 years

S/N L/CM t/s T=t/o/S LogT/S LogL/cm 1 80.0 17.90 1.79 0.252 1.903 2 70.0 17.30 1.73 0.238 1.845 3 60.0 15.34 1.53 0.185 1.775 4 50.0 13.90 1.39 0.145 1.698 5 40.0 13.06 1.31 0.155 1.602 Slope (s) = \(\frac{\bigtriangleup \text{logT}_{2}}{\bigtriangleup\text{logL}_{2}}\) - \(\frac{\bigtriangleup \text{logT}_{1}}{\bigtriangleup\text{logL}_{1}}\) = \(\frac{0.15 - 0.05}{0.164 - 0.148}\) = \(\frac{0.100}{0.016}\) = 6.25 PRECAUTIONS 1. avoided zero error in the reciding of the stop watch 2. avoided Parallax in reading the metre rule and the stop watch 3. ensured that the support of the bifilar pendulum is rigid 4. ensured smooth and regular oscillations in a horizontal plane. b. i. Simple harmonic motion is defined as the motion of an object whose acceleration is proportional to the distance from a fixed point and is always directed towards that point see graph above

bo = 102^\circc S/N M/g bi/ ^\circc \(\theta_{1}\)=(bi-bo)/^\circ 1.0 10.0 104.00 104 - 102 = 2.00 2.0 20.0 106.00 106 - 102 = 4.00 3.0 30.0 108.00 108 - 102 = 6.00 4.0 40.0 110.00 110 - 102 = 8.00 5.0 50.0 112.00 112 - 102 = 10.00 Slope, s = \(\frac{\bigtriangleup M_{2} - \bigtriangleup M_{1}}{\bigtriangleup\theta_{2} - \bigtriangleup\theta_{1}}\) = \(\frac{40}{8} - \frac{10}{2} = \frac{30}{6}\) S = 5 PRECAUTIONS 1. stirred the water gently to ensure uniform temperature. 2. avoided splashing of the water 3. avoided error due to parallax in reading the thermometer and the stop clock. (b) i. Boiling point is defined as the point when liquid changes to vapour when the saturation vapour pressure is equal to the atmospheric pressure. ii. Impurities affect the boiling of a liquid because their presence inicreases the boiling rboint of the liquidelcom- pared to the boiling point of a pure solvent.

\(\frac{g}{R^2}\) = \(\frac{g}{4^2}\) = \(\frac{g}{16}\)

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

The temperature difference between the heated liquid and the surrounding environment (the enclosure) is a crucial factor in determining the rate of heat transfer. According to Newton's Law of Cooling, the rate of cooling is proportional to the temperature difference between the object and its surroundings.

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

F = Ma + mg = 60 x 10 x 60 x 10 = 1200N

Clockwise moment = Anti-clockwise moment 10 x 30 = 15 x X X = 20cm

The total time of flight is given as \(T = \frac{2usin\theta}{g}\) Hence, time to maximum height is half of the total time of flight, t = \(\frac{u sin\theta}{g}\) = \(\frac{300 \text {sin} 30}{10}\) = 15s

t = \(\frac{v - u}{a}\) = \(\frac{20 - 0}{2}\) = 10s

T = \(\frac{PV}{nR}\) = \(\frac{7.6 x 10^6 x 10^-3}{6 x 8.3}\) = 153oC

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.

a manometer is used to measure pressure in liquids and gases. Manometer gauges measure the pressure of gases and vapors from atmospheric to about 1 Pa by balancing the force exerted by the gas or vapor against the weight of a column of liquid, usually mercury, water, or oil.