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(a)(i) Electric field intensity = Electric force per unit charge OR As negative potential i.e.E = \(\frac{-V}{d}\) OR E = \(\frac{F}{Q}\). where F is the force, Q is the charge and E electric field intensity. (ii) Electric potential =- work done per unit charge to bring a (positive) charge from infinity to a point in an electric field. (b) (c) (i) A dielectric substance is any insulating material placed between the plates of a capacitor (ii) Factors Effect on capacitor common area of plates Distance between plates Nature of dielectric capacitance increases with area capacitance decreased with increasing distance capacitance increases with dielectric constant C = \(\frac{KE_oA}{A}\) (iii) C\(_{2.3}\) = C\(_2\) + C\(_3\) = 20 + 20 = 40\(\mu\) F \(\frac{1}{C_T} = \frac{1}{C_{2.3}} + \frac{1}{C_1}\) \(C_T = \frac{\frac{C_{2.3} \times C_1}{C_2.3 + C_1} = \frac{40 \times 40}{40 \times 40}\) = 20\(\mu F\)

(i) Electrical \(to\) thermal \(to\) mechanical (kinetic energy) \(to\) thermal \(to\) electromagnetic (ii) Distinction between hard and soft X-rays Hard X-rays Soft x-rays (i) High (target) voltage (ii) short wavelength (iii) High penetration (iv) produced by high voltage Low (target) voltage Long wavelength Low penetration Produced by low voltage (iii) Uses: - Used to detect hidden cracks in materials - Used to detect alteration to work of art - To show broken bones in human body - used in crystallography i.e. to study crystals - used in diagnosis of ulcer (iv) Hazards: genetic mutations - skin burn - leukemia - Tissue damage - Tissue damage - baldness - graying - cataracts (v) Safety precautions: use of dosimeter, badges of GM counters - wearing of lead coats - Periodic medical counters - wearing of lead coats - periodic medical check up (b) \(^2_1H + ^2_1H \to ^3_1H + ^1_1H + Q\) Let atomic mass unit of tritium \(^3_1H\) be X 2.01410 x 2 x 931 = (x + 1.00783) 931 + 4.03 \(\frac{3750.254 - 4.03}{931}\) = x + 1.00783 4.0238 - 1.0078 = x 3.106 = x

M(g) \(W_1\)(g) \(W_2\)(g) \(W_3\)(g) U=(\(W_1\) - \(W_2\))g V=(\(W_1\)-\(W_3\))g 5.0 5.00 4.20 4.70 0.80 0.80 10.0 10.00 7.90 8.60 2.10 1.40 15.0 15.00 12.75 13.50 2.25 1.50 20.0 20.00 17.10 16.40 2.90 3.60 25.0 25.00 22.10 21.40 2.90 3.60 Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{2.35-1.00}{3.25-1.65} = \frac{1.35}{1.60}\) = 0.84 Precautions: I avoided parallax error in reading spring balance. I noted zero error of spring balance. l avoided splashing of liquid. I avoided object touching the bottom or side beaker. I cleaned the mass before dipping it into the liquid. (bi) Weight of kerosine displaced = volume of object x density of kerosine x g \(\frac{\text {Mass of object}}{\text {density of of object}}\) x density of kerosine x g = \(\frac{20 \times 10^{-3}kg \times 8.0 \times 10^{2} kgm^{-3} \times 10ms^{-2}}{ 8.0 \times 10^{-3} kgm^{-3}}\) = 0.02N = upthrust of the weight of object in air = mass x g = 20 x 10\(^{-3}\) X 10 = 0.2N Tension in the spring = weight of an object in the air - upthrust = 0.2N - 0.02N=0.18N The reading on the spring balance= 0.18 ii. Archimede's principle states that when a body is wholly or partially immersed in a liquid, it experiences an upthrust equal to the weight of the liquid displaced. On the other hand, the law of floatation states that a floating body displaces its own weight of the liquid in which it floats.

Let the forces be F1, and F2. F1 = m1 a1 = 3 x 2 = 6N F2 = m2 a2 = 1 x 9 = 9N F1 : F2 = 6:9 = 2:3 Thus ratio of forces is 2: 3

Centripetal Force (F\(_c\)) = \(\frac{mv ^2}{r}\) = \(m\omega^2r\) (Since V = \omegar) F = 0.01 x \(100^2 \times 0.2 \) F = 20 N

From principle of conservation of linear momentum, (0.05 x 200) + (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic). 10 + 0 = V Thus V = 10m/s. Recall Kinetic Energy = \(\frac{1}{2} mv^2\) \(\therefore\) K.E = 1/2 (0.05 + 0.95) x 10\(^2\) K.E = 1/2 (1 x 100) = 50 J.

Total Energy = Kinetic Energy + Potential Energy P.E = mgh = 0.1 x 10 x 10= 10 J K.E = \(\frac{1}{2}\) MV\(^2\) = \(\frac{1}{2}\) x 0.1 x 10\(^2\) = 0.5 x 0.1 x 100 = 5J T.E = K.E + P.E = 5 + 10 = 15J

To calculate the power developed in the engine, we need to determine the work done and the time taken. 1. Calculate the final kinetic energy (KE) of the car: \(KE = \frac{1}{2}mv^2 \) where \( m = 800 \text{ kg} \) and \( v = 25 \text{ m/s} \). \( KE = \frac{1}{2} \times 800 \, \text{kg} \times (25 \, \text{m/s})^2 \) \( KE = 0.5 \times 800 \times 625 = 250000 \text{ J} \, (\text{joules})\) 2. Calculate the power (P) developed in the engine using the formula: \(P = \frac{W}{t} \) where \( W \) is the work done (in this case, equal to the kinetic energy calculated) and \( t \) is the time taken. \(P = \frac{250000 \, \text{J}}{20 \, \text{s}} = 12500 \text{ W} \, (\text{watts}) \) Thus, the power developed in the engine is 12500 watts (or 12.5 kW).

When the brakes in a car are applied, the frictional force on the tires acts in the opposite direction of the car's motion, slowing down the vehicle.

R.D = \(\frac{\text{loss in weight in liquid X}}{\text{ loss in weight in water}}\) R.D = \(\frac{(10 - 7)}{(10 - 6)}\) = \(\frac{3}{4}\)

As temperature rises, the kinetic energy of the molecules increases, reducing the cohesive forces at the surface and leading to a decrease in surface tension.

At constant T, P1 V1 = P2 V2. Let P1 = P0 and V1 = V0, then solve accordingly. As pressure is varies inversely as volume according to: Boyle's law, also referred to as the Boyle–Mariotte law, or Mariotte's law, is an experimental gas law that describes how the pressure of a gas tends to decrease as the volume of the container increases. As volume is reduced by one-fifth; pressure will increase by five times.

Energy at the height = mgh This energy is converted to heat energy - mc\Deltat ∴ From the principle of energy conversion, mc\Deltat = mgh m x 450 x \Deltat = m x 10 x 20 ∴ \Deltat = \(\frac{m \times 10 \times 20}{(m \times 450)}\) = \(\frac{200}{450}\) = \(\frac{4}{9}\)^\circC

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

The general wave equation is given by; y = A Sin(wt - 2\pix/Wavelength) where w = angular velocity Thus, comparing this equation with the equation y = 0.25 x 10-3 Sin (500t - 0.025x) That Amplitude = 0.25 x 10-3m The angular velocity(w) = 500 (the coefficient of t in both equations) ∴ w = 500 rad s\(^{-1}\) = 5.00 x 10\(^2\)rad s\(^{-1}\)