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(a) The apparent cubic expansivity of a liquid is defined as the apparent increase in volume per unit volume per unit rise in temperature. (b)(i) Procedure: Density bottle dried and weighed, M,, fill bottle with liquid, replace the stopper and wipe the bottle dry. Weigh bottle and content M\(_2\), suspend the bottle with content in a water bath. Record the initial temperature of the water in the bath T\(_1\). Then heat water gradually up to boiling point. Notice that some liquid is expelled through the orifice of the stopper and the water in the bath is allowed to continue boiling until no more liquid is expelled through the orifice of the stopper.Record the final temperature T\(_2\) of the water bath. Remove the bottle from the bath, allow it to cool, wipe, dry it and reweigh M\(_3\). Calculation: Mass of the liquid expelled from the bottle = (M\(_2\) - M\(_3\)) Mass of the liquid left in the bottle = (M\(_3\) - M\(_1\) Temperature change, (T\(_2\) - T\(_1\)) Apparent cubic expansivity \(\alpha\) a = \(\frac{\text{Mass of the liquid expelled}}{\text{ Mass of remaining liquid \times change in temperature}}\) \(\alpha\) a = \(\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)}\) (ii) Precautions; (1) Avoid parallax in reading thermometer (2) Bottle should not be submerged (c) Apparent cubic expansivity \(\alpha\) a = \(\frac{M_2 - M_3}{(M_3 - M_1) - (T_2 - T_1)} = \frac{44.25 - 42.02}{42.02 \times 50}\) \(\alpha\) a = 1.06 x 10\(^{-3} k^{-1}\) But real cubic expansivity = \(\alpha\) a =+ \(\alpha\) g Where \(\alpha\) g is the cubic expansivity of the containing vessel i.e. glass \(\alpha\) a = 1.09 x 10\(^3 K^{-1}\)

3(a) This means that the ratio of the charge on the capacitor to the potential difference between the plates is 4LF. (b)(i) (1) Area of plates (2) Dielectric between plates (3) Distance between plates (b)(ii) (1) To separate a.c from d.c. (2) To control current in a.c circuit (3) To store charges or energy (c) Workdone, W = QV Average work done is given by \(\frac{1}{2}\) x QV .........(i) But V = \(\frac{Q}{V}\) ...............(ii) Substituting (ii) in (i) W = \(\frac{1}{2}\) x Q x \(\frac{Q}{C}\) Energy stored W = \(\frac{1}{2} \times \frac{Q^2}{2} \frac{1}{2} \frac{Q^2}{C}\) (i) Q\(_2 = C_2V\) = 100 x 10\(^{-6}\) C = 100\(mu\)C \(Q_1 = C_1 V\) = 3 10\(^{-6}\) x 50 = 150\(\mu\)C (iii) P.d across each capacitor is 50V (iii) E = \(\frac{1}{2} CV^2\) Since capacitors are in parallel C = \(\frac{1}{2}\) x 5 \(\mu\)F x 50 x 50 = \(\frac{1}{2}\) x 5 x 10\(^{-6}\) x 50 x 50 = 6.25 x 10\(^{-3}\)J

(a)(i) Viscosity is the property of a fluid which tends to prevent or oppose motion of one layer or part over another. It varies with the temperature and nature of the fluid. (ii) A body falling through a fluid eventually attains a steady velocity called terminal velocity. At this point the weight of the body equals the upthrust and viscous force. (b)(i) Description: A tall glass jar is filled with glycerine anJ supported vertically. Bands (acting as markers) are attached to points A.B.C.D.E and F at equal intervals. The ball is allowed to fall from the top of the jar while the time is noted as the ball passed each marker in turn. It will be observed that the time intervals between A and B, B and C are different while the time intervals for C and D, D and E, E and F are equal. Showing that a steady terminal velocity is reached from C. This ;Weans that \(\frac{CD}{t} = \frac{DE}{t} = \frac{EF}{t}\) is the terminal velocity. (ii) Precautions: (1) Drop the ball gently. (2) Ball must fall centrally down the glycerine (c)(i) (1) Viscosity is responsible for the different rates of flow of fluid (2) Viscosity affects motion of bodies in fluids. (ii) (1) Viscous liquids are used as lubricants (2) The knowledge of viscous drag is applied in the design of ships and aircraft.

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

Mechanical Advantage = \(\frac{Load}{Effort}\) but Load = 30N and Effort = 20N M . A = \(\frac{30}{20}\) = 1.5

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

P\(_1\)V\(_1\) = P\(_2\)V\(_2\) When the bubble comes to the surface and pressure drops 3 times, the volume increases 3 times. V\(_2\) = 3 x 2 = 6 cm\(^3\)

To find the ratio of the final volume of the gas to its initial volume, we use the ideal gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) Rearranging gives: \(\frac{V_2}{V_1} = \frac{P_1 T_2}{P_2 T_1}\) Substituting the values: - \( P_1 = 200 \, N/m^2 \) - \( P_2 = 100 \, N/m^2 \) - \( T_1 = 400.15 \, K \) - \( T_2 = 200.15 \, K \) We have: \(\frac{V_2}{V_1} = \frac{200 \times 200.15}{100 \times 400.15}\) Calculating the ratio: \(\frac{V_2}{V_1} = \frac{40030}{40015} \approx 1.0:1\) Thus, the ratio of the final volume of the gas to its initial volume is: 1.0:1

Set initial total energy to zero: \(\frac{1}{2}mv^2 - \frac{GMm}{R} = 0\) \(\frac{1}{2}mv^2 = \frac{GMm}{R}\) \(v^2 = \frac{2GM}{R}\) \(v = \sqrt{\frac{2GM}{R}}\)

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

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

To calculate the reactance of a capacitor, we use the formula: \(X_C = \frac{1}{2 \pi f C}\) Given: C = 25\(\mu\)F = 25 \(\times 10^{-6}\), F = 200 Hz Substituting the values into the formula, we approximate \( \pi \) as 10: \(X_C = \frac{1}{2 \times 10 \times 200 \times (25 \times 10^{-6})}\) Calculating this: \(X_C = \frac{1}{2 \times 10 \times 200 \times 25 \times 10^{-6}} = \frac{1}{0.01} = 100 \Omega\) Thus, the capacitive reactance is: 100\( \Omega\)

\(E = \frac{CV^{2}}{2}\) \(V^{2} = \frac{2E}{C}\) \(V^{2} = \frac{2\times 40}{5 \times 10^{-6}}\) \(V^{2} = 16 \times 10^{6}\) \(V = 4 \times 10^{3} = 4000v\)

Pressure = heg = m x \(\frac{kg}{m^3}\) x \(\frac{m}{5^2}\) = ML-1T-2

s = 10-2 on earth g = 6 ms-2 on earth Earth on moon = \(\frac{10}{6}\) = 1.6ms-2