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\(\theta\) = tan-1\(\frac{3}{7}\)

Thrust = \(\frac{\text{mass x velocity}}{time}\) = 10 x 5 x 103 = 5.0 x 104N

Intensity = \(\frac{1.002 \times 102}{60}\) = 1.67Nkg-1

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

V.R = \(\frac{1}{sin\theta}\) = 10 sin = \(\frac{1}{10}\)

Pressure = density x acceleration due to gravity x height. P = 4 x 1000 x 10 = 4 x 10\(^4\)Nm\(^{-2}\)( since max. pressure is required)

volume of object = \(\frac{400 \times 10^{-2}}{600}\) = 0.00067m3 volume of paraffin displaced = \(\frac{1}{2}\) x 0.00067 = 0.00034m3 mass of paraffin = 900 x 34 x 10-5 = 0.3kg weight of paraffin = 0.3 x 10 = 3N = upthrust = tension in the string

\(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\) V1 = 2V2 and T2 = 2T1 \(\frac{P_1\times 2V_2}{T_1}\) = \(\frac{P_2\times V_2}{2T_1}\) P2 = 4P1

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.

Distance during acceleration = area of the first triangle = \(\frac{1}{2}\) x 10 x 10 = 50m Distance during retardation( deceleration) = area of second triangle = \(\frac{1}{2}\) x 5 x 10 = 25m Total Distance = 50 + 25 = 75m.

Applying the principle of moment Clockwise moment = anti-clockwise moment. But moment = force x perpendicular distance F(200 - 20) = 1800 x 20 F = \(\frac{1800 \times 20}{180}\) F = 10 x 20 = 200N

For capacitors in series C\(_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2}\) C\(_{\text{eq}} = \frac{5 \times 15}{5 + 15} = \frac{75}{20}\) = 3.75\muF Charge on the series branch Q = Ceq \times V =3.75 \times 24 = 90 \muC. Potential difference across the 5 µF capacitor V\(_{5} = \frac{Q}{C} = \frac{90}{5}\) = 18 V

(a)(i) A machine is a device used to overcome a load at one end by the application of an effort at another end. Examples of simple machine include the screw jack, can opener, pulley system. (ii) M.A. is the ratio of the load (L) to the Effort (E). \(M.A = \frac{Load}{Effort}\) V.R. is the ratio of the distance moved by the effort tc the distarice moved by the load (at the same time). \(V.R = \frac{\text{distance moved by effort}}{\text{distance moved by load}}\) Efficiency (E) is the ratio of the useful work done by the machine (work output) to the total work put into the machine (work input). \(Efficiency = \frac{\text{work done on load}}{\text{work done by effort}} \times 100%\) \(Efficiency = \frac{M.A}{V.R} \times 100%\); \(Work = Force \times Distance\) \(Efficiency = \frac{\text{work output}}{\text{work input}} \) = \(\frac{Load \times \text{distance moved by load}}{Effort \times \text{distance moved by effort}}\) = \(M.A \times \frac{1}{V.R} \times 100%\) (b) The efficiency of a simple machine is not 100% because of the work done in overcoming friction. (c) length of tommy bar. L = 70cm = 0.7m; Pitch = 4.4mm = 0.0044m. (i) \(V.R = \frac{2\pi L}{pitch} = \frac{2 \pi \times 0.7}{0.0044}\) = \(2 \times \frac{22}{7} \times \frac{0.7}{44 \times 10^{-4}}\) \(V.R = 1000\) (ii) \(Efficiency = \frac{M.A}{V.R} \times 100%\) \(\frac{80}{100} = \frac{M.A}{1000} \implies M.A = 800\) (iii) \(M.A = \frac{Load}{Effort}\) \(Effort = \frac{8000 \times 10}{800} = 100N\) (iv) Work done = \(\frac{load \times \text{distance moved by load}}{effort \times \text{distance moved by effort}}\) = \(\frac{8000 \times 10 \times 0.2}{0.8} = 20,000J\)

(a) A mass of dry air is entrapped in a uniform capillary tube, sealed at one end, by means of a mercury pellet. Both the tube and thermometer are then firmly attached a half-meter rule and placed in a water bath as shown above. The water is stirred and its temperature \(\theta_{1}\) at a steady state is noted. The corresponding length, \(L_{1}\) of the entrapped air is also noted and recorded. The water is gradually heated and stirred uniformly until another steady temperature, \(\theta_{2}\) is attained and recorded. The corresponding length, \(L_{2}\) of the air column in the capillary tube is again read and recorded. The experiment is repeated to obtain more sets of readings. A graph of L against \(\theta\) is plotted as shown. Deduction: Since the volume V of the tube is proportional to its length, L the graph shows that the volume of the entrapped air increases (directly) with increase in temperature at constant pressure. Precautions: Dry air must be used. Also, error of parallax in reading thermometer and meter rule avoided. (b) \(S = \frac{V_{1} - V_{0}}{V_{0} \theta}\) = \(\frac{18.7 - 13.7}{13.7 \times 100}\) = \(\frac{5}{1370}\) = \(3.65 \times 10^{-3} K^{-1}\) (c) As temperature increases the kinetic energies the molecules increase and they bombard the wall the container at a faster rate since their velocities has increased. To keep the rate of change of momentum per unit area (i.e pressure) the same, the molecules will have to travel longer distances betweon collision This implies increase in volume.