Question Library

(a)(i) Rotational Motion; Examples: (1) Motion of a wind mill. (2) Motion of a turntable or disc. (3) Rotation of the earth about its axis. (4) Movement of the blade of an electric fan, etc. (ii) Linear Motion; Examples are: (1) An athlete running on a straight track. (2) A car moving on a straight road. (3)A ball rolling on a level ground, etc. (b) In the laboratory, the mass(m) of an irregular shaped solid can be determined using a chemical or beam balance using a graduated measuring cylinder fill it partially with water and the initial volume V\(_2\) recorded, then completely immersed the irregular solid shape in the water and the final volume V recorded. Volume of the solid = V\(_2\) - \(V_1\) But density = \(\frac{Mass}{ Volume}\) = \(\frac{M}{V_2 - V_1}\) (c) Newton's second law of motion states that the time rate of change of momentum of a body is directly propotional to the applied force acting on it and takes place in the direction of the force. (d) Inertia is the reluctance of a body to move if it is at rest or to stop if it is already in motion. The more the mass of a body, the greater is its inertia. (e)(i) Net force = F - F\(_1\) = Ma 5.0 x 2.0 = 10N (ii) F = Ma + F\(_1\) = 10 + 5 = 15.0N (iii) µ = \(\frac{F_{fric}}{R} = \frac{F_{fric}}{mg} = \frac{5}{5 \times 10}\) = 0.1

(a) Heat capacity is the quantity of heat required to change the temperature of a body by 1K. The unit is JK\(^{-1}\) (b) Effects of heat on a substance are: (i) Increase in temperature. (ii) Change of state. (iii) Change in colour . (iv) Change in electrical resistance, etc. (c) To remove a tightly fitted glass cork from a reagent bottle, warm the bottle neck using warm cloth to make the glass bottle to expand creating space for easy removal of the cork. (d)(i) Pt = MC\(\Delta \theta\) \(\Delta \theta = \frac{pt}{Mc} = \frac{1200 \times 3 \times 60}{0.8 \times 4000}\) =67.5^\circ = final temp \(\theta\) = 67.5 + 30^\circ = 97.5^\circC (ii) Containers with tight-fitted lids are not suitable because with increase in temperature the steam molecules from the container are confined in the container, thereby increasing the rate of collission of the steam with the walls of the container, hence in pressure inside the container. Thus the high pressure can burst the tightly-fitted container and the hot content released. (iii) The observed dent on the container is due to the condensation of the steam in the container which-will lead to a decrease in pressure in the greater container making the atmospheric pressure greater than the pressure in the container.

(a) Characteristics of sound and their corresponding Factors Characteristics factors - Pitch - Timber or quality - Loudness or intensity - Frequency - Harmonics or Overtone - Amplitude (b) Resonance is a phenomenon which occurs when an external vibrating source of sound such as a sounding tunning fork set another system such as molecules of column into vibration; at a stage the system vibrates with maximum amplitude producing, a loud sound. (c) Echo enables the construction engineers to line the interior wall with sound absorbing materials to prevent echo. (d)(i) Period T = \(\frac{1}{f} = \frac{1}{2400} = 4.2 \times 10^{-7}\) = 1.2 x 10\(^{-3}\)ms\(^{-1}\) (ii) Velocity V = \(\omega r = 2 \pi fr\) = 2 x 3.142 x 2400 x 0.8 x 10\(^{-7}\) = 1.2 x 10\(^{-3}\)ms\(^{-1}\) (iii) Acceleration a = \(\omega^2r\) = (\(2 \pi f)^2r = (2 \times 3.14 \times 2400)^2 \times 0.8 \times 10^{-7}\) = 18.2 ms\(^{-2}\) (iv) p = \(\frac{F}{A}\) force F = PA = 3.6 x 10\(^{-5} \times 3.14 \times (2.1 \times 10^{-3})^2\) = 5.0 x 10\(^{-11}\) N

(a) Elecromotive force is the energy needed to take a unit positive charge round a circuit in which a cell is connected. Or The energy per unit charge transformed in the cell. Or e.m.f is the p.d across the terminals of the cell or battery or generator in an open circuit. (b)(i) The principle of operation of a potentiometer is based on the fact that the p.d across the wire is directly proportional to the length of the wire.V \(\alpha\) L where V = p.d across the wire. L = length of the wire. (ii) Advantages of potentiometer over voltmeter in measuring p.d are: (1) It is more sensitive. (2) It is more accurate. (3) It eliminates error due to internal resistance of the cell. (c)(i) Labelled sketch of a Gold leaf electroscope: (ii) Uses of Gold leaf electroscope are: (1) To detetct the nature of electrostatic charge. (2) To detect electrostatic charge. (3) To determine the magnitude of the induced charges. (d)(i) The mode of action of magnetic relay relies on the principle of electromagnetic induction. A small current passing through one circuit controls another circuit containing devices which require larger current. (ii) Factors affecting the magnitude of induced e.m.f in a coil are: (1) the number of turns in the coil. (2) the magnetic field strength. (3) the cross-sectional area of the coil. (4) the speed of rotation of the coil. (iii) F = IBLsin\(\theta\) force per unit length \(\frac{F}{L}\) = IBsin\(\theta\) = 5 x 2.0 x 10\(^{-3}\) x sin30 = 5.0 x 10\(^{-3}\)Nm\(^{-1}\)

(a) Einstein's photoelectric equation is E = W + K Emax, where E = Energy of incident radiation W = Work function K.E\(_{\text{max}}\) = Maximum kinetic energy of photoelectron. (b) For photocell factors responsible for: (i) Emission \(\to\) frequency or wavelength of incident radiation. (ii) Rate of emission \(\to\) intensity of incident radiation or number of photons. (iii) Energy of photoelectrons ---> frequency or wavelength of incident radiation. (c)(i) (\(\alpha\)) 218 + q = 222 q = 4 (\(\alpha\)) P = 84 + 2 = 86, m = 214 — 214 = 0 n = 83 — 84 = —1 x is an electron or beta particle. ii) Needs to safely dispose radioactive waste: (1) it emits harmful radiations. (2) it is toxic (d)(i) E = \(\frac{hc}{\lambda}\) = \(\frac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{2.4 \times 10^{8}}\) = 8.25 x 10\(^{-19}\)J (ii) E\(_4\) ----------------- - 6.0 x 10\(^{-19}\)J E\(_3\) ---------------- - 8.2 x 10\(^{-19}\)J E\(_2\) ---------------- - 8.8 x 10\(^{-19}\)J E\(_1\) ---------------- - 16.7 x 10\(^{-19}\)J The energy level transitions which cause the emission of the photon in (d)(i) is. E\(_3\)

\(\frac{t_{1}}{s}\) \(\frac{t_{2}}{s}\) mean \(\frac{t}{s}\) w=\(\frac{D}{t}\) v=dw 140.0 4.50 5.00 4.750 29.474 58.948 120.0 4.20 4.40 4.300 27.907 55.814 100.0 4.00 3.80 3.980 25.641 51.282 80.0 3.80 3.50 3.650 21.918 43.836 60.00 3.20 3.40 3.300 18.182 36.364 Slope = \(\frac{70-32}{5.6-2.3} = \frac{38}{3.3}\) = 11.51 X = Slope 's" signifies acceleration due to gravity. Precaution; i ensured that l avoided parallax error when taking readings on the metre rule. ensured that i took care of zero when recording the period of oscillation. ensured that I took repeated readings in order to avoid random errors. (b)i. Velocity ratio (VR) of an inclined plane VR = \(\frac{1}{sin \theta}\), where \(\theta\) is the inclination to the horizontal. ii. Mass of an object = 5kg Angle of inclination = 30^\circ Acceleration due to gravity 10m/s\(^{2}\) F = mgsin \(\theta\) F = 5 x 10 x sin 30^\circ = 5 x 10 x 0.5 = 25N

a = 1.5cm u/cm \(\frac{a}{cm}\) M=\(\frac{a}{d}\) m\(^{-1}\) 30.0 2.2 1.467 0.682 35.0 1.5 1.000 1.000 40.0 1.2 0.800 1.250 45.0 0.9 0.600 1.667 50.0 0.7 0.467 2.141 Slope (s) = \(\frac{\bigtriangleup {m}^{-1}}{\bigtriangleup {y}}\) = \(\frac{2.4 - 0.4}{54 - 24} = \frac{2}{30}\) = 0.06 \(X_1\) = when m\(^{-1}\) = 0 u = 23cm Precautions; i - avoided error due to parallel when taking readings on the metre rule. - made sure that, I took the height of the image when it is sharp. - ensured that the intensity of the source of light was maintained. - took repeated readings to avoid random errors. (b)i. When U iss 37cm 37 = 1.01m\(^{-1}\) 37 x \(\frac{1.01}{m}\) m = \(\frac{1.01}{37}\) m = 0.03 ii. Object beyond 2F\(^{1}\)

Electricity I = 0.36A \(\frac{x}{cm}\) \(\frac{I}{A}\) \(\frac{X^{-1}}{cm^{-1}}\) 20.0 1.14 0.050 35.0 0.90 0.029 45.0 0.84 0.022 60.0 0.78 0.017 80.0 0.66 0.013 Slope (s) = \(\frac{\bigtriangleup {X}^{-1}}{\bigtriangleup {I}}\) = \(\frac{0.08-0.01}{1.68-0.644} = \frac{0.07}{1.036}\) From the graph, When X\(^{-1}\) = 0 \(I_0\) of \(I_1\) = 0.448A x. Evaluate \(\frac{I_{0}}{I}\) = \(\frac{0.448}{0.36}\) = 1.244 Precaution; The key was opened when readings were not taken. I ensured tight connections of the connecting wire I avoided parallax errors when reading the potentiometer. (bi) Define emf of a battery: The e.m.f of a cell is the potential difference across its terminals when it is in an open circuit. i.e. not Supplying current to an external circuit. ii. A cell x emf is 1.018V. Lengt\(h_1\) on the potentiometer x = 50.0cm Lengt\(h_2\) on the potentiometer, y,= 75.0cm A cell Y emf is? Solution By formula; \(\frac{E_{1}}{E_{2}}\) = \(\frac{L_{2}}{L_{1}}\) \(\frac{1.018}{E_{2}} = \frac{75.0}{50.0}\) 75.\(E_2\) = 1.018 x 50 \(E_2\) = \(\frac{1.018\times 50}{75}\) \(E_2\) = 0.612V. see graph above

F + Ma = P a = \(\frac{P - F}{m}\) = \(\frac{40 - 12}{5} = 5.6ms^{-2}\)

M = zit I = \(\frac{M}{Zt}\) = \(\frac{0.66}{30 \times 60 \times 33 \times 10^{-5}}\) = 1.11A Error = 1.11 - 1 = 0.11A

The photoelectric effect is dependent upon various factors including frequency of light, intensity of light, nature of material, energy of light and potential difference.

The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons emitted in this manner are called photoelectrons.

-We see that as it passes through centre of curvature it reflects back its path -Incident to the mirror at 90º - and with the normal at 0º

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

Lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) where f, d\(_o\) and d\(_i\) are the focal length, object distance and image distance respectively. The distance from the lens to the focal point is called the focal length. For converging lenses, the focal length is always positive, while diverging lenses always have negative focal lengths. ⇒ \(\frac{-1}{30} = \frac{1}{d_o} + \frac{1}{20}\) \(\frac{1}{d_o} = \frac{-1}{30} - \frac{1}{20}\) \(\frac{1}{d_o} = \frac{-2 + -3}{60}\) → \frac{-5}{60}\) = \frac{-1}{12}\) \(d_o = - 12 cm\) The object distance -12cm = 12cm on the other side of the lens.