Question Library

a) Effects to show the existence of matter waves: - diffraction of electrons — diffraction on neutron from crystal surface — diffraction pattern by reflecting hydrogen and helium molecules from crystal surface. (b) Effects to show that radiation behaves like waves: — compton effect — photoelectric effect — black body radiation — emission/absorption spectra — quantization of radiation energy proposed by Planck.

(a) Tensile stress is the force per unit cross-sectional area of a wire or rod. (b) Tensile strain is the extension per unit length. (c) Yield point is a point on a load versus extension graph that represents the least force beyond which plastic deformation occurs.

(a) The sum of the clockwise moments about a point = the sum of the anticlockwise moments about the same point (b)(i) Take a moment about the fulcrum F 60 x 24 = m x 18 m = \(\frac{60 \times 24}{18}\) = 80g (ii) Take a momen about F, 60 x (\(x\) - 13) = 80 x (48 - \(x\)) 3(\(x\) - 13) = 4(48 - \(x\)) 3\(x\) - 39 = 192 - 4\(x\) 7\(x\) = 231 \(x = \frac{231}{7}\) = 33cm (c)(i) (ii) (i) Minimum effort E = F + mgsin\(\theta\) = 1000 + 700 sin \(\theta\) = 1350N (ii) V.R = \(\frac{1}{sin \theta} = \frac{1}{sin 30}\) = 2 (ii) M.A = \(\frac{L}{E} = \frac{700}{1350}\) = 0.52

(a) Boiling point of a liquid is the temperature at which the saturated vapour pressure of the liquid is equal to the external atmospheric pressure. (b) Add and trap a few drops of the liquid in the closed limb of the J-tube above the mercury. Insert into a beaker of water the J-tube, stirrer and thermometer. Heat the beaker and content and stir continuously. Then read and record the temperature at which the mercury levels in the two limbs are the same. Heat the beaker a little further and remove heat source. Also read and record the temperature at which the mercury levels are once again the same. The average of the two temperature values is the boiling point of the liquid. (c) Heat lost by copper = Heat gained by water 0.3 x 400 (950 - 100) = 0.25 x 4200 x (100 - 25) + 2.26 x 10\(^6\) x m 102000 = 78750 + 2.26 x 10\(^6\)m m = \(\frac{102000 — 78750}{2.26 \times 10^6}\) = 0.0102876 kg or 10.3g

(a)(i) A wave motion is a process of transferring a disturbance in form of K.E from one point to another in a medium without any transfer of the particles of the medium. (ii) Radio wave Sound wave It is electro-magnetic wave. It is a transverse wave Requires no medium for propagation. Has velocity equal to that of light. It is a mechanical wave It is longitudinal wave Requires medium for propagation. Its velocity is less than that of light. (b) (i) Fill the burette with water. Set the prongs of the fork into vibration and bring it near the open end of the burette, then gradually open the tap to run out some water, there-by increasing the length of the air column until a loud sound is heard. Measure and record the length L\(_2\) of the air column. Further run out the water while le vibrating tuning fork is still held over open end of the burette until a second loud sound is heard, then read and record the length L\(_2\) of the air column. Last repeat the experiment to obtain L L\(_1\) and L L\(_2\) again and find the mean values of L L\(_1\) and L L\(_2\). Obtain the frequency of the tuning fork using f = \(\frac{v}{2(L_2 - L_1)}\) ii) Precautions: - repeat the experiment to obtain nean values - tap tuning fork on a pad not on a hard surface - hold tuning fork over the burette to avoid contact - obtain two lengths of air column to remove rnd correction. (c) \(f_o = \frac{v}{4L} = \frac{340}{4 \times 100 \times 10^{-2}}\) = 851Hz

(a) If the primary coil is connected to an A.C source, the A.C passing through the coil sets up a changing ternating magnetic flux in the soft iron core. The secondary coil is linked with the changing flux which Juces an e.m.f in it. Since the number of turns in a secondary is less than that in the primary, the induced e.m.f is less than e.m.f in the primary. For s reason the A.C output is less than the A.0 in primary. (b)(i) Ways in which heat is lost in a transformer: - Eddy current - Flux leakages - Hysteresis losses - Copper or joules losses (ii) Losses Minimizing method Eddy current losses. Flux leakage, Hysteresis losses. Copper or joules losses. Laminating the core. Design an efficient core. Core made of soft magnetic material of soft iron. Using low resistance coil. (c) (i) \(\frac{N_s}{N_p} = \frac{E_s}{E_p} = \frac{220}{4400}\) \(N_p : N_s = 20 : 1\) (ii) Efficiency = \frac{output}{input} = \frac{V_s \times I_s}{V_p \times I_p}\) = \(\frac{95}{100} = \frac{60}{4400 \times I_p}\) \(I_p = \frac{100 \times 60}{95 \times 440} = 0.014A\)

(a)(i) Thermally generated electrons from the filament are accelerated through high P.d and focused by the focusing cathode to a target. The energy of the impending electrons is converted mostly into heat which conducted through the copper anode. A small percentage of the energy is converted into X-rays which are radiated from the target through the window. (ii) Energy transformation: (b)(i) Hardness is the strength or penetrating power of the x-ray. (ii) Intensity is the energy radiated per unit time per unit area. (c)(i) Uses of x-rays: - Cure of cancer - Radio therapy - Detection of fracture in bones - Radiography - In metals to detect defect and fractures - In agriculture to kill germs - Analytical tool in crystallography - To detect metallic weapons (ii) Hazzards of over-exposure to x-rays; causes - leukemia - tissue damage - cataracts - skin burn or cancer - genetic mutations with undesirable heredity - baldness - graying.

Table of value/ Observation S/N L(cm) \(t_1\)(sec) T T\(^{2}\) L 1 130 45.70 2.29 5.22 100 2 110 42.00 2.10 4.41 80 3 90 38.00 1.90 3.61 60 4 70 33.10 1.66 2.74 40 5 50 28.70 1.44 2.06 20 Slope (s) = \(\frac{\bigtriangleup {T}^{2}}{\bigtriangleup {L}} = \frac{5.2-2.0}{100-18}\) S = \(\frac{3.2}{82}\) = 0.039 Intercept (c) = 1.3 Evaluation \(K_1\) = \(\frac{4 \pi ^{2}}{s}\) = \(\frac{4{(\frac{22}{7})^{2}}}{0.039}\) = \(\frac{4 \times 9.878}{0.039} = (\frac{39.51}{0.039})\)g = 1013g \(K_2\) = \(\frac{c}{s} = \frac{1.3}{0.039}\) = 33.33 Precautions: l avoided conical oscillations. l ensured firm suspension of pendulum ( l ensured reading is repeated (shown on the table). I avoided draught. I corrected for zero error of metre rule/stop clock I avoided parallax errors in reading the metre rule/stop clock. (b)i. The period of oscillation of a body is the time taken for the body to make one complete to and fro movement. ii. When a body is falling under its weight (alone), the rate of increase of its velocity with time (as a result of the earth's gravitational pull) is the acceleration of free fall due to gravity.

Table of values (Observation) R (\(\Omega\) \(V_1\)(v) V\(_{1}^{-1}\) 1 1.50 0.667 2 1.30 0.769 3 1.15 0.870 4 1.10 0.909 5 0.90 1.111 Slope (s) = \(\frac{\bigtriangleup {V}_{1}^{-1}}{DR} = \frac{1.22 - 0.6}{6.5 - 0.3}\) = \(\frac{0.62}{6.20}\) = 0.1 Intercept (c) = 0.57 Evaluation: y = \(\frac{1}{S} = \frac{1}{0.1}\) = 10 Precautions: I removed the key when not taking readings I ensured tight connections/clean terminals I avoided parallax error in reading the voltmeter I corrected for zero error on the voltmeter (b)i. The potential difference (pd) (in volts) between two points in a circuit is the work done (in joules) when one unit of positive charge (coulomb) moves from one point to the other ii. E = IR + Ir, 3 =0.5 x 4 + 0.5r, 3 = 2 +0.5r r = \(\frac{3 - 2}{0.5}\) = 2\(\Omega\)

For the uniform cube 10cm volume = 10 x 10 x 10 = 103 cm3 Mass = 700g ∴ density = (Mass)/Volume = (700)/100 = 0.7gcm-3 since the density is less than the density of water, the cube will float

R.D = (density of the subs)/density of water 19.2 =(P)/103; ∴P = 19.2 X 103 But Vol. = (Mass)/density = (2.4)/19.2 x 103 = 1.25 x 10-4 M3

The thermometric property of a constant volume thermometer is change in pressure with temperature

Electrical power = IV or I2R or (V)/R With V = 240v, R = 960Ω Power (V2)/R = (240 X 240)/960 =60Watts

In a capacitive circuit, current leads voltage by 90°, or voltage lags current by 90°

Power of heater = lvt = 2 x 240 x t heat required to boil the water = Mct = 2.5 x 4.2 x 103 x (100.52) = 2.5 x 4. 2 x 103 x 48 2 x 240 x t = 2.5 x 4.2 x 103 x 48 ∴ t = (2.5 x 4.2 x 103 x 48)/2 x 240 = 1050s or 1.05 x 103s