Solutions to SQA examination

Higher Grade Physics

2005

Return to past paper index page.

Section A

1. A 11. E 2. E 12. C 3. C 13. B 4. B 14. B 5. E 15. A 6. B 16. A 7. C 17. D 8. D 18. D 9. E 19. B 10.C 20. D

Section B

21.a.i. (A) Mean = sum/N Mean = (0.015 + 0.013 + 0.014 + 0.019 + 0.017 + 0.018)/6 Mean = 0.096/6 Mean = 0.016s (B) RND Uncertainty = (Max - Min)/N RND Uncertainty = (0.019 - 0.013)/6 RND Uncertainty = 0.006/6 RND Uncertainty = 0.001s a.ii. u = 0m/s v = 1.25m/s [ v = card width/t = 0.020/0.016 = 1.25m/s] s = 0.60m Use: v2 = u2 +2as a = (v2 -u2)/2s a = (1.252 - 0)/2x0.6 a = 1.5625/1.2 a = 1.30m/s2 b.i. Photoconducting mode. ii. When the beam is broken the photodiode is in darkness and does not conduct. This increases the voltage at point Y. When this voltage reaches 2.0V or higher the MOSFET conducts and starts the timer. 22.a.i. FV(total) = 2FRcos21o FV(total) = 2 x 4.5x103cos21o FV(total) = 8402.22N a.ii. Fun = FV(total) - mg Fun = 8402.22 - (236 x 9.8) Fun = 6089.42N a = Fun/m a = 6089.42/236 a = 25.8m/s2 b. The tension in the cords decrease as the capsule rises. This means the total upward force and therefore the unbalanced force decreases, reducing the acceleration. c. When there is no upward tension force the seat and occupants will both be projectiles in free fall. Both will be accelerating downwards at a rate of 9.8m/s2. 23.a. m = 12g = 12x10-3kg v = (0.5x0.3x0.1)m3 = 0.015m3 r = m/v r = 12x10-3kg/0.015m3 r = 0.8kg/m3 b. Pressure is directly proportional to the depth. This means that the lower surface has a greater pressure acting on it than the upper surface. The resulting forces (F = PA) are shown in the diagram. Fbottom is greater than Ftop. Fupthrust = Fbottom - Ftop c. The initial accelerating force is given by: Fun = Fupthrust - weight(w) However, as the float accelerates upwards and its speed increases the water resistance force also increases. This means the acceleration of the float decreases. Fun = Fupthrust - (weight(w) + water resistance) d. As the density is greater the mass and weight of the float is greater. The upward unbalanced force: Fun = Fupthrust - weight(w) ...is less. The acceleration : a = Fun/m ...is also less. 24.a.i.
Temperature/oC 25 50 75 100
Temperature/k 298 323 348 373
Volume/ml 20.6 22.6 24.0 25.4
(Volume/Temperature)/(ml/k) 0.0691 0.0700 0.0690 0.0681
Volume/Temperature(k) = constant Volume = constant x Temperature(k) Volume varies directly with the kelvin temperature. a.ii. T(k) = (65 +273)k = 338k Constant = 0.69ml/k Volume = constant x Temperature(k) Volume = 0.069 x 338 Volume = 23.3ml a.iii. As the temperature of the gas increases the kinetic energy and speed of the gas particles increase. This means that the average force, per collision, of the gas particles with the container walls increases. This produces an upward force on the syringe causing it to move upward until the pressure inside is the same as the air pressure outside. b.i. This is a wheatstone bridge circuit and is balanced when the voltmeter reads 0V. General equation: R1/R2 = R3/R4 Rvariable/500 = 2000/1000 Rvariable = 500 x 2 Rvariable = 1000W b.ii. 25.a. e.m.f The open circuit voltage. OR The energy supplied by the battery to each coulomb of charge. b.i. (A) The e.m.f. is found at the y-intercept. e.m.f. = 6.0V. (B) The internal resistance(r) is equal to the negative gradient (-m) m = (y2 - y1)/(x2 - x1) m = (6.0 - 1.0)/(0 - 1.0) m = -5 r = -m = -(-5) = 5W b.ii. E = I(R + r) R = (E/I) - r R = (6/0.3) - 5 R = 20 - 5 = 15W ...as required. c. 1/Rp = 1/R1 + 1/R2 1/Rp = 1/15 + 1/30 1/Rp = 3/30 Rp = 10W I = E/(Rp + r) I = 6.0/(10 + 5) I = 6.0/15 I = 0.4A 26.a.i. a.ii. The charging time is longer because less charge is transferred per second with a higher value resistor. a.iii. Q = CVC Vsupply = VC + VR VC = Vsupply - VR VC = 9 - 4 VC = 5V Q = 2200x10-6 x 5 Q = 11000x10-6C (11mC) b.i. Maximum energy is stored when: VC = Vsupply = 9V Emax = 1/2(CVC2) Emax = 1/2(2200x10-6 x 92) Emax = 1100x10-6 x 81 Emax = 89100x10-6J (89.1mJ) b.ii. The maximum discharge current occurs when the maximum voltage across the capacitor is applied across the resistor. Imax = V(R)max/R Imax = 9/100x103 Imax = 0.00009A (90mA) 27.a. x = source y = gate z = drain b.i. Vout = Rf/R1(V2 - V1) 2 = 1x106/500x103(V2 - V1) (V2 - V1) = 2/2 (V2 - V1) = 1V b.ii. V1 = VQ V2 = VP V2 = V1 + 1 VP = VQ + 1 The voltage at Q can be solved by proportion. VQ =[R20kW/(R20kW + R100kW)]Vsupply VQ =[20kW/(20kW + 100kW)]Vsupply VQ =[20kW/120kW]Vsupply VQ =[(1/6)]12 VQ = 2V VP = VQ + 1 VP = 2 + 1 VP = 3V (voltage across thermistor) VP/V75kW = Rthermistor/R75kW 3/V75kW = Rthermistor/75kW V75kW = 9V [as Vsupply = VP + V75kW] 3/9 = Rthermistor/75kW Rthermistor = 1/3(75kW) Rthermistor = 25kW 28.a.i. n =singlass = sinqair/sinqglass nglass = sin47o/sin29o nglass = 1.51 a.ii. sinqair = nglasssinglass sinqair = 1.51xsin31o sinqair = 0.778 qair = 51.1o b.i. When the path difference between adjacent slits to a point on the screen is zero or a whole number of wavelengths a bright spot is produced because of constructive interference. b.ii. dsinq = nl 300 lines/mm => 300000 lines/m d = 1/300000 = 3.3333x10-6m n = 2 (second order) sinq = nl/d sinq = 2 x 650x10-9/3.3333x10-6 sinq = 0.390 q = 22.95o b.iii. The blue fringes are closer together because shorter wavelength blue light is diffracted less than the longer wavelength red light. 29.a. Above the threshold frequency the photoelectric current is directly proportional to the intensity of radiation. b.i. Ek(max) = Ephoton - Ework function fphoton = v/l fphoton = 3x108/400x10-9 fphoton = 7.5x1014Hz Ephoton = hf Ephoton = 6.63x10-34 x 7.5x1014 Ephoton = 4.9725-19J Ek(max) = 4.9725x10-19 - 3.11x10-19 Ek(max) = 1.8625x10-19J b.ii. The photons may have a kinetic energy greater than the work the electric field does trying to stop them. Ek(max) > eV [V = potential difference applied by battery e = electronic charge] 30.a.i. At 200cpm material thickness = 1mm At 100cpm material thickness = 6mm 5mm of material is required to half the count. This is the half value thickness. a.ii. Three half value thicknesses are required. Thickness = 3 x 5 = 15mm b. Htotal/h = Hg/h + Hthermal neutrons/h + Hfast neutrons/h Hg/h = QgDg/h Hg/h = 1 x 2.0mGy/h Hg/h = 2.0mSv/h Hthermal neutrons/h = Qthermal neutronsDthermal neutrons/h Hthermal neutrons/h = 3 x 400mGy/h Hthermal neutrons/h = 1200mSv/h = 1.2mSv/h Hfast neutrons/h = Qfast neutronsDfast neutrons/h Hfast neutrons/h = 10 x 80mGy/h Hfast neutrons/h = 800mSv/h = 0.8mSv/h Htotal/h = (2.0 + 1.2 + 0.8)mSv/h Htotal/h = 4.0mSv/h thours = 500/4 thours = 125hours

END OF QUESTION PAPER

Return to past paper index page.