SRJC H2 PHY P3 SOLUTION

SRJC 2011 9646/Prelim/2011 SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 PHYSICS 9646 Preliminary Examination Paper 3 Longer Structured Questions 2 hours Candidates answer on the Question Paper. No additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, civics group and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer any two questions. At the end of the examination, fasten all your work securely together. The number of marks is given in bracket [ ] at the end of each question or part question. This document consist of 24 printed pages and no blank page For Examiner’s Use Section A 1 2 3 4 5 Section B 6 7 8 Total NAME CG INDEX NO.

2 SRJC 2011 9646/Prelim/2011 DATA AND FORMULAE Data speed of light in free space, c = 3.00  108 m s1 permeability of free space, 0 = 4   107 H m1 permittivity of free space, 0 = 8.85  1012 F m1 (1 / (36 π)) x 109 F m1 elementary charge, e = 1.60  1019 C the Planck constant, h = 6.63  1034 J s unified atomic mass constant, u = 1.66  1027 kg rest mass of electron, me = 9.11  1031 kg rest mass of proton, mp = 1.67  1027 kg molar gas constant, R = 8.31 J K 1 mol1 the Avogadro constant, NA = 6.02  1023 mol1 the Boltzmann constant, k = 1.38  1023 J K1 gravitational constant, G = 6.67  1011 N m2 kg2 acceleration of free fall, g = 9.81 m s 2 Formulae uniformly accelerated motion, s = ut + ½ at 2 v2 = u 2 + 2as work done on/by a gas, W = p V hydrostatic pressure, p = gh gravitational potential,  = – r Gm displacement of particle in s.h.m., x = x 0 sin t velocity of particle in s.h.m., v = v o cost v =  22 0 xxω  resistors in series, R = R 1 + R2 + … resistors in parallel, 1/ R = 1/R1 + 1/R2 + … electric potential, V = Q / 4or alternating current/voltage, x = x 0sin t transmission coefficient, T  exp(2kd) where k = 2 2 h E)m(U8 π radioactive decay, x = x0 exp(t) decay constant,  = 2 1 693.0 t

3 SRJC 2011 9646/Prelim/2011 [Turn Over For Examiner’s Use Section A Answer all the questions in this section 1 (a) Define gravitational potential at a point. ........................................................................................................................... .................. ......................................................................................................... .................. .....................................................................................................[1] (b) Two masses m1 and m2 are placed at positions as shown in the diagram, where m1 = 25 000 kg and m2 = 15 000 kg. (i) Determine the gravitational potential at t he origin due to m1 and m2 respectively. cm0.50.40.3r 22 1  cm12.200.90.18r 22 2  [M1] 15 11 1 1 Jkg10335.3050.0 )25000)(1067.6( r GM    [A1] 11 612 2 (6.67 10 )(15000) 4.973 100.2012 GM Jkgr       [A1] gravitational potential due to m1 = ............................................... J kg-1 [0] gravitational potential due to m2 = ............................................... J kg-1 [3] x/cm y/cm 0 4.0 18.0 9.0 3.0 m1 m2

4 SRJC 2011 9646/Prelim/2011 For Examiner’s Use (ii) Hence, or otherwise, det ermine the total gravitati onal potential energy of a 1000 kg mass placed at the origin. 1565 T Jkg1083.3)10973.4(10335.3   [M1] J1083.3)1083.3)(1000(mU 25   [A1] total gravitational potential energy = ........ ........................................... J [2] (iii) State the work required to move the 1000 kg mass from the origin to infinity. J1083.3W 2 work done = ...... ......................................... J [1] 2 The figure below shows the variation of pr essure with volume for a fixed mass of ideal gas. The gas is taken from state A to st ate C through two different paths ABC and path ADC. D 1.89  107 P/ Pa V/ 105 m3 1.50  106 7.0 50.0 C 1200 K A B 95.2 K

5 SRJC 2011 9646/Prelim/2011 [Turn Over For Examiner’s Use (a) Calculate the number of moles in the fixed mass of gas. 75 Consider point C, 1.89 10 7.0 10 8.31 1200 0.133 [A1 ] pV nRT pVn RT      number of moles of gas = …………………… [1] (b) (i) During process BC, the temperature of the gas rises from 95.2 K to 1200 K at constant volume. The molar heat capacity of the gas at constant volume is 12.5 J K 1 mol1. During process AB, 1613 J of thermal energy is lost. By showing your working clearly, calc ulate the change in internal energy from A to C. [Note: Molar heat capacity is the am ount of thermal en ergy required to raise the temperature of a mole of gas by one degree celcius]   64 5 1613 1613 1.5 10 5.0 10 7.0 10 1613 645 968 [M1 ] 0.133 1200 95.2 12.5 0 1837 [M1 ] 968 1837 869J [A1] AB AB AB AB BC BC BC AC AB BC UQ W pV UQ W UU U                          change in internal energy = …………………….. J [3]

6 SRJC 2011 9646/Prelim/2011 For Examiner’s Use (ii) Hence, calculate the thermal energy lost during process ADC.   74 5869 0 1.89 10 5.0 10 7.0 10 [C1 ] 869 8127 7260J [A1] AC AC AC AC AC AD DC UQ W QU W W            thermal energy lost = …………………….. J [2] (c) To bring the gas from stat e A to state C, less heat is needed for process ADC, compared to process ABC. Explain why. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………. [2] 3 A cell of e.m.f. E with an internal resistance r is connected in an electrical circuit consisting four resistors and a resistance wire W as shown below. When the current flowing through the cell is 12.0 A, the potential difference across AC is 6.0 V. The change in internal energy for both processes are the same as the change in internal energy is independent of the path taken [B1]. Since the work done on the gas for process ADC is higher than that for process ABC, the amount of heat supplied for process ADC is less [B1]. A r B B E C r 15.0 Ω 5.0 Ω A 10.0 Ω 20.0 Ω W

7 SRJC 2011 9646/Prelim/2011 [Turn Over For Examiner’s Use (a) 1.875  1019 electrons pass through the cell in t seconds. (i) Show that t = 0 . 2 5 s . [ 1 ] t Ne t QI     sx I Net 25.012 106.1 10 x 1.875 1919   (ii) Calculate the amount of electrical energy converted by the cell between the points A and C during this time period.     JxxNeVQVW 186106.110875.1 1919   energy = ………………….. J [1] (b) The resistivity of W is 1.36  10-5 Ω m, its length is 0.5 m, and its diameter is 4.0  10-3 m. Show that the effective re sistance across AC due to the four resistors and W is 0.5 Ω.