november 2021 physics paper 2 worked solutions

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linkage Your answer A D Monochromatic light from a laser is incident normally on a diffraction grating. A series of bright dots are formed on a distant screen. Which two terms can be used to explain these bright dots? diffraction, interference reflection, interference refraction, diffraction D refraction, reflection B C You should spend a maximum of 30 minutes on this section. Write your answer to each question in the box provided. Answer all the questions. Your answer 2 SECTION A ⒸOCR 2021 induced in a circuit is directly proportional to the rate of change of A [1] [1] PMT 3 4 The current in a lamp is 2.0mA. The potential difference across the lamp is 6.0V. What is the energy transfer in the lamp over a period of 3.0 hours? A B 0.036 J C 2.16J D 0.012 J Your answer D The diagram below shows two oppositely charged spheres. 130J B Which statement is correct? C -V The magnitude of the charge on each sphere is the same. The point P is on the line joining the centres of the spheres and is the same distance from the centre of each sphere. D A A negatively charged particle at P will move to the right. The direction of the electric field at P is to the left. X The electric potential at P is zero. The magnitude of the electric field strength at is zero. 3 P ⒸOCR 2021 Your answer [1] Turn over [1] PMT 4 5 A circuit with four light-emitting diodes (LEDs) P, Q, R and S is shown below. Two LEDs are lit in this circuit. Which two LEDs are lit? A B C D P and Q ⒸOCR 2021 P and R Q and R Q and S Your answer P D [1] PMT 5 6 A student is modelling the decay of charge for a capacitor discharging through a resistor using the equation AQ At The student decides to use At = 0.5 s. The initial charge on the capacitor is 1000 μC. 2.5 Part of the modelling spreadsheet from the student is shown below. t/s 0 0.5 1.0 1.5 A 700 B 720 What is the value of Q in uC at t = 1.5s? C 729 = -0.2Q. D 800 ⒸOCR 2021 Your answer Charge Q left on capacitor at time t/µC 1000 900 810 729 C -0.2(900) Charge AQ decaying in the next 0.5 s/µC 100 90 81 Turn over [1] PMT 7 A circuit with a thermistor is shown below. B The resistance of the resistor is R and the resistance of the thermistor is 2.5R. The potential difference (p.d.) across the thermistor is 5.0 V. What is the total p.d. across both components? A 2.0V 5 = 2.5 7.0V C D K Your answer L A 12.5 V B 17.5V B 8 A double-slit is used in an interference experiment to independently investigate the light from two sources K and L. The light from the sources have different wavelengths. The table below shows some data. Light source Wavelength of light C ⒸOCR 2021 0.48 mm 1.2 mm 25R 1.9 mm D 3.0mm Your answer What is the separation between adjacent bright fringes for source L? f= на 6 2 0.80 A 2.5+1 X V - ах D Separation between adjacent bright fringes 1.2mm to 1.2 Distance between screen and double-slit D 0.50D 2 [1] 0.8) xo.5D > [1] PMT 7 9 A stationary sound wave formed in a tube is shown below. 51=3,5 I A B C D The tube is closed at one end. The length of the tube is 3.5m. The speed of sound is 340ms-¹. What is the frequency of the sound wave? 97 Hz 120 Hz ⒸOCR 2021 240 Hz 486 Hz Your answer tube 3.5m 4 =2.8 K 3403 f(2.8) Turn over [1] PMT 8 10 A battery of electromotive force (e.m.f) 6.0V and of negligible internal resistance is used in the circuit below. 6.0 V The current from the battery is 30 mA. The current in the resistor X is 10mA. The resistors Y and Z are identical. A What is the power dissipated in the resistor Z? 30 mW B C D 60 mW ⒸOCR 2021 120 mW 30 mA 180 mW Your answer 10 mA Y 53 X B 0.06 [1] PMT 9 11 The diagram below shows the graphs of displacement x against time t for two waves S and T. x/mm B C D 3.01 ⒸOCR 2021 2.0- Ap 5.0 1.0+ 0- 0 -1.0- -2.0- -3.0- The waves meet at a point in space. The superposition of these two waves produces a resultant wave. What is the frequency fand the amplitude A of the resultant wave? A f = 100 Hz, A = 2.0mm f = 100 Hz, A= 4.0mm f = 200 Hz, A = 2.0mm f = 200 Hz, A = 4.0mm Your answer T C S. 10.0 t/ms Turn over [1] PMT 10 12 A flat coil has 200 turns and a cross-sectional area of 1.20 x 10-4 m². A The coil is placed horizontally in a uniform magnetic field. The magnetic flux density is 0.050T. The magnetic field is at an angle of 30.0° to the vertical. Q=200 x D cos 30 B What is the magnetic flux linkage for this coil? 3.00 x 10-6Wb 5.20 x 10-6Wb 6.00 x 10-4Wb 1.04 x 10-3Wb C D Your answer A 30° V B C 13 The nucleus of thorium-232 (232Th) emits two alpha particles and two beta-minus particles to become a nucleus of an isotope of radium. What is the nucleon number A and the proton number Z for the nucleus of this radium isotope? A = 224, Z = 88 D ⒸOCR 2021 A = 228, Z = 86 -magnetic field line A = 224, Z = 84 coil A = 228, Z = 88 Your answer A 232-4 ar 2 88 - 4 =224 1 + 1 [1] 2+ [1] PMT 11 14 A beam of ultrasound is incident normally at a boundary between two tissues F and G. The table below shows some data on the two tissues. Density of tissue Speed of ultrasound in tissue A 0.83% B C What percentage of the intensity of the ultrasound is reflected at the boundary? z opc 2=1.2px 1.50 8 D 8.2% 9.1% 29% Your answer C A exponential B linear ⒸOCR 2021 B 15 A radiation detector is placed in front of a beta-emitting source. The count-rate is measured and recorded every 10 minutes. The results are shown below. 10 311s-1 309 S-1 299s-¹ What term can be used to describe the data shown? random D spontaneous Your answer Tissue F I to C P C Tissue G 1.2p 1.5c (1.8-1)2 (111.8) ² 2 307 S-1 321 S-1 Turn over [1] [1] PMT 12 SECTION B 16 (a) The normal frequency range of hearing for young people is from 20 Hz to 20kHz. (i) The speed of sound in air is 340ms-1. Calculate the shortest wavelength a young person can hear. 340= 340 20000 ⒸOCR 2021 Answer all the questions. YJ wavelength = m [2] (ii) Describe how you can use an oscilloscope, and other additional laboratory equipment, to determine the actual upper limit of the frequency range for a young person. 1.7×10-2 use loudspeaker with signal generator. increase frequency until limit of hearing. frequency = 1/T on oscilloscope [3] PMT 13 (b) State one difference and one similarity between the oscillations of a stationary sound wave and a progressive sound wave. Difference:...stationary.was.has.nades.. Similarity: .........oscillations.are.longitudinal.. ⒸOCR 2021 (c) You are provided with a ray-box, a semi-circular block of plastic and other normal laboratory equipment. The outline of the block is shown below. protractor ...is.total.internal.refraction......... Describe how you could measure the refractive index n of the block using the critical angle method. Draw on the diagram and label it to make your answer clear. ....increase the incident.angle.until the.ray..of.light.runs.along the boundary....there......... ....measure.this.angle.using.protractor... ....n...1./..sinc.. -block [2] Turn over [3] PMT 17 (a) Define the work function of a metal. ...minimum.energy.required.to.liberate an electron from the surface.of.a.metal... ⒸOCR 2021 (b) The work function of potassium is 2.3 eV. (i) Potassium emits electrons from its surface when blue light is incident on it. Extremely intense red light produces no electrons. Explain these observations in terms of photons and their energy. blue light has a higher frequency, E-hf so has a greater energy. blue light has energy greater than 2.3eV. 14 energy of photon is independent of intensity one photon interacts with one electron . (ii) Light from a laser is incident on some potassium in a vacuum. Electrons are emitted. The wavelength of the light is 320 nm. Calculate the shortest de Broglie wavelength of the emitted electrons. h > x = mv = KEmax he 320x10-9 =2.54 x 10-19 V²2x2.54 x18-19 9.11×10-31 V=7.46x105 12 Mv² -2.3×1.6×1019 de Broglie wavelength = 9.8×10-10 [1] 9.\lx10°347.46x105 . [4] m [4] PMT 15 18 (a) The circuit diagram of an electrical circuit is shown below. 2.1V S ⒸOCR 2021 A + + The switch S is closed. 4.5V 2.4 V 0.80 Ω 0.50 92 The positive terminals of the batteries are connected together. One battery has electromotive force (e.m.f.) 4.5V and internal resistance 0.80 92. 21.2.12 R The other battery has e.m.f. 2.4 V and internal resistance 0.50 22. R is a coil of insulated wire of resistance 1.29 at room temperature. (i) On the diagram, draw an arrow to show the direction of the conventional current. [1] (ii) Calculate the current / shown by the ammeter. R₁ = 2.5 2.1=2.5T 1= 0.84 A [3] Turn over PMT 16 (iii) The insulated wire has diameter 4.6 x 10-4m. The number density of charge carriers in R is 4.2 x 1028 m-³. Calculate the mean drift velocity v of the charge carriers in R. 0.84-(4.2x1028) 23x10-4) ²T (1.6×10") 75x10 ms-¹ [2] (iv) The current measured by the ammeter is smaller than that calculated in (ii). This is because the temperature of R increased due to heating by the current. ⒸOCR 2021 V= Without any changes to the circuit itself, state and explain what practically can be done to make the measured current the same as the calculated current.. ..close.switch.only when taking readings to lower temperature... and resistance.of ..R. [2] PMT 1.2 1.5 17 (b)* A student is doing an experiment to determine the e.m.f. E of a cell and its internal resistance r. The circuit diagram of the arrangement is shown below. A ⒸOCR 2021 The student changes the resistance of the variable resistor. The potential difference Vacross the variable resistor and the current / in the circuit are measured. The Vagainst / graph plotted by the student is shown below. V/V -0.8 1.5 1.0- 0.5- 0 E 0 V 0.5 1.0 //A y intercept is E. gradient is -r. E = 1.2V r = 0.802 1.5 V/V 0.20 0.40 0.60 0.80 1.00 ΠΑ R/Ω 1.25 0.16 1.00 0.4 0.75 0.8 0.50 1.60 0.40 0.25 4.00 0.25 Use the graph to determine E and r. Explain your reasoning. Calculate R and P to complete the table. Describe how P depends on R. E = V + Ir P/W 0.25 0.4 There is an incomplete table next to the graph. R is the resistance of the variable resistor and P is the power dissipated by the variable resistor. 0.45 [6] R increases as V increases. P increases then decreases. maximum power when internal resistance = R = 0.8 Turn over PMT ⒸOCR 2021 Additional answer space if required 18 PMT 19 19 (a) The diagram below shows the top-view of a long current-carrying wire. The direction of the current in the wire is into the plane of the paper. Draw at least three field lines to indicate the magnetic field pattern around this wire. [2] (b) The arrangement shown in the diagram below is used to determine the magnetic flux density between the poles of a permanent magnet. ⒸOCR 2021 bench ● magnet wire 82.2 g top view current-carrying wire -digital balance The magnet is placed on the digital balance. The current-carrying wire is horizontal and at right angles to the magnetic field between the poles of the magnet. The wire is fixed. The following results are collected. length of the wire in the uniform field of the magnet = 6.0 ± 0.2 cm balance reading with no current in wire = 80.0g balance reading with current in wire = 82.2g current in wire = 5.0+ 0.1A Turn over PMT ⒸOCR 2021 20 Calculate the magnetic flux density B, including the absolute uncertainty. Ignore the absolute uncertainty in the balance readings. Write your value for B to 2 significant figures and the absolute uncertainty to 1 significant figure. 0.1 F=BIL F=2.2×10-³x9.81 B= 2.2,109.81 Sx0.06 = 0.072 B= 0.2 0.6 0.072 = 0.053 хо. 072 PMT =0.0038 I............... T [4] 21 20 (a) The diagram below shows a circuit to charge a capacitor. 1- 1.48 V ⒸOCR 2021 2000μF 1200005 The electromotive force (e.m.f.) E of the cell is 1.48 V and it has negligible internal resistance. The resistance of the resistor is 120k and the capacitance of the capacitor is 2000 μF. At time t=0 the capacitor is uncharged. The switch is closed at time t = 0. Calculate the time t when the potential difference across the capacitor is 1.00 V. + 120x10x2000x106 1=1.248 (1- 1.48 -1.126 = e CR t ce t = 270 s [4] Turn over PMT (b) A capacitor of capacitance C is connected across a strip of conductive paper. 6.00 V ⒸOCR 2021 The switch is moved from X to Y, and the time t for the potential difference across the capacitor to halve is measured. The time t is given by the expression t = (Ck In2) x L where k is the resistance of the conductive paper per unit length and L is the length of the conductive paper. The value of C is 1.2 x 10-³F. In an experiment, L is changed and t measured. The data points are plotted on a t against L grid as shown below. t/s 20 10- 4% 0 0.1 8.11 0.2 22 20. 4 053-0.11 0.3 0.4 - conductive paper = 38.1 0.5 %53 L/m 0.6 PMT ⒸOCR 2021 23 Draw a straight line of best fit through the data points, and use the gradient of this line to determine k. دیا 38=t=Ckinz 38-1.2x10-³ kln2 ... 4.6×10ª k= 92 m-¹ [4] Turn over PMT 24 21 (a) A high-energy X-ray photon interacts with an electron of an atom through the Compton effect. Describe this effect. electron.is ejected from the atom. photon is scattered with increased.wavelength. (b) The diagram below shows a beam of X-rays incident normally on some soft tissue. beam of X-rays ⒸOCR 2021 soft tissue I = 4.6x10²³ e =771.866 2.1 cm The attenuation (absorption) constant of the soft tissue is 0.85 cm-¹. The intensity of the beam is 4.6 x 10³ Wm-². P=1A P= 772 x 3.4 x 10-4 = 0.262 There is a small cyst 2.1 cm from the surface of the soft tissue. The cross-sectional area of the cyst normal to the beam is 3.4 x 10-4m². The beam is switched on for 30 s. Calculate the X-ray energy incident on the cyst in a period of 30s. P=IA E=Pt E=0.262x30 -cyst -0.85x2.1 7.9 [2] energy = ..... J [4] PMT 25 (c) The attenuation coefficients of the cyst and the soft tissues in (b) were similar. This prevented imaging the cyst using a two-dimensional X-ray image. Name a different X-ray technique that could be used to image the cyst. Explain the advantage of this technique. CAT scan. can provide a 3D image 22 The medical tracer fluorine-18 is used in positron emission tomography (PET). Fluorine-18 is a beta-plus emitter with a short half-life. ⒸOCR 2021 Describe how the fluorine-18 nuclei are located in a patient using a PET scanner. ....positrons from F-18 annihilate electrons within the patient. .... two gamma photons are produced and travel in opposite electrons.. difference in arrival times at the detectors can be used to locate point of annihilation..... [2] Turn over [4] PMT 26 23 (a) Explain the function of the control rods and the moderator in a nuclear fission reactor. control rods absorb neutrons to prevent a chain reaction. moderator slows down the neutrons [2] (b)* Some nuclear fission reactors use uranium-235 as fuel. In the future, there is possibility of using hydrogen-2 as fuel in fusion reactors. Here is some information and data on fission and fusion reactions. Typical reaction Approximate energy produced in each reaction Molar mass of fuel material Fission reactor 144. on + 235U→ 146Ba + 30Kr + 3 n 36 200 MeV ⒸOCR 2021 fusion E = 4 x 106 x 1.6x10-19 = 6₂4x10² 6.4x10-13 x 3.01x10²6 2 Fusion reactor uranium-235: 0.235 kg mol-¹ 4.255 mol = 1 kg =2.56×1024 nuclei Describe the similarities and the differences between fission and fusion feactions. Explain with the help of calculations, which fuel produces more energy per kilogram. ...fission.... reaction E = 200x106 x 1.6x10-19 = 3.2×10¹ J 1 ..fo... 3.2×10 ¹³ x 2.56x1013=8.2x1013 J for lkg.. 13 H+H³H+H energy is produced in both reactions. more energy is produced per reaction in fission. the total binding energy of products is greater in fusion nuclei repel each other so require high temperatures. fission reactions are triggered by thermal neutrons. hydrogen-2: 0.002 kg mol-¹ 500ml=1kg 9.6×10¹s roughly the same energy per kg 4 MeV [6] PMT Additional answer space if required ⒸOCR 2021 27 Turn over PMT 28 24 (a) Fig. 24 shows two horizontal metal plates in a vacuum. path of electrons ⒸOCR 2021 The diagram is not drawn to scale. Electrons travelling horizontally enter the space between the charged plates and are deflected vertically. Fig. 24 The potential difference between the plates is 4000 V. The distance between the plates is 0.08 m. The initial speed of the electrons is 6.0 x 107 ms-1. The vertical deflection of the electrons at the far end of the plates is x. P = (i) Show that the vertical acceleration a of an electron between the plates is 8.8 x 1015 m s-2. E = 4000 0.08 9 + 4000 V 2000x16x10-19 OOV S=0.12 V.=6x10² 008 =8x10-15 8x10-15 = 9.11x10-31 a a = 8.78×10¹5 ~ (ii) The length of each plate is 0.12 m. Show that the time t taken by the electron to travel this length is 2.0 x 10-⁹ s. 0.12=6x10²4 t = 2x10-9 [3] [1] PMT 29 (iii) Calculate the vertical deflection x of the electron. S= S = u=0 ⒸOCR 2021 a=8.78×1015 t = 2x10-9 +(8.78×10¹5) (2×10-7)2 1.8×10-2 m [2] (b) The arrangement shown in Fig. 24 is now used to investigate positrons emitted from a radioactive source. The speed of the positrons is also 6.0 x 107 ms-1 The initial path of the positrons is the same as that of the electrons in Fig. 24. On Fig. 24, sketch the path of the positrons between the plates. (c) Beta-minus particles (electrons) emitted from a radioactive source have a range of speeds. Describe and explain how a uniform magnetic field can be applied in the space between the charged plates to select beta-minus particles with a specific speed. No calculations are required. ....apply magnetic field at right angles to the electric field. ... electric force = magnetic force.…... ....no resultant vertical force, so beta particles travel with specific horizontal speed....... BQv = EQ V= E X = Turn over [2] [3] PMT 30 25 (a) A researcher is doing an experiment on a radioactive solution in a thin glass tube. The solution has two radioactive materials X and Y. The table below shows some data on these two materials. Half-life Particles emitted Daughter nuclei Material X ⒸOCR 2021 10 minutes Alpha Stable Material Y 10 hours Beta-minus Stable The solution has the same number of nuclei of X and Y at the start. (i) State and explain which material has the greatest activity at the start. ..material.X.as.it has a shorter.half.life.. (ii) State why it is dangerous for the researcher to handle the test tube with bare hands. beta particles are not stopped by glass so researcher exposed to radiation........... [1] (b) Carbon-14 (¹4C) is produced in the upper atmosphere of the Earth by collisions between nitrogen nuclei and fast-moving neutrons. The nuclear transformation equation below shows the formation of a single carbon-14 nucleus. 14N+n→ ¹4C + X (i) State the proton number of particle X. [1] proton number = 1 [1] PMT 31 (ii) Use the data below to determine the binding energy per nucleon of the ¹4C nucleus. Write your answer to 3 significant figures. mass of neutron = 1.675 x 10-27 kg mass of proton = 1.673 x 10-27 kg mass of ¹4C nucleus = 14.000 u 1 u = 1.66 x 10-27 kg mass of nucleons = (constituents) (6 × 1.673×10-²3) + (8×1.675×10-27) 2.3438×10-26 ⒸOCR 2021 (nucleus rest mass) Am = 2.3x10-26 - (14 x 1.66×10-27) -1.98x10-28 binding energy per nucleon = SE = (1.98×10-28) (3x108)² = 1.782 x 10-11 14 per nucleon 1.27x10-12 END OF QUESTION PAPER J per nucleon [4] PMT 32 ADDITIONAL ANSWER SPACE If additional space is required, you should use the following lined page(s). The question number(s) must be clearly shown in the margin(s). OCR Oxford Cambridge and RSA Copyright Information OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. 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