Boyle Law - Physics - Solved Paper, Exams of Physics

Download Boyle Law - Physics - Solved Paper and more Exams Physics in PDF only on Docsity! LEAVING CERTIFICATE EXAMINATION 2003: PHYSICS – HIGHER LEVEL 2003 Question 1 In an experiment to verify Boyle’s law, a student measured the volume V of a gas at different values of the pressure p. The mass of the gas was not allowed to change and its temperature was kept constant. The table shows the data recorded by the student. p/ kPa 120 180 220 280 320 380 440 V/cm3 9.0 6.0 5.0 4.0 3.5 3.0 2.5 (i) Describe with the aid of a diagram how the student obtained this data. See diagram. Note the pressure of the gas from the pressure-gauge and the volume from the graduated scale. Turn the screw to decrease the volume and increase the pressure. Note the new readings and repeat to get about seven readings. (ii) Draw a suitable graph on graph paper to show the relationship between the pressure of the gas and its volume. p/ kPa 120 180 220 280 320 380 440 1/V/cm-3 0.111 0.167 0.200 0.250 0.286 0.333 0.400 Axes labelled 6 points plotted correctly Straight line Good fit (iii) Explain how your graph verifies Boyle’s law. A straight line through the origin verifies that pressure is inversely proportional to volume (iv) Describe how the student ensured that the temperature of the gas was kept constant. Release the gas pressure slowly, allow time between readings. 2003 Question 2 In an experiment to measure the specific latent heat of vaporisation of water, cold water was placed in a copper calorimeter. Steam was passed into the cold water until a suitable rise in temperature was achieved. The following results were obtained. Mass of the calorimeter........................... = 73.4 g Mass of cold water .................................. = 67.5 g Initial temperature of water..................... = 10 °C Temperature of the steam........................ = 100 °C Mass of steam added ............................... = 1.1 g Final temperature of water ...................... = 19 °C (i) Describe how the mass of the steam was found. Final mass of (calorimeter + water + condensed steam) – Initial mass of (calorimeter + water) (ii) Using the data, calculate a value for the specific latent heat of vaporisation of water. (ml) steam + (mc∆ϑ) steam = (mc∆ϑ) water + (mc∆ϑ) cal ∆ϑwater = 90C, ∆ϑcal= 90C ∆ϑ) steam = 810C Answer = 2.2 × 106 J kg-1 (iii) Why is the rise in temperature the least accurate value? Read only to one significant figure {the concept of significant figures is not on the syllabus and shouldn’t have got asked. It hasn’t appeared since.] (iv) Give two ways of improving the accuracy of this value. Use a digital thermometer, use more steam, use less water, insulation, cover, stirring, steam trap 2003 Question 3 The following is part of a student’s report of an experiment to measure the focal length of a converging lens. “I found the approximate focal length of the lens to be 15 cm. I then placed an object at different positions in front of the lens so that a real image was formed in each case.” The table shows the measurements recorded by the student for the object distance u and the image distance v. u/cm 20.0 25.0 35.0 45.0 v/cm 66.4 40.6 27.6 23.2 (i) How did the student find an approximate value for the focal length of the lens? Focus the image of a distant object on a screen. The distance from the lens to screen corresponds to the focal length. (ii) Describe, with the aid of a labelled diagram, how the student found the position of the image. Set up as shown. Adjust the position of the screen until a sharp image is seen. (iii) Using the data in the table, find an average value for the focal length of the lens. (iv) 1/u+ 1/v = 1/f Average = 15.4 cm (v) Give two sources of error in measuring the image distance and state how one of these errors can be reduced. Image not sharp / parallax error in reading distance / not measuring to centre of lens / zero error in metre stick 4 In an experiment to verify Joule’s law, a heating coil was placed in a fixed mass of water. The temperature rise Δθ produced for different values of the current I passed through the coil was recorded. In each case the current was allowed to flow for a fixed length of time. The table shows the recorded data. I /A 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Δθ / °C 3.5 7.0 10.8 15.0 21.2 27.5 33.0 (i) Describe, with the aid of a labelled diagram, how the apparatus was arranged in this experiment. See diagram below. (ii) Using the given data, draw a suitable graph on graph paper and explain how your graph verifies Joule’s law. I /A 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Δθ / °C 3.5 7.0 10.8 15.0 21.2 27.5 33.0 I2 /A2 2.25 4.0 6.25 9.0 12.25 16.0 20.25 Label axes At least 6 correct points Straight line Good fit A straight line through origin shows that ∆ϑ ∝ I2 which verifies Joule’s Law. (iii) Explain why the current was allowed to flow for a fixed length of time in each case. You can only investigate the relationship between two variables at a time and time is a third variable. (iv) Apart from using insulation, give one other way of reducing heat losses in the experiment. Start with cold water, change the water for each run, use a lid, shorter time interval, polish calorimeter 5 (a) Stat Hooke’s law. Hooke’s Law states that when an object is stretched the restoring force is directly proportional to the displacement, provided the elastic limit is not exceeded. (b) What is the relationship between the acceleration due to gravity g and the distance from the centre of the earth? g is proportional to 1/d2 u/cm 20.0 25.0 35.0 45.0 v/cm 66.4 40.6 27.6 23.2 f/cm 15.4 15.5 15.4 15.3 The final kinetic energy gained by the electron equals the initial (electrical) potential energy = 4 KeV = (4000)(1.6 × 10–19) = 6.4×10−16 J (v) What is the speed of the electron at the anode? (Assume that the speed of the electron leaving the cathode is negligible.) E = ½ mv2 ⇒ 6.4 ×10-16 = ½ (9.1 × 10-31)(v2) ⇒ v =3.75 × 107 m s-1 (vi) After leaving the anode, the electron travels at a constant speed and enters a magnetic field at right angles, where it is deflected. The flux density of the magnetic field is 5 × 10–2 T. Calculate the force acting on the electron. F = Bev = (5 × 10–2)(1.6 × 10–19)( 3.75 × 107) = 3.0 ×10−13 N (vii) Calculate the radius of the circular path followed by the electron, in the magnetic field. F = mv2/r ⇒ 3.0 ×10−13 = (9.1 × 10–31)( 3.75 × 107)2/r ⇒ r = 4.3×10−3 m (viii) What happens to the energy of the electron when it hits the screen of the CRT? It gets converted to light. 10 (a) (i) Leptons, baryons and mesons belong to the “particle zoo”. Give (i) an example, (ii) a property, of each of these particles. LEPTONS; electron, positron, muon , tau, neutrino Not subject to strong force BARYONS; proton, neutron Subject to all forces, three quarks MESONS pi(on), kaon Subject to all forces, mass between electron and proton, quark and antiquark (ii) The following reaction represents pair production. γ → e+ + e– Calculate the minimum frequency of the γ-ray photon required for this reaction to occur. E = (2)mc2 = hf 2(9.1 × 10–31)( 3.0 × 108)2 = (6.6 × 10–34)f ⇒ f = 2.5×1020 Hz (iii) What is the effect on the products of the reaction if the frequency of the γ-ray photon exceeds the minimum value? The electrons which were created would move off with greater speed. There may also be more particles produced. (iv) The reverse of the above reaction is known as pair annihilation. Write a reaction that represents pair annihilation. e+ + e- → 2γ (v) Explain how the principle of conservation of charge and the principle of conservation of momentum apply in pair annihilation. Total charge on both sides is zero Momentum of positron + electron = momentum of photons 11 Read the following passage and answer the accompanying questions. Irish Times: Monday, January 11,1999 Radioactive decay helps to determine exact dates. Radioactive decay occurs with such precision that it is often used as a clock. Carbon dating has been invaluable to archaeologists, historians and anthropologists. The method is based on the measurement of 14C, a radioactive isotope of carbon with a half-life of 5730 years. 14C occurs to a small extent in the atmosphere together with the much more common 12C. Living organisms constantly exchange carbon with the atmosphere and the ratio of 14C to 12C in living tissue is the same as it is in the atmosphere. This ratio is assumed to have remained the same since prehistoric times. When an organism dies, it stops exchanging carbon with the atmosphere, and its 14C nuclei keep disintegrating while the 12C in the dead tissue remains undisturbed. (a) What is radioactive decay? Radioactivity is the breakup of unstable nuclei with the emission of one or more types of radiation. (b) What is an isotope? Isotopes are atoms which have the same Atomic Number but different Mass Numbers. (c) Apart from “carbon dating”, give two other uses of radioactive isotopes. Medical imaging, (battery of) heart pacemakers, sterilization, tracers, irradiation of food, killing cancer cells, measuring thickness, smoke detectors, nuclear fuel (d) How many neutrons are in a 14C nucleus? Eight (e) 14C decays to 14N. Write an equation to represent this nuclear reaction. 14 6C → 714N + -10e (f) How much of a 14C sample remains after 11 460 years? 11,460 corresponds to two half lives, and after two half lives one quarter remains. (g) Calculate the decay constant of 14C. T1/2 = ln 2 /λ ⇒ λ = 0.693/5730 = 1.21 × 10−4 y-1 = 3.8×10−12 s-1 (h) Why does the 12C in dead tissue remain “undisturbed”? It is not radioactive, it is not exchanging with the atmosphere, it is stable 12 (a) (i) State Newton’s second law of motion. Newton’s Second Law of Motion states that the rate of change of an object’s momentum is directly proportional to the force which caused it, and takes place in the direction of the force. (ii) A skydiver falls from an aircraft that is flying horizontally. He reaches a constant speed of 50 m s–1 after falling through a height of 1500 m. Calculate the average vertical acceleration of the skydiver. v 2= u2+2as ⇒ 502 = 0 + (2)(a)(1500) ⇒ a = 0.83 m s-2 (iii) If the mass of the skydiver is 90 kg, what is the magnitude and direction of the average resultant force acting on him? F = ma = 90 × 0.83 = 75 N Down (iv) Use a diagram to show the forces acting on the skydiver and explain why he reaches a constant speed. Weight acting down on diagram Air resistance / friction / buoyancy acting up on diagram Air resistance = weight, therefore resultant force = 0 Therefore acceleration = 0 12 (b) (i) What is the difference between heat and temperature? Heat is a form of energy, temperature is a measure of hotness. (ii) The emf of a thermocouple can be used as a thermometric property. Explain the underlined terms. An emf is a voltage applied to a full circuit. A thermometric property is any property which changes measurably with temperature. (iii) Name a thermometric property other than emf. Length, pressure, volume, resistance, colour (iv) Explain why it is necessary to have a standard thermometer. Two different types of thermometer will give slightly different readings at the same temperature 12 (c) (i) State Coulomb’s law of force between electric charges. Coulomb’s Law states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. (ii) Define electric field strength and give its unit. Electric field strength at a point is the force per unit charge at that point. The unit of electric field strength is the Newton per Coulomb (NC-1). (iii) How would you demonstrate an electric field pattern? Oil and semolina or seeds High tension / high voltage Lines of semolina show field (iv) The diagram shows a negative charge – Q at a point X. Copy the diagram and show on it the direction of the electric field strength at Y. Arrow towards X 12 (d) (i) State the laws of electromagnetic induction. Faraday’s Law states that the size of the induced emf is proportional to the rate of change of flux. Lenz’s Law states that the direction of the induced emf is always such as to oppose the change producing it. (ii) A small magnet is attached to a spring as shown in the diagram. The magnet is set oscillating up and down. Describe the current flowing in the circuit. Alternating current. (iii) If the switch at A is open, the magnet will take longer to come to rest. Explain why. There is no longer a full circuit, so even though there is an induced potential difference there is no (induced) current, therefore no induced magnetic field in the coil therefore no opposing force.