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2004 Leaving Cert Physics Solutions (Higher Level) 1 In an experiment to measure the acceleration due to gravity g by a free fall method, a student measured the time t for an object to fall from rest through a distance s. This procedure was repeated for a series of values of the distance s. The table shows the data recorded by the student. s /cm 30 40 50 60 70 80 90 t /ms 244 291 325 342 371 409 420 (i) Describe, with the aid of a diagram, how the student obtained the data. The clock starts as sphere is released and stops when the sphere hits the trapdoor. S is the distance from solenoid to trap-door. Record distance s and the time t
(ii) Calculate a value for g by drawing a suitable graph. Calculation of t^2 (at least five correct values) Axes s and t^2 labelled At least five points correctly plotted Straight line with good fit Method for slope Correct substitution g = 10.0 ± 0.2 m s− (iii) Give two precautions that should be taken to ensure a more accurate result. measure from bottom of sphere; avoid parallax error; for each value of s take several values for t / min t reference; ensure no external force (e.g. draughts, etc.) acts on sphere (during release); adjust ‘sensitivity’ of trap door; adjust ‘sensitivity’ of electromagnet (using paper between sphere and core); use large values for s (to reduce % error); use millisecond timer
2 In an experiment to measure the wavelength of monochromatic light, the angle θ between a central bright image ( n =
- and the first and second order images to the left and the right was measured_._ A diffraction grating with 500 lines per mm was used. The table shows the recorded data. n 2 1 0 1 2 θ /degrees 36.2 17.1 0 17.2 36. (i) Describe, with the aid of a diagram, how the student obtained the data.
See diagram, plus metre stick.
Measure distance x from central fringe for n = ±1, ±
Measure distance D from grating to screen and calculate θ in each case
using tan θ = x/D
(ii) Use all of the data to calculate a value for the wavelength of the light.
n λ = d sinθ
d = 1/500000 ⇒ d = 2 × 10-
n =1, λL= 588.1 nm, λR= 591.4 nm
n =2, λL= 590.6 nm, λR= 592.0 nm
Calculated average wavelength = 590 nm.
(iii) Explain how using a diffraction grating with 100 lines per mm leads to a less accurate result. It would result in a smaller value for θ which would mean larger percentage errors. (iv) The values for the angles on the left of the central image are smaller than the corresponding ones on the right. Suggest a possible reason for this.
The grating may not be perpendicular to the incident light
s /cm 30 40 50 60 70 80 90 t /ms 244 291 325 342 371 409 420 t^2 /s^2 0.060 0.085 0.106 0.117 0.138 0.167 0.
A student investigated the variation of the fundamental frequency f of a stretched string with its length l. (i) Draw a labelled diagram of the apparatus used in this experiment. Indicate on the diagram the points between which the length of the wire was measured.
The student drew a graph, as shown, using the data recorded in the experiment, to illustrate the relationship between the fundamental frequency of the string and its length. (ii) State this relationship and explain how the graph verifies it. f is proportional to 1/l. A straight line through the origin verifies this. The student then investigated the variation of the fundamental frequency f of the stretched string with its tension T. The length was kept constant throughout this investigation. (iii) How was the tension measured?
Using a newton-balance / pan with weights / suspended weights
(iv) What relationship did the student discover? Frequency is proportional to √Tension. (v) Why was it necessary to keep the length constant? Because length is a third variable and you can only investigate the relationship between two variables at a time. (vi) How did the student know that the string was vibrating at its fundamental frequency? The paper rider on the string falls off.
The following is part of a student’s report of an experiment to measure the resistivity of nichrome wire. “The resistance and length of the nichrome wire were found. The diameter of the wire was then measured at several points along its length.” The following data was recorded. Resistance of wire = 32.1 Ω Length of wire = 90.1 cm Diameter of wire = 0.19 mm, 0.21 mm, 0.20 mm, 0.21 mm, 0.20 mm (i) Name an instrument to measure the diameter of the wire and describe how it is used. Digital callipers Place the wire between the jaws Tighten the jaws Read the callipers (ii) Why was the diameter of the wire measured at several points along its length? To get an average because the material is not of uniform density. (iii) Using the data, calculate a value for the resistivity of nichrome. Average diameter = 0.202 mm A = π r^2 = 3.2 ×10−8^ m^2 ρ =RA/L ρ = (32.1)(3.2 × 10-8)/0.901) ρ = 1.1×10−6^ Ω m (iv) Give two precautions that should be taken when measuring the length of the wire. Ensure no kinks in wire, only measure length whose R value was measured, avoid parallax error, etc.
(i) Define specific heat capacity. The specific heat capacity of a substance is the heat energy needed to change one kilogram of the substance by one Kelvin. (ii) Define specific latent heat. The Specific Latent of a substance is the amount of heat energy need to change the state of 1 kg of the substance without a change in temperature. (iii) 500 g of water at a temperature of 15 0 C is placed in a freezer. The freezer has a power rating of 100 W and is 80% efficient. Calculate the energy required to convert the water into ice at a temperature of –20 oC. Cooling from 15 0 C to 0 0 C: Q = mcΔθ = (0.5)(4200)(15) = 31500 J Change of state: Q = ml = (0.5)(3.3 × 105) =165000 J Cooling ice from 0 oC to -20 oC: Q = (0.5)(2100)(20) =21000 J Total energy required = Qt = Q 1 +Q 2 +Q 3 =217500 = 2.2 × 10^5 J
(iv) How much energy is removed every second from the air in the freezer? 80% efficiency ⇒ 80 W ⇒ 80 J (per second) (v) How long will it take the water to reach a temperature of –20 oC? Power = Q ÷ time t = (217500 ÷ 80) = 2700 s (vi) Allowing a liquid to evaporate in a closed pipe inside the freezer cools the air in the freezer. The vapour is then pumped through the pipe to the outside of the freezer, where it condenses again. Explain how this process cools the air in the freezer. This change of state requires energy (latent heat) which is taken from inside the freezer and this lowers the temperature. (vii) The freezer causes the room temperature to rise. Explain why. Condensation (vapour to liquid) releases latent heat
(i) Define potential difference. The Potential difference (p.d.) between two points is the work done in bringing a charge of 1 Coulomb from one point to the other. (ii) Define capacitance. The Capacitance of a conductor is the ratio of the charge on the conductor to its potential. (iii) Describe an experiment to demonstrate that a capacitor can store energy.
- Set up as shown.
- Close the switch to charge the capacitor.
- Remove the battery and connect the terminals together to ‘short’ the circuit.
- The bulb will flash as the capacitor discharges, showing that it stores energy.
(iv) The circuit diagram shows a 50 μF capacitor connected in series with a 47 kΩ resistor, a 6 V battery and a switch. When the switch is closed the capacitor starts to charge and the current flowing at a particular instant in the circuit is 80 μA. Calculate the potential difference across the resistor and hence the potential difference across the capacitor when the current is 80 μA. V = IR V (across 47 kΩ resistor) = (80 ×10−6^ )(47 ×10^3 ) = 3.76 V V (across the capacitor) = 6 − 3.76 = 2.24 V (v) Calculate the charge on the capacitor at this instant.
C = Q/V ⇒ Q = CV = (50 ×10−6^ )(2.24) = 1.12 × 10-4^ C
(vi) Calculate the energy stored in the capacitor when it is fully charged. E = ½ CV^2 = ½ (50 ×10− 6)(6 )^2 = 9 ×10−4^ J (vii) Describe what happens in the circuit when the 6 V d.c. supply is replaced with a 6 V a.c. supply. The current will flow continually.
(i) Distinguish between photoelectric emission and thermionic emission. Photoelectric Effect: Emission of electrons when light of suitable frequency falls on a metal. Thermionic Emission: Emission of electrons from the surface of a hot metal. (ii) A freshly cleaned piece of zinc metal is placed on the cap of a negatively charged gold leaf electroscope and illuminated with ultraviolet radiation. Explain why the leaves of the electroscope collapse. Photoelectric emission occurs (electrons get emitted from the surface of the metal). The leaves become uncharged and therefore collapse. (iii) Explain why the leaves do not collapse when the zinc is covered by a piece of ordinary glass. Ordinary glass does not transmit UV light (iv) Explain why the leaves do not collapse when the zinc is illuminated with green light. The energy associated with photons of green light is too low for the photoelectric effect does not occur, so no electrons are emitted from the electroscope. (v) Explain why the leaves do not collapse when the electroscope is charged positively. Any electrons emitted are attracted back to the positive electroscope. (vi) The zinc metal is illuminated with ultraviolet light of wavelength 240 nm. The work function of zinc is 4. eV. Calculate the threshold frequency of zinc. φ = (4.3) eV = (4.3)( 1.6 × 10–19) J E = hf 0 ⇒ f 0 = (4.3)( 1.6 × 10–19)/ 6.6 × 10–34^ = 1.04 × 10^15 Hz
(vii) Calulate the maximum kinetic energy of an emitted electron. c = fλ ⇒ f = c/λ = (3.0 × 10^8 )/(240 × 10^9 ) = 1.25 × 10^15 Hz Ek = hf - φ Ek = (6.6 × 10–34)[( 1.25 × 10^15 ) - 1.04 × 10^15 )] = 1.39 × 10-
10 (a) (i) Beta decay is associated with the weak nuclear force. List two other fundamental forces of nature and give one property of each force. Strong: acts on nucleus/protons + neutrons/hadrons/baryons/mesons, short range Gravitational: attractive force, inverse square law/infinite range, all particles Electromagnetic: acts on charged particles, inverse square law/infinite range (ii) In beta decay, a neutron decays into a proton with the emission of an electron. Write a nuclear equation for this decay.
(iii) Calculate the energy released during the decay of a neutron. Mass lost = mass before – mass after = (mass of neutron) – [(mass of proton + electron)] = (1.6749 × 10–27) – [(1.6726 × 10–27^ + 9.1094 × 10–31)] = 1.3891 × 10-30^ kg E = mc^2 = (1.3891 × 10-30)(2.9979 × 10^8 )^2 = 1.25 × 10^13 J (iv) Momentum and energy do not appear to be conserved in beta decay. Explain how the existence of the neutrino, which was first named by Enrico Fermi, resolved this. Momentum and energy are conserved when the momentum and energy of the (associated) neutrino are taken into account.
During the late 1930s, Fermi continued to work on the nucleus. His work led to the creation of the first nuclear fission reactor in Chicago during 1942. The reactor consisted of a ‘pile’ of graphite moderator, uranium fuel with cadmium control rods. (v) What is nuclear fission? Fission is the splitting of a large nucleus into two smaller nuclei with the release of energy. (vi) What is the function of the moderator in the reactor? It slows down fast neutrons.
(vii) How did the cadmium rods control the rate of fission? They absorbed neutrons which would otherwise cause fission.
12 (b) (i) Give two reasons why the telecommunications industry uses optical fibres instead of copper conductors to transmit signals. Less interference, boosted less often, cheaper raw material, occupy less space, more information carried in the same space, flexible for inaccessible places, do not corrode, etc (ii) Explain how a signal is transmitted along an optical fibre. Light ray introduced at one end of fibre and strikes the interface at an angle greater than the internal angle so total internal reflection occurs. This continues all along the fibre. (iii) An optical fibre has an outer less dense layer of glass. What is the role of this layer of glass? Total internal reflection will only occur if the outer medium is of greater density. It also prevents damage to the surface of the core. (iv) An optical fibre is manufactured using glass of refractive index of 1.5. Calculate the speed of light travelling through the optical fibre. ng = ca /cg 1.5 = 3 × 10^8 / vg vg = 2.0 × 10^8 (m s-1)
12 (c) (i) What is electromagnetic induction? Electromagnetic Induction occurs when an emf is induced in a coil due to a changing magnetic flux. (ii) Describe an experiment to demonstrate electromagnetic induction****. Set up as shown. Move the magnet in and out of the coil and note the deflection in the galvanometer. (iii) A light aluminium ring is suspended from a long thread as shown in the diagram. When a strong magnet is moved away from it, the ring follows the magnet. Explain why. Current flows in the ring in such a direction as to oppose the change which caused it. Therefore the ring follows the magnet. (iv) What would happen if the magnet were moved towards the ring? The ring would be repelled.
12 (d) (i) A p-n junction is formed by taking a single crystal of silicon and doping separate but adjacent layers of it. A depletion layer is formed at the junction. What is doping? Doping is the addition of a small amount of atoms of another element to a pure semiconductor to increase its conductivity. (ii) Explain how a depletion layer is formed at the junction. Electrons from n-type and holes from p-type cross the common junction and cancel out with charge carriers on the other side. As a result a narrow insulating region is formed which now acts as a ‘barrier’ or depletion layer. (iii) The graph shows the variation of current I with potential difference V for a p-n junction in forward bias. Explain, using the graph, how the current varies with the potential difference. Very little current flows between 0 V and 0.6 V If the potential difference is greater than 0.6 V a large current flows. (iv) Why does the p-n junction become a good conductor as the potential difference exceeds 0. Volts? The depletion layer is overcome and as a result a large current flows.
