Good Distribution - Physics - Solved Paper, Exams of Physics

Download Good Distribution - Physics - Solved Paper and more Exams Physics in PDF only on Docsity! 2006 Leaving Cert Physics Solutions (Higher Level) 1 In investigating the relationship between the period and the length of a simple pendulum, a pendulum was set up so that it could swing freely about a fixed point. The length l of the pendulum and the time t taken for 25 oscillations were recorded. This procedure was repeated for different values of the length. The table shows the recorded data. l/cm 40.0 50.0 60.0 70.0 80.0 90.0 100.0 t/s 31.3 35.4 39.1 43.0 45.5 48.2 50.1 The pendulum used consisted of a small heavy bob attached to a length of inextensible string. (i) Explain why a small heavy bob was used. To reduce air resistance and to keep the string taut (ii) Explain why the string was inextensible. So that length remains constant because length would be another variable. (iii) Describe how the pendulum was set up so that it swung freely about a fixed point. The string was placed between two coins (or a split cork). (iv) Give one other precaution taken when allowing the pendulum to swing. Make sure that there are no draughts / make sure it oscillates in one plane only. (v) Draw a suitable graph to investigate the relationship between the period of the simple pendulum and its length. Values of t divided by 25 to get T Axes correctly labelled T2 vs. l At least six points plotted correctly Straight line drawn Good distribution (about straight line) (vi) What is this relationship? T2 is proportional to l (vii) Justify your answer. The graph resulted in a straight line through the origin 2 In an experiment to measure the wavelength of monochromatic light, a narrow beam of the light fell normally on a diffraction grating. The grating had 300 lines per millimetre. A diffraction pattern was produced. The angle between the second order image to the left and the second order image to the right of the central bright image in the pattern was measured. The angle measured was 40.60. (i) Describe, with the aid of a labelled diagram, how the data was obtained. The apparatus was set up as shown. To get a value for θ the distance x was measured between the centre image and the second order image, then the distance D between grating and screen was measured. θ = Tan-1 (x/D) We did the same for the other side and got an average value for θ. (ii) How was a narrow beam of light produced? Use a laser. (iii) Use the data to calculate the wavelength of the monochromatic light. nλ = d sin θ n = 2 d = 1/(3.00 x105) m = 3.33 x 10-6 m = 3.33 x 10-3 cm = 1/300 mm θ = 20.30 λ = 5.78 x 10-7 m (= 578 ≈ 580 nm) (iv) Explain how using a diffraction grating of 500 lines per mm leads to a more accurate result. Time for 25 swings /s 31.3 35.4 39.1 43.0 45.5 48.2 T/s 1.25 1.42 1.56 1.72 1.82 1.93 T 2 /s 2 1.57 2.01 2.45 2.96 3.31 3.72 l/m .40 .50 .60 .70 .80 .90 This would result in a greater angle for each order image and therefore a smaller percentage error in measuring the angle. (v) Give another way of improving the accuracy of this experiment. Repeat and get average value for the wavelength , repeat for higher orders. 3 A cylindrical column of air closed at one end and three different tuning forks were used in an experiment to measure the speed of sound in air. A tuning fork of frequency f was set vibrating and held over the column of air. The length of the column of air was adjusted until it was vibrating at its first harmonic and its length l was then measured. This procedure was repeated for each tuning fork. Finally, the diameter of the column of air was measured. The following data was recorded. Diameter of column of air = 2.05 cm (i) Describe how the length of the column of air was adjusted. The inner pipe was raised while immersed in water. (ii) Describe how the frequency of the column of air was measured. The frequency was read from the tuning fork which caused the vibration. (iii) Describe how the diameter of the column of air was measured. Using a digital calipers (iv) How was it known that the air column was vibrating at its first harmonic? The inner tube was raised until a loud sound could be heard. (v) Using all of the data, calculate the speed of sound in air. v = f λ λ = 4( l +0.3 d) v1 = 340(.3) m s -1; v2 = 342(.0) m s-1 ; v3 = 341(.1) m s-1 v ave = 341(.13) ms-1 4 In an experiment to verify Joule’s law a student passed a current through a heating coil in a calorimeter containing a fixed mass of water and measured the rise in temperature Δθ for a series of different values of the current I. The student allowed the current to flow for three minutes in each case. (i) Describe, with the aid of a labelled diagram, how the student arranged the apparatus. See diagram. (ii) Why was a fixed mass of water used throughout the experiment? The mass of water would be a third variable and you can only investigate the relationship between two variables at a time. (iii) The student drew a graph, as shown. Explain how this graph verifies Joule’s law. Straight line graph through origin ⇒ Δθ α I2 ⇒ P α I2 (iv) Given that the mass of water in the calorimeter was 90 g in each case, and assuming that all of the electrical energy supplied was absorbed by the water, use the graph to determine the resistance of the heating coil. The specific heat capacity of water is 4200 J kg–1 K–1. Electrical energy in = Heat energy out RI2 t = mcΔθ Rt = mc(Δθ/ I2) ⇒ Rt = mc(slope) ⇒ R = mc(slope)/t = (.09)(4200)(3.8)/180 R = (7.8 ↔ 8.2) Ω 5 (a) State Newton’s third law of motion. Newton’s First Law of Motion states that every object will remain in a state of rest or travelling with a constant velocity unless an external force acts on it. f/Hz 512 480 426 l/cm 16.0 17.2 19.4 (i) alternating current (ii) direct current. (i) axes labelled (I and t),sinusoidal curve (at least one full wave) (ii) (axes labelled) correct curve (v) The peak voltage of the mains electricity is 325 V. Calculate the rms voltage of the mains? V max = (√2)(Vrms) Vrms= 325/√2 = 230 V (vi) What is the resistance of the filament of a light bulb, rated 40 W, when it is connected to the mains? P = V2/ R 40 = (229.81)2/ R R = 1320 Ω (vii) Explain why the resistance of the bulb is different when it is not connected to the mains. Cold filament has lower resistance 10 (a) During a nuclear interaction an antiproton collides with a proton. Pair annihilation takes place and two gamma ray photons of the same frequency are produced. (i) What is a photon? A photon is a discrete amount of electromagnetic radiation. (ii) Calculate the frequency of a photon produced during the interaction. m [= mass of proton + mass of antiproton ] = 2(1.673 × 10-27) = 3.346 × 10-27 kg E = mc2 = (3.346 × 10-27 )(2.998 × 108)2 = 3.0074 × 10-10 Energy for one photon = 1.5037 × 10-10 J E = hf ⇒ f = E/h / = 1.5037 × 10-10 / 6.626 × 10-34 = 2.2694 × 1023 Hz (iii) Why are two photons produced? So that momentum is conserved. (iv) Describe the motion of the photons after the interaction. They move in opposite directions. (v) How is charge conserved during this interaction? Total charge before = +1-1 = 0 Total charge after = 0 since photons have zero charge (vi) After the annihilation, pairs of negative and positive pions are produced. Explain why. The energy of the photons is converted into matter . (vii) Pions are mesons that consist of up and down quarks and their antiquarks. Give the quark composition of (i) a positive pion, (ii) a negative pion. π+ = up and anti-down π- = down and anti-up (viii) List the fundamental forces of nature that pions experience. Electromagnetic, strong, weak , gravitational (ix) A neutral pion is unstable with a decay constant of 2.5 × 1012 s–1. What is the half-life of a neutral pion? λ T1/2 = ln 2 (= 0.693) T1/2 = 0.693 / 2.5 × 10 12 T1/2 = 2.8 ×10 -13 s 11 Read the following passage and answer the accompanying questions. The growth of rock music in the 1960s was accompanied by a switch from acoustic guitars to electric guitars. The operation of each of these guitars is radically different. The frequency of oscillation of the strings in both guitars can be adjusted by changing their tension. In the acoustic guitar the sound depends on the resonance produced in the hollow body of the instrument by the vibrations of the string. The electric guitar is a solid instrument and resonance does not occur. Small bar magnets are placed under the steel strings of an electric guitar, as shown. Each magnet is placed inside a coil and it magnetises the steel guitar string immediately above it. When the string vibrates the magnetic flux cutting the coil changes, an emf is induced causing a varying current to flow in the coil. The signal is amplified and sent to a set of speakers. Jimi Hendrix refined the electric guitar as an electronic instrument. He showed that further control over the music could be achieved by having coils of different turns. (Adapted from Europhysics News (2001) Vol. 32 No. 4) (a) How does resonance occur in an acoustic guitar? Energy is transferred from the strings to the hollow body and both vibrate at the same frequency. (b) What is the relationship between frequency and tension for a stretched string? Frequency is proportional to the square root of tension. (c) A stretched string of length 80 cm has a fundamental frequency of vibration of 400 Hz. What is the speed of the sound wave in the stretched string? v = f λ ⇒ v = 400(1.6) = 640 m s-1 (d) Why must the strings in the electric guitar be made of steel? Because only metal strings can be magnitised. (e) Define magnetic flux. Magnetic flux is the product of magnetic flux density multiplied by area. (f) Why does the current produced in a coil of the electric guitar vary? Because the induced emf varies due to the amplitude of the vibrating string. (g) What is the effect on the sound produced when the number of turns in a coil is increased? A louder sound is produced. (h) A coil has 5000 turns. What is the emf induced in the coil when the magnetic flux cutting the coil changes by 8 × 10–4 Wb in 0.1 s? E=−N(dφ / dt) E = 5000(8 × 10-4 /0.1) = 40 V 12 (a) (i) Define pressure. Pressure = Force divided by area. (ii) Is pressure a vector quantity or a scalar quantity? Justify your answer. It is a scalar because it has no direction. (iii) State Boyle’s law. Boyle’s Law states that pressure is inversely proportional to volume if temperature is constant. (iv) A small bubble of gas rises from the bottom of a lake. The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 1.01 × 105 Pa. The temperature of the lake is 4 oC. Calculate the pressure at the bottom of the lake. Pressure at bottom = 3 × pressure at top = 3.03 × 105 Pa (v) Calculate the depth of the lake. Pressure at bottom due to water = 2.02 × 105 Pa P= hρg h = P/ρg = 2.02 × 105 / (1.0 × 103)( 9.8 ) = 20.61 m 12 (b) (i) List the factors that affect the capacitance of a parallel plate capacitor. Common area of plates, distance apart, permittivity of dielectric between plates. (ii) The plates of an air filled parallel plate capacitor have a common area of 40 cm2 and are 1 cm apart. The capacitor is connected to a 12 V d.c. supply. Calculate the capacitance of the capacitor. C=εA/d C = [(8.85 × 10-12)(40 × 10-4)] / (0.01) C = 3.54 × 10-12 F (iii) Calculate the magnitude of the charge on each plate. Q = C V Q = (3.54 x 10-12)(12) = 4.2(5) x 10-11 C (iv) What is the net charge on the capacitor? zero (v) Give a use for a capacitor. blocks d.c. /smoothing /tuning circuits / timing circuits / flash guns for cameras. 12 (c) (i) Define power. Power is defined as energy divided by time. (ii) 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. (iii) 400 g of water at a temperature of 15 oC is placed in an electric kettle. The power rating of the kettle is 3.0 kW. Calculate the energy required to raise the temperature of the water to 100 oC. E = m cΔθ E = (0.40)(4200)(85) = 1.428 x 105 J (iv) Calculate the energy supplied by the kettle per second. 3000 J per second = 3000 W (v) Calculate the least amount of time it would take to heat the water to 100 oC. Time taken = 1.428 × 105 / 3000 = 47.6 s (vi) In reality, the time taken to heat the water will be greater. Explain why. Energy will be lost to the surroundings. 12 (d) The first Nobel Prize in Physics was awarded in 1901 for the discovery of X-rays. (i) What are X-rays? High frequency electromagnetic radiation. (ii) Who discovered them? Rontgen (iii) In an X-ray tube electrons are emitted from a metal cathode and accelerated across the tube to hit a metal anode. How are the electrons emitted from the cathode? By thermionic emission. (iv) How are the electrons accelerated? By the high voltage between the anode and cathode. (v) Calculate the kinetic energy gained by an electron when it is accelerated through a potential difference of 50 kV in an X-ray tube. Ek (= W) = q V = (1.6 × 10-19)(50 × 103) = 8.0 × 10-15 J (vi) Calculate the minimum wavelength of an X-ray emitted from the anode. E = h c/λ λ = [6.6 × 10-34 × 3.0 × 108] /(8.0 × 10-15)