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第10届国际物理奥林匹克竞赛试题及解答(英文)

时间:2011-09-19


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10th International Physics Olympiad 1977, Hradec Kr?alov?e, Czechoslovakia
Problem 1. The compression ratio of a four-stroke internal combustion engine is
o

The

engine draws in air and gaseous fuel at a temperature 27 C at a pressure 1 atm = 100 kPa. Compression follows an adiabatic process from point 1 to point 2, see Fig. 1. The pressure in the cylinder is doubled during the mixture ignition (2–3). The hot exhaust gas expands adiabatically to the volume V2 pushing the piston downwards (3–4). Then the exhaust valve opens and the pressure gets back to the initial value of 1 atm. All processes in the cylinder are supposed to be ideal. The Poisson constant (i.e. the ratio of specific heats Cp=CV ) for the mixture and exhaust gas is (The

compression ratio is the ratio of the volume of the cylinder when the piston is at the bottom to the volume when the piston is at the top.)

a) Which processes run between the points 0–1, 2–3, 4–1, 1–0? b) Determine the pressure and the temperature in the states 1, 2, 3 and 4. c) Find the thermal efficiency of the cycle. d) Discuss obtained results. Are they realistic? Solution: a) The description of the processes between particular points is the following: 0–1 : intake stroke isobaric and isothermal process 1–2 : compression of the mixture adiabatic process
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2–3 : mixture ignition isochoric process 3–4 : expansion of the exhaust gas adiabatic process 4–1 : exhaust isochoric process 1–0 : exhaust isobaric process Let us denote the initial volume of the cylinder before induction at the point 0 by V1, after induction at the point 1 by V2 and the temperatures at the particular points by T0, T1, T2, T3 and T4. b) The equations for particular processes are as follows. 0–1 : The fuel-air mixture is drawn into the cylinder at the temperature of T0 = T1 = 300 K and a pressure of p0 = p1 = 0:10 MPa. 1–2 : Since the compression is very fast, one can suppose the process to be adiabatic. Hence:

From the first equation one obtains

and by the dividing of both equations we arrive after a straightforward calculation at

For given values

,

, p1 = 0:10 MPa, T1 = 300 K we have p2 =

2:34 MPa and T2 = 738 K(t2 = 465oC). 2–3 : Because the process is isochoric and p3 = 2p2 holds true, we can write

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c) Thermal efficiency of the engine is defined as the proportion of the heat supplied that is converted to net work. The exhaust gas does work on the piston during the expansion 3–4, on the other hand, the work is done on the mixture during the compression 1–2. No work is done by/on the gas during the processes 2–3 and 4–1. The heat is supplied to the gas during the process 2–3. The net work done by 1 mol of the gas is

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d) Actually, the real pV -diagram of the cycle is smooth, without the sharp angles. Since the gas is not ideal, the real efficiency would be lower than the calculated one. Problem 2. Dipping the frame in a soap solution, the soap forms a rectangle film of length b and height h. White light falls on the film at an angle (measured with respect to the

normal direction). The reflected light displays a green color of wavelength a) Find out if it is possible to determine the mass of the soap film using the laboratory scales which has calibration accuracy of 0.1 mg. b) What color does the thinnest possible soap film display being seen from the perpendicular direction? Derive the related equations.

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b) If the light falls at the angle of 30o then the film seen from the perpendicular direction cannot be colored. It would appear dark. Problem 3. An electron gun T emits electrons accelerated by a potential difference U in a vacuum in the direction of the line a as shown in Fig. 2. The target M is placed at a distance d from the electron gun in such a way that the line segment connecting the points T and M and the line a subtend the angle as shown in Fig. 2. Find the magnetic

induction B of the uniform magnetic field

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Solution: a) If a uniform magnetic field is perpendicular to the initial direction of motion of an electron beam, the electrons will be deflected by a force that is always perpendicular to their velocity and to the magnetic field. Consequently, the beam will be deflected into a circular trajectory. The origin of the centripetal force is the Lorentz force, so

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b) If a uniform magnetic field is neither perpendicular nor parallel to the initial direction of motion of an electron beam, the electrons will be deflected into a helical trajectory. Namely, the motion of electrons will be composed of an uniform motion on a circle in the plane perpendicular to the magnetic field and of an uniform rectilinear motion in the direction of the magnetic field. The component initial velocity of the

, which is perpendicular to the magnetic field (see Fig. 4), will

manifest itself at the Lorentz force and during the motion will rotate uniformly around the line parallel to the magnetic field. The component field will remain parallel to the magnetic

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Hence we can calculate the radius of the circular trajectory

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