امتحانات الشهادة الثانوية العامة فرع العلوم العامة مسابقة في مادة الفيزياء المدة ثالث ساعات

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1 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات امتحانات الشهادة الثانوية العامة فرع العلوم العامة مسابقة في مادة الفيزياء المدة ثالث ساعات االسم: الرقم: دورة سنة 008 العادية This exam is formed of four exercises in four pages numbered from to 4 The use of non-programmable calculator is recommended First exercise (75 points) ompound pendulum A A compound pendulum is formed of a rod AB of negligible mass, which can rotate without friction in a vertical plane around a horizontal axis () passing through a point O of the rod so that OB = d A particle of mass M is fixed at point B and another particle of mass m < M, which can slide on the part OA of the rod is placed at a distance O = x of adjustable value Let a = OG be the distance between O and the center of gravity G of the pendulum (Fig) The gravitational potential energy reference is the horizontal plane containing O g = 0m/s ; = 0 ; sin = and cos = θ,( θ in rad ) for < 0 0 A- Theoretical study Md mx - Show that the position of G is given by: a = (M m) - Find the expression of the moment of inertia of the pendulum about the axis ( ) in terms of m, x, M and d m Fig 3- The pendulum thus formed is deviated by an angle 0 from its equilibrium position and then released from rest at the instant t0 = 0 The pendulum then oscillates around the stable equilibrium position At an instant t, the position of the pendulum is defined by the angular abscissa, the angle that the vertical through O makes with OG, and its angular velocity is dθ θ a) Write, at the instant t, the expression of the kinetic energy of the pendulum in terms of andθ b) Show that the expression of the gravitational potential energy of the system (pendulum, Earth) is PE = - (M + m) g a cos c) Write the expression of the mechanical energy of the system (pendulum, Earth) in terms of M, m, g, a,, and θ d) Derive the second order differential equation in that governs the motion of the pendulum e) Deduce that the expression of the proper period, for oscillations of small amplitude, has the form: T = (M m)ga f) Find the expression of the period T, in terms of M, m, d, g and x B- Application: metronome A metronome is an instrument that allows adjusting the speed at which music is played The compound pendulum studied in part A represents a metronome where M = 50 g, m = 5 g, and d = cm The graph of figure represents the variations of the period T of this metronome as a function of the distance x ) Find, in this case, the expression of the period T of the metronome as a function of x ) The leader of the orchestra (conductor), using a metronome to play a distribution, changes the position of along OA, to follow the rhythm of the musical piece The rhythm is indicated by terms inherited from talian for the classical distribution: x d O G B a M

2 Name ndication Period (in s) Grave very slow T = 5 Lento Slow T Moderato Moderate 06 T Prestissimo very fast 08 T 04 Determine, using a method of your choice, the positions between which the leader of the orchestra may move to adjust the speed to the rhythm Lento Second exercise (75 points) Determination of the capacitance of a capacitor n order to determine the capacitance of a capacitor, we consider two experiments A- First experiment i D We place the capacitor in series, with a coil of inductance L = 03 H, q a resistor of resistance R = 00 Ω and a low frequency generator G (LFG) that delivers across its terminals an alternating sinusoidal A voltage: ug = udb = 8sin (00 t π/3) (ug in V ; t in s) (Fig) As a result, the circuit carries an alternating sinusoidal current of value: i = m sin (00 t) (i in A; t in s) G An oscilloscope is connected so as to display, on channel Y, M the voltage ucoil= uam across the coil and, on channel Y, the voltage ur = umb across the resistor The knob «nv» (inverse) on channel Y is pushed Fig B On the screen of the oscilloscope, we observe the waveforms () and () represented in figure () () The vertical sensitivity Sv is the same on the two channels: Sv= V/div Take 03 = ) Why did we push in the knob «nv»? ) Referring to figure : a) Determine the horizontal sensitivity Sh that is selected on the oscilloscope b) Determine the phase difference between ub and ur c) Which of the two voltages leads the other? d) Deduce that the coil has a negligible resistance e) Determine the value of m 3) Determine the expression of ucoil as a function of time t Fig 4) Show that expression of the voltage u = uda across the m capacitor is given by u = - cos00π t 00π 5) Applying the law of addition of voltages, determine the value of by giving t a particular value, u L R Y Y

3 B- Second experiment The capacitor, initially charged, is now connected across the coil of inductance L = 03 H (Fig3) q The oscilloscope, adjusted on the horizontal sensitivity Sh = ms/div, allows to display the voltage u across the capacitor (Fig4) ) a) Show that the voltage u is sinusoidal of period T b) Determine T in terms of L and ) alculate the value of i L Fig3 Fig4 Third exercise (75 points) ndex of refraction of a piece of glass onsider a glass sheet of thickness e = 5 μm and of index of refraction n, and a source S of white light having a filter so that Young s apparatus receives monochromatic light of wavelength λ in air of adjustable value The object of this exercise is to study how the index n varies with λ A- Light interference nterfringe distance Young s slits apparatus is formed of two very thin slits F and F, parallel and separated by a distance a = 0 mm, and a screen of observation (E) placed parallel to the plane of the slits at a distance D = m from this plane ) F and F are illuminated with a monochromatic radiation of wavelength λ issued from S that is placed at equal distances from F and F a) F and F must have two basic properties for the phenomenon of interference to be observed What are they? b) Describe the system of fringes observed on (E) c) At the point O of the screen,equidistant from F and F, we observe a bright fringe Why? ) We admit that for a point M of (E), such that OM = x, the optical path difference in air or in ax vacuum is given by FM FM (E) D a) Determine the expression of xk corresponding to the center of the k th bright fringe b) Deduce the expression of the interfringe distance i in terms of λ, D and a B- ntroducing the sheet The glass sheet is put now just behind the slit F c and v are the speeds of light in vacuum (and practically in air) and in the glass sheet respectively ) Light crosses the glass sheet of thickness e during a time interval τ Give the expression of τ in terms of e and v ) Give the expression of the distance d, covered by light in air during the time interval τ, in terms of n and e 3) Deduce that the new optical path difference at point M is given by: ax - e(n ) FM FM D - Measurement of n NB : ntroducing the sheet does not affect the expression of the interfringe distance i n this question the calculation of n must include 3 decimal places ) F and F are illuminated with a red radiation, of wavelength λ = 768 nm, issued from S The center of the central fringe is formed at O, position that was occupied by the center of the 4 th bright fringe in the absence of the sheet Determine the value of n, the index of the sheet ) Fand F are illuminated with a violet radiation of wavelength λ = 434 nm, issued from S The center of the central fringe is now formed at O, position that was occupied by the center of the 8 th dark fringe in the absence of the sheet Determine the value of n, the index of the sheet 3 S F F M x O

4 3) an we consider the value of the index of refraction of a transparent medium without taking into account the radiation used? Why? Fourth exercise (75 points) Nuclear Fission The object of this exercise is to show evidence of certain properties of nuclear fission Given: Masses of nuclei: m ( 35 U) = u; m( 9 r) = 987 u; m( 4 Te ) = 4869 u; m( n ) = 008u; u = Kg ; c = ms - ; A- Energy of fission One of the fission reactions of the uranium 35, in a nuclear power plant may be written in the form : 9U 0n 40r Te x 0n ) Determine and x specifying the laws used ) alculate the energy produced by the fission of one nucleus of uranium 35 3) Determine the mass of uranium 35 used in the power plant during one year, knowing that its useful electric power is 900 MW, and that its efficiency is 30 % B- Products of fission Among the products of fission, we find, in the core of the reactor, the radioelements: 37 55s and 87 Rb of periods 30 years and years respectively These radioelements are placed in a pool called cooler The nuclei s and 37Rb have the masses 37u and 87u respectively ) Suppose that g of each of the radioelements is introduced into the pool at the instant t0 = 0 a) alculate the number of nuclei of each of the radioelements at the instant t0 = 0 b) Deduce, for each radioelement, the number of nuclei remaining after 3 years stay in the pool c) Determine, for each radioelement, the number of decays per day at the moment of taking them out of the pool (3 years later) ) Assuming that the danger of a radioelement on man depends on the radiations accumulated per day, which, of the two radioelements, is more dangerous? Justify - Probability of fission n a physics dictionary, we read that the probability of a nucleus A X to become fissionable is proportional to the ratio, called the stability factor of a nucleus This probability is no more A zero when this ratio exceeds 35 ) What do each of and A of the nuclide A X represent? ( 35) ) Show that a nucleus must contain a number of neutrons N such that N, so that 35 the probability to become fissionable is not zero 3) Find the maximum number of nucleons that must be contained in a uranium nucleus, of = 9, so that the probability to undergo fission is not zero 4

5 008 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات مشروع معيار التصحيح امتحانات الشهادة الثانوية العامة الفرع : علوم عامة مسابقة في مادة الفيزياء المدة ثالث ساعات االسم: الرقم: الدورة العادية للعام Part of the Q Answer Mark First exercise (75 points) A Md mx (M-m) OG = M OB + m O a = M m A = M + m = Md + mx 050 A3a E = ½ 050 EPP = -(M+m)gh = -(M+m)gacos O h G A3b 00 A3c Em = E + EPP =½ -(M+m)gacos 05 A3d de m 0 (M m)ga sin = 0 00 for small, sin = (M m)ga 0 A3e the proper angular frequency is: ( M m)ga ; 5 A3f the proper period T = T= (M m) ga T = Md mx g(md mx) B T = 008 0x 5x 050 B graphically or by calculus: For T = s, x = 3 cm For T = s, x = 3 cm 00 3 < x(cm) < 3 Second exercise (75 points) A To eliminate the phase opposition obtained from the way with which the coil and the resistor are connected to the oscilloscope (or: the oscilloscope, as it is connected, displays the voltage ubm, but as we want to display umb, then we have to push the knob inversion) 05 The angular frequency of the voltage is = 00 rad/s; or the period is T = = 0,0s = 0ms ; Aa T covers 4 divisions on the screen Sh= 0 5ms/div 4 Ab T covers 4 divisions that correspond to an angle of rad, the phase 5

6 difference is represented by division = rad 4 Ac ub leads ur 05 Because the voltage across the coil of zero resistance leads by the current Ad 050 that flows through it Ae 4 Rm = 4 div V / div = 4V m = 004A A3 di ub = L cos(00 t) = 4cos(00 t) A4 dq du i = u = primitiveof i = - m 00 cos( 00 t) 4 80 u = - cos(00 t) ug = u + ub + ur A sin(00 t- ) = - 3 cos(00 t) + 4cos(00 t) + 4 sin(00 t) 4 80 For t = 0 we have : = = 70-6 F di u = ub u = L( ) - L q = - L u Ba u + u= 0 the solution L 050 of such form of differential equation is sinusoidal of period T Bb The angular frequency of motion is such that = ; L L the period is T = L 050 From the waveform of figure 4 we have: T = 6 div ms / div = ms = 0,0s B T = = 50 F 4 L 5 Thierd exercise (75 points) Aa F and F become synchronous and coherent 050 Ab The interference fringes are bands that are equidistant, alternately dark and bright These fringes are parallel to the slits Ac We have = 0 then the radiations from Fand F arrive at O in phase thus they form at O a bright fringe 050 Aa For bright fringes: = k ax k D = k xk = ; D a 050 Ab D i = xk+ xk = a 050 B B d = c = B3 e 05 v 5 e c v = ne 050 the increase in the optical path for the light crossing the plate is: ne e = e(n-) = FM (FM + e(n ) = ax D 00 6

7 ax 0 For the central fringe: = 0 - e(n ) = 0 ; D so that x0 = 4i = 4 D 4 n = + = 64 a e 5 75 x0 = 75 i n = + e = 65 3 Non : n n Yes : n n Fourth exercise (75 points) A U 0n 40r Te x 0n onservation of mass number: 35 + = x x = 00 onservation of charge number: 9 = 40 + = 5 A Δm = = 05 u or kg E = Δmc = 30 - J E = 30 J / s 03 The energy for one year : 30 A = J The number of nuclei undergoing fission: nuclei 30 ts mass : = 494 kg N0( s ) = 4,40 nuclei 4 37,66 0 Ba 050 N0 ( Rb ) = 6,90 nuclei 4 87, t Bb N = N0 e - T N( s) = 4 0 nuclei N ( Rb) = nuclei N( Br ) = 0 number of disintegrations per day = N (:d - ) For s : 6 0 Bc For ( Rb) : B The more dangerous product is s, since its rate of disintegrations is greater 05 ca is the charge number, A is the mass number A N ( 35 ) ( 35) 35 N N 35 3 (9) 35 A A