Exam Sg Physics 2009

2009 ‫الدورة العادية للعام‬

‫امتحانات الشهادة الثانوية العامة‬ ‫ علوم عامة‬: ‫الفرع‬

‫وزارة التربية والتعليم العالي‬ ‫المديرية العامة للتربية‬ ‫دائرة المتحانات‬

‫مسابقة في مادة الفيزياء‬ ‫المدة ثلث ساعات‬

:‫السم‬ :‫الرقم‬

This exam is formed of four exercises in four pages numbered from 1 to 4. The use of non-programmble calculator is recommended, Answer the four following questions: First question (7 points) Torsion pendulum The object of this exercise is to determine through two experiments, and using theoretical study, the moment of inertia I of a homogeneous rod AB with respect to an axis perpendicular to the rod at its midpoint and the torsion constant C of a metallic wire OO'. The rod AB has a mass M = 50 g and a length ℓ = 60 cm. A torsion pendulum [P] is obtained by welding the midpoint of AB to one

end O of the wire while the other end O’ is fixed to a roof, thus AB can oscillate in a horizontal plane around an axis through OO’. AB is shifted, from its equilibrium position, by an angle θm radian in the horizontal plane and then released from rest at an instant taken as origin of time t0 = 0 . At an instant t during motion, AB makes an angle θ with the equilibrium position and has an d angular velocity & . dt The level of AB is taken as a gravitational potential energy reference. we neglect all friction and take π2 = 10. A- Theoretical study 1) Give, at the instant t, the expression of the mechanical energy ME of the system [P, Earth] in terms of I, C, θ and  . 2) Write the expression of ME: a) when θ = θm? b) when AB passes through the equilibrium position? 1

3) Derive the second order differential equation governing the motion of the pendulum [P]. 4) Is the motion of the pendulum [P] sinusoidal? Why? 5) Show that the natural period of the pendulum is given by T1 = 2π B- Experimental study1) By means of a stopwatch, we measure the duration t 1 of 20 oscillations we obtain t1 = 20 s. Verify that the relation between I and C is C = 40 I. 2) We fix two point masses each of 25 g at A and B. We thus obtain a new torsion pendulum [P’] that can perform also a rotational sinusoidal motion of natural period T2. a) Show that the moment of inertia I' of the system (rod + point masses) with respect to the axis OO’ is given l2 by: I' = I + m . 2 b) Write down the expression of T2 in

terms of I’ and C. c) By means of a stopwatch, we measure the duration t 2 of 20 oscillations and we obtain t2 = 40 s. Find a new relation between I and C. 3) Using both obtained relations between I and C, calculate I and C.

Second question (7 points) The phenomenon of self-induction The set up represented by the figure below consists of an ideal generator G of emf E = 12 V, of a coil of resistance r = 10  and of inductance L = 40 mH, of a resistor of resistance R = 40  and two switches K1 and K2.

AAt the instant t0 = 0, we close the switch K1 and we leave K2 open. The circuit carries a current i1 in the transient state. 1) Derive the differential equation verified by i1. 2) I0 is the current in the steady state. Determine the expression of I0 in terms of E,r and R then calculate its value. t

− ). e τ a) Determine the expression of τ in terms of L, r and R and calculate its value.

3) The solution of the differential equation is of the form: i1 = I0(1 –

2

b) Give the physical meaning of τ . 4) a) Determine the expression of the induced emf e as a function of time t. b) Calculate the value of the induced emf. e1 at the instant t0 = 0. B. After a few seconds, the steady state being established, we open K1 and we close at the same instant K2. We consider the instant of closing K2 as a new origin of time t0 = 0. The circuit (L, R, r) carries an induced current i2 at an instant t. 1) Determine the direction of i2 . 2) Derive the differential equation giving the variations of i2 as a function of time 3) Verify that i2 = I0E

−t τ

is the solution of this equation.

4) Calculate the value of the induced emf e2 at the instant t0 = 0. C. Compare e1 and e2 and deduce the role of the coil in each of the two previous circuits.

Third question (7 points) Characteristics of an (R, L, C) circuit

3

In order to determine the characteristics of an (R, L, C) circuit, we connect up the circuit represented in figure 1. This circuit is formed of a resistor of resistance R= 650 Ω, a coil of inductance L and of negligible resistance and a capacitor of capacitance C, all connected in series across a function generator (LFG) delivering a sinusoidal alternating voltage of the form: ug= uAM = Um cos(2 π f)t . A- We choose the value f1 for the frequency of the LFG and we display the variation of the voltage uAM across the generator on the channel (Y1) and the variation of the voltage uDM across the resistor on the channel (Y2) using an oscilloscope.

k

i

TheA waveforms obtained are represented in figure 2. Vertical sensitivity on both channels is: 2 V/div.

L

Horizontal sensitivity is: 0.1 ms /div. 1) Redraw figure (1) showing on it the connections of the

G

oscilloscope that allows to display the voltages uAM q us B

C

and uDM.

2) Referring to the waveforms, determine: a) The value of the frequency f1.

R

M

D

Fig. 1 b) The value uofDMφ1 the phase difference between uAM and uDM. 3) The current i carried by the circuit has the form: i = Im cos (2πf1 t - φ1 ). uAM down the expressions of the voltages: uAB, uBD and uDM as a function of time. a) Write b) The relation: uAM = uAB + uBD + uDM is valid for any t. Show, by giving to t a given value,

that:

Fig.2 L(2f1 )  tg1 

1 C  2f1 

R

B- Starting from the value f1, we decrease continuously the frequency f of the (LFG). We notice 4

that for f0 = 500 Hz the circuit is in a state of resonance. Deduce then a relation between L, C and f0. C- We keep decreasing the frequency f of the (LFG). For a value f2 of f we find that the phase difference between uAM and uDM is φ2 so that φ2 = - φ1. 1) Find the relation between f1, f2 and f0 2) Deduce the value of f2. D- Deduce from the preceding steps the values of L and of C.

Fourth question (7 points) Energy levels of the hydrogen atom The energies of the various levels of the hydrogen atom are given by the relation: E En = − 20 , where E0 is a constant and n is a whole positive number. n Given: Planck’s constant h = 6.63 × 10-34 J.s; 1 eV = 1.6 × 10-19 J; 1 nm = 10-9 m. Speed of light in vacuum: c = 3 × 108 m/s 1)-a- The energies of the hydrogen atom are quantized. What do we mean by quantized energy? b- Explain why the spectrum (absorption or emission) of hydrogen consists of lines. 2) A hydrogen atom, initially excited, undergoes a downward transition from the energy level E2 to the energy level E1. It then emits the radiation of wavelength: λ2,1 = 1.216 × 10-7 m. Determine, in J, the value: a- of the constant E0; b- of the ionization energy of the hydrogen atom taken in its ground state. 3) For the hydrogen we define several series to which the names of researchers who took part in their study are assigned. Among these series we consider that of Balmer, which is characterized by the downward transition from p > 2 to n = 2. To each transition p → 2 corresponds a line of wave λp,2. a- What do the term "series of lines" represent?  1 1 1  1.096 102   2 . b- Show that λp,2, expressed in nm, is given by the relation:  p,2 p   4 c- The analysis of the emission spectrum of the hydrogen atom shows four visible lines. Among which we consider the three lines: Hα = 656.28 nm ; Hβ = 486.13 nm and Hγ = 434.05 nm. These lines belong to the Balmer’s series. To which transition does each of these radiations correspond? d- Show that the wavelengths of the radiations corresponding to the series of lines studied tend when p → ∞ , towards a limit λ0. Calculate its value. 4) Balmer, in 1885, knew only the lines of the hydrogen atom that belong to the visible spectrum p2 λ = λ and he wrote the law of determination of these lines in the form: . 0 p2 − 4 a- Using the expression of the energy of the hydrogen atom and the energy of a photon emitted during a transition; determine the expression of λ0, in the formula of Balmer, in terms of E0, h and c. b- Show that the formula of Balmer makes it possible to find the result obtained in 3)-d.

5

Answer the four following questions: First question (7 points) Part of the Q A.1 A.2.a A.2.b A.3

A.4 A.5 B.1 B.2.a B.2.b B.2.c B.3

Answer

Mark

M.E = K.E + P.Eg + P.Ee = ½ Iθ’ 2+ 0 + ½ Cθ2 For maximum deviation , θ = θm and θ’ = 0. thus M.E = ½ C θ2m. In equilibrium position , θ = 0 and θ’ is maximum thus M.E = ½ Iθ’2m. M.E is conserved since no friction thus the derivative of M.E w.r.t time is zero I θ’ θ” + C θ θ’ = 0 θ’ ≠ 0 thus θ” + θ = 0 Equation has the form y” + ω2y = 0 has a solution sinusoidal provided ω2 = ω = thus T1 = 2π t = 20T = 20s thus T1 = 1 = 2π and 40 = 1 then C = 40 I l2 2 I’ = I + 2m () = I + m = I + 0.0045 2 Same law of motion thus T2 = 2π T2 = 2 thus 10 =1 or C = 10 I' = 10(I + 0.0045) = 10 I + 0.045 C = 40 I = 10I + 0.045 ⇒I = 1.5×10-3 kg.m2 and C = 0.06 N×m/rd

0.75 0.50 0.50 1.00

0.25 0.75 1.00 0.75 0.50 0.75 0.75

Second question (7 points) Part of the Q A.1

Answer E = ri1 + L

di1 dt

A.2

I0 =

di1 dt

A.4.a

i1

.

di1 dt

= 0; the current is then I0 such that: E = (r+R)I0 12 40 + 10

= I0/

1.00

τ

I0 =

⇒

E R+ r

.

= 0.24 A t eτ

(

−

)

E = (r + R)I0(1 -

⇒

t eτ −

) + LI0/

τ

(

t eτ −

)

⇒

L/

τ

=

L 0.04 = = 0.8 ms R+r 50 The time constant characterizes the duration of the current establishment in a (R+r, L) component

(r + R) A.3.b

⇒

When the steady state mode is established, i1 becomes constant and di1 dt

A.3.a

0.50

+ Ri1

E = (r+R) i1 + L

Mark

e=-L

⇒

di1 dt

τ

1.00

=

= -L I0/

τ

(

t eτ −

)=-E

t eτ −

6

.

0.25 0.50

A.4.b B.1

B.2

uAC = uAB + uBC L

⇒

B.3

B.4

C

0.25 0.50

For t =0, e = e1= -E = - 12 V. From the Lenz law, the coil will oppose the canceling of the current, It will create an induced current i2 of same direction as i1.

di 2 dt

di 2 dt

di 2 dt

0.75

+ Ri2

+ (R+ r ) i2 = 0

= - I0/

e=-L

0 = ri2 + L

⇒

i2

di 2 dt

τ

t eτ −

⇒

- L I0/

= - L( - I0/

τ

t eτ −

τ

t eτ −

)=E

−

+(R+r) I0 t eτ −

1.00

t

eτ

=0

. At t0 = 0, we have e = e2= E = 12

V e1 = - e2. When K1 is closed, the self-induced emf opposes the establishment of the current in the circuit e1 < 0 (the coil plays the role of a generator in opposition). When K2 is closed, the self-induced emf opposes the cancellation of the current in the circuit e2 > 0 (the coil plays the role of a generator).

0.75

1.00

⇒

⇒

Third question (7 points) Part of the Q A.1

Answer Connections of the oscilloscope.

Mark 0.25

7

k

i

A

Y1

L G

B C

q

D

R

M

Y2

A.2.a

T1 → 8 div ⇒ T1 = 0.8 ms

0.50

A.2.b A.3.a

f1= 1/T1= 1/ 0.8 × 10-3 = 1250 Hz | φ1| = 2 π 1/8 = π/4 rad. i = Imcos(2πf1⋅t - φ1)

0.25 1.00

uAB = L di/dt = -LIm(2πf1) sin(2πf1 t - φ1) uc= ∫i dt = ∫ cos (2 π f1 t - φ1) dt uc = () sin (2πf1 t - φ1) A.3.b

uR = Ri = R Im cos(2πf1 t - φ1) Um cos 2πf1 t = R Im cos(2πf1 t - φ1) +(() sin (2πf1 t - φ1)

0.75

– LIm (2 π f1) sin (2 π f1 t - φ1) 2πf1 t = π/2 ⇒ 0 = R Im sin φ1 + () cos φ1 – L Im (2πf1) cos φ1 R sin φ1 = [L(2πf1) – ] Im cosφ1

⇒

1 C  2f1 

L(2f1 )  tg1 

B

R Current resonance

φ=0

⇒

tgφ = 0

⇒

L2πf0 – = 0

⇒

0.75

LC4π2f02 = 1.

⇒

C.1 tgφ1 = - tg φ2 ⇒

⇒

1 1  L2f 2 C  2f1  = C2f 2 R R

L(2f1 ) 

1.50

L2πf1 + L2πf2 = 1/C [1/ (2πf1) + 1/ (2πf2)]

LC= 1/ 4π2 f1 f2 = 1/ 4π2 f02 f02 = f1 f2 f2 = (5002)/ 1250 = 250000/1250 = 200 Hz φ1= π/4 L2π(1250) – 1/(C 2π 1250) = 650 ⇒

C.2 D

⇒

×

LC = = 10-7 LC ⇒

×

4π2

×

×

(π2= 10) 12502 – 1= 650

C = 5.25 /(650

×

2π

×

C

×

×

2π

×

1250

1250) = 10-6 F = 1 μF

L = 10-7/ 10-6 = 10-1 H =0.1 H

8

0.50 2.00

Fourth question (7 points) Part of the Q 1.a 1.b

2.a

2.b 3.a

Answer

Mark

The energies of the hydrogen atom can take only particular values (well defined). For an electronic transition p → n the emitted photon (or absorbed) has hc a wavelength: λ p,n = . As Ep and EN are quantized then (Ep Ep − En EN) is too; which means that the λp,n has a well determined value, which corresponds to a line. 3E 0 hc E = E 2 = − 0 et E1 = − E 0 ⇒ E 2 − E1 = 4 λ 2,1 4 4  6,62 1034  3  108  E0   2,177  1018 J. 7 3 1, 216  10 E i  E   E1  E 0 = 2.177  10-18 J. It is the whole set of lines formed at the time of the electronic transitions of the atom from the level p to level n with p > n.

0.50

3.b

0.75

1.25

0.50 0.50 1.50

E 0 E 0 hc 1 E  1 1      0  2 2 p 4  p,2  p,2 hc  4 p  1 1 2,177 1018  109  1 1  1  2  1  2; = = 1.096  10   2  34 8   p,2  p,2 6,62  10  3 10  4 p   4 p 

Ep  E2  

3.c

3.d 4.a

4.b

Hα= 656.28 nm ⇒ p= 3, then it is the downward transition 3 → 2. Hβ; 4 → 2 et Hγ ; 5 → 2. 1.096 102 When p → ∞ ⇒ λ → λ0 =  364.96 nm 4  1 1  hc 4hc  p 2  4hc Ep  E2  E0   2     2   0   E0  p  4  E0  4 p    0

p2 1 4hc  0 2  364,96 nm 4 when p       0  p 4 1 2 E0 p

9

0.75

0.50 0.75

0.50