PHYSICS 212 FINAL EXAM 21 March 2003

Transcription

1 PHYSIS INAL EXAM Marh 00 Eam is losed book, losed notes. Use only the provided formula sheet. Write all work and answers in eam booklets. The baks of pages will not be graded unless you so ruest on the front of the page. Show all your work and eplain your reasoning (eept on #. Partial redit will be given (not on #. No redit will be given if no work is shown (not on #. If you have a question, raise your hand or ome to the front. harges labeled +q are positive and those labeled -q are negative.. (5 points or eah of these multiple hoie questions, indiate the orret response (A, B, or on the page for problem in your eam booklet. i A boat made entirely of wood floats in a large swimming pool. Does the water level in the pool move upward, move downward, or remain the same when the boat is disassembled and all the wooden piees are floating individually in the pool? A Upward. B Downward. emains the same. ii An ideal gas undergoes an isobari (onstant pressure epansion. Does the temperature of the gas inrease, derease, or remain the same? A Inreases. B Dereases. emains the same. iii A apaitor is harged up using the iruit shown at right. If the resistane of the resistor is inreased, does the time for the apaitor to aquire a ertain harge after the swith is losed inrease, derease, or remain the same? A Inreases. B Dereases. emains the same. iv The uations be desribe three waves. Whih wave has the largest speed? A y A os( - 4t B y B os(4 + t y 4 os( - t v Two harged partiles are fied in plae on a line as shown at right. In whih region on the line an a proton be plaed suh that the proton will remain at rest? A egion A B egion B egion vi A thin insulating rod has a uniformly distributed harge +q on its left half and -q on its right half. onsider a point P on the perpendiular bisetor of the rod, as shown at right. If the eletri potential is defined to be zero at infinity, then is the eletri potential at point P positive, negative, or zero? A Positive B Negative Zero vii Three idential light bulbs are onneted in the iruit shown at right. When the swith is losed, does the brightness of bulb inrease, derease, or remain the same? A Inreases. B Dereases. emains the same. A B +q +q P

2 . (5 points ive moles of an ideal monatomi gas are taken through the yle aba shown at right. Proess ab is adiabati and proess b is isothermal. Assume P b Pa, b 0.50 m, 6 b and 8. J/molK. a How muh work is done by the gas during the omplete yle? b What is the effiieny of this heat engine? Is the effiieny found in part (b larger, smaller, or the same as the effiieny of an ideal engine operating between the highest and est temperatures that our in the yle? d ind the entropy hange of the gas during the proess a. P b a. (5 points ive idential light bulbs are onneted in a iruit as shown at right. Eah light bulb dissipates 5 W of eletrial power when a 0 eletrial potential differene is plaed aross the bulb. The swith is open for parts (a and (b. a ind the urrent through light bulbs,, and 5. b Light bulb 5 is replaed by a light bulb that is normally brighter (i.e., when plaed aross 0. Does bulb get brighter, dimmer, or remain the same, and why? When the swith is losed, light bulb turns off. What is the value of the resistor? (0 points Two apaitors ( µ, 6 µ are onneted in a iruit as shown at right. (Swith is set to for (a and (b. a ind the magnitude of the voltage aross eah apaitor. a ind the magnitude of the voltage aross eah apaitor when a dieletri material of dieletri onstant κ 4 is inserted in apaitor. With the dieletri in plae, move the swith to z to onnet the resistor 5 Ω. ind the urrent through the resistor immediately after moving the swith and then a long time later. 5. (5 points A thin hoop of radius and uniform linear harge density λ has its enter at the origin of the -ais and its ais oinident with the -ais, as shown at right. Answer all parts in terms of the parameters given and any other physial onstants ruired. a ind the eletri potential on the -ais at a distane d from the origin (assuming the potential is zero at infinity. b ind the eletri field on the -ais at a distane d from the origin. ind the loation of an additional point harge -Q (oppositely harged ompared to the hoop that makes the eletri field at the position in (b beome zero. z d 6. (0 points A hol styrofoam (density ρ S 0. g/m ylinder has length L 0 m, inner radius r.0 m, and outer radius r 5.0 m. A small weight of mass m 0 60 g and volume 0 0 m is attahed to the bottom of the ylinder and the system floats in water (density ρ W.0 g/m. Assume that the ylinder floats with the ais vertial as shown. a ind the distane from the water level to the top of the ylinder. b Oil (density ρ oil 0.5 g/m is then added to the system (same level inside and outside ylinder until a tuning fork of frueny f 5 Hz is resonant with the third harmoni of the air olumn in the ylinder. ind the depth of the oil layer. Assume the speed of sound in air is 40 m/s.

3 PHYSIS INAL EXAM SOLUTIONS Marh 00. i When the boat is floating, the displaed water has the same weight as the total weight of the boat. When the individual piees float, the same weight must be supported, so the same amount of water must be displaed. ii A During an isobari proess the pressure is onstant. The ideal gas law (p nt then implies that T is proportional to. Thus if the gas epands, it must get hotter. iii A Inreasing the resistane dereases the urrent, whih means it takes longer for the harge to aumulate on the apaitor. iv The waves are written in the form y y ma os(k ± ωt. The speed of a wave is given by v ω/k. This ratio is largest for wave. The sign tells us whih way the wave travels. v B A positively harged proton is repelled by the +q harge and by the +q harge. In regions A and these repulsions will add to fore the proton to infinity. In region B the repulsions an anel at the appropriate plae to give no net fore. vi Sine P is ually spaed from the two sides and the potential is proportional to dq/r for eah harge element, the two sides will anel due to their opposite harges. Note that the eletri field is not zero, just the potential. vii A The swith short iruits bulb, reduing the total resistane of the iruit from to. The urrent (i / total thus inreases, whih auses the power dissipated in eah resistor (P i to inrease...0

4 PHYSIS INAL EXAM SOLUTIONS Marh 00. a The work done in a yle is the area enlosed by the yle in the p- diagram. There is no ontribution from a, we integrate along the isotherm b, and we use the st law of thermodynamis along the adiabat ab. To find the temperatures, relate the points using ideal gas law and adiabat definition. W Wab + Wb Wab Eab n Tab n( Tb Ta d Wb pd ntb ntb ln b b b W ntb ln n( Tb Ta b P Pa m b b Tb Pb b ntb Tb 5 0 ( 0. 5 : 40K T ( isotherm n 5mol 8. J molk γ γ ab: P P P P b γ a a b b a b Pb 6 T a W 5mol 8. J molk 40K ln6 40K 70K γ 5 5 a a b b a ( 6 ( 6 b ( P n P 6 n 6 T 70K W 74. 6kJ b Heat is transferred into the gas during the proess b, whih is isothermal ( E b 0. Qb Wb ntb ln b Qb ntb ln 6 5mol( 8. J molk 40K( kJ W 74. 6kJ ε Q 79kJ in ε 4% The highest and est temperatures in the yle our at b and a, respetively. Thus the effiieny of an ideal engine would be: T Ta 70K ε ideal TH Tb 40K ε 70% ε < ε ideal ideal d Proess a is a onstant volume proess, so the heat transferred is Q n T. The entropy hange is thus: a T dq a n dt a dt Ta S n n T T T T ln T K S n 70 ln 5mol ( 8. J molk( 9. 40K S 74. J K T T..0

5 PHYSIS INAL EXAM SOLUTIONS Marh 00. a irst find the resistane of the idential bulbs. Then find the uivalent resistanes of the various groups, alling the er 4 group "." P P + 8Ω 0 5W Ω Ω total up 4Ω 0 itotal 79 total 5.6Ω. A i i. 79A i. 79A( 4Ω 7. 6 total i A 4Ω 45 i i A 45 8Ω i 79. A i 07. A i5 06. A b If the new bulb is normally brighter, then its resistane must be smaller, sine P /. So replae 5 by a smaller resistane in the above alulations. A smaller 5 will give a smaller and hene a smaller total. This will mean a larger total urrent, whih implies a larger voltage drop aross and a orrespondingly smaller voltage drop aross the er set. Hene, lightbulb has a smaller voltage drop and will thus burn less brightly sine P /. Bulb gets dimmer If light bulb turns off, then its urrent and voltage drop must be zero. All the urrent (I fs through and. Apply Kirhoff's loop rule as shown. I I 0 I 0 I I I 0. Ω 0 6. Ω 0 I..0

6 PHYSIS INAL EXAM SOLUTIONS Marh a ind the uivalent apaitane of the two apaitors in series. Then find the harge on the uivalent apaitor and note that the same harge resides on eah of the apaitors in series. + + µ µ 6 µ 6 µ q q q µ ( 4µ q 4µ 8 µ q 4µ 4 6µ 8 4 b The dieletri auses the apaitane to inrease by the fator κ 4. ol the same alulation to get: + κ 4µ 6 q q q 4µ ( 48µ q 48µ 4 κ µ q 48µ 8 6µ 4 + µ µ 8 4µ Sine the apaitors are fully harged, the voltage aross the resistor immediately after moving the swith is. This voltage will drive a urrent through the resistor, whih will deplete the harges on the apaitors. Thus after a long time the urrent must stop, at whih point the voltage aross is 0. it ( 0 5Ω it ( 0. 4A it ( 0A..0 4

7 PHYSIS INAL EXAM SOLUTIONS Marh a onsider the hoop as omposed of infinitesimal harge elements dq. Eah element produes an eletri potential d at the point P on the -ais. To find the total potential, integrate over the length of the hoop. Sine eah element of the hoop is the same distane from the ais, the denominator an be pulled out of the integral. d k dq e r dq ke k r e kq + d dq + d k πλ + d e hoop e keπλ + d ke + d b The field on the ais has only an -omponent, by symmetry, and an be found by differentiating the potential, if we replae d by the variable for the derivative: keπλ + ke ke E π λ π λ + + dq keπλd E ( + d A new harge -Q must produe a field that eatly anels the field above. Sine -Q is opposite in sign to Q hoop, the fore will be attrative. Hene Q must be plaed on the -ais at a position < d. E keπλd ( + d Q + d ( d πλd kq e d 0 d Q ( + d πλd..0 5

8 PHYSIS INAL EXAM SOLUTIONS Marh a Let the distane from the water level to the top of the ylinder be. Arhimedes' priniple states that the buoyant fore is ual to the weight of the displaed fluid. Sine the ylinder is floating, the buoyant fore is ual to the weight of the ylinder and the attahed weight. buoyant mtotalg ρwg + ρwgπ( L ( r r m g + ρsgπl r r ρwπ( L ( r r m0 + ρsπl( r r ρw0 m0 + ρsπl( r r ρw0 ( L ρwπ( r r L m 0 + ρsπ L ( r r ρw 0 ρ π r r 0 0 w ( r r ( 5m ( m 5m 9m 6m 60g 0. g m 0m 6m. 0 g m 0m 0m π. 0 gm 6m +π 0m 5. 0m 5m b An air olumn with one end open and one end losed has a quarter-wave fundamental and odd integer harmonis. Let the air olumn height from the oil level to the top of the ylinder be d, and the thikness of the oil level be z. v 40 ms d λ 0. m m 4 4 f 4 5Hz buoyant mtotalg ρwg + ρwgπ( L d z ( r r + ρoilgπz( r r m g+ ρsgπl r r ( ρw ρoil πzr ( r ρw0 + ( ρw ρsπlr ( r ρwπdr ( r m ( [( ] w oil ( +π( ( ( π ρw m +π r r ρw ρs L ρwd z ( ρ ρ π r r [ ] 0g 60g 6m 0. 8g m 0m. 0 g m m z 05. gm 6m z 60. m