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Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 1 Q1. A sinusoidal wave with a speed of 22.4 m/s is travelling on a string of linear density 15.0 g/m. If the maximum transverse speed of a particle of the string is 7.05 m/s, then the average power transmitted by this wave is: A) B) C) D) E) 8.35 W 4.62 W 5.21 W 1.30 W 9.62 W Ans: 1 1 Pav = μω2 vym 2 = μv(ωym )2 2 2 1 = × 15.0 × 10−3 × 22.4 × (7.05)2 = 8.35 W 2 Q2. A car, travelling with a speed of 30.0 m/s, follows a police car travelling at 50.0 m/s. The siren of the police car emits sound with a frequency of 500 Hz. What frequency is heard by the driver of the car? The speed of sound is 340 m/s. A) B) C) D) E) 474 Hz 527 Hz 397 Hz 534 Hz 468 Hz Ans: Speed of car = x ; speed of police car = y f ′ = f0 . v + v0 v+x 340 + 30 = f0 . = (500). = 474 Hz v + vs v+y 340 + 50 Q3. For an ideal gas, which of the following statements is CORRECT? A) B) C) D) E) Ans: A In an isothermal expansion: W = 𝑄. 𝛥𝐸𝑖𝑛𝑡 = 0 for isothermal In an isothermal expansion: W =ΔEint. In an isothermal expansion: Q =ΔEint. In an isochoric process: W =ΔEint. W = 0 for isochoric In an isochoric process: W = Q. Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 2 Q4. Two moles of an ideal gas are in a 6.0 L container at a pressure of 5.0 × 105 Pa. Find the average translational kinetic energy of a single molecule. A) B) C) D) E) 3.7 × 10-21 J 1.2 × 10-21 J 7.5 × 10-21 J 1.9 × 10-21 J 9.3 × 10-21 J Ans: 3 3k pV kT = . 2 2 nR K avg = 3 5.0 × 105 × 6.0 × 10−3 −23 = × 1.38 × 10 × = 3.7 × 10−21 J 2 2 × 8.31 Q5. Five moles of an ideal monatomic gas are allowed to expand isobarically from 40.0 cm3 to 100 cm3. What is the change of entropy of the gas in this process? A) B) C) D) E) + 95.2 J/K + 57.1 J/K – 95.2 J/K – 57.1 J/K + 19.1 J/K Ans: ΔS = ∫ nCp dT dQ Tf Vf =∫ = n. Cp . ln ( ) = n. Cp . ln ( ) T T Ti Vi 5 100 = 5 × × 8.31 × ln ( ) = +95.2 J/K 2 40.0 King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 3 Q6. Two positive point charges are fixed on the x-axis as follows: q1 = 12 µC at the origin, and q2 = 3.0 µC at x = + 3.0 m. At what coordinate, on the x-axis, is the electric field due to q1 and q2 zero? A) B) C) D) E) + 2.0 m – 1.5 m + 3.0 m + 1.0 m – 1.0 m Ans: The requested point is between the two charges, a distance x from q1. magnitude: E1 = E2 kq1 kq 2 = (3 − x)2 x2 3−x 2 q2 3 1 ( ) = = = x q1 12 4 3−x 1 3 1 = ⇒ − 1= x 2 x 2 3 3 ⇒ = ⇒ x = +2.0 m x 2 Q7. In Figure 1, what is the ratio of the electric flux that penetrates surface S1 to the flux that penetrates surface S2? A) B) C) D) E) 3 0 2 1 1/3 Figure 1 Ans: Φ1 = q enc 3q + 2q + 4q 9q = = ε0 ε0 ε0 Φ2 = q enc 3q − 2q + 2q 3q = = ε0 ε0 ε0 Φ1 9q = =3 Φ2 3q King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 4 Q8. Three charges are arranged as shown in Figure 2. Calculate the work required by an external agent to remove the negative charge (– e) on the right end to infinity. A) B) C) D) E) 3.8 × 10-19 J 7.7 × 10-19 J 1.5 × 10-18 J 1.9 × 10-19 J 1.3 × 10-19 J Figure 2 Ans: ke2 ke2 ke2 2ke2 ke2 3ke2 Ui = − + − = − + = − a 2a a a 2a 2a Uf = − Wext ke2 a ke2 3ke2 ke2 = ∆U = Uf − Ui = − + = = + 3.8 × 10−19 J a 2a 2a Q9. A particle (m = 2.0 × 10-9 kg, q = – 5.0 µC) has a speed of 30 m/s at point A and moves, under the effect of an electric field opposite to its velocity, to point B where its speed becomes 80 m/s. What is the potential difference VB – VA? A) B) C) D) E) + 1.1 V + 3.5 V + 6.3 V – 2.4 V zero Ans: Ui + K i = Uf + K f 1 1 qVA + mvA 2 = qVB + mvB 2 2 2 m (v 2 − vB 2 ) ⇒ VB − VA = 2q A B A ⃗ E ⃗V a⃗ 2.0 × 10−9 VB − VA = . (900 − 6400) −1.0 × 10−5 = +1.1 V King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 5 Q10. The space between the plates of an isolated parallel-plate capacitor is filled with plastic of dielectric constant κ1 = 2.10. The potential difference between the plates is 600 V. The plastic is replaced with glass whose dielectric constant is κ2 = 3.40. What is the potential difference between the plates after inserting the glass? A) B) C) D) E) 371 V 247 V 325 V 199 V 462 V Ans: Qi = C1 . V1 = (k1 . C0 )(600) = 1260 C0 Qf = C2 . V2 = (k1 . C0 )(Vf ) = 3.40 C0 Vf isolated: Qi = Qf ⇒ 1260C0 = 3.40C0 Vf ⇒ Vf = 371 V Q11. Two resistors, A and B, are made of the same material of cylindrical wires having the same length. When a potential difference of 110 V is applied to each wire, the powers dissipated in A and B are 400 W and 100 W, respectively. Ignoring the variation of resistance with temperature, the ratio of the diameter of A to that of B is: A) B) C) D) E) 2 1 1/2 1/4 4 Ans: R= ρL ρL ρL 4ρL = = = 2 2 A πr π(D/2) πD2 P= V2 V2 ⇒R= R P 4ρL V 2 4ρLP = ⇒ D2 = 2 πD P πV 2 ⇒ D = K. √P (K = constant) ⇒ DA PA 400 =√ = √ =2 DB PB 100 King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 6 Q12. A light bulb has a tungsten filament with a resistance of 18.0 Ω at 20.0 oC. It is connected to a 120 V source. When operational, it dissipates a power of 100 W. What is the operational temperature of the filament? Assume that the dimensions of the filament do not change. The temperature coefficient of resistivity of tungsten is 4.50 × 10-3 (oC)-1. A) B) C) D) E) 1850 K 2880 K 1110 K 1280 K 2930 K Ans: (120)2 V2 = = 144 Ω P 100 ∆R 144 − 18 ∆R = α. R 0 . ∆T ⇒ ∆T = = = 1555.56 K α. R 0 4.5 × 10−3 × 18 Rf = Tf = To + ∆T = 293.15 + 1555.56 = 1848.71 → 1850 K Q13. A 45 Ω resistor is connected across the terminals of a 10 V battery. If a current of 0.20 A flows through the 45 Ω resistor, what is the internal resistance of the battery? A) B) C) D) E) 5Ω 3Ω 9Ω 8Ω 7Ω i r R ε Ans: i= ε ε 10 ⇒ r+R= = = 50 Ω r+R i 0.20 r = 50 − R = 50 − 45 = 5 Ω King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 7 Q14. The absolute value of the potential difference between points a and b in the Figure 6 is: Figure 6 A) 14 V B) 21 V C) 41 V D) 11 V E) 19 V Ans: R eq = 2 kΩ εnet = 12 + 12 − 4 = 20 V εnet 20 = = 1 × 10−2 A R eq 2 × 103 Va + 12 − (1 × 103 × 1 × 10−2 ) + 12 = Vb ⇒ Va − Vb = −24 + 10 = −14 V i= |Va − Vb | = 14 V Q15. Three resistors of resistances 2 Ω, 6 Ω and 12 Ω are connected to a battery, as shown in Figure 7. If the total current through the circuit is I = 5 A, what is the current through the 12 Ω resistor? Figure 7 2Ω A) 0.6 A B) 0.3 A C) 0.1 A 6Ω D) 0.2 A E) 0.9 A 12 Ω Ix I Ans: 1 1 1 1 6+2+1 9 = + + = = R eq 2 6 12 12 12 Type equation here. Ω V 12 4 ⇒ R eq = = Ω 9 3 5 × 4 20 ε = I. R eq = = V 3 3 Now, consider the loop containing 12 Ω and ε: +ε − 12 Ix = 0 ⇒ Ix = ε 1 20 20 = × = = 0.555 A ⇒ 0.6 A 12 12 3 36 King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 8 Q16. In the multi-loop circuit shown in Figure 8, the current through the 2.0 kΩ resistor is: Figure 8 A) B) C) D) E) 1.2 mA 3.1 mA 1.9 mA 0.7 mA 2.5 mA 𝑖2 6.0 kΩ 3.0 kΩ 𝑖1 A i 2.0 kΩ 6.0 V 6.0 V Ans: 3.0 kΩ Take the big outer loop: + 6.0 − 600i2 + 6000i1 − 6.0 = 0 ⟹ i1 = i2 At junction A: i = i1 + i2 = 2i2 ⟹ i2 = Now, take the left loop: 6 − 6000 i2 − 2000 i = 0 6 − 3000 i − 2000 i = 0 ⇒i= i 2 6 = 1.2 × 10−3 A 5000 Q17. In the circuits shown in Figure 9, all the capacitors have the same capacitance and initial charge, and all the resistors have the same resistance. At time t = 0, switch S is closed such that each capacitor discharges through its resistors network. Rank the time taken for discharging the capacitor to have its initial charge, shortest time first. Figure 9 A) B) C) D) E) B, C, A C, B, A A, C, B C, A, B B, A, C Ans: τ = RC ⟶ shortest τ ⇒ smallest R A: R = 2R R B: R = 2 ∴ B has the shortest time then C then A 3R C: R = 2} King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 9 Q18. A particle (mass = 6.0 ×10-6 kg) moves in the xy plane with a speed of km/s in a direction that makes an angle of 37o above the positive x axis. the instant it enters a magnetic field of (5.0 î ) mT, it experiences acceleration of (8.0 k̂ ) m/s2. What is the charge of the particle? Ignore gravitational force on the particle. A) B) C) D) E) 4.0 At an the – 4.0 µC – 4.8 µC + 4.0 µC + 4.8 µC – 5.0 µC Ans: ⃗F = ma⃗ = 6.0 × 10−6 × 8.0 k̂ = 48 × 10−6 k̂ (N) ⃗ = −(4.0 × 103 × 5.0 × 10−3 × sin37°)k̂ = −12 k̂ (N) ⃗v × B ⃗ = q (v ⃗ ): thus q = (48× 10−6 ) / (– 12) = – 4.0 C F ⃗ × B Q19. A region of space contains perpendicular uniform electric and magnetic fields, with the electric field pointing in the positive y direction. In this region, a proton travels in the positive x direction with constant velocity. What is the direction of the magnetic field? A) B) C) D) E) + z direction – z direction + y direction – y direction – x direction ⃗e F + ⃗ v Ans: A ⃗FB Q20. A proton moves around a circular path (radius = 2.0 mm) in a uniform 0.25 T magnetic field. What total distance does this proton travel during a 1.0-s time interval? A) B) C) D) E) 48 km 82 km 71 km 59 km 7.5 km Ans: mv 2 qBR ⇒v= R m qBRt 1.6 × 10−19 × 0.25 × 2.0 × 10−3 × 1.0 S = v. t = = = 48 km m 1.67 × 10−2 qvB = King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 10 Q21. A straight 20-cm wire is bent at its midpoint so as to form an angle of 90o, as shown in Figure 10, and carries a current of 10 A. It lies in the xy plane in a region where the magnetic field is in the positive z direction and has a constant magnitude of 3.0 mT. What is the magnitude of the total magnetic force on this wire? Figure 10 A) B) C) D) E) 4.2 × 10-3 N 3.2 × 10-3 N 5.3 × 10-3 N 2.1 × 10-3 N 6.0 × 10-3 N Ans: Portion on the x − axis: ⃗F1 = (iL1 B)ĵ = 10 × 0.1 × 3.0 × 10−3 ĵ = 3.0 × 10−3 ĵ (N) Portion on the y − axis: ⃗F2 = (iL2 B)î = 3.0 × 10−3 î (N) ⃗Fnet = ⃗F1 + ⃗F2 = (î + ĵ) × 3.0 × 10−3 (N) Fnet = √1 + 1 × 3.0 × 10−3 = 4.2 × 10−3 N Q22. A loop of radius r = 5.00 cm, carrying a current I = 0.200 A, is placed inside a magnetic field B = 0.300 ˆi (T). The normal to the loop is parallel to a unit vector n̂ 0.600 ˆi 0.800ˆj . Calculate the magnitude of the torque on the loop. A) B) C) D) E) 3.78 × 10-4 N.m 1.13 × 10-4 N.m 0.600 × 10-4 N.m 4.72 × 10-4 N.m Zero Ans: ⃗ = N iAn̂ = N iπr 2 n̂ = 1 × 0.200 × π × 25 × 10−4 × (−0.6î − 0.80ĵ) μ = −9.42 × 10−4 î − 1.26 × 10−3 ĵ (A. m2 ) τ⃗ = μ ⃗ × ⃗B = +3.78 × 10−4 k̂ (N. m) King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 11 Q23. Figure 11 shows three circuits consisting of straight radial lengths and concentric circular arcs (either half- or quarter-circles of radii r or 2r). The circuits carry the same current in the indicated direction. Rank the circuits according to the magnitude of the magnetic field produced at the center of curvature (the dot), greatest first. Figure 11 A) B) C) D) E) B2, B1, B3 B1, B2, B3 B3, B2, B1 B2, B3, B1 B1, B3, B2 i i i 1 2 3 Ans: 1: Contribution from the big circle only 2: Contribution from both circles 3. Inner circle cancels some of the outer Q24. A wire is bent as shown in FIGURE 12, and carries a current i = 15 mA along the indicated directions. What is the magnitude of the magnetic field at the center of the loop, C, if the radius of the loop is R = 5.0 cm? Figure 12 A) B) C) D) E) Ans: 1.3 × 10−7 T 3.3 × 10−6 T 1.1 × 10−8 T 3.1 × 10−7 T 5.3 × 10−8 T 2 μ0 i 4π × 10−7 × 15 × 10−3 Straight: B1 = = = 6.0 × 10−8 T 2πR 2π × 5.0 × 10−2 Out of page −7 −3 μ0 iϕ 4π × 10 × 15 × 10 × 2π Circle: B2 = = = 18.8 × 10−8 T 4πR 4π × 5.0 × 10−2 Into page Bc = B2 − B1 = 12.8 × 10 −8 T → into the page King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 12 Q25. Three long wires have currents flowing perpendicular to the page with directions as indicated in FIGURE 13. Wire 1 is at y = 2.0 m on the y-axis, wire 2 is located at the origin, and wire 3 is at x = 2.0 m on the x-axis. If I1 = 1.0 A, I2 = 2.0 A, and I3 = 3.0 A, what is the magnitude of the net force per unit length on wire 2 due to the other two wires? Figure 13 A) B) C) D) E) −7 6.3 × 10 N/m 5.3 × 10−5 N/m 5.3 × 10−9 N/m 3.6 × 10−7 N/m 1.3 × 10−7 N/m f21 f23 Ans: f = force per unit lenght f21 μ0 I1 I2 4π × 10−7 × 1.0 × 2.0 N = = = 2.0 × 10−7 2πd 2π × 2.0 m μ0 I2 I3 4π × 10−7 × 2.0 × 3.0 N f23 = = = 6.0 × 10−7 2πd 2π × 2.0 m N f2,net = √4.0 + 36 × 10−7 = 6.3 × 10−7 m Q26. Each of the four wires in Figure 14 carries a 2.0 A current into or out of the page. What is the value of the line integral B.d s for the indicated path of integration? Figure 14 A) B) C) D) E) −2.5 × 10−6 T.m +2.5 × 10−6 T.m −1.5 × 10−7 T.m +1.5 × 10−7 T.m +5.5 × 10−6 T.m Ans: ienc = +2.0 − 2.0 − 2.0 = −2.0 A ⃗ . ds = μ0 . ienc = −2.5 × 10−6 (T. m) ∮B King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 13 Q27. A solenoid with N turns carries a current of 12 A and has a length of 43 cm. If the magnitude of the magnetic field generated at the center of the solenoid is 90 mT, what is the value of N? A) B) C) D) E) 2.6 × 103 1.9 × 103 2.2 × 103 1.1 × 103 1.7 × 103 Ans: B = μ0 ni = μ0 Ni L BL 90 × 10−3 × 43 × 10−2 = = 2.6 × 103 μ0 i 4π × 10−7 × 12 ⇒N= Q28. A flexible loop has a radius of 12.0 cm and is placed with its plane perpendicular to a uniform magnetic field of magnitude 0.150 T, as shown in Figure 3. The loop is grasped at points A and B and stretched at constant rate until its area is zero. If it takes 0.300 s to stretch the loop, what is the magnitude of the induced emf during this time interval? Figure 3 A) B) C) D) E) 22.6 mV 6.31 mV 2.76 mV 63.1 mV 16.3 mV Ans: εind = dϕB d dA ∆A (BA) = B = = B. dt dt dt ∆t (B)(πr 2 ) 0.150 × π × 144 × 10−4 = = = 22.6 mV ∆t 0.300 King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0 Phys102 Coordinator: xyz Final-123 Monday, July 29, 2013 Zero Version Page: 14 Q29. A conducting loop is placed near a long straight wire carrying a constant current I, as shown in Figure 4. Which of the following actions will induce a counterclockwise current in the loop? Figure 4 A) Moving it up away from the wire B) Moving it toward the wire clockwise C) Moving it to the right (parallel to the wire) No change D) Moving it to the left (parallel to the wire) No change E) Rotating it through an axis that coincides with the wire No change Ans: A Q30. Figure 5 shows a metallic bar being moved to the right on two conducting parallel rails in a uniform magnetic field of magnitude 2.5 T, directed into the page. If R = 6.0 Ω, and L = 1.2 m, what is the magnitude of the applied force required to move the bar at a constant speed of 2.0 m/s? Figure 5 A) B) C) D) E) 3.0 N 5.0 N 2.0 N 4.0 N 6.0 N Ans: εind = BLv = 2.5 × 1.2 × 2.0 = 6.0 V i= εind 6.0 = = 1.0 A R 6.0 F = iLB = 1.0 × 1.2 × 2.5 = 3.0 N King Fahd University of Petroleum and Minerals Physics Department c-20-n-30-s-0-e-1-fg-1-fo-0