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### Phy102&05-s-ms2

Course: PHY 101, Spring 2007
School: Kuwait University
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Word Count: 755

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University Physics Kuwait Department Physics 102 Second Summer Midterm Examination July 24, 2005 Time: 5:00 pm 6:45 pm NameStudent No. Instructors: Bhatia, Rakshani, Farhan, Lajko, Marafi, Razee &amp; Sharma Fundamental constants k= 1 = 9.0 10 9 N m 2 / C 2 4 o (Coulomb constant) (Permittivity of free space) (Elementary unit of charge) (Avogadros number) (Acceleration due to gravity) (Conversion from...

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University Physics Kuwait Department Physics 102 Second Summer Midterm Examination July 24, 2005 Time: 5:00 pm 6:45 pm NameStudent No. Instructors: Bhatia, Rakshani, Farhan, Lajko, Marafi, Razee & Sharma Fundamental constants k= 1 = 9.0 10 9 N m 2 / C 2 4 o (Coulomb constant) (Permittivity of free space) (Elementary unit of charge) (Avogadros number) (Acceleration due to gravity) (Conversion from eV to J) (Electron mass) (Proton mass) n = 10-9 G = 109 p = 10-12 T = 1012 o = 8.85 10-12 C2 / (Nm2) e = +1.60 10-19 C NA = 6.022 1023 g = 9.8 m/s2 1 eV = 1.6 x 10-19 J me= 9.11 10-31 kg mp = 1.67 10-27 kg Prefixes of units m = 10-3 k = 10 3 = 10-6 M = 1 06 For use by Instructors only Prob. Marks 1 2 3 4 5 6 7 8 9 10 Total 0 1. An isolated 50 nF air-filled parallel plate capacitor has 360 nC charge on its plates. A dielectric material ( = 2.6) is inserted filling half the space between the plates, as shown. The work don by an external agent to insert the dielectric is a. b. c. d. e. 576 nJ. 2.3 J. 648 nJ. 213 nJ. 765 J. C eq = C o + C = 25 + 2.6 25 = 90 nF Ui 12 q, 2C i Uf = 12 q 2C eq 1 1 1 1 1 2 1 W = U = q 2 = ( 360 ) C eq C i 2 2 90 50 W = 576 nJ 2. A coaxial cable has inner and outer radii of 0.1 mm and 0.6 mm respectively. The capacitance per unit length for this cable is a. b. c. d. e. 49 pF/m. 31 pF/m. 2.810-19 nF/m. 9.710-18 nF/m. 1.010-17 nF/m. C = 2 o L b ln a 1 2 8.85 1012 C = 2 o = = 31pF 6 b L ln ln 1 a 3. For the circuit shown, the charge on the plates of C2 is 12.4 C. If C1=C2=C3=16 F and C4 = 36 F, the charge on the plates of C4 is a. b. c. d. e. 20.6 C. 12.4 C. 18.6 C. 24.8 C. 37.2 C. 1 q1 = q 2 = 12.4C C 12 = 16 16 = 8 F 32 q C q1 16 =3 q 3 = q1 3 = 12.4 = 24.8 C C 12 C 3 C 12 8 q 4 = q1 + q 3 = 12.4 + 24.8 = 37.2 C 4. Two cables of the same conduction material and the same length are connected to an EMF source shown. as If the diameter of conductor (2) is twice that of conductor (1), and the drift speed of the electrons in (1) is 4x10-7 m/s, the drift speed in conductor (2) is a. b. c. d. e. 410-7 m/s. 210-7 m/s. 810-7 m/s. 1610-7 m/s. 110-7 m/s. I1 = I 2 J 1A1 = J 2 A 2 nev d 1 a 2 = nev d 2 b 2 // / // / vd2 a2 1 = v d 1 2 = v d 1 = 1107 m/s b 4 5. Compared to the situation when switch (S) is open, the current in the EMF source when (S) is closed is three times greater. The value of the resistance R is a. b. c. d. e. 1 7 . 4 0 . 2 0 . 1 5 . 1 0 . 3I open = I closed 3 V 3 V = ( 20 + R ) 20 20R R = 10 2 6. For the circuit shown (r = 0.2 , R = 5 , 1 = 20V and 2 = 6V), the terminal voltage of 2 source is a. b. c. d. e. 9.2 V. 6.5 V. 5.5 V. 6 V. 2.6 V. R +r +r V 2 = 2 + Ir = 6.5 V I= 1 2 = 2.6 A 7. For the circuit shown (1=9 V, 2=12 V, R1= 15 , R2= 20 and R3= 40 ), the current passing through R3 is a. b. c. d. e. 0.35 A. 0.65 A. 0.38 A. 0.26 A. 0.2 A. 9 15I 1 + 20I 2 = 0 12 20I 2 40I 3 = 0 I 3 = I1 + I 2 SOLVING FOR I 3 I 3 = 0.35 A 8. For the circuit shown, the charge on the plates of the capacitor in the steady state situation is a. b. c. d. e. 3.3 C. 2.5 C. 5 C. 8.2 C. 9.6 C. 3 STEADY STATE: 10 1 = A, R 60 6 q = 3.3C I= = VC = 40 q q == 6 C 0.5 9. An electron moves enters a region, where a uniform magnetic field ( B = 0.8i mT ) exists, with a velocity given v = ( 4i 6 ) 104 m/s . The pitch of the resulting j helical path is r r a. b. c. d. e. 2.7 mm. 1.8 mm. 3.2 mm. 5.3 mm. 9.1 mm. 2 m 2 4 10 4 9.1 10 31 p = v xT = v x = qB 1.6 1019 0.8 10 3 p = 18 mm 10. A 15 A current passes through the wire shown. If the magnitude of the external magnetic field in the region is 20 mT, the magnitude of the force on the wire is a. b. c. d. e. 4.6 N. 1.2 N. 3.9 N. 8.3 N. 1.7 N. F = ILB sin L =4 2 F = 15 4 2 20sin / 2 = 1.7 N 4
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Kuwait University - PHY - 101
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Kuwait University - PHY - 101
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Kuwait University - PHY - 101
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Kuwait University - PHY - 101
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Kuwait University - PHY - 101
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