prac exam 3B - Physics 1112 Spring 2009 University of...

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Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler Solution to PHYS 1112 In-Class Exam #3B Thu. April 9, 2009, 2:00pm-3:15pm Conceptual Problems Problem 1: If a wire of some length L and with a circular cross-section of diameter D has a resistance of 60 kΩ, what will be the resistance of a wire made from the same material, at the same temperature, of length 24 L and diameter 6 D ? (A) 10 kΩ (B) 40 kΩ (C) 90 kΩ (D) 120 kΩ (E) 240 kΩ Answer: (B) Using R = ρL/A and A = π ( D/ 2) 2 , we get R L/D 2 . Increasing L L ± = 24 L and D D ± =6 D thus changes R R ± = 24 × R/ 6 2 = (24 / 36) × R = (2 / 3) × 60Ω = 40Ω. Problem 2: Three diFerent circuits, X , Y and Z , are built with the same three resistors, R 1 > 0, R 2 > 0, and R 3 > 0, and the same battery of battery voltage E , as shown in ±ig. 3.04. Compare and rank the magnitude of the currents I 3 through R 3 , observed in the three diFerent circuits. (Hint: I 3 = V 3 /R 3 .) Fig. 3.04 (X) R 1 I 2 R 3 R 2 I 3 I 1 E R 3 I 2 R 2 R 1 I 1 I 3 E I 3 R 3 R 1 I 1 R 2 I 2 E (Z) (Y) I o I o I o 1
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Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler (A) I 3 ( Z ) >I 3 ( Y ) 3 ( X ) (B) I 3 ( X ) 3 ( Z ) 3 ( Y ) (C) I 3 ( Y ) 3 ( X ) 3 ( Z ) (D) I 3 ( Y ) 3 ( Z ) 3 ( X ) (E) I 3 ( Z ) 3 ( X ) 3 ( Y ) Answer: (A) In circuit Z : R 3 is directly connected to battery, hence I 3 ( Z )= E /R 3 . In circuit Y : R 3 and R 2 in series are connected to battery. Hence I 3 ( Y E / ( R 3 + R 2 ) < I 3 ( Z ), since R 3 + R 2 >R 3 . In circuit X : R 3 , R 2 and R 1 in series are connected to battery. Hence I 3 ( X E / ( R 3 + R 2 + R 1 ) <I 3 ( Y ) 3 ( Z ), since R 3 + R 2 + R 1 3 + R 2 . Problem 3: A molecular ion beam containing four diFerent types of ions, called P , Q , R and S here, enters a uniform magnetic ±eld, as shown in ²ig. 3.15 below, with B ± = 0 above the lower horizontal line, and 8 B perpendicular to, and pointing out of, the plane of the drawing. The incident beam, below the lower horizontal line, is in the plane of the drawing. B (out) Fig. 3.15 P S Q R B=0 The diameters of the semicircular ion trajectories in the 8 B -±eld, denoted by d P , d Q , d R and d S , respectively are observed to be in a ratio of d P : d Q : d R : d S = 1 : 2 : 3 : 4 , as indicated in ²ig. 3.15. Assume all four ion types carry the same amount of charge per 2
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Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler ion, | q | , and they all enter the 8 B -Feld with the same speed v . What is the ratio of the four ion masses, denoted by m P , m Q , m R and m S , respectively ? (A) m P : m Q : m R : m S = 1 : 2 : 3 : 4 . (B) m P : m Q : m R : m S = 1: 2: 3: 4 . (C) m P : m Q : m R : m S = 1 1 : 1 2 : 1 3 : 1 4 . (D) m P : m Q : m R : m S = 1 1 : 1 2 : 1 3 : 1 4 . (E) m P : m Q : m R : m S = 1 : 4 : 9 : 16 . Answer: (A) ±rom d =2 r mv/ ( | q | B ), we get d m for Fxed v , B and | q | . So, the masses m must be in the same ratios as diameters d if all ions have the same v and | q | and travel in the same B . Problem 4: Two very thin circular rings, labeled 1 and 2 in the following, with radii R 1 and R 2 , respectively, and R 1 >R 2 , are both centered at the coordinate origin O (0 , 0 , 0), as shown in ±ig. 3.23.
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This note was uploaded on 04/16/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at University of Georgia Athens.

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prac exam 3B - Physics 1112 Spring 2009 University of...

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