Electromagnetism - RLC Circuit

A series RLC circuit is used in a radio to tune an FM station broadcasting at 103.7 MHz. The resistance in the circuit is 10 ohms and the inductance is 2.0 microhenries. What is the best estimate of the capacitance that should be used?

A. 200 pF
B. 50 pF
C. 1 pF
D. 0.2 pF
E. 0.02 pF

(GR9677 #26)

Solution:

X_L = X_C
ωL = 1 / ωC
C = 1 / ω²L 

Given:
f  = 103.7 × 10⁶ Hz
ω = 2πf  = 2π × 103.7 × 10⁶ ≈ 2π × 10² × 10⁶ = 2π × 10⁸
ω² = (2π × 10⁸)² ≈ 4π² × 10¹⁶ 
L  = 2.0 microhenries = 2 × 10⁻⁶ H

C = 1 / (4π² × 10¹⁶ × 2 × 10⁻⁶) 

= 1 / (8π² × 10¹⁰)

Take π² ≈ 10

C  ≈ 1/8 × 10⁻¹¹ 
= 1.25 × 10⁻¹² 
= 1.25 pF

Answer: C

Nuclear & Particle Physics - Radioactive

A radioactive nucleus decays, with the activity shown in the graph above. The half-life of the nucleus is

A. 2 min
B. 7 min
C. 11 min
D. 18 min
E. 23 min

(GR9277 #26)

Solution:

From the graph: at t = 0, N ≈ 6 × 10³

Half life, N_½ = 3 × 10³ → t = 7

Answer: B

Nuclear & Particle Physics - X-rays

A nickel target (Z = 28) is bombarded with fast electron. The minimum electron kinetic energy needed to produce X-rays in the K-series is most nearly

A. 10 eV
B. 100 eV
C. 1000 eV
D. 10,000 eV
E. 100,000 eV

(GR8677 #26)

Solution:
Moseley’s law:

K-series refers to a transition from some outer state, n_i to the inner-most shell, n_f = 1.
(The order from inner to outer → K, L, M, N).

As the electrons come in from n_i = ∞,


Answer: D

Classical Mechanics - Rotational Motion

A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point, as shown in the figure, and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?

A. 
B. 
C. 
D. 
E. 

(GR0177 #26)

Solution:


 


Conservation of energy of the system before and after the rod falls onto the ground:
Mgy = ½Iω²

y = L/2 (center of the mass of the rod is at the center of the rod)
Moment inertia of uniform rod: I = ¹/₃ML² 
v = rω 
v_f = ωL → ω = v_f /L 

Mg(L/2) = ½(¹/₃ML²)(v_f /L)²
gL = ¹/₃ v_f ²
v_f  = (3gL)^1/2 

Answer: C