Electromagnetism - Capacitor

The capacitor shown in Figure 1 above is charged by connecting switch S to contact a. If switch S is thrown to contact b at time t = 0, which of the curve in Figure 2 above represents the magnitude of the current through the resistor R as a function of time?

A. A
B. B
C. C
D. D
E. E

(GR9677 #01)

Solution:

The capacitor is charged by connecting switch S to contact a, so the current after connecting switch to contact b (t = 0) must start with I = V/R.

Only plot A and B are right.

The current can not be the same all the time and since the voltage of a capacitor follows an exponential decay:

V(t) = V₀e^−t/RC 
 I(t) = V(t)/R = V₀e^−t/RC/ R

Only plot B is right.

Answer: B

Electromagnetism - Faraday’s law

The circuit shown is in a uniform magnetic field that is into the page and is decreasing in magnitude at rate of 150 tesla/second. The ammeter reads

A. 0.15 A
B. 0.35 A
C. 0.50 A
D. 0.65 A
E. 0.80 A

(GR9677 #02)

Solution:

V − IR − ɛ = 0
I = (V − ɛ)/R

ɛ = − dΦ/dt = −AdB/dt

Given:
dB/dt = −150 t/s (minus sign because it’s decreasing)
A = (0.1 m)² = 0.01 m²
R = 10 Ω
V = 5 V

ɛ = − (0.01)(−150) = 1.5 V
I = (5 − 1.5)/10 = 3.5/10 = 0.35 A

Answer: B

Electromagnetism - Electric Potential

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

  

The electric potential at a point P, which is located on the axis of symmetry a distance x from the center of the ring, is given by

A. Q / (4πɛ₀x)
B. Q / [4πɛ₀(R² + x²)^1/2]
C. Qx / [4πɛ₀(R² + x²)]
D. Qx / [4πɛ₀(R² + x²)^3/2]
E. QR / [4πɛ₀(R² + x²)]

(GR9677 #03)

Solution:

Electric Potential, V = kQ/r 
with k = 1/4πɛ₀

The distance r of P from the charged ring is r² = R² + x²

V = Q / [4πɛ₀(R² + x²)^1/2]

Answer: B

Electromagnetism - Oscillation

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

A small particle of mass m and charge –q is placed at point P and released. If R ≫ x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to:

(GR9677 #04)

Solution:

F_electric = kqQ/r²
F_centripetal = mv²/r = mω²r

F_e = F_c
kqQ/r² = mω²r
ω² = kqQ/mr³

with
k = 1/4πɛ₀
r²  = R² + x²
R ≫ x
r² ∼ R²  → r³ ∼ R³

ω = √(^qQ/_4πɛ0mR³)

Answer: A

Notes: see problem GR9277 #65

Classical Mechanics - Circular Motion

A car travels with constant speed on a circular road on level ground. In the diagram above, F_air is the force of air resistance on the car. Which of the other force shown best represents the horizontal force of the road on the car's tires?

A. F_A 
B. F_B
C. F_C
D. F_D
E. F_E

(GR9677 #05)

Solution:

F_A = Centripetal force
F_C = Frictional force of the road, equals to force exerted by tires exert in the backward direction so that the car moves in the forward direction (Newton's third law Action-Reaction).

The horizontal force on the car’s tires, F_A + F_C = F_B

Answer: B

Classical Mechanics - Friction Force

A block of mass m sliding down an incline at constant speed is initially at height h above the ground as shown in the figure. The coefficient of kinetic friction between the mass and the incline is µ. If the mass continues to side down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reached the bottom of the incline?

A. mgh/µ
B. mgh
C. µmgh/sin θ 
D. mgh sin θ 
E. 0

(GR9677 #06)

Solution:

mg sin θ − F_r  = ma
At a constant speed, a = 0
F_r  = mg sin θ 

Energy dissipated = Work done by the frictional force
W = F_r ⋅ s

s = length of the inclined surface = h/sin θ

W = mg sin θ (h/sin θ) = mgh

Answer: B

Classical Mechanics - Linear Momentum

As shown in the picture, a ball of mass m suspended on the end of a wire, is released from height h and collides elastically, when it is at its lowest point, with a block of mass 2m at rest on a frictionless surface. After the collision, the ball rises to a final height equal to

A. 1/9 h
B. 1/8 h
C. 1/3 h
D. 1/2 h
E. 2/3 h

(GR9677 #07)

Solution:

Conservation of momentum of the system:

m_av_a + m_bv_b = m_av_a' + m_bv_b' 

Given:
m_a = m
m_b = 2m
v_b = 0

mv_a + 0 = mv_a' + 2mv_b'  
v_a  = v_a' + 2v_b'    (Eq.1)

Conservation of kinetic energy of the system:

½ m_av_a² + ½ m_bv_b² = ½ m_av_a'² + ½ m_bv_b'²
mv_a² + 0 = mv_a'² + 2mv_b'²
v_a² = v_a'² + 2v_b'²  (Eq.2)

(Eq.1) → (Eq.2)
(v_a' + 2v_b')² = v_a'² + 2v_b'²
v_a'² + 4v_a'v_b' + 4v_b'² = v_a'² + 2v_b'²
4v_a'v_b'  = 2v_b'² − 4v_b'²
2v_a' = − v_b'  
v_b'  = −2v_a'  (Eq.3)

(Eq.3) →  (Eq.1)
v_a  = v_a' + 2v_b' 
v_a  = v_a' + 2(−2v_a')
v_a  = − 3v_a'  (Eq. 4)

For the pendulum, conservation of energy:

at the moment of the collision
U = T → m_agh = ½ m_av_a² → v_a = (2gh)^½

after the collision
U'  = T' → m_agh' = ½ m_av_a'² → v_a' = (2gh')^½

Thus, (Eq. 4):
v_a  = − 3v_a'  
(2gh)^½  = − 3(2gh')^½ 
[(2gh)^½]² = [− 3(2gh')^½]²

h  = 9h'
h' = ¹⁄₉h 

Answer: A

Classical Mechanics - Harmonic Oscillator

A particle of mass m undergoes harmonic oscillation with period T₀. A force f proportional to the speed v of the particle, f = −bv, is introduced. If the particle continues to oscillate, the period with f acting is

A. Larger than T₀
B. Smaller than T₀
C. Independent of b
D. Dependent linearly on b
E. Constantly changing

(GR9677 #08)

Solution:

 f = −bv → minus sign means f is restoring force (damped oscillation).
The oscillation is getting slower (larger period) before it finally comes to stop.

Answer: A

Nuclear & Particle Physics - Hydrogen Spectrum

In the spectrum of Hydrogen, what is the ratio of the longest wavelength in the Lyman series (n_f  = 1) to the longest wavelength in the Balmer series (n_f  = 2)?

A. 5/27
B. 1/3
C. 4/9
D. 3/2
E. 3

(GR9677 #09)

Solution:

Rydberg formula:



λ_vac = the wavelength of the light emitted in vacuum
R_H = Rydberg constant for Hydrogen
n₁ and n₂ are integers such that n₁  n₂

For the longest wavelength take n₂ = ∞

For Lyman-radiation (n₂ = ∞ → n₁ = 1):

1/λ_L = R_H (1/1 − 0) = R_H

For Balmer-radiation (n₂ = ∞ → n₁ = 2):

1/λ_B = R_H (¼ − 0) = ¼R_H

The Ratio:

^λ_L/_λB = (¹/_RH)(^R_H/₄) = ¼ ≈ 5/27

Answer: A

Nuclear & Particle Physics - Internal Conversion

Internal conversion is the process whereby an excited nucleus transfers its energy directly to one of the most tightly bound atomic electrons, causing the electron to be ejected from the atom and leaving the atom in an excited state. The most probable process after an internal conversion electron is ejected from an atom with a high atomic number is that the

1. atom returns to its ground state through inelastic collisions with either atoms
2. atom emits one or several X-rays
3. nucleus emits a gamma-ray
4. nucleus emits an electron
5. nucleus emits a positron

(GR9677 #10)

Solution:

Electron transitions in atom (internal conversion) = X-ray production
→ An orbital electron is absorbed and ejected along with an X-ray

compared to:

Nuclear transitions = Gamma, γ Ray production

→ The excited nucleus jumps to a lower level and emits a photon γ

Answer: B

Note: 

1. (C), (D), and (E) are products of radioactive decay which are results of unstable nuclei.

2. In internal conversion:

• For low atomic number, it will produce the Auger effect and ionize the outside electron.
• For high atomic number, it will only emit X-rays.

Nuclear & Particle Physics - Stern-Gerlach Experiment

A beam of neutral hydrogen atoms in their ground state is moving into the plane of this page and passes through a region of a strong inhomogeneous magnetic field that is directed upward in the plane of the page. After the beam passes through this field, a detector would find that it has been

A. deflected upward
B. deflected to the right
C. undeviated
D. split vertically into two beams
E. split horizontally into three beams

(GR9677 #11)

Solution:

The Stern–Gerlach experiment → spin discovery

A beam of neutral atom passes through inhomogeneous magnetic field will split vertically into 2 beams representing spin-up and spin-down electrons.

Answer: D

Nuclear & Particle Physics - Positronium

The ground-state energy of positronium is most nearly equal to

A. − 27.2 eV
B. − 13.6 eV
C. − 6.8 eV
D. − 3.4 eV
E. 13.6 eV

(GR9677 #12)

Solution:

Energy levels of Positronium is half those of Hydrogen (See GR8677 #99)

E_n(H)  = − 13.6 / n²
E_n(Ps) = ½ E_n(H) 

For the ground-state n = 1 → E_Ps = − ½ × 13.6 eV = − 6.8 eV

Answer: C

Thermal Physics - Power

A 100-watt electric heater element is placed in a pan containing one liter of water. Although the heating element is on for a long time, the water, though close to boiling does not boil. When the heating element is removed, approximately how long will it take the water to cool by 1 degree Celsius? (Assume that the specific heat for water is 4.2 kJ/kg^oC)

A. 20 s
B. 40 s
C. 60 s
D. 130 s
E. 200 s

(GR9677 #13)

Solution:

P = 100 W
V = 1 L = 1 m³ → m = 1 kg (STP)
ΔT = 1^oC
c = 4.2 kJ/kg^oC

Q = mcΔT = Pt
1 × 4200 × 1 = 100t
t = 42 s

Answer: B

Thermal Physics - Specific Heat

Two identical 1 kg blocks of copper metal one initially at a temperature T₁ = 0^oC and the other initially at a temperature T₂ = 100^oC are enclosed in a perfectly insulating container. The two blocks are initially separated. When the blocks are placed in contact, they come to equilibrium at a final temperature T_f. The amount of heat exchanged between the two blocks in this process is equal to which of the following? (The specific heat of copper metal is equal to 0.1 kilocalorie/kg^oK)

A. 50 Kcal
B. 25 Kcal
C. 10 Kcal
D. 5 Kcal
E. 1 Kcal

(GR9677 #14)

Solution:

Q = mcΔT
|Q|_gain = |Q|_lost 
m₁c₁ΔT₁  = m₂c₂ΔT₂

m₁ = m₂ = 1 kg
c₁ = c₂ = c_copper = 0.1 kcal/kg^oK

T₁ = 0^oC = 273^oK

T₂ = 100^oC = 373^oK

ΔT₁ = ΔT₂
T_f   − T₁ = T₂ − T_f 
T_f  = (T₂ + T₁)/2 = (100 + 0)/2 = 50^oC = 323^oK

|Q|_gain = |Q|_lost  = m₁c₁ΔT₁  = 1 × 0.1 × (323 − 273) = 5kcal

Answer: D

Thermal Physics - Isothermal

Suppose one mole of an ideal gas undergoes the reversible cycle ABCA shown in the P-V diagram above, where AB is an isotherm. The molar heat capacities are C_p at constant pressure and C_v at constant volume. The net heat added to the gas during the cycle is equal to

A. RT_h (V₂/V₁)
B. −C_p(T_h − T_c)
C. C_p(T_h − T_c)
D. RT_h ln (V₂/V₁) − C_p(T_h − T_c)
E. RT_h ln (V₂/V₁) − R(T_h − T_c)

(GR9677 #15)

Solution:

AB Isotherm → T Constant = T_h

Ideal Gas: PV = nRT 
P = nRT / V

n = 1 mole,

W_AB = _V1∫^V₂ P dV = RT_h V1∫^V₂ (1/V) dV  = RT_h ln (V₂/V₁)

BC Isobaric → P constant = P₂

W_BC = _V2∫^V₁ P dV = P₂ (V₁− V₂) =  P₂V₁ −  P₂V₂

From the diagram:

P₂V₁ = nRT_c 

P₂V₂ = nRT_h

W_BC = nR(T_c − T_h)  = R(T_c − T_h)

W_CA = 0  since V constant

Total W = W_AB + W_BC = RT_h ln (V₂/V₁) + R(T_c − T_h)

or

W = RT_h ln (V₂/V₁) − R(T_h − T_c)

Answer: E

Thermal Physics - Mean Free Path

The mean free path for the molecules of a gas is approximately given by 1/ησ, where η is the number density and σ is the collision cross section. The mean free path for air molecules at room conditions is approximately

A. 10⁻⁴ m
B. 10⁻⁷ m
C. 10⁻¹⁰ m
D. 10⁻¹³ m
E. 10⁻¹⁶ m

(GR9677 #16)

Solution:

Mean free path = 1/ησ

Number density, η = N/V  
Cross section area, σ = πr² 

For Ideal Gas: PV = NkT

1/ησ = V / Nπr²  = NkT / PNπr² = kT / Pπr²   

k = 1.38 × 10⁻²³ Joule/K
Radius of atom is in order of Angstrom: 10⁻¹⁰ m
P (STP) = 1 atm = 10⁵ Newton/meter²
T (STP) = 0 °C = 32 °F = 273.15 K ≈ 2 × 10² K

kT / Pπr² = (1.38 × 10⁻²³ × 2 ×10²) / (π ×10⁵ × 10⁻²⁰) 
= (1.38 × 2 / π) 10^{−23+2−5+20} ≈ 10⁻⁷

Answer: B

Quantum Mechanics - Probability

The wave function for a particle constrained to move in one dimension is shown in the graph (Ψ = 0 for x ≤  0 and x ≥ 5). What is the probability that the particle would be found between x = 2 and x = 4?

A. 17/64
B. 25/64
C. 5/8
D. v(5/8)
E. 13/16

(GR9677 #17)

Solution:

Probability, P ∼ ψ²

Probability to find the particle between x = 2 and x = 4 (unnormalized probability):
ψ² = 2²  + 3² = 13

Total probability (normalized probability):
ψ² = 1²  +  1²  + 2²  + 3²  +  1² = 16

P = unnormalized probability / normalized probability = 13/16

Answer: E

Quantum Mechanics - Potential Wall

Consider a potential of the form

V(x) = 0, x ≤ a
V(x) = V_0, a < x < b
V(x) = 0, x ≥ b

As shown in the figure above. Which of the following wave functions is possible for a particle incident from the left with energy E < V₀.

(GR9677 #18)

Solution:

A. Classic not QM potential → FALSE
B. No decrease in amplitude → FALSE
C. Decay exponentially inside the wall, decrease amplitude (fits E < V₀) →  TRUE

D. QM Oscillator harmonics → FALSE
E. Cosine wave function not QM potential → FALSE

Answer: C

Nuclear & Particle Physics - Alpha/Rutherford Scattering

When alpha particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atoms and are scattered at large angles. If an alpha particles with an initial kinetic energy of 5 MeV happens to be scattered through an angle of 180^o, which of the following must have been its distance of the closest approach to the scattering nucleus? (Assume that the metal foil is made of silver, with Z = 50.)

A. 1.22 × 50^1/3 fm
B. 2.9 × 10⁻¹⁴ m
C. 1.0 × 10⁻¹² m
D. 3.0 × 10⁻⁸ m
E. 1.7 × 10⁻⁷ m

(GR9677 #19)

Solution:

"When alpha particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atoms and are scattered at large angles."→ Rutherford Scattering: the discovery of nucleus.

Rutherford estimated the radius of a silver nucleus to be 2 × 10⁻¹⁴ m, by observing the angular dependence of alpha-particle scattering (source).

Answer: B

Calculation:

Conservation of Energy: U = T

U = kq_α q_s / r = T
r =  kq_α q_s / T

Given:
T = 5 MeV = 5 × 10⁶ eV
Z_s = 50 → q_s = Z_αe = 50e
Z_α = 2 → q_α = Z_αe = 2e 

k = 1/4πɛ₀ = 1/(4 × 3.14 × 8.85 × 10⁻¹²)  ≈ 10¹⁰  Nm²/C²

r = (10¹⁰ Nm²/C²× 50e × 2e) / (5 × 10⁶ eV)
= 2 × 10⁵ e Nm²/VC² 

With e = 1.60 × 10⁻¹⁹ C

1 Volt = 1 Nm/C

r = 2 × 10⁵ × 1.60 × 10⁻¹⁹ m ≈ 3 × 10⁻¹⁴ m

Classical Mechanics - Linear Momentum

A helium atom, mass 4u travels with non relativistic speed v normal to the surface of a certain material, makes an elastic collision with an (essentially free) surface atom, and leaves in the opposite direction with speed 0.6v. The atom on the surface must be an atom of

A. Hydrogen, mass 1u
B. Helium, mass 4u
C. Carbon, mass 12u
D. Oxygen, mass 16u
E. Silicon, mass 28u

(GR9677 #20)

Solution:

m_a = 4u
v_a  = v
v_b  = 0
v_a' = − 0.6v

Conservation of momentum of the system:

m_av_a + m_bv_b = m_av_a' + m_bv_b' 
4uv = 4u(− 0.6v) + m_bv_b' 
4uv = − 2.4uv  + m_bv_b' 
m_bv_b' = 6.4uv (Eq.1)

Conservation of kinetic energy of the system:

½ m_av_a² + ½ m_bv_b² = ½ m_av_a'² + ½ m_bv_b'²
4uv² = 4u(− 0.6v)² + m_bv_b'²
4uv² = 4u(0.36v²) + m_bv_b'²
m_bv_b'² =  4uv²  − 1.44uv²
m_bv_b'² =  2.56uv² (Eq.2)

(Eq.1) → (Eq.2)
6.4 (v_b') = 2.56v
v_b' =  (2.56/6.4)v = 0.4v  (Eq.3)

(Eq.3) → (Eq.1)
m_b = 6.4uv / 0.4v = 16u

Answer: D