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

Quantum Mechanics - Laser

The approximate number of photons in a femtosecond (10⁻¹⁵ s) pulse of 600 nanometers wavelength light from a 10-kilowatt peak-power dye laser is

A. 10³
B. 10⁷
C. 10¹¹
D. 10¹⁵
E. 10¹⁸

(GR9677 #59)

Solution:

Number of photon, n = E_total / E_single

Energy of a single photon,  E_single = hc/λ 

h = 6.63 × 10⁻³⁴ Js
c = 3 × 10⁸ m/s
λ  =  600 nanometers = 600 × 10⁻⁹ m

→ E_single = 3 × 10⁻¹⁹  J

From Power, P = W/t = E_total/t

→ Total energy, E_total =  Pt

P = 10 kW = 10⁴ W
t = 10⁻¹⁵ s

→ E_total = 10⁻¹¹ J

n = E_total / E_single = 10⁻¹¹  / (3 × 10⁻¹⁹)  = 0.3 × 10⁸ = 3 × 10⁷  J

Answer: B

Electromagnetism - EM Radiation

A charged particle, A, moving at a speed much less than c, decelerates uniformly. A second particle, B, has one-half the mass, twice the charge, three times the velocity, and four times the acceleration of particle A. According to classical electrodynamics, the ratio P_B/P_A of the powers radiated is

A. 16
B. 32
C. 48
D. 64
E. 72

(GR9677 #70)

Solution:

Larmor formula:
Power radiated by an accelerating non-relativistic point charge particle,
P = q²a² ⁄ 6πɛ₀c³

P does not depend on m and v.

P_B/P_A =  (2q)² (4a)² ⁄q²a²  =  64

Answer: D

Electromagnetism - EM Radiation

An electron oscillates back and forth along the + and − x-axes, consequently emitting electromagnetic radiation. Which of the following statements concerning the radiation is NOT true?

1. The total rate of radiation of energy into all directions is proportional to the square of the electron's acceleration.
2. The total rate of radiation of energy into all directions is proportional to the square of the electron's charge.
3. Far from the electron, the rate at which radiated energy crosses a perpendicular unit area decreases as the inverse square of the distance from the electron
4. Far from the electron, the rate at which radiated energy crosses a perpendicular unit area is a maximum when the unit area is located on the + or − x-axes
5. Far from the electron, the radiated energy is carried equally by the transverse electric and the transverse magnetic fields.

(GR9277 #84)

Solution:

Power radiated, P = ∮ S ∙ dA
S is poynting vector, where S = (1/μ₀) E × B  = (1/cμ₀) |E|²
Electric field, E = − (kq/r)(a_⊥/c²)

(A) TRUE
P ∝ a²  since |E|² ∝  a_⊥²

(B) TRUE,
P ∝ q² since |E|² ∝ q²

(C) TRUE
P ∝ 1/r² since |E|² ∝ 1/r²

(D) FALSE
The electron oscillates back and forth along the + and − x-axes. Thus, the direction of acceleration is along the + and − x axes. P_max is when the area of electric flux perpendicular, ⊥ to the acceleration of the particle (not located on the + or − x-axes).
P_max ∝ a_⊥² since |E|² ∝  a_⊥²

(E) TRUE, 
Power radiated, P = ∮ S ∙ dA
S is poynting vector, where S = (1/μ₀) E × B
P is carried equally by E and B field.

Answer: D

Thermal Physics - Blackbody Radiation

A blackbody at temperature T₁ radiates energy at a power level of 10 milliwatts (mW). The same blackbody, when at a temperature 2T₁, radiates energy at a power level of 

A. 10 mW
B. 20 mW
C. 40 mW
D. 80 mW
E. 160 mW

(GR8677 #46)

Solution:

Blackbody radiation (Stefan-Boltzmann's law): u = σT⁴

Constant σ = u / T⁴

u₁T₁⁴ = u₂T₂⁴
10 / T₁⁴ = u₂ / (2T₁)⁴
u₂ = (10)(2T₁)⁴ / T₁⁴ = (10)(16) =160

Answer: E

Electromagnetism - EM Radiation

A charge particle oscillates harmonically along the x-axis as shown above. The radiation from the particle is detected at a distant point P which lies in the xy-plane. The electric field at P is in the

A. ±z direction and has a maximum amplitude at θ = 90^o
B. ±z direction and has a minimum amplitude at θ = 90^o
C. xy-plane and has a maximum amplitude at θ = 90^o
D. xy-plane and has a minimum amplitude at θ = 90^o
E. xy-plane and has a maximum amplitude at θ = 45^o

(GR8677 #53)

Solution:

Since point P lies in the xy-plane, the electric field at P also lies in the xy-plane, and its maximum amplitude is at θ = 90^o (perpendicular to the acceleration of particle).

Answer: C

Electromagnetism - Impedance

An alternating current electrical generator has a fixed internal impedance R_g + jX_g and is used to supply power to a passive load that has an impedance R_g + jX₁, where j =√(−1). R_g ≠ 0 and X_g ≠ 0. For a maximum power transfer between the generator and the load, X₁ should be equal to

A. 0
B. X_g
C. −X_g
D. R_g
E. −R_g

(GR8677 #64)

Solution:

Maximum power transfer theorem (Impedance matching): 

Maximum power is transferred from a source to a load when

• Load resistance = the internal resistance of the source, or 
• Load impedance = complex conjugate source impedance (Z_l = Z_s^*)

Z_s = R_g + jX_g  with  j =√(−1)

Z_s^* = R_g − jX_g

For max power transferred: 
Z_l  = Z_s^*
X₁ = −X_g

Answer: C

Optics - Polarization

Unpolarized light is incident on two ideal polarizers in series. The polarizers are oriented so that no light emerges through the second polarizer. A third polarizer is now inserted between the first two and its orientation direction is continuously rotated through 180^o. The maximum fraction of the incident power transmitted through all three polarizers is

A. Zero
B. 1/8
C. 1/2
D. 1/√2
E. 1

(GR8677 #74)

Solution:

Power ~ Intensity

The maximum fraction of intensity is: I = I₀/8 (see GR0177 #51)

→ P = P₀/8

Answer: B

Nuclear & Particle Physics - Gamma Rays Detector

An 8-cm diameter by 8-cm long NAI(TI) detector detects gamma rays of a specific energy from a point source of radioactivity. When the source is placed just next to the detector at the center of the circular face, 50 percents of all emitted gamma rays at that energy are detected. If the detector is moved to 1 meter away, the fraction of detected gamma rays drops to

A. 10⁻⁴
B. 2 ×10⁻⁴
C. 4 ×10⁻⁴
D. 8 ×10⁻⁴
E. 16 ×10⁻⁴

(GR0177 #14)

Solution:

The net power radiated by the source: P = I A
I = intensity
A = area of the surface

No power loss mentioned: P₁ = P₂
I₁ A₁ = I₂ A₂

A₁ = area of circle with diameter 8 cm
r₁  = 4 × 10⁻² m

A₂ = area of sphere with radius 1 m

The problem asks for the fraction, not how much the intensity is detected.
Thus, the fraction is just the ratio of the areas:



Answer: C

Classical Mechanics - Efficiency

An electric sander has a continuous belt that rubs against a wood surface as shown schematically above. The sander is 100 percent efficient and draws current of 9 amperes from a 120-volt line. The belt speed is 10 meters per second. If the sander is pushing against the wood with a normal force of 100 newtons, the coefficient of friction is most nearly

A. 0.02
B. 0.2
C. 0.4
D. 1.1
E. 10

(GR8677 #93)

Solution:

Efficiency:









Answer: D

Electromagnetism – Gauss' Law

A uniformly charged sphere of total charge Q expands and contract between radii R₁ and R₂  at a frequency f. The total power radiated by the sphere is

A. proportional to Q
B. proportional to f²
C. proportional to f⁴
D. proportional to (R₁ / R₂)
E. zero

(GR0177 #96)

Solution:

Power, P = VI
I = dq / dt
Q is constant → dq = 0 → I = 0  → P = 0

Answer: E