Physics Problems & Solutions: #62

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Electromagnetism - Capacitor

Two capacitors of capacitance 1.0 microfarad and 2.0 microfarad are each charged by being connected across a 5.0-Volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0 microfarad capacitor?

A. 0 V
B. 0.6 V
C. 1.7 V
D. 3.3 V
E. 5.0 V

(GR9677 #62)

Solution:

The charges stored at each capacitor:

Q₁ = C₁× V_t = 1 × 5 = 5 μF
Q₂ = C₂× V_t = 2 × 5 = μF

The plates of opposite charge are connected together (series connection):

Q₁' = Q₂' = Q_t
1/C_t = 1/C₁ + 1/C₂= 1 + 1/2 = 3/2
C_t= 2/3
Since V_t = 5
Q₂' = Q_t= C_t× V_t = 2/3 × 5 = 10/3

→ V₂ = Q₂' / C₂= 10/3 × 1/2 = 10/6 = 1.66 V

Answer: C

Thermal Physics - Isothermal

A mole of ideal gas initially at temperature T₀ and volume V₀ undergoes a reversible isothermal expansion to volume V₁. If the ratio of specific heats is c_p/c_v = γ and if R is the gas constant, the work done by the gas is

A. Zero
B. RT₀ (V₁/V₀ )^γ
C. RT₀ (V₁/V₀ − 1)
D. c_v T₀ [1 − (V₁/V₀ )^(γ−1)]
E. RT₀ ln (V₁/V₀ )

(GR9277 #62)

Solution:

W = ∫ P dV

Ideal Gas: PV = nRT → P = nRT / V

W = _{_V₀}∫^{^V₁} nRT (1/V) dV

1 mole → n = 1
Isothermal, constant T = T₀
→ W = RT₀ _{_V₀}∫^{^V₁} (1/V) dV = RT₀ ln (V₁/V₀ )

Answer: E

Classical Mechanics - Equation of Motion

Questions 61-62

The Equation of motion of a rocket in free space can be written where m is the rocket’s mass, v is its velocity, t is time, and u is a constant.

$m \frac{dv}{dt}+u \frac{dm}{dt}=0$

The equation can be solved to give v as a function of m. If the rocket has m = m₀ and v = 0 when it starts, what is the solution?

A. u m₀/m
B. u exp(m₀/m)
C. u sin (m₀/m)
D. u tan (m₀/m)
E. None of the above

(GR8677 #62)

Solution:

$\begin{aligned} m \frac{dv}{dt}+u \frac{dm}{dt}&=0 \\m\frac{dv}{dt}&=-u \frac{dm}{dt} \\dv&=-u \frac{dm}{m} \\\int_{v_0}^{v}dv&=-u \int_{m_0}^{m}\frac{dm}{m} \\v&=-u \ln \frac{m_0}{m} \end{aligned}$

Answer: E

Electromagnetism - Cyclotron

A nonrelativistic particle with a charge twice that of an electron moves through a uniform magnetic field. The field has a strength of π/4 tesla and is perpendicular to the velocity of the particle. What is the particle’s mass if it has a cyclotron frequency of 1,600 hertz?

A. 2.5 × 10^-23 kg
B. 1.2 × 10^-22 kg
C. 3.3 × 10^-22 kg
D. 5.0 × 10^-21 kg
E. 7.5 × 10^-21 kg

(GR0177 #62)

Solution:

$F_\text{lorentz}=F_\text{centripetal}$

$qvB=\frac{mv^2}{r}$

Angular velocity: $\omega=2\pi f=\frac{v}{r}$

$qvB=mv2\pi f \rightarrow m=\frac{qB}{2\pi f}$

$m=\frac{2(1.6 \times 10^{-19})\pi /4}{2\pi(1,600)}=\frac{1.6\times 10^{-19}}{(1.6\times 10^3)4}=2.5\times 10^{-23}$

Answer: A