Constants

TABLE OF INFORMATION
(Printed in the test booklet)

me = 9.11 × 10−31 kg
= 1.60 × 1019 coulomb
N= 6.02 × 1023 per mole
R = 8.31 joules/(mole K)
k = 1.38 × 1023 joule/K = 1.38 ×  10−16 erg/K
= 3 × 108 m/s
= 6.63 × 1034 Joule second = 4.1 × 1015 eV.second
ħ h/2π
ɛ= 8.85 × 10−12 coulomb2/(newton meter2)  
μ= 4π × 10−7 weber /(ampere meter)  
= 6.67 × 10−11 meter3/(kilogram second2)  
= 9.80 m/s2
1 atm = 1.0 × 105 newton/meter2 = 1.0 × 105 pascals (Pa)
1 Å = × 10−10 meter
1 weber/meter2 = 1 tesla = 104 gauss


Moments of inertia about center of mass

Rod : 1/12Ml
Disc : 1/2MR2
Sphere: 2/5MR2

Electromagnetism - Capacitor

 photo Screenshots_2014-11-06-17-07-52_zps4e3e803e.png
 photo Screenshots_2014-11-06-17-08-03_zpsc70fbde2.png


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)/RV0e−t/RC/ R

Only plot B is right.

Answer: B

Electromagnetism - Faraday’s law

 photo Screenshots_2014-11-06-17-08-16_zps1a3f486a.png
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:

IR − ɛ = 0
= (V − ɛ)/R

ɛ = − dΦ/dt = −AdB/dt

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

ɛ = − (0.01)(−150) = 1.5 V
= (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.

  photo Screenshots_2014-11-06-17-08-30_zpsa12ff792.png

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(Rx2)1/2]
C. Qx / [4πɛ0(Rx2)]
D. Qx / [4πɛ0(Rx2)3/2]
E. QR / [4πɛ0(Rx2)]
(GR9677 #03)

Solution:

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

The distance r of P from the charged ring is r2 = Rx2

Q / [4πɛ0(Rx2)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.

  photo Screenshots_2014-11-06-17-08-30_zpsa12ff792.png
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:

A. (qQ/4πɛ0mR³)½
B. (qQx/4πɛ0mR⁴)½
C. qQ/4πɛ0mR³
D. qQx/4πɛ0mR
E. (qQx/4πɛ0m)½ [1/(R² + x²)]
(GR9677 #04)

Solution:

Felectric = kqQ/r²
Fcentripetal = mv²/r = mω²r

FFc
kqQ/r² = mω²r
ω² = kqQ/mr³

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

ω = (qQ/4πɛ0mR³)½

Answer: A

Notes:
see problem GR9277 #65

Classical Mechanics - Circular Motion

 photo Screenshots_2014-11-06-17-08-43_zpscdab5748.png
A car travels with constant speed on a circular road on level ground. In the diagram above, Fair 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. FA 
B. FB
C. FC
D. FD
E. FE
(GR9677 #05)

Solution:

FA = Centripetal force
F= 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, FA + FFB

Answer: B

Classical Mechanics - Friction Force

 photo Screenshots_2014-11-06-17-08-51_zps70db792b.png
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 − Fr  ma
At a constant speed, a = 0
Fr  mg sin θ 

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

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

W = mg sin θ (h/sin θ) = mgh

Answer: B

Classical Mechanics - Linear Momentum

 photo Screenshots_2014-11-06-17-09-06_zpse8dbbe63.png

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:

mavmbvb = mavambvb

Given:
mm
m= 2m
vb = 0

mv+ 0 = mva+ 2mvb 
v = va+ 2vb   (Eq.1)

Conservation of kinetic energy of the system:

½ mava² + ½ mbvb² = ½ mava'² + ½ mbvb'²
mva² + 0 = mva'² + 2mvb'²
va² = va'² + 2vb'²  (Eq.2)

(Eq.1) → (Eq.2)
(va+ 2vb')² = va'² + 2vb'²
va'² + 4va'vb+ 4vb'² = va'² + 2vb'²
4va'vb = 2vb'² − 4vb'²
2va' = − vb'  
vb = −2va'  (Eq.3)

(Eq.3) →  (Eq.1)
v = va+ 2vb
v = va+ 2(−2va')
v = − 3va'  (Eq. 4)

For the pendulum, conservation of energy:

at the moment of the collision
U T → magh = ½ mava² → va = (2gh)½

after the collision
U'  T→ magh' = ½ mava'² → va= (2gh')½

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

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, fbv, 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 = 1) to the longest wavelength in the Balmer series (n = 2)?

A. 5/27
B. 1/3
C. 4/9
D. 3/2
E. 3
(GR9677 #09)

Solution:

Rydberg formula:



λvac = the wavelength of the light emitted in vacuum
RH = Rydberg constant for Hydrogen
n1 and n2 are integers such that n n2

For the longest wavelength take n2 = ∞

For Lyman-radiation (n2 = ∞ → n= 1):

1/λL = RH (1/1 − 0) = RH

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

1/λB = RH (¼ − 0) = ¼RH

The Ratio:

λL/λB = (1/RH)(RH/4) = ¼ ≈ 5/27

Answer: A