Sample items “Mechanics”


1- Which of the following cannot be negative?

A) elastic potential energy
B) gravitational potential energy
C) mechanical energy
D) velocity
E) work

The elastic potential energy cannot be negative since
EPe =  1 / 2  kx2
where k and x2 are always positive


2- A small plane is used for training flights with constant speed v = 60m/s on a circular path of radius 1,000 m. The centripetal force exerted on the plane to keep it on its path is F = 4 x 104 N.
What is the mass of the plane?

A) 66.6 x 103 kg
B) 11.11 x 103 kg
C) 4 x 103 kg
D) 2 x 103 kg
E) 900 kg

The centripetal force exerted on the plane is directed towards the center of the circular trajectory and its magnitude is given by the relation:
F = m  v2 / r
m =  F r / v2
m =  4 × 104 × 1000 / 3600  = 11.11 x103 kg


3- A particle of mass m = 0.5kg undergoes a simple harmonic motion on a horizontal support with period T = π/5 s and amplitude 5 cm.
What is the maximum speed for this particle?

    A) 10 cm/s
    B) 50 cm/s
    C) 5 cm/s
    D) 25 cm/s
    E) 100 cm/s

The simple harmonic motion is represented by:
x = xm sin (ωt)
The speed is the derivative of x with respect to time.
So, v = xm ω cos (ωt)
Thus, vm = xm ω = 0.05 ( / T ) = 0.05 ( / π ) x 5 = 0.5 m/s = 50 cm/s
Or Second way:
The maximum speed is attained for x = 0;
So, sin (ωt) = 0 ; thus ωt= π or 2π
The time where v is maximum is t =  π / ω ; or t =  / ω
Replace t in v expression to get vm;
Thus, vm = xm ω = 0.05 ( / T ) = 0.05 ( / π ) x 5 = 0.5 m/s = 50 cm/s



Sample items “Electricity”


4- Two resistors of resistances R1 and R2 are connected in parallel. The equivalent resistance of the grouping is 10 Ω. Which of the following statements about the resistances is correct?

    A) Both R1 and R2 are greater than 10 Ω
    B) Both R1 and R2 are equal to 10 Ω
    C) Both R1 and R2 are less than 10 Ω
    D) The sum of R1 and R2 is 10 Ω
    E) R1 is greater than 10 Ω and R2 is smaller than 10 Ω

In parallel grouping of two resistors, the equivalent resistor is smaller than the smallest resistor. Thus, both R1 and R2 are greater than Req = 10 Ω.


5- A long wire in a DC circuit carries 2A electric current. What is the magnitude of the magnetic field at a point in space at a distance 40 cm from the center of the wire? (µ0 = 4π x 10-7 T.m/A)

    A) 10-6 T
    B) 0.5 x 10-6 T
    C) 0.5 π x 10-6 T
    D) π x 10-6 T
    E) 2π x 10-6 T

The magnitude of the magnetic field at a point in space at a distance d from the center of the wire is given by:
B =  µ0 × I / 2π × d  =  2 × 10-7 × I / d  =  2 × 10-7 × 2 / 0.4  = 10-6 T


6- Three resistors connected in parallel have individual values of 4.0 Ω, 6.0 Ω, and 10.0 Ω. If this combination is connected in series with a 12.0 V battery and a 2.0 Ω resistor as shown below, what is the current in the 10.0 Ω resistor?

A) 0.59 A
B) 1 A
C) 3.33 A
D) 11 A
E) 16 A

Let R’ be the resistance of the equivalent resistor for the three parallel resistors
1 / R’  =  1 / 4  +  1 / 6  +  1 / 10  
R’ = 1.93 Ω
The Re = 1.93 + 2 = 3.93Ω
The main current I =   12 / 3.93  = 3.05 A
The voltage across the parallel branches is v’ = 3.05 x 1.93 = 5.88 V
Thus, the current in the 10.0 Ω resistor is I’ =  5.88 / 10   = 0.588 A



Sample items “Waves and Particle Nature of Light”


7- The phenomenon of the total internal reflection is the main principle for the functioning of the fiber optics. This phenomenon appears if:

A) the index of refraction of the medium is less than 1.
B) the angle of incidence is greater than the critical angle.
C) the index of refraction of the medium is greater than 1.
D) the angle of incidence is smaller than the critical angle.
E) the angle of incidence is zero.

Total internal reflection appears if the angle of incidence i is greater than the critical angle il.


8- A convergent lens of focal length f = 20 cm gives, for an object of size 2 cm placed 60 cm in front of it:

A) a virtual image formed on a screen 30 cm behind the lens.
B) a real and inverted image of size 1 cm.
C) a real image of a size of 4 cm formed 30 cm behind the lens.
D) an erect image of a size of 4 cm formed 40 cm behind the lens.
E) a real image formed on a screen 60 cm behind the lens.

Apply the formula of position
1 / p  + 1p’ =  1 / f
1 / 60  +  1 / p’  =  1 / 20
P’ = +30 cm (real)
Magnification formula:
A’B’ / 2  = -  1 / 2
So the size of the image is A’B’ = 1 cm ( inverted )


9- Water drops fall at a rate of 120 drops per minute on a free surface of still water, causing ripples whose crests are 20 cm apart. The speed of propagation of the waves is:

A) 0.3 m/s
B) 0.4 m/s
C) 0.75 m/s
D) 1.5 m/s
E) 2.4 m/s

The frequency is f =  120 / 60  = 2 Hz
The distance between two consecutive crests represents the wave length. So, λ = 0.2 m
The speed of propagation of the waves is given by the relation
v = λ f = 0.2 x 2 = 0.4 m/s



Sample items “Thermo-dynamics”


10- The latent heat of fusion of water is 3.35 x 105 J/kg. The amount of energy required to change 500 g of ice into water is:

A) 1.49 x 103 J
B) 1.49 x 10-3 J
C) 6.75 x 105 J
D) 1.675 x 105 J
E)1.675 x 107 J

Q = m L
= 0.5 x 3.35 x 105 = 167500 J


11- The pressure of a gas inside a cylinder is 300 kPa. If the gas is compressed to half its original volume and the temperature rises from 23 °C to 323 °C, the new pressure will be:

A) 1200 Pa
B) 1200 kPa
C) 2400 Pa
D) 600 kPa
E) 2400 kPa

The gas will be considered as an ideal gas. The equation of state could be written in the form:
P1 V1 / T1  =  P2 V2 / T2
300 × 103 × V / 23 + 273  =  P2 × 0.5V / 323 + 273
P2 = 1208108.1 Pa ≅ 1200 kPa.


12- A gasoline engine does 50 J of work in one cycle and releases 35 J as heat loss. What is the engine’s efficiency?

A) 0.15
B) 0.30
C) 0.65
D) 0.45
E) 0.59

The efficiency of an engine is defined as ratio of the useful work done to the total energy provided.
The efficiency is given by the relation:
η =  Energy use / Eergy total  =  50 / 50 + 35  =  50 / 85  = 0.588



Sample items “Modern Physics”


13- Which of the following is the correct order from least to most massive fundamental particles?

A) electron = beta particle < proton < neutron < alpha particle
B) neutron < proton < alpha particle < electron < alpha particle
C) alpha particle < neutron < proton < electron < beta particle
D) alpha particle < neutron < proton < electron = beta particle
E) neutron < proton < electron < alpha particle < beta particles

The mass of the electron is the smallest mass. The correct answer is (A).
me = 5.5 x 10-4 u < mp = 1.00727 u < mn = 1.00866 u < mHe = 4.0015 u


14- The energy level in the diagram below shows the first six energy levels for an electron in a hydrogen atom that drops from energy level E3 to energy level E2. What is the frequency of the emitted photon? (h = 6 × 10-34 J.s)

A) 2.52 x 1014 Hz
B) 5.04 x 1014 Hz
C) 7.85 x 1014 Hz
D) 4 x 1013 Hz
E) 1 x 1015 Hz

The energy of the emitted photon is equal to the energy loss due to the transition between the two levels
Ei – Ef = h υ
[-1.51 – (-3.4)] x 1.6 x 10-19 = 6 x 10-34 υ
υ = 5.04 x 1014 Hz



Sample items “General Knowledge”


15- A custodian tries to turn on the light upon entering the room. The bulb does not light up. His hypothesis is that a circuit breaker is open. However, the custodian finds that the room's breaker is not open. What is the next best step to solve the problem?
I. Check whether other bulbs are plugged in
II. Check whether other breakers are open.
III. Develop a new hypothesis for why the lights do not turn on

A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III

Bulbs are connected in parallel and checking other lamps or other circuit breakers is useless.
So, it is recommended to develop a new hypothesis for why the lights do not turn on.


16- Which among the following is the fundamental equation of the simple harmonic motion (S.H.M) along the horizontal axis x’ox for a mass m connected to the second extremity of horizontal spring of stiffness constant k?

A) x’’ + (k / m) x’ = 0
B) x + ω2x’’ = 0
C) x’’ + (k/ m) x = 0
D) x2 + ωx’2 = 0
E) x’’ + (m/ k)2 x = 0

For the given oscillator, the mechanical energy at any time is:
EM = ½ mv2 + ½ kx2 = constant
Derive with respect to time to get:
mvv' + kxx’ = 0 ; x’ = v ≠ 0 and v’ = x”
so,
mx” + kx = 0  divide by m to get
x” +  k / m  x = 0
Differential equation of form: x” + ω2 x = 0 where k / m  = ω2 that represents a SHM.