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King Fahd University of Petroleum and Minerals – Physics Department

Physics 102 Recitation – Term 172 – Spring 2018 – Section 7

Quiz # 1

Name: ID # _______ __

A transverse sinusoidal wave is travelling to the right (+ -axis) on a stretched

string. The general form of the wave is ( , ) = sin ( ± + ). The

amplitude of the wave is 0.800 cm, its wavelength is 0.550 m, and its phase

constant is zero. The speed of waves on the string is 24.0 m/s. What is the speed

of the particle of the string located at = 0.300 m when the time is 1.50 s?

SOLUTION: = 2 = = 2 / ( , ) = sin( − ) ( , ) = − cos( − ) ( , ) = −(2 / ) cos (2 / ) − (2 / ) ∴ = − 2 × 24.00.550 × 8.00 × 10 cos 2 × 0.3000.550 − 2 × 24.0 × 1.500.550 = − . /



Quiz # 1

Name: ID # _______ __

Two identical sinusoidal traveling waves, each with a frequency of 100 Hz, are moving

in the same direction along a stretched string. If they are interfering and the combined

wave has an amplitude that is 0.25 times that of the amplitude of each wave, calculate

the phase difference between the two waves in radians and in meters. The speed of

waves on the string is 25 m/s.

SOLUTION:

The amplitude of the resultant wave is : = 2 cos /2 .

Thus, the phase difference is : = 2 cos /2 = 2 cos 0.25 /2 = 2.89 rad

The wavelength is : = = = 0.25 m.

Therefore, the phase difference in meters is : ∆ = = 0.115 m.



Quiz # 1

Name: ID # _______ __

A standing wave is set up on a string of length 1.2 m that is fixed at its ends. The string vibrates

according to the equation: , = 0.50 sin 2.5 cos 40 , where and are in

meters and is in seconds.

(a) What is the speed of waves on this string?

(b) What is the harmonic number of the standing wave?

SOLUTION:

(a) = = . = 16 m/s.

(b) = ⟹ = = = = × . × . = 3



Quiz # 2

Name: ID #

The difference between two consecutive resonant frequencies in a tube closed at

one end is 80 Hz. What is the fifth harmonic frequency of this tube?

SOLUTION:

The resonant frequencies of the tube are given by: = , where n = 1, 3, 5, …

The difference between two consecutive resonant frequencies is: ∆ = − = − = .

The fifth harmonic frequency of the tube is = = = × 80 = 200 Hz.

DELL

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Quiz # 2

Name: ID #

The intensity of sound is 3.20 × 10 W/m2 at a distance of 5.00 m from a

source. What is the sound level at a distance of 10.0 m from the source?

SOLUTION:

Recall that the intensity ( ) of sound at a distance from a source is: = . Let be the intensity at 5.00 m, and be the intensity at 10.0 m, then: = 4 × 100 ÷ 4 × 25 = 0.25

Thus, = 0.25 = 8.00 × 10 W/m .

Therefore, the sound level at a distance of 10.0 m is: = 10 log = 10 × log 8.00 × 10 10 = 89 dB

DELL

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Quiz # 2

Name: ID #

A stationary speaker sends sound waves of frequency 1000 Hz toward a car

moving away at a speed of 35 m/s. What is the frequency of the waves reflected

back to the speaker? [speed of sound in air = 343 m/s]

SOLUTION:

= speed of sound, = speed of car

From the speaker to the car:

Source = speaker, frequency =

Observer = car, frequency = = −

From the car to the speaker:

Source = car, frequency =

Observer = speaker, frequency = = + = − + = −+ = 1000 × 343 − 35343 + 35 = 815 Hz

DELL

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Quiz # 3

Name: ID #

An ice cube, of mass 20 g and initial temperature 0 °C, is placed in an aluminum

container whose initial temperature is 80 oC. The system comes to an equilibrium

temperature of 30 °C. What is the mass of the container? [cAl = 900 J/kg.K]

SOLUTION: For ice:

Melting: = = 20 × 10 × 333 × 10 = 6660 J Heating: = ∆ = 20 × 10 × 4190 × 30 = 2514 J For aluminum:

Cooling: = ∆ = × 900 × −50 = − 45000

Conservation of energy: + + = 0 6660 + 2514 − 45000 = 0

Thus: mAl = 0.204 kg



Quiz # 3

Name: ID #

A metal rod has a length of 7.50 m at 15 °C and a length of 7.60 m at 95 °C. What

is the temperature of the rod when its length is 7.45 m?

SOLUTION:

In the first change (15 → 95 °C): = ∆ ∆ = 7.60 − 7.507.50 × (95 − 15) = 1.67 × 10 (℃)

In the second change (15 °C → ??): ∆ = ∆ = 7.45 − 7.501.67 × 10 × 7.50 = − 40 C = + ∆ = 15 − 40 = −25 ℃



Quiz # 3

Name: ID #

The figure below shows a steel bar 20.0 cm long welded end to end to a copper

bar 40.0 cm long. The two bars have the same cross sectional area. The free end

of steel is maintained at 100 oC, and the free end of copper is maintained at 0 oC.

Find the temperature (Tx) at the junction between the two bars, assuming steady

state. [Thermal conductivity: ksteel = 50.2 W/m.K, kcopper = 385 W/m.K]

SOLUTION: In steady state: ( ) = ( ) ( ) = (100 − )

= 100 − 3.835 = 100 −

Thus: Tx = 20.7 oC



Quiz # 4

Name: ID #

An ideal gas, initially occupying a volume of 0.400 m3 at a pressure of 1.00 atm,

expands isothermally to double its initial volume. How much heat is absorbed by

the gas in this process?

SOLUTION:

Isothermal process: = constant; thus = constant; therefore ∆ = 0.

From the 1st law of thermodynamics: ∆ = − = 0.

Thus: = = ln = ln = 0.400 × 1.01 × 10 × ln 2 = 28 kJ



Quiz # 4

Name: ID #

An ideal monatomic gas, initially occupying a volume of 1.00 L at a pressure of

1.00 atm, absorbs 500 J of heat and expands isobarically. What is the final volume

of the gas?

SOLUTION: For an ideal monatomic gas: = 3 /2, = 5 /2. Isobaric process: = ∆ = 5 /2 ∆ .

Thus: ∆ = 2 /5. = ∆ = ∆ = 2 /5

Thus: ∆ = = × × . × = 1.98 × 10 m = 1.98 L.

Therefore: = 2.98 L.



Quiz # 4

Name: ID #

An ideal monatomic gas ( = 2.00), initially occupying a volume of 1.00 L at

300 K, expands adiabatically to double its initial volume. How much work is done

by the gas in this process?

SOLUTION: For an ideal monatomic gas: = 3 /2, = 5 /2, = / = 5/3. Adiabatic process: = 0, and = .

= = 12 / × 300 = 189 K

From the 1st law of thermodynamics: ∆ = − .

Thus: = −∆ = − ∆ = − 3 /2 − = −2.00 × 3 × 8.312 189 − 300 = 2.77 kJ



Quiz # 5

Name: ID #

An ideal diatomic gas is the working substance in an engine that operates on the

cycle shown in the figure below, where V3 = 4V1. What is the engine’s efficiency?

SOLUTION:

Consider process 2→3:

Thus: 3 0.431

Consider process 4→1:

Thus: 0.144 ∆ 5 /2 ∆ 2.5 ∆ 2.5 ∆ ∴ 2.5 4 0.144 0.431 2.87 ( ⟹ ) ∆ 5 /2 ∆ 2.5 ∆ 2.5 ∆ ∴ 2.5 3 5.00 ( ⟹ ) 0 | | 1 | | 1 2.875.00 0.426



Quiz # 5

Name: ID #


cycle shown in the figure below. Processes BC and DA are reversible and

adiabatic. What is the efficiency of this engine?

SOLUTION:

Consider process BC:

Thus: / / /32 / 2 23.78

Consider process DA:

Thus: / / /32 / 11.89 ∆ 7 /2 ∆ 3.5 ∆ 3.5 ∆ ∴ 3.5 2 3.5 (+ ⟹ ) ∆ 7 /2 ∆ 3.5 ∆ 3.5 ∆ ∴ 3.5 11.89 23.78 1.30 ( ⟹ ) 0 | | 1 | | 1 1.33.5 0.629



Quiz # 5

Name: ID #


cycle shown in the figure below. What is the engine’s efficiency?

SOLUTION: ln 3 ln 2 3 ln 2 ( ⟹ ) ∆ 5 /2 ∆ 2.5 3 5 ( ⟹ ) ln ln 1/2 ln 2 ( ⟹ ) ∆ 5 /2 ∆ 2.5 3 5 ( ⟹ ) 3 ln 2 5 ln 2 5 | | 1 | | 1 ln 2 53 ln 2 5 0.196



Quiz # 6

Name: ID #

Two small identical conducting spheres, initially uncharged are separated by a

distance of 0.50 m. Find the number of electrons that must be transferred from

one sphere to the other in order to produce an attractive force of 2.0 × 104 N

between the spheres.

SOLUTION: The magnitude of the charge on each sphere is the same; let’s call it Q. = = ⟹ = = 2.0 × 109.0 × 10 × 0.50 = 7.5 × 10 C

=

Thus: = = . × . × = 47 × 10 electrons



Quiz # 6

Name: ID #

Two point charges are placed on the x-axis as follows: q1 = – 6.00 × 10-6 C at the origin,

and q2 = + 2.00 × 10-6 C at x = 10.0 cm. A third positive point charge q3 is to be located

somewhere, on the x axis, such that the net electrostatic force on it due to q1 and q2 is

zero. What is the x-coordinate of q3?

SOLUTION: Since q1 and q2 are of opposite signs, the point of equilibrium (zero force) must be: (1) on the

x-axis, (2) outside the region between q1 and q2, (3) closer to the weaker charge.

Thus, q3 must be placed at a distance d to the right of q2.

Let the distance between q1 and q2 be L, then: = ∶ + = ⟹ + =

Thus: 1 + = ± ⟹ = −1 ± = 0.732

Therefore, = .. = 13.7 cm, and the x-coordinate of q3 is x = + 23.7 cm.



Quiz # 6

Name: ID #

The figure below shows an equilateral triangle ABC. A point charge q = + 1.0 μC is located at

each of the three corners. A point charge Q is placed at the midpoint between B and C.

Determine the magnitude and sign of Q so that the net force on the charge at A is zero.

SOLUTION: Note the following:

(1) Q must be negative.

(2) The forces on A due to B and C are equal in magnitude ( = ), and each of them

makes an angle of 60o with the horizontal.

(3) The distance d between Q and A is such that: = − = .

(4) The force on A due to Q must be equal to the sum of the y-components of the forces

due to B and C.

Thus: = + = 2 = 2 sin 60 = 2 √32 ⟹ = √3 = √3 × 34 × 1.0 = 1.3 μC

Therefore: Q = – 1.3 μC.



Quiz # 7

Name: ID #

A neutral conducting spherical shell has an outer radius of 2.0 m. A point charge

is placed at the center of the shell. If the charge induced on the inner surface of

the shell is + 5.0 μC, what are the magnitude and direction (inward/outward) of

the electric field at a distance of 3.0 m from the center of the sphere?

SOLUTION:

The shell is neutral, thus: shell = 0. = + ⟹ = − = 0 − 5.0 = − 5.0 μC.

The requested point is outside the shell; thus, the only charge that will contribute

to the electric field is the charge on the outer surface of the shell.

Since that charge is negative, the electric field is radially inward. = = × × . × . = / .



Quiz # 7

Name: ID #

Two infinite parallel lines of charge are shown in the figure below, and are separated

by 5.0 cm. Their uniform linear charge densities are = + 2.0 × 10 C/m and = − 3.0 × 10 C/m. Find the net electric field at a point that is 1.0 cm to the left

of . Express your answer in unit vector notation.

SOLUTION: The electric fields due to the lines are: = − ̂ = − × × × . × . × ̂ = − 3.6 × 10 ̂ (N/C) = + ̂ = + × × × . × . × ̂ = + 9.0 × 10 ̂ (N/C)

The net electric field is: = + = − 3.6 × 10 ̂ + 9.0 × 10 ̂ = − 2.7 × 10 ̂ (N/C)



Quiz # 7

Name: ID #

A cube of edge length 2.00 m is placed in a region in which the electric field is given

by = 20.0 − 28.0 (N/C). Determine the net charge contained within the cube.

SOLUTION: The area of each face is 4.0 m2.

Since the electric field is in the z direction, there will be electric flux only through

the top ( ) and bottom ( ) faces.

At the top (z = 2.0 m): Φ = . = − 36.0 . + 4.00 = − 144 N. m /C.

At the bottom (z = 0): Φ = . = + 20.0 . − 4.00 = − 80.0 N. m /C.

Thus, the net flux is Φ = Φ + Φ = −224 N. m /C.

Gauss’ law: Φ = .

Therefore: = Φ = −224 × 8.85 × 10 = − 1.98 × 10 C ≈ −2 nC .



Quiz # 8

Name: ID #

In the rectangle shown below, the sides have lengths of 2.0 cm and 4.0 cm. The charges

are 1.0μC and 4.0μC. How much work is required to a move a third

charge 3.0μC from A to B along the diagonal of the rectangle?

SOLUTION: 9 10 1.0 100.04 4.0 100.02 1575kV

9 10 1.0 100.02 4.0 100.04 450kV ∆ 3.0 10 450 1575 10 3.375J



Quiz # 8

Name: ID #

Two isolated concentric conducting thin spherical shells have radii = 0.50 m and = 1.0 m; uniform charges = + 3.0 μC and = − 4.0 μC. With V = 0 at infinity,

what is the electric potential at a point that is 0.80 m from the center?

SOLUTION: = + = 9 × 10 3.00.80 − 4.01.0 × 10 = − 2250 V



Quiz # 8

Name: ID #

In the figure below, two parallel charged plates are separated by distance 5.00mm.

The plate potentials are 50.0V and 100V. An electron is released from

rest at the midpoint between the plates. Which plate will it strike? With what speed will

it strike that plate?

SOLUTION: The electron will move toward the higher potential, i.e. to plate 1.

The electric field is uniform, so the potential varies linearly in the region between

the plates.

Thus, the potential at the midpoint is – 75.0 V.

Apply the conservation of mechanical energy: ⟹ { 0} 12

2 2 1.60 109.11 10 50.0 75.0 8.78 10

Therefore, 2.96 10 m/s.



Quiz # 9

Name: ID #

In the circuit shown below, each capacitor has a capacitance of 3.0μF and 12V.

How much total energy is stored by this combination of capacitors?

SOLUTION: and are connected in parallel: 6.0μF

and are connected in series: 2.0μF

The stored energy is : 2.0 10 144 0.144mJ



Quiz # 9

Name: ID #

In the circuit shown below, each capacitor has a capacitance of 3.0μF and 12V.

How much total energy is stored by this combination of capacitors?

SOLUTION:

and are connected in series: 1.5μF

and are connected in parallel: 4.5μF

The stored energy is : 4.5 10 144 0.324mJ



Quiz # 9

Name: ID #

A 15.0 μF capacitor is charged using a 12.0 V battery. The battery is removed, and an

insulator of dielectric constant κ = 5.00 is inserted filling the space between the plates of

the capacitor. Calculate the energy stored in the capacitor after insertion of the insulator.

SOLUTION: =

This charge is maintained after removing the battery.

After inserting the dielectric, the stored energy is = 2 = 2 = 12 1 = 2 = 15.0 × 10 × 1442 × 5.00 = 0.216 mJ



Quiz # 10

Name: ID #

A potential difference of 0.500 V is applied to a conducting wire of length 3.00 m and

resistivity of 9.68 × 10-8 Ω.m. Calculate the magnitude of the current density in the wire.

SOLUTION: = = 1 × = × = = 0.5009.68 × 10 × 3.00 = 1.72 × 10 A/m



Quiz # 10

Name: ID #

A conducting wire is 25.0 m long, has a radius of 0.200 mm, and is made of a material

with resistivity of 1.50 × 10-6 Ω.m. If it carries a current of 0.300 A, how much power

is dissipated in it?

SOLUTION: = = = 1.50 × 10 × 25.0 × 4.00 × 10 = 298 Ω = = 0.0900 × 298 = 26.9 W



Quiz # 10

Name: ID #

A conducting wire is 25.0 m long, has a radius of 0.200 mm, and is at 20.0 oC. If the

temperature is increased to 340 oC, what is the resistance of the wire? Ignore the change

in the dimensions of the wire. [ρo = 1.50 × 10-6 Ω.m, α = 0.400 × 10-3 ( oC )-1]

SOLUTION: = 1 + ∆ = 1.50 × 10 1 + 0.400 × 10 × 320 = 1.692 × 10 Ω. m = = = 1.692 × 10 × 25.0 × 4.00 × 10 = 337 Ω



Quiz # 11

Name: ID #

Three resistors are connected as shown in the figure. The potential difference between

points A and B is VA – VB = + 26 V. How much current flows through the 2.0-Ω resistor?

SOLUTION:

The equivalent resistance of the 2.0 Ω and 4.0 Ω resistors (connected in parallel) is

(4×2)/(4+2) = 4/3 Ω.

The total equivalent resistance is (4/3) + 3 = 13/3 Ω.

Thus, the current flowing the 3 Ω resistor is 26/(13/3) = 6 A.

Now, consider the path going from A through the 3 Ω resistor, the 2 Ω resistor,

then to B: − 3 × 6 − 2 =

Therefore: = = = 4 A



Quiz # 11

Name: ID #

In the circuit shown below, determine the power dissipated by the 40-Ω resistor?

SOLUTION: Reduce the circuit to a single-loop:

The 10 Ω and 40 Ω resistors are connected in parallel, their equivalent resistance is

(40×10)/(40+10) = 8 Ω.

The total equivalent resistance is 8 + 2 = 10 Ω.

Thus, the current from the battery is i = 18/10 = 1.8 A.

Now, consider the left loop, and call ix the current through the 40 Ω resistor:

Start from the battery and go clockwise

+18 – (40 ix) – (2 × 1.8) = 0.

Therefore, ix = 0.36 A.

The power dissipated in it is P = (0.36)2 × 40 = 5.2 W



Quiz # 11

Name: ID #

Five resistors, each of resistance 6.0 Ω, are connected as shown in the diagram.

The potential difference between points A and B is 32 V. What is the current in

resistor R?

SOLUTION: Number the resistors as shown below:

R23 = 12 Ω

R234 = (12×6)/(12+6) = 4 Ω

Req = 6 + 4 + 6 = 16 Ω

The current through resistor 1 is 32/16 = 2 A.

Now, consider the path that goes from A to B through R, and let the current through R be i : − 6 × 2 − 6 − 6 × 2 =

Thus: = = = 1.3 A

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