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WAEC: PHYSICS 2023 Physics WAEC (WASSCE) Past Questions 2023 Physics WAEC (WASSCE) Essay Past Questions (Paper 2)
Lesson 3, Topic 1
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2023 Physics WAEC (WASSCE) Essay Past Questions (Paper 2)

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Content:

  • 2023 Physics WAEC (WASSCE) Essay Past Questions (Paper 2)
  • SECTION A (1 – 7) – Will comprise seven short-structured questions. Candidates will be required to answer any five questions for a total of 15 marks.
  • SECTION B (8 – 12) – Will comprise five essay questions out of which candidates will be required to answer any three for 45 marks.
  • Duration – 1½ hours

SECTION A (1 – 7)

Answer any five (5)

Question 1:

The force, F, acting on the wings of an aircraft moving through the air of velocity, v, and density, ρ, is given by the equation F = kvxρyAz, where k is a dimensionless constant and A is the surface area of the wings of the aircraft. Use dimensional analysis to determine the values of x, y, and z.

View Solution

Question 2:

a. Define strain energy.

b. Write an expression for the energy stored, E, in a stretched wire of original length, l , cross-sectional area, A, extension, e, and Young’s modulus, Y, of the material of the wire.

View Solution

Question 3:

a. A projectile is fired at an angle, θ, to the horizontal with velocity, u. Show that at any time, t, during the motion, the:

i. horizontal component of the velocity is independent of t;

ii. vertical component of the velocity depends on t.

b. State the assumption on which projectile motion is based.

View Solution

Question 4:

State three differences between geostationary satellites and polar satellites.

View Answer

Question 5:

a. Using the kinetic theory, explain the term diffusion of fluid molecules.

b. Name one phenomenon that demonstrates that light behaves as a:

 i. wave
ii. particles

View Answers

Question 6:

a. State one difference between an intrinsic and an extrinsic semiconductor.

b. Draw a circuit diagram to illustrate full wave smoothing rectification.

View Answers

Question 7:

a. What is fibre optics?

b. State two reasons why optical fibres are preferred to copper cables in the telecommunication industry.

View Answers

SECTION B (8 – 12)

Answer any three (3)

Question 8:

a.

i. State the reason why simple harmonic motion is periodic.

ii. State two factors that affect the period of oscillation of a simple pendulum.

iii. Sketch a graph of the total mechanical energy, E, against displacement, y, for the motion of a simple pendulum from one extreme position to the other.

b. 

Screenshot 2025 01 28 at 20.33.27

The diagram above illustrates an oscillatory pendulum.

Calculate the work done in raising the pendulum to point B, if the mass of the bob is 50 g.

c. A spiral spring of spring constant, k, and natural length, l, has a scale pan of mass 0.04 kg hanging on its lower end while the upper end is firmly fixed to a support. When an object of mass 0.20 kg is placed on the scale pan, the length of the spring becomes 0.055 m and when the object is replaced with another object of mass 0.28 kg, the length of the spring becomes 0.065 m. Calculate the values of k and l.
[g = 10 ms-2]

View Answers

Question 9:

a. Define each of the following terms used with simple machines:

i. Pivot
ii. Load
iii. Efficiency.

b. A truck of mass 1.2 × 103 kg is pulled from rest by a constant horizontal force of 25.2 N on a levelled road. If the maximum speed attainable in the process is 60 km/h.

Calculate the:

i. work done by the force;
ii. distance travelled by the truck in reaching the maximum speed.

c. State two differences between absolute zero temperature and ice point.

d. An uncalibrated liquid-in-glass thermometer was used in determining a Celsius temperature. The readings are tabulated below:

Temperature °C-60100
Length of column / cmL2.015.0

Determine the value of L.

View Answers

Question 10:

a. i. Why are parabolic mirrors suitable for use in headlamps of vehicles?

ii.  Draw a ray diagram to illustrate the answer in 10(a)(i).

b. i. State two applications of echoes.

ii. An observer standing at a point, P, on the same horizontal ground as the foot, H, of a tower, shouts, and 1.20 s later, he hears the echo. He then moved to another point, Q, 40 m from P, and shouted again but the echo was heard after 1.45 s. Calculate the:
I. distance between P and H;
II. speed of sound in air.

c. i. Define the term absolute refractive index of a medium.

ii. A piece of coin falls accidentally into a tank containing two immiscible liquids A and B as illustrated in figure below:

Screenshot 2025 01 29 at 08.41.57

Calculate the displacement of the coin when viewed vertically from above. [refractive index of A = 1.3, refractive index of B = 1.4]

View Answers

Question 11:

a. i. Define the electric potential at a point in an electric field.

ii.  An uncharged body, A, was charged electrostatically by a test charge, B, using the method of induction and the method of contact. State two differences between the two methods.

b. An important precaution during an electricity experiment is to open the circuit when no readings are being taken. Give two reasons for the stated precaution.

c. i. The Figure below is a circuit diagram in which a coil of inductance, L, and a resistor of resistance, R, are connected to a variable alternating source of frequency, f.

Screenshot 2025 01 29 at 09.28.41

The table shows the square of the impedance, Z2, corresponding to each value of f2;

f2/Hz2198.8400.0600.3800.9900.0
Z2/Ω2249.6400.0552.3702.3800.0

(i) Write down the equation for Z in terms of f2, R2, and L2.

(ii) Plot a graph of Z2 against f2 and use it to deduce:

(I) L
(II) R

[π2 = 10]

View Answers

Question 12:

a. State the function of each of the following parts of a modern x-ray tube:

i. heater;
ii. high tension source;
iii. cooling fins.

b. State one reason for each of the following design features of a modern x-ray tube:

i. the glass envelope is highly evacuated; 
ii. the target is a metal of very high melting point;
iii. the cooling fins are located outside the glass envelope.

c. In a nuclear fission reaction, a nuclide \( \scriptsize ^{235}_{92}U \) is bombarded with a neutron to produce \( \scriptsize ^{93}_{36}Kr \: and \: ^{141}_{56}Ba\)with additional neutrons, the energy involved in the process is Q.

  • \( \scriptsize mass \: of \: ^{235}_{92}U = 235.044\: u\)
  • \( \scriptsize mass \: of \: ^{93}_{36}Kr = 91.898 \:u\) 
  • \( \scriptsize mass \: of \: ^{141}_{56}Ba = 140.914 \:u \)
  • mass of neutron = 1.009 u
  • 1 u = 1.66 × 1027 kg
  • c = 3.0 × 108 ms-1

i. Write down the balanced nuclear reaction equation for the process.
ii. State with reason whether Q is absorbed or released in the process.
iii. Calculate the value of Q in joules.
iv. State three differences between nuclear fusion and nuclear fission.

View Answers
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Question 1

The force, F, acting on the wings of an aircraft moving through the air of velocity, v, and density, ρ, is given by the equation F = kvxρyAz, where k is a dimensionless constant and A is the surface area of the wings of the aircraft. Use dimensional analysis to determine the values of x, y, and z.

Solution

F = k [v]x [ρ]y [A]z

  • F = Force (N or kg ms-2) = \( \scriptsize MLT^{-2} \)
  • k = constant
  • v = velocity (ms-1) = \( \scriptsize LT^{-1} \)
  • ρ = density (kg/m3) = \( \scriptsize ML^{-3} \)
  • A = surface area (m2) = \( \scriptsize L^{2} \)

∴  \( \scriptsize  \left(MLT^{-2}\right) = k \left(LT^{-1}\right)^x \left(ML^{-3}\right)^y \left(L^{2}\right)^z \)

Since k is dimensionless,

⇒ \( \scriptsize  \left(MLT^{-2}\right) = \left(LT^{-1}\right)^x \left(ML^{-3}\right)^y \left(L^{2}\right)^z \)

⇒ \( \scriptsize  \left(MLT^{-2}\right) = L^x\: \times \: T^{(-1)x} \: \times \:  M^y\: \times \: L^{(-3)^y} \: \times \: L^{(2)z} \)

⇒ \( \scriptsize  \left(MLT^{-2}\right) = L^x\: \times \: T^{-x} \: \times \:  M^y\: \times \: L^{-3y} \: \times \: L^{2z} \)

⇒ \( \scriptsize  \left(MLT^{-2}\right) = L^{x\:-\:3y\:+\: 2z}\: \times \: T^{-x} \: \times \:  M^y  \)

⇒ \( \scriptsize  M^1L^1T^{-2} = L^{x\:-\:3y\:+\: 2z}\: \times \: T^{-x} \: \times \:  M^y  \)

For M:

y = 1

For L:

x – 3y +  2z = 1 ……(i)

For T:

-x = -2

∴ x = 2

To find z, substitute x and y into equation (i)

(2) – 3(1) +  2z = 1

2 – 3 +  2z = 1

-1 +  2z = 1

2z = 1 + 1

2z = 2

z = \( \frac{2}{2} \)

z = 1

Hence, x = 2, y = 1 and z = 1.

Question 2

a. Define strain energy.

Answer:

Strain energy is the potential energy stored within an elastic material as a result of its deformation when an external force is applied. It is the ratio of extension produced in a solid/wire to the original length.

 

b. Write an expression for the energy stored, E, in a stretched wire of original length, l, cross-sectional area, A, extension, e, and Young’s modulus, Y, of the material of the wire.

Solution

Strain energy , E = \( \frac{1}{2} \scriptsize Fe \)

where F = Force, e = extension

⇒ \(\scriptsize Y = \normalsize \frac{Stress}{Strain}\)

⇒ \(\scriptsize Strain = \normalsize \frac{Extension}{Original \: Length} = \frac{e}{L}\)

⇒ \( \scriptsize Stress, \: \sigma \) = \( \frac{Force}{Area} \\ = \frac{F}{A} \)

∴ Y = \(\normalsize \large \frac{\frac{F}{A}}{\frac{e}{L}}\)

Y = \( \frac{FL}{Ae}\)

∴ F = \( \frac{YAe}{L} \)

Substituting F into the strain energy equation, we have:

E = \( \normalsize \frac{1}{2} \scriptsize \: \times\: \normalsize \frac{YAe}{L} \scriptsize\: \times \:  e \)

E = \( \normalsize \frac{1}{2} \scriptsize \: \times\: \normalsize \frac{YAe^2}{L}  \)

E = \(\frac{YAe^2}{2L}  \)

Question 3

a. A projectile is fired at an angle, θ, to the horizontal with velocity, u.

Show that at any time, t, during the motion, the:

i. horizontal component of the velocity is independent of t;

Solution

At any point the projectile is rising;

Vx = ux – gt

ux = u cos θ, g = 0

∴  Vx = u cos θ – 0

∴  Vx = u cos θ

∴ horizontal component of the velocity is independent of t

 

ii. Show that at any time, t, during the motion, the: vertical component of the velocity depends on t.

Solution

Vy = uy – gt

uy = u sin θ

∴  Vy = u sin θ – gt

∴  the vertical component of the velocity depends on t

 

b. State the assumption on which projectile motion is based.

Answer:

It assumes that the only force acting on the projectile is gravity, and all other forces such as air resistance or drag, affecting its motion, are negligible.

Question 4

State three differences between geostationary satellites and polar satellites.

Answer:

Geostationary Satellites: Polar Satellites:
Geostationary satellites orbit the globe in an east-west orbit. Polar satellites orbit the globe in a north-south orientation.
The period of revolution of a geostationary satellite is 24 hours. The period of revolution of a polar satellite is approximately 100 minutes
They are positioned at an altitude of approximately 35,786 kilometres above the Earth’s surface. They orbit at low altitudes closer to Earth’s surface,
at a distance of 500-800 km
They have an inclination of 0°, meaning their orbital plane aligns with the equatorial plane. They have a high inclination angle (usually around 90°) to achieve polar orbits.
Geostationary satellites provide continuous coverage of a specific region because they orbit the Earth at the same speed as the Earth rotates. A polar satellite provides global coverage because it orbits the Earth from pole to pole, passing over every part of the planet during each orbit
They primarily collect data for meteorological observation weather forecasting, and telecommunications. They collect data for a wide range of applications, including weather monitoring, climate research, Environmental monitoring, and scientific research.

Question 5

a. Using the kinetic theory, explain the term diffusion of fluid molecules.

Answer:

According to kinetic theory, “diffusion of fluid molecules” refers to the random movement and mixing of fluid particles (molecules) due to their inherent kinetic energy, causing them to collide with each other and spread out from areas of high concentration to areas of low concentration, resulting in a gradual mixing of the fluid throughout a container; essentially, the molecules are constantly in motion and will naturally distribute themselves evenly through the available space.

 

b. Name one phenomenon that demonstrates that light behaves as a:

i. wave

Answers:

  • reflection;
  • refraction;
  • diffraction;
  • interference;
  • and polarization.

Any (1) one

 

ii. particle

Answers:

  • the absorption of light;
  • the photoelectric effect;
  • the Compton effect;
  • thermionic emission;
  • and radiation of energy from heated bodies.

Any (1) one

Question 6

a. State one difference between an intrinsic and an extrinsic semiconductor.

Answer:

Intrinsic Semiconductor Extrinsic Semiconductor
An intrinsic semiconductor is a pure material
with no intentional impurities.		 An extrinsic semiconductor is a semiconductor that has been intentionally doped with impurities to modify its electrical properties.
Intrinsic semiconductors have low conductivity
\high resistivity.		 Extrinsic semiconductors have higher conductivity
\low resistivity
In an intrinsic semiconductor, the number of electrons and holes (positive charge carriers) are equal. In an extrinsic semiconductor, the number of electrons and holes are not equal depending on the type of dopant added (n-type or p-type)
It has thermally generated current carriers. In an extrinsic semiconductor, the charge carriers are created by the process of doping.
Intrinsic semiconductors are not further subdivided n-type and p-type semiconductors are two categories of extrinsic semiconductors.

 

b. Draw a circuit diagram to illustrate full wave smoothing rectification.

Question 7

a. What is fibre optics?

Answer:

This is a technology that transmits information/data using pulses of light through thin transparent fibres made of high-quality glass/plastic.

 

b. State two reasons why optical fibres are preferred to copper cables in the telecommunication industry.

Answer:

  • Optical fibres have greater bandwidth
  • They transmit signals faster
  • cover long distances
  • Greater efficiency
  • High-quality transmission
  • Low resistance

Question 8

a.

i. State the reason why simple harmonic motion is periodic.

Answer:

Simple harmonic motion is considered periodic because of the restoring force which is always acting in the opposite direction to the displacement and is directly proportional to it. This causes the object to oscillate back and forth around an equilibrium position, creating a repeating pattern of motion.
ii. State two factors that affect the period of oscillation of a simple pendulum.

 

ii. State two factors that affect the period of oscillation of a simple pendulum.

Answer:

  • Length of the pendulum
  • Acceleration due to gravity

 

iii. Sketch a graph of the total mechanical energy, E, against displacement, y, for the motion of a simple pendulum from one extreme position to the other.

Answer:

b.

The diagram above illustrates an oscillatory pendulum.

Calculate the work done in raising the pendulum to point B, if the mass of the bob is 50 g.

Solution:

mgl – mgl cos(θ)” represents the change in potential energy of a mass “m” hanging from a length “l” at an angle “theta” from the vertical, where “g” is the acceleration due to gravity; essentially, it calculates the difference in potential energy between the object at its lowest point (theta = 0) and at an angle theta.

Workdone = mgl(1 – cosθ)

= 0.05 × 10  × 0.6(1 – cos60)

= 0.05 × 10  × 0.6(1 – 0.5)

= 0.05 × 10  × 0.6 × 0.5

= 0.15 J

or

Workdone = mgh

= 0.05 × 10 × 10

= 5 J

 

c. A spiral spring of spring constant, k, and natural length, l, has a scale pan of mass 0.04 kg hanging on its lower end while the upper end is firmly fixed to a support. When an object of mass 0.20 kg is placed on the scale pan, the length of the spring becomes 0.055 m and when the object is replaced with another object of mass 0.28 kg, the length of the spring becomes 0.065 m. Calculate the values of k and l.
[g = 10 ms-2]

Solution:

From Hooke’s law:

F = ke

F = kΔl

where k is the spring constant, l is the new length, lo is the original length, and F = mg.

Total mass = mass of pan (m) + added mass (mo)

∴ (m + mo) g = k (l – lo)

For 0.20 kg:

(0.20 + 0.04) × 10 = k (0.055 – lo)

k (0.055 – lo) = 2.4  ………(1)

For 0.28 kg:

(0.28 + 0.04) × 10 = k (0.065 – lo)

k (0.065 – lo) = 3.2 ………(2)

Divide equation 2 by 1

⇒ \( \frac{k(0.065 \: – \: l_o)}{k(0.055 \: – \: l_o)} = \frac{3.2}{2.4} \)

2.4(0.065 – lo) =  3.2(0.055 – lo)

0.156 – 2.4lo = 0.176 – 3.2lo

3.2lo – 2.4lo = 0.176 – 0.156

0.8lo = 0.02

lo = \(\frac{0.02}{0.8}\\ = \scriptsize 0.025\: m\)

from eq. 1

k (0.055 – lo) = 2.4

∴ k = \(\frac{2.4}{(0.055 \: – \: l_o)}\)

substitute lo into the equation

∴ k = \(\frac{2.4}{(0.055 \: – \: 0.025)}\)

⇒ k = \(\frac{2.4}{(0.03)}\)

∴ k = 80 Nm-1

Question 9

a. Define each of the following terms used with simple machines:

i. Pivot

Answer:

A pivot, often referred to as a fulcrum, is the point or axis around which a simple machine, such as a lever or a seesaw, rotates. It would be observed that at both the load and effort ends, linear motion is produced.

ii. Load

Answer:

A force E (Effort) is applied at one end of the bar to overcome load L at the other end.

iii. Efficiency

Answer:

This is the ratio of work output to the work input of a machine expressed as a percentage.

 

b. A truck of mass 1.2 × 103 kg is pulled from rest by a constant horizontal force of 25.2 N on a levelled road. If the maximum speed attainable in the process is 60 km/h.

Calculate the:

i. work done by the force;

Solution

m = 1.2 × 103 kg, F = 25.2 N, v = 60 km/h

Convert v in km/h to m/s

v = 60 km/hr = \( \frac{60 \: \times \: 1000}{3600} \\ = \scriptsize 16.67\: ms^{-1} \)

Work done = Change in Kinetic energy

Work done = \( \frac{1}{2} \scriptsize mv^2 \)

Work done = \( \frac{1}{2} \scriptsize \: \times \: 1.2 \: \times \: 10^3 \: \times (16.67)^2\\ = \scriptsize 1.67 \: \times \: 10^5 \: J \)

 

ii. distance travelled by the truck in reaching the maximum speed.

Solution

Work done = Force × displacement

Work done = F × d

25.2 × d = 1.67 × 105 J

d = \( \frac{1.67\:\times\:10^5}{25.2} \)

d = \( \scriptsize 6.63 \: \times \: 10^3\:m \)

d = \( \scriptsize 6.6 \:km \)

 

c. State two differences between absolute zero temperature and ice point.

Answer:

Absolute Zero: Ice Point:
Temperature at which all molecules are at rest. Temperature at which the vibratory molecular
motion still occurs.
Lowest temperature on the absolute scale. Temperature lower than 0°C, exists on the
Celsius scale.
Not a fixed point on the absolute scale. Lower fixed point on the Celsius scale.
Has a value of 0K / -273.15°C Has a value of 0°C / 273.15 K

 

d. An uncalibrated liquid-in-glass thermometer was used in determining a Celsius temperature. The readings are tabulated below

Temperature °C -6 0 100
Length of column / cm L 2.0 15.0

Determine the value of L.

Solution

⇒ \( \frac{L_{\theta} \: – \: L_0}{L_{100} \: – \: L_0}= \frac{\theta}{100} \)

⇒ \( \frac{L \: – \: 2}{15 \: – \: 2}= \frac{-6}{100} \)

⇒ \( \frac{L \: – \: 2}{13}=\scriptsize -0.06 \)

⇒ L – 2 = -0.78

⇒ L = 2 – 0.78

⇒ L = 1.22 cm

Question 10

a. i. Why are parabolic mirrors suitable for use in headlamps of vehicles?

Answer:

They produce a parallel beam of light with the same intensity over a long distance.

Parabolic mirrors form strong reflected parallel beams of constant intensity over a long distance.

 

ii.  Draw a ray diagram to illustrate the answer in 10(a)(i).

 

b. i. State two applications of echoes.

Answer:

1. Measuring the Speed of Sound in Air: Measuring the time taken for the observer to hear the echo and the distance between the observer and the obstacle can be used to determine the speed of sound.

2. Oil and Gas Exploration: This is achieved by sending ultrasonic signals into the earth which are reflected by different solids and liquids under the earth. Geologists and Geophysicists interpret this to determine the types of mineral deposits at a particular location.

3. Echolocation: Bats use echolocation when they fly around in the dark. The echoes help the bats to know what objects are around them, therefore preventing them from flying into trees and other objects and helping them hunt and identify prey. Echolocation is used by other animals such as dolphins, toothed whales and some nocturnal birds.

4. Detection of a Submarine: Warships use sonar sound systems or devices to detect objects underwater. For example, during conflict, warships can detect underwater enemy submarines. Fishermen can also use sonar to detect fish.

 

i. An observer standing at a point, P, on the same horizontal ground as the foot, H, of a tower, shouts, and 1.20 s later, he hears the echo. He then moved to another point, Q, 40 m from P, and shouted again but the echo was heard after 1.45 s. Calculate the:

I. distance between P and H;

Solution

At point P,

v = \( \frac{2PH}{t} \\ = \frac{2d}{t} \)

t = 1.20 s

∴ v = \( \frac{2d}{1.2} \) …..(1)

 

At Q,

v = \( \frac{2d \: + \: 40}{1.45} \) ……(2)

equate (1) and (2)

⇒ \( \frac{2d}{1.2}  = \frac{2d \: + \: 40}{1.45} \)

⇒ \( \frac{1.2}{1.45}   = \frac{2d}{2d \: + \: 40}\)

⇒ \( \frac{1.2}{1.45}   = \frac{\not{2}d}{\not{2}d \: + \: 40}\)

⇒ \( \frac{1.2}{1.45}   = \frac{d}{d \: + \: 40}\)

⇒ \( \scriptsize 1.45d = 1.2d \: + \: 48\)

⇒ \( \scriptsize 1.45d \: -\: 1.2d = 48\)

⇒ \( \scriptsize 0.25d = 48\)

⇒ \( \scriptsize d = \normalsize \frac{48}{0.25} \)

⇒ \( \scriptsize d =192 \: m \)

∴ distance between P and H = 192 m

 

II. speed of sound in air.

Solution

v = \(  \frac{2d}{t} \)

v = \(  \frac{2 \: \times \: 192}{1.2} \)

v = \(  \frac{384}{1.2} \)

v = \(  \scriptsize 320\: ms^{-1} \)

 

c. i. Define the term absolute refractive index of a medium.

Answer:

The absolute refractive index of a medium is the ratio of the speed of light in a vacuum/air to the speed of light in that medium or the ratio of the sine of angle of incidence to the ratio of the sine of angle of reflection in the medium.

 

ii. A piece of coin falls accidentally into a tank containing two immiscible liquids A and B as illustrated in the figure below:

Calculate the displacement of the coin when viewed vertically from above. [refractive index of A = 1.3, refractive index of B = 1.4]

Solution

Displacement, d = real depth – apparent depth

but refractive index, \( n = \frac{real \: depth}{apparent\: depth}\)

∴ apparent depth = \(  \frac{real \: depth}{n}\)

∴ d = real depth – \(  \frac{real \: depth}{n}\)

∴ d = \( \scriptsize real\: depth\left(\scriptsize 1 \: – \: \normalsize \frac{1}{n}\right) \)

d = \( \scriptsize t \left( \scriptsize 1 \: – \:  \normalsize \frac{1}{n}\right) \)

With respect to liquid A

  • real depth = 40 cm = 0.4 m
  • refractive index of A = 1.3

dA = \( \scriptsize 0.4\left( \scriptsize 1 \: – \:  \normalsize \frac{1}{1.3}\right)  \\ =\scriptsize 0.092 \: m = 9.2 \: cm \)

With respect to liquid B

  • real depth = 8 cm = 0.08 m
  • refractive index of B = 1.4

dB = \( \scriptsize 0.08\left( \scriptsize 1 \: – \:  \normalsize \frac{1}{1.4}\right)  \\ =\scriptsize 0.023 \: m = 2.3 \: cm \)

∴ The total displacement from the bottom of the tank:

dT = dA + dB

dT = 0.092 + 0.023 = 0.115 m

∴ dT = 0.115 m or 11.5 cm

Question 11

a. i. Define the electric potential at a point in an electric field.

Answer:

Electric potential (V) at a point P is the work done in bringing a unit positive charge from infinity to that point in the electric field of another charge.

 

ii.  An uncharged body, A, was charged electrostatically by a test charge, B, using the method of induction and the method of contact. State two differences between the two methods.

Answer:

Induction Contact
B never touches A B touches A
Polarity of B is opposite to that of A
eventually.		 Polarity of B is the same as A
eventually.
Charge of B is repelled to the opposite
end of A.		 Charge on B is transferred to same
size of contact.
Requires earthing Does not require earthing.

 

b. An important precaution during an electricity experiment is to open the circuit when no readings are being taken. Give two reasons for the stated precaution.

Answer:

  • To prevent/minimize overheating (in the load)
  • To prevent the cell from draining quickly.
  • To allow the depolarizer to recover.

 

c. i. The Figure below is a circuit diagram in which a coil of inductance, L, and a resistor of resistance, R, are connected to a variable alternating source of frequency, f.

The table shows the square of the impedance, Z2, corresponding to each value of f2;

f2/Hz2 198.8 400.0 600.3 800.9 900.0
Z2/Ω2 249.6 400.0 552.3 702.3 800.0

Write down the equation for Z in terms of f2, R2, and L2.

Solution

⇒ ZRL =  \( \scriptsize \sqrt{R^2 \; + \; X_L^2} \)

but \( \scriptsize X_L = 2\pi f L \)

⇒ ZRL =  \( \scriptsize \sqrt{R^2 \; + \; \left(2\pi f L\right)^2} \)

⇒ ZRL =  \( \scriptsize \sqrt{R^2 \; + \; 4\pi^2 f^2 L^2} \)

 

(ii) Plot a graph of Z2 against f2 and use it to deduce:

(I) L
(II) R

Given ⇒ \(\scriptsize \pi^2 = 10 \)

Working:

 

slope, m  =  \(\frac{Z_2^2 \: – \: Z_1^2}{f_2^2 \: – \: f_1^2}\\ =\frac{700 \: – \: 200}{800 \: – \: 130}\\ = \frac{500}{670}\\ = \scriptsize 0.746 \)

⇒ ZRL =  \( \scriptsize \sqrt{R^2 \; + \; 4\pi^2 f^2 L^2} \)

∴  \( \scriptsize Z_{RL}^2 = R^2 \; + \; 4\pi^2 f^2 L^2 \)

∴  \( \scriptsize Z_{RL}^2 =  4\pi^2 L^2 f^2  \; + \; R^2\)

(I) L

Compare with:

⇒ \( \scriptsize y = mx \:+\: c\)

m = \( \scriptsize 4\pi^2 L^2 \)

⇒ \( \scriptsize L^2 = \normalsize \frac{m}{4\pi^2 } \)

but from the question, \(\scriptsize \pi^2 = 10 \)

∴  \( \scriptsize L^2 = \normalsize \frac{m}{4(10)} \)

⇒ \( \scriptsize L^2 = \normalsize \frac{m}{40} \)

⇒ \( \scriptsize L = \normalsize \sqrt{ \frac{m}{40}} \)

from the graph m = 0.746

∴  \( \scriptsize L = \normalsize \sqrt{ \frac{0.76}{40}} \\ = \scriptsize \sqrt{0.019} \\ = \scriptsize 0.138 \:H \: or \: 138 \:mH \)

 

(II) R = \( \scriptsize \sqrt{intercept} \)

R = \( \scriptsize \sqrt{100} \\ = \scriptsize  10 \: \Omega \)

Question 12

a. State the function of each of the following parts of a modern x-ray tube:

i. heater;

Answer:

  • Heats up the filament to release thermo-electrons.

 

ii. high tension source;

Answer:

  • It accelerates the liberated thermo-electrons towards the target.

 

iii. cooling fins;

Answer:

  • It reduces the intense heat produced at the target by radiation.

 

b. State one reason for each of the following design features of a modern x-ray tube:

i. the glass envelope is highly evacuated;

Answer:

  • To prevent gaseous ionization in the tube.

 

ii. the target is a metal of very high melting point;

Answer:

  • To withstand the high energy generated as a result of the collisions of the thermo-electrons.
  • To prevent it from melting.

 

iii. the cooling fins are located outside the glass envelope;

Answer:

  • To prevent a possible break up/ crack of the glass envelope due to heat generated.
  • To increase the rate of cooling.

 

c. In a nuclear fission reaction, a nuclide \( \scriptsize ^{235}_{92}U \) is bombarded with a neutron to produce \( \scriptsize ^{93}_{36}Kr \: and \: ^{141}_{56}Ba\) with additional neutrons, the energy involved in the process is Q.

  • \( \scriptsize mass \: of \: ^{235}_{92}U = 235.044\: u\)
  • \( \scriptsize mass \: of \: ^{93}_{36}Kr = 91.898 \:u\)
  • \( \scriptsize mass \: of \: ^{141}_{56}Ba = 140.914 \:u \)
  • mass of neutron = 1.009 u
  • 1 u = 1.66 × 1027 kg
  • c = 3.0 × 108 ms-1

i. Write down the balanced nuclear reaction equation for the process.

Answer:

⇒ \( \scriptsize \underset{uranium}{_{92} ^{235} \textrm {U}}\: + \: \underset{neutron}{_{0} ^{1} \textrm {n}} \: \rightarrow \: \underset{barium}{_{56} ^{141} \textrm {Ba}} \: + \: \underset{krypton}{_{36} ^{93} \textrm {Kr}} \: + \: \underset{neutrons}{2\: _{0} ^{1} \textrm {n}} \:+ \:Q \)

 

ii. State with reason whether Q is absorbed or released in the process.

Solution/Answer:

Sum of reactant masses:

= (235.044 + 1.099) = 236.053 u

 

Sum of product masses:

= [91.898 + 140.914)  + 2(1.009) = 234.830 u

Q is released because the sum of the masses of reactants is greater than the product masses.

 

iii. Calculate the value of Q in joules.

Solution

Δm = 236.053 u – 234.830 u

Δm = 1.223 u

Q = \( \scriptsize mc^2 \\ \scriptsize = 1.223 \: \times \: 1.66 \: \times \: 10^{-27} \: \times \: \left( 3.0 \: \times \: 10^8 \right)^2\\ \scriptsize  = 1.83 \: \times \: 10^{-10} \: J  \)

 

iv. State three differences between nuclear fusion and nuclear fission.

Answer:

Nuclear Fission: Nuclear Fusion:
This is the breaking up of the nucleus of a heavy
atom or element into nearly two equal parts.		 Two or more light nuclei combine, at
high temperatures, to form a heavier nucleus.
Less energy produced. Greater energy produced.
By-products are radioactive. By-products are non-radioactive.
The raw material for fission is more expensive. The raw material for fusion is cheaper.
Lower initiation temperature. Higher initiation temperature.
Fission process is more dangerous. Fusion process is less dangerous.
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