Quiz 5 of 13

2018 Physics WAEC Theory Past Questions

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(1) (a) Define strain.

Solution

Strain is defined as the ratio of the extension to the original length (e/l).

(b) A rubber band is stretched to twice its original length. Calculate the strain on the rubber band.

Solution

Let original length be X

new length be 2X

extension = 2X - X= X

Strain = \( \frac{e}{l} = \frac{x}{x} \)

Strain =

(2) State three materials used for making optical fibres

Solution
- Silica
- Germanra
- Fluoroaluminate

(3) Name three classes of magnetic materials

Solution

- Diamagnetism
- Ferromagnetism
- Paramagnetism

(4) (a) What is an intrinsic semi-conductor?

Solution
An intrinsic semiconductor is an undoped semiconductor, pure, without any significant dopant species present.

 

(b) Distinguish between the p-type and n-type semiconductors

Solution

P-type semiconductors are these produced by doping with indium or boron so as to produce an impure semiconductor with more holes or positive charges than electrons WHILE n-type semiconductors are those produced by doping with germanium or arsenic so as to produce an impure semiconductor with more free electrons or negative charges than positive charges.

(5) A missile is projected so as to attain its maximum range. Calculate the maximum height attained if the initial velocity of projection is 200 ms-¹ (Leaving the answer in whole number)

Solution

V = 200ms-1

g = 10ms-2

H = \(  \frac{u^2 sin^2 \theta}{2g} \)

θ = 45º (since maximum range is obtained at this angle)

H = \( \frac{2w^2 sin^2 45}{2g} \)

= \(  \frac{400 \: \times \: 0.5}{20} \)

= 1000m

(6) A black body radiates maximum energy when its surface temperature T and the corresponding wavelength λmax are related by the equation λmax T = constant Given the values of the constant and surface temperature as 2.9 x 10-3 mK and 57ºC respectively. Calculate the frequency of the energy radiated

Solution

λmax T = constant

T = 57ºC = 330K

This question was unpopular among candidates. Many candidates used the formula λmax T = constant velocity to solve the question and the rest failed because of their failure to convert 57ºC to Kelvin

The expected answer is

λmax  = \( \frac{2.9 \: \times \: 10^{-3}}{330} \)

= \( \scriptsize 8.8 \: \times \: 10^{-6}m\)

c = fλ

or

f = \( \frac{c}{\lambda} \)

where c = velocity of light = 3.0 x 108m/s

λ = \( \scriptsize 8.8 \: \times \: 10^{-6}m\)

f = \(\frac{3.0 \: \times \: 10^{8}}{8.8 \: \times \: 10^{-6}} \)

= \( \scriptsize 3.4 \: \times \: 10^{13}Hz\)

(7) (a) What does the acronym LASER stands for?

Solution

Light Amplification by Stimulated Emission of Radiation

(b) What is a laser?

Solution
A laser is a device that emits light through a process of optical amplification based
on the stimulated emission of electromagnetic radiation.

(8) (a) Define uniform acceleration

Solution

Uniform acceleration can be defined as when the velocity of a moving body
increases by equal amounts in equal time intervals.

(b) (i) Forces act on a car in motion. List the: horizontal forces and their directions;

Solution

Frictional force - opposite driving force
Driving force – forward

b(ii) List the: vertical forces and their directions

Solution

Gravitational force - Downward
Reactive force – Upward

(8) (c) (i) A car starts from rest and accelerates uniformly for 20 s to attain a speed of 25 ms-2. It maintains this speed for 30 s before decelerating uniformly to rest. The total time for the journey is 60 s.
Sketch a velocity-time graph for the motion.

Solution

(8) (c) (ii) a. Use the graph to determine the total distance travelled by the car (Leaving the answer in whole number)

Solution

Distance travelled = Area of the trapezium
= ½(RS + PQ) × h
=½(60 + 30) × 25
=½ × 90 × 25
= 1125m

(8) (c) (ii) b. deceleration of the car. (Leaving the answer in three decimal place)

Solution

Deceleration = \( \frac{QT}{TS}= \frac{25}{10}\)

= 2.5 ms-1

 

(d)

The figure above illustrates force - extension graph for a stretched spiral spring. Determine the work done on the spring. (Leaving the answer in three decimal places)

Solution

Work done = Area of Triangle
= ½ × base × height
= ½ × 0.5 × 12
= 3Ncm-¹ or 0.003Nm-¹

(a) List two factors each that affect heat loss by:

(i) radiation;

Solution

- Surface area
- Rate of atomic activity

a (ii) convection.

Solution

-Nature/type of fluid
-Density/viscosity of fluid
-Thermal conductivity of fluid
=Specific heat capacity of fluid
=Exposed surface area

 

(b) State two factors that determine the quantity of heat in a body.

Solution

- Specific heat capacity of the body
- Temperature change of the body

 

(c) Explain the statement: The specific latent heat of vaporization of mercury is 2.72 x 10⁵ J kg-1

Solution

It means 2.72 × 10⁵ J is required to vaporise a unit mass of mercury without a change in temperature.

 

(d) (i) A jug of heat capacity 250 JK-1 contains water at 28°C. An electric heater of resistance 35 ohms connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surrounding, calculate the: Mass of water in the jug before heating

Solution

Heat supplied by heater = Heat gained by water +  Heat gained by jug

Heat supplied by heater = ivt = \( \frac{v^2t}{R} \)

Heat gained by water = \(  \scriptsize M_w C_w (\theta_2 \: - \: \theta_1)\)

Where Mw = mass of water

Cw = specific heat capacity of water

θ2 = final temp. of water

θ1 = initial temp. of water

Heat gained by Jug = \(  \scriptsize C_j (\theta_2 \: - \: \theta_1)\)

Cj = specific heat capacity of jug

\( \frac{v^2t}{R} =   \scriptsize M_w C_w (\theta_2 \: - \: \theta_1) \: + \: C_j (\theta_2 \: - \: \theta_1)\)

\( \frac{220^2 \: \times \: 4 \: \times \: 60}{35}  =  \scriptsize M_w \: \times \: 4200 (100 \: - \: 28) \: + \: 250 (100 \: - \: 28)\)

Mw = 1.038kg

 

(d) (ii) A jug of heat capacity 250 JK-1 contains water at 28°C. An electric heater of resistance 35 ohms connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surrounding, calculate the:  specific latent heat of vaporization

Solution

\( \scriptsize M_v L_v = \normalsize \frac{v^2t}{R} \)

\( \scriptsize  L_v = \normalsize \frac{200^2 \: \times \: 5 \: \times \: 60}{35 \: \times \: 0.3} \)

= \( \scriptsize 1.38  \: \times \: 10^6 Jkg^{-1} \)

(10) (a) Define diffraction

Solution

Diffraction is the ability of waves to bend around obstacles in their path.

(b) (i) Explain critical angle

Solution

Critical angle is the angle of incidence in the denser medium when the angle of refraction in the less dense medium is 90°

(b) (ii) (a)

The diagram above illustrates a ray of light passing through a rectangular transparent plastic block.

Determine the value of the critical angle.

Solution

Critical angle = Critical angle = 90° - 44° = 46°

(b) (ii) (b)

The diagram above illustrates a ray of light passing through a rectangular transparent plastic block.

Calculate the refractive index of the block.

Solution

Refractive index = \( \frac{1}{sin C} \)

Refractive index =\( \frac{1}{sin 46} \)

= 1.39

 

(c) (i) A pipe closed at one end has a fundamental frequency of 200 Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe.

Calculate the: fundamental frequency of the open pipe
(Leaving the answer in whole number)

Solution

fo = 200Hz

First overtone = 3fo= 3 × 200 = 600Hz

1st overtone (open pipe) = 600Hz

Fundamental overtone = \( \frac{1st \: overtone}{2} \)

\( \frac{600}{2} \)

= 300Hz

 

(c) (ii) length of the closed pipe

(Leaving the answer in three decimal places)

Solution

fo = \( \frac{V}{4l} \)

200 = \( \frac{330}{4l} \)

l = \( \frac{330}{200}  \scriptsize \: \times \: 4\)

l = 0.413m

 

(c) (iii) length of the open pipe
(Leaving the answer in two decimal places)

Solution

f1 = \( \frac{V}{l} \)

600 = \( \frac{330}{l} \)

l = \( \frac{330}{600} \)

l = 0.55m

(11) (a) (i) Define reactance

Solution

Reactance can be defined as the opposition to the flow of a.c offered by a capacitor or an inductor or both.

(a) (ii) Define impedance in an a.c circuit

Solution

Impedance is the overall opposition of a mixed circuit containing a resistor an inductor and/or a capacitor.

(b) (i)

The diagram above illustrates an a.c. generator. When the coil is rotated, e.m.f. is induced in the coil. Explain why an e.m.f. is induced

Solution

An e.m.f is induced because the plane or the coil is parallel to the field, thereby causing a change of flux at maximum.

 

(b) (ii)

The diagram above illustrates an a.c. generator. When the coil is rotated, e.m.f. is induced in the coil.
State the purpose of the slip-rings

Solution

The purpose of the slip rings is to obtain an alternating voltage.

 

(b) (iii)

The diagram above illustrates an a.c. generator. When the coil is rotated, e.m.f. is induced in the coil. Name and state the law used to determine the direction of the induced current

Solution

Fleming's Right-Hand Rule: It states that if the thumb, forefinger, and middle finger of the right hand are held mutually at right angles to each other, with the fore-finger pointing in the direction of motion, then the middle finger will point in the induced e.m.f or current direction.

 

(b) (iv) State two ways to increase the induced e.m.f.

Solution

- Increase the number of turns in the coil
- Increase the area of the coil.

 

(c) A lamp is rated 12 V, 6 W. Calculate the amount of energy transformed by the lamp in 5 minutes.

Solution

E= Power × time

∴ = 6 × 5 ×60

= 1800J

(12) (a) Define binding energy in an atom

Solution

Binding energy is the minimum work/energy required to seperate the nucleons of an atom.

 

(b) List three pieces of evidence to support the claim that X-rays are electromagnetic waves.

Solution

Required Evidence:

- Not affected by electric/magnetic fields
- Can travel through a vacuum
- Travel with the speed of light
- Can be polarized
- Cause Fluorescence

 

(c) List three peaceful uses of nuclear energy

Solution

- It is used to generate electricity
- Treatment of tumor
- Power submarine/rockets

(d)Light of wavelength 4.5 x 10-7in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10-9J, calculate the:

(i) frequency of the incident light;

(ii) energy of the incident light;

(iii) energy of the photoelectrons. [Speed of light = 3.0 x 108ms-1, h = 6.6 x 10-34 Js]

Solution d(i)

c = \( \scriptsize f \lambda \)

or

f = \( \frac{c}{\lambda} \)

f = \( \frac{3.0 \: \times \: 10^8}{4.5 \: \times \: 10^{-7}} \)

f = \( \scriptsize 6.67 \: \times \: 10^{14} Hz\)

 

Solution d(ii)

E = hf

= \( \scriptsize 6.6 \: \times \: 10^{-34}\: \times \:  6.67 \: \times \: 10^{14} \)

= \( \scriptsize 4.4 \times \: 10^{-19} J\)

 

Solution d(ii)

K.E = E - W

= \( \scriptsize 4.4 \: \times \: 10^{-19}\: \times \:  3.0 \: \times \: 10^{-19} J\)

=  = \( \scriptsize 1.4 \: \times \: 10^{-19} J\)

(1) (a) Define strain.

Solution

Strain is defined as the ratio of the extension to the original length (e/l).

(b) A rubber band is stretched to twice its original length. Calculate the strain on the rubber band.

Solution

Let original length be X

new length be 2X

extension = 2X - X= X

Strain = \( \frac{e}{l} = \frac{x}{x} \)

Strain =

(2) State three materials used for making optical fibres

Solution
- Silica
- Germanra
- Fluoroaluminate

(3) Name three classes of magnetic materials

Solution

- Diamagnetism
- Ferromagnetism
- Paramagnetism

(4) (a) What is an intrinsic semi-conductor?

Solution
An intrinsic semiconductor is an undoped semiconductor, pure, without any significant dopant species present.

 

(b) Distinguish between the p-type and n-type semiconductors

Solution

P-type semiconductors are these produced by doping with indium or boron so as to produce an impure semiconductor with more holes or positive charges than electrons WHILE n-type semiconductors are those produced by doping with germanium or arsenic so as to produce an impure semiconductor with more free electrons or negative charges than positive charges.

(5) A missile is projected so as to attain its maximum range. Calculate the maximum height attained if the initial velocity of projection is 200 ms-¹ (Leaving the answer in whole number)

Solution

V = 200ms-1

g = 10ms-2

H = \(  \frac{u^2 sin^2 \theta}{2g} \)

θ = 45º (since maximum range is obtained at this angle)

H = \( \frac{2w^2 sin^2 45}{2g} \)

= \(  \frac{400 \: \times \: 0.5}{20} \)

= 1000m

(6) A black body radiates maximum energy when its surface temperature T and the corresponding wavelength λmax are related by the equation λmax T = constant Given the values of the constant and surface temperature as 2.9 x 10-3 mK and 57ºC respectively. Calculate the frequency of the energy radiated

Solution

λmax T = constant

T = 57ºC = 330K

This question was unpopular among candidates. Many candidates used the formula λmax T = constant velocity to solve the question and the rest failed because of their failure to convert 57ºC to Kelvin

The expected answer is

λmax  = \( \frac{2.9 \: \times \: 10^{-3}}{330} \)

= \( \scriptsize 8.8 \: \times \: 10^{-6}m\)

c = fλ

or

f = \( \frac{c}{\lambda} \)

where c = velocity of light = 3.0 x 108m/s

λ = \( \scriptsize 8.8 \: \times \: 10^{-6}m\)

f = \(\frac{3.0 \: \times \: 10^{8}}{8.8 \: \times \: 10^{-6}} \)

= \( \scriptsize 3.4 \: \times \: 10^{13}Hz\)

(7) (a) What does the acronym LASER stands for?

Solution

Light Amplification by Stimulated Emission of Radiation

(b) What is a laser?

Solution
A laser is a device that emits light through a process of optical amplification based
on the stimulated emission of electromagnetic radiation.

(8) (a) Define uniform acceleration

Solution

Uniform acceleration can be defined as when the velocity of a moving body
increases by equal amounts in equal time intervals.

(b) (i) Forces act on a car in motion. List the: horizontal forces and their directions;

Solution

Frictional force - opposite driving force
Driving force – forward

b(ii) List the: vertical forces and their directions

Solution

Gravitational force - Downward
Reactive force – Upward

(8) (c) (i) A car starts from rest and accelerates uniformly for 20 s to attain a speed of 25 ms-2. It maintains this speed for 30 s before decelerating uniformly to rest. The total time for the journey is 60 s.
Sketch a velocity-time graph for the motion.

Solution

(8) (c) (ii) a. Use the graph to determine the total distance travelled by the car (Leaving the answer in whole number)

Solution

Distance travelled = Area of the trapezium
= ½(RS + PQ) × h
=½(60 + 30) × 25
=½ × 90 × 25
= 1125m

(8) (c) (ii) b. deceleration of the car. (Leaving the answer in three decimal place)

Solution

Deceleration = \( \frac{QT}{TS}= \frac{25}{10}\)

= 2.5 ms-1

 

(d)

The figure above illustrates force - extension graph for a stretched spiral spring. Determine the work done on the spring. (Leaving the answer in three decimal places)

Solution

Work done = Area of Triangle
= ½ × base × height
= ½ × 0.5 × 12
= 3Ncm-¹ or 0.003Nm-¹

(a) List two factors each that affect heat loss by:

(i) radiation;

Solution

- Surface area
- Rate of atomic activity

a (ii) convection.

Solution

-Nature/type of fluid
-Density/viscosity of fluid
-Thermal conductivity of fluid
=Specific heat capacity of fluid
=Exposed surface area

 

(b) State two factors that determine the quantity of heat in a body.

Solution

- Specific heat capacity of the body
- Temperature change of the body

 

(c) Explain the statement: The specific latent heat of vaporization of mercury is 2.72 x 10⁵ J kg-1

Solution

It means 2.72 × 10⁵ J is required to vaporise a unit mass of mercury without a change in temperature.

 

(d) (i) A jug of heat capacity 250 JK-1 contains water at 28°C. An electric heater of resistance 35 ohms connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surrounding, calculate the: Mass of water in the jug before heating

Solution

Heat supplied by heater = Heat gained by water +  Heat gained by jug

Heat supplied by heater = ivt = \( \frac{v^2t}{R} \)

Heat gained by water = \(  \scriptsize M_w C_w (\theta_2 \: - \: \theta_1)\)

Where Mw = mass of water

Cw = specific heat capacity of water

θ2 = final temp. of water

θ1 = initial temp. of water

Heat gained by Jug = \(  \scriptsize C_j (\theta_2 \: - \: \theta_1)\)

Cj = specific heat capacity of jug

\( \frac{v^2t}{R} =   \scriptsize M_w C_w (\theta_2 \: - \: \theta_1) \: + \: C_j (\theta_2 \: - \: \theta_1)\)

\( \frac{220^2 \: \times \: 4 \: \times \: 60}{35}  =  \scriptsize M_w \: \times \: 4200 (100 \: - \: 28) \: + \: 250 (100 \: - \: 28)\)

Mw = 1.038kg

 

(d) (ii) A jug of heat capacity 250 JK-1 contains water at 28°C. An electric heater of resistance 35 ohms connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surrounding, calculate the:  specific latent heat of vaporization

Solution

\( \scriptsize M_v L_v = \normalsize \frac{v^2t}{R} \)

\( \scriptsize  L_v = \normalsize \frac{200^2 \: \times \: 5 \: \times \: 60}{35 \: \times \: 0.3} \)

= \( \scriptsize 1.38  \: \times \: 10^6 Jkg^{-1} \)

(10) (a) Define diffraction

Solution

Diffraction is the ability of waves to bend around obstacles in their path.

(b) (i) Explain critical angle

Solution

Critical angle is the angle of incidence in the denser medium when the angle of refraction in the less dense medium is 90°

(b) (ii) (a)

The diagram above illustrates a ray of light passing through a rectangular transparent plastic block.

Determine the value of the critical angle.

Solution

Critical angle = Critical angle = 90° - 44° = 46°

(b) (ii) (b)

The diagram above illustrates a ray of light passing through a rectangular transparent plastic block.

Calculate the refractive index of the block.

Solution

Refractive index = \( \frac{1}{sin C} \)

Refractive index =\( \frac{1}{sin 46} \)

= 1.39

 

(c) (i) A pipe closed at one end has a fundamental frequency of 200 Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe.

Calculate the: fundamental frequency of the open pipe
(Leaving the answer in whole number)

Solution

fo = 200Hz

First overtone = 3fo= 3 × 200 = 600Hz

1st overtone (open pipe) = 600Hz

Fundamental overtone = \( \frac{1st \: overtone}{2} \)

\( \frac{600}{2} \)

= 300Hz

 

(c) (ii) length of the closed pipe

(Leaving the answer in three decimal places)

Solution

fo = \( \frac{V}{4l} \)

200 = \( \frac{330}{4l} \)

l = \( \frac{330}{200}  \scriptsize \: \times \: 4\)

l = 0.413m

 

(c) (iii) length of the open pipe
(Leaving the answer in two decimal places)

Solution

f1 = \( \frac{V}{l} \)

600 = \( \frac{330}{l} \)

l = \( \frac{330}{600} \)

l = 0.55m

(11) (a) (i) Define reactance

Solution

Reactance can be defined as the opposition to the flow of a.c offered by a capacitor or an inductor or both.

(a) (ii) Define impedance in an a.c circuit

Solution

Impedance is the overall opposition of a mixed circuit containing a resistor an inductor and/or a capacitor.

(b) (i)

The diagram above illustrates an a.c. generator. When the coil is rotated, e.m.f. is induced in the coil. Explain why an e.m.f. is induced

Solution

An e.m.f is induced because the plane or the coil is parallel to the field, thereby causing a change of flux at maximum.

 

(b) (ii)

The diagram above illustrates an a.c. generator. When the coil is rotated, e.m.f. is induced in the coil.
State the purpose of the slip-rings

Solution

The purpose of the slip rings is to obtain an alternating voltage.

 

(b) (iii)

The diagram above illustrates an a.c. generator. When the coil is rotated, e.m.f. is induced in the coil. Name and state the law used to determine the direction of the induced current

Solution

Fleming's Right-Hand Rule: It states that if the thumb, forefinger, and middle finger of the right hand are held mutually at right angles to each other, with the fore-finger pointing in the direction of motion, then the middle finger will point in the induced e.m.f or current direction.

 

(b) (iv) State two ways to increase the induced e.m.f.

Solution

- Increase the number of turns in the coil
- Increase the area of the coil.

 

(c) A lamp is rated 12 V, 6 W. Calculate the amount of energy transformed by the lamp in 5 minutes.

Solution

E= Power × time

∴ = 6 × 5 ×60

= 1800J

(12) (a) Define binding energy in an atom

Solution

Binding energy is the minimum work/energy required to seperate the nucleons of an atom.

 

(b) List three pieces of evidence to support the claim that X-rays are electromagnetic waves.

Solution

Required Evidence:

- Not affected by electric/magnetic fields
- Can travel through a vacuum
- Travel with the speed of light
- Can be polarized
- Cause Fluorescence

 

(c) List three peaceful uses of nuclear energy

Solution

- It is used to generate electricity
- Treatment of tumor
- Power submarine/rockets

(d)Light of wavelength 4.5 x 10-7in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10-9J, calculate the:

(i) frequency of the incident light;

(ii) energy of the incident light;

(iii) energy of the photoelectrons. [Speed of light = 3.0 x 108ms-1, h = 6.6 x 10-34 Js]

Solution d(i)

c = \( \scriptsize f \lambda \)

or

f = \( \frac{c}{\lambda} \)

f = \( \frac{3.0 \: \times \: 10^8}{4.5 \: \times \: 10^{-7}} \)

f = \( \scriptsize 6.67 \: \times \: 10^{14} Hz\)

 

Solution d(ii)

E = hf

= \( \scriptsize 6.6 \: \times \: 10^{-34}\: \times \:  6.67 \: \times \: 10^{14} \)

= \( \scriptsize 4.4 \times \: 10^{-19} J\)

 

Solution d(ii)

K.E = E - W

= \( \scriptsize 4.4 \: \times \: 10^{-19}\: \times \:  3.0 \: \times \: 10^{-19} J\)

=  = \( \scriptsize 1.4 \: \times \: 10^{-19} J\)