2016 Physics WAEC Theory Past Questions

Question 1

(a) State the dimension of impulse.

Answer

Impluse = mass × velocity

Impluse = mass × \(\frac{displacement}{time}\)

Impluse = M × \(\frac{L}{T}\)

= M × LT^-1

= MLT^-1

(b) State the dimension of acceleration

Answer

Acceleration = \(\frac{velocity}{time}\)

Acceleration = \(\frac{displacement}{time} \: \times \: \frac{1}{time}\)

Acceleration = \(\frac{L}{T} \: \times \: \frac{1}{time}\)

Acceleration = \(\frac{L}{T^2} \)

∴ Acceleration = LT^-2

(c) State the dimension of work

Answer

Work = Force × distance

Work = Mass × Acceleration × distance

Work = M × LT^-2 × L

∴ Work = ML² T^-2

Question 2

A projectile is fired with a velocity of 20 ms^-1 at an angle of 40° to horizontal. Determine the components of the velocity of the projectile at its maximum height.

Solution

V_x = U cos θ = 20 cos 40° = 15.32 ms^-1

The vertical component is 0 at max. height,

i.e. V_y = 0

Question 3

State three different materials that can be used to demonstrate Brownian motion.

Answer

• smoke particles
• carbon particles
• dust particles
• pollen particles
• chalk particles
• potassium permanganate crystals

Question 4

An electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12 T . This results in a magnetic force of 9.6 × 10^-12 N on the electron. Calculate the speed of the electron as its enters the magnetic field.

(e = 1.6 ×10^-19 C ).

Solution

F = e VB sin θ

9.6 × 10^-12 = 1.6 × 10^-19 × V × 0.12× sin 90^o

V = \(\frac{9.6 \: \times \: 10^{-12}}{1.6 \: \times \: 10^{-19} \: \times \: 0.12\: \times \: 1}\)

V  = 5.0 × 10⁸ ms^-1

Question 5

List three uses of rockets.

Answer

• space exploration
• space travel
• warfare
• fireworks/entertainment
• launching of (artificial) satellites (into orbits)

Question 7

The diagram above illustrates a cathode ray tube.
Identify the components X, Y, and Z

Answer

X – filament/heater/cathode
Y – anode
Z – plate(s)/deflector(s)

Question 8

(a) Explain the term net force.

Answer

Net force is the effective force resulting from the actions of a system of forces on a body.

OR

Net force is an unbalanced force that produces an acceleration of a body.

(b) Define the principle of conservation of linear momentum and state one example of it.

Answer

In an isolated/closed system of colliding bodies, the total linear momentum in a fixed direction remains unchanged/constant.

Example;

• rocket propulsion;
• recoil of gun
• colliding trolleys.

(c) A ball of mass 200 g released from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 m. Calculate the impulse received by the floor.
(g = 10 ms^-2)

Solution

Impulse = mass × [(Final velocity) – (Initial velocity)]

• Initial velocity is the velocity with which the ball hits the ground
• Final velocity is the velocity of the rebound

Let’s calculate the velocity with which the ball hits the ground.

• h₁ = 2 m
• g = 10 m/s²
• u = 0

v₁² = u₁² + 2gh₁

v₁²  = 0 + (2 × 10 × 2)

v₁² = 40

v₁ = \(\scriptsize \sqrt{40}\)

v₁ = 6.325 ms^-1

Next, calculate the velocity of the rebound;

• h₂ = 1.8 m
• v = 0
• (-ve g because the direction is upwards)

v²₂ = u₂² – 2gh₂

0  = u₂² – 2 × 10 × 1.8

u₂² = 36

u₂ = \(\scriptsize \sqrt{36}\)

u₂ = 6.000 ms^-1

Impulse = change in momentum

= mv₁ – (-mu₂) = m(v₁ + u₂)

=0.2(6.325 + 6.000)

= 2.46 Ns or 2.46 kgm/s

(d) (i) A body of mass 20 g performs a simple harmonic motion at a frequency of  5 Hz . At a distance of 10 cm from the mean position, its velocity is 200 cms^-1. Calculate its maximum displacement from the mean position;
(g = 10 ms^-2 ; π = 3.14)

Solution

v =  \(\scriptsize \omega \sqrt{r^2 \: – \: y^2}\)

= \(\scriptsize 2 \pi f \sqrt{r^2 \: – \: y^2}\)

2 = \(\scriptsize 2 \:  \times \: 3.14 \: \times \: 5 \: \times \:   \sqrt{r^2 \: – \: 0.1^2}\)

r = 0.12 m

(ii) A body of mass 20 g performs a simple harmonic motion at a frequency of 5 Hz . At a distance of 10 cm from the mean position, its velocity is 200 cms^-1.

Calculate its maximum velocity;
(g = 10 ms^-2 ; π = 3.14)

Solution

V_max = ωr = 2πfr

= 2 × 3.14 × 5 × 0.12

= 3.77 ms^-1

(iii) A body of mass 20g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms^-1. Calculate its maximum potential energy.
(g = 10ms^-2 ; π = 3.14)

Solution

P.E = \(\frac{1}{2} \scriptsize mw^2r^2\)

P.E = \(\frac{1}{2} \scriptsize m(2 \pi f)^2r^2\)

P.E = \(\frac{1}{2} \scriptsize \: \times \:  0.02  \: \times \:   (2 \: \times \: 3.14 \: \times \: 5)^2\: \times \: (0.12)^2\)

= 1.42 × 10^-1 J

Question 9

(a) Explain the term:

(i) thermal equilibrium;

Solution

Thermal equilibrium is a condition in which two bodies in contact have no net flow of heat between them

(ii) Explain the term: fundamental interval.

Answer

Fundamental interval is the difference in temperature between the lower fixed point (temperature of melting ice at standard atmospheric pressure) and the upper fixed point (temperature of steam at standard atmospheric pressure).

(b) List two uses of hydraulic press.

Answer

– lifting of objects;
– compressing metal sheets;
– compressing soft materials into bales;
– compressing paper in printing industries.

(c) Name the material used to reset the steel index in the Six’s maximum and minimum thermometer.

Answer

Magnet

(d) (i) A nursing mother prepared her baby’s milk mixture at 85°C, in a feeding bottle. In order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity 90 J K ^-1 containing 500 g of water at 26°C. If the mass of the mixture is 300 g, calculate the specific heat capacity of the mixture.

[Neglect the heat losses and heat capacity of the bottle;
specific heat capacity of water = 4200 J kg^-1 K^-1]

Solution

Heat lost by mixture = Heat gained by water + Heat gained by aluminium

M_m c_m∆θ_m = M_w c_w∆θ_w + c_a∆θ_a

0.3 × c_m × (85 – 40) = 0.5 × 4200 × (40 – 26) + 90 (40 – 26)

13.5c_m = 29400 + 1260

c_m = \(\frac{29400+1260}{13.5}\)

c_m = 2271.1 JKg^-1K^-1

(ii) A nursing mother prepared her baby’s milk mixture at 85°C, in a feeding bottle. In order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity 90 J K ^-1 containing 500 g of water at 26°C. If the mass of the mixture is 300 g, calculate the specific heat capacity of the mixture.

(*) Name two ways through which the bottle loses heat

Answer

– Conduction;
– Convection.

(**) Name two industrial processes in which heat exchange is used

Answer

– incubator;
– cooling by radiators;
– generation of electricity;
– refrigeration;
– air – conditioning;
– sewage treatment.

Question 11

(a) Explain briefly dielectric strength.

Answer

Dielectric strength is the maximum potential gradient/electric field strength an insulator can withstand without breaking down. Measured in Vmm^-1/Vm^-1

(b) An electromagnetic wave has its wavelength shorter than those of radiowaves and microwaves but longer than that of visible light.

(i) Identify the wave.

Answer

The wave is infrared.

(ii) Name one suitable detector for the wave.

Answer

Suitable detectors can be:

• Infra-red camera;
• Thermopile
• Bolometer
• Photographic film

(iii) Name one source of the wave.

Answer

Sources of the wave can be:

• Hot bodies
• Sun

(c) An oil drop carrying a charge of 1.0 × 10^-19 C is found to remain at rest in a uniform electric field of intensity 1200 N C ^-1. Calculate the weight of the oil drop.

Solution

E =  \(\frac{F}{q} = \frac{W}{q}\)

OR

W = mF = Eq

= 1200 × 1.0 × 10^-19

= 1.2 × 10^-16 N

(d) (i) An RLC series circuit consists of a 100 Ω resistor, 0.05 H inductor, and a 25 μF capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the impedance.

Solution

X_L = 2πfL

= 2 × 3.14 × 50 × 0.05

= 15.7 Ὠ

X_C = \(\frac{1}{2 \pi f C}\)

X_C = \(\frac{1}{2 \: \times \: 3.14 \: \times \: 50 \: \times \: 25 \: \times \: 10^{-6}}\)

X_C = 127.4 Ὠ

i. Z = \(\scriptsize \sqrt{R^2 \: + \: (X_L \: -\: X_C)^2}\)

Z = \(\scriptsize \sqrt{100^2 \: + \: (12.74 \: -\: 15.7)^2}\)

Z = \(\scriptsize 149.9 \:  \Omega\)

(ii) An RLC series circuit consists of a 100 Ω resistor, 0.05 H inductor and a 25 μF capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the current.

Solution

I = \(\frac{V}{Z}\)

I = \(\frac{220}{149.9}\)

I = 1.47 A

Question 12

(a) (i) Explain the following term; mass defect;

Answer

Mass defect is the difference between the mass of a nucleus and the total mass of its nucleons.

– expressed in atomic mass unit ;
– total mass of nucleons – mass of nucleons;
– it is a measure of binding energy

(ii) Explain the following term: binding energy of a nucleus.

Answer

Binding energy is the work done/energy needed to separate the nucleus.

– measured in eV or J.  

(b) (i) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its;

(*) wavelength: 

Answer

wavelength increases

(**) energy:

Answer

energy decreases  

(ii) A particle of mass 4.4 × 10^-23 kg moves with a velocity of 10⁵ ms^-1, calculate its wavelength.

(h = 6.6 × 10^-34 Js)

Solution

λ = \(\frac{h}{mv}\)

λ = \(\frac{6.6 \: \times \: 10^{-34}}{4.4 \: \times \: 10^{-23} \: \times \: 10^5}\)

λ = 1.5 × 10^-16 m  

(c)

The diagram above shows part of a radioactive decay series. Use it to answer the following questions;

(i) Name a pair of isotopes.

Answer

⇒ \(\scriptsize ^{212}_{84} Po \: and \: ^{216}_{84} Po\)

⇒ \(\scriptsize ^{288}_{90}Th \: and \: ^{232}_{90}Th\)

⇒ \(\scriptsize ^{288}_{82}Pb \: and \: ^{212}_{82}Pb\)

Any one  

(ii) Name the isotope with which the series starts.

Answer

⇒ \(\scriptsize ^{232}_{90}Th\)

and

⇒ \(\scriptsize ^{228}_{90}Th\)  

(iii) Write down a nuclear equation for two examples of each of

(*) alpha decay;

Answer

⇒ \(\scriptsize ^{232}_{90}Th \: \rightarrow \: ^{228}_{88}Ra \: + \: ^{4}_{2}He \: + \: energy\)

⇒ \(\scriptsize ^{212}_{84}Po \: \rightarrow \: ^{208}_{82}Pb \: + \: ^{4}_{2}He \: + \: energy\)  

(**) beta decay;

Answer

⇒ \(\scriptsize ^{212}_{83}Bi \: \rightarrow \: ^{212}_{84}Po\: + \: ^{0}_{-1}e \: + \: energy\)

⇒ \(\scriptsize ^{224}_{88}Ra \: \rightarrow \: ^{224}_{89}Ac\: + \: ^{0}_{-1}e \: + \: energy\)