Quiz 7 of 16

2017 Physics WAEC Theory Past Questions

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(1) A particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with t².

Solution

Recall: h = ut + ½ gt² ——— (1)

Now, for an object which drops from rest, i.e, u = 0

h = ½ gt² ———- (2)

or

h = 0 + ½ gt² ——– (3)

Equate (3) to y = mx + c

h = 1/2 gt² + 0

y = mx + c

On the y-axis, we have h; and on the x – axis, we have t²; and m (slope) = ½ g

 

(2) A particle is projected horizontally at 15ms-1 from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting the ground. [g= 10ms-2]

(Leaving the answer in whole number)

Solution

Ux = 15ms-1 (horizontal velocity)
H = 20m
R = Unknown
Diagrammatically, we have

Recall:

H = Uyt + ½ gt²

Uy = 0

H = ½ gt² —————- *

20 = 1/2 × 10 × t²

40 = 10t²

t² = 4

t = 2s

Recall:

R = U x T

R = (15)(2) = 30m

(3) List three phenomena which can be explained by the molecular theory of matter.

Solution

The following phenomena can be explained by the molecular theory of matter:

i. Brownian movement
ii. Diffusion
iii. Effusion
iv. Osmosis
v. Viscosity
vi. Evaporation

(4) (a) State two factors on which surface tension depends.

Solution

The following factors affect surface tension:
i. Nature of liquid
ii. Temperature of liquid
iii. Presence/Absence of Impurities

 

(b) How can mosquito larvae be made to sink in stagnant water?

Solution

It should be noted that mosquito larva float on water, due to the surface tension of water.
Surface tension can be reduced by any of the following methods

i. Alcohol
ii. Soap/Detergent
iii. Camphor
iv. Application of heat

It implies that the above-listed methods will reduce the surface tension of water and hence cause the mosquito larva to sink

(5) List three advantages of fluorescent tubes over filament bulbs.

Solution

Fluorescent tubes are advantageous compared to fluorescent bulbs in the sense that, fluorescent tubes.

i. produce less heat

ii. are brightening

iii. use up less energy (electrical)

(6) List three advantages of p – n junction diode over diode valve.

Solution

The advantages of p – n junction over diode valve include:

1. It is cheaper

2. It requires low voltage in its operations

3. It operates at a very fast rate, i.e, it does not require time to warm up.

(7) (a) State two deductions that can be made from a displacement-time graph.

Solution

The following deductions can be made from a displacement time graph

i. Instantaneous velocity
ii. Uniform velocity

 

(b) If the distance between two equal masses is doubled and their individual masses are also doubled, what would happen to the force between them? Support your answer quantitatively.

Solution

From Newton’s Law of Universal Gravitation

F = \( \frac{GM_1M_2}{d^2} \)

G = \( \frac{Fd^2}{M_1M_2} \)

It is stated in the question that the bodies have equal masses, therefore, M1 = M2 = m

G = \( \frac{Fd^2}{m} \)

G1= G2

\( \frac{F_1d_1^2}{m_1^2} = \frac{F_2d_2^2}{m_2^2} \)

d2 = 2d1 ———-

m2 = 2m1 ———-*

\( \frac{F_1d_1^2}{m_1^2} = \frac{F_2 (2d_1)^2}{(2m_1)^2} \) \( \frac{F_1d_1^2}{m_1^2} = \frac{F_2 4d_1^2}{4m_1^2} \)

 

7 (c) State two factors that affect the maximum height attained by a bullet fired from a gun.

Solution

An external/environmental factor which affects the maximum height is air resistance.

Other factors which affect the value of the maximum height are:

i. Angle of projection (θ)
ii. Initial speed (U)
iii. Acceleration due to gravity (g)
i, ii and iii can be deduced from:

Hm = \( \frac{U^2sin^2 \theta}{2g} \) _________***

 

7 (d) State two practical examples of mechanical resonance.

Solution

Examples of mechanical resonance include:

i. the collapse of a bridge caused by soldiers marching on it.
ii. Car bodies rattle
iii. the breaking of glass by high-frequency vibrations

 

7 (e) A body is released from rest at the top of a plane inclined at 30° to the horizontal and 4.0m high. If the coefficient of friction between the body and the plane is 0.3, calculate the time the body takes to reach the bottom of the plane.

(Leaving the answer in one decimal place)

Solution

Diagrammatically, the information in the question can be represented as

μ = 0.3 (coefficient of static friction)

N.B: Two horizontal forces act on this body.

* The force acting on the body and which makes it move down the plane = mgsinθ

* The opposing frictional force on the body = μmgcosθ

Since mgsinθ > μmgcosθ, the net force on the body is given as:

F = mgsinθ – μmgcosθ ————— (1)

Recall: F = ma

ma = mgsinθ – μmgcosθ

ma = m(gsinθ – μgcosθ)

a = gsinθ – μgcosθ ————– (2)

a = 10(sin30°) – 0.3(10cos30°)

a = 10(0.5) – 0.3(10 × 0.8660)

a = 5 – 2.6

a = 2.4m/s2

Recall: s = ut + ½ at2 ———– (3)

S is the distance travelled down the plane.

Sin30° =  \( \frac{4}{S} \)

S =  \( \frac{4}{sin 30^o} \)

= 8m

Equation (3) becomes

8 = ut + ½ (2.4)t²

since the body starts its motion from rest, therefore, u = 0

8 = 0(t) + ½ (2.4)t²

8 = 1.2t²

t² = \( \frac{8}{1.2} \)

t² = 6.67

t = √6.67 = 2.6s

(8) (a) Define stable equilibrium as applied to a rigid body.

Solution

A body is said to be in stable equilibrium if after a slight displacement the body returns to its original position.

 

(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.

Solution

A block and tackle system with a velocity ratio of 3 is one that has three (3) pulleys.

 

(c) At the beginning of a race, a tyre of volume 8.0 × 10-4m3 at 20°C has a gas pressure of 4.5 × 105 Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to 4.6 × 105 Pa.

(Leaving the answer in one decimal place)

Solution

This question borders on general gas equation.

V1 = 8.0 × 10-4m3

T1 = 20°C = 293K

P1 = 4.5 × 105pa

T2 = unknown

P2 = 4.6 × 105pa

The question does not say anything about the volume changing, therefore, let us assume it is constant, i.e V1 = V2

\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) Pressure law

\( \frac{4.5 \: \times \: 10^5}{293} = \frac{4.6 \: \times \: 10^5}{T_2} \) \( \scriptsize T_2 = \normalsize \frac{4.6 \: \times \: 293}{4.5}  \) \( \scriptsize T_2 = 299.5K \: or \: 26.5^oC \)

 

(8) (d) (i)

Ice point

273K

Steam point

375K

Resistance/Ω5.677.75
Pressure/Pa7.13 x 1049.74 x 104

The table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the

(*) resistance thermometer
(*) gas thermometer

Solution

Resistance Thermometer

\( \frac{a}{b} = \frac{c}{d} \) \( \frac{7.43 \: – \: 5.67}{7.75 \: – \: 5.67} = \frac{R \: – \: 273}{375 \: – \: 273} \)

R = 359.3K

Gas Thermometer

\( \frac{a}{b} = \frac{c}{d} \) \( \frac{(9.33 \: – \:  7.13) \: \times \: 10^4}{(9.74 \: – \:  7.13) \: \times \: 10^4} = \frac{G \: – \: 273}{375 \: – \: 273} \)

G = 359.0K

 

(d) (ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?
(Leaving the answer in three decimal place)

Solution

Error = 359.3 – 359.0

= 0.3

Percentage error = \( \frac{0.3}{359.0}\scriptsize  \: \times \: 100 \% \)

= 0.084%

(9) (a) What is a wavefront?

Solution

A wavefront is a surface connecting points of equal phase on all waves in a geometric plane.

 

(b) (i) State two practical uses of glass prisms.

Solution

The practical uses of glass prisms include:
1. Binoculars (for viewing distant objects)
2. Periscope
3. For the dispersion of light rays (beam)

 

(ii) List two factors that determine the deviation of a ray of light travelling from air into a triangular glass prism.

Solution

1.  The refractive index of the prism in use
2.  The angle of incidence
3.  The refracting angle of the prism.

 

(iii) Sketch a graph to illustrate the variation of the angle of deviation d, with that of incidence i, for a ray of light travelling from air into a triangular glass prism.

Indicate on the graph the point at which the angle of incidence i is equal to the angle of emergence e.

N.B: dm = Minimum deviation
i = angle of incidence
e = angle of emergence

 

(c) (i) Draw and label a diagram of an astronomical telescope in normal adjustment.

Solution

(ii) The angular magnification of an astronomical telescope in normal adjustment is 5. If the focal length of the objective is 100cm, calculate the:
(*) focal length of the eyepiece;
(**) length of the telescope.

Solution

Angular magnification = 5
Fo (objective focal length) = 100cm
* Fe (eye piece focal length ) = ?
** LT (length of telescope) = ?

Recall:

Angular magnification = \( \frac{objective \: focal \: length }{eye \: piece\: length} \)

5 = \( \frac{100}{F_e} \)

Fe  = 20cm

LT = Fo  + Fe

LT = 100cm + 20cm = 120cm

(10) (a) (i) What is dielectric?

Solution

A dielectric is a non-conducting material such as rubber, glass, or waxed paper separating the two plates of a capacitor.

 

(ii) A parallel plate capacitor consists of two plates each of area 9.6 × 10-2m2, separated by a dielectric of thickness 2.25 × 10-3m and dielectric constant 900. Calculate the capacitance of the capacitor.
o = permittivity of free space = 8.85 × 10--12Fm-1]

Solution

Area, A = 9.6 × 10-2m2
Distance d, between the plates = 2.25 × 10-3m
εr = 900
εo = 8.85 × 10-¹²Fm-¹

Recall: C = \( \frac{A \varepsilon_o\varepsilon _r}{d} \)——–(1)

C = \( \frac{9.6 \: \times \: 10^{-2} \: \times \: 8.86 \: \times \: 10^{-2} \: \times \: 900}{2.25 \: \times \: 10^{-3}} \)

C = 3.4 × 10-7F

 

(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.

Solution

A voltmeter should have a higher resistance. This is because an ammeter measures current, therefore it should not be constructed in such a way that the opposition to the flow of current in it will be very high.

This is why in a connection, the voltmeter is connected in parallel to allow a minimum passage of current through it.

(ii)

The resistance of the voltmeter in the circuit diagram illustrated above is 800Ω. Calculate the voltmeter reading.
(Leaving the answer in whole number)

Solution

Two 400Ω resistors are in parallel, the effective resistance, Re, can be gotten by:

\( \frac{1}{R_e} = \frac{1}{400} \: + \:  \frac{1}{400} \) \( \frac{1}{R_e} = \frac{1 \: + \: 1}{400}  \) \( \frac{1}{R_e} = \frac{2}{400}  \) \( \scriptsize  2R_e = 400 \) \( \scriptsize  R_e = \normalsize \frac{400}{2}\) \( \scriptsize  R_e = 200 \Omega \)

 

The question says the resistance of the voltmeter V is 800Ω, so we redraw the diagram B

Two 800Ω resistors are in parallel, the effective resistance, Re, can be gotten by:

\( \frac{1}{R_e} = \frac{1}{800} \: + \:  \frac{1}{800} \)

=  \( \frac{1}{R_e} = \frac{1 \: + \: 1}{800}  \)

=  \( \frac{1}{R_e} = \frac{2}{400}  \)

= \( \scriptsize  2R_e = 800 \)

= \( \scriptsize  R_e =  \normalsize\frac{800}{2}\)

= \( \scriptsize  R_e = 400 \Omega \)

N.B: Resistors in parallel arrangement will have equal voltage (p.d). Therefore, V800Ω = V800Ω

6V = VA + VB ———————- (1)
VA = Ie × 200 ————- (2)
VA = IB × 400 ———- (3)

First calculate the total current in the circuit

Ve = Ie × Re

6V = Ie × (200 + 400)

Ie = 0.01A

Since the two resistors are now in series, Ie = IA = IB = 0.01A

VA = 0.01 × 200 = 2V

VB = 0.01 × 400 = 4V

The voltmeter reads 4V

 

(c)

A battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.
(Leaving the answer in whole number)

Solution

\( \frac{1}{R_e} = \frac{1}{2 \Omega} \: + \:  \frac{1}{2 \Omega} \)

=  \( \frac{1}{R_e} = \frac{1 \: + \: 1}{2}  \)

=  \( \frac{1}{R_e} = \frac{2}{2}  \)

= \( \scriptsize  2R_e = 2 \)

= \( \scriptsize  R_e =  \normalsize \frac{2}{2}\)

= \( \scriptsize  R_e = 1 \Omega \)

The two resistors in parallel becomes 1Ω

The equivalent resistance therefore is

Re = 2Ω + 2Ω + 1Ω

Re = 5Ω

Now, the diagram becomes

(11) (a) (i) What is nuclear fission?

Solution

Nuclear fission is a radioactive process which involves the disintegration (splitting) of heavy nucleus into lighter nuclei accompanied by energy release.

 

(ii) State the function of each of the following materials in a nuclear fission reactor;
(*) graphite;
(**) boron rods
(***) liquid sodium

Solution

– Graphite helps to slow down neutrons in reaction chambers
– Boron rods absorb completely the energetic neutrons so as to control the fission rate
– Liquid sodium serves as a thermal conductor which conducts excess heat away from the reaction chamber.

 

(b) The table below gives some of the energy levels of a hydrogen atom.

n12345
En/eV-13.60-3.39-1.51-0.85-0.540.00

(i) Draw the energy level diagram for the atom.

Solution

En: n = α ————– 0eV
E4: n = 5 ————- -0.54eV
E3: n = 4 ————- -0.85eV
E2: n = 3 ————— -1.51eV
E1; n = 2 —————- -3.39eV
E0; n = 1 ————— -13.60 eV

 

(ii)The table below gives some of the energy levels of a hydrogen atom.

n12345
En/eV-13.60-3.39-1.51-0.85-0.540.00

Determine the wavelength of the photon emitted when the atom goes from the energy state n = 3 to the ground state.
[h = 6.6 × 10-34Js, c = 3.0 × 108ms-1, e = 1.6 × 10-19C]

Solution

Planck’s constant, h = 6.6 × 10-34Js
Velocity of light, c = 3.0 × 108ms-1
Charge of electron, e = 1.6 × 10-19C
Wavelength, λ = ?

Recall:

λ =  \( \frac{hc}{\delta e} \)

λ = \( \frac{6.6 \: \times \: 10^{-34} \: \times \: 3.0 \: \times \: 10^8}{[-1.51 \: – \: (13.60)] \: \times \: 1.6 \: \times \: 10^{-19}} \)

λ = 1.02 × 10-7m

 

(c) A piece of ancient bone from an excavation site showed 146C activity of 9.5 disintegrations per minute per 1.0 × 10-3kg. If a bone specimen from a living creature shows 146C activity of 12.0 disintegrations per minute per 1.0 × 10-3kg, determine the age of the ancient bone.

[Half-life of 146C = 5572 years]

Solution

λ = \( \frac{0.693}{T^{\frac{1}{2}}} \)

λ = \( \frac{0.693}{5572} \)

= 0.0001244 years-1

N = \( \scriptsize Noe^{-\lambda t} \)

\( \scriptsize e^{-0.0001244t} = 1.2632 \)

0.0001244t = 0.234

t = 1878 years

(12) A spiral spring has a length of 14cm when a force of 4N is hung on it. A force of 6N extends the spring by 4cm. Calculate the unstretched length of the spring.
(Leaving the answer in one decimal place)

Solution

Initial length = L1 = 14cm

Initial Force = F1 = 4N

When a force of 6N is applied, the extension, e = 4cm, i.e L2 = 18cm

Unstretched length, Lo = Unknown

N.B: Unstretched length = Original length

Recall;

\( \frac{F_1}{L_1 \: – \: L_0} = \frac{F_2}{L_2 \: -\: L_1} = \frac{F_n}{L_n \: -\: L_{n\: -\:1}}  \) \( \frac{4}{14 \: – \: L_0} = \frac{6}{18 \: -\: 14}  \) \( \frac{4}{14 \: – \: L_0} = \frac{6}{4}  \)

cross multiply

\( \scriptsize 16 = 6(14 \: – \: L_0 )\) \(\frac{16}{6} = \scriptsize(14 \: – \: L_0 )\) \(\scriptsize(14 \: – \: L_0 ) = 2.67 \) \(\scriptsize – L_0  = 2.67 \: – \: 14 \) \(\scriptsize L_0  =  14 \: – \: 2.67  \) \(\scriptsize L_0  =  11.33cm \)
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