WAEC: PHYSICS Free WAEC Physics Past Questions Free WAEC Physics Theory Past Questions

Lesson 1, Topic 1

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Free WAEC Physics Theory Past Questions

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Question 1 ( No. 12. 2022 Physics WAEC Theory Past Question)

(a)(i) What is meant by the term artificial radioactivity?

(ii) Complete the table below.

Emission	Nature	Charge	Ionizing-ability
	High speed electron		Moderately ionizing
		Neutral	Negligible ionizing ability
Alpha particle		Positive

(b) In an x-ray tube, an electron is accelerated from rest towards a metal target by a 30 kV source. Calculate the kinetic energyEnergy is the ability to do work. Energy exists in several forms such as heat, kinetic or mechanical energy, light, potential energy, and electrical energy. Units of Energy: The SI unit... More of the electron. [e = 1.6 × 10^-19C]

(c) The table below shows the frequencies of radiations incident on a certain metal and the corresponding kinetic energies of the photoelectrons.

Frequency × 10¹⁴ (Hz)	6.8	8.0	9.2	10.0	11.0
Kinetic energy × 10^-19 (J)	0.8	1.6	2.4	2.9	3.8

(i) Plot a graph of kinetic energy, K.E, on the vertical axis and frequency, f, on the horizontal axis starting both axes from the origin (0,0).

(ii) From the graph, determine the:

I. Planck’s constant;
II. Threshold frequency of radiations;
III Work function of the metal.

Question 2 ( No. 9. 2022 Physics WAEC Theory Past Question)

(a)(i) State Coulomb’s law of electrostatics.

(ii) The electron and proton of a hydrogen atom are separated by a mean distance of \( \scriptsize 5.2 \: \times \: 10^{-11} \: m\)

Calculate the magnitude of the electrostatic force between the particles.

[e = 1.6 × 10^-19 C, (4πε₀)^-1= 9.0 × 10⁹ mF^-1]

(b)(i) The diagram below shows a potential divider circuit.

phy theory 9b

I. Show that Vout = \(\scriptsize V_{in} \left( \normalsize \frac{R_1}{R_1 \:+\: R_2}\right) \)

II. If \( \frac{V_{in}}{V_{out}}\) = 2.5 and R₁ = 30Ω, calculate R₂.

iii. Define the volt.

(c) Explain why wood is not suitable for use as the core of transformers.

(d) State one application for the cathode ray tube.

Responses

2022 Physics WAEC Theory Past Question 9

(a)(i) State Coulomb’s law of electrostatics.

Answer: Coulomb’s law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

(ii) The electron and proton of a hydrogen atom are separated by a mean distance of \( \scriptsize 5.2 \: \times \: 10^{-11}\: m\)

Calculate the magnitude of the electrostatic force between the particles.

[e = 1.6 × 10^-19 C, (4πε₀)^-1= 9.0 × 10⁹ mF^-1]

Solution

Given values: r = 5.2 × 10^-11m, e = q = 1.6 × 10^-19 C, k = (4πε₀)^-1= 9.0 × 10⁹ mF^-1

Using the formula: F = \( \frac{kq_1q_2}{r^2} \)

Using the formula: F = \( \frac{9\: \times \: 10^9 \: \times \: 1.6 \: \times \: 10^{-19}\: \times \: 1.6 \: \times \: 10^{-19}}{\left(5.2 \: \times \: 19^{-11}\right)^2} \)

F = \( \frac{2.304 \: \times \: 10^{-28}}{2.704 \: \times \: 10^{-21}} \)

F = \( \scriptsize 0.8521 \: \times \: 10^{-28 + 21} \)

F = \( \scriptsize 0.8521 \: \times \: 10^{-7} \)

F = \( \scriptsize 8.521 \: \times \: 10^{-8}\: N \)

(b)(i) The diagram below shows a potential divider circuit.

I. Show that Vout = \(\scriptsize V_{in} \left( \normalsize \frac{R_1}{R_1 \:+\: R_2}\right) \)

Proof:

Assuming that the current through the circuit is I

∴ \( \scriptsize V_{in} = IR_1 \: + \: IR_2 \)

⇒ \( \scriptsize V_{in} = I(R_1 \: + \: R_2)\: …..(1) \)

⇒ \( \scriptsize V_{out} = IR_2 \: …..(2) \)

divide equation (2) by (1)

⇒ \(\frac{V_{out}}{V_{in}} = \frac{\not{I}R_2}{\not{I}(R_1 \: + \: R_2)} \)

⇒ \(\frac{V_{out}}{V_{in}} = \frac{R_2}{R_1 \: + \: R_2} \)

∴ \(\scriptsize V_{out} = V_{in} \left( \frac{R_2}{R_1 \: + \: R_2} \right) \)

II. If \( \frac{V_{in}}{V_{out}}\) = 2.5 and R₁ = 30Ω, calculate R₂.

Solution:

⇒\( \frac{V_{in}}{V_{out}} \scriptsize = 2.5\)

from the proof above

⇒ \(\frac{V_{out}}{V_{in}} = \frac{R_2}{R_1 \: + \: R_2} \)

If we invert the equation we can use the value given to calculate R₂

⇒ \(\frac{V_{in}}{V_{out}} = \frac{R_1 \: + \: R_2}{ R_2} \)

⇒ \(\scriptsize 2.5 = \normalsize \frac{30 \: + \: R_2}{ R_2} \)

cross multiply

⇒ \(\scriptsize 2.5 R_2 = 30 \: + \: R_2\)

collect like terms

⇒ \(\scriptsize 2.5 R_2 \: -\: R_2 = 30 \)

⇒ \(\scriptsize 1.5 R_2 = 30 \)

⇒ \(\scriptsize R_2 = \normalsize \frac{30}{1.5} \)

⇒ \(\scriptsize R_2 = 20 \: \Omega \)

iii. Define the volt.

Answer:

The Volt is equal to the difference in electric potentials between two given points in a wire carrying an electric current of magnitude 1 ampere and dissipating one watt of power between those two points.
It can also be defined as the potential difference that exists between two points and imparts one Joule of energy to each Coulomb of charge that passes through it.

(c) Explain why wood is not suitable for use as the core of transformers.

Answer: Generally magnetic materials are used in transformer core because it has the tendency to carry the flux. If we use wooden pieces in place of magnetic materials then Transformer will not work at all because the wooden core can not provide the path to the flux so flux will not link either in the secondary side or primary side.

(d) State one application for the cathode ray tube.

Answer:

The working of CRT is based upon the movement of fast-moving electrons which strikes the photosensitive screen and helps in the production of moving images. The two main uses of cathode-ray tubes are as given below:

1) They are used in oscilloscopes which are used in physics laboratories.
2) They are also used in imaging equipment for space exploration and laboratory findings.

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