Free WAEC Physics Theory Past Questions

2022 Physics WAEC Theory Past Question 12

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

Answer:

Induced radioactivity, also called artificial radioactivity or man-made radioactivity, is the process of using radiation to make a previously stable material radioactive.

It is the process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and energy.

(ii) Complete the table below.

Answer:

(b) In an x-ray tube, an electron is accelerated from rest towards a metal target by a 30 kV source. Calculate the kinetic energy of the electron. [e = 1.6 × 10^-19 C]

K.E = eV

= 1.6 × 10^-19 × 30 × 10³

= 4.8 × 10^-15 J

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

(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).

Solution:

Scale
x-axis - 2 cm represents 2 Hz
y-axis - 2cm represents 1 J

(ii) From the graph, determine the:

I. Planck's constant;

Solution:

Planck's constant. h = slope of the graph

Slope = \(\frac{y_2 \: - \: y_1}{x_2\: - \: x_1 }\)

Slope = \(\frac{(3.8 \: - \: 1.6) \: \times \: 10^{-19}}{(11\: - \: 8) \: \times \: 10^{14}}\)

Slope = \(\frac{2.2 \: \times \: 10^{-19}}{3 \: \times \: 10^{14}}\)

Slope = \(\scriptsize 0.733 \: \times \: 10^{-19 \: - \: 14}\)

Slope = \(\scriptsize   0.733 \: \times \: 10^{-33} \: Js\)

Slope = \(\scriptsize  7.33 \: \times \: 10^{-34} \: Js\)

II. Threshold frequency of radiations;

Solution:

Threshold frequency  f₀ is the intercept on the frequency axis from the graph.

∴ f₀ = \(\scriptsize 5.8 \: \times \: 10^{14}\: Hz\) (see graph)

III. Work function of the metal.

Solution:

The negative intercept on the K.E axis is the work function of the metal, W₀. (see graph)

From the graph,  W₀ =\(\scriptsize -3.8 \: \times \: 10^{-19} J\)

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

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

⇒ \(\scriptsize 1.5 R_2 = 30\)

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

⇒ \(\scriptsize R_2 = 20 \: \Omega\)

ii. 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 JouleJoule is the SI (International System of Units) unit of energy and work. It is equal to the amount of work done when a force of 1 newton displaces a mass... More 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.