WAEC SSCE - Physics - 2024 (Essay)
Question 1 Report
(a) State the condition for light ray incident on a concave mirror to reflect through the principal focus of the mirror.
(b) Explain spherical aberration as a defect of curved mirrors
(c) The diagram above illustrates an image formed when an object is placed in front of a particular lens. Redraw the diagram and indicate the center of curvature, C, principal focus, F, and the principal axis. ( SEE THE DIAGRAM ABOVE)
(d) What is meant by accommodation in connection to the human?
(e) State quantitatively, the (i) values of the near and far point of a normal eye. (ii) Use the answer in 10c(i) to determine the change in optical power of the normal human eye when reading a book and viewing the sky.[lens-to-retina distance = 2.5 cm].
(a) The incident light rays must be parallel and close to the principal axis
(b) Spherical aberration as a defect of curved mirrors: When a wide parallel beam is incident on a spherical mirror, it is not brought to the same focus. This makes the image distorted.
(c) SEE THE DIAGRAM ABOVE
(d) Accommodation is the ability of the eyes to alter its focal length to form a clear image
(e) Quantitatively, (i) far point = infinity, Near point = 25 cm
P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
(ii) for near point, u = 0.25 m
P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{0.25}\) + \(\frac{1}{v}\)
similarly, for far point, u = infinity(∞)
P\(_{far point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
P\(_{far point}\) = \(\frac{1}{∞}\) + \(\frac{1}{v}\)
= 0 + \(\frac{1}{v}\)
So, change in optical power
ΔP = P\(_{near point}\) - P\(_{far point}\)
ΔP = 4 + \(\frac{1}{v}\) - \(\frac{1}{v}\)
ΔP = 4 dioptres.
Answer Details
(a) The incident light rays must be parallel and close to the principal axis
(b) Spherical aberration as a defect of curved mirrors: When a wide parallel beam is incident on a spherical mirror, it is not brought to the same focus. This makes the image distorted.
(c) SEE THE DIAGRAM ABOVE
(d) Accommodation is the ability of the eyes to alter its focal length to form a clear image
(e) Quantitatively, (i) far point = infinity, Near point = 25 cm
P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
(ii) for near point, u = 0.25 m
P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{0.25}\) + \(\frac{1}{v}\)
similarly, for far point, u = infinity(∞)
P\(_{far point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
P\(_{far point}\) = \(\frac{1}{∞}\) + \(\frac{1}{v}\)
= 0 + \(\frac{1}{v}\)
So, change in optical power
ΔP = P\(_{near point}\) - P\(_{far point}\)
ΔP = 4 + \(\frac{1}{v}\) - \(\frac{1}{v}\)
ΔP = 4 dioptres.
Question 2 Report
(a) Derive the dimension of surface tension.
(b) Name the instrument used to measure the force of gravity at a place
(a) SURFACE TENSION = \(\frac{\text{force}}{\text{length}}\)
FORCE = mass x accel. due to gravity
Dimension of force = M.LT\(^{-2}\)
Dimension of length = L
Dimension of surface area = \(\frac{MLT^{-2}}{L}\) = MT\(^{-2}\)
(b) The instruments used for measuring the force of gravity at a place are:
(1) Gravimeter, (2) Spring balance, (3) Gravitational accelerometer.
Answer Details
(a) SURFACE TENSION = \(\frac{\text{force}}{\text{length}}\)
FORCE = mass x accel. due to gravity
Dimension of force = M.LT\(^{-2}\)
Dimension of length = L
Dimension of surface area = \(\frac{MLT^{-2}}{L}\) = MT\(^{-2}\)
(b) The instruments used for measuring the force of gravity at a place are:
(1) Gravimeter, (2) Spring balance, (3) Gravitational accelerometer.
Question 3 Report
(a)(i) What is a thermometric liquid?
(ii) State the reason for the following design features of a clinical thermometer. I. Narrow bore: II. Thin wall of the bulb
(b) Distinguish between heat and temperature of an object in terms of the energy of a particle
(c) Explain why evaporation leads to cooling
(d) A kettle rated 2000W, contains water at 20ºC. The kettle is switched on and after two minutes, the water starts boiling. After another six minutes, 45% of the water in the kettle boils away. (i) Determine the specific latent heat of the vaporization of the water (ii) State one assumption made in your calculation 9d(i) above
(a) A thermometric liquid is a liquid used in thermometers to measure temperature. It expands and contracts uniformly with changes in temperature, allowing for accurate readings.
(ii) I. Narrow Bore
Reason: Allows precise measurement by controlling the movement of the thermometric liquid, ensuring quick and accurate readings
II. Thin Wall of the Bulb
Reason: Enhances thermal conductivity, enabling faster heat transfer for quicker and more accurate temperature readings.
(b) Heat is a measure of the change in total internal energy in a body while temperature is a measure of the average kinetic energy of a molecule of the body.
(c) Evaporation leads to cooling for these reasons:
Energy Absorption: Molecules at the surface absorb energy to break free, often from the liquid and its surroundings.
Loss of High-Energy Molecules: Higher-energy molecules evaporate, reducing the average kinetic energy of the remaining liquid.
Temperature Decrease: As the average kinetic energy drops, the temperature of the liquid decreases, resulting in cooling.
(d)(i) Given: P = 2000W, \(\theta\) = 20ºC, t = 2 mins = 120 secs.
P x t = mc\(\Delta\)\(\theta\)
m = \(\frac{P \times t}{c \times \Delta \theta}\)
m = \(\frac{2000 \times 120}{4200 \times (100 - 20)}\) = 0.714 kg
Also, Pt = ml
l = \(\frac{Pt}{m}\)
where m = 45% of 0.714, t = 6mins = 360secs
l = \(\frac{2000 \times 60 \times 6}{0.714 \times 0.45}\) = 2.29 x 10\(^6\)Jkg\(^{-1}\)
(d)(i) Assumption made: there is no loss of heat to the surroundings, and the heat capacity of the material of the kettle is negligible.
Answer Details
(a) A thermometric liquid is a liquid used in thermometers to measure temperature. It expands and contracts uniformly with changes in temperature, allowing for accurate readings.
(ii) I. Narrow Bore
Reason: Allows precise measurement by controlling the movement of the thermometric liquid, ensuring quick and accurate readings
II. Thin Wall of the Bulb
Reason: Enhances thermal conductivity, enabling faster heat transfer for quicker and more accurate temperature readings.
(b) Heat is a measure of the change in total internal energy in a body while temperature is a measure of the average kinetic energy of a molecule of the body.
(c) Evaporation leads to cooling for these reasons:
Energy Absorption: Molecules at the surface absorb energy to break free, often from the liquid and its surroundings.
Loss of High-Energy Molecules: Higher-energy molecules evaporate, reducing the average kinetic energy of the remaining liquid.
Temperature Decrease: As the average kinetic energy drops, the temperature of the liquid decreases, resulting in cooling.
(d)(i) Given: P = 2000W, \(\theta\) = 20ºC, t = 2 mins = 120 secs.
P x t = mc\(\Delta\)\(\theta\)
m = \(\frac{P \times t}{c \times \Delta \theta}\)
m = \(\frac{2000 \times 120}{4200 \times (100 - 20)}\) = 0.714 kg
Also, Pt = ml
l = \(\frac{Pt}{m}\)
where m = 45% of 0.714, t = 6mins = 360secs
l = \(\frac{2000 \times 60 \times 6}{0.714 \times 0.45}\) = 2.29 x 10\(^6\)Jkg\(^{-1}\)
(d)(i) Assumption made: there is no loss of heat to the surroundings, and the heat capacity of the material of the kettle is negligible.
Question 4 Report
An object projected at an angle to a ground level has a time of flight 4 seconds to move through still air. Calculate the maximum height attained by the object.[g = 10ms\(^{-2}\)]
Given: time of flight T = 4s, g = 10ms\(^{-2}\), H\(_{max}\)
\(\theta\) = ?
T = \(\frac{2u sin\theta}{g}\)
4 = \(\frac{2 \times u sin \theta}{10}\)
40 = 2u sin\(\theta\)
sin\(\theta\) = \(\frac{20}{u}\)
But, H\(_{max}\) = \(\frac{u^2 sin^2 \theta}{2g}\)
H\(_{max}\) = \(\frac{u^2 \times (\frac{20}{u})^2}{2 \times 10}\)
H\(_{max}\) = \(\frac{20 \times 20}{20}\) = 20 metres.(u\(^2\)s cancel out)
Thus H\(_{max}\) attained = 20metres.
Answer Details
Given: time of flight T = 4s, g = 10ms\(^{-2}\), H\(_{max}\)
\(\theta\) = ?
T = \(\frac{2u sin\theta}{g}\)
4 = \(\frac{2 \times u sin \theta}{10}\)
40 = 2u sin\(\theta\)
sin\(\theta\) = \(\frac{20}{u}\)
But, H\(_{max}\) = \(\frac{u^2 sin^2 \theta}{2g}\)
H\(_{max}\) = \(\frac{u^2 \times (\frac{20}{u})^2}{2 \times 10}\)
H\(_{max}\) = \(\frac{20 \times 20}{20}\) = 20 metres.(u\(^2\)s cancel out)
Thus H\(_{max}\) attained = 20metres.
Question 5 Report
(a) State two uses of a polar satellite.
(b) What does the slope of a graph of tensile stress against tensile strain represent?
(a) The uses of polar satellites include:
(i) for location identification
(ii) to collect data on climate change
(iii) observe and predict a natural disaster
(iv) use for communication.
(b) The slope of a graph of tensile stress against tensile strain represents the modulus of elasticity, also known as Young's modulus. Young's modulus quantifies the stiffness of a material. A steeper slope indicates a stiffer material, while a flatter slope indicates a more flexible material.
Answer Details
(a) The uses of polar satellites include:
(i) for location identification
(ii) to collect data on climate change
(iii) observe and predict a natural disaster
(iv) use for communication.
(b) The slope of a graph of tensile stress against tensile strain represents the modulus of elasticity, also known as Young's modulus. Young's modulus quantifies the stiffness of a material. A steeper slope indicates a stiffer material, while a flatter slope indicates a more flexible material.
Question 6 Report
An electron of mass, m, and charge, e moves through the electric field of potential difference V\(_o\) with a speed, v. Show that de Broglie wavelength associated with the electron is given as \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\), where h is the Plank's constant.
To show that \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
The de broglie wavelength \(\lambda\), is given by
\(\lambda\) = \(\frac{\text{h}}{\text{p}}\)
where p = mv
\(\lambda\) = \(\frac{\text{h}}{\text{mv}}\)
When an electron is accelerated through a potential difference V\(_0\), it gains kinetic energy given K.E.= eV\(_0\)
The kinetic energy can also be expressed in terms of the mass m of the electron and its speed v: K.E = \(\frac{1}{2}\)mv\(^2\)
K.E = eV\(_0\) = \(\frac{1}{2}\)mv\(^2\)
v = \(\sqrt{\frac{2eV_o}{m}}\)
Substituting v = \(\sqrt{\frac{2eV_o}{m}}\) into the de Broglie equation
\(\lambda\) = \(\frac{\text{h}}{m \times \sqrt{\frac{2eV_o}{m}}}\) = \(\frac{\text{h}}{\sqrt{\frac{2eV_o m^2}{m}}}\)
\(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
Note: use the fact that \(\sqrt{m}\) x \(\sqrt{m}\) = m
Answer Details
To show that \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
The de broglie wavelength \(\lambda\), is given by
\(\lambda\) = \(\frac{\text{h}}{\text{p}}\)
where p = mv
\(\lambda\) = \(\frac{\text{h}}{\text{mv}}\)
When an electron is accelerated through a potential difference V\(_0\), it gains kinetic energy given K.E.= eV\(_0\)
The kinetic energy can also be expressed in terms of the mass m of the electron and its speed v: K.E = \(\frac{1}{2}\)mv\(^2\)
K.E = eV\(_0\) = \(\frac{1}{2}\)mv\(^2\)
v = \(\sqrt{\frac{2eV_o}{m}}\)
Substituting v = \(\sqrt{\frac{2eV_o}{m}}\) into the de Broglie equation
\(\lambda\) = \(\frac{\text{h}}{m \times \sqrt{\frac{2eV_o}{m}}}\) = \(\frac{\text{h}}{\sqrt{\frac{2eV_o m^2}{m}}}\)
\(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
Note: use the fact that \(\sqrt{m}\) x \(\sqrt{m}\) = m
Question 7 Report
State three uses of ferromagnetic materials
The uses of ferromagnetic materials are:
(i) Electromagnets
(ii) Magnetic Storage
(iii) Magnetic Sensors
(iv) Magnetic Shielding
(v) Permanent Magnets
(vi) Inductive Components, others are transformers, electric bells, electric generators, etc.
Answer Details
The uses of ferromagnetic materials are:
(i) Electromagnets
(ii) Magnetic Storage
(iii) Magnetic Sensors
(iv) Magnetic Shielding
(v) Permanent Magnets
(vi) Inductive Components, others are transformers, electric bells, electric generators, etc.
Question 8 Report
(a) Mention three facts about photoelectric effect
(b) An electric magnetic radiation source of power 6 x 10\(^{-3}\) W emits 1 x 10\(^{16}\) photons per second. The most energetic photo electron ejected from a metal surface is stopped by a potential difference of 2.2V. Calculate the work function of the metal. [ mass of (photon) electrons 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C]
(c) State two factors on which the activity of a radioactive sample depends
(d) Cobalt-60 source has an activity of 2.0 x 10\(^6\) Bq and a half-life of 1.8 x 10\(^8\)s. Calculate the number of radioisotope nuclei in the source.
(a) electrons are only emitted when the photon frequency exceeds the threshold frequency,
- the energy of each photon is proportional to its frequency,
- the emission of photoelectrons depends on the frequency of incident light on the metal,
- the intensity of light does not affect the energy of the electrons emitted/photoelectrons,
- the energy of the photoelectron depends on the frequency of incident radiation.
(b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ?
Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\)
= \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\)
= 6 x 10\(^{-19}\)J
But K.E = E + W\(_o\)
W\(_o\) = E - K.E = E - eV
W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J
(c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc
(d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s
A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity.
But \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\)
\(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\)
Recall, A = \(\lambda\)N
N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei
Answer Details
(a) electrons are only emitted when the photon frequency exceeds the threshold frequency,
- the energy of each photon is proportional to its frequency,
- the emission of photoelectrons depends on the frequency of incident light on the metal,
- the intensity of light does not affect the energy of the electrons emitted/photoelectrons,
- the energy of the photoelectron depends on the frequency of incident radiation.
(b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ?
Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\)
= \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\)
= 6 x 10\(^{-19}\)J
But K.E = E + W\(_o\)
W\(_o\) = E - K.E = E - eV
W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J
(c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc
(d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s
A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity.
But \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\)
\(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\)
Recall, A = \(\lambda\)N
N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei
