WAEC SSCE - Physics - 2024 (Essay)
Pergunta 1 Relatório
(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.
Detalhes da Resposta
(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.
Pergunta 2 Relatório
PART2
(a) State the effect of increasing temperature on the viscosity of a: (i) liquid, (ii) gas
(b) State two factors that determine the magnitude of a moment of a force
(c) A uniform stick AB of length, L, and mass, m, is balanced horizontally on a knife edge 10.0cm from A when an object of 400 g is suspended at A. When the knife edge is moved 5 cm further, the object has to be moved to a point 9.00 cm from A for the stick to balance.
(i) Represent the balance system with a suitable diagram
(ii) Determine the: I. mass, m of the stick; II. length, L.
(d) Explain in terms of air molecules why pressure at the top of a high mountain is less than at sea level.
(e) Mercury of density 13.6 x 10\(^3\)kgm\(^{-3}\) is poured in a container of uniform cross-sectional area 40cm \(^2\) to a height of 20 cm. The total pressure exerted on the base of the container is 1.42 x 10\(^5\)P. Calculate the: (i) mass of the mercury in the container and (ii) pressure exerted at the surface of the mercury. [g = 10ms\(^{-2}\)
(ai) Increasing the temperature of a liquid decreases the viscosity of the liquid. (aii) Increasing the temperature of a gas increases the viscosity of the gas
(b) Factors that determine the magnitude of the moment of a force about a point are: 1. Angle between the distance from the turning point and the line of action of the force. 2. magnitude of the applied force. 3. perpendicular distance from the pivot to the line of action of the applied force.
(c) SEE THE DIAGRAM ABOVE.
(ii) Taking a moment about the support: clockwise moment = anti-clockwise moment
400 x 10 = m x (\(\frac{L}{2}\) - 10) -- ------- from figure 1 above
4000 = m x (\(\frac{L}{2}\) - 10) - - - -- - - (i)
400 x 6 = m x (\(\frac{L}{2}\) - 15) - - - - -- - - -(ii)
divide eqn i by ii
\(\frac{4000}{2400}\) = \(\frac{m \times \frac{L}{2} - 10)}{m \times \frac{L}{2} - 15)}\)
\(\frac{5}{3}\) = \(\frac{\frac{L}{2} - 10)}{\frac{L}{2} - 15)}\)
3 x (\(\frac{L}{2}\) - 10) = 5 x (\(\frac{L}{2}\) - 15)
\(\frac{3L}{2}\) - 30) = \(\frac{5L}{2}\) - 75)
\(\frac{5L}{2}\) - \(\frac{3L}{2}\) = 75 - 30 = 45
L = \(\frac{2L}{2}\) = 45 cm
From equation (ii) 400 x 6 = m x (\(\frac{L}{2}\) - 15)
2400 = m x (\(\frac{45}{2}\) - 15)
2400 = m x 7.5
m = \(\frac{2400}{7.5}\) = 320 g
Therefore, I. m = 320g and II. L = 45 cm
(d) Reasons for less pressure at the top of a mountain:
The lower pressure at the top of a mountain is due to the reduced number of air molecules in the atmosphere above that elevation, resulting in a lighter air column and fewer molecular collisions, which together create lower atmospheric pressure compared to sea level.
(e)(i) Density = \(\rho\) = \(\frac{m}{v}\)
m = v x \(\rho\), but v = A x h
m = A x h x \(\rho\) = 13.6 x 10\(^3\) x 40 x 20 x 10\(^{-6}\) = 10.88kg
(e)(ii) pressure exerted by the surface of the mercury
P\(_t\) = P\(_{atm}\) + pressure of mercury
P\(_{atm}\) = P\(_t\) - \(\rho\) x h x g = 1.42 x 10\(^5\) - 13.6 x 10\(^3\) x 0.2 x 10 = 1.148 x 10\(^5\)Pa
Detalhes da Resposta
(ai) Increasing the temperature of a liquid decreases the viscosity of the liquid. (aii) Increasing the temperature of a gas increases the viscosity of the gas
(b) Factors that determine the magnitude of the moment of a force about a point are: 1. Angle between the distance from the turning point and the line of action of the force. 2. magnitude of the applied force. 3. perpendicular distance from the pivot to the line of action of the applied force.
(c) SEE THE DIAGRAM ABOVE.
(ii) Taking a moment about the support: clockwise moment = anti-clockwise moment
400 x 10 = m x (\(\frac{L}{2}\) - 10) -- ------- from figure 1 above
4000 = m x (\(\frac{L}{2}\) - 10) - - - -- - - (i)
400 x 6 = m x (\(\frac{L}{2}\) - 15) - - - - -- - - -(ii)
divide eqn i by ii
\(\frac{4000}{2400}\) = \(\frac{m \times \frac{L}{2} - 10)}{m \times \frac{L}{2} - 15)}\)
\(\frac{5}{3}\) = \(\frac{\frac{L}{2} - 10)}{\frac{L}{2} - 15)}\)
3 x (\(\frac{L}{2}\) - 10) = 5 x (\(\frac{L}{2}\) - 15)
\(\frac{3L}{2}\) - 30) = \(\frac{5L}{2}\) - 75)
\(\frac{5L}{2}\) - \(\frac{3L}{2}\) = 75 - 30 = 45
L = \(\frac{2L}{2}\) = 45 cm
From equation (ii) 400 x 6 = m x (\(\frac{L}{2}\) - 15)
2400 = m x (\(\frac{45}{2}\) - 15)
2400 = m x 7.5
m = \(\frac{2400}{7.5}\) = 320 g
Therefore, I. m = 320g and II. L = 45 cm
(d) Reasons for less pressure at the top of a mountain:
The lower pressure at the top of a mountain is due to the reduced number of air molecules in the atmosphere above that elevation, resulting in a lighter air column and fewer molecular collisions, which together create lower atmospheric pressure compared to sea level.
(e)(i) Density = \(\rho\) = \(\frac{m}{v}\)
m = v x \(\rho\), but v = A x h
m = A x h x \(\rho\) = 13.6 x 10\(^3\) x 40 x 20 x 10\(^{-6}\) = 10.88kg
(e)(ii) pressure exerted by the surface of the mercury
P\(_t\) = P\(_{atm}\) + pressure of mercury
P\(_{atm}\) = P\(_t\) - \(\rho\) x h x g = 1.42 x 10\(^5\) - 13.6 x 10\(^3\) x 0.2 x 10 = 1.148 x 10\(^5\)Pa
Pergunta 3 Relatório
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.
Detalhes da Resposta
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.
Pergunta 4 Relatório
(a)i State the condition for a charged particle to experience a force in magnetic field
(ii) State the expression for the magnetic force, F acting on a charged particle, Q in a magnetic field of flux density, \(\beta\) with speed, V
(iii) Prove quantitatively that there is no magnetic force moving along the direction of a magnetic field.
(b)i Mention the main parts of an electrical transformer.
(ii) Explain how an alternating potential difference applied to the primary coil gives rise to an induced e.m.f in the secondary coil of a transformer.
(c) Proton of mass 1.7 x 10(^{-27}\) kg enters a magnetic field of flux density 0.207 normally and followed a quarter circle path before existing with a constant speed of 4.5 x 10\(^6\)m/s
(i) Explain why the speed of the proton remains constant.
(ii) Calculate the
I. radius of the circular path
II. time taken by the proton to move through the magnetic field [e = 1.6 x 10\(^{-19}\) C, \(\pi\) = 3.14]
(a)i The charged particle must be in motion or have velocity and the particle's velocity must be inclined to the magnetic field at an angle \(\theta\)
(ii) F = Qv\(\beta\)sin\(\theta\)
(iii) for a charged particle moving parallel to the magnetic field,
\(\theta\) = 0 and F = Qv\(\beta\)sin\(\theta\)
F = Qv\(\beta\)sin 0 = 0
Thus, F = 0
(b)The main parts of an electrical transformer are the coil, Soft iron core
(ii) An alternating current in the primary coil produces a changing magnetic field in the primary coil. The changing magnetic field around the core gives rise to a changing magnetic flux linking the secondary coil thereby inducing an alternating e.m.f in the secondary coil as stated in Faraday's law.
c(i) Why the speed of the proton remains constant: The speed of the proton remains constant because the magnetic force acts perpendicular to its velocity, providing centripetal acceleration without doing work on it. This means the force changes the direction of the proton's motion but not its speed, resulting in constant kinetic energy and a uniform speed throughout its circular path.
(ii) I. radius of curved path
qv\(\beta\) = \(\frac{mv^2}{r}\)
r = \(\frac{mv}{q\beta}\)
r = \(\frac{1.7 \times 10^{-27}}{1.6 \times 10^{-19}}\) = 0.24 m.
II. time taken by the proton to move through the magnetic field,
t = \(\frac{\pi r}{2v}\) = \(\frac{3.14 \times 0.24}{2 \times 4.5 \times 10^6}\) = 8.37 x 10\(^{-8}\)s
Detalhes da Resposta
(a)i The charged particle must be in motion or have velocity and the particle's velocity must be inclined to the magnetic field at an angle \(\theta\)
(ii) F = Qv\(\beta\)sin\(\theta\)
(iii) for a charged particle moving parallel to the magnetic field,
\(\theta\) = 0 and F = Qv\(\beta\)sin\(\theta\)
F = Qv\(\beta\)sin 0 = 0
Thus, F = 0
(b)The main parts of an electrical transformer are the coil, Soft iron core
(ii) An alternating current in the primary coil produces a changing magnetic field in the primary coil. The changing magnetic field around the core gives rise to a changing magnetic flux linking the secondary coil thereby inducing an alternating e.m.f in the secondary coil as stated in Faraday's law.
c(i) Why the speed of the proton remains constant: The speed of the proton remains constant because the magnetic force acts perpendicular to its velocity, providing centripetal acceleration without doing work on it. This means the force changes the direction of the proton's motion but not its speed, resulting in constant kinetic energy and a uniform speed throughout its circular path.
(ii) I. radius of curved path
qv\(\beta\) = \(\frac{mv^2}{r}\)
r = \(\frac{mv}{q\beta}\)
r = \(\frac{1.7 \times 10^{-27}}{1.6 \times 10^{-19}}\) = 0.24 m.
II. time taken by the proton to move through the magnetic field,
t = \(\frac{\pi r}{2v}\) = \(\frac{3.14 \times 0.24}{2 \times 4.5 \times 10^6}\) = 8.37 x 10\(^{-8}\)s
Pergunta 5 Relatório
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
Detalhes da Resposta
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
Pergunta 6 Relatório
(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.
Detalhes da Resposta
(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.
Pergunta 7 Relatório
The fractional change in length produced in an elastic material of spring constant 680Nm\(^{-1}\) when a force of 306N is applied to stretch it is 1.5. Calculate the original length of the material.
From Hooke's law, F = kx where x = extension, k = force constant
K = 680Nm\(^{-1}\), F = 306N
x = \(\frac{\text{F}}{\text{x}}\) = \(\frac{306}{680}\) ≈ 0.450m
Fractional Change in Length: Given the fractional change is 1.5:
1.5 = \(\frac{\text{x}}{\text{L}}\)
L = \(\frac{0.450}{1.5}\) = 0.30m.
Detalhes da Resposta
From Hooke's law, F = kx where x = extension, k = force constant
K = 680Nm\(^{-1}\), F = 306N
x = \(\frac{\text{F}}{\text{x}}\) = \(\frac{306}{680}\) ≈ 0.450m
Fractional Change in Length: Given the fractional change is 1.5:
1.5 = \(\frac{\text{x}}{\text{L}}\)
L = \(\frac{0.450}{1.5}\) = 0.30m.
Pergunta 8 Relatório
(a) State the S.I base unit of the following quantities: (i) a stress, (ii) force constant (iii) Plank's constant
(i) Stress: The base unit of stress is the Pascal (Pa), which is equivalent to kg⋅m\(^{−1}\)⋅s\(^{−2}\)
(ii) Force Constant (Spring Constant,k): The base unit of the spring constant is also in Newtons per meter (N/m), which is equivalent to kg⋅s\(^{−2}\)
(iii) Planck's Constant (h): The base unit of Planck's constant is Joule seconds (J·s), which is equivalent to kg⋅m\(^2\)s\(^{−1}\)
.
Detalhes da Resposta
(i) Stress: The base unit of stress is the Pascal (Pa), which is equivalent to kg⋅m\(^{−1}\)⋅s\(^{−2}\)
(ii) Force Constant (Spring Constant,k): The base unit of the spring constant is also in Newtons per meter (N/m), which is equivalent to kg⋅s\(^{−2}\)
(iii) Planck's Constant (h): The base unit of Planck's constant is Joule seconds (J·s), which is equivalent to kg⋅m\(^2\)s\(^{−1}\)
.
