WAEC SSCE - Physics - 2023 (Essay)
Swali 1 Ripoti
a. Define each of the following terms used with simple machines:
i. Pivot ii.Load iii. Efficiency.
b. A truck of mass 1.2 × \(10^3\) kg is pulled from rest by a constant horizontal force of 25.2N on a leveled road. If the maximum speed attainable in the process is 60 km/h.
Calculate the: i. work done by the force; ii. distance traveled by the truck in reaching the maximum speed.
c. State two differences between absolute zero temperature and ice point.
d. An uncalibrated liquid-in-glass thermometer was used in determining a Celsius temperature. The readings are tabulated below
| Temperature/°C | -6 | 0 | 100 |
| Length of column/ cm | L | 2.0 | 15.0 |
a
i. A pivot, often referred to as a fulcrum, is the point or axis around which a simple machine, such as a lever or a seesaw, rotates.
ii. A load is the object or weight that is being moved, lifted, or supported by the machine.
iii. Efficiency refers to how effectively a machine can perform its intended task while minimizing energy loss
b.
i. Given: F = 25.2N, m = 1200kg, u = 0m/s , v = 60km/h = 16.67m/s
Workdone = ΔK.E = \(K.E_2 - K. E_1\)
Workdone = \(\frac{1}{2}K.E_2 - \frac{1}{2}K.E_1\)
Workdone = \(\frac{1}{2}m(v^2 - u^2)\)
Workdonk = \(\frac{1}{2}\times 1200 \times 16.67^2\) ( since u = 0)
Workdone = 600 x 277.89 = 166733.34J
Therefore, Workdone = \(1.67 \times 10^5\)J
ii. F = ma → a = \(\frac{F}{m} = \frac{25.2}{1.2 \times10^3} = 0.021ms^2\)
recall, \(v^2 = u^2 +2aS\)
S = \(\frac{v^2 - u^2}{ 2a} = \frac{16.67^2 - 0^2}{ 2\times 0.021}\)
S = 6616.4m = 6.62km.
c. Absolute zero temperature is the lowest possible temperature that theoretically represents the complete absence of thermal energy while the ice point is the temperature at which water coexists with ice in thermal equilibrium.
- Absolute zero temperature is defined as 0 Kelvin (0 K) or approximately -273.15°C while ice point is typically defined as 0°C at sea level
d. \(\frac{0 - ( - 6)}{ 2 - L} = \frac{ 100 - 0 }{ 15 - 2}\)
\(\frac{ 6}{ 2 - L} = \frac{100 }{ 13}\)
6 x 13 = 100( 2 - L )
78 = 200 - 100L
100L = 200 - 78
100L = 122
L = \(\frac{122}{100}\) = 1.22cm
Maelezo ya Majibu
a
i. A pivot, often referred to as a fulcrum, is the point or axis around which a simple machine, such as a lever or a seesaw, rotates.
ii. A load is the object or weight that is being moved, lifted, or supported by the machine.
iii. Efficiency refers to how effectively a machine can perform its intended task while minimizing energy loss
b.
i. Given: F = 25.2N, m = 1200kg, u = 0m/s , v = 60km/h = 16.67m/s
Workdone = ΔK.E = \(K.E_2 - K. E_1\)
Workdone = \(\frac{1}{2}K.E_2 - \frac{1}{2}K.E_1\)
Workdone = \(\frac{1}{2}m(v^2 - u^2)\)
Workdonk = \(\frac{1}{2}\times 1200 \times 16.67^2\) ( since u = 0)
Workdone = 600 x 277.89 = 166733.34J
Therefore, Workdone = \(1.67 \times 10^5\)J
ii. F = ma → a = \(\frac{F}{m} = \frac{25.2}{1.2 \times10^3} = 0.021ms^2\)
recall, \(v^2 = u^2 +2aS\)
S = \(\frac{v^2 - u^2}{ 2a} = \frac{16.67^2 - 0^2}{ 2\times 0.021}\)
S = 6616.4m = 6.62km.
c. Absolute zero temperature is the lowest possible temperature that theoretically represents the complete absence of thermal energy while the ice point is the temperature at which water coexists with ice in thermal equilibrium.
- Absolute zero temperature is defined as 0 Kelvin (0 K) or approximately -273.15°C while ice point is typically defined as 0°C at sea level
d. \(\frac{0 - ( - 6)}{ 2 - L} = \frac{ 100 - 0 }{ 15 - 2}\)
\(\frac{ 6}{ 2 - L} = \frac{100 }{ 13}\)
6 x 13 = 100( 2 - L )
78 = 200 - 100L
100L = 200 - 78
100L = 122
L = \(\frac{122}{100}\) = 1.22cm
Swali 2 Ripoti
State three differences between geostationary satellites and polar satellites.
| Geostationary satellites | Polar satellites | |
| 1 | They orbit the earth directly above the equator in geosynchronous orbit. they remain fixed relative to a specific point on the Earth's surface | They orbit the Earth, passing over or near the north and south poles on each orbit. |
| 2 | They have an inclination of 0°, meaning their orbital plane aligns with the equatorial plane | They have a high inclination angle( usually around 90°) to achieve polar orbits. |
| 3 | They observe the Earth from a fixed position, providing a continuous view of a specific region. | They observe the Earth from a changing perspective as the orbit, resulting in different views of the Earth's surface with each pass |
| 4 | They primarily collect data for meteorological observation weather forecasting, and telecommunications. | They collect data for a wide range of applications, including weather monitoring, climate research, Environmental monitoring, and scientific research. |
| 5 | They are positioned at an altitude of approximately 35,786 kilometres above the Earth's surface. | They operate at a lower altitudes, typically between 700 and 1500 kilometers |
| 6 | They provide a continuous stream of data and observations for a specific region, allowing for real-time monitoring. | They have a lower data refresh rate since they pass over a given less frequently. However, they provide a broader coverage. |
pick any three according to the given instructions.
Maelezo ya Majibu
| Geostationary satellites | Polar satellites | |
| 1 | They orbit the earth directly above the equator in geosynchronous orbit. they remain fixed relative to a specific point on the Earth's surface | They orbit the Earth, passing over or near the north and south poles on each orbit. |
| 2 | They have an inclination of 0°, meaning their orbital plane aligns with the equatorial plane | They have a high inclination angle( usually around 90°) to achieve polar orbits. |
| 3 | They observe the Earth from a fixed position, providing a continuous view of a specific region. | They observe the Earth from a changing perspective as the orbit, resulting in different views of the Earth's surface with each pass |
| 4 | They primarily collect data for meteorological observation weather forecasting, and telecommunications. | They collect data for a wide range of applications, including weather monitoring, climate research, Environmental monitoring, and scientific research. |
| 5 | They are positioned at an altitude of approximately 35,786 kilometres above the Earth's surface. | They operate at a lower altitudes, typically between 700 and 1500 kilometers |
| 6 | They provide a continuous stream of data and observations for a specific region, allowing for real-time monitoring. | They have a lower data refresh rate since they pass over a given less frequently. However, they provide a broader coverage. |
pick any three according to the given instructions.
Swali 3 Ripoti
a. Using the kinetic theory, explain the term diffusion of fluid molecules.
b. Name one phenomenon that demonstrates that light behaves as a:
i. wave ii. particles
a. According to the kinetic theory, diffusion in fluids occurs because of the random and continuous motion of molecules. This motion results in a net movement of the molecules from regions of higher concentration to regions to regions of lower concentration ultimately leading to the mixing of substances in the fluid i.e. the process continues until the concentration becomes uniform throughout the fluid. It's driven by the kinetic energy of the molecules and leads to the mixing of substances in fluids.
b.
i. Interference of light waves
ii .– Photo-electric effect
–Compton Effect
Maelezo ya Majibu
a. According to the kinetic theory, diffusion in fluids occurs because of the random and continuous motion of molecules. This motion results in a net movement of the molecules from regions of higher concentration to regions to regions of lower concentration ultimately leading to the mixing of substances in the fluid i.e. the process continues until the concentration becomes uniform throughout the fluid. It's driven by the kinetic energy of the molecules and leads to the mixing of substances in fluids.
b.
i. Interference of light waves
ii .– Photo-electric effect
–Compton Effect
Swali 4 Ripoti
a. Define strain energy.
b. Write an expression for the energy stored, E, in a stretched wire of original length, l , cross-sectional area, A, extension, e, and Young's modulus, Y, of the material of the wire.
a. Strain energy is the potential energy stored within an elastic material as a result of its deformation caused by external forces.
b. Energy stored E = \(\frac{1}{2}Fe\)
Young Modulus = \(\frac{ Stress}{ Strain}\) = Y
Stress = \(\frac{Force}{Area} = \frac{(F)}{(A)}\)
Strain = \(\frac{ Extension}{Original Length} = \frac{(e)}{(l)}\)
Y = \(\frac{F}{A}\div \frac{e}{l}\)
Y = \(\frac{F}{A} \times \frac{l}{e}\)
F = \(\frac{YAe}{l}\)
Therefore, Energy stored E = \(\frac{1}{2}\frac{YAe}{l}\times e\)
E = \(\frac{YAe^2}{2l}\)
Maelezo ya Majibu
a. Strain energy is the potential energy stored within an elastic material as a result of its deformation caused by external forces.
b. Energy stored E = \(\frac{1}{2}Fe\)
Young Modulus = \(\frac{ Stress}{ Strain}\) = Y
Stress = \(\frac{Force}{Area} = \frac{(F)}{(A)}\)
Strain = \(\frac{ Extension}{Original Length} = \frac{(e)}{(l)}\)
Y = \(\frac{F}{A}\div \frac{e}{l}\)
Y = \(\frac{F}{A} \times \frac{l}{e}\)
F = \(\frac{YAe}{l}\)
Therefore, Energy stored E = \(\frac{1}{2}\frac{YAe}{l}\times e\)
E = \(\frac{YAe^2}{2l}\)
Swali 5 Ripoti
a. State one difference between an intrinsic and an extrinsic semiconductor.
b. Draw a circuit diagram to illustrate full wave smoothing rectification.
a. – Extrinsic semiconductors are produced by adding impurities into pure semiconductors, while intrinsic semiconductors are always present in their most pure state.
- Extrinsic semiconductors have a high electrical conductivity in comparison to other materials at room temperature, while intrinsic semiconductors have a low electrical conductivity.
- In intrinsic semiconductors, the number of electrons and holes is equal, however, this is not the case in extrinsic semiconductors.
- Temperature alone determines the behavior of intrinsic semiconductors, whereas extrinsic semiconductors are influenced by both temperature and the number of impurities present.
- n-type and p-type semiconductors are two categories of extrinsic semiconductors, while intrinsic semiconductors are not further subdivided.
b. diagram above
Maelezo ya Majibu
a. – Extrinsic semiconductors are produced by adding impurities into pure semiconductors, while intrinsic semiconductors are always present in their most pure state.
- Extrinsic semiconductors have a high electrical conductivity in comparison to other materials at room temperature, while intrinsic semiconductors have a low electrical conductivity.
- In intrinsic semiconductors, the number of electrons and holes is equal, however, this is not the case in extrinsic semiconductors.
- Temperature alone determines the behavior of intrinsic semiconductors, whereas extrinsic semiconductors are influenced by both temperature and the number of impurities present.
- n-type and p-type semiconductors are two categories of extrinsic semiconductors, while intrinsic semiconductors are not further subdivided.
b. diagram above
Swali 6 Ripoti
ai. Define the electric potential at a point in an electric field.
ii. An uncharged body, A, was charged electrostatically by a test charge, B, using the method of induction and the method of contact. State two differences between the two methods.
b. An important precaution during an electricity experiment is to open the circuit when no readings are being taken. Give two reasons for the stated precaution.
ci. Fig. 11.0 is a circuit diagram in which a coil of inductance, L, and a resistor of resistance, R, are connected to a variable alternating source of frequency, f.
The table shows the square of the impedance, \(Z^2\); corresponding to each value of \(ƒ^2\).
| \(ƒ^2\)/ \(Hz^2\) |
198.80 | 400.00 | 600.30 | 800.90 | 900.00 |
| \(Z^2\)/ \(Ω^2\) |
249.60 | 400.00 | 550.30 | 702.30 | 800.90 |
Write down the equation for Z in terms of \(f^2\), \(R^2\), and \(L^2\).
ii. Plot a graph \(Z^2 against \(f^2\) of and use it to determine the values of:
i. L
ii. R
[\(π^2\) = 10]
ai. Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.
ii. Process of Charging:
Induction: Involves redistributing charges within an uncharged body without direct contact with a charged object.
Contact: Requires direct physical contact with a charged object to transfer charges.
- Nature of Final Charge:
Induction: Results in a charge of opposite polarity to the charged object initially brought close.
Contact: Results in a charge of the same polarity as the charged object with which contact is made.
b. - Safety: Opening the circuit prevents current from flowing, which reduces the risk of electric shock. This is especially important when working with high voltages or currents.
- Conservation of battery power: If the circuit is left closed, the current will continue to flow even when no readings are being taken. This can drain the battery and shorten its lifespan.
- Accurate readings: If the circuit is left closed, the current flowing through the circuit can cause the components to heat up. This can affect the accuracy of the readings.
- Protection of equipment: If the circuit is left closed, a surge of current could flow through the circuit if there is a fault. This could damage the equipment or even cause a fire.
c. \(Z^2 = R^2 + X{_L}^2\)
\(Z^2 = R^2 + (2πf_L)^2
\(Z^2 = R^2 + 4π^2f^2L^2\)
\(Z^2 = 4π^2f^2L^2 + R^2\)
∴ \(Z = \sqrt{4π^2L^2f^2 + R^2}\)
ii. from the equation
\(Z^2 = 4π^2f^2L^2 + R^2\)
using y = mx + c = y = \(Z^2\), x = \(f^2\)
m = \(4π^2L^2\) = slope of the grape
m = \(\frac{ Z_2^2 - Z_1^2}{f_2^2 - f_1^2} = \frac{ 700 - 200 }{ 800 - 140 }\)
\(4π^2L^2 = \frac{ 500}{ 660}\)
\(L^2 = \frac{ 500}{ 660} \times \frac{1}{ 4π^2} =\frac{ 500}{ 660} \times \frac{1}{4 \times 10}\) ( since \(π^2\) = 10)
\(L^2 = 0.0189\)
∴ L = \(\sqrt{ 0.0189}\) = 0.138H = 138mH
Maelezo ya Majibu
ai. Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.
ii. Process of Charging:
Induction: Involves redistributing charges within an uncharged body without direct contact with a charged object.
Contact: Requires direct physical contact with a charged object to transfer charges.
- Nature of Final Charge:
Induction: Results in a charge of opposite polarity to the charged object initially brought close.
Contact: Results in a charge of the same polarity as the charged object with which contact is made.
b. - Safety: Opening the circuit prevents current from flowing, which reduces the risk of electric shock. This is especially important when working with high voltages or currents.
- Conservation of battery power: If the circuit is left closed, the current will continue to flow even when no readings are being taken. This can drain the battery and shorten its lifespan.
- Accurate readings: If the circuit is left closed, the current flowing through the circuit can cause the components to heat up. This can affect the accuracy of the readings.
- Protection of equipment: If the circuit is left closed, a surge of current could flow through the circuit if there is a fault. This could damage the equipment or even cause a fire.
c. \(Z^2 = R^2 + X{_L}^2\)
\(Z^2 = R^2 + (2πf_L)^2
\(Z^2 = R^2 + 4π^2f^2L^2\)
\(Z^2 = 4π^2f^2L^2 + R^2\)
∴ \(Z = \sqrt{4π^2L^2f^2 + R^2}\)
ii. from the equation
\(Z^2 = 4π^2f^2L^2 + R^2\)
using y = mx + c = y = \(Z^2\), x = \(f^2\)
m = \(4π^2L^2\) = slope of the grape
m = \(\frac{ Z_2^2 - Z_1^2}{f_2^2 - f_1^2} = \frac{ 700 - 200 }{ 800 - 140 }\)
\(4π^2L^2 = \frac{ 500}{ 660}\)
\(L^2 = \frac{ 500}{ 660} \times \frac{1}{ 4π^2} =\frac{ 500}{ 660} \times \frac{1}{4 \times 10}\) ( since \(π^2\) = 10)
\(L^2 = 0.0189\)
∴ L = \(\sqrt{ 0.0189}\) = 0.138H = 138mH
Swali 7 Ripoti
a. A projectile is fired at an angle, θ, to the horizontal with velocity, u. Show that at any time, t, during the motion, the: i. horizontal component of the velocity is independent of t;
ii. vertical component of the velocity depends on t.
b. State the assumption on which projectile motion is based.
i. From v = u + gt:
\(u_x\) = ucosθ, g = 0 (horizontal component is not affected by gravity)
\(v_x\) = ucosθ + (0)t
∴ \(v_x\) = ucosθ
ii. \(v_y = u_y + gt\)
∴ \(v_y = usinθ + gt\)
b. It assumes that the only force acting on the projectile is gravity, and there are no other forces, such as air resistance or drag, affecting its motion.
Maelezo ya Majibu
i. From v = u + gt:
\(u_x\) = ucosθ, g = 0 (horizontal component is not affected by gravity)
\(v_x\) = ucosθ + (0)t
∴ \(v_x\) = ucosθ
ii. \(v_y = u_y + gt\)
∴ \(v_y = usinθ + gt\)
b. It assumes that the only force acting on the projectile is gravity, and there are no other forces, such as air resistance or drag, affecting its motion.
Swali 8 Ripoti
ai. State the reason why simple harmonic motion is periodic.
ii. State two factors that affect the period of oscillation of a simple pendulum.
iii. Sketch a graph of the total mechanical energy, E, against displacement, y, for the motion of a simple pendulum from one extreme position to the other.
b. The diagram above illustrates an oscillatory pendulum. Calculate the work done in raising the pendulum to point B, if the mass of the bob is 50 g.
[g = \(10 ms^2\)] see the figure above
c. A spiral spring of spring constant, k, and natural length, l, has a scale pan of mass 0.04 kg hanging on its lower end while the upper end is firmly fixed to a support. When an object of mass 0.20 kg is placed on the scale pan, the length of the spring becomes 0.055 m and when the object is replaced with another object of mass 0.28 kg, the length of the spring becomes 0.065 m. Calculate the values of k and l.
[g = \(10 ms^2\)]
ai. Simple harmonic motion is periodic because of the restoring force which is proportional to the displacement, acts in the opposite direction to the motion
ii. ♦ Length of the pendulum ♦ Strength of the gravitational field
iii. See the figure above.
b. m = 50g = 0.05kg, g = 10 \(ms^2\), h = 10cm = 0.1m, P.E = ?
P.E = mgh = 0.05 × 10 × 0.1
∴ P.E = 0.05 J
c. Let the extension be ∆l
F = k ∆l = k(\(l_n\) - l )
where k is the spring constant, ln is the new length, l is the natural length, and F = mg.
Total mass = mass of pan + added mass
In the first case,
\(m_1\) = 0.04 + 0.20 = 0.24 kg
\(F_1\) = 0.24 × 10 = 2.4 N
∆l = 0.055 - l
Since F = k ∆l,
2.4 = k(0.055 - l )
2.4 = 0.055k - l k --- (i)
In the second case,
\(m_2\) = 0.04 + 0.28 = 0.32 kg
\(F_2\) = 0.32 × 10 = 3.2 N
∆l = 0.065 - l
Since F = k ∆l,
3.2 = k(0.065 - l )
3.2 = 0.065k - l k --- (ii)
Subtracting equation (i) from (ii)
⇒ 0.8 = 0.01k
k = \(\frac{0.8}{ 0.01}\)
Substitute (80) for k in equation (i)
⇒ 2.4 = 0.055(80) - 80 l
⇒ 2.4 = 4.4 - 80 l
⇒ 80 l = 4.4 - 2.4
⇒ 80 l = 2
l = \(\frac{2}{80}\)
Therefore, the value of k = 80 N/m and l = 0.025 m
Maelezo ya Majibu
ai. Simple harmonic motion is periodic because of the restoring force which is proportional to the displacement, acts in the opposite direction to the motion
ii. ♦ Length of the pendulum ♦ Strength of the gravitational field
iii. See the figure above.
b. m = 50g = 0.05kg, g = 10 \(ms^2\), h = 10cm = 0.1m, P.E = ?
P.E = mgh = 0.05 × 10 × 0.1
∴ P.E = 0.05 J
c. Let the extension be ∆l
F = k ∆l = k(\(l_n\) - l )
where k is the spring constant, ln is the new length, l is the natural length, and F = mg.
Total mass = mass of pan + added mass
In the first case,
\(m_1\) = 0.04 + 0.20 = 0.24 kg
\(F_1\) = 0.24 × 10 = 2.4 N
∆l = 0.055 - l
Since F = k ∆l,
2.4 = k(0.055 - l )
2.4 = 0.055k - l k --- (i)
In the second case,
\(m_2\) = 0.04 + 0.28 = 0.32 kg
\(F_2\) = 0.32 × 10 = 3.2 N
∆l = 0.065 - l
Since F = k ∆l,
3.2 = k(0.065 - l )
3.2 = 0.065k - l k --- (ii)
Subtracting equation (i) from (ii)
⇒ 0.8 = 0.01k
k = \(\frac{0.8}{ 0.01}\)
Substitute (80) for k in equation (i)
⇒ 2.4 = 0.055(80) - 80 l
⇒ 2.4 = 4.4 - 80 l
⇒ 80 l = 4.4 - 2.4
⇒ 80 l = 2
l = \(\frac{2}{80}\)
Therefore, the value of k = 80 N/m and l = 0.025 m
