WAEC SSCE - Physics - 2023 (Essay)
Frage 1 Bericht
a. What is fibre optics?
b. State two reasons why optical fibres are preferred to copper cables in the telecommunication industry.
a. Fiber optics is a technology that uses light to transmit data over long distances. Fiber optic cables are made of thin, flexible strands of glass or plastic that can carry a large amount of data at very high speeds.
b. - Higher bandwidth: Optical fibers can carry a much larger amount of data than copper cables. This is because light travels much faster than electricity and optical fibers can carry more light signals than copper cables can carry electrical signals.
- Longer distances: Optical fibers can transmit data over longer distances than copper cables without signal loss. This is because light signals are less attenuated in optical fibers than electrical signals are in copper cables.
- Immunity to interference: Optical fibers are immune to electrical interference, which can cause problems with copper cables. This is because optical fibers transmit light signals, which are not affected by electrical fields.
- Security: Optical fibers are more secure than copper cables because they are difficult to tap. This is because light signals cannot be detected without damaging the optical fiber.
- Durability: Optical fibers are more durable than copper cables. They are less susceptible to damage from moisture, temperature fluctuations, and physical impact.
- Cost-effectiveness: Optical fibers are becoming more cost-effective than copper cables, especially for high-bandwidth applications. This is because the cost of manufacturing optical fibers has decreased significantly in recent years.
Antwortdetails
a. Fiber optics is a technology that uses light to transmit data over long distances. Fiber optic cables are made of thin, flexible strands of glass or plastic that can carry a large amount of data at very high speeds.
b. - Higher bandwidth: Optical fibers can carry a much larger amount of data than copper cables. This is because light travels much faster than electricity and optical fibers can carry more light signals than copper cables can carry electrical signals.
- Longer distances: Optical fibers can transmit data over longer distances than copper cables without signal loss. This is because light signals are less attenuated in optical fibers than electrical signals are in copper cables.
- Immunity to interference: Optical fibers are immune to electrical interference, which can cause problems with copper cables. This is because optical fibers transmit light signals, which are not affected by electrical fields.
- Security: Optical fibers are more secure than copper cables because they are difficult to tap. This is because light signals cannot be detected without damaging the optical fiber.
- Durability: Optical fibers are more durable than copper cables. They are less susceptible to damage from moisture, temperature fluctuations, and physical impact.
- Cost-effectiveness: Optical fibers are becoming more cost-effective than copper cables, especially for high-bandwidth applications. This is because the cost of manufacturing optical fibers has decreased significantly in recent years.
Frage 2 Bericht
The force, F, acting on the wings of an aircraft moving through the air of velocity, v, and density, ρ, is given by the equation F = \(kv^xρ^yA^z\), where k is a dimensionless constant and A is the surface area of the wings of the aircraft. Use dimensional analysis to determine the values of x, y, and z.
F = \(kv^xρ^yA^z\)
For the left-hand side:
F = mass × acceleration = \(MLT^{-2}\)
For the right-hand side:
v = \(LT^{-1}\), ρ = \(ML^{-3}\) and A = \(L^2\)
So,
\(MLT^{-2} = k(LT^{-1})^x(ML^{-3})^y(L^2)^z\)
Since k is dimensionless,
\(MLT^{-2} = (LT^{-1})^x(ML^{-3})^y(L^2)^z\)
\(MLT^{-2} = L^x × T^{(-1)x} \times M^y × L^{(-3)y} \times L^{(2)z}\)
\(MLT^{-2} = L^x × T^{-x} \times M^y \times L^{-3y} \times L^{2z}\)
\(MLT^{-2} = L^{(x - 3y + 2z)} \times T^{-x} \times M^y\)
\(M^1L^1T^{-2}= L^{(x - 3y + 2z)} \times T^{-x} \times M^{y}\)
Comparing the powers:
For M,
y = 1
For L,
x - 3y + 2z = 1 ---- (i)
For T,
-x = -2
∴ x = 2
Substitute (2) for x and (1) for y in equation (i)
⇒ 2 - 3(1) + 2z = 1
⇒ 2 - 3 + 2z = 1
⇒ -1 + 2z = 1
⇒ 2z = 1 + 1
⇒ 2z = 2
∴ z = 1
Hence, x = 2, y = 1 and z = 1.
Antwortdetails
F = \(kv^xρ^yA^z\)
For the left-hand side:
F = mass × acceleration = \(MLT^{-2}\)
For the right-hand side:
v = \(LT^{-1}\), ρ = \(ML^{-3}\) and A = \(L^2\)
So,
\(MLT^{-2} = k(LT^{-1})^x(ML^{-3})^y(L^2)^z\)
Since k is dimensionless,
\(MLT^{-2} = (LT^{-1})^x(ML^{-3})^y(L^2)^z\)
\(MLT^{-2} = L^x × T^{(-1)x} \times M^y × L^{(-3)y} \times L^{(2)z}\)
\(MLT^{-2} = L^x × T^{-x} \times M^y \times L^{-3y} \times L^{2z}\)
\(MLT^{-2} = L^{(x - 3y + 2z)} \times T^{-x} \times M^y\)
\(M^1L^1T^{-2}= L^{(x - 3y + 2z)} \times T^{-x} \times M^{y}\)
Comparing the powers:
For M,
y = 1
For L,
x - 3y + 2z = 1 ---- (i)
For T,
-x = -2
∴ x = 2
Substitute (2) for x and (1) for y in equation (i)
⇒ 2 - 3(1) + 2z = 1
⇒ 2 - 3 + 2z = 1
⇒ -1 + 2z = 1
⇒ 2z = 1 + 1
⇒ 2z = 2
∴ z = 1
Hence, x = 2, y = 1 and z = 1.
Frage 3 Bericht
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
Antwortdetails
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
Frage 4 Bericht
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
Antwortdetails
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
Frage 5 Bericht
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.
Antwortdetails
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.
Frage 6 Bericht
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
Antwortdetails
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
Frage 7 Bericht
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
Antwortdetails
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
Frage 8 Bericht
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.
Antwortdetails
| 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.
