WAEC SSCE - Physics - 2023 (Essay)
Question 1 Report
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
Answer Details
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
Question 2 Report
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 = \(\frac{YAe^2}{2l}\)
Answer Details
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 = \(\frac{YAe^2}{2l}\)
Question 3 Report
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
Answer Details
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
Question 4 Report
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
Answer Details
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
Question 5 Report
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.
Answer Details
| 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.
Question 6 Report
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.
Answer Details
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.
Question 7 Report
ai. Why are parabolic mirrors suitable for use in the headlamps of vehicles?
ii. Draw a ray diagram to illustrate the answer in 10(a)(i).
bi. State two applications of echoes.
ii. An observer standing at a point, P, on the same horizontal ground as the foot, H, of a tower, shouts, and 1.20 s later, he hears the echo. He then moved to another point, Q, 40 m from P, and shouted again but the echo was heard after 1.45 s. Calculate the:
I. distance between P and H;
II. speed of sound in air.
ci. Define the term absolute refractive index of a medium.
ii. A piece of coin falls accidentally into a tank containing two immiscible liquids A and B as illustrated in Fig. 10.0 above.
Calculate the displacement of the coin when viewed vertically from above.
[refractive index of A = 1.3, refractive index of B = 1.4]
ai. Parabolic mirrors, due to their ability to efficiently collect and focus incoming parallel light rays, are suitable for use in vehicle headlamps, ensuring a concentrated and directed beam of light for improved visibility and reduced glare.
ii. See diagram above.
bi. – Sonar Systems: Sonar (Sound Navigation and Ranging) systems use echoes to determine the depth and location of objects underwater. A sound signal is emitted, and the time it takes for the signal to bounce off an object and return as an echo is used to calculate the distance to the object. Sonar is commonly used in marine navigation, fishing, and underwater mapping.
- Medical Imaging: In medical ultrasound imaging, echoes are used to create images of the interior of the body. High-frequency sound waves are directed into the body, and the echoes that bounce back from different tissues are used to generate detailed images of organs, blood vessels, and developing fetuses during pregnancy.
- Architectural Design: Architects and engineers use echoes to evaluate the acoustics of buildings and design spaces with optimal sound qualities. The controlled use of echoes can enhance the sound in concert halls, auditoriums, and theaters, providing a more enjoyable listening experience for audiences.
- Geological Surveys: Seismic surveys use echoes generated by controlled explosions or mechanical sources to map subsurface geological features. By analyzing the time it takes for seismic waves to reflect off different rock layers, geologists can infer the composition and structure of the Earth's subsurface.
- Surveying and Distance Measurement: Echoes are used in surveying and distance measurement equipment like total stations and laser rangefinders. A laser or light signal is emitted, and the time it takes for the light to bounce off a target and return is used to measure distances accurately.
- Echo Sounding in Navigation: In addition to sonar, echo sounding is used in navigation to determine water depth, especially in shallow or coastal areas. A sound signal is sent to the seabed, and the time it takes for the echo to return helps sailors and navigators avoid underwater hazards.
- Musical Effects: Musicians and audio engineers often use echoes and reverb effects in music production. These effects add depth and dimension to music by simulating the reflection of sound in different environments. Echoes and reverb are commonly used in recording studios, live concerts, and digital audio effects.
See diagram above.
ii. V = \(\frac{2d}{t}\)
Case1: V = \(\frac{2d}{1.2}\)
Case2: V = \(\frac{2(40 + d)}{1.45}\)
Equating case 1 and 2
\(\frac{2d}{1.2}\) = \(\frac{2(40 + d)}{1.45}\)
2.9d = 96 + 2.4d
2.9d - 2.4d = 96
0.5d = 96
d = \(\frac{96}{0.5}\)
= 192m
∴ Distance between P and H = 192m
put d = 192m into case(1)
V = \(\frac{2d}{1.2}\)
V = \(\frac{2\times192}{1.2}\) = 320m/s
ci. The absolute refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in the medium. It is a measure of how much a medium slows down light.
ii. Displacement = real depth - \(\frac{real depth}{refractive index }\) ( n = \(\frac{real depth}{apparentdepth }\))
= real depth( 1 - \(\frac{1}{n}\))
For liquid A, displacement,d = 8 (1 - \(\frac{1}{1.3}\)) = 9.23cm
For liquid B, displacement,d = 40( 1 - \(\frac{1}{1.4}\)) = 2.29cm
∴ Total displacement = 9.23 + 2.29 = 11.52cm
∴ the apparent position of the coin is 11.52cm from the bottom.
Answer Details
ai. Parabolic mirrors, due to their ability to efficiently collect and focus incoming parallel light rays, are suitable for use in vehicle headlamps, ensuring a concentrated and directed beam of light for improved visibility and reduced glare.
ii. See diagram above.
bi. – Sonar Systems: Sonar (Sound Navigation and Ranging) systems use echoes to determine the depth and location of objects underwater. A sound signal is emitted, and the time it takes for the signal to bounce off an object and return as an echo is used to calculate the distance to the object. Sonar is commonly used in marine navigation, fishing, and underwater mapping.
- Medical Imaging: In medical ultrasound imaging, echoes are used to create images of the interior of the body. High-frequency sound waves are directed into the body, and the echoes that bounce back from different tissues are used to generate detailed images of organs, blood vessels, and developing fetuses during pregnancy.
- Architectural Design: Architects and engineers use echoes to evaluate the acoustics of buildings and design spaces with optimal sound qualities. The controlled use of echoes can enhance the sound in concert halls, auditoriums, and theaters, providing a more enjoyable listening experience for audiences.
- Geological Surveys: Seismic surveys use echoes generated by controlled explosions or mechanical sources to map subsurface geological features. By analyzing the time it takes for seismic waves to reflect off different rock layers, geologists can infer the composition and structure of the Earth's subsurface.
- Surveying and Distance Measurement: Echoes are used in surveying and distance measurement equipment like total stations and laser rangefinders. A laser or light signal is emitted, and the time it takes for the light to bounce off a target and return is used to measure distances accurately.
- Echo Sounding in Navigation: In addition to sonar, echo sounding is used in navigation to determine water depth, especially in shallow or coastal areas. A sound signal is sent to the seabed, and the time it takes for the echo to return helps sailors and navigators avoid underwater hazards.
- Musical Effects: Musicians and audio engineers often use echoes and reverb effects in music production. These effects add depth and dimension to music by simulating the reflection of sound in different environments. Echoes and reverb are commonly used in recording studios, live concerts, and digital audio effects.
See diagram above.
ii. V = \(\frac{2d}{t}\)
Case1: V = \(\frac{2d}{1.2}\)
Case2: V = \(\frac{2(40 + d)}{1.45}\)
Equating case 1 and 2
\(\frac{2d}{1.2}\) = \(\frac{2(40 + d)}{1.45}\)
2.9d = 96 + 2.4d
2.9d - 2.4d = 96
0.5d = 96
d = \(\frac{96}{0.5}\)
= 192m
∴ Distance between P and H = 192m
put d = 192m into case(1)
V = \(\frac{2d}{1.2}\)
V = \(\frac{2\times192}{1.2}\) = 320m/s
ci. The absolute refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in the medium. It is a measure of how much a medium slows down light.
ii. Displacement = real depth - \(\frac{real depth}{refractive index }\) ( n = \(\frac{real depth}{apparentdepth }\))
= real depth( 1 - \(\frac{1}{n}\))
For liquid A, displacement,d = 8 (1 - \(\frac{1}{1.3}\)) = 9.23cm
For liquid B, displacement,d = 40( 1 - \(\frac{1}{1.4}\)) = 2.29cm
∴ Total displacement = 9.23 + 2.29 = 11.52cm
∴ the apparent position of the coin is 11.52cm from the bottom.
Question 8 Report
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
Answer Details
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
