WAEC SSCE - Physics - 2022 (Essay)

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(a)() Meaning of artificial radioactivity;

The process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and. energy.

(b) The kinetic energy of the electron can be calculated using the formula: KE = qV, where q is the charge of the electron and V is the potential difference. Substituting the given values, we get:

K.E = eV

KE = (1.6 x 10^-19 C)(30,000 V)

KE = 4.8 x 10^-15 J (c)

(i)

To calculate the slope of the graph, you need to determine the change in the dependent variable (kinetic energy) divided by the change in the independent variable (frequency). In this case, you can choose any two points on the graph and calculate the slope using the following formula:

slope = (kinetic_energy2 - kinetic_energy1) / (frequency2 - frequency1)

Let's take two points from the given data, for example:

Point 1: (frequency1, kinetic_energy1) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J)
Point 2: (frequency2, kinetic_energy2) = (8.0 x 10^14 Hz, 1.6 x 10^-19 J)

Now, we can calculate the slope:

slope = (1.6 x 10^-19 J - 0.8 x 10^-19 J) / (8.0 x 10^14 Hz - 6.8 x 10^14 Hz)

slope = 1 x 10^-5 J Hz^(-1).

To determine Planck's constant from the given graph and slope, we can use the equation:

slope = h / e

where h is Planck's constant and e is the elementary charge (1.602176634 x 10^-19 C).

From the previous calculation, the slope of the graph is 1 x 10^-5 J Hz^(-1).

Let's substitute the values into the equation to solve for Planck's constant:

1 x 10^-5 J Hz^(-1) = h / (1.602176634 x 10^-19 C)

To isolate h, we can rearrange the equation:

h = slope * e

Substituting the values:

h = (1 x 10^-5 J Hz^(-1)) * (1.602176634 x 10^-19 C)

Evaluating the expression:

h ≈ 1.602176634 x 10^-24 J·s

Therefore, from the given graph and slope, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s.

(ii) To determine the threshold frequency of radiation from the given information, we need to use the concept of the photoelectric effect and the relationship between the kinetic energy of photoelectrons and the frequency of incident radiation.

According to the photoelectric effect, electrons are ejected from a metal surface when illuminated by electromagnetic radiation of sufficient energy. The minimum frequency of radiation required to eject electrons is known as the threshold frequency.

The relationship between the kinetic energy of photoelectrons and the frequency of incident radiation is given by the equation:

K.E. = h * (frequency - threshold_frequency)

where K.E. is the kinetic energy of the photoelectrons, h is Planck's constant, frequency is the frequency of incident radiation, and threshold_frequency is the threshold frequency.

From the graph, we have the slope, which is equal to h, and the kinetic energy corresponding to each frequency. We can select any point on the graph where the kinetic energy is non-zero and solve for the threshold frequency.

Let's choose the point (frequency, kinetic energy) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J) from the given data.

0.8 x 10^-19 J = slope * (6.8 x 10^14 Hz - threshold_frequency)

Substituting the slope value:

0.8 x 10^-19 J = 1.602176634 x 10^-24 J·s * (6.8 x 10^14 Hz - threshold_frequency)

To solve for the threshold frequency, we can rearrange the equation:

threshold_frequency = 6.8 x 10^14 Hz - (0.8 x 10^-19 J / (1.602176634 x 10^-24 J·s))

Calculating the threshold frequency:

threshold_frequency = 6.8 x 10^14 Hz - 4.992706701 x 10^4 Hz

threshold_frequency ≈ 6.799500729 x 10^14 Hz

Therefore, the threshold frequency of radiation is approximately 6.799500729 x 10^14 Hz.

(iii)

To determine the work function of the metal, we can use the equation:

Work function = h * threshold_frequency

where h is Planck's constant and threshold_frequency is the threshold frequency of radiation.

From the previous calculations, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s and the threshold frequency is approximately 6.799500729 x 10^14 Hz.

Substituting these values into the equation, we can calculate the work function:

Work function = (1.602176634 x 10^-24 J·s) * (6.799500729 x 10^14 Hz)

Work function ≈ 1.090589631 x 10^-9 J

Therefore, based on the given information, the approximate value of the work function of the metal is 1.090589631 x 10^-9 J.

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Semiconductors are materials that have an electrical conductivity between that of a conductor and an insulator. Three properties of a semiconductor are:

1. Conductivity: Semiconductors have intermediate electrical conductivity, which means they are not good conductors like metals but not as poor as insulators. They can conduct electricity under certain conditions, such as when exposed to light or heat, or when impurities are added to them (a process called doping).
2. Energy gap: Semiconductors have a small energy gap between their valence band (the highest energy level occupied by electrons) and their conduction band (the lowest energy level that electrons can occupy and still be free to move). This energy gap determines the conductivity of the material and can be manipulated through doping or other processes.
3. Temperature sensitivity: Semiconductors are sensitive to changes in temperature, which can affect their electrical conductivity. As the temperature of a semiconductor increases, the energy of its electrons increases and they become more likely to move into the conduction band, increasing the material's conductivity. However, if the temperature gets too high, the semiconductor may begin to behave like a conductor and lose its unique properties.

Overall, these properties make semiconductors useful in a wide range of electronic devices, including transistors, diodes, and solar cells.

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(a)

(i) One necessary condition for constructive interference of waves is that the waves have the same frequency, and the crests and troughs of the waves align with each other.

(ii) One necessary condition for total internal reflection is that the angle of incidence is greater than the critical angle for the boundary between two media, and the wave travels from a denser medium to a less dense medium.

(iii) One necessary condition for the production of beats is that two waves with slightly different frequencies interfere with each other, and their amplitudes vary periodically in time.

(b) The velocity of sound in air can be calculated as follows:

- The distance between the first and second position of resonance is 100 cm - 35 cm = 65 cm = 0.65 m.

- The wavelength of the sound wave is twice the distance between the first and second position of resonance, which is 2 x 0.65 m = 1.3 m.

- The frequency of the tuning fork is 256 Hz. - Using the equation v = fλ, where v is the velocity of sound, f is the frequency, and λ is the wavelength, we can calculate the velocity of sound as

v = 256 Hz x 1.3 m = 332.8 m/s.

(c) (i) The three classifications of musical instruments are:

- Stringed instruments

- Wind instruments

- Percussion instruments

(ii) Examples of each classification are:

- Stringed instruments: guitar, violin

- Wind instruments: flute, trumpet

- Percussion instruments: drums, xylophone

(d) The critical angle for light traveling from glass to air can be calculated as follows:

- The refractive index of glass is given as 1.5.

- Using the formula sin ?c = 1/n, where ?c is the critical angle and n is the refractive index, we can calculate the critical angle as sin θc = 1/1.5 = 0.67.

- Taking the inverse sine of 0.67, we can find the critical angle as θc = 42.3 degrees.

(e) The speed of sound in a medium is directly proportional to the square root of the temperature of the medium. Using this relationship, we can calculate the temperature of the medium as follows:

- Let T1 be the temperature of the medium where the speed of sound is 240 m/s, and T2 be the temperature of the medium where the speed of sound is 310 m/s.

- The ratio of the speeds of sound is 310/240 = 1.29.

- The ratio of the square roots of the temperatures is √(T2/T1) = 1.29. - Solving for T2, we get T2 = T1 x (1.29)^2 = T1 x 1.6641. - Substituting T1 = 102 + 273 = 375 K, we get T2 = 625 K. -

Therefore, the temperature of the medium where the speed of sound is 310 m/s is 625 - 273 = 352 °C.

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(a)(i) Coulomb's law of electrostatics states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Mathematically, F = kQ1Q2/r^2, where F is the electrostatic force, Q1 and Q2 are the charges on the objects, r is the distance between them, and k is Coulomb's constant.

(ii) Using Coulomb's law, the magnitude of the electrostatic force between the electron and proton of a hydrogen atom can be calculated as follows:

F = k(Q1Q2/r^2) = (9.0 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)(1.6 x 10^-19 C)/(5.2 x 10^-11 m)^2 = 8.2 x 10^-8 N.

(b)(I) Proof of Vout

Let the current through the circuit be I

Vin

)

OR

I = VinR1+R2

II. Calculation of R1

;

Vout

)

OR

VoutVin=R1R1+R2

(ii) The Volt The unit of potential difference between two points when one joule of work is done in taking one coulomb of charge between the points

(c) Wood is not suitable for use as the core of transformers because it is not a good conductor of electricity. Transformers require a material with a high magnetic permeability, which allows for efficient transfer of energy between coils. Wood does not have the necessary magnetic properties and can also be damaged by the heat generated during operation.

(d) One application for the cathode ray tube (CRT) is in television and computer displays. A CRT works by producing a beam of electrons that is directed towards a phosphorescent screen, causing it to emit light and create an image. By controlling the path of the electron beam, the image can be manipulated and displayed on the screen. Although CRT technology has largely been replaced by LCD and LED displays, it is still used in some specialized applications such as medical imaging and oscilloscopes.

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(a) The S.I. unit of intensity of blackbody radiation is watt per square metre, \(\mathrm{W\,m^{-2}}\).

(b) Intensity-wavelength curves for a perfect blackbody at two different temperatures are shown below.

1. As the temperature increases, the maximum intensity increases. Thus, the curve for \(T_2\) has a higher peak than that for \(T_1\).
2. As the temperature increases, the wavelength at which maximum intensity occurs decreases. Thus, \(\lambda_{\max}\) shifts towards shorter wavelengths, in accordance with \(\lambda_{\max}T=\text{constant}\).

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Magnetic and non-magnetic materials are two distinct categories of materials. Here are three differences between these two: 1. Magnetic materials are attracted to a magnet, while non-magnetic materials are not. This means that if you hold a magnet near a magnetic material, it will be pulled towards the magnet, but the same won't happen with a non-magnetic material. 2. Magnetic materials can be magnetized, whereas non-magnetic materials cannot. Magnetization is the process of aligning the magnetic domains of a material, and this is only possible with magnetic materials. 3. Magnetic materials have magnetic domains, while non-magnetic materials do not. Magnetic domains are regions within a material where the magnetic moments of the atoms are aligned in the same direction, creating a magnetic field. In non-magnetic materials, the magnetic moments are randomly oriented, resulting in no net magnetic field. These are the three main differences between magnetic and non-magnetic materials.

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|PQ| = Range = Ux.T

= UT Cosθ $\mathrm{<br>}$

= 250 x 40 Cos 42º

 = 7431 m (7.4 km)

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(a) Results and graph

The external diameter of the boiling tube measured with the slide vernier caliper is:

\[D=2.41\ \text{cm}\]

Hence,

\[A=\frac{\pi D^2}{4}=\frac{3.14(2.41)^2}{4}=4.56\ \text{cm}^2.\]

The initial depth of immersion of the loaded boiling tube is:

\[y_0=3.000\ \text{cm}.\]

Table of readings

Mass, \(m\) (g)	Depth, \(y\) (cm)	\(h=y-y_0\) (cm)	\(\log h\)	\(\log m\)
2	3.439	0.439	−0.358	0.301
5	4.097	1.097	0.040	0.699
7	4.535	1.535	0.186	0.845
10	5.193	2.193	0.341	1.000
12	5.632	2.632	0.420	1.079

The graph of \(\log m\) against \(\log h\) is shown below.

Using two widely separated points on the best-fit line, \((-0.358,\ 0.301)\) and \((0.420,\ 1.079)\),

\[\text{gradient}=\frac{1.079-0.301}{0.420-(-0.358)}=\frac{0.778}{0.778}=1.00.\]

Thus, \(\log m=\log h+0.659\), showing that \(m\propto h\).

(b)(i)

Law of flotation: A body floating in a fluid displaces a quantity of the fluid whose weight is equal to the weight of the body.

(b)(ii)

Mass of rod:

\[m=\rho V=720(0.1)(0.63)=45.36\ \text{kg}.\]

Weight of rod:

\[W=mg=45.36\times10=453.6\ \text{N}.\]

Density of liquid:

\[\rho_l=1.26\times1000=1260\ \text{kg m}^{-3}.\]

If \(y\) is the depth of immersion, then the upthrust is

\[U=\rho_l(0.1y)g=1260(0.1y)(10)=1260y.\]

For flotation, upthrust equals weight:

\[1260y=453.6.\]

\[y=\frac{453.6}{1260}=0.36\ \text{m}.\]

Depth of immersion = \(0.36\ \text{m}\).

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(a) Two factors that affect the rate of evaporation of a liquid are: 1. Temperature: Higher temperatures generally result in faster rates of evaporation, as the heat energy increases the kinetic energy of the liquid molecules, causing them to move faster and escape into the air more readily. 2. Surface area: A larger surface area of the liquid exposed to the air can increase the rate of evaporation, as there are more liquid molecules available to escape into the air. (b) Latent heat refers to the amount of heat energy that is required to change the phase of a substance without changing its temperature. For example, when ice is heated, it melts and changes from a solid to a liquid, but its temperature remains constant at 0°C until all the ice has melted. The energy absorbed during this phase change is known as the latent heat of fusion. (c) (i) On a dry day, water in a clay pot is cooler than water in a rubber container because the water in the clay pot evaporates faster due to the porous nature of the clay, which allows air to circulate through it. As the water evaporates, it removes heat from the remaining water, causing it to cool down. (ii) Cooking of food is faster in a pressure cooker than in an ordinary pot because the pressure inside the cooker increases the boiling point of water. This means that the food can be cooked at a higher temperature than in an ordinary pot, resulting in faster cooking times. (d) (i) Latent heat of fusion can be calculated using the formula Q = mL, where Q is the heat energy absorbed, m is the mass of the substance, and L is the latent heat of fusion. In this case, Q = (40 V x 12 A x 1400 s) = 672,000 J, and m = 1.5 kg. Assuming that the temperature of the substance was constant during melting, L can be calculated as L = Q/m = 672,000 J / 1.5 kg = 448,000 J/kg. (ii) Specific heat capacity can be calculated using the formula c = Q/(mΔT), where c is the specific heat capacity, ΔT is the change in temperature, and all other variables are the same as in part (i). In this case, Q = (40 V x 12 A x 72 s) = 34,560 J and ΔT = 60°C. Substituting these values into the formula, we get c = 34,560 J / (1.5 kg x 60°C) = 384 J/(kg·°C). (e) Two differences between evaporation and boiling are: 1. Evaporation occurs at the surface of a liquid, while boiling occurs throughout the liquid. 2. Evaporation occurs at a range of temperatures below the boiling point of the liquid, while boiling occurs only at the boiling point. Evaporation is a slower process than boiling, as it only occurs at the surface of the liquid, while boiling involves the rapid formation of bubbles throughout the liquid.

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(a)(i) The law of inertia states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by an external force.

(a)(ii) Wearing a safety belt in a moving vehicle could reduce the possibilities of severe injuries when the vehicle is involved in a collision because of the law of inertia. When a car suddenly stops in a collision, the passengers in the car tend to continue moving forward at the same speed the car was moving before the collision. If they are not restrained by a seat belt, they will keep moving forward and could collide with the dashboard, steering wheel, or windshield. However, if they are wearing a seat belt, the belt applies a force to the passenger in the opposite direction, keeping them from moving forward, and reducing the risk of injury.

(b)(i) The moment of a force is the turning effect of the force about a pivot and is given by the product of the force and the perpendicular distance from the pivot to the line of action of the force.

(b)(ii)
i. The force diagram for the arrangement is a diagram that shows all the forces acting on an object in a particular situation. For this arrangement, the force diagram will show the weight of the plank acting downwards at the center, the tension T acting upwards at point A, and the reaction force F acting upwards at the knife edge.

ii. To determine the tension, T, in the string, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge. Thus,

54 N x 0.50 m = T x 1.50 m.

Therefore, T = 18 N.

iii. To determine the force, F, acting on the knife edge, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge and at equilibrium, the sum of upward forces = the sum of forces acting downward

T + F = 54

T = 18 N

18 + F = 54

F = 36 N

(c) Three differences between solid friction and viscosity are:
1. Solid friction is the force that opposes the motion of an object along a solid surface, whereas viscosity is the force that opposes the motion of an object through a fluid.
2. Solid friction depends on the nature of the surfaces in contact, whereas viscosity depends on the properties of the fluid, such as its density and viscosity.
3. Solid friction is independent of velocity, whereas viscosity is directly proportional to velocity.

(d) One method of increasing the velocity ratio of a pulley system is to increase the number of pulleys in the system. The velocity ratio of a pulley system is the ratio of the distance moved by the effort to the distance moved by the load. The more pulleys there are in the system, the greater the distance the effort can move for a given distance moved by the load, thus increasing the velocity ratio.

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(a) A geostationary satellite is an artificial satellite that orbits the Earth at the same rate as the Earth's rotation. This means that it stays fixed at a particular point in the sky over the Earth's equator. From the ground, it appears as though the satellite is stationary. (b) Two types of lasers are solid-state lasers and gas lasers. Solid-state lasers use a solid material as the active medium, while gas lasers use a gas or a mixture of gases as the active medium. Solid-state lasers are commonly used in industry, medicine, and research, while gas lasers are used for applications such as cutting, welding, and laser printers.

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(a) Observations, table and graph

The external diameter of the boiling tube is

\[D=1.20\ \text{cm}\]

Hence, the cross-sectional area is

\[A=\frac{\pi D^2}{4}=\frac{3.14(1.20)^2}{4}=1.13\ \text{cm}^2.\]

The initial depth of immersion of the loaded boiling tube is

\[y_0=3.0\ \text{cm}.\]

Mass added, \(m\) (g)	Depth of immersion, \(y\) (cm)	\(h=y-y_0\) (cm)	\(\log h\)	\(\log m\)
2	4.8	1.8	0.255	0.301
5	7.4	4.4	0.643	0.699
7	9.2	6.2	0.792	0.845
10	11.9	8.9	0.949	1.000
12	13.6	10.6	1.025	1.079

The graph of \(\log m\) against \(\log h\) is shown below.

From the line of best fit,

\[\text{gradient}=\frac{\Delta(\log m)}{\Delta(\log h)}\approx1.01\approx1.\]

Thus, \(m\propto h\). This agrees with the flotation relation \(m=\rho Ah\), where \(\rho\) and \(A\) are constant.

(b)(i) Law of flotation

A floating body displaces a weight of fluid equal to its own weight.

(b)(ii) Depth of immersion of the rod

Density of liquid:

\[\rho_l=1.26\times1000=1260\ \text{kg m}^{-3}.\]

For equilibrium of the floating rod,

\[\rho_rALg=\rho_lAyg,\]

where \(y\) is the immersed length. Therefore,

\[y=\frac{\rho_rL}{\rho_l}=\frac{720\times0.63}{1260}=0.36\ \text{m}.\]

Depth of immersion = \(0.36\ \text{m}\).

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(a) Potentiometer experiment

(i) The e.m.f. of the battery is:

\[E=2.0\ \text{V}\]

With the jockey at \(U\), the ammeter reading is:

\[i=1.00\ \text{A}\]

(ii) Table of results

Distance, \(d\) (cm)	Current, \(I\) (A)	\(I^{-1}\) (A−1)
30.0	0.769	1.30
40.0	0.714	1.40
50.0	0.667	1.50
60.0	0.625	1.60
70.0	0.588	1.70

As \(d\) increases, \(I\) decreases.

(iii) Graph of \(d\) against \(I^{-1}\)

(iv) Slope of the graph

Using two widely separated points on the straight line, \((1.20\ \text{A}^{-1},20.0\ \text{cm})\) and \((1.70\ \text{A}^{-1},70.0\ \text{cm})\):

\[s=\frac{70.0-20.0}{1.70-1.20}=\frac{50.0}{0.50}=100\ \text{cm A}.\]

(v) Value of \(I^{-1}\) when \(d=0\)

From the intercept on the \(I^{-1}\)-axis,

\[I^{-1}=1.00\ \text{A}^{-1}.\]

Therefore,

\[I=\frac{1}{1.00}=1.00\ \text{A},\]

which agrees with the current \(i\) when the jockey is at \(U\).

(vi) Resistance per unit length, \(\delta\)

Given that

\[s=\frac{E}{\delta},\]

\[\delta=\frac{E}{s}=\frac{2.0}{100}=0.020\ \Omega\,\text{cm}^{-1}.\]

(vii) Precautions

1. The key was closed only while taking a reading, to prevent heating of the potentiometer wire and consequent change in resistance.
2. The jockey was touched lightly on the wire and not pressed hard, to ensure good contact without damaging the wire.

(b)

(i) The resistance of a uniform wire is given by

\[R=\rho\frac{l}{A}.\]

\(R\) is the resistance of the wire, \(\rho\) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is its cross-sectional area.

(ii)

\[P=VI=240\times0.75=180\ \text{W}=0.180\ \text{kW}.\]

Energy used in 10 hours:

\[E=Pt=0.180\times10=1.80\ \text{kWh}.\]

Cost of energy:

\[\text{Cost}=1.80\times\$0.50=\boxed{\$0.90}.\]

Awọn alaye Idahun

(a) Results and graph

Mass of unloaded metre rule, \(M_0=75.0\text{ g}\).

Mass of weight hanger, \(m=20.0\text{ g}\).

Balance point of unloaded metre rule: \(d=AG=50.0\text{ cm}\).

For each load, \(y=z-5\), \(x=d-z\), and \(v=\dfrac{x}{y}\).

\(M\) (g)	\(z=AK\) (cm)	\(y=z-5\) (cm)	\(x=d-z\) (cm)	\(v=x/y\)
20	34.3	29.3	15.7	0.536
40	30.0	25.0	20.0	0.800
60	26.8	21.8	23.2	1.064
80	24.3	19.3	25.7	1.332
100	22.3	17.3	27.7	1.601

The plotted graph of \(M\) against \(v\), with both axes beginning at the origin, is shown below.

Using two widely separated points on the line of best fit, \((v_1,M_1)=(0.536,20)\) and \((v_2,M_2)=(1.601,100)\):

\[s=\frac{M_2-M_1}{v_2-v_1}=\frac{100-20}{1.601-0.536}=75.1\text{ g}\approx75.0\text{ g}.\]

The vertical intercept is \(c\approx-20.0\text{ g}\).

Thus, within experimental accuracy, \(s=M_0\) and \(c=-m\).

Precautions

1. The metre rule was allowed to come to rest and was balanced horizontally before each reading was taken.
2. All scale readings were taken with the eye vertically above the mark to avoid parallax error.

(b)

(i) An object is in stable equilibrium if, when slightly displaced, its centre of gravity rises and a restoring moment acts to return it to its original position.

(ii) Let \(y\) be the distance of the support from the \(100\text{ N}\) body. Taking moments about the support:

\[100y=50(6-y)\]

\[100y=300-50y\]

\[150y=300\]

\[y=2.0\text{ m}.\]

Therefore, the support should be placed \(\boxed{2.0\text{ m}}\) from the \(100\text{ N}\) body.

Awọn alaye Idahun

Dimensions of the angular momentum, K =

[K]2[m1][r]2=[G][m1][m2][r] $\frac{<br>}{[}=\frac{<br>}{}$

[k]2ML2=M−1L3T−2M2L $\frac{\mathrm{<br>\; <br>}}{\mathrm{<br>}}=\frac{{\mathrm{<br>}}^{}}{}$

[K] = (M2 ${}^2$L4T−2 ${}^{-2}$)½

∴ $\therefore$ [k] = ML2T−1

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(a) Potentiometer Experiment

The e.m.f. of the battery was measured as:

\(E=3.0\text{ V}\)

With the jockey at \(U\), \(d=0\) and the ammeter reading was:

\(i=1.50\text{ A}\)

The readings obtained are shown below. The reciprocal current was evaluated from \(I^{-1}=1/I\).

Distance, \(d\) (cm)	Current, \(I\) (A)	\(I^{-1}\) (A−1)
30.0	1.154	0.867
40.0	1.111	0.900
50.0	1.071	0.933
60.0	1.034	0.967
70.0	1.000	1.000

A graph of \(d\) against \(I^{-1}\), with both axes beginning at the origin, is plotted below.

Using two widely separated points on the straight line, \((0.867\text{ A}^{-1},30.0\text{ cm})\) and \((1.000\text{ A}^{-1},70.0\text{ cm})\):

\[ s=\frac{70.0-30.0}{1.000-0.867} =\frac{40.0}{0.133} \approx 3.00\times10^2\text{ cm A}. \]

Hence, the slope of the graph is \(3.00\times10^2\text{ cm A}\).

Extrapolating the straight line to \(d=0\),

\(I^{-1}=0.667\text{ A}^{-1}\).

Therefore, \(I=1/0.667=1.50\text{ A}\), which agrees with the current when the jockey is at \(U\).

Given that \(s=E/\delta\),

\[ \delta=\frac{E}{s}=\frac{3.0}{3.00\times10^2} =1.00\times10^{-2}\ \Omega\text{ cm}^{-1}. \]

Precautions

1. The key was closed only while a reading was being taken, to prevent heating and consequent change in resistance of the wire.
2. The jockey was touched lightly on the potentiometer wire and not dragged along it, to avoid damaging the wire or altering its resistance.

(b)

(i) The resistance of a uniform wire is given by

\[R=\frac{\rho l}{A}.\]

• \(R\) is the resistance of the wire in ohms \((\Omega)\).
• \(\rho\) is the resistivity of the material in \(\Omega\,\text{m}\).
• \(l\) is the length of the wire in metres.
• \(A\) is the cross-sectional area of the wire in \(\text{m}^2\).

(ii)

\[ P=VI=240\times0.75=180\text{ W}=0.180\text{ kW}. \] \[ \text{Electrical energy}=0.180\times10=1.80\text{ kWh}. \] \[ \text{Cost}=1.80\times\$0.50=\$0.90. \]

Therefore, the cost of operating the fan for 10 hours is \(\$0.90\).

Awọn alaye Idahun

Calculation of workdone by the force to scratch the string

W = ½ ke\(^{2}\) = \(\frac{f^{2}}{2k}\)

W = \(\frac{50^{2}}{2 \times 100}\)

W = 12.5 J

OR

e = \(\frac{F}{K} = \frac{50}{100}\) = 0.5m

w = ½ = ke\(^{2}\)

w = ½ x 100 x 0.5\(^{2}\)

W = 12.5 J