WAEC SSCE - Physics - 2022

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r = resistance of the galvanometer

l = current through galvanometer = 20mA or 0.02A

V1 = p.d across the galvanometer = I X r = 0.02 X r

: V1 = 0.02r

V2 =  p.d across the  multiplier = 8 - 0.02r

R = resistance of the multiplier = 395Ω

where R = VI $\frac{\mathrm{<br>}}{}$ 

--> 395 = 8−0.02r0.02 $\frac{\mathrm{<br>}}{}$

CROSS MULTIPLY

8−0.02r = 395 * 0.02

8−0.02r = 7.9

--> 0.02r = 8 - 7.9

∴ r = 0.10.02 $\frac{0.1}{0.02}$

 = 5Ω

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The periodic rise and fall in the intensity of sound produced when two notes of nearly equal frequencies are sounded together is called a "beat". When two sound waves of slightly different frequencies interfere with each other, they create a periodic variation in the sound intensity, resulting in a pulsating sound. This pulsation is the beat. The beat frequency is equal to the difference between the frequencies of the two notes. For example, if one note has a frequency of 440 Hz and the other note has a frequency of 442 Hz, the beat frequency would be 2 Hz. The other options, "doppler effect", "interference", and "resonance", are different phenomena in acoustics. The Doppler effect is the change in frequency of a sound wave due to the motion of the source or the observer. Interference is the combination of two or more waves that results in a new wave pattern. Resonance is the tendency of a system to vibrate with maximum amplitude at a specific frequency called its natural frequency. Therefore, the correct answer to the question is "beat".

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A freely suspended compass needle on the earth's surface settles in a plane called magnetic meridian. This is because the needle of a compass always points towards the earth's magnetic north pole due to the magnetic field of the earth. The plane containing the magnetic north and south poles of the earth is called the magnetic meridian. This is different from the geographic meridian, which is the imaginary line on the earth's surface passing through the geographic poles. The magnetic declination is the angle between the magnetic meridian and the geographic meridian at a particular location, while isoganals are lines connecting points with the same magnetic declination.

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Two conductors with current flowing in the same direction will attract. 

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The formula for the period of a wave is given by T = 1/f, where T is the period and f is the frequency. In this case, the frequency of the radio wave is 10 kHz, which means that f = 10,000 Hz. We can use the formula v = λf, where v is the velocity of the wave, λ is the wavelength, and f is the frequency. Solving for λ, we get λ = v/f. Substituting the given values, we get λ = 10 m/s / 10,000 Hz = 0.001 m. Now, we can use the formula T = λ/v to find the period. Substituting the given values, we get T = 0.001 m / 10 m/s = 0.0001 s. Therefore, the period of the 10 kHz radio wave traveling at 10 m/s is 0.0001 seconds or 1.0 x 10^-4 s. The answer is the second option.

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Joule as the unit of work done or energy used.

Force = mass × acceleration
 Work = Force × Distance
 Work = mass x acceleration x distance

Joule = kg × ms2 × m

→ kg m2 ${}^2$ s−2 ${}^{-2}$ 

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Given Data: Emf = 1v, r = 2 ohms, R = 3 ohms, I = ?

3 resistance in parallel = 1r1 + 1r2 $\frac{\mathrm{<br>}}{}$ + 1r3 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);">3</span>}}{}$

1rT= 12 $\frac{1}{2}$ + 12 $\frac{1}{2}$ + 12 $\frac{1}{2}$

1rT = 1+1+12 $\frac{1+1+1}{2}$

1rT $\frac{\mathrm{<br>}}{}$ = 32 $\frac{3}{2}$

cross multiply

rT ${}_{\mathrm{<br>}}$ = 23 $\frac{\mathrm{<br>}}{}$ or 0.67

E = I(R+r) 

1 = I(3+0.67)

1 = I(3.67)

1 ÷ 3.67 = I

0.27 = I

: I ≈ 0.30A

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The viscosity of a fluid refers to its resistance to flow. It is affected by various factors such as the relative motion between the layers of the fluid, the nature of the material of the fluid, and the normal reaction in the fluid. However, the surface area of the fluid in contact is not a factor that affects viscosity. In simple terms, viscosity is affected by things like how thick or thin a fluid is and how easily its layers slide past each other. The larger the normal reaction in the fluid or the stickier the fluid, the higher the viscosity.

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Given Data: L = 10, V = 50, F = 100, I = ?

Inductive Reactance [XL ${}_{\mathrm{<br>}}$] = 2π $\mathrm{<br>}$FL → 2 * 3.14 * 100 * 10

XL
 = 6,280

Current[I] = VXL $\frac{\mathrm{<br>}}{}$ → 506280 $\frac{50}{6280}$

I = 0.0079 

≈ 0.008A 

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The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. 

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A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. A marble S on a flat horizontal surface is an example.

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The period of a simple pendulum is doubled, when its length is made four times.

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When the particles of a medium vibrate perpendicular to the direction of wave propagation, the wave is called a transverse wave. In a transverse wave, the particles move up and down or side to side, but the wave itself moves forward. This type of wave is often seen in water waves, where the water moves up and down while the wave moves forward, or in light waves, where the electric and magnetic fields oscillate up and down perpendicular to the direction of the wave.

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An example of a mechanical wave is "water waves". A mechanical wave is a type of wave that requires a medium (a substance or material) to travel through. When the medium is disturbed, it creates a disturbance that propagates through the medium, carrying energy with it. This disturbance can be in the form of oscillations or vibrations of particles in the medium. Water waves are a type of mechanical wave that propagates through water as a medium. When wind or a disturbance creates a ripple in the water, the ripple spreads out in a circular pattern, with the particles of water moving up and down as the wave passes through them. This motion of the water particles is the oscillation that carries the energy of the wave. On the other hand, radio waves, X-rays, and light rays are all examples of electromagnetic waves, which do not require a medium to travel through. Electromagnetic waves consist of oscillating electric and magnetic fields that propagate through space. They can travel through a vacuum and do not need a material medium to carry their energy. Therefore, the correct answer to the question is "water waves".

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Vernier calipers commonly used in industry provide a precision to 0.01 mm (10 micrometres), or 0.01cm or one thousandth of an inch.

 Usually, it is the total length of the main scale. Vernier calipers, generally, have a range of 300 mm

300 millimeters equal 30.0 centimeters (300mm = 30.0cm).

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In a nuclear reactor, neutrons are used to induce and sustain a chain reaction. Fast-moving neutrons, however, are not very efficient at inducing fission, so they need to be slowed down to increase the likelihood of fission. This process is called moderation. One way to moderate neutrons is by using materials with light nuclei that can scatter neutrons without absorbing them, such as graphite. Therefore, the correct answer is "graphite rods".

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The basic principle of operation of a beam balance is the principle of moments. This principle states that when an object is placed on one side of a beam balance, it creates a moment (or a turning force) around the pivot point, which is the center of the beam. This moment is counteracted by an equal moment created by a weight placed on the other side of the beam. The weight on the other side is moved along a calibrated scale until the beam is in a horizontal position, indicating that the moments on both sides are equal. This means that the weight of the object can be determined by reading the position of the weight on the calibrated scale.

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 friction, force that resists the sliding or rolling of one solid object over another.

Examples of friction in our daily life

• Driving of a a vehicle on a surface.
• Applying brakes to stop a moving vehicle.
• Skating.
• Walking on the road.

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The kinetic-molecular theory of gases assumes that ideal gas molecules:

(1) are constantly moving;

(2) have negligible volume;

(3) have negligible intermolecular forces;

(4) undergo perfectly elastic collisions; and

(5) have an average kinetic energy proportional to the ideal gas's absolute temperature.

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If a body moves with uniform retardation, the velocity-time graph is a straight line.

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To calculate the effort required to lift a load of mass 20kg using a wheel and axle machine with 100% efficiency and radii ratio of 5:1, we need to use the principle of moments. The principle of moments states that the sum of the moments acting on a system in equilibrium is equal to zero. In this case, we can assume that the system is in equilibrium since the wheel and axle are not rotating, and the load is not moving. Let R1 and R2 be the radii of the wheel and axle, respectively. Since the ratio of their radii is 5:1, we have R1 = 5R2. Let E be the effort required to lift the load and G be the weight of the load. We have G = 20kg × 10 m/s² = 200N. The moment of the effort E about the axle is ER1, and the moment of the weight G about the wheel is GR2. Since the machine is in equilibrium, we have: ER1 = GR2 Substituting R1 = 5R2 and G = 200N, we get: E(5R2) = 200N(R2) Simplifying the equation, we get: E = 40N Therefore, the effort required to lift the load of mass 20kg using this machine is 40N.

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Archimede's Principle states that a body immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced, and this is fundamental to the equilibrium of a body floating in still water.

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The answer is "melting point" because the total internal energy of a liquid depends on various factors, such as its temperature, mass, and specific heat capacity. However, the melting point of a substance is not related to the total internal energy of the liquid, but rather to the temperature at which the solid form of the substance changes into a liquid state. Therefore, "melting point" is the odd one out, and does not affect the total internal energy of the liquid.

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We can solve this problem using the principle of conservation of energy. When the object is at point A, it has some gravitational potential energy given by mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above some reference level. When the object falls to point B, it loses some potential energy, which is converted into kinetic energy. We can write this as: mgh = (1/2)mv^2 where v is the velocity of the object at point B. Rearranging the equation, we get: v = √(2gh) Substituting the given values, we get: v = √(2 × 10 × 30) = √600 = 24.5 m/s Therefore, the velocity of the object when it reaches point B is 24.5 m/s. So, the correct answer is "24.5 m/s".

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A plane mirror forms a virtual image of an object. The image is always erect (upright), of the same size as the object, and laterally inverted (left and right are switched). Therefore, the correct option is: "erect and bigger than the object". This statement is not correct as the image formed by a plane mirror is of the same size as the object and not bigger.

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Sensitivity of a galvanometer increases when number of turns, area of coil and magnetic field increase

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Gamma rays are produced when energy changes occur within the nuclei of atoms. Gamma rays are high-energy electromagnetic radiation that are emitted by atomic nuclei during radioactive decay or nuclear reactions. These changes within the nuclei of atoms can result in the release of excess energy in the form of gamma rays. This process is known as gamma decay. Gamma rays have the highest energy and shortest wavelength in the electromagnetic spectrum, and they can penetrate most materials, making them useful in a wide range of applications, including medical imaging and cancer treatment.

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The statement that is not correct is "The electric potential energy of a negatively charged particle increases when it moves to a point of lower potential." This statement contradicts the definition of electric potential energy. Electric potential energy is the energy that a charged particle possesses due to its position in an electric field. The potential energy of a charged particle is directly proportional to the potential difference between two points in the electric field, where the potential difference is defined as the change in potential energy per unit charge. Therefore, a negatively charged particle would move towards a point of higher potential energy since it moves towards a region of higher electric field. When a positively charged particle moves towards a point of higher potential energy, it gains electric potential energy, and when it moves towards a point of lower potential energy, it loses electric potential energy. The work done in taking a charged particle around a closed path in an electric field is zero because the potential difference between the starting point and ending point is zero.

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The greatest force is experienced when the conductor is at right angle to the field.

I.e at perpendicular or Ø = 90°

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The ionization of air refers to the process of creating charged particles by adding or removing electrons from atoms or molecules in the air. The ability of a radiation to ionize air depends on its energy and type. Out of the given options, the correct order of increasing ionization of air is "II < III < I". Gamma radiation has the highest energy and is the most penetrating type of radiation, but it has the lowest ionizing power because it interacts weakly with matter. It can only ionize air when it directly hits an atom or molecule. Beta radiation has a medium energy and a higher ionizing power than gamma radiation. It can ionize air when it collides with an atom or molecule and transfers its energy to it, causing it to eject an electron. Alpha radiation has the lowest energy and the highest ionizing power among the three types of radiation. It can ionize air when it collides with an atom or molecule, transfers its energy to it, and ejects one or more electrons. Therefore, the correct order of increasing ionization of air is "II < III < I", where "II" represents gamma radiation, "III" represents beta radiation, and "I" represents alpha radiation.

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When a sound is produced, it travels through the air as a sound wave. The speed at which sound waves travel through air is constant and depends on the air temperature. Assuming a temperature of 20°C, the speed of sound in air is approximately 340 m/s. Now, let's consider the situation in the problem. The hunter fires a gun and hears an echo 2.4 s later. The echo is the result of the sound waves reflecting off the cliff and traveling back to the hunter's ear. We know that the distance between the hunter and the cliff is 408 m. The sound wave has to travel this distance twice, once from the gun to the cliff and then from the cliff back to the hunter. So, the total distance traveled by the sound wave is 2 × 408 = 816 m. We also know that the time taken for the sound wave to travel this distance is 2.4 s. Using the formula speed = distance / time, we can calculate the speed of the sound wave as: speed = distance / time = 816 m / 2.4 s ≈ 340 m/s Therefore, the speed of the sound wave is approximately 340 m/s, which is the first option in the list.

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Radiation is a method of heat transfer that does not require a material medium. It is the process of emission of energy in the form of electromagnetic waves from a warm body to its surrounding environment. Radiation can travel through a vacuum and can also travel through transparent media like air, water, and glass. This means that heat can be transferred from one object to another through radiation even if they are not in contact with each other or if there is no physical medium between them.

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cubic and linear expansivity are related by this equation. γ=3α.

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A mercury-in-glass thermometer measures temperature based on the expansion of the mercury in the bulb as the temperature changes. The length of the mercury column in the glass tube rises or falls as the temperature increases or decreases. In this question, the distance between the fixed points of the thermometer is given as 30 cm. The lower fixed point is the temperature at which the mercury bulb is completely immersed in melting ice and the upper fixed point is the temperature at which the mercury bulb is completely immersed in steam at standard atmospheric pressure. If the mercury level is 10.5 cm above the lower fixed point, then the length of the mercury column above the lower fixed point is 10.5 cm. To find the temperature corresponding to this length, we can use the following formula: (T - T_lfp)/(T_ufp - T_lfp) = l/l_fp where T is the temperature, T_lfp is the temperature of the lower fixed point (0 °C), T_ufp is the temperature of the upper fixed point (100 °C), l is the length of the mercury column above the lower fixed point (10.5 cm), and l_fp is the distance between the fixed points (30 cm). Solving for T, we get: (T - 0)/(100 - 0) = 10.5/30 T = (10.5/30) x 100 T = 35 °C Therefore, the temperature when the mercury level is 10.5 cm above the lower fixed point is 35 °C. The correct answer is: 35.0 ºC

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The power of a lens, P is defined as the reciprocal of the focal length of the lens, f in meters, expressed in diopters (D). Therefore, we can use the formula P = 1/f to find the focal length of the lens. For a thin lens, the radius of curvature, R of each surface of the lens is related to the focal length, f, by the formula: 1/f = (n - 1) × (1/R1 - 1/R2) Where n is the refractive index of the lens medium, R1 is the radius of curvature of one surface and R2 is the radius of curvature of the other surface. For a single curved surface of a thin lens, the formula reduces to: 1/f = (n - 1) × (1/R) where R is the radius of curvature of the curved surface. Rearranging the formula, we get: R = 1 / ((n-1) x f) Substituting P = 2.5D = 2.5m⁻¹ in place of f, and assuming the refractive index of the lens medium is 1, we get: R = 1 / ((1-1) x 2.5) = undefined Therefore, it is not possible to determine the radius of curvature of the lens from the given information. The answer is undefined.

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As the temperature of water rises, its density changes due to thermal expansion or contraction. Water is most dense at 4°C, and its density decreases both below and above this temperature. As a result, when water is heated from 0°C to 30°C, its density will decrease with each degree of temperature rise. Therefore, the correct option is "fall steadily."

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The statement that is not correct is: "They are longitudinal". Electromagnetic waves are actually transverse in nature, which means that the electric and magnetic fields oscillate perpendicular to the direction of wave propagation. Longitudinal waves, on the other hand, have oscillations parallel to the direction of wave propagation. The other three statements are correct: electromagnetic waves carry energy as they travel through space, they travel with the speed of light, and the electric and magnetic fields are at right angles to each other.

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Acceleration[a] = [forward−resistive]forcemass

a = 1240−8001220 $\frac{1240-800}{1220}$ → 440−8001220 $\frac{440-800}{1220}$

a = 0.36m/s2 ${}^2$,  v = 4m/s, s = ?

Second equation of motion:

s is distance
u is initial velocity
v is final velocity
a is acceleration

v2 ${}^2$ = u2 ${}^2$ + 2as

42 ${}^2$ = 02 ${}^2$ + 2 * 0.36 * s

16 =  0.72s 

22.2m = s

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The correct relationship between the displacement, s, of a particle initially at rest in a linear motion and the time, t, is s α t2. This means that the displacement of a particle is directly proportional to the square of the time elapsed when the particle is moving in a straight line with a constant acceleration. In other words, if the time is doubled, the displacement will be four times greater; if the time is tripled, the displacement will be nine times greater, and so on. This relationship is also known as the second equation of motion, which is commonly used in physics to describe the motion of objects under constant acceleration.

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The problem involves the concept of linear expansivity. The diameter of the brass ring increases with an increase in temperature. We can use the formula for linear expansivity to solve the problem: ΔL = αLΔT where ΔL is the change in length, L is the original length, α is the linear expansivity and ΔT is the change in temperature. Since we are given the initial diameter, we can use it to find the original length (L) of the ring. L = πd/2 where d is the diameter. Substituting the values, we get: L = π × 50.0/2 = 78.54 cm The change in length (ΔL) is given by the difference in the final and initial lengths: ΔL = 50.29/2 - 50.0/2 = 0.145 cm Substituting the given values of α and L, we get: ΔL = αLΔT 0.145 = (1.9 × 10^-5) × 78.54 × ΔT Solving for ΔT, we get: ΔT = 0.145/(1.9 × 10^-5 × 78.54) = 116.3 °C Therefore, the temperature to which the ring must be heated is: 30 + 116.3 = 146.3 °C Rounding off to one decimal place, we get: 146.3 °C Hence, the correct option is 152.6 °C (Option A).

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Since the current flowing through a parallel resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its maximum value.

 (Z=R )

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Refraction occurs when light travels from one medium to another which changes the speed at which the light travels.

This causes light to bend upon incidence with the interface of a new material.

 Angle of refraction is smaller than angle of incidence when light ray travels from rarer medium to denser medium.

I.e  light can either refract towards the normal(when slowing down while crossing the boundary) or away from the normal (when speeding up while crossing the boundary).

When light undergoes refraction, its frequency remains the same.

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de broglie wavelength formula in terms of kinetic energy:

 λ = h2mE√ $\frac{\mathrm{<br>}}{}$

 λ = 6.6∗10−342∗9.1∗10−31∗100√ $\frac{\mathrm{<br>}}{}$

 λ = 5.50 x 10−14 ${}^{<br>}$m

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Objects at a distance greater than two times of the focal length produces a real image, inverted and diminished in size.

→ 2 * 15 = 30cm

: At >30cm this is applicable  

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P = 60w, V = 240v, R = ?

P = V2R $\frac{{\mathrm{<br>}}^{}}{}$

: 60 = 2402R $\frac{{\mathrm{<br>}}^{}}{}$

→ R * 60 = 57,600

R = 5 $\frac{57,600}{60}$

R = 960Ω

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It is a graph that represents the force that acts on a body in a certain amount of time.
Force on the object is represented by the y-axis and time taken by the object is represented by the y axis.
The area under the curve indicates the impulse, the change in momentum that the object undergoes.

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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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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) 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) 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:

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

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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

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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) 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 \(\$0.90\).

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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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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.