WAEC SSCE - Physics - 2022

The basic principle of operation of a beam balance is------------------

Awọn alaye Idahun

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.

An example of a mechanical wave is---------

Awọn alaye Idahun

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".

.An inductor of inductance 10 H is connected across an a.c circuit source of 50 V, 100 Hz. What is the current in the circuit? [π $�$ = 3.14]

Awọn alaye Idahun

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 

The speed of fast-moving neutrons in a nuclear reactor can be reduced by using?

Awọn alaye Idahun

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".

Which of the following concepts is a method of heat transfer that does not require a material medium?

Awọn alaye Idahun

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.

From the principle of flotation, a body sinks in a fluid until it displace a quantity equal to its own?

Awọn alaye Idahun

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.

A hunter fires a gun at a point 408 m away from a cliff. If he hears an echo 2.4 s later, determine the speed of the sound wave

Awọn alaye Idahun

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.

The correct relationship between the displacement, s, of a particle initially at rest in a linear motion and the time, t, is?

Awọn alaye Idahun

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.

Which of the illustrated graphs above represents a body moving with uniform retardation?

Awọn alaye Idahun

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

Awọn alaye Idahun

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.

The engine of a car provides a forward force of 1240 N and the total resistive force on the car is 800N. If the mass of the car is 1220 kg, determine the distance the car has to travel from the rest before acquiring a speed of 4 m s−1 ${}^{-1}$.

Awọn alaye Idahun

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

In which of the following situations is friction not useful?

Awọn alaye Idahun

 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.

The linear expansivity, α $\mathrm{<br>}$, and cubic expansivity, γ $\mathrm{<br>}$, of a material are related by the equation?

Awọn alaye Idahun

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

The periodic rise and fall in the intensity of sound produced whn two notes of nearly equal frequencies are sounded together is called?

Awọn alaye Idahun

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".

Arrange the following radiations in order of increasing ionization of air.

I. Alpha

Il. Gamma

IIl. Beta

Awọn alaye Idahun

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.

The length of a simple pendulum is increased by a factor of four. By what factor is its period increase?

Awọn alaye Idahun

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

Which of the following statements about a straight current-carrying wire placed in a uniform magnetic field is correct? The wire experiences ---------------

Awọn alaye Idahun

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

I.e at perpendicular or Ø = 90°

A small object of mass 50 g is released from a point A. Determine the velocity of the object when it reaches a point B, a vertical distance of 30m below A. [g = 10 ms−2

Awọn alaye Idahun

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".

The efficiency of a wheel and axle is 100% and the ratio of their radii is 5:1. calculate the effort required to lift a load of mass 20kg using this machine, [g = 10 m s−2 ${}^{-2}$]

Awọn alaye Idahun

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.

When the direction of vibration of the particles of a medium is perpendicular to the direction of propagation of a wave, the wave is said to be--------

Awọn alaye Idahun

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.

The area under a force-time graph represents--------

Awọn alaye Idahun

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.

The joule is equivalent to?

Awọn alaye Idahun

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 ${}^{-2}$ 

Which of the folowing statements about light travelling from one material medium to another is not correct?

Awọn alaye Idahun

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.

Which of the following statements-about electromagnetic waves is not correct?

Awọn alaye Idahun

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.

Awọn alaye Idahun

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

In doping an intrinsic semiconductor to produce a p-type semiconductor,

Awọn alaye Idahun

Three cells each of emf, 1.0 V, and internal resistance, 2 Ω $\mathrm{\Omega }$, are connected in parallel across a 3Ω $\mathrm{\Omega }$  resistor. Determine the current in the resistor.

Awọn alaye Idahun

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

Using venier calipers, which of the following readings gives the correct measurement for the length of a rod?

Awọn alaye Idahun

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).

The reason for having a large number of turns in the coil of a moving coil galvanometer is to-----

Awọn alaye Idahun

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

Suppose three identical steel balls Q, R, and S, are placed on an undulating ground as illustrated in the diagram above. Which of the balls is/are in neutral equilibrium?

Awọn alaye Idahun

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.

In a series R-L-C circuit at resonance, impedance is----

Awọn alaye Idahun

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 )

A galvanometer with a full-scale deflection of 20 mA is Converted to read 8 K by connecting a 395 Ω $\mathrm{\Omega }$ resistor in series with it. Determine the internal resistance of the galvanometer

Awọn alaye Idahun

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

The power of the lens is +2.5D. What is the radius of the curvature?

Awọn alaye Idahun

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.

If the kinetic energy of an electron is 100 eV, what is the wavelength of the de-Broglie wave associated with it? [h = 6.6 x 10−34 ${}^{-34}$ J s, e = 1.6 x 1019 ${}^{19}$C, me ${}_{�}$, = 9.1 x 10−31 ${}^{-31}$ kg]

Awọn alaye Idahun

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

Which of the following statements about electric potential energy is not correct?

Awọn alaye Idahun

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.

Which of the following statement(s) is/are correct about a fixed mass of gas compressed in an inexpansible container;

I. The average speed of the molecules increases

II. The temperature of the gas increases

III. The molecules hit the walls of the container more often than before the compression 

Awọn alaye Idahun

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.

The viscosity of a fluid depends on the following factors except the?

Awọn alaye Idahun

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.

An object is placed at different distances, u, from a converging lens of focal length, 15.0cm. For what value of u does the lens act as a microscope?

Awọn alaye Idahun

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  

The period of a 10 kHz radio wave traveling at 10
${}^8$ ms−1 ${}^{<br>}$ is--

Awọn alaye Idahun

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.

Gamma rays are produced when?

Awọn alaye Idahun

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.

Molecules move in random motion within a liquid. The total internal energy of the liquid depends on all of the following except its?

Awọn alaye Idahun

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.

A body of mass, M, moving with velocity, V, has a wavelength, X, associated with it. This phenomenon is called----------

Awọn alaye Idahun

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

A lamp is rated 240 V, 60 W. Determine the resistance of the lamp when lit?

Awọn alaye Idahun

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

The distance between the fixed points of a mercury-in-glass thermometer is 30 cm. Determine the temperature when the mercury level is 10.5 cm above the lower fixed point

Awọn alaye Idahun

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

A freely suspended compass needle on the earth's surface settles in a plane called----------

Awọn alaye Idahun

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.

The diameter of a brass ring at 30 °C is 50.0 cm. To what temperature must this ring be heated to increase its diameter to 50.29 cm? [ linear expansivity of brass = 1.9 x 10−5 ${}^{<br>}$ K−1 ${}^{<br>}$]

Awọn alaye Idahun

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).

The half-life of a radioactive substance is 15 hours. If at some instance, the sample has a mass of 512 g, calculate the time it will take 78 $\frac{7}{8}$ of the sample to decay

Awọn alaye Idahun

A quantity of water at 0 °C is heated to 30 °C. For each degree rise in temperature, its density will---------

Awọn alaye Idahun

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."

You are provided with a battery of e.m.f, E, a standard resistor, R, of resistance 2 Ω $\mathrm{\Omega }$, a key, K, an ammeter, A, a jockey, J, a potentiometer, UV, and some connecting wires.

(i) Measure and record the emf, E, of the battery.

(ii) Set up the circuit as shown in the diagram above with the key open.

(iii) Place the jockey at the point, U, of the potentiometer wire. Close the key and record the reading, i, of the ammeter.

(iv) Place the jockey at a point T on the potentiometer wire UV such that d = UT = 30.0 cm.

(v) Close the circuit, read and record the current, I, on the ammeter,

(vi) Evaluate I1 ${}^1$.

(vi) Repeat the experiment for four other values of d = 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm. In each case, record I and evaluate I1 ${}^1$.

(vii) Tabulate the results

(ix) Plot a graph with d on the vertical axis and I on the horizontal axis stalling both axes from the origin (0,0).

(x) Determine the slope, s, of the graph.

(xi) From the graph determine the value I1 ${}_1$, of I when d = 0. (ci) Given that=s, calculate 8.

(xii) State two precautions taken to ensure accurate results.

(xii) Given that Eδ $\frac{�}{�}$ = s, calculate δ $�$.

(b)(i) Write down the equation that connects the resistance, R, of a wire and the factors on which it depends. State the meaning of each of the symbols.

(ii) An electric fan draws a current of0.75 A in a 240 V circuit. Calculate the cost of using, the fan for 10 hours if the utility rate is $ 0.50 per kWh.

Awọn alaye Idahun

(a) OBSERVATIONS

(i) Value of E correctly read and recorded to at least 1 d.p in volts.

(ii) Value of i correctly read and recorded to at least 1 d.pin amperes.

(iii) Five values of d correctly determined and recorded to at least 1 d.p in cm.

(iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend.
Trend;  As d increases, I decreases

(v) Five values of I1 ${}^1$ correctly evaluated to at least 3 s.f.

(vi) Composite table showing at least d, I, and I−1 ${}^{-1}$.

(ii) P = IV

Power P = 240×0.75100 $\frac{240\times 0.75}{100}$

0.18kw

Energy consumed in 10hr =  0.18 x 10

= 1.8kWh

Cost of energy = 1.8 x 0.5 = $0.9

OR 

Cost of energy = P x cost x time100 $\frac{\text{<br>}}{}$

0.7×240×101000 $\frac{\mathrm{<br>}}{}$

=  $0.9

State three differences between magnetic and non-magnetic materials.

Awọn alaye Idahun

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.

You are provided with a loaded boiling tube with a centimeter scale fixed inside it, a transparent vessel filled with water, standard masses 2 g, 5g and 10 g, and a slide vernier caliper. Use the diagram above as a guide to perform the experiment.

(i) Use the slide vernier caliper to measure and record the external diameter, D, of the boiling tube.

(ii) Evaluated A = 0.25? $\mathrm{<br>}$D2 ${}^2$, where ? $\mathrm{<br>}$ = 3.14.

(iii) Place the loaded boiling tube gently in the water in the transparent vessel such that it floats vertically.

(iv) Read and record the depth of immersion, y, from the zero mark of the scale fixed inside, the boiling tube.

(v) Add a mass, m = 2g, to the boiling tube. Read and record the new depth of immersion, y, from the zero mark of the scale.

(vi) Evaluate h = (y - y0 ${}_0$), log h and log m.

(vii) Repeat the experiment for four other values of m = 5g, 7g, 10 g, and 12g. In each case, record y and evaluate h, log h, and log m.

(viii) Tabulate the results.

(ix) Plot a graph with log m on the vertical axis and log h on the horizontal axis starting both axes from the origin (0,0).

(b)(i) State in full the law on which the experiment in (a) is based.

(ii) A uniform cylindrical rod is 0.63 m long and it has a cross-sectional area of 0.1 m2 ${}^2$. Calculate the depth of immersion of the rod if it floats vertically in a liquid of relative density 1.26. [density of rod =720 kg m?3 ${}^{?3}$, g = 10 m s?2 ${}^{?2}$].

Awọn alaye Idahun

The law on which this experiment is based is the Archimedes' principle, which states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. The buoyant force arises because of the difference in pressure acting on the top and bottom surfaces of the immersed object. This principle helps in determining the density of the liquid in which the object is immersed.

To calculate the depth of immersion of a uniform cylindrical rod, we can use the Archimedes' principle by considering the weight of the liquid displaced by the rod, which is equal to the weight of the rod. Given the length, cross-sectional area, density of the rod, and the relative density of the liquid, we can determine the depth of immersion of the rod.

(a) OBSERVATIONS

(i) Value of D correctly measure and recorded to at least 2 d.p in cm

(ii) Value of A = 0.25? $\mathrm{<br>}$D2 ${}^2$ correctly evaluated

(iii) Value of y0 ${}_0$ correctly read and recorded to at least 1 d.p in cm.

(iv) Five values of m correctly recorded in grammes.

(v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend.

(vi) Five values of h = (y - yo) correctly evaluated.

(vii) Five values of log h correctly evaluated to at least 3 d.p

(viii) Five values of log m correctly evaluated to 3d.

(ix) Composite table showing at least m, y, h, log h and log m.

(b)(i) The Law of Flotation; 

A floating body displaces its own weight of fluid in which it floats.

(ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6

Let y be the depth of immersion 

? $?$ Weight of water displaced

= (0.1y) x 1.26 x 1000 x 10

1260y = 453.6 y=453.6=9.36

= 1260y = 453.6

y = 453.61260 $\frac{453.6}{1260}$ = 9.36

You are provided with a loaded boiling tube with a centimeter scale fixed inside it, a transparent vessel filled with water, standard masses 2 g, 5g and 10 g, and a slide vernier caliper. Use the diagram above as a guide to perform the experiment.

(i) Use the slide vernier caliper to measure and record the external diameter, D, of the boiling tube.

(ii) Evaluated A = 0.25? $\mathrm{<br>}$D2 ${}^{\mathrm{<br>}}$, where ? $\mathrm{<br>}$ = 3.14.

(iii) Place the loaded boiling tube gently in the water in the transparent vessel such that it floats vertically.

(iv) Read and record the depth of immersion, y, from the zero mark of the scale fixed inside, the boiling tube.

(v) Add a mass, m = 2g, to the boiling tube. Read and record the new depth of immersion, y, from the zero mark of the scale.

(vi) Evaluate h = (y - y0 ${}_0$), log h and log m.

(vii) Repeat the experiment for four other values of m = 5g, 7g, 10 g, and 12g. In each case, record y and evaluate h, log h, and log m.

(viii) Tabulate the results.

(ix) Plot a graph with log m on the vertical axis and log h on the horizontal axis starting both axes from the origin (0,0).

(b)(i) State in full the law on which the experiment in (a) is based.

(ii) A uniform cylindrical rod is 0.63 m long and it has a cross-sectional area of 0.1 m2 ${}^2$. Calculate the depth of immersion of the rod if it floats vertically in a liquid of relative density 1.26. [density of rod =720 kg m?3 ${}^{<br>}$, g = 10 m s?2 ${}^{<br>}$].

Awọn alaye Idahun

(a) OBSERVATIONS

(i) Value of D correctly measure and recorded to at least 2 d.p in cm

(ii) Value of A = 0.25π $\mathrm{<br>}$D2 ${}^{\mathrm{<br>}}$ correctly evaluated

(iii) Value of y0 ${}_0$ correctly read and recorded to at least 1 d.p in cm.

(iv) Five values of m correctly recorded in grammes.

(v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend.

(vi) Five values of h = (y - yo) correctly evaluated.

(vii) Five values of log h correctly evaluated to at least 3 d.p

(viii) Five values of log m correctly evaluated to 3d.

(ix) Composite table showing at least m, y, h, log h and log m.

(b)(i) The Law of Flotation; 

A floating body displaces its own weight of fluid in which it floats.

(ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6

Let y be the depth of immersion 

∴ $\therefore$ Weight of water displaced

= (0.1y) x 1.26 x 1000 x 10

1260y = 453.6 y=453.6=9.36

= 1260y = 453.6

y = 453.61260 $\frac{\mathrm{<br>}}{}$ = 9.36

You are provided with a battery of e.m.f, E, a standard resistor, R, of resistance 2 ? $\mathrm{?}$, a key, K, an ammeter, A, a jockey, J, a potentiometer, UV, and some connecting wires.

(i) Measure and record the emf, E, of the battery.

(ii) Set up the circuit as shown in the diagram above with the key open.

(iii) Place the jockey at the point, U, of the potentiometer wire. Close the key and record the reading, i, of the ammeter.

(iv) Place the jockey at a point T on the potentiometer wire UV such that d = UT = 30.0 cm.

(v) Close the circuit, read and record the current, I, on the ammeter,

(vi) Evaluate I1 ${}^1$.

(vi) Repeat the experiment for four other values of d = 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm. In each case, record I and evaluate I1 ${}^1$.

(vii) Tabulate the results

(ix) Plot a graph with d on the vertical axis and I on the horizontal axis stalling both axes from the origin (0,0).

(x) Determine the slope, s, of the graph.

(xi) From the graph determine the value I1 ${}_1$, of I when d = 0. (ci) Given that=s, calculate 8.

(xii) State two precautions taken to ensure accurate results.

(xii) Given that E? $\frac{\mathrm{<br>\; <span\; style="font-size:\; var(--bs-body-font-size);\; font-weight:\; var(--bs-body-font-weight);\; word-spacing:\; normal;"><br></span>}}{}$ = s, calculate ??
.

(b)(i) Write down the equation that connects the resistance, R, of a wire and the factors on which it depends. State the meaning of each of the symbols.

(ii) An electric fan draws a current of0.75 A in a 240 V circuit. Calculate the cost of using, the fan for 10 hours if the utility rate is $ 0.50 per kWh.

Awọn alaye Idahun

(a) OBSERVATIONS

(i) Value of E correctly read and recorded to at least 1 d.p in volts.

(ii) Value of i correctly read and recorded to at least 1 d.pin amperes.

(iii) Five values of d correctly determined and recorded to at least 1 d.p in cm.

(iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend.
Trend;  As d increases, I decreases

(v) Five values of I1 ${}^1$ correctly evaluated to at least 3 s.f.

(vi) Composite table showing at least d, I, and I−1 ${}^{-1}$.

(ii) P = IV

Power P = 240×0.75100 $\frac{240\times 0.75}{100}$

0.18kw

Energy consumed in 10hr =  0.18 x 10

= 1.8kWh

Cost of energy = 1.8 x 0.5 = $0.9

OR 

Cost of energy = P x cost x time100 $\frac{\text{P x cost x time}}{100}$

0.7×240×101000 $\frac{0.7\times 240\times 10}{1000}$

=  $0.9

The effective potential energy, E, of a lunar satellite of mass, m, moving in an. elliptical orbit around the moon of mass, m, is given by

E = K22m1r2−Gm1m2r $\frac{{\mathrm{<br>}}^{}}{\mathrm{<br>}{1}_{}{\mathrm{<br>}}^{}}<br>\frac{\mathrm{<br>}}{}$ where r is the distance of the satellite from the mooń and G is the universal gravitational constant of dimensions, M−1 ${}^{-1}$L3 ${}^{}$T22
.

Ďetermine the dimensions of the angular momentum, K, of the satellite using dimensional analysis.