WAEC SSCE - Physics - 2009

Akọwa Nkọwa

Akọwa Nkọwa

Electric energy is measured in kilowatt-hour (kWh). A kilowatt-hour is a unit of energy equivalent to the energy used by a load of one kilowatt in one hour. This unit is commonly used by electric utility companies to measure the amount of energy consumed by households or businesses. An ampere is a unit of electric current, while a coulomb is a unit of electric charge. A kilowatt is a unit of power, which is the rate at which energy is consumed or produced. Therefore, none of these units are suitable for measuring electric energy.

Akọwa Nkọwa

Akọwa Nkọwa

When an object is immersed in a fluid, it experiences an upward force called buoyant force or upthrust. This force is equal to the weight of the fluid displaced by the object. In this case, the volume of the object submerged in the fluid is \(\frac{3}{4}\) of its total volume. So, the volume of fluid displaced by the object is \(\frac{3}{4}\) x 0.046m³ = 0.0345m³ The weight of this displaced fluid can be calculated as: mass = density x volume = 980kgm^-3 x 0.0345m³ = 33.81kg weight = mass x g = 33.81kg x 10ms^-2 = 338.1N Therefore, the upthrust on the body is 338.1N. So, the correct answer is (C) 338.10N.

Akọwa Nkọwa

When the stone is dropped, it gains potential energy due to its position above the ground. This potential energy is converted into kinetic energy as the stone falls towards the ground. The maximum kinetic energy of the stone will occur just before it hits the ground when all its potential energy has been converted to kinetic energy. The potential energy of the stone is given by the formula P.E. = mgh, where m is the mass of the stone, g is the acceleration due to gravity and h is the height from which it is dropped. Substituting the given values into the formula, we get: P.E. = (0.5 kg)(10 ms^-2)(12 m) = 60 J Since the maximum kinetic energy is equal to the potential energy at the maximum height, the maximum kinetic energy of the stone is also 60 J. Therefore, the correct option is (d) 60.0 J.

Akọwa Nkọwa

Akọwa Nkọwa

The increase in length of the rod can be calculated using the formula: ΔL = αLΔT where: ΔL = change in length α = linear expansivity coefficient L = initial length ΔT = change in temperature Substituting the given values, we have: ΔL = (4.0 x 10^-3 K^-1) x (2m) x (80°C - 25°C) ΔL = 0.44m Therefore, the increase in length of the rod is 0.44m. Note: When a material is heated, its length increases due to the expansion of its particles. The linear expansivity coefficient (α) represents the fractional change in length per unit change in temperature.

Akọwa Nkọwa

To calculate the magnitude of the impulse received by the body, we need to use the impulse-momentum theorem, which states that the impulse applied to a body is equal to the change in momentum of the body. The momentum of an object is defined as the product of its mass and velocity, i.e., P = mv, where P is momentum, m is mass, and v is velocity. The change in momentum (ΔP) of the body can be calculated as the difference between the final momentum (P_f) and the initial momentum (P_i), i.e., ΔP = P_f - P_i. In this case, the initial momentum of the body is given by P_i = mv = 5 kg x 30 m/s = 150 kg m/s (due east), and the final momentum is given by P_f = mv = 5 kg x 50 m/s = 250 kg m/s (due east). Therefore, the change in momentum (ΔP) of the body is: ΔP = P_f - P_i = 250 kg m/s - 150 kg m/s = 100 kg m/s (due east). The magnitude of the impulse (J) received by the body is equal to the change in momentum (ΔP), i.e., J = ΔP = 100 kg m/s. Therefore, the answer is option A: 100 Ns.

Akọwa Nkọwa

Akọwa Nkọwa

The property that will have a different value if measured in space is (i) weight. Weight is the force with which an object is attracted towards the center of the earth due to gravity. In space, there is no gravity or the gravitational force is very weak compared to that on earth. Therefore, the weight of an object measured on earth will be different from the weight of the same object measured in space. On the other hand, specific heat capacity and density are intrinsic properties of a substance that do not depend on the gravitational force, so they will have the same value whether measured on earth or in space. Thus, the answer is (i) only.

Akọwa Nkọwa

Alpha rays are not part of the electromagnetic spectrum. The electromagnetic spectrum includes all forms of electromagnetic radiation that differ in wavelength and frequency, including radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Alpha rays are actually alpha particles, which consist of two protons and two neutrons and are emitted by some types of radioactive materials. Alpha particles have a positive charge and are relatively large compared to other types of radiation, so they are not part of the electromagnetic spectrum.

Akọwa Nkọwa

Sound interference is necessary to produce the phenomenon of beats. Beats are produced when two sound waves of slightly different frequencies are superimposed on each other. The interference of these sound waves causes the sound intensity to oscillate, resulting in a periodic variation of loudness known as beats. These beats can be heard as a throbbing sound. The number of beats produced per second is equal to the difference in frequency between the two sound waves. The phenomenon of beats is used in tuning musical instruments to ensure they are in tune with each other.

Akọwa Nkọwa

In a hydraulic press, the pressure exerted is constant throughout the system. The pressure exerted is given by the formula: Pressure = Force / Area The force is directly proportional to the product of the pressure and area. F ∝ P.A If F₁ is the force applied on the smaller piston of area A₁, and F₂ is the force on the larger piston of area A₂, then: F₁ / A₁ = F₂ / A₂ Rearranging the formula above to solve for F₂: F₂ = F₁ x (A₂ / A₁) Substituting the values given in the question, we get: F₂ = 40N x (200cm² / 10cm²) = 800N Therefore, the force obtained is 800N. The answer is (a) 800N.

Akọwa Nkọwa

Akọwa Nkọwa

When table salt (sodium chloride) is added to ice, it lowers the melting point of the ice. This is because salt dissolves in the thin layer of water on the surface of the ice, forming a saltwater solution. The saltwater solution has a lower freezing point than pure water, so it requires a lower temperature to freeze. As a result, the ice melts at a lower temperature when salt is added, leading to a lowering of the melting point. This phenomenon is called freezing point depression.

Akọwa Nkọwa

The maximum displacement of a vibrating tuning fork is its amplitude. In a vibrating system, amplitude refers to the maximum displacement of the vibrating object from its equilibrium or rest position. For a tuning fork, the amplitude refers to the maximum distance that the prongs of the fork move away from their resting position when they are vibrating. The amplitude of a tuning fork determines the intensity or loudness of the sound it produces. A larger amplitude produces a louder sound, while a smaller amplitude produces a softer sound. Therefore, the correct option is "amplitude".

Akọwa Nkọwa

The characteristic of light that determines its color is wavelength. Wavelength is the distance between consecutive peaks or troughs of a wave. Different colors of light have different wavelengths, with longer wavelengths corresponding to colors like red and orange, and shorter wavelengths corresponding to colors like blue and violet. So, when light interacts with an object and some wavelengths are absorbed while others are reflected or transmitted, the color that we perceive is determined by the wavelengths that are reflected or transmitted to our eyes. Therefore, the wavelength is responsible for the color of light we see.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

When an electric current flows through a straight wire, it generates a magnetic field around the wire. The direction of the magnetic field is dependent on the direction of the current flow. If the direction of the current flow is reversed, then the direction of the magnetic field will also be reversed. Therefore, the correct answer is: the magnetic field is oppositely directed.

Akọwa Nkọwa

The wavelength of the associated wave of a moving particle can be determined using the de Broglie equation, which relates the particle's momentum to its wavelength: λ = h / p where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. We can determine the momentum of the electron using the formula: p = m * v where p is the momentum, m is the mass, and v is the velocity. Substituting the given values, we get: p = (9.1 x 10^-31 kg) * (10^7 ms^-1) = 9.1 x 10^-24 kgms^-1 Substituting p into the de Broglie equation, we get: λ = h / p = (6.6 x 10^-34 Js) / (9.1 x 10^-24 kgms^-1) = 7.25 x 10^-11 m Therefore, the wavelength of the associated wave is 7.25 x 10^-11 m. The correct option is (c) 7.25 x 10^-11 m.

Akọwa Nkọwa

Akọwa Nkọwa

When two forces act on a body, their resultant force can be calculated using vector addition. The equilibrant is the force required to balance the two forces and bring the body into a state of equilibrium. It has the same magnitude as the resultant force but acts in the opposite direction. To calculate the equilibrant, we need to find the resultant force of the two given forces. We can do this by drawing a vector diagram or using trigonometry. Using trigonometry, we can find the magnitude of the resultant force as follows: \begin{align*} F_{\text{resultant}} &= \sqrt{(6\text{N})^2 + (8\text{N})^2} \\ &= \sqrt{36\text{N}^2 + 64\text{N}^2} \\ &= \sqrt{100\text{N}^2} \\ &= 10\text{N} \end{align*} The equilibrant has the same magnitude as the resultant force but acts in the opposite direction. Therefore, the magnitude of the equilibrant is also 10N. So, the correct answer is 10N.

Akọwa Nkọwa

The resistance of a wire can be calculated using the formula R = \(\frac{\rho L}{A}\), where R is the resistance, \(\rho\) is the resistivity of the wire, L is the length of the wire, and A is the cross-sectional area of the wire. In this problem, we are given the resistivity of the wire and the cross-sectional area of the wire. We are also told that the wire needs to have a resistance of 4.0\(\Omega\). We can rearrange the formula to solve for L: L = \(\frac{AR}{\rho}\) Substituting the values given in the question, we have: L = \(\frac{(7.50 \times 10^{-4} cm^{2}) (4.0 \Omega)}{4.40 \times 10^{-5} \Omega cm}\) = 68.18 cm Therefore, the length of the wire required to make a 4.0\(\Omega\) resistor is 68.18 cm. The correct option is: 68.18cm.

Akọwa Nkọwa

Akọwa Nkọwa

If the current is doubled, the heating power will also double, which means that the coil will generate twice as much heat per unit of time. Since the amount of water to be boiled is the same, the amount of heat needed to raise its temperature to boiling point will also be the same. However, with twice the heating power, the time required to deliver that amount of heat will be halved. Therefore, the time it will take to boil the water using the doubled current will be half of the time it took with the original current, which is 2 minutes. So the answer is 2 minutes.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The internal resistance of a cell can be calculated using the equation: r = (E - V)/I where r is the internal resistance of the cell, E is the e.m.f. of the cell, V is the voltage across the cell, and I is the current passing through the circuit. In this case, the e.m.f. of the cell is given as 1.5V, and the resistance of the resistor is 3.0\(\Omega\). Since the cell and the resistor are connected in series, the current passing through them will be the same. We can calculate the current using Ohm's law: I = V/R = 0.9V/3.0\(\Omega\) = 0.3A Now we can use the formula for the internal resistance: r = (E - V)/I = (1.5V - 0.9V)/0.3A = 2.0\(\Omega\) Therefore, the internal resistance of the cell is 2.0\(\Omega\). Answer option A is correct.

Akọwa Nkọwa

The rate at which the boat receives the waves is equal to the frequency of the waves. The distance between successive crests of the waves is 10m, which is also the wavelength of the wave. Therefore, the formula for the speed of the wave can be written as: speed = wavelength x frequency We are given the speed of the wave as 30m/s and the wavelength as 10m. Rearranging the formula above to solve for frequency gives: frequency = speed / wavelength Substituting the values we have: frequency = 30m/s / 10m = 3s^-1 Therefore, the rate at which the boat receives the waves is 3s^-1, which is option (C).

Akọwa Nkọwa

Akọwa Nkọwa

When a pair of equal and opposite non-collinear parallel forces act on a body, a rotational motion will be produced. This is because the two forces will cause a torque or moment about the center of mass of the body, which will result in the body rotating around that point. The magnitude of the torque is proportional to the product of the forces and the perpendicular distance between them, which is also known as the lever arm. Therefore, the greater the force or the longer the lever arm, the greater the torque and the faster the body will rotate. The other options, such as vibrational motion, random motion, and translational motion, do not apply in this scenario because the forces are not acting in such a way as to cause those types of motion.

Akọwa Nkọwa

The numbers 238 and 235 represent the nucleon numbers of the two isotopes of uranium, namely \(^{238}U\) and \(^{235}U\). The nucleon number represents the total number of protons and neutrons in the nucleus of an atom. Therefore, \(^{238}U\) has a nucleus consisting of 238 protons and neutrons, and \(^{235}U\) has a nucleus consisting of 235 protons and neutrons. Atomic number, on the other hand, represents the number of protons in the nucleus of an atom, and proton number and neutron number are not commonly used scientific terms.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

In an AC circuit, the current is constantly changing direction and amplitude, and we use the root mean square (rms) value to measure the effective current flowing in the circuit. The peak value is the maximum value the current can reach, and it is equal to the square root of 2 times the rms value. So, in this case, we can use the formula: Peak value = sqrt(2) x rms value Given that the ammeter records 5.5, which is the rms value, we can calculate the peak value as follows: Peak value = sqrt(2) x 5.5 Peak value = 7.77 Rounded to one decimal place, the answer is 7.8A. Therefore, option A is the correct answer.

Akọwa Nkọwa

The period of a simple pendulum depends on its length. The relationship between the period (T) of a simple pendulum and its length (L) is given by the formula: T = 2π√(L/g), where g is the acceleration due to gravity. If the period of the pendulum doubles, then the new period (T') is twice the original period (T): T' = 2T = 4π√(L/g). If the length of the pendulum is increased by X, then the new length (L') is: L' = L + X. Substituting this into the formula for the new period gives: T' = 2π√((L+X)/g) Since T' = 2T, we can equate the two expressions for T' and simplify: 4π√(L/g) = 2π√((L+X)/g) Squaring both sides and canceling the common factor of 4π²/g, we get: L = (L+X)/2 Solving for X gives: X = L Therefore, the length of the pendulum needs to be increased by 80.0 cm to double its period, so X = 80.0 cm. The answer is option D: 240.0 cm.

Akọwa Nkọwa

A concave mirror forms an image of an object when light rays from the object are reflected off the mirror. The image formed can either be magnified or reduced in size, and it can be erect or inverted depending on where the object is placed in relation to the mirror. In this question, we are told that the concave mirror forms a magnified and erect image. For this to happen, the object must be placed between the focus and the pole of the mirror. This is because, in a concave mirror, light rays coming parallel to the principal axis are reflected and converge at the focus. When an object is placed between the focus and the pole of the mirror, the reflected rays will still converge and form a real image that is magnified and erect. If the object is placed at the focus, the reflected rays will be parallel, and no image will be formed. If the object is placed beyond the radius of curvature, the image formed will be real, but it will be inverted. If the object is placed at the center of curvature, the image formed will be real, inverted, and the same size as the object.

Akọwa Nkọwa

Akọwa Nkọwa

A satellite in circular motion around the earth does not have a uniform velocity. Even though its speed is constant, the direction of its velocity is constantly changing due to the centripetal force provided by the gravitational force of the earth. Therefore, it has an acceleration towards the center of the circle, which is equal in magnitude to the centripetal force acting on it.

Akọwa Nkọwa

The instrument used to measure relative humidity is called a hygrometer. A hygrometer is a device that measures the amount of water vapor in the air or the humidity level. It measures the amount of moisture in the air by using two thermometers. One of the thermometers is dry while the other is kept wet. The difference in temperature between the two thermometers is then measured, and this difference is used to calculate the relative humidity of the air. Therefore, the correct option is (d) hygrometer.

Akọwa Nkọwa

Akọwa Nkọwa

When a conductor mounted on an insulating stand is charged and left for some time, the conductor eventually loses all its charges. This happens because the charges on the conductor attract opposite charges from the surrounding air, which neutralize the charge on the conductor. This process is called electrostatic discharge or simply discharge. As a result, the charges are dissipated, and the conductor becomes neutral again. Therefore, the correct option is "charges on the conductor are neutralized by opposite charges from the surrounding air".

Akọwa Nkọwa

Akọwa Nkọwa

The phenomenon that causes light travelling through a small pinhole to not make a shadow with a distinct sharp edge is called diffraction. Diffraction is the bending of waves around an obstacle or through an opening. When light waves pass through a small pinhole, they bend and spread out after passing through the hole. As a result, the edges of the shadow become blurred, and a distinct sharp edge cannot be seen. This effect is also observed when sound waves pass through small openings, such as in the case of musical instruments. Therefore, the correct option is "diffraction".

Akọwa Nkọwa

(a) A clinical thermometer has a small temperature range, approximately \(35\,^{\circ}\mathrm{C}\) to \(43\,^{\circ}\mathrm{C}\). In boiling water, the mercury expands excessively and may produce enough pressure to crack or burst the glass thermometer.

(b)

1. An increase in pressure increases the boiling point of water.
2. An increase in pressure decreases the melting point of ice.

(c) A liquid boils when its saturated vapour pressure is equal to the external atmospheric pressure. Thus, at \(100\,^{\circ}\mathrm{C}\), the saturated vapour pressure of pure water is \(760\ \mathrm{mmHg}\).

(d) The arrangements of the trapped air and mercury thread are represented below.

Since the capillary tube has uniform cross-sectional area, volume is proportional to length. Hence, Boyle's law may be written as:

\[P_1L_1=P_2L_2\]

Initially, with the open end uppermost, the pressure of the trapped air is

\[P_1=76+20=96\ \mathrm{cmHg}\]

and \(L_1=15\ \mathrm{cm}\). Therefore,

\[P_1L_1=96\times15=1440\ \mathrm{cmHg\,cm}.\]

(i) Tube held horizontally

The pressure of the trapped air is atmospheric pressure:

\[P_2=76\ \mathrm{cmHg}\]

\[96\times15=76L_2\]

\[L_2=\frac{1440}{76}=18.95\ \mathrm{cm}\approx\boxed{18.9\ \mathrm{cm}}\]

(ii) Tube held vertically with the open end underneath

The pressure of the trapped air is

\[P_3=76-20=56\ \mathrm{cmHg}\]

\[96\times15=56L_3\]

\[L_3=\frac{1440}{56}=25.7\ \mathrm{cm}\]

\[\boxed{L_3=25.7\ \mathrm{cm}}\]

Akọwa Nkọwa

(a) Cations are positively charged ions. During electrolysis, they are attracted to and discharged at the cathode.

(b) A labelled electrolytic cell is shown below.

The anode is connected to the positive terminal of the d.c. supply, while the cathode is connected to the negative terminal. Cations move towards the cathode and anions move towards the anode through the electrolyte.

Akọwa Nkọwa

Unpolarized light (in which the vibrations occur in all planes perpendicular to the direction of travel) can be polarized by the following methods:

• Selective absorption (using a Polaroid): passing the light through a Polaroid sheet, which transmits vibrations in only one plane and absorbs those in all other planes.
• Reflection: light reflected from a smooth non-metallic surface (such as glass or water) at the polarizing (Brewster) angle becomes plane polarized.
• Double refraction (using a crystal): passing the light through a doubly refracting crystal such as calcite or tourmaline (or a Nicol prism), which splits it into two plane-polarized rays.

(Scattering, for example by dust or air molecules, also produces partially polarized light.)

Akọwa Nkọwa

For a particle dropped from rest, the initial velocity, \(u=0\). Hence,

\[h=ut+\frac{1}{2}gt^2=\frac{1}{2}gt^2.\]

Taking \(g=9.8\ \text{m s}^{-2}\),

\[h=4.9t^2.\]

(a) Graph of \(h\) against \(t\)

The graph is a parabola passing through the origin. Thus, \(h\) varies parabolically with \(t\), since \(h\propto t^2\). Its gradient increases as time increases.

(b) Graph of \(h\) against \(t^2\)

This is a straight line through the origin. Therefore,

\[h\propto t^2.\]

The gradient of the graph is

\[\frac{h}{t^2}=4.9=\frac{g}{2}.\]

Hence, \(g=2\times4.9=9.8\ \text{m s}^{-2}\).

Akọwa Nkọwa

(a) Tensile stress

Tensile stress is the stretching force acting per unit cross-sectional area of a material when it is subjected to a pull along its length.

Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\).

(b) Young's modulus

Young's modulus (the modulus of elasticity) is the ratio of tensile stress to tensile strain for a material, within the limit of proportionality (the region where Hooke's law holds).

where \(F\) is the stretching force, \(A\) the cross-sectional area, \(l\) the original length and \(e\) the extension. Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\). A large value of \(E\) indicates a stiff material that is difficult to stretch.

Akọwa Nkọwa

(a) Two properties of cathode rays

• They travel in straight lines and cast sharp shadows of objects placed in their path.
• They are streams of negatively charged particles (electrons); hence they are deflected by both electric and magnetic fields (towards the positive plate).

(Other valid properties: they possess kinetic energy and momentum, they can produce heat, they cause certain substances to fluoresce, and they can produce X-rays on striking a metal target.)

(b) Increasing the intensity and energy of cathode rays

Intensity: The intensity (number of electrons emitted per second) is increased by raising the temperature of the cathode/filament, that is by increasing the heating (filament) current. A hotter cathode emits more electrons per second by thermionic emission, giving a more intense beam.

Energy: The energy (speed) of the cathode rays is increased by raising the accelerating potential difference (the anode voltage) between the cathode and anode. Each electron gains energy \(eV\), so a larger \(V\) gives faster, more energetic electrons.

Akọwa Nkọwa

De Broglie proposed that a moving particle of momentum \(p = mv\) has an associated wavelength \(\lambda = \dfrac{h}{mv}\). The following observations support this wave nature of moving particles:

• Electron diffraction (Davisson and Germer): a beam of electrons directed at a crystal (nickel) produced a diffraction pattern of maxima and minima, just as X-rays do. Diffraction is a wave property, so the electrons behaved as waves.
• Thomson's experiment: when a beam of fast electrons was passed through a thin metal foil, it produced concentric diffraction rings on a photographic plate, similar to those given by X-rays passing through the same foil.
• The electron microscope: its successful operation relies on the very short wavelength associated with fast-moving electrons; the electron beam can be focused and diffracted like a wave, which is why it resolves far finer detail than a light microscope.

(A further supporting observation is the diffraction of other particles such as neutrons and protons by crystals.)

Akọwa Nkọwa

(a) Diffusion

Diffusion is the gradual movement or spreading of the molecules of a substance from a region of higher concentration to a region of lower concentration until the molecules are uniformly distributed. It occurs because the molecules are in continuous random motion, and it is faster in gases than in liquids because gas molecules move faster and are farther apart.

(b) Two applications of electrical conduction through gases

• Fluorescent tubes and neon (discharge) lamps - light is produced when an electric current is passed through a gas at low pressure.
• Advertising / sodium and mercury vapour street lamps - conduction through the vapour produces the characteristic coloured light used in signs and lighting.

(Other acceptable examples: the cathode-ray tube / X-ray tube, and the lightning discharge, which is conduction through air.)

Akọwa Nkọwa

(a) Differences between a moving-coil galvanometer and a d.c. generator

Moving-coil galvanometer	D.C. generator
It detects or measures small electric currents.	It generates electrical energy.
It converts electrical energy to mechanical deflection; its coil turns through a limited angle and is controlled by hair springs.	It converts mechanical energy to electrical energy; its coil rotates continuously and uses a split-ring commutator and carbon brushes.

(b) Eddy currents
Eddy currents are circulating currents induced in the body of a conductor when the magnetic flux through it changes. They flow in closed paths and oppose the change producing them. They are applied in an induction furnace and an eddy-current brake.

(c) Principle of a potentiometer
When a steady current flows through a uniform wire of constant cross-sectional area, the potential difference across the wire is directly proportional to its length:

\[V \propto l.\]

(d) Vector diagram

Taking the current, and hence \(V_R\), as the horizontal reference, \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).

\[V^2=V_R^2+(V_L-V_C)^2\]

\[110^2=80^2+(40-V_C)^2\]

\[(40-V_C)^2=12100-6400=5700\]

Since the circuit is net capacitive, \(V_C>V_L\):

\[V_C=40+\sqrt{5700}=40+75.5=115.5\text{ V}\]

(i) Potential difference across the capacitor:

\[\boxed{V_C=115.5\text{ V}}\]

(ii) Capacitive reactance:

\[X_C=\frac{V_C}{I}=\frac{115.5}{2}=57.75\ \Omega\]

\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\times3.14\times60\times57.75}=4.59\times10^{-5}\text{ F}\]

\[\boxed{C\approx46\ \mu\text{F}}\]

(iii) Inductive reactance:

\[X_L=\frac{V_L}{I}=\frac{40}{2}=20\ \Omega\]

\[L=\frac{X_L}{2\pi f}=\frac{20}{2\times3.14\times60}=0.0531\text{ H}\]

\[\boxed{L\approx0.053\text{ H}}\]

Akọwa Nkọwa

(a)

Sound wave	Radio wave
It is a mechanical wave and requires a material medium for propagation.	It is an electromagnetic wave and can travel through a vacuum.
It is longitudinal in air.	It is transverse.

(b)
The vibrating tuning fork forces the table top to vibrate. The table top has a much larger vibrating surface than the prongs of the fork and therefore sets a larger volume of air into vibration. More sound energy is transmitted to the air per second, so the sound is louder.

(c)

1. Production of stationary waves:
   • Two waves of the same frequency and equal or nearly equal amplitude must be present.
   • The waves must travel in opposite directions along the same line and superpose.
2. Formation of interference patterns:
   • The sources must be coherent, that is, have the same frequency and a constant phase difference.
   • The waves must overlap in the same region with equal or nearly equal amplitudes.
3. Occurrence of total internal reflection:
   • The wave must travel from an optically denser medium to an optically less dense medium.
   • The angle of incidence in the denser medium must be greater than the critical angle.

(d)(i) At minimum deviation, the path through the equilateral prism is symmetrical and the refracted ray is parallel to the base.

(d)(ii)
For an equilateral prism,

\[A=60^\circ, \qquad D_m=41^\circ\]

\[n=\frac{\sin\left(\frac{A+D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\]

\[n=\frac{\sin\left(\frac{60^\circ+41^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}=\frac{\sin 50.5^\circ}{\sin 30^\circ}\]

\[n=\frac{0.7716}{0.5000}=1.54\]

Therefore, the refractive index of the glass is \(1.54\).

Akọwa Nkọwa

(a) Faraday's second law of electrolysis

When the same quantity of electricity is passed through different electrolytes, the masses of the different substances liberated (or deposited) are directly proportional to their chemical equivalents (that is, to the ratio of their relative atomic mass to valency).

(b) Value of the electronic charge

The substance is monovalent, so each atom is discharged by one electron. One mole contains \(N = 6.0\times10^{23}\) atoms and is liberated by a charge \(Q = 9.6\times10^{4}\ \text{C}\).

The charge carried by one electron (one atom) is:

The value of the electronic charge is \(1.6\times10^{-19}\ \text{C}\).

Akọwa Nkọwa

(a) Determination of the centre of gravity of the cardboard

1. Make three small, well-spaced holes near the edge of the irregular cardboard.
2. Clamp the stout pin horizontally in the retort stand. Suspend the cardboard freely from the pin through one hole.
3. Hang the simple pendulum from the same pin, with its string very close to the surface of the cardboard.
4. Allow the cardboard and pendulum to come to rest. Trace the vertical line indicated by the pendulum string on the cardboard with the pencil.
5. Suspend the cardboard successively from each of the other two holes and trace the vertical line in each case.
6. The point of intersection of the three vertical lines is the centre of gravity of the cardboard.

Precautions: The pin must be firmly clamped; the cardboard must swing freely; and the pendulum must be at rest before each line is traced.

(b) Velocity-time graph

(i) Acceleration

Acceleration is the gradient of the velocity-time graph. For the straight-line section joining points A and B,

\[a=\frac{\text{change in velocity}}{\text{time taken}}=\frac{v_B-v_A}{t_B-t_A}.\]

(ii) Total distance covered

The total distance covered is the area between the velocity-time graph and the time axis. Thus, for the graph shown, the distance is the area under the graph from the starting time to the final time.

(c) Velocity-time graph of the motion

For the first 20 s:

\[v_1=0+(6.0\times20)=120\ \text{m s}^{-1}.\]

For the next 10 s:

\[v_2=120+(4.0\times10)=160\ \text{m s}^{-1}.\]

The body then continues at \(160\ \text{m s}^{-1}\) for a further 30 s.

The graph consists of straight-line sections through \((0,0)\), \((20,120)\), \((30,160)\), and \((60,160)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).

(i) Maximum speed

\[\boxed{160\ \text{m s}^{-1}}\]

(ii) Total distance travelled during the first 30 s

This is the area under the graph from 0 s to 30 s:

\[\begin{aligned}s_1&=\frac12\times20\times120=1200\ \text{m},\\s_2&=\frac12(120+160)\times10=1400\ \text{m}.\end{aligned}\]

\[s=1200+1400=\boxed{2600\ \text{m}}.\]

(iii) Average speed during the first 30 s

\[\text{Average speed}=\frac{\text{total distance}}{\text{total time}}=\frac{2600}{30}=86.7\ \text{m s}^{-1}.\]

\[\boxed{86.7\ \text{m s}^{-1}}\]

Akọwa Nkọwa

The particle is projected horizontally, so its vertical motion is free fall from rest and its horizontal velocity stays constant.

Time to reach the ground (vertical fall of \( h = 20\,\text{m} \)): \[ h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4} = 2\,\text{s}. \]

Horizontal distance (range): \[ x = u \times t = 15 \times 2 = 30\,\text{m}. \]

The horizontal distance covered before hitting the ground is 30 m.

Akọwa Nkọwa

Whether a liquid wets a solid surface depends on the relative sizes of two types of intermolecular force:

• Cohesion - the force of attraction between molecules of the same substance (liquid to liquid).
• Adhesion - the force of attraction between molecules of different substances (liquid to solid).

Water on glass: The adhesive force between water molecules and glass molecules is greater than the cohesive force between the water molecules themselves. Water molecules are therefore pulled towards the glass, spread over it, and the water forms a concave meniscus with an acute angle of contact. Because it clings to and spreads on the surface, water is said to wet glass.

Mercury on glass: The cohesive force between mercury molecules is greater than the adhesive force between mercury and glass. The mercury molecules are pulled more strongly towards one another than towards the glass, so the mercury draws itself together into rounded drops, forms a convex meniscus with an obtuse angle of contact, and does not spread. Hence mercury does not wet glass.

In short: a liquid wets a solid when adhesion exceeds cohesion (water and glass) and does not wet it when cohesion exceeds adhesion (mercury and glass).

Akọwa Nkọwa

(a)(i) Emission line spectrum: This consists of separate bright lines of definite wavelengths on a dark background. It is produced when excited atoms emit radiation as their electrons fall from higher to lower energy levels.

(ii) Line absorption spectrum: This consists of dark lines on a continuous bright spectrum. It is obtained when continuous light passes through a cool gas, which absorbs particular wavelengths corresponding to its energy transitions.

(b) Simple photocell

When radiation of sufficiently high frequency falls on the photosensitive cathode, photoelectrons are emitted from its surface. The anode is maintained positive relative to the cathode, so it attracts and collects the emitted electrons. The resulting movement of electrons through the external circuit constitutes a photocurrent, which is indicated by the microammeter.

(c)(i) Evidence that X-rays are waves:

1. They undergo diffraction, for example by crystals.
2. They undergo interference.

(ii) Uses of X-rays other than in medicine:

1. Detecting cracks and faults in metal castings and welded joints.
2. Examining crystal structures.

(d)

For an electron falling from [?25l[?25h\,[?25l[?25h-1.6\,\text{eV}\) to \(-10.4\,\text{eV}\), the energy emitted is

\[\begin{aligned}\Delta E&=E_i-E_f\\&=(-1.6)-(-10.4)\\&=8.8\,\text{eV}.\end{aligned}\]

Since \(1\,\text{eV}=1.6\times10^{-19}\,\text{J}\),

\[E=8.8\times1.6\times10^{-19}=1.408\times10^{-18}\,\text{J}\approx1.41\times10^{-18}\,\text{J}.\]

Using \(E=\dfrac{hc}{\lambda}\),

\[\begin{aligned}\lambda&=\frac{hc}{E}\\&=\frac{(6.6\times10^{-34})(3.00\times10^8)}{1.408\times10^{-18}}\\&=1.41\times10^{-7}\,\text{m}.\end{aligned}\]

Energy of radiation = \(1.41\times10^{-18}\,\text{J}\) (or \(8.8\,\text{eV}\)); wavelength = \(1.41\times10^{-7}\,\text{m}\).