WAEC SSCE - Physics - 2013

Bayanin Amsa

The rising of a liquid in an open and ended glass tube of narrow bore is called capillarity. Capillarity is the result of the combination of adhesive and cohesive forces between the liquid molecules and the molecules of the tube material. In a narrow tube, the adhesive forces between the liquid and the tube are stronger than the cohesive forces between the liquid molecules themselves, causing the liquid to rise in the tube. The narrower the tube, the higher the liquid rises due to capillary action.

Bayanin Amsa

As the pump handle is slowly pushed downwards, the volume of the pump is reduced. This means that the same amount of air inside the pump is now occupying a smaller space. Since the temperature is constant, the increase in pressure results from an increase in the frequency of collisions of air molecules with the walls of the pump. The air molecules are being compressed into a smaller volume, leading to an increase in pressure. Therefore, the correct answer is: the frequency of collision of the air molecules with the walls of the pump is increased.

Bayanin Amsa

Bayanin Amsa

The heat gained by the water can be calculated using the formula: Heat = Power × Time where Power = Current × Voltage and Time = 2 minutes = 120 seconds The voltage can be calculated using Ohm's law: Voltage = Current × Resistance Substituting the values given: Voltage = 4.0A × 50\(\Omega\) = 200V Therefore, Power = 4.0A × 200V = 800W Substituting the values in the formula: Heat = Power × Time = 800W × 120s = 96,000J Therefore, the heat gained by the water is 96,000J. The correct option is (d) 96,000J.

Bayanin Amsa

The phenomenon that supports the theory that waves have a particle nature is the "photoelectric effect." This effect demonstrates that light behaves as a stream of particles, called photons, which transfer their energy to electrons in a metal surface. When the energy of a photon exceeds a certain threshold value, an electron is ejected from the metal surface. This effect cannot be explained by wave theory alone, but it can be explained by assuming that light has a particle nature. Therefore, the photoelectric effect provides strong evidence for the wave-particle duality of light.

Bayanin Amsa

Bayanin Amsa

The velocity ratio of an inclined plane is a measure of the effectiveness of the plane in reducing the force required to move an object to a certain height. It is defined as the ratio of the distance moved by the load to the distance moved by the effort. The velocity ratio of an inclined plane is dependent on the angle of inclination. As the angle of inclination increases, the velocity ratio also increases. This means that a steeper inclined plane is more effective at reducing the force required to move an object to a certain height. Therefore, the correct option is: increases with increase in the angle of inclination.

Bayanin Amsa

The minimum energy required to cause photoelectric emission when the surface of a metal is irradiated with electromagnetic radiation of suitable frequency is called the work function. It is the amount of energy required to remove an electron from the metal surface, overcoming the attractive forces holding the electron in the metal. The energy of the incident photon must be greater than or equal to the work function for photoelectric emission to occur. If the energy of the incident photon is less than the work function, no electrons will be emitted.

Bayanin Amsa

To calculate the acceleration of the block, we need to apply Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. Mathematically, this is expressed as a = F_net/m, where a is the acceleration, F_net is the net force, and m is the mass of the object. In this problem, the net force acting on the block is the difference between the applied force and the frictional force, i.e., F_net = F_applied - F_friction = 40N - 12N = 28N. Therefore, the acceleration of the block is a = F_net/m = 28N/5kg = 5.6ms^-2. Therefore, the correct option is 5.6ms^-2.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The general equation of a progressive transverse wave is given by: y = A sin (kx - ωt) where A is the amplitude, k is the wave number, x is the position of the particle, ω is the angular frequency, and t is the time. Comparing this equation with the given equation, we have: A = 2 cm k = 2π/30 cm^(-1) = π/15 cm^(-1) ω = 2π/T, where T is the period Thus, we have: ω = 2π/T = 2π/(0.01 s) Substituting the value of ω, we get: 2π/(0.01 s) = π/15 cm^(-1) Simplifying this equation, we get: T = 0.01 s Therefore, the period of the wave is 0.01 s.

Bayanin Amsa

The energy required to eject an electron from a metal is given by: Kinetic energy of ejected electron = Energy of photon – Work function We can rearrange this equation to find the energy of the photon: Energy of photon = Kinetic energy of ejected electron + Work function The kinetic energy of the ejected electron is given as 3.0 eV. We can convert this to joules using the conversion factor 1 eV = 1.6 x 10^-19 J: Kinetic energy of ejected electron = 3.0 eV x 1.6 x 10^-19 J/eV = 4.8 x 10^-19 J Substituting this value and the given work function into the equation for the energy of the photon: Energy of photon = 4.8 x 10^-19 J + 2.56 x 10^-19 J = 7.36 x 10^-19 J We can use the formula E = hf to relate the energy of the photon to its frequency f: E = hf where h is Planck's constant, h = 6.6 x 10^-34 J s. Rearranging the equation gives: f = E/h Substituting the values we obtained: f = (7.36 x 10^-19 J) / (6.6 x 10^-34 J s) ≈ 1.12 x 10¹⁵ Hz Therefore, the frequency of the photon required to eject an electron with a kinetic energy of 3.0 eV from a metal with a work function of 2.56 x 10^-19 J is approximately 1.12 x 10¹⁵ Hz.

Bayanin Amsa

The slope of a linear distance-time graph represents the object's speed or rate of change of distance with respect to time. In other words, the slope is a measure of how fast the object is moving. A steeper slope indicates a faster speed, while a shallower slope indicates a slower speed. Therefore, the correct answer is speed.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The critical angle is the angle of incidence at which the angle of refraction is 90 degrees. Using Snell's law, we can relate the critical angle to the refractive index. At the critical angle, the sine of the angle of refraction is 1, so we can write: sin(critical angle) = 1/n where n is the refractive index. Substituting n=1.5 gives: sin(critical angle) = 1/1.5 = 0.67 Taking the inverse sine of both sides gives: critical angle = sin^-1(0.67) = 42^o Therefore, the correct answer is 42^o.

Bayanin Amsa

Bayanin Amsa

To determine the speed of projection of a body projected vertically upwards in a vacuum, we can use the principles of kinematics and the given information. Here are the steps:

1. Understand the total time of flight: The total time taken for the body to go up and come back down is 1.2 seconds.

2. Determine the time to reach the maximum height: Since the time to go up is equal to the time to come down, the time to reach the maximum height is half of the total time of flight.

   $t_{\text{up}}=\frac{1.2\text{ s}}{2}=0.6\text{ s}t\_\{\backslash text\{up\}\}\; =\; \backslash frac\{1.2\; \backslash text\{\; s\}\}\{2\}\; =\; 0.6\; \backslash text\{\; s\}$

3. Use the kinematic equation for velocity: The velocity at the maximum height is 0 (since the body momentarily stops before falling back down). We can use the equation:

   $v=u-gtv\; =\; u\; -\; gt$

   where $vv$ is the final velocity (0 m/s at the maximum height), $uu$ is the initial velocity (speed of projection), $gg$ is the acceleration due to gravity (10 m/s²), and $tt$ is the time taken to reach the maximum height (0.6 s).

   Setting $v=0v\; =\; 0$:

   $0=u-g\cdot t0\; =\; u\; -\; g\; \backslash cdot\; t$

4. Solve for $uu$:

   $u=g\cdot t=10{\text{ m/s}}^2\cdot 0.6\text{ s}=6\text{ m/s}u\; =\; g\; \backslash cdot\; t\; =\; 10\; \backslash text\{\; m/s\}^2\; \backslash cdot\; 0.6\; \backslash text\{\; s\}\; =\; 6\; \backslash text\{\; m/s\}$

Therefore, the speed of projection is $\boxed{{\displaystyle 6.0\text{ m/s}}}\backslash boxed\{6.0\; \backslash text\{\; m/s\}\}$6.0 m/s

Bayanin Amsa

Bayanin Amsa

The electric field intensity (E) is defined as the force per unit charge. Therefore, we can find E by dividing the force (F) by the charge (q). E = F/q Substituting the given values, we have: E = 80N / (2.0 x 10\(^{-5}\) C) E = 4.0 x 10\(^6\) NC\(^{-1}\) Therefore, the magnitude of the electric field intensity is 4.0 x 10\(^6\) NC\(^{-1}\). So, the correct option is (B) 4.0 x 10⁶ NC^-1.

Bayanin Amsa

Radio waves have frequencies lower than infrared radiation. Radio waves have the lowest frequency among all the electromagnetic radiations, ranging from a few hertz to several gigahertz. On the other hand, infrared radiation has a frequency range of about 430 THz to 300 GHz. Therefore, radio waves have a lower frequency than infrared radiation.

Bayanin Amsa

Bayanin Amsa

When a body of mass m is released on an inclined plane at an angle θ to the horizontal, the component of its weight acting down the plane is mgsinθ. The force acting parallel to the plane that causes the body to slide down is given by F = mgsinθ. The acceleration of the body down the plane is given by a = F/m = gsinθ. In this case, m = 2kg and θ = 60°. Therefore, the acceleration of the body down the plane is given by: a = gsinθ = 10ms\(^{-2}\) x sin(60°) = 8.7ms\(^{-2}\) Therefore, the correct answer is 8.7ms\(^{-2}\).

Bayanin Amsa

The type of thermometer used for the calibration of other thermometers is the constant volume gas thermometer. This thermometer operates on the principle of the ideal gas law, which states that the pressure and volume of a gas are proportional to its absolute temperature. In a constant volume gas thermometer, a gas is trapped in a container with a fixed volume, and its pressure is measured at different temperatures to obtain a calibration curve. This curve can then be used to calibrate other types of thermometers, such as liquid-in-glass thermometers or thermocouples. Therefore, the correct answer is constant volume gas thermometer.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The instrument used to determine the accurate value of the electromotive force (emf) of a cell is called a potentiometer. A potentiometer measures the potential difference across a known length of a uniform wire and compares it with the potential difference across the unknown emf source, which is connected in parallel with the wire. By adjusting the length of the wire until the potential difference across the known length is equal to the potential difference across the cell, the emf of the cell can be determined accurately. Therefore, the correct option is a potentiometer.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The statement that radioisotopes are used to induce mutations in plants and animals to obtain new and improved varieties is not correct. Radioisotopes are unstable isotopes of an element that emit radiation as they undergo decay, and they have several applications in various fields. However, inducing mutations in plants and animals is not one of them. Instead, mutagenic agents like chemicals or radiation are used for this purpose. Radioisotopes are used to trace the paths of metabolic processes in plants and animals, estimate the age of rocks, locate broken bones, and several other applications in fields such as medicine, agriculture, and industry.

Bayanin Amsa

When salt is dissolved in water, it lowers the freezing point of the water. This is because adding salt lowers the temperature at which water freezes, a process known as freezing point depression. When salt is added to water, it lowers the freezing point of the water because the salt ions interfere with the formation of ice crystals. The result is that the saltwater mixture will remain in a liquid state at a lower temperature than pure water would. Therefore, the correct option is "decreases".

Bayanin Amsa

The maximum displacement on either side of the equilibrium position of an object in simple harmonic motion represents its **amplitude**. In simple harmonic motion, the object oscillates back and forth around a central point, which is its equilibrium position. The amplitude is the maximum displacement of the object from this equilibrium position, in either direction. It represents the "strength" or "size" of the oscillation. The other options are not correct: the **period** is the time it takes for one complete oscillation, the **wavelength** is the distance between two adjacent points in a wave that are in phase, and the **frequency** is the number of oscillations per unit time (usually measured in Hertz).

Bayanin Amsa

Solid friction is in opposition to motion. When two solid surfaces are in contact, there is a force that resists the relative motion between them, known as friction. In the case of solid friction, this force acts perpendicular to the surfaces in contact and depends on the normal reaction between them, which is the force exerted by one surface on the other perpendicular to the contact area. Friction is also independent of the surface area in contact and dependent on the relative motion between the layers.

Bayanin Amsa

When a person holds a negatively charged metal rod and stands bare-footed on the ground, the negative charges will flow to the ground through the person's body. This is because the human body is a good conductor of electricity, allowing charges to flow easily through it. When a person is standing on the ground, they are in contact with a large conductive surface (the earth), which provides a pathway for the charges to flow away from the person's body and into the ground. This process is called grounding or earthing, and it helps to prevent the buildup of static electricity on the person's body.

Bayanin Amsa

The frequency of a note produced by a string under tension is directly proportional to the square root of the tension. This relationship is given by the equation: f = (1/2L) √(T/μ) Where f is the frequency, T is the tension, L is the length of the string and μ is the linear mass density of the string. Since the length and mass density of the string are constant, we can say that the frequency is directly proportional to the square root of the tension: f ∝ √T So, if we quadruple the tension, the frequency will double. Therefore, the frequency produced by the string when the tension is quadrupled will be: f' = 2f = 2 × 14Hz = 28Hz Therefore, the correct answer is 28Hz.

Bayanin Amsa

Bayanin Amsa

To determine the potential energy of the body at the maximum height of its motion, we need to follow these steps:

1. Calculate the speed of projection (initial velocity): From the previous solution, we determined that the speed of projection $uu$ is 6 m/s.

2. Determine the maximum height: Use the kinematic equation for the vertical motion to find the maximum height:

   $v^2=u^2-2ghv^2\; =\; u^2\; -\; 2gh$

   At the maximum height, the final velocity $vv$ is 0, so:

   $0=u^2-2gh⟹h=\frac{u^2}{2g}0\; =\; u^2\; -\; 2gh\; \backslash implies\; h\; =\; \backslash frac\{u^2\}\{2g\}$

   Substituting $u=6\text{ m/s}u\; =\; 6\; \backslash text\{\; m/s\}$ and $g=10{\text{ m/s}}^{2}g\; =\; 10\; \backslash text\{\; m/s\}^2$:

   $h=\frac{(6\text{ m/s}{)}^2}{2\times 10{\text{ m/s}}^2}=\frac{36{\text{ m}}^2\mathrm{/}{\text{s}}^2}{20{\text{ m/s}}^2}=1.8\text{ m}h\; =\; \backslash frac\{(6\; \backslash text\{\; m/s\})^2\}\{2\; \backslash times\; 10\; \backslash text\{\; m/s\}^2\}\; =\; \backslash frac\{36\; \backslash text\{\; m\}^2/\backslash text\{s\}^2\}\{20\; \backslash text\{\; m/s\}^2\}\; =\; 1.8\; \backslash text\{\; m\}$

3. Calculate the potential energy (PE): The potential energy at the maximum height is given by:

   $\text{PE}=mgh\backslash text\{PE\}\; =\; mgh$

   Convert the mass from grams to kilograms:

   $m=20\text{ g}=0.02\text{ kg}m\; =\; 20\; \backslash text\{\; g\}\; =\; 0.02\; \backslash text\{\; kg\}$

   Now substitute $m=0.02\text{ kg}m\; =\; 0.02\; \backslash text\{\; kg\}$, $g=10{\text{ m/s}}^{2}g\; =\; 10\; \backslash text\{\; m/s\}^2$, and $h=1.8\text{ m}h\; =\; 1.8\; \backslash text\{\; m\}$:

   $\text{PE}=0.02\text{ kg}\times 10{\text{ m/s}}^2\times 1.8\text{ m}=0.36\text{ J}\backslash text\{PE\}\; =\; 0.02\; \backslash text\{\; kg\}\; \backslash times\; 10\; \backslash text\{\; m/s\}^2\; \backslash times\; 1.8\; \backslash text\{\; m\}\; =\; 0.36\; \backslash text\{\; J\}$

Therefore, the potential energy of the body at the maximum height of its motion is $\boxed{{\displaystyle 0.36\text{ J}}}\backslash boxed\{0.36\; \backslash text\{\; J\}\}$

Bayanin Amsa

We can use the formula for the current in an AC circuit with an inductor and a resistor: \begin{equation*} I = \frac{V}{\sqrt{R^2 + (\omega L)^2}} \end{equation*} where V is the voltage, R is the resistance, L is the inductance, and \(\omega\) is the angular frequency. We can find \(\omega\) using the formula: \begin{equation*} \omega = 2\pi f \end{equation*} where f is the frequency. Substituting the given values, we get: \begin{align*} \omega &= 2\pi \times 50\text{ Hz} \\ &= 100\pi\text{ rad/s} \end{align*} Now we can substitute all the values into the first formula: \begin{align*} I &= \frac{V}{\sqrt{R^2 + (\omega L)^2}} \\ &= \frac{240\text{ V}}{\sqrt{(4\Omega)^2 + (100\pi \text{ rad/s} \times 0.12\text{ H})^2}} \\ &\approx 6.3\text{ A} \end{align*} Therefore, the current through the coil is approximately 6.3A. Answer is correct.

Bayanin Amsa

Images formed by a convex mirror are always behind the mirror, real, and upright. However, the images are always smaller than the actual object. This is due to the fact that convex mirrors cause light rays to diverge, making the image appear farther away and smaller than it actually is. Therefore, the correct option in this case would be "behind the mirror".

Bayanin Amsa

The correct statement about pressure in a liquid is "the pressure in a liquid increases with depth." This means that as the depth of a liquid increases, the pressure at that point also increases. This is because the weight of the liquid above a certain depth creates a force that increases the pressure at that depth. The higher the density of a liquid, the higher the pressure it exerts, which means that the second statement is incorrect. The third statement is also incorrect because pressure in a liquid acts in all directions, not just perpendicular to the sides of the containing vessel. Finally, the fourth statement is also incorrect because pressure in a liquid is affected by the acceleration due to gravity, as the weight of the liquid above a certain depth creates a force that increases the pressure at that depth.

Bayanin Amsa

The correct option is: "the final image is inverted". This statement is incorrect. The Galilean telescope consists of a converging lens (the objective lens) and a diverging lens (the eyepiece) placed at a distance equal to the sum of their focal lengths. The objective lens produces a real and inverted image of the object being observed, which is then magnified and made upright by the eyepiece lens. So, the final image is upright. The other options are: - "it is shorter than terrestial telescope" - This statement is correct. The Galilean telescope is shorter than the terrestrial telescope, as it uses a combination of lenses to produce a magnified image rather than a long focal length objective lens. - "it has a small field of view" - This statement is generally correct. The Galilean telescope has a smaller field of view compared to the terrestrial telescope due to its design, which results in a narrow angle of view. - "the final image is erect" - This statement is correct. As mentioned earlier, the final image produced by a Galilean telescope is upright or erect.

Bayanin Amsa

In a neutral atom, the number of positively charged protons in the nucleus is equal to the number of negatively charged electrons surrounding the nucleus. The nucleons in an atom refer to the particles present in the nucleus, which are protons and neutrons. Therefore, the correct answer is (a) protons and neutrons.

Bayanin Amsa

The correct statement is (i and ii only). The magnitude of expansion or contraction of a substance depends on the temperature change and the nature of the substance. The size of the substance does not affect the magnitude of the expansion or contraction. When a substance is heated, its temperature increases, and the particles in the substance gain kinetic energy, leading to an increase in the space between the particles. This increase in space between the particles results in expansion of the substance. The nature of the substance also plays a role in determining the magnitude of expansion or contraction. Different substances have different coefficients of thermal expansion, which is a measure of how much a substance expands or contracts when its temperature changes. Some substances expand more than others when they are heated and contract more than others when they are cooled.

Bayanin Amsa

Bayanin Amsa

The unit of power is the watt (W), which is defined as the rate at which work is done or energy is transferred. In terms of fundamental units, power can be expressed as the product of force and velocity, which can be further expressed as mass times acceleration times velocity. Simplifying this expression, we get the unit of power as kgm²s^-3, which is option B.

Bayanin Amsa

(a) Boiling point

The boiling point of a liquid is the constant temperature at which the liquid boils, i.e. at which its saturated vapour pressure equals the external atmospheric pressure.

(b) Boiling water without heating it

Put warm water in a round-bottomed flask, cork it, and connect it to a vacuum pump. As the pump lowers the pressure above the water, the boiling point falls; when the reduced pressure equals the water's vapour pressure at that temperature, the water boils without any further heating. (Alternatively, boil the water, cork the flask, invert it and pour cold water over the top: the vapour inside condenses, pressure drops, and the water boils again.)

(c) Three applications of expansion of metals

• The bimetallic strip in thermostats and fire alarms.
• Fitting (shrink-fitting) of metal tyres/rims onto wheels and rivets.
• Expansion gaps/rollers left in railway lines and bridges.

(d) Cooling the room air from 34°C to 24°C

Floor area \(= 7\times10 = 70\,\text{m}^{2}\); air mass \(= 250\,\text{kg}\); \(c = 1010\,\text{Jkg}^{-1}\text{K}^{-1}\); density \(= 1.25\,\text{kg m}^{-3}\).

(i) Height of the room: volume of air \(= \dfrac{m}{\rho} = \dfrac{250}{1.25} = 200\,\text{m}^{3}\).

\[ \text{height} = \frac{\text{volume}}{\text{floor area}} = \frac{200}{70} = 2.86\,\text{m} \]

(ii) Energy extracted:

\[ Q = mc\,\Delta\theta = 250\times1010\times(34-24) = 250\times1010\times10 = 2.525\times10^{6}\,\text{J} \]

(iii) Which is higher? The actual energy that must be extracted is higher than the calculated value. The calculation accounts only for cooling the air; in practice heat must also be removed from the walls, furniture and occupants, and heat continually leaks in from the warmer surroundings, so more energy is needed than the value found above.

Bayanin Amsa

The diagram is a cathode-ray tube (CRT), and the three labelled sections are its three basic components:

• P - the electron gun: the assembly at the narrow end containing the heated filament/cathode and the anodes (focusing and accelerating electrodes). It produces, focuses and accelerates a narrow beam of electrons.
• Q - the deflecting system (deflecting plates): the pairs of parallel plates (the X- and Y-plates) through which the beam passes. They deflect the electron beam horizontally and vertically by means of the electric fields between the plates.
• R - the fluorescent screen: the flared front face of the tube, coated on the inside with a fluorescent (phosphor) material. It glows to produce a visible spot where the electron beam strikes it.

Bayanin Amsa

(a) Two particles used in explaining the wave nature of matter:

• The electron.
• The neutron (a proton is also acceptable).

(b) One device based on the wave nature of matter:

• The electron microscope.

Bayanin Amsa

Uses of the LASER

• (a) Communication: transmission of signals (voice, data, TV) through optical fibre cables using laser light.
• (b) Medicine: laser surgery, for example bloodless cutting of tissue, welding a detached retina, or removing tumours.
• (c) Security: laser beams used in burglar-alarm/intruder-detection systems (and in barcode/scanner and printing security).

Bayanin Amsa

(a) Triangle law of vector addition

The triangle law of vector addition states that if two vectors are represented in both magnitude and direction by the two adjacent sides of a triangle taken in order, then their resultant (sum) is represented in magnitude and direction by the third side of the triangle taken in the reverse order.

(b) Physical quantities associated with the equations of linear motion

The four physical quantities are:

• Displacement (or distance), \(s\)
• Velocity/speed (initial \(u\) and final \(v\))
• Acceleration, \(a\)
• Time, \(t\)

(c) Variation of potential energy and kinetic energy with time in S.H.M.

In simple harmonic motion the total mechanical energy \(E\) stays constant, so the potential energy (P.E.) and kinetic energy (K.E.) continuously interchange. When the body is at the equilibrium position the K.E. is maximum and the P.E. is zero; at the extreme positions the P.E. is maximum and the K.E. is zero. Both curves are \(\sin^2\)/\(\cos^2\) shapes that always add up to the constant total energy \(E\).

Key features of the sketch: both curves lie between \(0\) and \(E\); K.E. starts at its maximum \(E\) (body passing through equilibrium) while P.E. starts at zero; wherever one curve peaks the other is zero, and for every instant \(\text{K.E.} + \text{P.E.} = E\).

(d) The spring-and-block problem

Given: compression \(e = 5\,\text{cm} = 0.05\,\text{m}\); energy stored \(W = 10\,\text{J}\); mass of P, \(m_P = 2.0\,\text{kg}\); mass of Q, \(m_Q = 4.0\,\text{kg}\); time after release \(t = 0.25\,\text{s}\).

(i) Value of the force constant \(k\)

The energy stored in a compressed spring is \(W = \tfrac{1}{2}k e^{2}\), so

\[ k = \frac{2W}{e^{2}} = \frac{2 \times 10}{(0.05)^{2}} = \frac{20}{0.0025} = 8000\ \text{N m}^{-1}. \]

(ii) Force used to compress the spring

\[ F = k e = 8000 \times 0.05 = 400\ \text{N}. \]

(iii) Acceleration of block P after release

Using Newton's second law with the spring force acting on P:

\[ a = \frac{F}{m_P} = \frac{400}{2.0} = 200\ \text{m s}^{-2}. \]

(iv) Common speed after the inelastic collision

First find the velocity of P just before impact. Taking the acceleration to act over \(t = 0.25\,\text{s}\):

\[ u_P = a t = 200 \times 0.25 = 50\ \text{m s}^{-1}. \]

The collision is inelastic, so the blocks move together. Applying conservation of linear momentum:

\[ m_P u_P + m_Q u_Q = (m_P + m_Q)v \] \[ (2.0 \times 50) + (4.0 \times 0) = (2.0 + 4.0)\,v \] \[ 100 = 6v \quad\Rightarrow\quad v = \frac{100}{6} = 16.7\ \text{m s}^{-1}. \]

The common speed of the two blocks after collision is \(16.7\ \text{m s}^{-1}\).

Bayanin Amsa

None

Bayanin Amsa

Projectile motion and time of flight

A body is projected from ground level with speed \(u\) at an angle \(\theta\) to the horizontal. The diagram shows the parabolic path and the initial velocity resolved into its horizontal and vertical components.

Resolving the initial velocity:

Horizontal component: \(u_x = u\cos\theta\) (this stays constant because there is no horizontal force).

Vertical component: \(u_y = u\sin\theta\) (this is decelerated by gravity \(g\) acting downward).

Time to reach maximum height

Taking upward as positive, the vertical velocity at time \(t\) is

\[ v_y = u\sin\theta - g t \]

At the maximum height the vertical velocity is momentarily zero, \(v_y = 0\). Let \(t\) be the time taken to reach this point:

\[ 0 = u\sin\theta - g t \]\[ t = \frac{u\sin\theta}{g} \]

Time of flight

By the symmetry of the path, the time taken to fall from the maximum height back to the projection level equals the time taken to rise to it. Hence the total time of flight \(T\) is twice the time to reach maximum height:

\[ T = 2t = \frac{2u\sin\theta}{g} \]

This is the required result: the time of flight \(T = \dfrac{2u\sin\theta}{g}\), where \(g\) is the acceleration of free fall.

Bayanin Amsa

None

Bayanin Amsa

(a) Purpose of earthing an electrical appliance

Earthing means connecting the metal casing of an appliance to the ground through a low-resistance earth wire. Its purpose is safety: if a live wire becomes loose and touches the metal casing (a fault), the casing would otherwise become live and give a fatal shock to anyone who touches it. The earth wire provides a path of very low resistance straight to the ground, so the large fault current flows to earth (rather than through the user's body) and blows the fuse, cutting off the supply. Thus earthing protects the user from electric shock.

(b) Why the bulb of a simple cell dims and goes off

The light produced by a bulb connected to a simple (voltaic) cell gradually dims and finally goes off because of two defects of the simple cell:

• Polarisation: hydrogen gas liberated during the chemical reaction collects as a layer of bubbles on the copper (positive) plate. This layer increases the internal resistance and also sets up a back e.m.f. that opposes the current, so the current falls.
• Local action: impurities in the zinc plate form tiny local cells with the zinc, so the zinc is eaten away (and current is wasted) even when the cell is not in use, weakening the cell.

Together these defects steadily reduce the current, so the bulb dims and eventually goes out.

(c) Series R-L-C circuit on an a.c. source

Given: coil \(L = 0.007\,\text{H}\), resistor \(R = 8\,\Omega\), capacitor \(C = 0.001\,\text{F}\), frequency \(f = \dfrac{500}{\pi}\,\text{Hz}\). The r.m.s voltages are: across the coil \(V_L = 30\,\text{V}\), across the resistor \(V_R = 20\,\text{V}\), across the capacitor \(V_C = 70\,\text{V}\).

(i) Vector (phasor) diagram of the voltages

Taking the current \(I\) as the reference (horizontal) direction: \(V_R\) is in phase with the current, \(V_L\) leads the current by \(90^\circ\) (drawn upward) and \(V_C\) lags the current by \(90^\circ\) (drawn downward). The source voltage \(V\) is the resultant.

(ii) Calculations

(\(\alpha\)) r.m.s voltage of the source

From the phasor diagram, the resultant of the perpendicular components gives

\[ V = \sqrt{V_R^{2} + (V_L - V_C)^{2}} = \sqrt{20^{2} + (30 - 70)^{2}} \]\[ V = \sqrt{400 + 1600} = \sqrt{2000} \approx 44.7\,\text{V} \]

(\(\beta\)) r.m.s current in the circuit

First find the reactances and the impedance.

\[ X_L = 2\pi f L = 2\pi \times \frac{500}{\pi} \times 0.007 = 1000 \times 0.007 = 7\,\Omega \]\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times \frac{500}{\pi} \times 0.001} = \frac{1}{1000 \times 0.001} = 1\,\Omega \]\[ Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = \sqrt{8^{2} + (7 - 1)^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10\,\Omega \]\[ I = \frac{V}{Z} = \frac{44.7}{10} \approx 4.47\,\text{A} \]

(\(\gamma\)) power dissipated in the circuit

Power is dissipated only in the resistor:

\[ P = I^{2}R = (4.47)^{2} \times 8 = 20 \times 8 = 160\,\text{W} \]

(iii) Sinusoidal equation for the source voltage

The peak (maximum) voltage is

\[ V_o = V\sqrt{2} = 44.7 \times \sqrt{2} \approx 63.2\,\text{V} \]

The angular frequency is

\[ \omega = 2\pi f = 2\pi \times \frac{500}{\pi} = 1000\,\text{rad s}^{-1} \]

Therefore the instantaneous source voltage is

\[ V = V_o \sin(\omega t) = 63.2\,\sin(1000\,t)\ \text{volts} \]

Bayanin Amsa

(a) Faster on a hot or cold day?

Sound travels faster on a hot day. The speed of sound in air increases with temperature because the air molecules move faster and transmit the disturbance more rapidly (the speed is proportional to \(\sqrt{T}\), where \(T\) is the absolute temperature).

(b) Why megaphones are shaped like funnels

The funnel (conical) shape channels the sound energy and directs it forward in a narrow beam instead of allowing it to spread out in all directions. This concentrates the sound waves, so the sound is louder and carries farther in the intended direction.

(c) Refraction at the air-glass surface (refractive index \(n = 1.5\), angle of incidence read from the diagram \(= 30^{\circ}\))

(i) In the copied diagram: the angle of incidence (x) is between the incident ray and the normal in the air; the angle of reflection (y) is between the reflected ray and the normal (in air, equal to x); the angle of refraction (z) is between the refracted ray and the normal inside the glass.

(ii) Applying Snell's law from air into glass:

\[ n = \frac{\sin i}{\sin r} \;\Rightarrow\; \sin r = \frac{\sin i}{n} = \frac{\sin 30^{\circ}}{1.5} = \frac{0.5}{1.5} = 0.3333 \]

\[ r = \sin^{-1}(0.3333) = 19.47^{\circ} \approx 19^{\circ} \]

(d) Frequency of the vibrating sonometer wire (SHM)

Amplitude \(A = 1.0\ \text{cm} = 0.01\ \text{m}\); maximum acceleration \(a_{max} = 980\ \text{m s}^{-2}\). For SHM, \(a_{max} = \omega^{2} A\):

\[ \omega^{2} = \frac{a_{max}}{A} = \frac{980}{0.01} = 98000 \]

\[ \omega = \sqrt{98000} = 313.05\ \text{rad s}^{-1} \]

Since \(\omega = 2\pi f\), with \(\pi = \dfrac{22}{7}\):

\[ f = \frac{\omega}{2\pi} = \frac{313.05}{2 \times \tfrac{22}{7}} = \frac{313.05}{6.286} \approx 49.8\ \text{Hz} \approx 50\ \text{Hz} \]

Bayanin Amsa

The particle is projected at \(30^\circ\) to the horizontal, and its horizontal component of speed is \(u_x = 50\,\text{ms}^{-1}\).

(a) Initial speed \(u\)

The horizontal component is \(u_x = u\cos\theta\), so:

\[ u = \frac{u_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.866} = 57.7\,\text{ms}^{-1} \]

(b) Speed at maximum height

At the maximum height the vertical component of velocity is zero, so the speed there equals the (unchanged) horizontal component:

\[ v = u\cos\theta = 50\,\text{ms}^{-1} \]

Bayanin Amsa

An electron accelerated through a potential difference \(V\) gains kinetic energy equal to the electrical work done on it:

\[ eV = \frac{1}{2}m v^{2} \Rightarrow v = \sqrt{\frac{2eV}{m}} \]

Data: \(V = 2.5\,\text{kV} = 2.5\times10^{3}\,\text{V}\); \(e = 1.6\times10^{-19}\,\text{C}\); \(m = 9.1\times10^{-31}\,\text{kg}\).

\[ v = \sqrt{\frac{2(1.6\times10^{-19})(2.5\times10^{3})}{9.1\times10^{-31}}} = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} \] \[ v = \sqrt{8.79\times10^{14}} = 2.97\times10^{7}\,\text{ms}^{-1} \]

The maximum speed of the accelerated electrons is about \(3.0\times10^{7}\,\text{ms}^{-1}\).

Bayanin Amsa

(a) What is a polarizer?

A polarizer is an optical device (for example a Polaroid sheet or a tourmaline crystal) that has a single characteristic direction called its transmission axis. When light falls on it, it transmits only the component of the light vibrations that is parallel to this transmission axis and absorbs (or blocks) all the other components. It therefore converts an unpolarized beam, whose vibrations occur in all planes perpendicular to the direction of travel, into a plane-polarized beam whose vibrations lie in only one plane.

(b) How a polarizer polarizes a beam of unpolarized light

In an unpolarized beam the electric-field vibrations take place in every direction in the plane perpendicular to the direction of propagation. When this beam strikes the polarizer, only the vibration component that is parallel to the polarizer's transmission axis is allowed through; every component perpendicular to that axis is absorbed. The light emerging from the polarizer therefore vibrates in one plane only, that is, it is plane-polarized. Because only one component is transmitted, the emerging intensity is about half that of the incident unpolarized light \(\left(I = \tfrac{1}{2}I_0\right)\).

The action is shown in the diagram below. The unpolarized light (vibrating in all directions) passes through the first polarizer and emerges polarized in the vertical plane. When a second polarizer (the analyser) is placed with its transmission axis crossed at \(90^\circ\) to the first, it blocks these vertical vibrations and no light passes, which confirms that the beam leaving the first polarizer is indeed plane-polarized.

Bayanin Amsa

By Faraday's law of electrolysis, the mass deposited is \(m = zIt\), where \(z\) is the electrochemical equivalent, \(I\) the true current and \(t\) the time.

Data: \(m = 0.66\,\text{g}\), \(z = 0.00033\,\text{g C}^{-1}\), \(t = 30\,\text{min} = 1800\,\text{s}\).

True charge that passed:

\[ Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]

True (actual) current:

\[ I_{true} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]

Error in the ammeter reading:

\[ \text{Error} = I_{true} - I_{reading} = 1.11 - 1.00 = 0.11\,\text{A} \]

The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of about \(\dfrac{0.11}{1.11}\times100 \approx 10\%\)).

Bayanin Amsa

(a) Brownian motion

Brownian motion is the continuous, random (zig-zag) movement of tiny particles suspended in a fluid (such as smoke particles in air or pollen grains in water), caused by their being bombarded unevenly by the fast-moving molecules of the surrounding fluid.

(b) Two inferences from the Brownian-motion experiment

• Matter (the fluid) is made up of tiny particles (molecules).
• These molecules are in constant, random motion and possess kinetic energy (the motion becomes more vigorous as temperature rises).