WAEC SSCE - Physics - 2012 (Essay)
Question 1 Report
(a) Explain the term electrodes in electric cells.
b) An electric current passing through an electrolyte for 2 minutes deposited 200 g of a substance. If the electrochemical equivalent of the substance is 8.33 x 10\(^{-4}\)g C\(^{-1}\), calculate the current passed.
(a) Electrodes in electric cells
Electrodes are the two conducting plates or rods through which electric current enters and leaves the electrolyte in a cell. The electrode by which conventional current enters the electrolyte is the anode (positive), and the one by which it leaves is the cathode (negative).
(b) Calculating the current
By Faraday's first law of electrolysis, the mass deposited is \( m = Z I t \), where Z is the electrochemical equivalent, I the current and t the time.
Given: \( m = 200\,\text{g} \), \( Z = 8.33 \times 10^{-4}\,\text{g C}^{-1} \), \( t = 2\,\text{min} = 120\,\text{s} \).
\[ I = \frac{m}{Z t} = \frac{200}{(8.33 \times 10^{-4}) \times 120} \]
\[ I = \frac{200}{0.09996} \approx 2.0 \times 10^{3}\,\text{A} \]
The current passed is about 2000 A.
Answer Details
(a) Electrodes in electric cells
Electrodes are the two conducting plates or rods through which electric current enters and leaves the electrolyte in a cell. The electrode by which conventional current enters the electrolyte is the anode (positive), and the one by which it leaves is the cathode (negative).
(b) Calculating the current
By Faraday's first law of electrolysis, the mass deposited is \( m = Z I t \), where Z is the electrochemical equivalent, I the current and t the time.
Given: \( m = 200\,\text{g} \), \( Z = 8.33 \times 10^{-4}\,\text{g C}^{-1} \), \( t = 2\,\text{min} = 120\,\text{s} \).
\[ I = \frac{m}{Z t} = \frac{200}{(8.33 \times 10^{-4}) \times 120} \]
\[ I = \frac{200}{0.09996} \approx 2.0 \times 10^{3}\,\text{A} \]
The current passed is about 2000 A.
Question 2 Report
(a) Explain wave-particle duality of light.
(b) Illustrate your answer in (a) with observable phenomena.
(a) Wave-particle duality of light
Wave-particle duality means that light has a dual nature: it can behave both as a wave and as a stream of particles (photons). In some experiments light shows wave properties, while in others it shows particle properties. No single model fully describes all the behaviour of light; the two descriptions complement each other.
(b) Observable phenomena illustrating this
Wave nature is shown by:
- Interference (for example Young's double-slit fringes).
- Diffraction of light around edges and through narrow slits.
- Polarization of light.
Particle nature is shown by:
- The photoelectric effect, where light ejects electrons only if its frequency is high enough (energy comes in photon packets \( E = hf \)).
- The Compton effect, where an X-ray photon collides with an electron like a particle.
Answer Details
(a) Wave-particle duality of light
Wave-particle duality means that light has a dual nature: it can behave both as a wave and as a stream of particles (photons). In some experiments light shows wave properties, while in others it shows particle properties. No single model fully describes all the behaviour of light; the two descriptions complement each other.
(b) Observable phenomena illustrating this
Wave nature is shown by:
- Interference (for example Young's double-slit fringes).
- Diffraction of light around edges and through narrow slits.
- Polarization of light.
Particle nature is shown by:
- The photoelectric effect, where light ejects electrons only if its frequency is high enough (energy comes in photon packets \( E = hf \)).
- The Compton effect, where an X-ray photon collides with an electron like a particle.
Question 3 Report
(a) (i) Explain why x-rays can be used to produce photographs of fractures in bones.
(ii) List four uses of x-rays other than in medicine.
(b) State the energy transformations which takes place during the operation of an x-ray tube.
(c) (i) Explain three named dangers to which human beings may be exposed when subjected to large doses of x-rays.
(ii) State two precautions that must be taken by persons working with x-rays.
(d) In an x-ray tube, an electron is accelerated from rest towards a tungsten target biased at a potential of 33 kV. Calculate, for the electron, the
(i) kinetic energy;
(ii) velocity. [h = 6.6 x 10\(^{-14}\) Js; Me = 9.1 x10\(^{-31}\) kg; c = 3.0 x 10\(^4\) ms\(^{-1}\); e = 1.6 x 10\(^{-19}\) C.]
(a)(i) Why X-rays can photograph bone fractures
X-rays are highly penetrating. They pass easily through soft tissue (flesh) but are strongly absorbed by the denser bone. When X-rays are directed through a limb onto a photographic film or detector, the bones cast dark shadows while the flesh lets the rays through. A break or crack in a bone shows up as a discontinuity (a line or gap) in the bone shadow, so fractures can be seen.
(a)(ii) Four uses of X-rays other than in medicine
- Detecting flaws, cracks and cavities in metal castings and welds (industrial radiography).
- Studying the arrangement of atoms in crystals (X-ray crystallography/diffraction).
- Screening luggage and cargo for hidden or dangerous objects at airports and borders.
- Checking the authenticity of paintings and examining the internal structure of sealed goods.
(b) Energy transformations in an X-ray tube
Electrical energy is converted into the kinetic energy of the fast-moving electrons; when these electrons strike the target, their kinetic energy is converted mainly into heat (about 99%) and a small part into X-ray (electromagnetic) energy.
(c)(i) Three dangers of large doses of X-rays
- They destroy or damage living body cells and tissues, causing radiation burns.
- They can cause cancer, for example leukaemia.
- They can damage reproductive cells, causing genetic mutations or sterility (and can damage the eyes).
(c)(ii) Two precautions
- Shield the body with lead aprons and work behind lead-lined walls/screens.
- Limit the exposure time and dose, and wear a film badge (dosimeter) to monitor the radiation received.
(d)(i) Kinetic energy of the electron
\[ KE = eV = (1.6 \times 10^{-19}) \times (33 \times 10^{3}) = 5.28 \times 10^{-15}\,\text{J} \]
(d)(ii) Velocity of the electron
Using \( \tfrac{1}{2}m v^2 = KE \):
\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2 \times 5.28 \times 10^{-15}}{9.1 \times 10^{-31}}} \]
\[ v = \sqrt{1.16 \times 10^{16}} \approx 1.08 \times 10^{8}\,\text{m s}^{-1} \]
The electron reaches about \( 1.08 \times 10^{8}\,\text{m s}^{-1} \).
Answer Details
(a)(i) Why X-rays can photograph bone fractures
X-rays are highly penetrating. They pass easily through soft tissue (flesh) but are strongly absorbed by the denser bone. When X-rays are directed through a limb onto a photographic film or detector, the bones cast dark shadows while the flesh lets the rays through. A break or crack in a bone shows up as a discontinuity (a line or gap) in the bone shadow, so fractures can be seen.
(a)(ii) Four uses of X-rays other than in medicine
- Detecting flaws, cracks and cavities in metal castings and welds (industrial radiography).
- Studying the arrangement of atoms in crystals (X-ray crystallography/diffraction).
- Screening luggage and cargo for hidden or dangerous objects at airports and borders.
- Checking the authenticity of paintings and examining the internal structure of sealed goods.
(b) Energy transformations in an X-ray tube
Electrical energy is converted into the kinetic energy of the fast-moving electrons; when these electrons strike the target, their kinetic energy is converted mainly into heat (about 99%) and a small part into X-ray (electromagnetic) energy.
(c)(i) Three dangers of large doses of X-rays
- They destroy or damage living body cells and tissues, causing radiation burns.
- They can cause cancer, for example leukaemia.
- They can damage reproductive cells, causing genetic mutations or sterility (and can damage the eyes).
(c)(ii) Two precautions
- Shield the body with lead aprons and work behind lead-lined walls/screens.
- Limit the exposure time and dose, and wear a film badge (dosimeter) to monitor the radiation received.
(d)(i) Kinetic energy of the electron
\[ KE = eV = (1.6 \times 10^{-19}) \times (33 \times 10^{3}) = 5.28 \times 10^{-15}\,\text{J} \]
(d)(ii) Velocity of the electron
Using \( \tfrac{1}{2}m v^2 = KE \):
\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2 \times 5.28 \times 10^{-15}}{9.1 \times 10^{-31}}} \]
\[ v = \sqrt{1.16 \times 10^{16}} \approx 1.08 \times 10^{8}\,\text{m s}^{-1} \]
The electron reaches about \( 1.08 \times 10^{8}\,\text{m s}^{-1} \).
Question 4 Report
The diagram above illustrates an arrangement of a cathode ray from an electron gun and a bar magnet placed perpendicularly to the direction of the ray. Will the ray bend downward or upward? Explain.
From the diagram: the electron gun fires the cathode ray horizontally to the right, and the bar magnet stands vertically to the right of the beam with its S pole at the top and its N pole at the bottom, so its axis is perpendicular to the ray.
Direction of the magnetic field. In the region in front of the magnet the field lines run from the N pole to the S pole, that is, vertically upward (from the bottom N pole to the top S pole), in the plane of the paper.
Nature of the beam. A cathode ray is a stream of negatively charged electrons. Taking the electron velocity to the right as \(v\) and the field upward as \(B\), the magnetic force on each electron is
\[ \vec{F} = -e\,(\vec{v} \times \vec{B}) \]
Both \(\vec{v}\) (horizontal, to the right) and \(\vec{B}\) (vertical, upward) lie in the plane of the paper and are mutually perpendicular. Their cross product \(\vec{v}\times\vec{B}\) therefore points out of the paper, and because the electron charge is negative, the force \(\vec{F}\) points into the plane of the paper.
Conclusion. The ray is not simply pushed up or down within the plane of the diagram. It is deflected perpendicular to the paper (into the page, away from the reader), because the deflecting force is at right angles to both the ray and the vertical field. Had the ray moved in the same direction and the field been directed into the page, the electrons would then be deflected downward; with the field as drawn (upward, in the plane of the paper) the deflection is out of the plane.
Answer Details
From the diagram: the electron gun fires the cathode ray horizontally to the right, and the bar magnet stands vertically to the right of the beam with its S pole at the top and its N pole at the bottom, so its axis is perpendicular to the ray.
Direction of the magnetic field. In the region in front of the magnet the field lines run from the N pole to the S pole, that is, vertically upward (from the bottom N pole to the top S pole), in the plane of the paper.
Nature of the beam. A cathode ray is a stream of negatively charged electrons. Taking the electron velocity to the right as \(v\) and the field upward as \(B\), the magnetic force on each electron is
\[ \vec{F} = -e\,(\vec{v} \times \vec{B}) \]
Both \(\vec{v}\) (horizontal, to the right) and \(\vec{B}\) (vertical, upward) lie in the plane of the paper and are mutually perpendicular. Their cross product \(\vec{v}\times\vec{B}\) therefore points out of the paper, and because the electron charge is negative, the force \(\vec{F}\) points into the plane of the paper.
Conclusion. The ray is not simply pushed up or down within the plane of the diagram. It is deflected perpendicular to the paper (into the page, away from the reader), because the deflecting force is at right angles to both the ray and the vertical field. Had the ray moved in the same direction and the field been directed into the page, the electrons would then be deflected downward; with the field as drawn (upward, in the plane of the paper) the deflection is out of the plane.
Question 5 Report
A ray of light is incident on an air-glass boundary at an angle \(\theta\). If the angle between the partially reflected ray and the refracted ray is 90°, calculate \(\theta\), given that the refractive index of glass is 1.50.
Finding the angle of incidence \( \theta \)
The reflected ray makes an angle \( \theta \) with the normal (equal to the angle of incidence), and the refracted ray makes an angle \( r \) with the normal on the other side. If the angle between the reflected ray and the refracted ray is \( 90^\circ \), then, measuring from the normal on each side,
\[ \theta + 90^\circ + r = 180^\circ \quad\Rightarrow\quad \theta + r = 90^\circ \quad\Rightarrow\quad r = 90^\circ - \theta \]
Applying Snell's law at the air-glass boundary (\( n = 1.50 \)):
\[ \sin\theta = n\sin r = n\sin(90^\circ - \theta) = n\cos\theta \]
\[ \frac{\sin\theta}{\cos\theta} = n \quad\Rightarrow\quad \tan\theta = 1.50 \]
\[ \theta = \tan^{-1}(1.50) = 56.3^\circ \]
The angle of incidence is \( \theta \approx 56.3^\circ \). (This is the polarizing, or Brewster, angle, for which \( \tan\theta = n \).)
Answer Details
Finding the angle of incidence \( \theta \)
The reflected ray makes an angle \( \theta \) with the normal (equal to the angle of incidence), and the refracted ray makes an angle \( r \) with the normal on the other side. If the angle between the reflected ray and the refracted ray is \( 90^\circ \), then, measuring from the normal on each side,
\[ \theta + 90^\circ + r = 180^\circ \quad\Rightarrow\quad \theta + r = 90^\circ \quad\Rightarrow\quad r = 90^\circ - \theta \]
Applying Snell's law at the air-glass boundary (\( n = 1.50 \)):
\[ \sin\theta = n\sin r = n\sin(90^\circ - \theta) = n\cos\theta \]
\[ \frac{\sin\theta}{\cos\theta} = n \quad\Rightarrow\quad \tan\theta = 1.50 \]
\[ \theta = \tan^{-1}(1.50) = 56.3^\circ \]
The angle of incidence is \( \theta \approx 56.3^\circ \). (This is the polarizing, or Brewster, angle, for which \( \tan\theta = n \).)
Question 6 Report
(a) State the principle of conservation of linear momentum.
(b) Explain the mode of action of a propelled rocket.
(c) During a training session, two footballers pass a ball repeatedly between themselves. Give two reasons why the to and fro motion of the ball is not simple harmonic.
(d) A ball is dropped from a height, at the same time as another ball is projected horizontally from the same height.
(i) Would the balls hit the ground at the same time?
(ii) Explain your answer in (i).
(e) A ball of mass 0.10 kg is projected horizontally onto a vertical wall with a speed of 17 ms\(^{-1}\). The ball makes contact with the wall for 0.15 s and rebounds horizontally with a speed of 13 ms\(^{-1}\).
Calculate the:
(i) change in momentum of the ball;
(ii) average force exerted on the ball during its collision with the wall.
(a) Principle of conservation of linear momentum
In a closed (isolated) system on which no net external force acts, the total linear momentum of the bodies before an interaction (such as a collision) is equal to the total linear momentum after the interaction.
(b) Mode of action of a propelled rocket
A rocket burns fuel and expels a large mass of hot exhaust gases backwards at very high speed. By the conservation of momentum, the backward momentum given to the gases is balanced by an equal and opposite forward momentum gained by the rocket, so the rocket is driven forward. (This is also Newton's third law: action and reaction.)
(c) Two reasons the to-and-fro motion of the ball is not simple harmonic
- There is no restoring force that is proportional to, and directed opposite to, the displacement from a fixed equilibrium position.
- The motion is not periodic about a fixed point with a fixed period; the ball travels between the players at roughly constant speed rather than with sinusoidally varying speed.
(d)(i) Yes, both balls hit the ground at the same time.
(d)(ii) The vertical motion is independent of the horizontal motion. Both balls start with the same vertical velocity (zero) and fall under the same acceleration due to gravity g through the same height, so \( h = \tfrac{1}{2}g t^2 \) gives the same time of fall for each, regardless of the horizontal projection.
(e) Take the initial direction of motion as positive; the ball rebounds in the opposite direction.
(i) Change in momentum
\[ \Delta p = m(v - u) = 0.10 \times (-13 - 17) = 0.10 \times (-30) = -3.0\,\text{kg m s}^{-1} \]
The change in momentum is 3.0 kg m s\(^{-1}\) (directed away from the wall).
(ii) Average force
\[ F = \frac{\Delta p}{t} = \frac{3.0}{0.15} = 20\,\text{N} \]
The average force exerted on the ball is 20 N (directed away from the wall).
Answer Details
(a) Principle of conservation of linear momentum
In a closed (isolated) system on which no net external force acts, the total linear momentum of the bodies before an interaction (such as a collision) is equal to the total linear momentum after the interaction.
(b) Mode of action of a propelled rocket
A rocket burns fuel and expels a large mass of hot exhaust gases backwards at very high speed. By the conservation of momentum, the backward momentum given to the gases is balanced by an equal and opposite forward momentum gained by the rocket, so the rocket is driven forward. (This is also Newton's third law: action and reaction.)
(c) Two reasons the to-and-fro motion of the ball is not simple harmonic
- There is no restoring force that is proportional to, and directed opposite to, the displacement from a fixed equilibrium position.
- The motion is not periodic about a fixed point with a fixed period; the ball travels between the players at roughly constant speed rather than with sinusoidally varying speed.
(d)(i) Yes, both balls hit the ground at the same time.
(d)(ii) The vertical motion is independent of the horizontal motion. Both balls start with the same vertical velocity (zero) and fall under the same acceleration due to gravity g through the same height, so \( h = \tfrac{1}{2}g t^2 \) gives the same time of fall for each, regardless of the horizontal projection.
(e) Take the initial direction of motion as positive; the ball rebounds in the opposite direction.
(i) Change in momentum
\[ \Delta p = m(v - u) = 0.10 \times (-13 - 17) = 0.10 \times (-30) = -3.0\,\text{kg m s}^{-1} \]
The change in momentum is 3.0 kg m s\(^{-1}\) (directed away from the wall).
(ii) Average force
\[ F = \frac{\Delta p}{t} = \frac{3.0}{0.15} = 20\,\text{N} \]
The average force exerted on the ball is 20 N (directed away from the wall).
Question 7 Report
(a) Explain the term critical angle.
(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.
(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.
(d) Study the ray diagram below and use it to answer the questions that follow.
Calculate the:
(i) values of angles P,Q and R;
(ii) refractive index n of the glass prism;
(iii) value of e;
(iv) total deviation D.
(a) Critical angle. The critical angle is the angle of incidence in the optically denser medium for which the corresponding angle of refraction in the less dense medium is exactly \(90^\circ\). For an angle of incidence greater than this value, total internal reflection occurs.
(b) Two factors that determine the deviation by a triangular prism.
- The angle of incidence of the ray on the first face.
- The refracting (apex) angle of the prism, and the refractive index of the glass (which itself depends on the colour/wavelength of the light).
(c) Least deviation. The ray is least deviated when it passes symmetrically through the prism, so the two internal angles are equal to the given angle of refraction: \(r_1=r_2=r=29.0^\circ\). The refracting angle is then
\[A=r_1+r_2=2(29.0^\circ)=58.0^\circ.\]
At the first face, using \(n=\dfrac{\sin i}{\sin r}\) with \(n=1.5\):
\[\sin i = n\sin r = 1.5\times\sin 29.0^\circ = 1.5\times0.4848 = 0.7272,\]
\[i=\sin^{-1}(0.7272)=46.7^\circ.\]
The angle of minimum (least) deviation is
\[D_{min}=2i-A = 2(46.7^\circ)-58.0^\circ = 93.4^\circ-58.0^\circ \approx 35.4^\circ.\]
(d) From the ray diagram the apex angle \(A=60^\circ\), the angle of incidence on the left face is \(45^\circ\), and the ray is refracted through \(30^\circ\) inside the glass.
(i) Angles P, Q and R.
\(P\) is the angle of refraction at the first face, read directly from the diagram: \(P=r_1=30^\circ\).
Since \(r_1+r_2=A\), the angle of incidence at the second face is \(R=r_2=A-r_1=60^\circ-30^\circ=30^\circ\).
\(Q\) is the angle between the two normals (drawn at P and R) where they meet inside the prism. In the triangle formed by the internal ray PR and the two normals, \[P+R+Q=180^\circ,\] \[Q=180^\circ-30^\circ-30^\circ=120^\circ.\]
So \(P=30^\circ,\; Q=120^\circ,\; R=30^\circ.\)
(ii) Refractive index n. At the first face,
\[n=\frac{\sin 45^\circ}{\sin 30^\circ}=\frac{0.7071}{0.5000}=1.41.\]
(iii) Angle of emergence e. At the second face the angle of incidence inside is \(r_2=30^\circ\), so
\[\sin e = n\sin r_2 = 1.41\times\sin 30^\circ = 1.41\times0.5 = 0.707,\]
\[e=\sin^{-1}(0.707)=45^\circ.\]
(iv) Total deviation D.
\[D=(i_1+e)-A=(45^\circ+45^\circ)-60^\circ=30^\circ.\]
Because \(i_1=e=45^\circ\) and \(r_1=r_2=30^\circ\), the ray passes symmetrically, so this \(30^\circ\) is in fact the minimum deviation.
Answer Details
(a) Critical angle. The critical angle is the angle of incidence in the optically denser medium for which the corresponding angle of refraction in the less dense medium is exactly \(90^\circ\). For an angle of incidence greater than this value, total internal reflection occurs.
(b) Two factors that determine the deviation by a triangular prism.
- The angle of incidence of the ray on the first face.
- The refracting (apex) angle of the prism, and the refractive index of the glass (which itself depends on the colour/wavelength of the light).
(c) Least deviation. The ray is least deviated when it passes symmetrically through the prism, so the two internal angles are equal to the given angle of refraction: \(r_1=r_2=r=29.0^\circ\). The refracting angle is then
\[A=r_1+r_2=2(29.0^\circ)=58.0^\circ.\]
At the first face, using \(n=\dfrac{\sin i}{\sin r}\) with \(n=1.5\):
\[\sin i = n\sin r = 1.5\times\sin 29.0^\circ = 1.5\times0.4848 = 0.7272,\]
\[i=\sin^{-1}(0.7272)=46.7^\circ.\]
The angle of minimum (least) deviation is
\[D_{min}=2i-A = 2(46.7^\circ)-58.0^\circ = 93.4^\circ-58.0^\circ \approx 35.4^\circ.\]
(d) From the ray diagram the apex angle \(A=60^\circ\), the angle of incidence on the left face is \(45^\circ\), and the ray is refracted through \(30^\circ\) inside the glass.
(i) Angles P, Q and R.
\(P\) is the angle of refraction at the first face, read directly from the diagram: \(P=r_1=30^\circ\).
Since \(r_1+r_2=A\), the angle of incidence at the second face is \(R=r_2=A-r_1=60^\circ-30^\circ=30^\circ\).
\(Q\) is the angle between the two normals (drawn at P and R) where they meet inside the prism. In the triangle formed by the internal ray PR and the two normals, \[P+R+Q=180^\circ,\] \[Q=180^\circ-30^\circ-30^\circ=120^\circ.\]
So \(P=30^\circ,\; Q=120^\circ,\; R=30^\circ.\)
(ii) Refractive index n. At the first face,
\[n=\frac{\sin 45^\circ}{\sin 30^\circ}=\frac{0.7071}{0.5000}=1.41.\]
(iii) Angle of emergence e. At the second face the angle of incidence inside is \(r_2=30^\circ\), so
\[\sin e = n\sin r_2 = 1.41\times\sin 30^\circ = 1.41\times0.5 = 0.707,\]
\[e=\sin^{-1}(0.707)=45^\circ.\]
(iv) Total deviation D.
\[D=(i_1+e)-A=(45^\circ+45^\circ)-60^\circ=30^\circ.\]
Because \(i_1=e=45^\circ\) and \(r_1=r_2=30^\circ\), the ray passes symmetrically, so this \(30^\circ\) is in fact the minimum deviation.
Question 8 Report
A spiral spring with a metal extends by 10.5 cm in air. When the metal is fully submerged in water, the spring extends by 6.8 cm. Calculate the relative density of the metal. (Assume Hooke's law is obeyed)
Relative density from spring extensions
By Hooke's law the extension of the spring is proportional to the force (weight) it supports.
In air the spring supports the full weight of the metal, so its extension is proportional to the weight:
\[ e_1 = 10.5\,\text{cm} \propto W \]
In water the spring supports the apparent weight (weight minus upthrust), so:
\[ e_2 = 6.8\,\text{cm} \propto (W - \text{upthrust}) \]
Therefore the extension due to the upthrust is proportional to
\[ e_1 - e_2 = 10.5 - 6.8 = 3.7\,\text{cm} \]
The upthrust equals the weight of water displaced, so
\[ \text{relative density} = \frac{\text{weight of metal in air}}{\text{weight of equal volume of water}} = \frac{e_1}{e_1 - e_2} \]
\[ = \frac{10.5}{3.7} = 2.84 \]
The relative density of the metal is about 2.84.
Answer Details
Relative density from spring extensions
By Hooke's law the extension of the spring is proportional to the force (weight) it supports.
In air the spring supports the full weight of the metal, so its extension is proportional to the weight:
\[ e_1 = 10.5\,\text{cm} \propto W \]
In water the spring supports the apparent weight (weight minus upthrust), so:
\[ e_2 = 6.8\,\text{cm} \propto (W - \text{upthrust}) \]
Therefore the extension due to the upthrust is proportional to
\[ e_1 - e_2 = 10.5 - 6.8 = 3.7\,\text{cm} \]
The upthrust equals the weight of water displaced, so
\[ \text{relative density} = \frac{\text{weight of metal in air}}{\text{weight of equal volume of water}} = \frac{e_1}{e_1 - e_2} \]
\[ = \frac{10.5}{3.7} = 2.84 \]
The relative density of the metal is about 2.84.
Question 9 Report
(a) On what principle does lighting in a fluorescent tube operate?
(b) State two factors which determine the colour of light produced in a fluorescent tube.
(a) Principle of a fluorescent tube
A fluorescent tube works on the principle of fluorescence. An electric discharge is passed through mercury vapour in the tube, which emits ultraviolet (UV) radiation. This UV radiation strikes the fluorescent (phosphor) powder coating the inside wall of the tube, causing it to fluoresce and give out visible light.
(b) Two factors that determine the colour of the light produced
- The nature (chemical composition) of the fluorescent powder (phosphor) coating the inside of the tube.
- The type of gas or vapour used in the tube.
Answer Details
(a) Principle of a fluorescent tube
A fluorescent tube works on the principle of fluorescence. An electric discharge is passed through mercury vapour in the tube, which emits ultraviolet (UV) radiation. This UV radiation strikes the fluorescent (phosphor) powder coating the inside wall of the tube, causing it to fluoresce and give out visible light.
(b) Two factors that determine the colour of the light produced
- The nature (chemical composition) of the fluorescent powder (phosphor) coating the inside of the tube.
- The type of gas or vapour used in the tube.
Question 10 Report
Explain why sound waves cannot be plane Polarized.
Why sound waves cannot be plane polarized
Sound waves are longitudinal waves: the particles of the medium vibrate along the same direction in which the wave travels (backwards and forwards, forming compressions and rarefactions).
Polarization is the restriction of the vibrations of a wave to a single plane. This is only possible when the vibrations are perpendicular (transverse) to the direction of travel, because only then are there sideways vibrations that can be confined to one plane.
Since the vibrations of a sound wave are parallel to the direction of propagation and are the same in every plane containing that direction, there is no sideways component that can be selected out. Therefore a sound wave cannot be plane polarized. Only transverse waves (such as light) can be polarized.
Answer Details
Why sound waves cannot be plane polarized
Sound waves are longitudinal waves: the particles of the medium vibrate along the same direction in which the wave travels (backwards and forwards, forming compressions and rarefactions).
Polarization is the restriction of the vibrations of a wave to a single plane. This is only possible when the vibrations are perpendicular (transverse) to the direction of travel, because only then are there sideways vibrations that can be confined to one plane.
Since the vibrations of a sound wave are parallel to the direction of propagation and are the same in every plane containing that direction, there is no sideways component that can be selected out. Therefore a sound wave cannot be plane polarized. Only transverse waves (such as light) can be polarized.
Question 11 Report
(a) Define magnetic line of force.
(b) A wire of length 10 cm carrying a current of 4.0 A is placed between the poles of a powerful electromagnet of magnetic flux density 2.0 T. Calculate the:
(i) force on the wire when it is parallel to the field;
(ii) maximum force on the wire;
(iii) force on the wire when it makes an angle of 60° with the field.
(c) Describe how keepers can be used to preserve the magnetic strength of permanent bar magnets.
(d) A sailor observes that his mariners' compass reads N 10° W at a place where the angle of declination is N15° W. Calculate the true bearing of the place.
(a) Magnetic line of force
A magnetic line of force is a line (curve) drawn in a magnetic field such that the tangent to it at any point gives the direction of the magnetic field at that point; equivalently, it is the path along which a free (isolated) north pole would move if placed in the field.
(b) Force on the current-carrying wire
Data: \( L = 10\,\text{cm} = 0.10\,\text{m} \), \( I = 4.0\,\text{A} \), \( B = 2.0\,\text{T} \). The force is \( F = BIL\sin\theta \).
(i) Wire parallel to the field (\( \theta = 0^\circ \))
\[ F = BIL\sin 0^\circ = 0\,\text{N} \]
(ii) Maximum force (\( \theta = 90^\circ \))
\[ F = BIL = 2.0 \times 4.0 \times 0.10 = 0.80\,\text{N} \]
(iii) Wire at \( 60^\circ \) to the field
\[ F = BIL\sin 60^\circ = 0.80 \times 0.866 = 0.69\,\text{N} \]
(c) Use of keepers
Bar magnets are stored in pairs, laid side by side with the north pole of one next to the south pole of the other, and short bars of soft iron (keepers) are placed across the two ends. The keepers become magnetized by induction, and together with the magnets they form a closed loop of magnetic flux. This keeps the molecular magnets (domains) aligned and prevents self-demagnetization, so the magnets retain their strength.
(d) True bearing
The compass reads N 10° W relative to magnetic north, and magnetic north itself lies N 15° W of true north (angle of declination). Since both deviations are to the west, they add:
\[ 10^\circ + 15^\circ = 25^\circ \text{ west of true north} \]
The true bearing of the place is N 25° W.
Answer Details
(a) Magnetic line of force
A magnetic line of force is a line (curve) drawn in a magnetic field such that the tangent to it at any point gives the direction of the magnetic field at that point; equivalently, it is the path along which a free (isolated) north pole would move if placed in the field.
(b) Force on the current-carrying wire
Data: \( L = 10\,\text{cm} = 0.10\,\text{m} \), \( I = 4.0\,\text{A} \), \( B = 2.0\,\text{T} \). The force is \( F = BIL\sin\theta \).
(i) Wire parallel to the field (\( \theta = 0^\circ \))
\[ F = BIL\sin 0^\circ = 0\,\text{N} \]
(ii) Maximum force (\( \theta = 90^\circ \))
\[ F = BIL = 2.0 \times 4.0 \times 0.10 = 0.80\,\text{N} \]
(iii) Wire at \( 60^\circ \) to the field
\[ F = BIL\sin 60^\circ = 0.80 \times 0.866 = 0.69\,\text{N} \]
(c) Use of keepers
Bar magnets are stored in pairs, laid side by side with the north pole of one next to the south pole of the other, and short bars of soft iron (keepers) are placed across the two ends. The keepers become magnetized by induction, and together with the magnets they form a closed loop of magnetic flux. This keeps the molecular magnets (domains) aligned and prevents self-demagnetization, so the magnets retain their strength.
(d) True bearing
The compass reads N 10° W relative to magnetic north, and magnetic north itself lies N 15° W of true north (angle of declination). Since both deviations are to the west, they add:
\[ 10^\circ + 15^\circ = 25^\circ \text{ west of true north} \]
The true bearing of the place is N 25° W.
Question 12 Report
Explain why it is desirable to install an air conditioner near the ceiling of a room and not close to the floor.
Why an air conditioner is installed near the ceiling
An air conditioner cools the air passing through it. Cold air is denser (heavier) than warm air, so it tends to fall, while warm air is less dense and rises.
When the air conditioner is placed near the ceiling, the cold air it produces sinks downwards through the room, pushing the warmer air upwards towards the unit, where it is cooled in turn. This sets up a continuous convection current that circulates the air and cools the whole room evenly and quickly.
If it were fixed near the floor, the cold air produced would simply remain at floor level (having nowhere to fall to), the warm air would stay trapped near the ceiling, and there would be no effective circulation, so the room would be cooled poorly.
Answer Details
Why an air conditioner is installed near the ceiling
An air conditioner cools the air passing through it. Cold air is denser (heavier) than warm air, so it tends to fall, while warm air is less dense and rises.
When the air conditioner is placed near the ceiling, the cold air it produces sinks downwards through the room, pushing the warmer air upwards towards the unit, where it is cooled in turn. This sets up a continuous convection current that circulates the air and cools the whole room evenly and quickly.
If it were fixed near the floor, the cold air produced would simply remain at floor level (having nowhere to fall to), the warm air would stay trapped near the ceiling, and there would be no effective circulation, so the room would be cooled poorly.
Question 13 Report
a) Define surface tension.
(b) State two methods by which the surface tension of a liquid can be reduced.
(a) Surface tension
Surface tension is the force per unit length acting along (perpendicular to) a line drawn on the surface of a liquid, tending to make the surface behave like a stretched elastic skin and to contract to the smallest possible area. Its unit is the newton per metre (\( \text{N m}^{-1} \)).
(b) Two methods of reducing surface tension
- Raising the temperature of the liquid (surface tension falls as temperature rises).
- Adding a surface-active substance such as soap, detergent or oil (impurities) to the liquid.
Answer Details
(a) Surface tension
Surface tension is the force per unit length acting along (perpendicular to) a line drawn on the surface of a liquid, tending to make the surface behave like a stretched elastic skin and to contract to the smallest possible area. Its unit is the newton per metre (\( \text{N m}^{-1} \)).
(b) Two methods of reducing surface tension
- Raising the temperature of the liquid (surface tension falls as temperature rises).
- Adding a surface-active substance such as soap, detergent or oil (impurities) to the liquid.
Question 14 Report
(a)
In the diagram illustrated, a body of mass m slides on an inclined plane. Show that the coefficient Mg of friction between the surfaces in contact is tan \(\theta\).
A spiral spring with a metal extends by 10.5 cm in air. When the metal is fully submerged in water, the spring extends by 6.8 cm. Calculate the relative density of the metal. (Assume Hooke's law is obeyed)
(a) Showing that \(\mu = \tan\theta\)
The diagram shows a body of mass m on a plane inclined at angle \(\theta\). The forces on it are the weight \(Mg\) acting vertically downward, the normal reaction \(R\) perpendicular to the plane, and the frictional force \(F_p\) acting up the plane (opposing the tendency to slide down).
Resolve the weight into components parallel and perpendicular to the plane:
- Component along the plane (down the slope): \(Mg\sin\theta\)
- Component perpendicular to the plane: \(Mg\cos\theta\)
Perpendicular to the plane there is no motion, so
\[ R = Mg\cos\theta. \]When the body is just on the point of sliding (or slides down at constant velocity), the frictional force is limiting and balances the component of weight down the plane:
\[ F_p = Mg\sin\theta. \]But the limiting friction is \(F_p = \mu R\). Therefore
\[ \mu R = Mg\sin\theta. \]Substituting \(R = Mg\cos\theta\):
\[ \mu\,Mg\cos\theta = Mg\sin\theta \]\[ \mu = \frac{\sin\theta}{\cos\theta} = \tan\theta. \]Hence the coefficient of friction equals \(\tan\theta\), where \(\theta\) is the angle of repose.
Relative density of the metal
Because the spring obeys Hooke's law, the extension is proportional to the load (force) on it.
Weight of metal in air \(\propto\) extension in air \(= 10.5\,\text{cm}\).
Apparent weight in water \(\propto\) extension in water \(= 6.8\,\text{cm}\).
Upthrust = loss in weight \(\propto (10.5 - 6.8) = 3.7\,\text{cm}\).
The upthrust equals the weight of water displaced, so
\[ \text{Relative density} = \frac{\text{weight in air}}{\text{weight of water displaced}} = \frac{\text{extension in air}}{\text{loss in extension}}. \]\[ \text{R.D.} = \frac{10.5}{10.5 - 6.8} = \frac{10.5}{3.7} = 2.84. \]Answer Details
(a) Showing that \(\mu = \tan\theta\)
The diagram shows a body of mass m on a plane inclined at angle \(\theta\). The forces on it are the weight \(Mg\) acting vertically downward, the normal reaction \(R\) perpendicular to the plane, and the frictional force \(F_p\) acting up the plane (opposing the tendency to slide down).
Resolve the weight into components parallel and perpendicular to the plane:
- Component along the plane (down the slope): \(Mg\sin\theta\)
- Component perpendicular to the plane: \(Mg\cos\theta\)
Perpendicular to the plane there is no motion, so
\[ R = Mg\cos\theta. \]When the body is just on the point of sliding (or slides down at constant velocity), the frictional force is limiting and balances the component of weight down the plane:
\[ F_p = Mg\sin\theta. \]But the limiting friction is \(F_p = \mu R\). Therefore
\[ \mu R = Mg\sin\theta. \]Substituting \(R = Mg\cos\theta\):
\[ \mu\,Mg\cos\theta = Mg\sin\theta \]\[ \mu = \frac{\sin\theta}{\cos\theta} = \tan\theta. \]Hence the coefficient of friction equals \(\tan\theta\), where \(\theta\) is the angle of repose.
Relative density of the metal
Because the spring obeys Hooke's law, the extension is proportional to the load (force) on it.
Weight of metal in air \(\propto\) extension in air \(= 10.5\,\text{cm}\).
Apparent weight in water \(\propto\) extension in water \(= 6.8\,\text{cm}\).
Upthrust = loss in weight \(\propto (10.5 - 6.8) = 3.7\,\text{cm}\).
The upthrust equals the weight of water displaced, so
\[ \text{Relative density} = \frac{\text{weight in air}}{\text{weight of water displaced}} = \frac{\text{extension in air}}{\text{loss in extension}}. \]\[ \text{R.D.} = \frac{10.5}{10.5 - 6.8} = \frac{10.5}{3.7} = 2.84. \]Question 15 Report
(a) (i) What is a machine?
(ii) State two uses of gears.
(iii) Define the velocity ratio for a pair of gear wheels.
(iv) How can the mechanical advantage of a gear system be increased?
The diagram above illustrates the gears system of a bicycle.
(i) Determine its velocity ratio.
(ii) If the bicycle has an efficiency of 90%, calculate the effort required to overcome a load of 70N.
(iii) Why is the calculated effort less than the actual effort required?
(a)(i) What is a machine? A machine is a device in which an effort applied at one point is used to overcome a load at another point, thereby making work easier to do.
(a)(ii) Two uses of gears:
- To change the speed of rotation (and to change the direction of motion).
- To magnify force, i.e. to increase the turning effect (torque) transmitted from one shaft to another.
(a)(iii) Velocity ratio of a pair of gear wheels. It is the ratio of the number of teeth on the driven wheel to the number of teeth on the driving wheel:
\[ V.R. = \frac{\text{number of teeth on driven wheel}}{\text{number of teeth on driving wheel}} \](a)(iv) How the mechanical advantage is increased. Make the driving gear smaller, with fewer teeth than the driven gear; this raises the velocity ratio and hence the mechanical advantage.
(b)(i) Velocity ratio of the bicycle gear system. The driven gear has 12 teeth and the driving gear has 18 teeth, so:
\[ V.R. = \frac{12}{18} = \frac{2}{3} \](b)(ii) Effort to overcome a load of 70 N at 90% efficiency.
\[ \text{Efficiency} = \frac{M.A.}{V.R.} \Rightarrow M.A. = \text{Efficiency} \times V.R. = 0.90 \times \frac{2}{3} = 0.60 \]Since \(M.A. = \dfrac{\text{Load}}{\text{Effort}}\):
\[ \text{Effort} = \frac{\text{Load}}{M.A.} = \frac{70}{0.60} = 116.7\ \text{N} \](b)(iii) Why the calculated effort is less than the actual effort required. The calculation ignores the extra force lost to friction between the chain and the gears and between the tyres and the load; in practice this friction means a larger effort is actually needed.
Answer Details
(a)(i) What is a machine? A machine is a device in which an effort applied at one point is used to overcome a load at another point, thereby making work easier to do.
(a)(ii) Two uses of gears:
- To change the speed of rotation (and to change the direction of motion).
- To magnify force, i.e. to increase the turning effect (torque) transmitted from one shaft to another.
(a)(iii) Velocity ratio of a pair of gear wheels. It is the ratio of the number of teeth on the driven wheel to the number of teeth on the driving wheel:
\[ V.R. = \frac{\text{number of teeth on driven wheel}}{\text{number of teeth on driving wheel}} \](a)(iv) How the mechanical advantage is increased. Make the driving gear smaller, with fewer teeth than the driven gear; this raises the velocity ratio and hence the mechanical advantage.
(b)(i) Velocity ratio of the bicycle gear system. The driven gear has 12 teeth and the driving gear has 18 teeth, so:
\[ V.R. = \frac{12}{18} = \frac{2}{3} \](b)(ii) Effort to overcome a load of 70 N at 90% efficiency.
\[ \text{Efficiency} = \frac{M.A.}{V.R.} \Rightarrow M.A. = \text{Efficiency} \times V.R. = 0.90 \times \frac{2}{3} = 0.60 \]Since \(M.A. = \dfrac{\text{Load}}{\text{Effort}}\):
\[ \text{Effort} = \frac{\text{Load}}{M.A.} = \frac{70}{0.60} = 116.7\ \text{N} \](b)(iii) Why the calculated effort is less than the actual effort required. The calculation ignores the extra force lost to friction between the chain and the gears and between the tyres and the load; in practice this friction means a larger effort is actually needed.
