WAEC SSCE - Physics - 1995 (Essay)
Question 1 Report
(a) State Faraday's law of electromagnetic induction.
(b) Draw a labelled diagram of an induction coil and explain how it works.
(c) How is the effect of eddy currents minimized in the coil?
(d) State two reasons why a capacitor should be included in the primary circuit of the coil.
(e) State three uses of an induction coil.
(a) Faraday's law of electromagnetic induction
Whenever the magnetic flux linked with a circuit changes, an e.m.f. is induced in the circuit. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage:
\[|E|=N\left|\frac{d\Phi}{dt}\right|\]
where \(E\) is the induced e.m.f., \(N\) is the number of turns and \(\Phi\) is the magnetic flux through each turn.
(b) Labelled diagram of an induction coil
How it works: When key \(K\) is closed and the platinum contacts touch, current flows through the primary coil. The soft-iron core becomes magnetised and attracts the soft-iron armature. This separates the contacts and breaks the primary circuit. The core then loses its magnetism, so the spring returns the armature and remakes the contact. Thus, the primary current is made and broken rapidly.
At each break, the magnetic flux in the core collapses rapidly and induces an e.m.f. in the secondary coil. Since the secondary has very many turns, a very large e.m.f. is induced, sufficient to produce a spark across the spark gap.
(c) Eddy currents are minimised by making the core from a bundle of thin insulated soft-iron wires, or by laminating the soft-iron core.
(d) Reasons for connecting a capacitor across the contact points
- It reduces sparking at the platinum contacts and therefore prevents the contacts from burning away.
- It makes the break in the primary current more rapid, causing a faster collapse of flux and hence a larger induced e.m.f. in the secondary.
(e) Uses of an induction coil
- In the ignition system of petrol motor vehicles to produce sparks at spark plugs.
- For operating X-ray tubes.
- For studying electric discharge through gases, such as in discharge tubes.
Answer Details
(a) Faraday's law of electromagnetic induction
Whenever the magnetic flux linked with a circuit changes, an e.m.f. is induced in the circuit. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage:
\[|E|=N\left|\frac{d\Phi}{dt}\right|\]
where \(E\) is the induced e.m.f., \(N\) is the number of turns and \(\Phi\) is the magnetic flux through each turn.
(b) Labelled diagram of an induction coil
How it works: When key \(K\) is closed and the platinum contacts touch, current flows through the primary coil. The soft-iron core becomes magnetised and attracts the soft-iron armature. This separates the contacts and breaks the primary circuit. The core then loses its magnetism, so the spring returns the armature and remakes the contact. Thus, the primary current is made and broken rapidly.
At each break, the magnetic flux in the core collapses rapidly and induces an e.m.f. in the secondary coil. Since the secondary has very many turns, a very large e.m.f. is induced, sufficient to produce a spark across the spark gap.
(c) Eddy currents are minimised by making the core from a bundle of thin insulated soft-iron wires, or by laminating the soft-iron core.
(d) Reasons for connecting a capacitor across the contact points
- It reduces sparking at the platinum contacts and therefore prevents the contacts from burning away.
- It makes the break in the primary current more rapid, causing a faster collapse of flux and hence a larger induced e.m.f. in the secondary.
(e) Uses of an induction coil
- In the ignition system of petrol motor vehicles to produce sparks at spark plugs.
- For operating X-ray tubes.
- For studying electric discharge through gases, such as in discharge tubes.
Question 2 Report
(a) Explain what is meant by acceleration of free fall due to gravity, g.
(b) State two reasons why g varies on the surface of the earth
(c) A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower of height 100 m and it hits the ground at a point Q. If the initial velocity of projection is 100ms\(^{-1}\), calculate the
(i) maximum height of the stone above the ground;
(ii) time it takes to reach this height;
(iii) time of flight
(iv) horizontal distance from the foot of the tower to the point Q. (Neglect air resistance and take g as 10m\(^{-2}\))
(a) The acceleration of free fall due to gravity, g, is the constant acceleration produced in a body that is falling freely under the action of gravity alone (no air resistance). It is the rate of increase of velocity of a freely falling body, having an approximate value of \( 9.8\,\text{m s}^{-2} \) (taken as \( 10\,\text{m s}^{-2} \) here).
(b) Two reasons g varies over the earth:
- The earth is not a perfect sphere; its radius is greater at the equator than at the poles, so g is smaller at the equator (since \( g \propto 1/r^2 \)).
- Variation of altitude (height above sea level) - g decreases with increasing distance from the earth's centre.
(c) Resolve the initial velocity: \( u_x = 100\cos 30^\circ = 86.6\,\text{m s}^{-1} \), \( u_y = 100\sin 30^\circ = 50\,\text{m s}^{-1} \).
(i) Maximum height above the ground. Height risen above the tower top: \( H = \dfrac{u_y^2}{2g} = \dfrac{50^2}{2\times 10} = 125\,\text{m} \). Above the ground: \( 125 + 100 = 225\,\text{m} \).
(ii) Time to reach maximum height. \( t = \dfrac{u_y}{g} = \dfrac{50}{10} = 5\,\text{s} \).
(iii) Time of flight. Taking downward displacement to the ground as \(-100\,\text{m}\): \[ -100 = 50t - \tfrac{1}{2}(10)t^2 \] \[ 5t^2 - 50t - 100 = 0 \Rightarrow t^2 - 10t - 20 = 0 \] \[ t = \dfrac{10 + \sqrt{100 + 80}}{2} = \dfrac{10 + 13.42}{2} = 11.7\,\text{s}. \]
(iv) Horizontal distance to Q. \( x = u_x \times t = 86.6 \times 11.7 = 1.01 \times 10^{3}\,\text{m} \) (about 1014 m).
Answer Details
(a) The acceleration of free fall due to gravity, g, is the constant acceleration produced in a body that is falling freely under the action of gravity alone (no air resistance). It is the rate of increase of velocity of a freely falling body, having an approximate value of \( 9.8\,\text{m s}^{-2} \) (taken as \( 10\,\text{m s}^{-2} \) here).
(b) Two reasons g varies over the earth:
- The earth is not a perfect sphere; its radius is greater at the equator than at the poles, so g is smaller at the equator (since \( g \propto 1/r^2 \)).
- Variation of altitude (height above sea level) - g decreases with increasing distance from the earth's centre.
(c) Resolve the initial velocity: \( u_x = 100\cos 30^\circ = 86.6\,\text{m s}^{-1} \), \( u_y = 100\sin 30^\circ = 50\,\text{m s}^{-1} \).
(i) Maximum height above the ground. Height risen above the tower top: \( H = \dfrac{u_y^2}{2g} = \dfrac{50^2}{2\times 10} = 125\,\text{m} \). Above the ground: \( 125 + 100 = 225\,\text{m} \).
(ii) Time to reach maximum height. \( t = \dfrac{u_y}{g} = \dfrac{50}{10} = 5\,\text{s} \).
(iii) Time of flight. Taking downward displacement to the ground as \(-100\,\text{m}\): \[ -100 = 50t - \tfrac{1}{2}(10)t^2 \] \[ 5t^2 - 50t - 100 = 0 \Rightarrow t^2 - 10t - 20 = 0 \] \[ t = \dfrac{10 + \sqrt{100 + 80}}{2} = \dfrac{10 + 13.42}{2} = 11.7\,\text{s}. \]
(iv) Horizontal distance to Q. \( x = u_x \times t = 86.6 \times 11.7 = 1.01 \times 10^{3}\,\text{m} \) (about 1014 m).
Question 3 Report
(a) What is surface tension? Explain the phenomenon in terms of intermolecular forces
(b) Describe a simple experiment to demonstrate the surface tension of a liquid
(c) State three examples to illustrate the effects of surface tension
(d) Why does water wet a clean glass surface whereas mercury does not?
(e) State two methods by which the surface tension of a liquid may reduced.
(a) Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic skin (membrane) and tends to contract to occupy the smallest possible area.
In terms of intermolecular forces: A molecule deep inside the liquid is attracted equally in all directions by neighbouring molecules, so the net force on it is zero. A molecule at the surface has no liquid molecules above it, so it experiences a net inward (downward) cohesive pull. This inward pull on all surface molecules makes the surface behave as though it were under tension.
(b) Experiment: Tie a loop of thread loosely across a wire ring and dip the ring into soap solution to form a film. The thread lies in an irregular shape. Now break the film inside the loop (with a hot wire). The thread is at once pulled outward into a perfect circle, showing that the remaining film pulls equally on all sides - a demonstration of surface tension.
(c) Effects of surface tension:
- Small insects (e.g. pond skaters) can walk on the surface of water.
- Falling raindrops and small liquid drops are spherical.
- A greased needle can float on the surface of water although it is denser than water.
(d) Water wets clean glass because the force of adhesion between water and glass molecules is greater than the force of cohesion between water molecules, so the water spreads over the glass (concave meniscus). For mercury the cohesive force between mercury molecules is greater than the adhesive force between mercury and glass, so mercury does not wet the glass and instead forms rounded beads (convex meniscus).
(e) Methods of reducing surface tension:
- Raising the temperature of the liquid.
- Adding an impurity such as soap or detergent.
Answer Details
(a) Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic skin (membrane) and tends to contract to occupy the smallest possible area.
In terms of intermolecular forces: A molecule deep inside the liquid is attracted equally in all directions by neighbouring molecules, so the net force on it is zero. A molecule at the surface has no liquid molecules above it, so it experiences a net inward (downward) cohesive pull. This inward pull on all surface molecules makes the surface behave as though it were under tension.
(b) Experiment: Tie a loop of thread loosely across a wire ring and dip the ring into soap solution to form a film. The thread lies in an irregular shape. Now break the film inside the loop (with a hot wire). The thread is at once pulled outward into a perfect circle, showing that the remaining film pulls equally on all sides - a demonstration of surface tension.
(c) Effects of surface tension:
- Small insects (e.g. pond skaters) can walk on the surface of water.
- Falling raindrops and small liquid drops are spherical.
- A greased needle can float on the surface of water although it is denser than water.
(d) Water wets clean glass because the force of adhesion between water and glass molecules is greater than the force of cohesion between water molecules, so the water spreads over the glass (concave meniscus). For mercury the cohesive force between mercury molecules is greater than the adhesive force between mercury and glass, so mercury does not wet the glass and instead forms rounded beads (convex meniscus).
(e) Methods of reducing surface tension:
- Raising the temperature of the liquid.
- Adding an impurity such as soap or detergent.
Question 4 Report
(a) Define the boiling it of a liquid.
(b) Describe with the a d. labelled diagram, an experiment to determire the boiling point of a small quantity of a liquid.
(c) factors that may affect the boiling point of a liquid
(d) Using the kinetic theory of matter, explain why pure water changes to steam at S.T.P. without any change in temperature, although heat is being supplied to the water.
(a) The boiling point of a liquid is the temperature at which its saturated vapour pressure becomes equal to the external atmospheric pressure, so that the liquid changes to vapour throughout its bulk.
(b) Determination of the boiling point of a small quantity of a liquid
The apparatus is arranged as shown below.
Procedure
- Introduce a little mercury into a J-tube. Remove the air trapped in the shorter, closed limb.
- Put a small quantity of the liquid under test into the shorter limb, above the mercury.
- Place the J-tube in a beaker containing water. Heat the water bath gently and stir it continuously so that its temperature is uniform.
- Observe the mercury levels in the two limbs. When the mercury levels are at the same horizontal level, read and record the thermometer temperature.
- Allow the water bath to cool, still stirring it. Again record the temperature at which the mercury levels are equal.
- The mean of the heating and cooling readings is the boiling point of the liquid.
When the mercury levels are equal, the pressure of the liquid vapour in the closed limb is equal to atmospheric pressure in the open limb. Hence, at this temperature, the saturated vapour pressure of the liquid equals atmospheric pressure and the temperature is its boiling point.
(c) Factors affecting the boiling point of a liquid include:
- External pressure: an increase in external pressure raises the boiling point, while a decrease in pressure lowers it.
- Impurities: dissolved non-volatile impurities generally raise the boiling point of a liquid.
(d) At the boiling point, the heat supplied to pure water is absorbed as latent heat of vaporisation. This energy is used to overcome the intermolecular forces of attraction between water molecules and to separate the molecules further apart against atmospheric pressure as steam forms. Since the supplied heat does not increase the average kinetic energy of the molecules, the temperature of the water remains constant while it changes to steam.
Answer Details
(a) The boiling point of a liquid is the temperature at which its saturated vapour pressure becomes equal to the external atmospheric pressure, so that the liquid changes to vapour throughout its bulk.
(b) Determination of the boiling point of a small quantity of a liquid
The apparatus is arranged as shown below.
Procedure
- Introduce a little mercury into a J-tube. Remove the air trapped in the shorter, closed limb.
- Put a small quantity of the liquid under test into the shorter limb, above the mercury.
- Place the J-tube in a beaker containing water. Heat the water bath gently and stir it continuously so that its temperature is uniform.
- Observe the mercury levels in the two limbs. When the mercury levels are at the same horizontal level, read and record the thermometer temperature.
- Allow the water bath to cool, still stirring it. Again record the temperature at which the mercury levels are equal.
- The mean of the heating and cooling readings is the boiling point of the liquid.
When the mercury levels are equal, the pressure of the liquid vapour in the closed limb is equal to atmospheric pressure in the open limb. Hence, at this temperature, the saturated vapour pressure of the liquid equals atmospheric pressure and the temperature is its boiling point.
(c) Factors affecting the boiling point of a liquid include:
- External pressure: an increase in external pressure raises the boiling point, while a decrease in pressure lowers it.
- Impurities: dissolved non-volatile impurities generally raise the boiling point of a liquid.
(d) At the boiling point, the heat supplied to pure water is absorbed as latent heat of vaporisation. This energy is used to overcome the intermolecular forces of attraction between water molecules and to separate the molecules further apart against atmospheric pressure as steam forms. Since the supplied heat does not increase the average kinetic energy of the molecules, the temperature of the water remains constant while it changes to steam.
