WAEC SSCE - Physics - 1990 (Essay)
Question 1 Report
(a) Define the boiling point of a liquid.
(b) Describe an experiment to determine the boiling point of small quantity of a liquid.
(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred to a vessel of negligible thermal capacity containing 250 g of water at 25°C. If the final steady temperature of the mixture is 100°C, calculate the mass of the water that will boil away.
[Specific heat capacity of copper = \(4.0 \times 10^{2} Jkg^{-1} K^{-1}\); Specific heat capacity of water = \(4.2 \times 10^{3} Jkg^{-1} K^{-1}\); Specific latent heat of vaporization of steam = \(2.26 \times 10^{6} Jkg^{-1}\)
(d) State four other effects of heat on a substance other than expansion.
(a) Boiling point
The boiling point of a liquid is the constant temperature at which the liquid changes to vapour throughout its bulk, when its saturated vapour pressure equals the external (atmospheric) pressure.
(b) Experiment to determine the boiling point of a small quantity of liquid
- Put the small quantity of liquid in a narrow ignition tube.
- Attach a thermometer to the tube with a rubber band so the bulb is level with the liquid.
- Immerse the tube and thermometer in a beaker of water (a heating bath) clamped on a stand.
- Heat the bath steadily while stirring, and observe the thermometer.
- The temperature at which the liquid boils steadily (bubbles rise continuously and temperature stays constant) is read and recorded as the boiling point.
(c) Mass of water that boils away
Heat lost by copper cooling from 950°C to 100°C:
\[ Q_{Cu} = m_{Cu} c_{Cu}\,\Delta\theta = 0.300 \times 4.0\times10^{2} \times (950-100) = 0.300\times400\times850 = 102000\,\text{J} \]Heat used to raise the water from 25°C to 100°C:
\[ Q_{w} = 0.250 \times 4.2\times10^{3} \times (100-25) = 0.250\times4200\times75 = 78750\,\text{J} \]Heat left for vaporization:
\[ Q_{v} = 102000 - 78750 = 23250\,\text{J} \]Mass boiled away:
\[ m = \frac{Q_{v}}{L} = \frac{23250}{2.26\times10^{6}} = 1.03\times10^{-2}\,\text{kg} \approx 10.3\,\text{g} \](d) Four other effects of heat (besides expansion)
- Change of state (melting, boiling, evaporation).
- Rise in temperature of the substance.
- Chemical change (e.g. decomposition or burning).
- Change in electrical resistance (or thermionic/thermoelectric effects).
Answer Details
(a) Boiling point
The boiling point of a liquid is the constant temperature at which the liquid changes to vapour throughout its bulk, when its saturated vapour pressure equals the external (atmospheric) pressure.
(b) Experiment to determine the boiling point of a small quantity of liquid
- Put the small quantity of liquid in a narrow ignition tube.
- Attach a thermometer to the tube with a rubber band so the bulb is level with the liquid.
- Immerse the tube and thermometer in a beaker of water (a heating bath) clamped on a stand.
- Heat the bath steadily while stirring, and observe the thermometer.
- The temperature at which the liquid boils steadily (bubbles rise continuously and temperature stays constant) is read and recorded as the boiling point.
(c) Mass of water that boils away
Heat lost by copper cooling from 950°C to 100°C:
\[ Q_{Cu} = m_{Cu} c_{Cu}\,\Delta\theta = 0.300 \times 4.0\times10^{2} \times (950-100) = 0.300\times400\times850 = 102000\,\text{J} \]Heat used to raise the water from 25°C to 100°C:
\[ Q_{w} = 0.250 \times 4.2\times10^{3} \times (100-25) = 0.250\times4200\times75 = 78750\,\text{J} \]Heat left for vaporization:
\[ Q_{v} = 102000 - 78750 = 23250\,\text{J} \]Mass boiled away:
\[ m = \frac{Q_{v}}{L} = \frac{23250}{2.26\times10^{6}} = 1.03\times10^{-2}\,\text{kg} \approx 10.3\,\text{g} \](d) Four other effects of heat (besides expansion)
- Change of state (melting, boiling, evaporation).
- Rise in temperature of the substance.
- Chemical change (e.g. decomposition or burning).
- Change in electrical resistance (or thermionic/thermoelectric effects).
Question 2 Report
(a) Explain the terms nuclear fission and nuclear fusion.
(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages, fusion is not normally used for the generation of power.
(c) \(^{238} _{92} U\) is a long half- life alpha emitter and decays to thorium Th which, in turn decays by beta emission with a small decay constant to an isotope of protactinium Pa. The protactinium decay scheme of \(^{238} _{92} U \) as stated above.
(d) State three uses of radioisotopes.
Question 3 Report
(a) Define the capacitance of a capacitor.
(b) State three factors on which the capacitance of a parallel plate capacitor depends.
(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.
(d) Two capacitors of capacitance 4\(\mu F\) and 6\(\mu F\) are connected in series to a 100V d.c supply. Draw the circuit diagram and calculate the (i) charge on either plate of each capacitor (ii) p.d. across each capacitor; (iii) energy of the combined capacitors.
(a) Capacitance of a capacitor
The capacitance of a capacitor is the ratio of the magnitude of the charge \(Q\) on either plate to the potential difference \(V\) between the plates:
\[ C = \frac{Q}{V} \]where \(Q\) is the charge (coulomb) and \(V\) is the p.d. (volt). Its SI unit is the farad (F).
(b) Factors affecting the capacitance of a parallel-plate capacitor
- The common (overlapping) area \(A\) of the plates, with \(C \propto A\).
- The distance of separation \(d\) between the plates, with \(C \propto \dfrac{1}{d}\).
- The permittivity \(\varepsilon\) (nature) of the dielectric medium between the plates, with \(C \propto \varepsilon\).
These combine as \(C = \dfrac{\varepsilon A}{d}\).
(c) Energy \(W\) stored in a charged capacitor
When the capacitor already holds a charge \(q\), the p.d. across it is \(v = \dfrac{q}{C}\). To move a further small charge \(dq\) onto the plates the work done is
\[ dW = v\,dq = \frac{q}{C}\,dq \]The total work to charge the capacitor from \(0\) to the final charge \(Q\) is therefore
\[ W = \int_{0}^{Q} \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{Q^{2}}{2} = \frac{Q^{2}}{2C} \]Using \(Q = CV\), the same result may be written as
\[ W = \frac{Q^{2}}{2C} = \frac{1}{2}QV = \frac{1}{2}CV^{2} \](d) Two capacitors 4\(\mu\text{F}\) and 6\(\mu\text{F}\) in series across 100 V
Circuit diagram: the two capacitors are joined end to end (in series) in a single loop with the 100 V d.c. supply.
Effective (combined) capacitance of capacitors in series:
\[ \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12} \] \[ \Rightarrow C = \frac{12}{5} = 2.4\,\mu\text{F} \](i) Charge on either plate of each capacitor. In a series circuit the charge is the same on every capacitor and equals the charge supplied to the combination:
\[ Q = CV = 2.4\times10^{-6} \times 100 = 2.4\times10^{-4}\,\text{C} = 240\,\mu\text{C} \]Hence each capacitor carries \(Q = 240\,\mu\text{C}\) on either plate.
(ii) P.d. across each capacitor.
\[ V_{4} = \frac{Q}{C_{4}} = \frac{2.4\times10^{-4}}{4\times10^{-6}} = 60\,\text{V} \] \[ V_{6} = \frac{Q}{C_{6}} = \frac{2.4\times10^{-4}}{6\times10^{-6}} = 40\,\text{V} \]Check: \(V_{4} + V_{6} = 60 + 40 = 100\,\text{V}\), which equals the supply voltage.
(iii) Energy of the combined capacitors.
\[ W = \frac{Q^{2}}{2C} = \frac{(2.4\times10^{-4})^{2}}{2 \times 2.4\times10^{-6}} = \frac{5.76\times10^{-8}}{4.8\times10^{-6}} = 1.2\times10^{-2}\,\text{J} \]Equivalently \(W = \tfrac{1}{2}QV = \tfrac{1}{2}(2.4\times10^{-4})(100) = 0.012\,\text{J}\).
Therefore the energy stored in the combination is \(W = 1.2\times10^{-2}\,\text{J}\).
Answer Details
(a) Capacitance of a capacitor
The capacitance of a capacitor is the ratio of the magnitude of the charge \(Q\) on either plate to the potential difference \(V\) between the plates:
\[ C = \frac{Q}{V} \]where \(Q\) is the charge (coulomb) and \(V\) is the p.d. (volt). Its SI unit is the farad (F).
(b) Factors affecting the capacitance of a parallel-plate capacitor
- The common (overlapping) area \(A\) of the plates, with \(C \propto A\).
- The distance of separation \(d\) between the plates, with \(C \propto \dfrac{1}{d}\).
- The permittivity \(\varepsilon\) (nature) of the dielectric medium between the plates, with \(C \propto \varepsilon\).
These combine as \(C = \dfrac{\varepsilon A}{d}\).
(c) Energy \(W\) stored in a charged capacitor
When the capacitor already holds a charge \(q\), the p.d. across it is \(v = \dfrac{q}{C}\). To move a further small charge \(dq\) onto the plates the work done is
\[ dW = v\,dq = \frac{q}{C}\,dq \]The total work to charge the capacitor from \(0\) to the final charge \(Q\) is therefore
\[ W = \int_{0}^{Q} \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{Q^{2}}{2} = \frac{Q^{2}}{2C} \]Using \(Q = CV\), the same result may be written as
\[ W = \frac{Q^{2}}{2C} = \frac{1}{2}QV = \frac{1}{2}CV^{2} \](d) Two capacitors 4\(\mu\text{F}\) and 6\(\mu\text{F}\) in series across 100 V
Circuit diagram: the two capacitors are joined end to end (in series) in a single loop with the 100 V d.c. supply.
Effective (combined) capacitance of capacitors in series:
\[ \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12} \] \[ \Rightarrow C = \frac{12}{5} = 2.4\,\mu\text{F} \](i) Charge on either plate of each capacitor. In a series circuit the charge is the same on every capacitor and equals the charge supplied to the combination:
\[ Q = CV = 2.4\times10^{-6} \times 100 = 2.4\times10^{-4}\,\text{C} = 240\,\mu\text{C} \]Hence each capacitor carries \(Q = 240\,\mu\text{C}\) on either plate.
(ii) P.d. across each capacitor.
\[ V_{4} = \frac{Q}{C_{4}} = \frac{2.4\times10^{-4}}{4\times10^{-6}} = 60\,\text{V} \] \[ V_{6} = \frac{Q}{C_{6}} = \frac{2.4\times10^{-4}}{6\times10^{-6}} = 40\,\text{V} \]Check: \(V_{4} + V_{6} = 60 + 40 = 100\,\text{V}\), which equals the supply voltage.
(iii) Energy of the combined capacitors.
\[ W = \frac{Q^{2}}{2C} = \frac{(2.4\times10^{-4})^{2}}{2 \times 2.4\times10^{-6}} = \frac{5.76\times10^{-8}}{4.8\times10^{-6}} = 1.2\times10^{-2}\,\text{J} \]Equivalently \(W = \tfrac{1}{2}QV = \tfrac{1}{2}(2.4\times10^{-4})(100) = 0.012\,\text{J}\).
Therefore the energy stored in the combination is \(W = 1.2\times10^{-2}\,\text{J}\).
Question 4 Report
(a) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph (i) acceleration; (ii) retardation; (iii) total distance covered.
(b) Show that the displacement of a body moving with uniform acceleration a is given by \(s = ut + \frac{1}{2} at^{2}\) where u is the velocity of the body at time t = 0.
(c) A particle moving in a straight line with uniform deceleration has a velocity of 40ms\(^{-1}\) at a point P, 20ms\(^{-1}\) at a point Q and comes to rest at a point R where QR = 50m. Calculate the: (i) distance PQ; (ii) time taken to cover PQ; (iii) time taken to cover PR.
(a) Obtaining quantities from a velocity-time graph
On a velocity-time graph the velocity is plotted on the vertical axis and time on the horizontal axis. A typical graph rises from A to B, stays level from B to C, then falls from C to D as shown below.
(i) Acceleration is obtained from the gradient of the rising portion AB. If the velocity increases from \(v_1\) to \(v_2\) as time changes from \(t_1\) to \(t_2\) along AB, then \[ a = \text{gradient of AB} = \frac{v_2 - v_1}{t_2 - t_1} \quad (\text{a positive slope}). \]
(ii) Retardation (deceleration) is obtained from the gradient of the falling portion CD. Here the velocity decreases with time, so the slope is negative: \[ \text{retardation} = -\,(\text{gradient of CD}) = \frac{v_C - v_D}{t_D - t_C}. \]
(iii) Total distance covered is the total area enclosed between the graph line and the time axis (the shaded region O-A-B-C-D). It is found by splitting the region into triangles and a rectangle (or one trapezium) and adding the areas.
(b) Deriving \(s = ut + \tfrac{1}{2}at^{2}\)
The displacement equals the area under the velocity-time graph, which for uniform acceleration is a trapezium of parallel sides \(u\) (initial velocity) and \(v\) (final velocity) and width \(t\): \[ s = \text{average velocity} \times t = \left(\frac{u + v}{2}\right)t. \tag{1} \] For uniform acceleration \(a\), the final velocity is \[ v = u + at. \tag{2} \] Substituting (2) into (1): \[ s = \left(\frac{u + (u + at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t \] \[ \boxed{\,s = ut + \tfrac{1}{2}at^{2}\,}. \]
(c) Uniform deceleration problem
Along the straight line the velocities are: at P, \(u = 40\,\text{ms}^{-1}\); at Q, \(20\,\text{ms}^{-1}\); at R the particle is at rest (\(0\,\text{ms}^{-1}\)); and \(QR = 50\,\text{m}\).
Find the deceleration using the motion from Q to R, with \(v^{2} = u^{2} - 2as\): \[ 0 = 20^{2} - 2a(50) \;\Rightarrow\; 400 = 100a \;\Rightarrow\; a = 4\,\text{ms}^{-2}\ (\text{deceleration}). \]
(i) Distance PQ (from P to Q): \[ 20^{2} = 40^{2} - 2(4)(PQ) \;\Rightarrow\; 400 = 1600 - 8\,PQ \;\Rightarrow\; PQ = \frac{1200}{8} = 150\,\text{m}. \]
(ii) Time taken to cover PQ, using \(v = u - at\) from P to Q: \[ 20 = 40 - 4t \;\Rightarrow\; 4t = 20 \;\Rightarrow\; t = 5\,\text{s}. \]
(iii) Time taken to cover PR (from P to rest at R): \[ 0 = 40 - 4t \;\Rightarrow\; 4t = 40 \;\Rightarrow\; t = 10\,\text{s}. \]
Answer Details
(a) Obtaining quantities from a velocity-time graph
On a velocity-time graph the velocity is plotted on the vertical axis and time on the horizontal axis. A typical graph rises from A to B, stays level from B to C, then falls from C to D as shown below.
(i) Acceleration is obtained from the gradient of the rising portion AB. If the velocity increases from \(v_1\) to \(v_2\) as time changes from \(t_1\) to \(t_2\) along AB, then \[ a = \text{gradient of AB} = \frac{v_2 - v_1}{t_2 - t_1} \quad (\text{a positive slope}). \]
(ii) Retardation (deceleration) is obtained from the gradient of the falling portion CD. Here the velocity decreases with time, so the slope is negative: \[ \text{retardation} = -\,(\text{gradient of CD}) = \frac{v_C - v_D}{t_D - t_C}. \]
(iii) Total distance covered is the total area enclosed between the graph line and the time axis (the shaded region O-A-B-C-D). It is found by splitting the region into triangles and a rectangle (or one trapezium) and adding the areas.
(b) Deriving \(s = ut + \tfrac{1}{2}at^{2}\)
The displacement equals the area under the velocity-time graph, which for uniform acceleration is a trapezium of parallel sides \(u\) (initial velocity) and \(v\) (final velocity) and width \(t\): \[ s = \text{average velocity} \times t = \left(\frac{u + v}{2}\right)t. \tag{1} \] For uniform acceleration \(a\), the final velocity is \[ v = u + at. \tag{2} \] Substituting (2) into (1): \[ s = \left(\frac{u + (u + at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t \] \[ \boxed{\,s = ut + \tfrac{1}{2}at^{2}\,}. \]
(c) Uniform deceleration problem
Along the straight line the velocities are: at P, \(u = 40\,\text{ms}^{-1}\); at Q, \(20\,\text{ms}^{-1}\); at R the particle is at rest (\(0\,\text{ms}^{-1}\)); and \(QR = 50\,\text{m}\).
Find the deceleration using the motion from Q to R, with \(v^{2} = u^{2} - 2as\): \[ 0 = 20^{2} - 2a(50) \;\Rightarrow\; 400 = 100a \;\Rightarrow\; a = 4\,\text{ms}^{-2}\ (\text{deceleration}). \]
(i) Distance PQ (from P to Q): \[ 20^{2} = 40^{2} - 2(4)(PQ) \;\Rightarrow\; 400 = 1600 - 8\,PQ \;\Rightarrow\; PQ = \frac{1200}{8} = 150\,\text{m}. \]
(ii) Time taken to cover PQ, using \(v = u - at\) from P to Q: \[ 20 = 40 - 4t \;\Rightarrow\; 4t = 20 \;\Rightarrow\; t = 5\,\text{s}. \]
(iii) Time taken to cover PR (from P to rest at R): \[ 0 = 40 - 4t \;\Rightarrow\; 4t = 40 \;\Rightarrow\; t = 10\,\text{s}. \]
