WAEC SSCE - Physics - 2017 (Practical)
Swali 1 Ripoti
You are provided with a beaker, a thermometer, a stirrer Stopwatch/clock, measuring cylinder, table salt, water, and other necessary materials.
i. You Measure \(200\text{cm}^{3}\) of water into the beaker.
ii. Heat the water until it boils steadily for about 2 minutes.
iii. Read and record the boiling point \(b_{0}\).
iv. Add table salt of mass \(M = 10.0\text{ g}\) to the boiling water and stir continuously until another boiling point \(b_{1}\) is attained.
v. Read and record \(b_{i}\).
vi. Evaluate \(\theta_{i} = (b_{i} - b_{0})\)
vii. Using the same mixture, repeat the procedure four more times by adding 10.0 g of salt each time to give the cumulative mass \(M_{i}\) of salt as 20 g, 30g, 40g, and 50g.
viii. In each case allow the mixture to boil steadily for at least 2 minutes then read and record the boiling point b.
ix. Tabulate your readings.
x. Plot a graph with \(M_{i}\) on the vertical axis and \(\theta_{1}\) on the horizontal axis.
xi. Determine the slope, s, of the graph.
xii. State two precautions taken to ensure accurate results.
(b) i. Define the boiling point of a liquid.
ii. What effects do impurities have on the boiling point of a liquid?
Effect of dissolved salt on the boiling point of water
The boiling point of the water before salt was added was:
\[b_0=102.0^{\circ}\mathrm{C}\]
| S/N | Cumulative mass, \(M_i\) / g | Boiling point, \(b_i\) / \(^\circ\mathrm{C}\) | \(\theta_i=b_i-b_0\) / \(^\circ\mathrm{C}\) |
|---|---|---|---|
| 1 | 10.0 | 104.0 | \(104.0-102.0=2.0\) |
| 2 | 20.0 | 106.0 | \(106.0-102.0=4.0\) |
| 3 | 30.0 | 108.0 | \(108.0-102.0=6.0\) |
| 4 | 40.0 | 110.0 | \(110.0-102.0=8.0\) |
| 5 | 50.0 | 112.0 | \(112.0-102.0=10.0\) |
Graph of cumulative mass of salt against rise in boiling point
Using two widely separated points on the line, \((\theta_1,M_1)=(2.0,10.0)\) and \((\theta_2,M_2)=(10.0,50.0)\):
\[s=\frac{\Delta M}{\Delta\theta}=\frac{50.0-10.0}{10.0-2.0}=\frac{40.0}{8.0}=5.0\ \mathrm{g\,^{\circ}C^{-1}}.\]
Slope, \(s=5.0\ \mathrm{g\,^{\circ}C^{-1}}\).
Precautions
- Stir the solution gently and continuously so that its temperature is uniform throughout.
- Take each thermometer reading only after steady boiling has continued for at least 2 minutes, reading the scale at eye level to avoid parallax error.
(b)(i) Boiling point
The boiling point of a liquid is the temperature at which its saturated vapour pressure equals the external atmospheric pressure, so that vaporisation occurs throughout the liquid.
(b)(ii) Effect of impurities
Dissolved impurities raise the boiling point of a liquid. Thus, as more salt is dissolved in the water, a higher temperature is required for it to boil.
Maelezo ya Majibu
Effect of dissolved salt on the boiling point of water
The boiling point of the water before salt was added was:
\[b_0=102.0^{\circ}\mathrm{C}\]
| S/N | Cumulative mass, \(M_i\) / g | Boiling point, \(b_i\) / \(^\circ\mathrm{C}\) | \(\theta_i=b_i-b_0\) / \(^\circ\mathrm{C}\) |
|---|---|---|---|
| 1 | 10.0 | 104.0 | \(104.0-102.0=2.0\) |
| 2 | 20.0 | 106.0 | \(106.0-102.0=4.0\) |
| 3 | 30.0 | 108.0 | \(108.0-102.0=6.0\) |
| 4 | 40.0 | 110.0 | \(110.0-102.0=8.0\) |
| 5 | 50.0 | 112.0 | \(112.0-102.0=10.0\) |
Graph of cumulative mass of salt against rise in boiling point
Using two widely separated points on the line, \((\theta_1,M_1)=(2.0,10.0)\) and \((\theta_2,M_2)=(10.0,50.0)\):
\[s=\frac{\Delta M}{\Delta\theta}=\frac{50.0-10.0}{10.0-2.0}=\frac{40.0}{8.0}=5.0\ \mathrm{g\,^{\circ}C^{-1}}.\]
Slope, \(s=5.0\ \mathrm{g\,^{\circ}C^{-1}}\).
Precautions
- Stir the solution gently and continuously so that its temperature is uniform throughout.
- Take each thermometer reading only after steady boiling has continued for at least 2 minutes, reading the scale at eye level to avoid parallax error.
(b)(i) Boiling point
The boiling point of a liquid is the temperature at which its saturated vapour pressure equals the external atmospheric pressure, so that vaporisation occurs throughout the liquid.
(b)(ii) Effect of impurities
Dissolved impurities raise the boiling point of a liquid. Thus, as more salt is dissolved in the water, a higher temperature is required for it to boil.
Swali 2 Ripoti
You are provided with cells, a potentiometer, an ammeter, a voltmeter, a bulb, a key, a jockey, and other necessary materials.
i. Measure and record the e.m.f E of the battery.
ii. Set up a circuit as shown in the diagram above.
iii. Close the key K and use the jockey to make a firm.
iii. Contact at J on the potentiometer wire such that PJ = x= 10cm.
iv. Take and record the voltmeter reading V and the () Corresponding ammeter reading.
v. Evalute log V and log I.
vi. Repeat the procedure for five other values of x = 20 cm, 30 cm, 40 cm, 50 cm, and 60 cm.
vii. Tabulate your readings.
viii. Plot a graph with log I on the vertical axis and log V on the horizontal axis.
ix. Determine the slope s, of the graph.
x. Determine the intercept, c, on the vertical axis.
xi. State two precautions taken to ensure accurate results.
(b)i. How is the brightness of the bulb affected as x increases?
ii. List two electrical devices whose actions do not obey Ohm's law.
Test of Practical Knowledge - Potentiometer, Bulb, Ammeter and Voltmeter
(i) E.m.f of the battery: On open circuit the voltmeter connected directly across the battery terminals reads \( E = 3.0\,\text{V} \).
(ii) Circuit: The circuit is connected as shown below. The battery, key K and the potentiometer wire PQ form the driver loop. The section PJ of length \( x \) is tapped by the jockey J and supplies the bulb through the ammeter A, with the voltmeter V connected across the bulb.
(iii)-(vii) Readings: The key K is closed and the jockey pressed firmly at J so that \( PJ = x \). For each length \( x \) the voltmeter reading \( V \) and the corresponding ammeter reading \( I \) are taken, and \( \log V \) and \( \log I \) are evaluated. The results are tabulated below.
| S/N | x / cm | V / V | I / A | log V | log I |
|---|---|---|---|---|---|
| 1 | 10 | 0.30 | 0.15 | -0.523 | -0.824 |
| 2 | 20 | 0.45 | 0.19 | -0.347 | -0.721 |
| 3 | 30 | 0.60 | 0.23 | -0.222 | -0.638 |
| 4 | 40 | 0.75 | 0.26 | -0.125 | -0.585 |
| 5 | 50 | 0.90 | 0.29 | -0.046 | -0.538 |
| 6 | 60 | 1.05 | 0.32 | 0.021 | -0.495 |
(viii) Graph: A graph of \( \log I \) (vertical axis) against \( \log V \) (horizontal axis) is plotted below. The points lie on a straight line, confirming the relation \( I = kV^{s} \), i.e. \( \log I = s\,\log V + c \).
(ix) Slope of the graph: Two well-separated points on the line of best fit are taken, \( (\log V_1, \log I_1) = (-0.523,\,-0.824) \) and \( (\log V_2, \log I_2) = (0.021,\,-0.495) \):
\[ s = \frac{\Delta(\log I)}{\Delta(\log V)} = \frac{-0.495 - (-0.824)}{0.021 - (-0.523)} = \frac{0.329}{0.544} \]\[ \boxed{s \approx 0.60} \](x) Intercept on the vertical axis: Extending the line of best fit to \( \log V = 0 \), it cuts the vertical axis at
\[ c = \log I - s\,\log V = -0.824 - (0.60)(-0.523) = -0.824 + 0.314 \]\[ \boxed{c \approx -0.51} \](xi) Two precautions:
- Parallax error was avoided by viewing the ammeter and voltmeter pointers directly (line of sight perpendicular to the scale).
- The key K was opened whenever readings were not being taken, so that the cells did not run down and the wire did not heat up. (The jockey was also pressed firmly and not dragged along the potentiometer wire.)
(b)(i) Effect on brightness: As \( x \) increases, the potential difference tapped from the potentiometer wire increases, so both the voltage \( V \) across the bulb and the current \( I \) through it increase. The bulb therefore becomes brighter as \( x \) increases.
(b)(ii) Two electrical devices whose action does not obey Ohm's law:
- A semiconductor diode.
- A thermistor. (A filament lamp and a transistor are also acceptable - the slope \( s \approx 0.60 \ne 1 \) obtained above is itself evidence that the filament bulb used here is non-ohmic.)
Maelezo ya Majibu
Test of Practical Knowledge - Potentiometer, Bulb, Ammeter and Voltmeter
(i) E.m.f of the battery: On open circuit the voltmeter connected directly across the battery terminals reads \( E = 3.0\,\text{V} \).
(ii) Circuit: The circuit is connected as shown below. The battery, key K and the potentiometer wire PQ form the driver loop. The section PJ of length \( x \) is tapped by the jockey J and supplies the bulb through the ammeter A, with the voltmeter V connected across the bulb.
(iii)-(vii) Readings: The key K is closed and the jockey pressed firmly at J so that \( PJ = x \). For each length \( x \) the voltmeter reading \( V \) and the corresponding ammeter reading \( I \) are taken, and \( \log V \) and \( \log I \) are evaluated. The results are tabulated below.
| S/N | x / cm | V / V | I / A | log V | log I |
|---|---|---|---|---|---|
| 1 | 10 | 0.30 | 0.15 | -0.523 | -0.824 |
| 2 | 20 | 0.45 | 0.19 | -0.347 | -0.721 |
| 3 | 30 | 0.60 | 0.23 | -0.222 | -0.638 |
| 4 | 40 | 0.75 | 0.26 | -0.125 | -0.585 |
| 5 | 50 | 0.90 | 0.29 | -0.046 | -0.538 |
| 6 | 60 | 1.05 | 0.32 | 0.021 | -0.495 |
(viii) Graph: A graph of \( \log I \) (vertical axis) against \( \log V \) (horizontal axis) is plotted below. The points lie on a straight line, confirming the relation \( I = kV^{s} \), i.e. \( \log I = s\,\log V + c \).
(ix) Slope of the graph: Two well-separated points on the line of best fit are taken, \( (\log V_1, \log I_1) = (-0.523,\,-0.824) \) and \( (\log V_2, \log I_2) = (0.021,\,-0.495) \):
\[ s = \frac{\Delta(\log I)}{\Delta(\log V)} = \frac{-0.495 - (-0.824)}{0.021 - (-0.523)} = \frac{0.329}{0.544} \]\[ \boxed{s \approx 0.60} \](x) Intercept on the vertical axis: Extending the line of best fit to \( \log V = 0 \), it cuts the vertical axis at
\[ c = \log I - s\,\log V = -0.824 - (0.60)(-0.523) = -0.824 + 0.314 \]\[ \boxed{c \approx -0.51} \](xi) Two precautions:
- Parallax error was avoided by viewing the ammeter and voltmeter pointers directly (line of sight perpendicular to the scale).
- The key K was opened whenever readings were not being taken, so that the cells did not run down and the wire did not heat up. (The jockey was also pressed firmly and not dragged along the potentiometer wire.)
(b)(i) Effect on brightness: As \( x \) increases, the potential difference tapped from the potentiometer wire increases, so both the voltage \( V \) across the bulb and the current \( I \) through it increase. The bulb therefore becomes brighter as \( x \) increases.
(b)(ii) Two electrical devices whose action does not obey Ohm's law:
- A semiconductor diode.
- A thermistor. (A filament lamp and a transistor are also acceptable - the slope \( s \approx 0.60 \ne 1 \) obtained above is itself evidence that the filament bulb used here is non-ohmic.)
Swali 3 Ripoti
You are provided with two retort stands, two-metre rules, pieces of thread and other necessary apparatus.
i. Set up the apparatus as illustrated above ensuring the strings are permanently 10cm from either end of the rule.
ii. Measure and record the length L = 80 cm of the two strings.
iii. Hold both ends of the rule and displace the rule slightly, then release so that it oscillates about a vertical axis through its centre.
iv. Determine and record the time t for 10 complete oscillations.
v. Determine the period T of oscillations.
vi. Evaluate log T and L.
vii. Repeat the procedure for four other values of L= 70 cm, 60 cm, 50 cm, and 40 cm
viii. Tabulate your readings.
ix. Plot a graph with log T on the vertical axis and log L on the horizontal axis.
x. Determine the slope, s, and the intercept, c on the vertical axis.
xi. State two precautions taken to ensure accurate results.
(b)i. Define simple harmonic motion.
ii. Determine the value of L corresponding to t= 12 s from the graph in 1.
Bifilar pendulum: torsional oscillation of a suspended metre rule
The two threads of equal length \(L\) are fixed to the rigid horizontal support, each 10 cm from the ends of the metre rule, so that the rule hangs horizontally and can oscillate about the vertical axis through its centre.
For each length the period is obtained from the timing of ten complete oscillations:
\[ T = \frac{t}{10} \]and \(\log T\) and \(\log L\) are then evaluated for each reading.
(viii) Table of readings
| S/N | L /cm | t /s (10 osc.) | T = t/10 /s | log T | log L |
|---|---|---|---|---|---|
| 1 | 80.0 | 17.9 | 1.79 | 0.253 | 1.903 |
| 2 | 70.0 | 16.7 | 1.67 | 0.223 | 1.845 |
| 3 | 60.0 | 15.5 | 1.55 | 0.190 | 1.778 |
| 4 | 50.0 | 14.1 | 1.41 | 0.149 | 1.699 |
| 5 | 40.0 | 12.6 | 1.26 | 0.100 | 1.602 |
(ix) Graph of log T against log L
The points lie on a straight line, confirming that \( \log T = s\,\log L + c \).
(x) Slope and intercept
Taking two widely separated points on the line of best fit, \((1.602,\;0.100)\) and \((1.903,\;0.253)\):
\[ s = \frac{\Delta(\log T)}{\Delta(\log L)} = \frac{0.253 - 0.100}{1.903 - 1.602} = \frac{0.153}{0.301} = 0.51 \]Extending the line back to \(\log L = 0\) (or using \( c = \log T - s\log L = 0.253 - 0.51\times1.903 \)) gives the vertical intercept:
\[ c = -0.71 \]Hence \( \log T = 0.51\,\log L - 0.71 \), which corresponds to \( T \propto L^{1/2} \), the expected law for the bifilar pendulum.
(xi) Two precautions
- I avoided parallax error when reading the metre rule and when starting and stopping the stopwatch, by viewing each scale directly from the front.
- I ensured that the support was rigid and that the rule oscillated smoothly about a vertical axis through its centre with only a small angular displacement, counting the oscillations from a fixed reference mark.
(b)(i) Simple harmonic motion
Simple harmonic motion is the motion of a body whose acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point:
\[ a = -\omega^{2}x \](b)(ii) Value of L corresponding to t = 12 s
For \( t = 12\,\text{s} \):
\[ T = \frac{t}{10} = \frac{12}{10} = 1.2\,\text{s}, \qquad \log T = \log 1.2 = 0.079 \]Reading from \( \log T = 0.079 \) on the vertical axis across to the line of best fit and down to the horizontal axis (or solving \( 0.079 = 0.51\log L - 0.71 \)):
\[ \log L = \frac{0.079 + 0.71}{0.51} = \frac{0.789}{0.51} = 1.56 \]\[ L = 10^{1.56} = 36\,\text{cm} \]Therefore the length of the threads corresponding to \( t = 12\,\text{s} \) is \( L \approx 36\,\text{cm} \).
Maelezo ya Majibu
Bifilar pendulum: torsional oscillation of a suspended metre rule
The two threads of equal length \(L\) are fixed to the rigid horizontal support, each 10 cm from the ends of the metre rule, so that the rule hangs horizontally and can oscillate about the vertical axis through its centre.
For each length the period is obtained from the timing of ten complete oscillations:
\[ T = \frac{t}{10} \]and \(\log T\) and \(\log L\) are then evaluated for each reading.
(viii) Table of readings
| S/N | L /cm | t /s (10 osc.) | T = t/10 /s | log T | log L |
|---|---|---|---|---|---|
| 1 | 80.0 | 17.9 | 1.79 | 0.253 | 1.903 |
| 2 | 70.0 | 16.7 | 1.67 | 0.223 | 1.845 |
| 3 | 60.0 | 15.5 | 1.55 | 0.190 | 1.778 |
| 4 | 50.0 | 14.1 | 1.41 | 0.149 | 1.699 |
| 5 | 40.0 | 12.6 | 1.26 | 0.100 | 1.602 |
(ix) Graph of log T against log L
The points lie on a straight line, confirming that \( \log T = s\,\log L + c \).
(x) Slope and intercept
Taking two widely separated points on the line of best fit, \((1.602,\;0.100)\) and \((1.903,\;0.253)\):
\[ s = \frac{\Delta(\log T)}{\Delta(\log L)} = \frac{0.253 - 0.100}{1.903 - 1.602} = \frac{0.153}{0.301} = 0.51 \]Extending the line back to \(\log L = 0\) (or using \( c = \log T - s\log L = 0.253 - 0.51\times1.903 \)) gives the vertical intercept:
\[ c = -0.71 \]Hence \( \log T = 0.51\,\log L - 0.71 \), which corresponds to \( T \propto L^{1/2} \), the expected law for the bifilar pendulum.
(xi) Two precautions
- I avoided parallax error when reading the metre rule and when starting and stopping the stopwatch, by viewing each scale directly from the front.
- I ensured that the support was rigid and that the rule oscillated smoothly about a vertical axis through its centre with only a small angular displacement, counting the oscillations from a fixed reference mark.
(b)(i) Simple harmonic motion
Simple harmonic motion is the motion of a body whose acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point:
\[ a = -\omega^{2}x \](b)(ii) Value of L corresponding to t = 12 s
For \( t = 12\,\text{s} \):
\[ T = \frac{t}{10} = \frac{12}{10} = 1.2\,\text{s}, \qquad \log T = \log 1.2 = 0.079 \]Reading from \( \log T = 0.079 \) on the vertical axis across to the line of best fit and down to the horizontal axis (or solving \( 0.079 = 0.51\log L - 0.71 \)):
\[ \log L = \frac{0.079 + 0.71}{0.51} = \frac{0.789}{0.51} = 1.56 \]\[ L = 10^{1.56} = 36\,\text{cm} \]Therefore the length of the threads corresponding to \( t = 12\,\text{s} \) is \( L \approx 36\,\text{cm} \).
