WAEC SSCE - Physics - 2018 (Practical)
Question 1 Rapport
You are provided with a metre rule, lens, screen, ray box, and other necessary apparatus.
i. Set up the experiment as shown in the diagram above. Measure and record the diameter \(a_{0}\), of the illuminated object.
ii. Place the object at a distance \(x = 25\text{cm}\) from the lens. Adjust the screen until a sharp image is obtained on the screen.
iii. Measure and record the diameter, \(a\), of the image.
iv. Measure and record the distance \(v\) between the lens and the screen.
v. Evaluate \(y = P = \frac{1+y^{2}}{y}\) and \(T = x+v\).
vi. Repeat the procedure for \(x = 30\text{cm}\), \(35\text{cm}\), \(40\text{cm}\) and \(45\text{cm}\). In each case, determine the corresponding values of \(a,v,y, P\) and \(T\).
vii. Tabulate your results.
viii. Plot a graph of \(P\) on the vertical axis against \(T\) on the horizontal axis starting both axes from the origin \((0,0)\).
ix. Determine the slope, \(s\), of the graph.
x. Determine the intercept, \(c\), on the horizontal axis.
xi Evaluate \(K = \frac{c}{2}\)
xii. State two precautions taken to ensure accurate results.
(b)i. Explain the statement, the focal length of a converging lens is 20cm.
ii. An object is placed at a distance x from a converging lens of focal length 20cm. If the magnification of the real image is 5, calculate the value of x.
(a) Determination of image magnification with a converging lens
(i) Diameter of the illuminated object, \(a_{0} = 2.00\ \text{cm}\).
(vii) Table of results
| S/N | x (cm) | a (cm) | v (cm) | y = a/a₀ | P = (1+y²)/y | T = x+v (cm) |
|---|---|---|---|---|---|---|
| 1 | 25.0 | 2.60 | 49.0 | 1.30 | 2.07 | 74.00 |
| 2 | 30.0 | 2.40 | 46.0 | 1.20 | 2.03 | 76.00 |
| 3 | 35.0 | 2.10 | 45.0 | 1.05 | 2.00 | 80.00 |
| 4 | 40.0 | 1.80 | 43.0 | 0.90 | 2.01 | 83.00 |
| 5 | 45.0 | 1.60 | 40.0 | 0.80 | 2.05 | 85.00 |
Sample evaluations (S/N 1)
\[ y = \frac{a}{a_{0}} = \frac{2.60}{2.00} = 1.30 \] \[ P = \frac{1+y^{2}}{y} = \frac{1+(1.30)^{2}}{1.30} = \frac{1+1.69}{1.30} = \frac{2.69}{1.30} = 2.07 \] \[ T = x + v = 25.0 + 49.0 = 74.00\ \text{cm} \]Sample evaluations (S/N 3)
\[ y = \frac{2.10}{2.00} = 1.05,\qquad P = \frac{1+(1.05)^{2}}{1.05} = \frac{2.1025}{1.05} = 2.00,\qquad T = 35.0 + 45.0 = 80.00\ \text{cm} \](viii) Graph of P against T
(ix) Slope of the graph
Two points are read on the line of best fit:
\[ (T_{1},\,P_{1}) = (74.00\ \text{cm},\ 2.070) \qquad (T_{2},\,P_{2}) = (80.00\ \text{cm},\ 2.037) \] \[ s = \frac{T_{2}-T_{1}}{P_{2}-P_{1}} = \frac{80.00 - 74.00}{2.070 - 2.037} = \frac{6.00}{0.033} = 181.8\ \text{cm} \](x) Intercept on the horizontal (T) axis
Extending the line of best fit to the horizontal axis gives
\[ c = 81.0\ \text{cm} \](xi) Evaluation of K
\[ K = \frac{c}{2} = \frac{81.0}{2} = 40.5\ \text{cm} \](xii) Two precautions
- The object, the lens and the screen were kept coaxial, with their centres at the same height on a straight line.
- Parallax error was avoided when reading the metre rule, and the screen was adjusted until the sharpest possible image was obtained.
(b)(i) Meaning of “focal length of a converging lens is 20 cm”
It means that the distance between the optical centre of the lens and its principal focus is 20 cm; that is, a beam of light travelling parallel to the principal axis is converged (brought to a focus) at a point 20 cm from the optical centre of the lens.
(b)(ii) Value of x for a real image of magnification 5
For a real image the magnification is
\[ m = \frac{v}{u} = 5 \quad\Rightarrow\quad v = 5u \]Applying the lens formula with \(f = 20\ \text{cm}\):
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] \[ \frac{1}{20} = \frac{1}{u} + \frac{1}{5u} = \frac{5+1}{5u} = \frac{6}{5u} \] \[ 5u = 20 \times 6 = 120 \] \[ u = \frac{120}{5} = 24\ \text{cm} \]Therefore \(x = 24\ \text{cm}\).
Détails de la réponse
(a) Determination of image magnification with a converging lens
(i) Diameter of the illuminated object, \(a_{0} = 2.00\ \text{cm}\).
(vii) Table of results
| S/N | x (cm) | a (cm) | v (cm) | y = a/a₀ | P = (1+y²)/y | T = x+v (cm) |
|---|---|---|---|---|---|---|
| 1 | 25.0 | 2.60 | 49.0 | 1.30 | 2.07 | 74.00 |
| 2 | 30.0 | 2.40 | 46.0 | 1.20 | 2.03 | 76.00 |
| 3 | 35.0 | 2.10 | 45.0 | 1.05 | 2.00 | 80.00 |
| 4 | 40.0 | 1.80 | 43.0 | 0.90 | 2.01 | 83.00 |
| 5 | 45.0 | 1.60 | 40.0 | 0.80 | 2.05 | 85.00 |
Sample evaluations (S/N 1)
\[ y = \frac{a}{a_{0}} = \frac{2.60}{2.00} = 1.30 \] \[ P = \frac{1+y^{2}}{y} = \frac{1+(1.30)^{2}}{1.30} = \frac{1+1.69}{1.30} = \frac{2.69}{1.30} = 2.07 \] \[ T = x + v = 25.0 + 49.0 = 74.00\ \text{cm} \]Sample evaluations (S/N 3)
\[ y = \frac{2.10}{2.00} = 1.05,\qquad P = \frac{1+(1.05)^{2}}{1.05} = \frac{2.1025}{1.05} = 2.00,\qquad T = 35.0 + 45.0 = 80.00\ \text{cm} \](viii) Graph of P against T
(ix) Slope of the graph
Two points are read on the line of best fit:
\[ (T_{1},\,P_{1}) = (74.00\ \text{cm},\ 2.070) \qquad (T_{2},\,P_{2}) = (80.00\ \text{cm},\ 2.037) \] \[ s = \frac{T_{2}-T_{1}}{P_{2}-P_{1}} = \frac{80.00 - 74.00}{2.070 - 2.037} = \frac{6.00}{0.033} = 181.8\ \text{cm} \](x) Intercept on the horizontal (T) axis
Extending the line of best fit to the horizontal axis gives
\[ c = 81.0\ \text{cm} \](xi) Evaluation of K
\[ K = \frac{c}{2} = \frac{81.0}{2} = 40.5\ \text{cm} \](xii) Two precautions
- The object, the lens and the screen were kept coaxial, with their centres at the same height on a straight line.
- Parallax error was avoided when reading the metre rule, and the screen was adjusted until the sharpest possible image was obtained.
(b)(i) Meaning of “focal length of a converging lens is 20 cm”
It means that the distance between the optical centre of the lens and its principal focus is 20 cm; that is, a beam of light travelling parallel to the principal axis is converged (brought to a focus) at a point 20 cm from the optical centre of the lens.
(b)(ii) Value of x for a real image of magnification 5
For a real image the magnification is
\[ m = \frac{v}{u} = 5 \quad\Rightarrow\quad v = 5u \]Applying the lens formula with \(f = 20\ \text{cm}\):
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] \[ \frac{1}{20} = \frac{1}{u} + \frac{1}{5u} = \frac{5+1}{5u} = \frac{6}{5u} \] \[ 5u = 20 \times 6 = 120 \] \[ u = \frac{120}{5} = 24\ \text{cm} \]Therefore \(x = 24\ \text{cm}\).
Question 2 Rapport
You are provided with a pendulum bob, a metre rule, a stopwatch, a retort stand with clamp, and other necessary apparatus.
i. Suspend the pendulum bob from the clamp as illustrated in the diagram.
ii. Adjust the pendulum such that AC=L= 90 cm
iii. Displace the pendulum bob slightly such that it oscillates in a vertical plane.
iv. Measure and record the time t for 20 complete oscillations.
v. Evaluate T and \(\sqrt L\)
vi.Repeat the procedure for four others values of L= 80 cm,70 cm, 60 cm, and 50cm.
vii. Tabulate your readings
viii. Plot a graph with T on the vertical axis and \(\sqrt L\) on the horizontal axis.
ix. Determine the slope, s, of the graph.
x. Evaluate g= \(\frac{4\pi^{2}}{5^{2}}\)
xi. State two precautions taken to ensure accurate results.
(b) i. Determine from your graph, the period of the pendulum for L= 75 cm.
ii. A simple pendulum bob is set into simple harmonic motion. Sketch a diagram of the setup and indicate on it; the positions of:
(a) maximum velocity.
(b) maximum acceleration of the bob
Set-up and measurements
The pendulum bob is suspended from the clamp of the retort stand by a light thread. The length \(L = AC\) is measured from the point of suspension \(A\) to the centre of the bob \(C\). For each length the bob is displaced through a small angle and released so that it swings in a vertical plane, and the time \(t\) for 20 complete oscillations is taken with the stopwatch. The period is obtained from \(T = \dfrac{t}{20}\) and \(\sqrt{L}\) is evaluated with \(L\) in metres.
Table of readings
| S/N | L / cm | L / m | t (20 osc.) / s | T / s | √L / m1/2 |
|---|---|---|---|---|---|
| 1 | 90 | 0.90 | 38.1 | 1.91 | 0.949 |
| 2 | 80 | 0.80 | 35.9 | 1.79 | 0.894 |
| 3 | 70 | 0.70 | 33.6 | 1.68 | 0.837 |
| 4 | 60 | 0.60 | 31.1 | 1.55 | 0.775 |
| 5 | 50 | 0.50 | 28.4 | 1.42 | 0.707 |
Graph of T against √L
Plotting \(T\) (vertical axis) against \(\sqrt{L}\) (horizontal axis) gives a straight line passing through the origin.
Slope of the graph
Taking two widely separated points on the line of best fit, \((0.707,\,1.42)\) and \((0.949,\,1.90)\):
\[ s = \frac{\Delta T}{\Delta \sqrt{L}} = \frac{1.90 - 1.42}{0.949 - 0.707} = \frac{0.48}{0.242} = 1.98\ \text{s\,m}^{-1/2}. \]Evaluating g
For a simple pendulum \(T = 2\pi\sqrt{L/g} = \dfrac{2\pi}{\sqrt{g}}\sqrt{L}\), so the slope is \(s = \dfrac{2\pi}{\sqrt{g}}\) and therefore
\[ g = \frac{4\pi^{2}}{s^{2}} = \frac{4 \times (3.142)^{2}}{(1.98)^{2}} = \frac{39.48}{3.92} = 10.1\ \text{m s}^{-2}. \]Two precautions
- The bob was displaced through a small angle (less than about \(10^\circ\)) and made to oscillate in a vertical plane, so that conical (circular) oscillation was avoided and the motion remained simple harmonic.
- The timing was started and stopped as the bob passed the central rest position, and the experiment was shielded from draught, so that counting and timing errors were reduced.
(b)(i) Period for L = 75 cm
For \(L = 0.75\ \text{m}\), \(\sqrt{L} = 0.866\ \text{m}^{1/2}\). Reading up from the graph (equivalently \(T = s\sqrt{L}\)):
\[ T = 1.98 \times 0.866 = 1.72\ \text{s}. \](b)(ii) Positions of maximum velocity and maximum acceleration
The bob swings between the two extreme positions \(P\) and \(Q\) about the central (lowest) equilibrium position \(O\).
- Maximum velocity occurs at \(O\), the equilibrium position, where the displacement is zero and the bob moves fastest.
- Maximum acceleration occurs at the extreme positions \(P\) and \(Q\), where the displacement from \(O\) is greatest and the restoring force is largest.
Détails de la réponse
Set-up and measurements
The pendulum bob is suspended from the clamp of the retort stand by a light thread. The length \(L = AC\) is measured from the point of suspension \(A\) to the centre of the bob \(C\). For each length the bob is displaced through a small angle and released so that it swings in a vertical plane, and the time \(t\) for 20 complete oscillations is taken with the stopwatch. The period is obtained from \(T = \dfrac{t}{20}\) and \(\sqrt{L}\) is evaluated with \(L\) in metres.
Table of readings
| S/N | L / cm | L / m | t (20 osc.) / s | T / s | √L / m1/2 |
|---|---|---|---|---|---|
| 1 | 90 | 0.90 | 38.1 | 1.91 | 0.949 |
| 2 | 80 | 0.80 | 35.9 | 1.79 | 0.894 |
| 3 | 70 | 0.70 | 33.6 | 1.68 | 0.837 |
| 4 | 60 | 0.60 | 31.1 | 1.55 | 0.775 |
| 5 | 50 | 0.50 | 28.4 | 1.42 | 0.707 |
Graph of T against √L
Plotting \(T\) (vertical axis) against \(\sqrt{L}\) (horizontal axis) gives a straight line passing through the origin.
Slope of the graph
Taking two widely separated points on the line of best fit, \((0.707,\,1.42)\) and \((0.949,\,1.90)\):
\[ s = \frac{\Delta T}{\Delta \sqrt{L}} = \frac{1.90 - 1.42}{0.949 - 0.707} = \frac{0.48}{0.242} = 1.98\ \text{s\,m}^{-1/2}. \]Evaluating g
For a simple pendulum \(T = 2\pi\sqrt{L/g} = \dfrac{2\pi}{\sqrt{g}}\sqrt{L}\), so the slope is \(s = \dfrac{2\pi}{\sqrt{g}}\) and therefore
\[ g = \frac{4\pi^{2}}{s^{2}} = \frac{4 \times (3.142)^{2}}{(1.98)^{2}} = \frac{39.48}{3.92} = 10.1\ \text{m s}^{-2}. \]Two precautions
- The bob was displaced through a small angle (less than about \(10^\circ\)) and made to oscillate in a vertical plane, so that conical (circular) oscillation was avoided and the motion remained simple harmonic.
- The timing was started and stopped as the bob passed the central rest position, and the experiment was shielded from draught, so that counting and timing errors were reduced.
(b)(i) Period for L = 75 cm
For \(L = 0.75\ \text{m}\), \(\sqrt{L} = 0.866\ \text{m}^{1/2}\). Reading up from the graph (equivalently \(T = s\sqrt{L}\)):
\[ T = 1.98 \times 0.866 = 1.72\ \text{s}. \](b)(ii) Positions of maximum velocity and maximum acceleration
The bob swings between the two extreme positions \(P\) and \(Q\) about the central (lowest) equilibrium position \(O\).
- Maximum velocity occurs at \(O\), the equilibrium position, where the displacement is zero and the bob moves fastest.
- Maximum acceleration occurs at the extreme positions \(P\) and \(Q\), where the displacement from \(O\) is greatest and the restoring force is largest.
Question 3 Rapport
You are provided with an ammeter, resistor, key, metre bridge, and other necessary apparatus.
i. Connect a circuit as shown in the diagram above.
ii. Close the key and use the jockey to make contact with AB at N such that AN = d = 25cm
iii. Read and record the ammeter reading.
iv. Evaluate 1\(^{-1}\).
v. Repeat the procedures for values of d = 35cmm, 50cm, 65cm and 80cm. In each case, record I and determine 1\(^{-1}\)
vi. Tabulate your results.
vii. Plot a graph with log on the vertical axis and d on the horizontal axis.
viii. Determine the slope, s of the graph.
ix. State two precautions taken to obtain accurate results.
(b)i. Use your graph to determine the value of d = l = 1.5A.
ii. State two factors that affect the resistance of a wire.
Test of Practical Knowledge: Metre Bridge and Ammeter
The apparatus is connected so that the cell, key, ammeter and resistor \(R\) are joined in series to end \(A\) of the metre bridge wire, while the jockey is used to make contact with the wire at \(N\). Only the length \(AN = d\) of the bridge wire carries the current, so as \(d\) increases the resistance in the circuit rises and the ammeter reading \(I\) falls.
(ii) - (vi) Readings and table. For each contact length \(d = AN\) the key is closed, the ammeter reading \(I\) is taken, its reciprocal \(I^{-1}\) is evaluated and \(\log I^{-1}\) obtained. The results are tabulated below.
| S/N | d (cm) | I (A) | I-1 (A-1) | log I-1 |
|---|---|---|---|---|
| 1 | 25 | 0.65 | 1.54 | 0.19 |
| 2 | 35 | 0.55 | 1.82 | 0.26 |
| 3 | 50 | 0.44 | 2.50 | 0.40 |
| 4 | 65 | 0.30 | 3.33 | 0.52 |
| 5 | 80 | 0.23 | 4.35 | 0.64 |
(vii) Graph. \(\log I^{-1}\) is plotted on the vertical axis against \(d\) on the horizontal axis and the best straight line is drawn through the points.
(viii) Slope. Taking two widely separated points on the line, \((25,\;0.20)\) and \((80,\;0.64)\):
\[ s = \frac{\Delta(\log I^{-1})}{\Delta d} = \frac{0.64 - 0.20}{80 - 25} = \frac{0.44}{55} = 8.0\times10^{-3}\ \text{A}^{-1}\,\text{cm}^{-1}. \](ix) Two precautions.
- The key was opened between successive readings so that the bridge wire did not heat up and change its resistance.
- Clean, tight connections were made and parallax error was avoided by reading the ammeter and the metre rule with the eye directly in line with the pointer and the scale.
(b)(i) Value of d when I = 1.5 A.
\[ I^{-1} = \frac{1}{1.5} = 0.67\ \text{A}^{-1}, \qquad \log I^{-1} = \log(0.67) = -0.17. \]The best-fit line is \(\log I^{-1} = 8.0\times10^{-3}\,d - 0.01\). Producing this line downwards until it meets \(\log I^{-1} = -0.17\) gives
\[ d = \frac{-0.17 + 0.01}{8.0\times10^{-3}} \approx -20\ \text{cm}. \]Because this value falls below \(d = 0\) (off the lower end of the scale), a current as large as \(1.5\ \text{A}\) cannot be obtained with this circuit: even the shortest usable length of wire keeps the current below about \(0.65\ \text{A}\).
(b)(ii) Two factors that affect the resistance of a wire.
- The length of the wire (resistance increases with length).
- The cross-sectional area (thickness or diameter) of the wire (resistance decreases as the area increases).
(The resistivity or nature of the material and the temperature of the wire also affect its resistance.)
Détails de la réponse
Test of Practical Knowledge: Metre Bridge and Ammeter
The apparatus is connected so that the cell, key, ammeter and resistor \(R\) are joined in series to end \(A\) of the metre bridge wire, while the jockey is used to make contact with the wire at \(N\). Only the length \(AN = d\) of the bridge wire carries the current, so as \(d\) increases the resistance in the circuit rises and the ammeter reading \(I\) falls.
(ii) - (vi) Readings and table. For each contact length \(d = AN\) the key is closed, the ammeter reading \(I\) is taken, its reciprocal \(I^{-1}\) is evaluated and \(\log I^{-1}\) obtained. The results are tabulated below.
| S/N | d (cm) | I (A) | I-1 (A-1) | log I-1 |
|---|---|---|---|---|
| 1 | 25 | 0.65 | 1.54 | 0.19 |
| 2 | 35 | 0.55 | 1.82 | 0.26 |
| 3 | 50 | 0.44 | 2.50 | 0.40 |
| 4 | 65 | 0.30 | 3.33 | 0.52 |
| 5 | 80 | 0.23 | 4.35 | 0.64 |
(vii) Graph. \(\log I^{-1}\) is plotted on the vertical axis against \(d\) on the horizontal axis and the best straight line is drawn through the points.
(viii) Slope. Taking two widely separated points on the line, \((25,\;0.20)\) and \((80,\;0.64)\):
\[ s = \frac{\Delta(\log I^{-1})}{\Delta d} = \frac{0.64 - 0.20}{80 - 25} = \frac{0.44}{55} = 8.0\times10^{-3}\ \text{A}^{-1}\,\text{cm}^{-1}. \](ix) Two precautions.
- The key was opened between successive readings so that the bridge wire did not heat up and change its resistance.
- Clean, tight connections were made and parallax error was avoided by reading the ammeter and the metre rule with the eye directly in line with the pointer and the scale.
(b)(i) Value of d when I = 1.5 A.
\[ I^{-1} = \frac{1}{1.5} = 0.67\ \text{A}^{-1}, \qquad \log I^{-1} = \log(0.67) = -0.17. \]The best-fit line is \(\log I^{-1} = 8.0\times10^{-3}\,d - 0.01\). Producing this line downwards until it meets \(\log I^{-1} = -0.17\) gives
\[ d = \frac{-0.17 + 0.01}{8.0\times10^{-3}} \approx -20\ \text{cm}. \]Because this value falls below \(d = 0\) (off the lower end of the scale), a current as large as \(1.5\ \text{A}\) cannot be obtained with this circuit: even the shortest usable length of wire keeps the current below about \(0.65\ \text{A}\).
(b)(ii) Two factors that affect the resistance of a wire.
- The length of the wire (resistance increases with length).
- The cross-sectional area (thickness or diameter) of the wire (resistance decreases as the area increases).
(The resistivity or nature of the material and the temperature of the wire also affect its resistance.)
