WAEC SSCE - Physics - 2013 (Practical)
Swali 1 Ripoti
You are provided with a uniform metre rule of mass, M indicated on its reverse side, a knife-edge, a graduated measuring cylinder of known mass, M\(_{1}\) marked on it and other necessary apparatus.
- Read and record with values of M and m\(_{1}\).
- Balance the metre rule horizontally on the knife edge. Read and record the balance point as G.
- Tie a loop of thread around the neck of the measuring cylinder.
- Fill the cylinder with the sand provided to the 2cm\(^{3}\) mark. Record the volume, V, of the sand.
- Hang the cylinder at the 2 cm mark of the metre rule and adjust the position of the knife edge until the rule balances horizontally.
- Read and record the new balance position K.
- Determine the value of e and f.
- Determine the mass, m\(_{2}\), of the sand in the measuring cylinder. Hint: m\(_{2}\) = (\(\frac{\text {M x f}}{e}\)) - m\(_{1}\).
- Repeat the procedure by filling the measuring cylinder to the mark V = 4,6,8 and 10 cm\(^{3}\). In each case, ensure that the measuring cylinder is kept constant at the 2 cm mark on the metre rule.
- Tabulate your readings.
- Plot a graph with m\(_{2}\) on the vertical axis and V on the horizontal axis.
- Determine the slope, s, of the graph.
- State two precautions taken to ensure accurate results.
(b)i. Determine the mass of 7.5 cm\(^{3}\) of the sand using your graph.
ii. A gold coin of mass 102.0 g has a uniform cross-sectional area of 10.0 cm\(^{2}\). Calculate its thickness. [Density of gold=19.3 g cm\(^{-3}\)]
Test of Practical Knowledge (moments: mass of sand by balancing a metre rule)
Observations
- Mass of the uniform metre rule, \(M = 130\,\text{g}\)
- Mass of the empty measuring cylinder, \(m_1 = 20\,\text{g}\)
- Balance point of the bare rule, \(G = 50.4\,\text{cm}\)
The cylinder is hung at the 2 cm mark of the rule and the knife-edge is moved until the rule balances at position \(K\). Then \(e = K - 2\) is the distance from the knife-edge to the load, and \(f = G - K\) is the distance from the knife-edge to the rule's centre of gravity. Taking moments about the knife-edge:
\[ (m_1 + m_2)\,e = M\,f \quad\Rightarrow\quad m_2 = \frac{M f}{e} - m_1 \]For example, at \(V = 2\,\text{cm}^3\): \(K = 43.5\), \(e = 41.5\), \(f = 6.9\), so \(m_2 = \dfrac{130 \times 6.9}{41.5} - 20 = 21.61 - 20 = 1.61\,\text{g}\).
Table of readings
| \(V\)/cm\(^3\) | \(K\)/cm | \(e = K-2\)/cm | \(f = G-K\)/cm | \(m_2 = \frac{Mf}{e}-m_1\)/g |
|---|---|---|---|---|
| 2 | 43.5 | 41.5 | 6.9 | 1.61 |
| 4 | 42.7 | 40.7 | 7.7 | 3.44 |
| 6 | 41.9 | 39.9 | 8.5 | 6.37 |
| 8 | 40.1 | 39.1 | 9.3 | 9.42 |
| 10 | 40.4 | 38.4 | 10.0 | 12.18 |
Graph of \(m_2\) against \(V\)
Plotting \(m_2\) (vertical axis) against \(V\) (horizontal axis) and drawing the best straight line through the points:
Slope of the graph
Reading two well-separated points off the line of best fit, \((V_1, m_2{}_1) = (4.4,\,4.0)\) and \((V_2, m_2{}_2) = (10.4,\,12.8)\):
\[ s = \frac{\Delta m_2}{\Delta V} = \frac{12.8 - 4.0}{10.4 - 4.4} = \frac{8.8}{6.0} = 1.47\ \text{g cm}^{-3} \]The slope is the mass per unit volume (density) of the sand.
Two precautions
- Draught was avoided and the knife-edge was firmly clamped so that the rule balanced steadily and horizontally before each reading.
- Parallax error was avoided by reading the balance point on the rule and the sand volume in the cylinder with the eye placed directly above the scale.
(b)(i) Mass of 7.5 cm\(^3\) of sand
Reading the best-fit line at \(V = 7.5\,\text{cm}^3\) (shown by the dashed guide lines on the graph) gives
\[ m_2 = 8.4\,\text{g} \](b)(ii) Thickness of the gold coin
Mass \(= 102.0\,\text{g}\), cross-sectional area \(= 10.0\,\text{cm}^2\), density \(= 19.3\,\text{g cm}^{-3}\).
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{102.0}{19.3} = 5.28\,\text{cm}^3 \]\[ \text{thickness} = \frac{\text{Volume}}{\text{Area}} = \frac{5.28}{10.0} = 0.528\,\text{cm} \]Maelezo ya Majibu
Test of Practical Knowledge (moments: mass of sand by balancing a metre rule)
Observations
- Mass of the uniform metre rule, \(M = 130\,\text{g}\)
- Mass of the empty measuring cylinder, \(m_1 = 20\,\text{g}\)
- Balance point of the bare rule, \(G = 50.4\,\text{cm}\)
The cylinder is hung at the 2 cm mark of the rule and the knife-edge is moved until the rule balances at position \(K\). Then \(e = K - 2\) is the distance from the knife-edge to the load, and \(f = G - K\) is the distance from the knife-edge to the rule's centre of gravity. Taking moments about the knife-edge:
\[ (m_1 + m_2)\,e = M\,f \quad\Rightarrow\quad m_2 = \frac{M f}{e} - m_1 \]For example, at \(V = 2\,\text{cm}^3\): \(K = 43.5\), \(e = 41.5\), \(f = 6.9\), so \(m_2 = \dfrac{130 \times 6.9}{41.5} - 20 = 21.61 - 20 = 1.61\,\text{g}\).
Table of readings
| \(V\)/cm\(^3\) | \(K\)/cm | \(e = K-2\)/cm | \(f = G-K\)/cm | \(m_2 = \frac{Mf}{e}-m_1\)/g |
|---|---|---|---|---|
| 2 | 43.5 | 41.5 | 6.9 | 1.61 |
| 4 | 42.7 | 40.7 | 7.7 | 3.44 |
| 6 | 41.9 | 39.9 | 8.5 | 6.37 |
| 8 | 40.1 | 39.1 | 9.3 | 9.42 |
| 10 | 40.4 | 38.4 | 10.0 | 12.18 |
Graph of \(m_2\) against \(V\)
Plotting \(m_2\) (vertical axis) against \(V\) (horizontal axis) and drawing the best straight line through the points:
Slope of the graph
Reading two well-separated points off the line of best fit, \((V_1, m_2{}_1) = (4.4,\,4.0)\) and \((V_2, m_2{}_2) = (10.4,\,12.8)\):
\[ s = \frac{\Delta m_2}{\Delta V} = \frac{12.8 - 4.0}{10.4 - 4.4} = \frac{8.8}{6.0} = 1.47\ \text{g cm}^{-3} \]The slope is the mass per unit volume (density) of the sand.
Two precautions
- Draught was avoided and the knife-edge was firmly clamped so that the rule balanced steadily and horizontally before each reading.
- Parallax error was avoided by reading the balance point on the rule and the sand volume in the cylinder with the eye placed directly above the scale.
(b)(i) Mass of 7.5 cm\(^3\) of sand
Reading the best-fit line at \(V = 7.5\,\text{cm}^3\) (shown by the dashed guide lines on the graph) gives
\[ m_2 = 8.4\,\text{g} \](b)(ii) Thickness of the gold coin
Mass \(= 102.0\,\text{g}\), cross-sectional area \(= 10.0\,\text{cm}^2\), density \(= 19.3\,\text{g cm}^{-3}\).
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{102.0}{19.3} = 5.28\,\text{cm}^3 \]\[ \text{thickness} = \frac{\text{Volume}}{\text{Area}} = \frac{5.28}{10.0} = 0.528\,\text{cm} \]Swali 2 Ripoti
You are provided with a constantan wire, a2 \(\Omega\)standard resistor, an accumulator E, an ammeter A, a key K, and other necessary apparatus.
- Measure and record the emf of the accumulator provided
- Connect a circuit as shown in the diagram above.
- Close the key, read and record the ammeter reading l\(_{o}\) when the crocodile clip is not in contact with the Constantan wire.
- Open the key with the clip making contact with the wire, when d=90cm, close the key. Read and record the ammeter reading l.
- Evaluate d\(^{_1}\)
- Repeat the procedure for four other values of d=80, 70, 60, and 50 cm
- In each case, read and record the ammeter reading and evaluatre d\(^{_1}\).
- Tabulate your readings.
- Plot a graph with I on the vertical axis and d\(^{_1}\) on the horizontal aXIS.
- Determine the slope, S of the graph and its intercept, c, on the vertical axis.
- Evaluate k = \(\frac{c}{s}\)
- Using your graph, determine the current I when d=55cm.
- State two precautions taken to obtain accurate results.
(b)i. Explain what is meant by the potential difference between two points in an electric circuit.
ii. State two factors on which the resistance of a wire depends.
Test of practical knowledge: current through a length of constantan wire
The circuit
The accumulator \(E\), key \(K\), ammeter \(A\), standard resistor \(R = 2\,\Omega\) and the length \(d\) of constantan wire (tapped by the crocodile clip) are joined in series as shown.
Readings recorded before the main experiment
- e.m.f. of the accumulator, \(E = 2.0\ \text{V}\)
- ammeter reading with the clip off the wire, \(I_0 = 0.90\ \text{A}\)
Table of readings
| \(d/\text{cm}\) | \(I/\text{A}\) | \(d^{-1}/\text{cm}^{-1}\) |
|---|---|---|
| 90.0 | 0.21 | 0.0111 |
| 80.0 | 0.23 | 0.0125 |
| 70.0 | 0.25 | 0.0143 |
| 60.0 | 0.29 | 0.0167 |
| 50.0 | 0.34 | 0.0200 |
Graph of \(I\) against \(d^{-1}\)
Slope of the graph
Two points are taken on the line of best fit:
\((d^{-1}_1, I_1) = (0.0060\ \text{cm}^{-1},\ 0.13\ \text{A})\) and \((d^{-1}_2, I_2) = (0.0240\ \text{cm}^{-1},\ 0.40\ \text{A})\).
\[ S = \frac{\Delta I}{\Delta (d^{-1})} = \frac{0.40 - 0.13}{0.0240 - 0.0060} = \frac{0.27}{0.0180} = 15\ \text{A}\,\text{cm} \]Intercept on the vertical axis
Where the line of best fit cuts the \(I\)-axis (at \(d^{-1} = 0\)):
\[ c = 0.04\ \text{A} \]Evaluation of \(k\)
\[ k = \frac{c}{S} = \frac{0.04}{15} = 2.67\times 10^{-3}\ \text{cm}^{-1} \]Current when \(d = 55\ \text{cm}\)
\(d^{-1} = \dfrac{1}{55} = 0.0182\ \text{cm}^{-1}\). Reading up from \(0.0182\ \text{cm}^{-1}\) to the line of best fit and across to the \(I\)-axis:
\[ I = 0.31\ \text{A} \]Two precautions
- The key was opened immediately after each reading so that the wire did not heat up (which would change its resistance) and the accumulator did not run down.
- Clean, tight terminal connections were used and the zero error of the ammeter was noted and corrected before each reading.
(b)(i) Potential difference between two points
The potential difference between two points in an electric circuit is the work done (energy converted from electrical form to other forms) in moving one coulomb of positive charge from one point to the other. It is measured in volts, where \(1\ \text{V} = 1\ \text{J C}^{-1}\).
(b)(ii) Two factors on which the resistance of a wire depends
- The length of the wire (resistance increases with length).
- The cross-sectional area of the wire (resistance decreases as the area increases).
(It also depends on the material/resistivity of the wire and on its temperature.)
Maelezo ya Majibu
Test of practical knowledge: current through a length of constantan wire
The circuit
The accumulator \(E\), key \(K\), ammeter \(A\), standard resistor \(R = 2\,\Omega\) and the length \(d\) of constantan wire (tapped by the crocodile clip) are joined in series as shown.
Readings recorded before the main experiment
- e.m.f. of the accumulator, \(E = 2.0\ \text{V}\)
- ammeter reading with the clip off the wire, \(I_0 = 0.90\ \text{A}\)
Table of readings
| \(d/\text{cm}\) | \(I/\text{A}\) | \(d^{-1}/\text{cm}^{-1}\) |
|---|---|---|
| 90.0 | 0.21 | 0.0111 |
| 80.0 | 0.23 | 0.0125 |
| 70.0 | 0.25 | 0.0143 |
| 60.0 | 0.29 | 0.0167 |
| 50.0 | 0.34 | 0.0200 |
Graph of \(I\) against \(d^{-1}\)
Slope of the graph
Two points are taken on the line of best fit:
\((d^{-1}_1, I_1) = (0.0060\ \text{cm}^{-1},\ 0.13\ \text{A})\) and \((d^{-1}_2, I_2) = (0.0240\ \text{cm}^{-1},\ 0.40\ \text{A})\).
\[ S = \frac{\Delta I}{\Delta (d^{-1})} = \frac{0.40 - 0.13}{0.0240 - 0.0060} = \frac{0.27}{0.0180} = 15\ \text{A}\,\text{cm} \]Intercept on the vertical axis
Where the line of best fit cuts the \(I\)-axis (at \(d^{-1} = 0\)):
\[ c = 0.04\ \text{A} \]Evaluation of \(k\)
\[ k = \frac{c}{S} = \frac{0.04}{15} = 2.67\times 10^{-3}\ \text{cm}^{-1} \]Current when \(d = 55\ \text{cm}\)
\(d^{-1} = \dfrac{1}{55} = 0.0182\ \text{cm}^{-1}\). Reading up from \(0.0182\ \text{cm}^{-1}\) to the line of best fit and across to the \(I\)-axis:
\[ I = 0.31\ \text{A} \]Two precautions
- The key was opened immediately after each reading so that the wire did not heat up (which would change its resistance) and the accumulator did not run down.
- Clean, tight terminal connections were used and the zero error of the ammeter was noted and corrected before each reading.
(b)(i) Potential difference between two points
The potential difference between two points in an electric circuit is the work done (energy converted from electrical form to other forms) in moving one coulomb of positive charge from one point to the other. It is measured in volts, where \(1\ \text{V} = 1\ \text{J C}^{-1}\).
(b)(ii) Two factors on which the resistance of a wire depends
- The length of the wire (resistance increases with length).
- The cross-sectional area of the wire (resistance decreases as the area increases).
(It also depends on the material/resistivity of the wire and on its temperature.)
Swali 3 Ripoti
Using the diagram above as a guide, carry out the following instructions:
- Fix a plain sheet of paper on the Celotex board.
- Place the rectangular glass prism on the paper and trace its outline, ABCD. Remove the prism.
- Draw a normal NMP to meet AB and DC at M and P respectively such that \(|AM|=DP=\)2.0cm.
- Trace the ray PQ with two pins, \(P_{1}\) and \(P_{2}\) at P and Q respectively such that angle MPQ = i = 50º
- Replace the prism on its outline. Trace the emergent ray with two other pins \(P_{3}\) and \(P_{4}\) such that they lie in a straight line with \(P_{2}\) and the image of \(P_{1}\) viewed through the glass prism.
- Measure and record \(\theta\), the angle between the emergent ray and face AB of the glass prism.
- Evaluate \(\cos \theta\) and \(\sin\) i.
- Repeat the procedure for four other values of i = 10°, 15°,20° and25°.
- Tabulate your readings.
- Plot a graph with \(\cos \theta\) on the vertical axis and \(\sin\) i on the horizontal axis.
- Determine the slope, s, of the graph
- State two precautions taken to ensure accurate results. Attach your traces to your answer booklet
(b)i. State the laws of refraction of light
ii. Explain what is meant by the statement the refractive index of a material is 1.65
Test of Practical Knowledge: Refraction through a Rectangular Glass Prism
The outline ABCD is traced, the normal NMP drawn with \(|AM| = |DP| = 2.0\,\text{cm}\), and for each angle of incidence \(i\) the emergent ray is located with pins P3 and P4. The angle \(\theta\) between the emergent ray and face AB is measured, and \(\cos\theta\) and \(\sin i\) are computed.
Table of Readings
| \(i\,/\,^{\circ}\) | \(\theta\,/\,^{\circ}\) | \(\cos\theta\) | \(\sin i\) |
|---|---|---|---|
| 5.0 | 83.0 | 0.137 | 0.087 |
| 10.0 | 74.0 | 0.276 | 0.174 |
| 15.0 | 65.0 | 0.423 | 0.259 |
| 20.0 | 56.0 | 0.559 | 0.342 |
| 25.0 | 48.0 | 0.669 | 0.423 |
Graph of \(\cos\theta\) against \(\sin i\)
The points are plotted with \(\cos\theta\) on the vertical axis and \(\sin i\) on the horizontal axis, and a straight line of best fit is drawn through them.
Determination of the Slope
Two widely separated points are read from the line of best fit:
\(A = (\sin i,\ \cos\theta) = (0.150,\ 0.250)\) and \(B = (0.450,\ 0.720)\).
\[ s = \frac{\Delta(\cos\theta)}{\Delta(\sin i)} = \frac{0.720 - 0.250}{0.450 - 0.150} = \frac{0.470}{0.300} = 1.567 \]Slope, \(s = 1.57\).
Two Precautions
- The pins were fixed erect (vertical) and spaced about 4 cm apart so that the ray directions were accurately located.
- Parallax error was avoided by reading the protractor and metre rule with the eye directly above the scale, and sharp, neat traces were made with a well-sharpened pencil.
(b)(i) Laws of Refraction of Light
- First law: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
- Second law (Snell's law): The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media, \(\dfrac{\sin i}{\sin r} = n\), or \(n_1 \sin i = n_2 \sin r\), where the symbols have their usual meanings.
(b)(ii) Meaning of "the refractive index of a material is 1.65"
It means that the ratio of the speed of light in vacuum (air) to the speed of light in the material is 1.65, i.e. \(n = \dfrac{c}{v} = 1.65\). Equivalently, for light passing from air into the material, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is 1.65, so light travels 1.65 times more slowly in the material and bends towards the normal on entering it.
Maelezo ya Majibu
Test of Practical Knowledge: Refraction through a Rectangular Glass Prism
The outline ABCD is traced, the normal NMP drawn with \(|AM| = |DP| = 2.0\,\text{cm}\), and for each angle of incidence \(i\) the emergent ray is located with pins P3 and P4. The angle \(\theta\) between the emergent ray and face AB is measured, and \(\cos\theta\) and \(\sin i\) are computed.
Table of Readings
| \(i\,/\,^{\circ}\) | \(\theta\,/\,^{\circ}\) | \(\cos\theta\) | \(\sin i\) |
|---|---|---|---|
| 5.0 | 83.0 | 0.137 | 0.087 |
| 10.0 | 74.0 | 0.276 | 0.174 |
| 15.0 | 65.0 | 0.423 | 0.259 |
| 20.0 | 56.0 | 0.559 | 0.342 |
| 25.0 | 48.0 | 0.669 | 0.423 |
Graph of \(\cos\theta\) against \(\sin i\)
The points are plotted with \(\cos\theta\) on the vertical axis and \(\sin i\) on the horizontal axis, and a straight line of best fit is drawn through them.
Determination of the Slope
Two widely separated points are read from the line of best fit:
\(A = (\sin i,\ \cos\theta) = (0.150,\ 0.250)\) and \(B = (0.450,\ 0.720)\).
\[ s = \frac{\Delta(\cos\theta)}{\Delta(\sin i)} = \frac{0.720 - 0.250}{0.450 - 0.150} = \frac{0.470}{0.300} = 1.567 \]Slope, \(s = 1.57\).
Two Precautions
- The pins were fixed erect (vertical) and spaced about 4 cm apart so that the ray directions were accurately located.
- Parallax error was avoided by reading the protractor and metre rule with the eye directly above the scale, and sharp, neat traces were made with a well-sharpened pencil.
(b)(i) Laws of Refraction of Light
- First law: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
- Second law (Snell's law): The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media, \(\dfrac{\sin i}{\sin r} = n\), or \(n_1 \sin i = n_2 \sin r\), where the symbols have their usual meanings.
(b)(ii) Meaning of "the refractive index of a material is 1.65"
It means that the ratio of the speed of light in vacuum (air) to the speed of light in the material is 1.65, i.e. \(n = \dfrac{c}{v} = 1.65\). Equivalently, for light passing from air into the material, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is 1.65, so light travels 1.65 times more slowly in the material and bends towards the normal on entering it.
