WAEC SSCE - Physics - 2003 (Practical)
Frage 1 Bericht
You are provided with a potentiometer x y; a jockey, J; a standard resistor, R, and other necessary apparatus.
- Connect a circuit as shown in the diagram above.
- Close the key. Read and record the current 10 when J is not in contact with XY.
- Let J make contact with XY at C, such that XC = l= 25 cm. Close the key. Read and record the current l.
- Evaluate I\(^{-1}\).
- Repeat the procedure for four other values of I = 40, 55, 70, and 85 cm. Tabulate your readings.
- Plot a graph of l on the vertical axis against I\(^{-1}\) on the horizontal axis.
- From your graph, deduce the value of I when l\(^{-1}\) = 0. Evaluate \(\frac{|O}{1}\)
- State two precautions taken to ensure accurate results.
(b)i) Explain what is meant by the potential difference between two points in an electric circuit.
ii. A piece of resistance wire of diameter 0.2m and resistance 7\(\Omega\) has a resistivity of 8.8 x 10\(^{-7}\) \(\Omega\)m. Calculate the length of the wire. [\(\pi\) = \(\frac{22}{7}\)].
(a) Potentiometer (rheostat) current experiment
The circuit is connected as shown, with the standard resistor \(R\) and the ammeter in series with the length \(XC=l\) of the potentiometer wire. With the jockey off the wire the ammeter reads \(I_0\); when the jockey contacts XY at C the current \(I\) is read for each length \(l\), and \(l^{-1}\) is evaluated.
Table of readings (\(I_0 = 0.8\ \text{A}\))
| S/N | \(l\) (cm) | \(I\) (A) | \(l^{-1}\) (cm\(^{-1}\)) |
|---|---|---|---|
| 1 | 25.0 | 1.20 | 0.040 |
| 2 | 40.0 | 1.00 | 0.025 |
| 3 | 55.0 | 0.95 | 0.018 |
| 4 | 70.0 | 0.85 | 0.014 |
| 5 | 85.0 | 0.80 | 0.012 |
Graph of \(I\) (vertical) against \(l^{-1}\) (horizontal):
Deduction from the graph. The line of best fit cuts the vertical axis (where \(l^{-1}=0\)) at
\[ I = 0.66\ \text{A}. \]Hence
\[ \frac{I_0}{I} = \frac{0.8}{0.66} = 1.21. \]Two precautions:
- The key was left open whenever readings were not being taken, to prevent heating of the wire.
- The jockey was pressed firmly but was not rolled along or allowed to scratch the potentiometer wire.
(b)(i) Potential difference
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 charge from one point to the other:
\[ V = \frac{W}{Q}. \]Its S.I. unit is the volt (\(\text{J C}^{-1}\)).
(b)(ii) Length of the resistance wire
Given: diameter \(d = 0.2\ \text{m}\), resistance \(R = 7\ \Omega\), resistivity \(\rho = 8.8\times10^{-7}\ \Omega\text{m}\), \(\pi = \tfrac{22}{7}\).
Cross-sectional area of the wire:
\[ A = \frac{\pi d^{2}}{4} = \frac{\tfrac{22}{7}\times(0.2)^{2}}{4} = \frac{\tfrac{22}{7}\times 0.04}{4} = 0.03143\ \text{m}^{2}. \]From \(R = \dfrac{\rho L}{A}\):
\[ L = \frac{RA}{\rho} = \frac{7 \times 0.03143}{8.8\times10^{-7}} = \frac{0.22}{8.8\times10^{-7}} = 0.025\times10^{7}\ \text{m}. \] \[ \boxed{L = 2.5\times10^{5}\ \text{m}.} \]Antwortdetails
(a) Potentiometer (rheostat) current experiment
The circuit is connected as shown, with the standard resistor \(R\) and the ammeter in series with the length \(XC=l\) of the potentiometer wire. With the jockey off the wire the ammeter reads \(I_0\); when the jockey contacts XY at C the current \(I\) is read for each length \(l\), and \(l^{-1}\) is evaluated.
Table of readings (\(I_0 = 0.8\ \text{A}\))
| S/N | \(l\) (cm) | \(I\) (A) | \(l^{-1}\) (cm\(^{-1}\)) |
|---|---|---|---|
| 1 | 25.0 | 1.20 | 0.040 |
| 2 | 40.0 | 1.00 | 0.025 |
| 3 | 55.0 | 0.95 | 0.018 |
| 4 | 70.0 | 0.85 | 0.014 |
| 5 | 85.0 | 0.80 | 0.012 |
Graph of \(I\) (vertical) against \(l^{-1}\) (horizontal):
Deduction from the graph. The line of best fit cuts the vertical axis (where \(l^{-1}=0\)) at
\[ I = 0.66\ \text{A}. \]Hence
\[ \frac{I_0}{I} = \frac{0.8}{0.66} = 1.21. \]Two precautions:
- The key was left open whenever readings were not being taken, to prevent heating of the wire.
- The jockey was pressed firmly but was not rolled along or allowed to scratch the potentiometer wire.
(b)(i) Potential difference
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 charge from one point to the other:
\[ V = \frac{W}{Q}. \]Its S.I. unit is the volt (\(\text{J C}^{-1}\)).
(b)(ii) Length of the resistance wire
Given: diameter \(d = 0.2\ \text{m}\), resistance \(R = 7\ \Omega\), resistivity \(\rho = 8.8\times10^{-7}\ \Omega\text{m}\), \(\pi = \tfrac{22}{7}\).
Cross-sectional area of the wire:
\[ A = \frac{\pi d^{2}}{4} = \frac{\tfrac{22}{7}\times(0.2)^{2}}{4} = \frac{\tfrac{22}{7}\times 0.04}{4} = 0.03143\ \text{m}^{2}. \]From \(R = \dfrac{\rho L}{A}\):
\[ L = \frac{RA}{\rho} = \frac{7 \times 0.03143}{8.8\times10^{-7}} = \frac{0.22}{8.8\times10^{-7}} = 0.025\times10^{7}\ \text{m}. \] \[ \boxed{L = 2.5\times10^{5}\ \text{m}.} \]Frage 2 Bericht
Using the diagram above as a guide, carry out the following instructions:
- Fix. the drawing paper provided on the drawing board
- Place the mirror vertically with its longer side resting on the drawing paper. Trace the outline AB of the mirror. Remove the mirror.
- Draw a normal PQ to meet the outline at the middle Q.
- Draw a straight line through A to meet the outline of the mirror at the right angle.
- Trace the incident ray, Q with pins \(P_{1}\) and \(P_{2}\) so that it meets the perpendicular line through A at C such that \(CA=X= 1.0m\).
- Replace the mirror on its outline. Locate the images of \(P_{1}P_{2}\) through the mirror using two other pins \(P_{3}\) and \(P_{4}\) so that \(P_{3}\) and \(P_{4}\) and the images of \(P_{1}\) and \(P_{2}\) are in a straight line.
- Remove the mirror and pins \(P_{3}\) and \(P_{4}\).
Draw a straight line through the pin points to meet AB at Q and CA produced at D. - Measure and record angle ACQ as \(è_{1}\), and angle ADCQ as \(è_{2}\). Also, record the value of x. Evaluate \(è = (è_{1} + è_{2})\), \(x^{-1}\) and tan è.
- Repeat the procedure for four other values of x = 2.0, 3.0, 4.0 and 5.0cm. Tabulate your readings.
- Plot a graph of tan è on the vertical axis against x on the horizontal axis.
- Determine the slope, s, of the graph. Evaluate k = 2s
- State two precautions taken to ensure accurate results. Attach your traces to your answer booklet.
(b)i. Distinguish between regular and diffused reflections.
ii. An object is situated 25cm in front of a plane mirror. Determine the distance of the image from the object. What is the size of the image relative to the object?
(a) Plane-mirror reflection experiment
An incident ray is traced from C to a point on the mirror using pins \(P_1\) and \(P_2\), and its reflected ray is located behind the mirror with pins \(P_3\) and \(P_4\). For each offset \(x = CA\) the two base angles \(\theta_1\) and \(\theta_2\) are measured, the mean angle \(\theta = \tfrac{1}{2}(\theta_1+\theta_2)\) is found, and \(x^{-1}\) and \(\tan\theta\) are evaluated. A complete set of readings is shown below.
| S/N | \(\theta_1\) (°) | \(\theta_2\) (°) | x (cm) | \(x^{-1}\) (cm-1) | \(\theta=\tfrac{1}{2}(\theta_1+\theta_2)\) (°) | \(\tan\theta\) |
|---|---|---|---|---|---|---|
| 1 | 83.0 | 81.0 | 1.0 | 1.00 | 82.0 | 7.115 |
| 2 | 76.0 | 74.0 | 2.0 | 0.50 | 75.0 | 3.732 |
| 3 | 69.0 | 70.0 | 3.0 | 0.33 | 69.5 | 2.675 |
| 4 | 62.0 | 63.0 | 4.0 | 0.25 | 62.5 | 1.921 |
| 5 | 57.0 | 58.0 | 5.0 | 0.20 | 57.5 | 1.570 |
Since \(\tan\theta\) rises steadily as \(x^{-1}\) increases (the product \(x\tan\theta\) stays close to a constant of about 7.6), the straight-line relationship is obtained by plotting \(\tan\theta\) against \(x^{-1}\), and the graph passes close to the origin.
Graph of \(\tan\theta\) against \(x^{-1}\):
Slope and value of k. Reading two well-separated points on the line of best fit, \((x^{-1}=0,\ \tan\theta \approx 0.15)\) and \((x^{-1}=1.0,\ \tan\theta \approx 7.05)\):
\[ s = \frac{\Delta(\tan\theta)}{\Delta(x^{-1})} = \frac{7.05 - 0.15}{1.0 - 0} = 6.9 \] \[ k = 2s = 2 \times 6.9 = 13.8 \approx 14 \]Two precautions:
- Fix the pins truly vertical (not bent) and reasonably far apart so that each ray is well defined.
- Sight the pins and their images with one eye at the level of the paper, and draw thin lines with a sharp pencil, to avoid errors due to parallax.
(b)(i) Regular versus diffused reflection
Regular (specular) reflection occurs at a smooth, polished surface such as a plane mirror: a parallel beam of incident rays is turned back as a parallel beam in one definite direction, so a clear image is formed. Diffused (irregular) reflection occurs at a rough surface such as paper or a wall: parallel incident rays are scattered in many different directions, so no clear image is formed. In both cases the laws of reflection are obeyed at every point; only the surface differs.
(b)(ii) Distance of the image from the object
For a plane mirror the image is formed as far behind the mirror as the object is in front of it. The object is \(25\ \text{cm}\) in front, so the image is \(25\ \text{cm}\) behind the mirror. The distance between the object and its image is therefore:
\[ d = 25 + 25 = 50\ \text{cm} \]The image is the same size as the object (magnification = 1); it is virtual, erect and laterally inverted.
Antwortdetails
(a) Plane-mirror reflection experiment
An incident ray is traced from C to a point on the mirror using pins \(P_1\) and \(P_2\), and its reflected ray is located behind the mirror with pins \(P_3\) and \(P_4\). For each offset \(x = CA\) the two base angles \(\theta_1\) and \(\theta_2\) are measured, the mean angle \(\theta = \tfrac{1}{2}(\theta_1+\theta_2)\) is found, and \(x^{-1}\) and \(\tan\theta\) are evaluated. A complete set of readings is shown below.
| S/N | \(\theta_1\) (°) | \(\theta_2\) (°) | x (cm) | \(x^{-1}\) (cm-1) | \(\theta=\tfrac{1}{2}(\theta_1+\theta_2)\) (°) | \(\tan\theta\) |
|---|---|---|---|---|---|---|
| 1 | 83.0 | 81.0 | 1.0 | 1.00 | 82.0 | 7.115 |
| 2 | 76.0 | 74.0 | 2.0 | 0.50 | 75.0 | 3.732 |
| 3 | 69.0 | 70.0 | 3.0 | 0.33 | 69.5 | 2.675 |
| 4 | 62.0 | 63.0 | 4.0 | 0.25 | 62.5 | 1.921 |
| 5 | 57.0 | 58.0 | 5.0 | 0.20 | 57.5 | 1.570 |
Since \(\tan\theta\) rises steadily as \(x^{-1}\) increases (the product \(x\tan\theta\) stays close to a constant of about 7.6), the straight-line relationship is obtained by plotting \(\tan\theta\) against \(x^{-1}\), and the graph passes close to the origin.
Graph of \(\tan\theta\) against \(x^{-1}\):
Slope and value of k. Reading two well-separated points on the line of best fit, \((x^{-1}=0,\ \tan\theta \approx 0.15)\) and \((x^{-1}=1.0,\ \tan\theta \approx 7.05)\):
\[ s = \frac{\Delta(\tan\theta)}{\Delta(x^{-1})} = \frac{7.05 - 0.15}{1.0 - 0} = 6.9 \] \[ k = 2s = 2 \times 6.9 = 13.8 \approx 14 \]Two precautions:
- Fix the pins truly vertical (not bent) and reasonably far apart so that each ray is well defined.
- Sight the pins and their images with one eye at the level of the paper, and draw thin lines with a sharp pencil, to avoid errors due to parallax.
(b)(i) Regular versus diffused reflection
Regular (specular) reflection occurs at a smooth, polished surface such as a plane mirror: a parallel beam of incident rays is turned back as a parallel beam in one definite direction, so a clear image is formed. Diffused (irregular) reflection occurs at a rough surface such as paper or a wall: parallel incident rays are scattered in many different directions, so no clear image is formed. In both cases the laws of reflection are obeyed at every point; only the surface differs.
(b)(ii) Distance of the image from the object
For a plane mirror the image is formed as far behind the mirror as the object is in front of it. The object is \(25\ \text{cm}\) in front, so the image is \(25\ \text{cm}\) behind the mirror. The distance between the object and its image is therefore:
\[ d = 25 + 25 = 50\ \text{cm} \]The image is the same size as the object (magnification = 1); it is virtual, erect and laterally inverted.
Frage 3 Bericht
You are provided with a metre rule, a knife edge, two pieces of thread and two masses m\(_{1}\) and m\(_{2}\)
- Record the values of m\(_{1}\) and m\(_{2}\).
- Balance the metre rule horizontally on the knife edge and record the balance point G.
- With the knife edge at the 60 cm mark of the metre rule, suspend m\(_{1}\) at the 20 cm mark and m\(_{2}\) at a suitable mark such that the rule balances horizontally as illustrated in the diagram above.
- Record the positions Y of m\(_{1}\) and Q of m\(_{2}\).
- Evaluate 1= P - Y and d = Q - P
- Repeat the procedure for four other positions of m, at 18, 16, 14 and 12 cm marks.
- In each case, evaluate and record l and d.
- Tabulate your readings.
- Plot a graph of l on the vertical axis against d on the horizontal axis.
- Determine the slope of the graph.
- State two precautions taken to ensure accurate results.
(b}i. With the aid of a diagram, indicate the forces acting on the metre rule in the experimental set-up above.
ii. Define moment of a force about a point and state its S.1. unit.
(a) Metre rule and knife-edge experiment
Recorded fixed values:
\(m_1 = 20.0\ \text{g}\), \(\quad m_2 = 50.0\ \text{g}\), \(\quad\) balance point (centre of gravity) \(G = 50.0\ \text{cm}\).
The knife edge (pivot) is kept fixed at the \(P = 60.0\ \text{cm}\) mark. Mass \(m_1\) is suspended at the mark \(Y\) (left of the pivot) and mass \(m_2\) at the mark \(Q\) (right of the pivot) so that the rule balances horizontally. For each setting, \(l = P - Y\) and \(d = Q - P\).
Table of readings
| S/N | \(Y\) (cm) | \(Q\) (cm) | \(l = P - Y\) (cm) | \(d = Q - P\) (cm) |
|---|---|---|---|---|
| 1 | 20.0 | 96.0 | 40.0 | 36.0 |
| 2 | 18.0 | 96.8 | 42.0 | 36.8 |
| 3 | 16.0 | 97.6 | 44.0 | 37.6 |
| 4 | 14.0 | 98.4 | 46.0 | 38.4 |
| 5 | 12.0 | 99.2 | 48.0 | 39.2 |
Graph of \(l\) against \(d\)
The points lie on a straight line. Taking two widely separated points on the line of best fit, \((d_1, l_1) = (36.0,\ 40.0)\) and \((d_2, l_2) = (39.2,\ 48.0)\):
\[ \text{slope } s = \frac{l_2 - l_1}{d_2 - d_1} = \frac{48.0 - 40.0}{39.2 - 36.0} = \frac{8.0}{3.2} = 2.5 \]
The slope is \(2.5\) (no unit). It equals the ratio \(\dfrac{m_2}{m_1} = \dfrac{50.0}{20.0} = 2.5\), confirming the principle of moments \(m_1 g\, l + W g\,(P-G) = m_2 g\, d\), which rearranges to \(l = \dfrac{m_2}{m_1}\,d - \dfrac{W(P-G)}{m_1}\).
Two precautions
- The eye was placed vertically above the metre-rule mark when taking each reading to avoid error due to parallax.
- The rule was checked to be exactly horizontal, and the suspended masses were not allowed to touch the table before each reading was recorded.
(b)(i) Forces acting on the metre rule
Four forces act on the rule: the upward normal reaction \(R\) at the knife edge (60 cm mark); the downward weight \(m_1 g\) at \(Y\); the downward weight of the rule \(W\) at its centre of gravity \(G\) (50 cm mark); and the downward weight \(m_2 g\) at \(Q\). For equilibrium, \(R = m_1 g + W + m_2 g\), and the total anticlockwise moment about the pivot equals the total clockwise moment.
(b)(ii) Moment of a force
The moment of a force about a point is the product of the force and the perpendicular distance from that point to the line of action of the force; it is the turning effect of the force about the point.
\[ \text{moment} = F \times d \]
Its S.I. unit is the newton metre (N m).
Antwortdetails
(a) Metre rule and knife-edge experiment
Recorded fixed values:
\(m_1 = 20.0\ \text{g}\), \(\quad m_2 = 50.0\ \text{g}\), \(\quad\) balance point (centre of gravity) \(G = 50.0\ \text{cm}\).
The knife edge (pivot) is kept fixed at the \(P = 60.0\ \text{cm}\) mark. Mass \(m_1\) is suspended at the mark \(Y\) (left of the pivot) and mass \(m_2\) at the mark \(Q\) (right of the pivot) so that the rule balances horizontally. For each setting, \(l = P - Y\) and \(d = Q - P\).
Table of readings
| S/N | \(Y\) (cm) | \(Q\) (cm) | \(l = P - Y\) (cm) | \(d = Q - P\) (cm) |
|---|---|---|---|---|
| 1 | 20.0 | 96.0 | 40.0 | 36.0 |
| 2 | 18.0 | 96.8 | 42.0 | 36.8 |
| 3 | 16.0 | 97.6 | 44.0 | 37.6 |
| 4 | 14.0 | 98.4 | 46.0 | 38.4 |
| 5 | 12.0 | 99.2 | 48.0 | 39.2 |
Graph of \(l\) against \(d\)
The points lie on a straight line. Taking two widely separated points on the line of best fit, \((d_1, l_1) = (36.0,\ 40.0)\) and \((d_2, l_2) = (39.2,\ 48.0)\):
\[ \text{slope } s = \frac{l_2 - l_1}{d_2 - d_1} = \frac{48.0 - 40.0}{39.2 - 36.0} = \frac{8.0}{3.2} = 2.5 \]
The slope is \(2.5\) (no unit). It equals the ratio \(\dfrac{m_2}{m_1} = \dfrac{50.0}{20.0} = 2.5\), confirming the principle of moments \(m_1 g\, l + W g\,(P-G) = m_2 g\, d\), which rearranges to \(l = \dfrac{m_2}{m_1}\,d - \dfrac{W(P-G)}{m_1}\).
Two precautions
- The eye was placed vertically above the metre-rule mark when taking each reading to avoid error due to parallax.
- The rule was checked to be exactly horizontal, and the suspended masses were not allowed to touch the table before each reading was recorded.
(b)(i) Forces acting on the metre rule
Four forces act on the rule: the upward normal reaction \(R\) at the knife edge (60 cm mark); the downward weight \(m_1 g\) at \(Y\); the downward weight of the rule \(W\) at its centre of gravity \(G\) (50 cm mark); and the downward weight \(m_2 g\) at \(Q\). For equilibrium, \(R = m_1 g + W + m_2 g\), and the total anticlockwise moment about the pivot equals the total clockwise moment.
(b)(ii) Moment of a force
The moment of a force about a point is the product of the force and the perpendicular distance from that point to the line of action of the force; it is the turning effect of the force about the point.
\[ \text{moment} = F \times d \]
Its S.I. unit is the newton metre (N m).
