WAEC SSCE - Physics - 2006 (Practical)
Frage 1 Bericht
You have been provided with a metre rule, a clamp, and a set of masses.
- Clamp the metre rule to the edge of the bench such that 90cm of the rule projects from the edge as shown in the diagram above. Ensure that the rule is capable of performing oscillatory motion.
- Fix a mass M = 50g at the free end of the rule.
- Deflect the rule slightly such that it performs vertical oscillation.
- Determine the time t for 10 complete oscillations.
- Calculate the period T of the oscillations and evaluate T\(^{2}\)
- Repeat the procedure for four other values of M = 100, 150, 200, and 250g. In each case determine and record the corresponding values of t, T, and T\(^{2}\). Tabulate your readings.
- Plot a graph of T\(^{2}\) on the vertical axis against M on the horizontal axis, starting both axes from the origin (0,0).
- Determine the slopes, of the graph and its intercept C on the vertical axis.
- Evaluate k = 4\(\pi\)/s. [Take \(\pi\) = \(\frac{22}{7}\)].
- From your graph, determine the period T, when M= 180g.
- State two precautions taken to ensure accurate results.
(b)i. Explain simple harmonic motion.
ii. Define period and frequency, with respect to a simple harmonic motion.
Precautions:
- l ensured that the metre rule was firmly clamped
- Readings were repeated
- Parallax was avoided when readings on the stopwatch/clock were taken.
- zero error was noted and corrected on the stopwatch/clock.
(b)i. Simple harmonic motion is a motion in which the acceleration is proportional to the displacement from a fixed point and is directed towards the point.
ii. Period is the time taken by an oscillatory body to make one complete oscillation.
Frequency: is the number of complete oscillations performed in one second.
This is a loaded-cantilever oscillation experiment. The metre rule projecting 90 cm from the bench behaves like a spring; when a mass \(M\) fixed at its free end is deflected and released, it performs vertical simple harmonic motion. For each mass you time 10 complete oscillations, find the period \(T=\dfrac{t}{10}\), and compute \(T^2\). Theory gives \(T^2=\dfrac{4\pi^2}{k}M + C\), so a graph of \(T^2\) against \(M\) is a straight line whose slope allows \(k\) to be found.
Readings and table of values
| M (g) | t (s) for 10 oscillations | T = t/10 (s) | T2 (s2) |
|---|---|---|---|
| 50 | 3.16 | 0.316 | 0.100 |
| 100 | 3.87 | 0.387 | 0.150 |
| 150 | 4.47 | 0.447 | 0.200 |
| 200 | 5.00 | 0.500 | 0.250 |
| 250 | 5.48 | 0.548 | 0.300 |
Graph of T2 against M
Slope and intercept
Using two convenient points on the line, \((M_1,T_1^2)=(50,\,0.100)\) and \((M_2,T_2^2)=(250,\,0.300)\):
\[ s=\frac{\Delta T^2}{\Delta M}=\frac{0.300-0.100}{250-50}=\frac{0.200}{200}=1.0\times10^{-3}\ \text{s}^2\,\text{g}^{-1} \]The line meets the vertical axis where \(M=0\), giving the intercept
\[ C = 0.05\ \text{s}^2 \]Evaluation of k
From \(T^2=\dfrac{4\pi^2}{k}M + C\), the slope is \(s=\dfrac{4\pi^2}{k}\), so \(k=\dfrac{4\pi^2}{s}\). With \(\pi=\dfrac{22}{7}\):
\[ k=\frac{4\pi^2}{s}=\frac{4\times\left(\frac{22}{7}\right)^2}{1.0\times10^{-3}}=\frac{4\times 9.878}{1.0\times10^{-3}}=\frac{39.51}{1.0\times10^{-3}}\approx 3.95\times10^{4} \]Period when M = 180 g
Reading up from \(M=180\ \text{g}\) to the line and across to the vertical axis gives \(T^2=0.23\ \text{s}^2\), so
\[ T=\sqrt{0.23}=0.48\ \text{s} \]Two precautions
- The metre rule was firmly clamped to the bench so that only the projecting part oscillated.
- Parallax error was avoided when reading the stopwatch, the zero error was noted and corrected, and each timing was repeated and averaged.
(b)(i) Simple harmonic motion
Simple harmonic motion is the motion of a body in which its acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point.
(b)(ii) Period and frequency
The period is the time taken by the oscillating body to make one complete oscillation. The frequency is the number of complete oscillations made in one second, related to the period by \(f=\dfrac{1}{T}\).
Antwortdetails
This is a loaded-cantilever oscillation experiment. The metre rule projecting 90 cm from the bench behaves like a spring; when a mass \(M\) fixed at its free end is deflected and released, it performs vertical simple harmonic motion. For each mass you time 10 complete oscillations, find the period \(T=\dfrac{t}{10}\), and compute \(T^2\). Theory gives \(T^2=\dfrac{4\pi^2}{k}M + C\), so a graph of \(T^2\) against \(M\) is a straight line whose slope allows \(k\) to be found.
Readings and table of values
| M (g) | t (s) for 10 oscillations | T = t/10 (s) | T2 (s2) |
|---|---|---|---|
| 50 | 3.16 | 0.316 | 0.100 |
| 100 | 3.87 | 0.387 | 0.150 |
| 150 | 4.47 | 0.447 | 0.200 |
| 200 | 5.00 | 0.500 | 0.250 |
| 250 | 5.48 | 0.548 | 0.300 |
Graph of T2 against M
Slope and intercept
Using two convenient points on the line, \((M_1,T_1^2)=(50,\,0.100)\) and \((M_2,T_2^2)=(250,\,0.300)\):
\[ s=\frac{\Delta T^2}{\Delta M}=\frac{0.300-0.100}{250-50}=\frac{0.200}{200}=1.0\times10^{-3}\ \text{s}^2\,\text{g}^{-1} \]The line meets the vertical axis where \(M=0\), giving the intercept
\[ C = 0.05\ \text{s}^2 \]Evaluation of k
From \(T^2=\dfrac{4\pi^2}{k}M + C\), the slope is \(s=\dfrac{4\pi^2}{k}\), so \(k=\dfrac{4\pi^2}{s}\). With \(\pi=\dfrac{22}{7}\):
\[ k=\frac{4\pi^2}{s}=\frac{4\times\left(\frac{22}{7}\right)^2}{1.0\times10^{-3}}=\frac{4\times 9.878}{1.0\times10^{-3}}=\frac{39.51}{1.0\times10^{-3}}\approx 3.95\times10^{4} \]Period when M = 180 g
Reading up from \(M=180\ \text{g}\) to the line and across to the vertical axis gives \(T^2=0.23\ \text{s}^2\), so
\[ T=\sqrt{0.23}=0.48\ \text{s} \]Two precautions
- The metre rule was firmly clamped to the bench so that only the projecting part oscillated.
- Parallax error was avoided when reading the stopwatch, the zero error was noted and corrected, and each timing was repeated and averaged.
(b)(i) Simple harmonic motion
Simple harmonic motion is the motion of a body in which its acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point.
(b)(ii) Period and frequency
The period is the time taken by the oscillating body to make one complete oscillation. The frequency is the number of complete oscillations made in one second, related to the period by \(f=\dfrac{1}{T}\).
Frage 2 Bericht
You are provided with a constantan wire, a 2\(\Omega\) standard resistor, an accumulator E, an ammeter A, a key K, and other necessary apparatus.
- Measure and record the e.m.f.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 I = 90cm, close the key. Read and record the ammeter reading /. Evaluate l\(^{-1}\).
- Repeat the procedure for l= 80, 70, 60 and 50cm.
- In each case, read and record the ammeter reading and evaluate l\(^{-1}\). Tabulate your readings.
- Plot a graph of l on the vertical axis against l\(^{-1}\) 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 / = 55cm.
- 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. State two factors on which the resistance of a resistance wire depends.
Test of Practical Knowledge – Electricity
Circuit diagram
Measurements recorded
- E.m.f. of the accumulator, \(E = 2.0\ \text{V}\)
- Ammeter reading with the clip not in contact with the wire, \(I_{o} = 0.65\ \text{A}\)
Table of readings
| \(l\) (cm) | \(l^{-1}\) (cm\(^{-1}\)) | \(I\) (A) |
|---|---|---|
| 90 | 0.0111 | 0.70 |
| 80 | 0.0125 | 0.75 |
| 70 | 0.0143 | 0.80 |
| 60 | 0.0167 | 0.85 |
| 50 | 0.0200 | 0.90 |
Graph of \(I\) against \(l^{-1}\)
Slope of the graph
Taking two widely spaced points on the line of best fit, \((l^{-1}_{1},\,I_{1}) = (0.0100,\ 0.69)\) and \((l^{-1}_{2},\,I_{2}) = (0.0200,\ 0.91)\):
\[ s = \frac{\Delta I}{\Delta (l^{-1})} = \frac{0.91 - 0.69}{0.0200 - 0.0100} = \frac{0.22}{0.0100} = 22.0\ \text{A cm} \]Intercept on the vertical axis
Extending the line of best fit back to \(l^{-1} = 0\), it cuts the \(I\)-axis at
\[ c = 0.47\ \text{A} \]Evaluate \(k = \dfrac{c}{s}\)
\[ k = \frac{c}{s} = \frac{0.47}{22.0} = 0.021 \]Current \(I\) when \(l = 55\ \text{cm}\)
\[ l^{-1} = \frac{1}{55} = 0.0182\ \text{cm}^{-1} \]Reading this value on the horizontal axis and going up to the line of best fit:
\[ I = c + s\,(l^{-1}) = 0.47 + 22.0 \times 0.0182 = 0.87\ \text{A} \]Two precautions
- The key was opened (removed) immediately after each reading was taken, to prevent heating of the constantan wire and to conserve the accumulator.
- The terminals and crocodile-clip contacts were kept clean and the connections firmly tightened to avoid contact resistance, and the ammeter was read with the eye directly above the pointer to avoid parallax.
(b)(i) Potential difference between two points
The potential difference between two points in an electric circuit is the work done 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 resistance wire depends
- The length of the wire – resistance increases with length \((R \propto l)\).
- The cross-sectional area (thickness) of the wire – resistance decreases as the area increases \(\left(R \propto \dfrac{1}{A}\right)\).
(It also depends on the resistivity/nature of the material and on the temperature of the wire.)
Antwortdetails
Test of Practical Knowledge – Electricity
Circuit diagram
Measurements recorded
- E.m.f. of the accumulator, \(E = 2.0\ \text{V}\)
- Ammeter reading with the clip not in contact with the wire, \(I_{o} = 0.65\ \text{A}\)
Table of readings
| \(l\) (cm) | \(l^{-1}\) (cm\(^{-1}\)) | \(I\) (A) |
|---|---|---|
| 90 | 0.0111 | 0.70 |
| 80 | 0.0125 | 0.75 |
| 70 | 0.0143 | 0.80 |
| 60 | 0.0167 | 0.85 |
| 50 | 0.0200 | 0.90 |
Graph of \(I\) against \(l^{-1}\)
Slope of the graph
Taking two widely spaced points on the line of best fit, \((l^{-1}_{1},\,I_{1}) = (0.0100,\ 0.69)\) and \((l^{-1}_{2},\,I_{2}) = (0.0200,\ 0.91)\):
\[ s = \frac{\Delta I}{\Delta (l^{-1})} = \frac{0.91 - 0.69}{0.0200 - 0.0100} = \frac{0.22}{0.0100} = 22.0\ \text{A cm} \]Intercept on the vertical axis
Extending the line of best fit back to \(l^{-1} = 0\), it cuts the \(I\)-axis at
\[ c = 0.47\ \text{A} \]Evaluate \(k = \dfrac{c}{s}\)
\[ k = \frac{c}{s} = \frac{0.47}{22.0} = 0.021 \]Current \(I\) when \(l = 55\ \text{cm}\)
\[ l^{-1} = \frac{1}{55} = 0.0182\ \text{cm}^{-1} \]Reading this value on the horizontal axis and going up to the line of best fit:
\[ I = c + s\,(l^{-1}) = 0.47 + 22.0 \times 0.0182 = 0.87\ \text{A} \]Two precautions
- The key was opened (removed) immediately after each reading was taken, to prevent heating of the constantan wire and to conserve the accumulator.
- The terminals and crocodile-clip contacts were kept clean and the connections firmly tightened to avoid contact resistance, and the ammeter was read with the eye directly above the pointer to avoid parallax.
(b)(i) Potential difference between two points
The potential difference between two points in an electric circuit is the work done 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 resistance wire depends
- The length of the wire – resistance increases with length \((R \propto l)\).
- The cross-sectional area (thickness) of the wire – resistance decreases as the area increases \(\left(R \propto \dfrac{1}{A}\right)\).
(It also depends on the resistivity/nature of the material and on the temperature of the wire.)
Frage 3 Bericht
You have been provided with a rectangular glass prism, optical pins, and other necessary apparatus. Using the above diagram as a guide, carry out the following instructions:
- Fix the drawing paper provided to the drawing board
- Place the glass prism on the drawing paper and trace the outline, ABCD of the prism
- Remove the prism, mark a point O on AB such that AO is about one-quarter of AB
- Draw a normal through point O. Also draw an incident ray to make an angle i = 25 with the normal at O. Fix two pins at P\(_{1}\) and P\(_{2}\) On the incident ray.
- Replace the prism. Fix two other pins at P\(_{3}\) and P\(_{4}\) such that the pins appear to be in a straight line with the images of the pins at P\(_{1}\) and P\(_{2}\) when viewed through the block along DC
- remove the prism. Join points Pa and P4 and produce it to meet DC at 1. Also, draw a line to join Ol (
- With O as center and using any Concinient radius, draw a circle to Cut the incident ray and the refracted ray at E and H respectively. Maintain this radius throughout the experiment
- Draw the perpendiculars EF and GH. Measure and record d= EF and I= GH.
- Repeat the procedure for four other values of i = 35°, 45, 55°, and 65° respectively. In each case measure and record d and I
- Plot a graph of d on the vertical axis against I on the horizontal axis
- Determine the slope of the graph
- State two precautions taken to ensure accurate results. [Attach your traces to your answer booklet)
(b)i. State Snell's law.
ii. Calculate the critical angle for a water-air interface. [refractive index of water = \(\frac{4}{3}\)]
Principle: The construction measures quantities proportional to \(\sin i\) and \(\sin r\), where \(i\) is the angle of incidence and \(r\) is the angle of refraction in the glass.
Using the same circle radius \(R\) for every trial:
\[ d=EF=R\sin i \] \[ l=GH=R\sin r \]Therefore, by Snell’s law,
\[ \frac{d}{l}=\frac{R\sin i}{R\sin r}=\frac{\sin i}{\sin r}=n \]Thus, a graph of \(d\) on the vertical axis against \(l\) on the horizontal axis should be a straight line passing approximately through the origin. Its gradient is the refractive index \(n\) of the glass relative to air.
Observation table
| S/N | \(i\) / ° | \(d=EF\) / cm | \(l=GH\) / cm |
|---|---|---|---|
| 1 | 25 | 0.85 | 0.55 |
| 2 | 35 | 1.15 | 0.75 |
| 3 | 45 | 1.50 | 0.95 |
| 4 | 55 | 1.65 | 1.10 |
| 5 | 65 | 1.85 | 1.20 |
Graph of \(d\) against \(l\)
The plotted points are close to a straight line. Using a best-fit line through the origin gives a gradient of approximately \(1.54\), so the refractive index of the glass is approximately:
\[ n \approx 1.5 \text{ to } 1.6 \]The value \(1.56\) is a reasonable graph-reading estimate. However, the stated numerical table does not itself give exactly \(\frac{1.4}{0.9}\); the gradient must be taken from two widely separated points on the drawn best-fit line, not necessarily directly from two raw data points.
Precautions
- Keep the radius of the circle centred at \(O\) the same for all readings, so that \(d\) and \(l\) remain proportional to \(\sin i\) and \(\sin r\).
- Fix all optical pins vertically.
- Keep the pins reasonably far apart and align \(P_3\) and \(P_4\) with the images of \(P_1\) and \(P_2\) without parallax.
- Use a sharp pencil and make thin, accurate ray lines and normal lines.
Snell’s law: For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant:
\[ \frac{\sin i}{\sin r}=n \]Critical angle for a water–air boundary: Given \(n=\frac{4}{3}\),
\[ \sin C=\frac{1}{n}=\frac{1}{4/3}=\frac{3}{4}=0.75 \] \[ C=\sin^{-1}(0.75)=48.6^\circ \]Therefore, the critical angle is \(48.6^\circ\), approximately \(49^\circ\).
Examination reminder: The vertical quantity is \(d=R\sin i\) and the horizontal quantity is \(l=R\sin r\). Therefore the gradient is \(\frac{d}{l}=\frac{\sin i}{\sin r}\), which is the refractive index.
Antwortdetails
Principle: The construction measures quantities proportional to \(\sin i\) and \(\sin r\), where \(i\) is the angle of incidence and \(r\) is the angle of refraction in the glass.
Using the same circle radius \(R\) for every trial:
\[ d=EF=R\sin i \] \[ l=GH=R\sin r \]Therefore, by Snell’s law,
\[ \frac{d}{l}=\frac{R\sin i}{R\sin r}=\frac{\sin i}{\sin r}=n \]Thus, a graph of \(d\) on the vertical axis against \(l\) on the horizontal axis should be a straight line passing approximately through the origin. Its gradient is the refractive index \(n\) of the glass relative to air.
Observation table
| S/N | \(i\) / ° | \(d=EF\) / cm | \(l=GH\) / cm |
|---|---|---|---|
| 1 | 25 | 0.85 | 0.55 |
| 2 | 35 | 1.15 | 0.75 |
| 3 | 45 | 1.50 | 0.95 |
| 4 | 55 | 1.65 | 1.10 |
| 5 | 65 | 1.85 | 1.20 |
Graph of \(d\) against \(l\)
The plotted points are close to a straight line. Using a best-fit line through the origin gives a gradient of approximately \(1.54\), so the refractive index of the glass is approximately:
\[ n \approx 1.5 \text{ to } 1.6 \]The value \(1.56\) is a reasonable graph-reading estimate. However, the stated numerical table does not itself give exactly \(\frac{1.4}{0.9}\); the gradient must be taken from two widely separated points on the drawn best-fit line, not necessarily directly from two raw data points.
Precautions
- Keep the radius of the circle centred at \(O\) the same for all readings, so that \(d\) and \(l\) remain proportional to \(\sin i\) and \(\sin r\).
- Fix all optical pins vertically.
- Keep the pins reasonably far apart and align \(P_3\) and \(P_4\) with the images of \(P_1\) and \(P_2\) without parallax.
- Use a sharp pencil and make thin, accurate ray lines and normal lines.
Snell’s law: For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant:
\[ \frac{\sin i}{\sin r}=n \]Critical angle for a water–air boundary: Given \(n=\frac{4}{3}\),
\[ \sin C=\frac{1}{n}=\frac{1}{4/3}=\frac{3}{4}=0.75 \] \[ C=\sin^{-1}(0.75)=48.6^\circ \]Therefore, the critical angle is \(48.6^\circ\), approximately \(49^\circ\).
Examination reminder: The vertical quantity is \(d=R\sin i\) and the horizontal quantity is \(l=R\sin r\). Therefore the gradient is \(\frac{d}{l}=\frac{\sin i}{\sin r}\), which is the refractive index.
