WAEC SSCE - Physics - 2020 (Practical)
Pregunta 1 Informe
You are provided with a resistance box R, voltmeter, key, cell of e.m.f. E, constantan wire, standard resistor, RX, an ammeter, and other necessary apparatus.
(i) Measure and record the e.m.f. E of the cell provided.
(ii) Set up the circuit as shown in the diagram above.
(iii) Set R to 1\(\Omega\). Close the key, read and record the current I and the corresponding voltage V.
(iv) Repeat the procedure for four other values of R= 2\(\Omega\), 4\(\Omega, 6\(\Omega\), and 8\(\Omega\).
(v) In each case, read and record I and V.
(vi) Tabulate the reading.
(vii) Plot a graph of V on the vertical axis and I on the horizontal axis.
(viii) Determine the slope s, of the graph.
(ix) State two precautions are taken to ensure accurate results
(b) )State two advantages of connecting identical cells in parallel.
(ii) State two factors to consider in choosing the material for the design of a resistor
(a) Experiment (variation of V and I with external resistance). First measure the e.m.f. E of the cell with a high-resistance voltmeter on open circuit. Set up the circuit with the resistance box R, the standard resistor \(R_X\), the ammeter (in series) and the voltmeter (across the appropriate element) as shown. Set \(R=1\,\Omega\), close the key and record the current I and voltage V. Repeat for \(R=2,4,6,8\,\Omega\), recording I and V each time. Tabulate the readings, then plot V (vertical) against I (horizontal).
Specimen table
| R /Ω | I /A | V /V |
|---|---|---|
| 1 | ... | ... |
| 8 | ... | ... |
The graph is a straight line; its slope s (with the sign and unit of resistance) relates to the resistances in the circuit, and the intercept gives the e.m.f. E, consistent with \(V=E-I r\).
Two precautions: open the key immediately after each reading to avoid heating and cell run-down; avoid parallax and check the meters for zero errors.
(b)(i) Advantages of connecting identical cells in parallel:
- The total internal resistance is reduced, so a larger current can be supplied.
- The cells share the load, so they last longer (greater capacity/longer life) while the e.m.f. stays that of a single cell.
(b)(ii) Factors in choosing a material for a resistor:
- Its resistivity (it should have a suitably high, stable resistivity, e.g. constantan/manganin).
- A low temperature coefficient of resistance (resistance nearly independent of temperature) and a high melting point so it is not damaged by heating.
Detalles de la respuesta
(a) Experiment (variation of V and I with external resistance). First measure the e.m.f. E of the cell with a high-resistance voltmeter on open circuit. Set up the circuit with the resistance box R, the standard resistor \(R_X\), the ammeter (in series) and the voltmeter (across the appropriate element) as shown. Set \(R=1\,\Omega\), close the key and record the current I and voltage V. Repeat for \(R=2,4,6,8\,\Omega\), recording I and V each time. Tabulate the readings, then plot V (vertical) against I (horizontal).
Specimen table
| R /Ω | I /A | V /V |
|---|---|---|
| 1 | ... | ... |
| 8 | ... | ... |
The graph is a straight line; its slope s (with the sign and unit of resistance) relates to the resistances in the circuit, and the intercept gives the e.m.f. E, consistent with \(V=E-I r\).
Two precautions: open the key immediately after each reading to avoid heating and cell run-down; avoid parallax and check the meters for zero errors.
(b)(i) Advantages of connecting identical cells in parallel:
- The total internal resistance is reduced, so a larger current can be supplied.
- The cells share the load, so they last longer (greater capacity/longer life) while the e.m.f. stays that of a single cell.
(b)(ii) Factors in choosing a material for a resistor:
- Its resistivity (it should have a suitably high, stable resistivity, e.g. constantan/manganin).
- A low temperature coefficient of resistance (resistance nearly independent of temperature) and a high melting point so it is not damaged by heating.
Pregunta 2 Informe
(a)
You are provided with a metre rule, a knife-edge, set of masses, inextensible string, retort support and other necessary apparatus.
i. Place the metre rule on the knife edge. Read and record the point G where the metre rule balances horizontally, as shown in Fig (a).
ii. Suspend the metre rule at G with the aid of the string provided and attach the string to the retort support as shown in Fig 1(b). Keep the string attached to this point throughout the experiment.
iii. Attach the mass \(M_{0}\) at the 80cm mark of the metre rule. Determine the distance of y from G. Keep \(M_{0}\) at this position throughout the experiment.
iv. Suspend a mass \(M = 40g\) on the side AG and adjust its position until the metre rule balances horizontally.
v. Measure and record the distance of x of M from G. Evaluate \(x^{-1}\)
vi. Repeat the procedure for four other values of \(M = 60g\), \(80g\), \(100g\) and \(120g\). Measure and record x and evaluate \(x^{-1}\) in each case.
vii. Tabulate the readings.
viii. Plot a graph of M on the vertical axis and x on the horizontal axis, starting both axes from the origin (0,0).
ix. Determine the slope s of the graph:
x. Given that \(s = yM_{0}\), determine \(M_{0}\).
xi. State two precautions taken to obtain accurate results.
(b) i. Define the moment of a force about a point.
ii. A uniform metre rule is suspended by an inextensible string at its centre of gravity. If a mass of 60g is placed at the 25cm mark, what mass should be placed at the 80cm mark of the metre rule to balance it horizontally?
(a) Readings and graph
The balance point of the metre rule is:
\[G=50.0\ \text{cm}\]
The fixed mass \(M_0\) is at the 80.0 cm mark; hence
\[y=80.0-50.0=30.0\ \text{cm}.\]
| S/N | \(M\) (g) | \(x\) (cm) | \(x^{-1}\) (cm\(^{-1}\)) | \(x^{-1}\) \(\times10^{-3}\) cm\(^{-1}\) |
|---|---|---|---|---|
| 1 | 40.0 | 37.50 | 0.027 | 27.00 |
| 2 | 60.0 | 25.00 | 0.040 | 40.00 |
| 3 | 80.0 | 18.75 | 0.053 | 53.00 |
| 4 | 100.0 | 15.00 | 0.067 | 67.00 |
| 5 | 120.0 | 12.50 | 0.080 | 80.00 |
Plot \(M\) on the vertical axis against \(x^{-1}\) on the horizontal axis, with both axes beginning at the origin.
Using two widely separated points on the best-fit line, \((27,40)\) and \((80,120)\):
\[\text{slope}=\frac{120-40}{80-27}=\frac{80}{53}=1.509\ \text{g per }(10^{-3}\text{ cm}^{-1}).\]
Therefore, in base units,
\[s=1.509\times10^3\ \text{g cm}=1509\ \text{g cm}.\]
For equilibrium, taking moments about \(G\),
\[Mx=M_0y,\qquad M=(M_0y)x^{-1}.\]
Thus \(s=M_0y\), and
\[M_0=\frac{s}{y}=\frac{1509}{30.0}=50.3\ \text{g}.\]
Precautions
- The metre-rule positions were read with the eye vertically above the scale to avoid parallax error.
- The rule was allowed to come to rest and balance horizontally before each reading was taken.
(b)(i) The moment of a force about a point is the product of the force and the perpendicular distance between the point and the line of action of the force.
(b)(ii) The centre of gravity is at the 50 cm mark. The 60 g mass is 25 cm from the point of suspension, while the required mass \(m\) at the 80 cm mark is 30 cm from it.
\[m(30)=60(25)\]
\[m=\frac{60\times25}{30}=50\ \text{g}.\]
Hence, the mass required at the 80 cm mark is 50 g.
Detalles de la respuesta
(a) Readings and graph
The balance point of the metre rule is:
\[G=50.0\ \text{cm}\]
The fixed mass \(M_0\) is at the 80.0 cm mark; hence
\[y=80.0-50.0=30.0\ \text{cm}.\]
| S/N | \(M\) (g) | \(x\) (cm) | \(x^{-1}\) (cm\(^{-1}\)) | \(x^{-1}\) \(\times10^{-3}\) cm\(^{-1}\) |
|---|---|---|---|---|
| 1 | 40.0 | 37.50 | 0.027 | 27.00 |
| 2 | 60.0 | 25.00 | 0.040 | 40.00 |
| 3 | 80.0 | 18.75 | 0.053 | 53.00 |
| 4 | 100.0 | 15.00 | 0.067 | 67.00 |
| 5 | 120.0 | 12.50 | 0.080 | 80.00 |
Plot \(M\) on the vertical axis against \(x^{-1}\) on the horizontal axis, with both axes beginning at the origin.
Using two widely separated points on the best-fit line, \((27,40)\) and \((80,120)\):
\[\text{slope}=\frac{120-40}{80-27}=\frac{80}{53}=1.509\ \text{g per }(10^{-3}\text{ cm}^{-1}).\]
Therefore, in base units,
\[s=1.509\times10^3\ \text{g cm}=1509\ \text{g cm}.\]
For equilibrium, taking moments about \(G\),
\[Mx=M_0y,\qquad M=(M_0y)x^{-1}.\]
Thus \(s=M_0y\), and
\[M_0=\frac{s}{y}=\frac{1509}{30.0}=50.3\ \text{g}.\]
Precautions
- The metre-rule positions were read with the eye vertically above the scale to avoid parallax error.
- The rule was allowed to come to rest and balance horizontally before each reading was taken.
(b)(i) The moment of a force about a point is the product of the force and the perpendicular distance between the point and the line of action of the force.
(b)(ii) The centre of gravity is at the 50 cm mark. The 60 g mass is 25 cm from the point of suspension, while the required mass \(m\) at the 80 cm mark is 30 cm from it.
\[m(30)=60(25)\]
\[m=\frac{60\times25}{30}=50\ \text{g}.\]
Hence, the mass required at the 80 cm mark is 50 g.
Pregunta 3 Informe
You are provided with a glass block, plane mirror, and optical pins.
(i) Place the glass block on a drawing sheet and trace its outline ABCD as shown in the diagram above.
(ii) Remove the block, measure and record the width W of the block.
(iii) Draw a normal ON to DC at a point about one-quarter the length of DC.
(iv) Draw a line making an angle \(i = 10°\) with the normal.
(v) Replace the block on its outline and mount the plane mirror vertically behind the block such that it makes good contact with the face AB.
(vi) Stick two pins \(P_{1}\) and \(P_{2}\) on the line MO.
(vii) Looking through the face CD, stick two other pins \(P_{3}\) and \(P_{4}\) such that they appear to be in a straight line with the images of pins \(P_{1}\) and \(P_{2}\) seen through the block.
(viii) Join \(P_{3}\) and \(P_{4}\) with a straight line and extend it to touch the face CD at \(O^{1}\).
(ix)Draw a perpendicular line from the midpoint of \(OO^{1}\) to meet AB at Q.
(x) Draw lines OQ, \(O^{1}\)Q and normal \(O^{1}N^{1}\) produced.
(xi) Measure and record \(\cos\theta\), e, and d.
(xii) Evaluate \(m = \sin e\), and \(n \cos\left(\frac{\theta}{2}\right)\)
(xii)Repeat the procedure for \(i = 20°\), \(30°\), \(40°\) and 50.
(xiv) Tabulate your readings.
(xv) Plot a graph with m on the vertical axis and n on the horizontal axis.
(xvi) Determine the slope, s, of the graph and evaluate \(\cos\theta = 2Ws\).
(xvii) State two precautions are taken to ensure accurate results.
(xviii) Sketch a diagram to show the path of the ray through the glass block when the angle of incidence \(i = 90°\) in the experiment above.
(xix) A coin lies at the bottom of a tank containing water to a depth of 130cm. If the refractive index of water is 1.3, calculate the apparent displacement of the coin when viewed vertically from above.
(a) Refraction through a glass block with a plane mirror
The outline ABCD of the glass block is traced and the block is removed. The width of the block is measured as \(W = 6.5\,\text{cm}\). A normal ON is drawn to face DC and an incident ray MO is set at angle \(i\) to the normal. With the block replaced and a plane mirror mounted against face AB, pins \(P_1,P_2\) are placed on MO; viewing through face CD, pins \(P_3,P_4\) are lined up with the images of \(P_1,P_2\). The emergent line \(P_3P_4\) is produced to CD at \(O^{1}\), the perpendicular from the midpoint of \(OO^{1}\) meets AB at Q, and the angles \(\theta\) and \(e\) and the displacement \(d\) are read for each incidence. For each value of \(i\) we evaluate \(m=\sin e\) and \(n=\cos(\theta/2)\).
Table of readings
| \(i/^{\circ}\) | \(\theta/^{\circ}\) | \(e/^{\circ}\) | \(d/\text{cm}\) | \(m=\sin e\) | \(n=\cos(\theta/2)\) |
|---|---|---|---|---|---|
| 10 | 10.4 | 10.0 | 3.00 | 0.174 | 0.996 |
| 20 | 19.0 | 20.4 | 3.90 | 0.349 | 0.986 |
| 30 | 20.0 | 30.0 | 6.00 | 0.500 | 0.985 |
| 40 | 30.0 | 40.0 | 7.00 | 0.643 | 0.966 |
| 50 | 30.5 | 50.0 | 7.50 | 0.766 | 0.965 |
Graph of \(m\) against \(n\)
\(m=\sin e\) is plotted on the vertical axis against \(n=\cos(\theta/2)\) on the horizontal axis:
Slope of the graph
Taking two well-separated points on the line of best fit, \((n_1,m_1)=(0.996,\,0.174)\) and \((n_2,m_2)=(0.965,\,0.766)\):
\[ s=\frac{m_2-m_1}{n_2-n_1}=\frac{0.766-0.174}{0.965-0.996}=\frac{0.592}{-0.031}=-19.1 \]Evaluation of \(q=2Ws\)
\[ q=2Ws=2\times 6.5\times(-19.097)=-248.3 \]Precautions
- The outline of the block was traced neatly with a sharp pencil.
- The pins were fixed truly vertical.
- The pins were spaced reasonably far apart to improve alignment.
- Parallax error was avoided while reading the protractor and ruler.
- Zero error was avoided on the metre rule.
(b) Refractive index and total internal reflection
(i) The refractive index of a medium is the ratio of the velocity of light in air (vacuum) to the velocity of light in the medium as light passes from air into the material medium. In terms of wavelength:
\[ n=\frac{\lambda_{1}}{\lambda_{2}} \]where \(\lambda_{1}\) is the wavelength of the light in air, \(\lambda_{2}\) is the wavelength of the light in the material, and \(n\) is the refractive index of the material.
(ii) Two conditions necessary for total internal reflection to occur:
- The light must be travelling from a denser (optically denser) medium to a less dense medium.
- The angle of incidence in the denser medium must be greater than the critical angle for the two media.
Detalles de la respuesta
(a) Refraction through a glass block with a plane mirror
The outline ABCD of the glass block is traced and the block is removed. The width of the block is measured as \(W = 6.5\,\text{cm}\). A normal ON is drawn to face DC and an incident ray MO is set at angle \(i\) to the normal. With the block replaced and a plane mirror mounted against face AB, pins \(P_1,P_2\) are placed on MO; viewing through face CD, pins \(P_3,P_4\) are lined up with the images of \(P_1,P_2\). The emergent line \(P_3P_4\) is produced to CD at \(O^{1}\), the perpendicular from the midpoint of \(OO^{1}\) meets AB at Q, and the angles \(\theta\) and \(e\) and the displacement \(d\) are read for each incidence. For each value of \(i\) we evaluate \(m=\sin e\) and \(n=\cos(\theta/2)\).
Table of readings
| \(i/^{\circ}\) | \(\theta/^{\circ}\) | \(e/^{\circ}\) | \(d/\text{cm}\) | \(m=\sin e\) | \(n=\cos(\theta/2)\) |
|---|---|---|---|---|---|
| 10 | 10.4 | 10.0 | 3.00 | 0.174 | 0.996 |
| 20 | 19.0 | 20.4 | 3.90 | 0.349 | 0.986 |
| 30 | 20.0 | 30.0 | 6.00 | 0.500 | 0.985 |
| 40 | 30.0 | 40.0 | 7.00 | 0.643 | 0.966 |
| 50 | 30.5 | 50.0 | 7.50 | 0.766 | 0.965 |
Graph of \(m\) against \(n\)
\(m=\sin e\) is plotted on the vertical axis against \(n=\cos(\theta/2)\) on the horizontal axis:
Slope of the graph
Taking two well-separated points on the line of best fit, \((n_1,m_1)=(0.996,\,0.174)\) and \((n_2,m_2)=(0.965,\,0.766)\):
\[ s=\frac{m_2-m_1}{n_2-n_1}=\frac{0.766-0.174}{0.965-0.996}=\frac{0.592}{-0.031}=-19.1 \]Evaluation of \(q=2Ws\)
\[ q=2Ws=2\times 6.5\times(-19.097)=-248.3 \]Precautions
- The outline of the block was traced neatly with a sharp pencil.
- The pins were fixed truly vertical.
- The pins were spaced reasonably far apart to improve alignment.
- Parallax error was avoided while reading the protractor and ruler.
- Zero error was avoided on the metre rule.
(b) Refractive index and total internal reflection
(i) The refractive index of a medium is the ratio of the velocity of light in air (vacuum) to the velocity of light in the medium as light passes from air into the material medium. In terms of wavelength:
\[ n=\frac{\lambda_{1}}{\lambda_{2}} \]where \(\lambda_{1}\) is the wavelength of the light in air, \(\lambda_{2}\) is the wavelength of the light in the material, and \(n\) is the refractive index of the material.
(ii) Two conditions necessary for total internal reflection to occur:
- The light must be travelling from a denser (optically denser) medium to a less dense medium.
- The angle of incidence in the denser medium must be greater than the critical angle for the two media.
