WAEC SSCE - Physics - 2001 (Practical)
Swali 1 Ripoti
You are provided with two wires marked P and C. a resistor \(R_s\) = \(1\Omega\) and other necessary apparatus.
- Connect \(R_s\) in the left-hand gap of the metre bridge, a length L = 100cm of wire P in the right-hand gap and the other apparatus as shown in the diagram: above
- Determine the balance point B on the bridge wire AC
- Measure and record AB =/s, and BC = /
- Evaluate \(R_1\) = \(\left(\frac{|_{p}}{|_{s}}\right)Rs\)
- Repeat the procedure for four other values of L = 90, 80, 70 and 60cm. In each case obtain and record the value of \(|_{s}\) and \(|_{p}\) and evaluate \(R_1\) = \(\left(\frac{|_{p}}{|_{s}}\right)Rs\)
- Repeat the experiment with the second wire, Q. Obtain the value of \(|_{s}\) and \(|_{Q}\) for equal lengths of wire as used in wire P.
- Evaluate \(R_1\) = \(\left(\frac{|_{p}}{|_{s}}\right)Rs\). In each case, tabulate your readings.
- Plot a graph of \(R_2\) on the Vertical axis against \(R_1\) on the horizontal axis.
- Determine the slope S, of the graph.
- Evaluate the k = \(\sqrt s\).
- State two precautions taken to ensure accurate results.
(b)i. Define the resistivity of the material of a wire.
ii. A galvanometer with a full-scale-deflection of 1.5 x \(10^{3}\). A has a resistance of \(50\Omega\). Determine the resistance required to convert it into a voltmeter reading up to 1.5V.
Metre-bridge determination of wire resistances
The standard resistor \(R_s = 1\,\Omega\) is connected in the left-hand gap and a length \(L\) of the test wire in the right-hand gap. The jockey is moved along the bridge wire AC until the galvanometer shows no deflection at the balance point B. Writing \(l_s = AB\) and \(l_p = BC\), the unknown resistance is
\[ R = \left(\frac{l_p}{l_s}\right)R_s .\]The reading is taken for \(L = 100, 90, 80, 70\) and \(60\ \text{cm}\), first with wire P (giving \(R_1\)) and then with wire Q at the same lengths (giving \(R_2\)).
Table of readings
| \(L\)/cm | \(l_s\)/cm (P) | \(l_p\)/cm | \(R_1\)/\(\Omega\) | \(l_s\)/cm (Q) | \(l_q\)/cm | \(R_2\)/\(\Omega\) |
|---|---|---|---|---|---|---|
| 100 | 5.3 | 94.7 | 17.90 | 21.6 | 78.4 | 3.60 |
| 90 | 6.5 | 93.5 | 14.40 | 22.1 | 77.9 | 3.50 |
| 80 | 7.0 | 93.0 | 13.30 | 23.4 | 76.6 | 3.30 |
| 70 | 8.5 | 91.5 | 10.80 | 24.3 | 75.7 | 3.10 |
| 60 | 9.5 | 90.5 | 9.50 | 25.3 | 74.7 | 2.95 |
Sample evaluation (L = 100 cm, wire P): \(R_1 = \dfrac{l_p}{l_s}R_s = \dfrac{94.7}{5.3}\times 1 = 17.90\,\Omega\); and for wire Q: \(R_2 = \dfrac{74.7\ \text{to}\ 78.4}{21.6}\times 1\), e.g. \(\dfrac{78.4}{21.6}\times 1 = 3.60\,\Omega\).
Graph of \(R_2\) against \(R_1\)
Plotting \(R_2\) (vertical axis) against \(R_1\) (horizontal axis) gives a straight line, showing that \(R_2\) increases uniformly with \(R_1\).
Slope of the graph
Taking two well-separated points on the line of best fit, \((R_1, R_2) = (9.5,\ 3.00)\) and \((17.9,\ 3.67)\):
\[ S = \frac{\Delta R_2}{\Delta R_1} = \frac{3.67 - 3.00}{17.9 - 9.5} = \frac{0.67}{8.4} = 0.080 .\]Hence
\[ k = \sqrt{S} = \sqrt{0.080} = 0.28 .\]Two precautions
- The key was opened when readings were not being taken, so that the wires did not heat up and change their resistance; all connections were kept clean and tight.
- The jockey was tapped gently (not slid or allowed to scratch the bridge wire) and the balance length was read with the eye directly above the wire to avoid parallax error.
(b)(i) Resistivity
The resistivity of the material of a wire is the resistance of a specimen of the material having unit length and unit cross-sectional area. From \(R = \dfrac{\rho L}{A}\),
\[ \rho = \frac{RA}{L},\]where \(A\) is the cross-sectional area, \(R\) the resistance and \(L\) the length of the wire. Its SI unit is the ohm-metre \((\Omega\,\text{m})\).
(b)(ii) Converting the galvanometer to a voltmeter
Full-scale-deflection current \(I_g = 1.5\times10^{-3}\,\text{A}\), coil resistance \(G = 50\,\Omega\), required full-scale reading \(V = 1.5\,\text{V}\). A multiplier resistance \(R\) is connected in series with the galvanometer:
\[ V = I_g(G + R) \;\Rightarrow\; G + R = \frac{V}{I_g} = \frac{1.5}{1.5\times10^{-3}} = 1000\,\Omega .\]\[ R = 1000 - 50 = 950\,\Omega .\]A resistance of \(\mathbf{950\,\Omega}\) must be connected in series with the galvanometer.
Maelezo ya Majibu
Metre-bridge determination of wire resistances
The standard resistor \(R_s = 1\,\Omega\) is connected in the left-hand gap and a length \(L\) of the test wire in the right-hand gap. The jockey is moved along the bridge wire AC until the galvanometer shows no deflection at the balance point B. Writing \(l_s = AB\) and \(l_p = BC\), the unknown resistance is
\[ R = \left(\frac{l_p}{l_s}\right)R_s .\]The reading is taken for \(L = 100, 90, 80, 70\) and \(60\ \text{cm}\), first with wire P (giving \(R_1\)) and then with wire Q at the same lengths (giving \(R_2\)).
Table of readings
| \(L\)/cm | \(l_s\)/cm (P) | \(l_p\)/cm | \(R_1\)/\(\Omega\) | \(l_s\)/cm (Q) | \(l_q\)/cm | \(R_2\)/\(\Omega\) |
|---|---|---|---|---|---|---|
| 100 | 5.3 | 94.7 | 17.90 | 21.6 | 78.4 | 3.60 |
| 90 | 6.5 | 93.5 | 14.40 | 22.1 | 77.9 | 3.50 |
| 80 | 7.0 | 93.0 | 13.30 | 23.4 | 76.6 | 3.30 |
| 70 | 8.5 | 91.5 | 10.80 | 24.3 | 75.7 | 3.10 |
| 60 | 9.5 | 90.5 | 9.50 | 25.3 | 74.7 | 2.95 |
Sample evaluation (L = 100 cm, wire P): \(R_1 = \dfrac{l_p}{l_s}R_s = \dfrac{94.7}{5.3}\times 1 = 17.90\,\Omega\); and for wire Q: \(R_2 = \dfrac{74.7\ \text{to}\ 78.4}{21.6}\times 1\), e.g. \(\dfrac{78.4}{21.6}\times 1 = 3.60\,\Omega\).
Graph of \(R_2\) against \(R_1\)
Plotting \(R_2\) (vertical axis) against \(R_1\) (horizontal axis) gives a straight line, showing that \(R_2\) increases uniformly with \(R_1\).
Slope of the graph
Taking two well-separated points on the line of best fit, \((R_1, R_2) = (9.5,\ 3.00)\) and \((17.9,\ 3.67)\):
\[ S = \frac{\Delta R_2}{\Delta R_1} = \frac{3.67 - 3.00}{17.9 - 9.5} = \frac{0.67}{8.4} = 0.080 .\]Hence
\[ k = \sqrt{S} = \sqrt{0.080} = 0.28 .\]Two precautions
- The key was opened when readings were not being taken, so that the wires did not heat up and change their resistance; all connections were kept clean and tight.
- The jockey was tapped gently (not slid or allowed to scratch the bridge wire) and the balance length was read with the eye directly above the wire to avoid parallax error.
(b)(i) Resistivity
The resistivity of the material of a wire is the resistance of a specimen of the material having unit length and unit cross-sectional area. From \(R = \dfrac{\rho L}{A}\),
\[ \rho = \frac{RA}{L},\]where \(A\) is the cross-sectional area, \(R\) the resistance and \(L\) the length of the wire. Its SI unit is the ohm-metre \((\Omega\,\text{m})\).
(b)(ii) Converting the galvanometer to a voltmeter
Full-scale-deflection current \(I_g = 1.5\times10^{-3}\,\text{A}\), coil resistance \(G = 50\,\Omega\), required full-scale reading \(V = 1.5\,\text{V}\). A multiplier resistance \(R\) is connected in series with the galvanometer:
\[ V = I_g(G + R) \;\Rightarrow\; G + R = \frac{V}{I_g} = \frac{1.5}{1.5\times10^{-3}} = 1000\,\Omega .\]\[ R = 1000 - 50 = 950\,\Omega .\]A resistance of \(\mathbf{950\,\Omega}\) must be connected in series with the galvanometer.
Swali 2 Ripoti
- Fix the 100g mass marked P at B, the 80 cm mark of the uniform metre rule, using an adhesive.
- Suspend another 100g mass marked Q at A, a distance V = 10.Ocm from the 0 cm mark of the metre rule.
- Balance the whole arrangement horizontally on a knife edge as shown in the diagram above.
- Measure and record the distance U of K from the 0 cm mark of the metre rule.
- Repeat the procedure for five other values of V = 15.0, 20.0, 25.0, 30.0 and 35.0 cm
- In each case, measure and record the Corresponding values of U. Tabulate your readings.
- Plot a graph of U on the vertical axis against V on the horizontal axis.
- determine the:
(1) slope, s, of the graph
(2) intercept c, on the vertical axis. - Evaluate (i) \(K_{1} = \frac{(1 - 2s)}{s} 100\):
(ii) \(K_{2} = \frac{2c}{s} - 160\) - State two precautions taken to ensure an accurate result
(b)i. State two conditions under which a rigid body at rest remains in equilibrium when acted upon by three non-parallel coplanar forces.
ii. Explain how the position of the centre or gravity of a body affects the equilibrium of the body.
Test of practical knowledge: balancing a metre rule (principle of moments)
The mass P = 100 g is fixed with adhesive at B, the 80.0 cm mark. The second mass Q = 100 g is suspended at A, a distance V from the 0 cm mark. For each value of V the knife edge K is moved until the rule balances horizontally, and its distance U from the 0 cm mark is read.
Table of readings
| V / cm | U / cm |
|---|---|
| 10.0 | 47.0 |
| 15.0 | 48.3 |
| 20.0 | 49.4 |
| 25.0 | 51.2 |
| 30.0 | 52.5 |
| 35.0 | 54.0 |
Graph of U against V
Plotting U on the vertical axis against V on the horizontal axis gives a straight line of best fit through the points.
(1) Slope of the graph
Taking two widely separated points on the line of best fit, \((V_1,U_1)=(10.0,\;47.0)\) and \((V_2,U_2)=(35.0,\;54.0)\):
\[s=\frac{\Delta U}{\Delta V}=\frac{54.0-47.0}{35.0-10.0}=\frac{7.0}{25.0}=0.28.\](2) Intercept on the vertical axis
Extending the line of best fit back to \(V=0\) gives
\[c=44\;\text{cm}.\]Evaluation
\[K_{1}=\frac{(1-2s)}{s}\times100=\frac{1-(2\times0.28)}{0.28}\times100=\frac{1-0.56}{0.28}\times100=\frac{0.44}{0.28}\times100=157.\]\[K_{2}=\frac{2c}{s}-160=\frac{2\times44}{0.28}-160=\frac{88}{0.28}-160=314.3-160=154.3.\]Two precautions
- Errors due to parallax were avoided by reading the position of the knife edge on the metre rule with the line of sight directly above the mark.
- Each suspended mass was checked to hang freely without touching the table, and the arrangement was shielded from draught so that it settled to a true balance.
(b)(i) Conditions for equilibrium under three non-parallel coplanar forces
- The three forces must be concurrent - their lines of action must all pass through a single point.
- The resultant (vector sum) of the three forces must be zero, so that they can be represented in magnitude and direction by the three sides of a triangle taken in order.
(b)(ii) Effect of the position of the centre of gravity on equilibrium
The position of the centre of gravity determines the type of equilibrium of a body. When the centre of gravity is low and the vertical line through it falls within the base of support, the body is in stable equilibrium: after a small tilt the weight provides a restoring moment that returns it to its original position. When the centre of gravity is high so that a small tilt makes the vertical line through it fall outside the base, the body is in unstable equilibrium and topples further. When the centre of gravity remains at the same height as the body is displaced, it is in neutral equilibrium. Hence lowering the centre of gravity and widening the base of support both increase the stability of a body.
Maelezo ya Majibu
Test of practical knowledge: balancing a metre rule (principle of moments)
The mass P = 100 g is fixed with adhesive at B, the 80.0 cm mark. The second mass Q = 100 g is suspended at A, a distance V from the 0 cm mark. For each value of V the knife edge K is moved until the rule balances horizontally, and its distance U from the 0 cm mark is read.
Table of readings
| V / cm | U / cm |
|---|---|
| 10.0 | 47.0 |
| 15.0 | 48.3 |
| 20.0 | 49.4 |
| 25.0 | 51.2 |
| 30.0 | 52.5 |
| 35.0 | 54.0 |
Graph of U against V
Plotting U on the vertical axis against V on the horizontal axis gives a straight line of best fit through the points.
(1) Slope of the graph
Taking two widely separated points on the line of best fit, \((V_1,U_1)=(10.0,\;47.0)\) and \((V_2,U_2)=(35.0,\;54.0)\):
\[s=\frac{\Delta U}{\Delta V}=\frac{54.0-47.0}{35.0-10.0}=\frac{7.0}{25.0}=0.28.\](2) Intercept on the vertical axis
Extending the line of best fit back to \(V=0\) gives
\[c=44\;\text{cm}.\]Evaluation
\[K_{1}=\frac{(1-2s)}{s}\times100=\frac{1-(2\times0.28)}{0.28}\times100=\frac{1-0.56}{0.28}\times100=\frac{0.44}{0.28}\times100=157.\]\[K_{2}=\frac{2c}{s}-160=\frac{2\times44}{0.28}-160=\frac{88}{0.28}-160=314.3-160=154.3.\]Two precautions
- Errors due to parallax were avoided by reading the position of the knife edge on the metre rule with the line of sight directly above the mark.
- Each suspended mass was checked to hang freely without touching the table, and the arrangement was shielded from draught so that it settled to a true balance.
(b)(i) Conditions for equilibrium under three non-parallel coplanar forces
- The three forces must be concurrent - their lines of action must all pass through a single point.
- The resultant (vector sum) of the three forces must be zero, so that they can be represented in magnitude and direction by the three sides of a triangle taken in order.
(b)(ii) Effect of the position of the centre of gravity on equilibrium
The position of the centre of gravity determines the type of equilibrium of a body. When the centre of gravity is low and the vertical line through it falls within the base of support, the body is in stable equilibrium: after a small tilt the weight provides a restoring moment that returns it to its original position. When the centre of gravity is high so that a small tilt makes the vertical line through it fall outside the base, the body is in unstable equilibrium and topples further. When the centre of gravity remains at the same height as the body is displaced, it is in neutral equilibrium. Hence lowering the centre of gravity and widening the base of support both increase the stability of a body.
Swali 3 Ripoti
Using the diagram as a out the guide carry out the following instructions:
- Place the pin O horizontally inside the cylinder provided. Pour some water on the pin in the cylinder such that the length of the water column \(l = \) SO \(= 10.0\text{cm}\), where S place the represents the water meniscus.
- lnsert another pin, P in the cork held by the boss of the retort stand.
- Adjust the position of P vertically upward or 0 formed by refraction at S.
- Read and record the distance \(h = PO\).
- Repeat the procedure for four other values of \(l = 15, 20, 25,\) and \(30\text{cm}\).
- In each case measure and record the corresponding value of h. Tabulate your readings.
- Plot a graph of h on the vertical axis against l on the horizontal axis.
- Determine the slope, \(s\), of the graph
- Evaluate (i) \(K_{1} = 1 - S\)
(ii) \(K_{2} = \frac{1}{k}\) - State two precautions taken to ensure accurate results.
(b)i. Explain the total internal reflection of light.
ii. A rectangular glass prism of thickness 6 cm and refractive index 1.5 is placed on the page of a book. The prints on the book are viewed vertically down Determine the apparent upward displacement of the print.
Test of practical knowledge: real and apparent depth (refraction at a water surface)
The pin O is placed horizontally at the bottom of the cylinder. Water is poured in so that the length of the water column \(l = SO\), where S is the position of the water meniscus. A second (search) pin P in the cork on the retort stand is adjusted vertically until, viewed from above, it coincides with no parallax with the image of O formed by refraction at S. The distance \(h = PO\) is read and recorded. The procedure is repeated for \(l = 15,\ 20,\ 25\) and \(30\ \text{cm}\).
Table of readings
| \(l\) (cm) | \(h\) (cm) |
|---|---|
| 10.0 | 7.50 |
| 15.0 | 11.25 |
| 20.0 | 15.00 |
| 25.0 | 18.75 |
| 30.0 | 22.50 |
Graph of \(h\) against \(l\)
Plotting \(h\) on the vertical axis against \(l\) on the 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, \((l_1, h_1) = (10.0,\ 7.50)\) and \((l_2, h_2) = (30.0,\ 22.50)\):
\[ s = \frac{\Delta h}{\Delta l} = \frac{22.50 - 7.50}{30.0 - 10.0} = \frac{15.00}{20.00} = 0.625 \]Evaluation
\[ K_1 = 1 - s = 1 - 0.625 = 0.375 \]\[ K_2 = \frac{1}{K_1} = \frac{1}{0.375} = 0.67 \]Two precautions
- The lower level of the water meniscus was read with the eye at the same horizontal level as the meniscus to avoid parallax error.
- The zero error of the metre rule was noted and allowed for when measuring \(h\) and \(l\).
(b)(i) Total internal reflection of light
Total internal reflection occurs when light travels from an optically denser medium into a less dense medium and the angle of incidence in the denser medium is greater than the critical angle. Under this condition no light is refracted into the less dense medium; all of it is reflected back into the denser medium, obeying the ordinary laws of reflection.
(b)(ii) Apparent upward displacement of the print
Real depth (thickness of glass) \(D = 6\ \text{cm}\), refractive index \(_a\mu_g = 1.5\).
\[ _a\mu_g = \frac{\text{real depth}}{\text{apparent depth}} \quad\Rightarrow\quad \text{apparent depth} = \frac{6}{1.5} = 4\ \text{cm} \]\[ \text{apparent upward displacement} = \text{real depth} - \text{apparent depth} = 6 - 4 = 2.0\ \text{cm} \]The print appears raised (displaced upward) by \(2.0\ \text{cm}\).
Maelezo ya Majibu
Test of practical knowledge: real and apparent depth (refraction at a water surface)
The pin O is placed horizontally at the bottom of the cylinder. Water is poured in so that the length of the water column \(l = SO\), where S is the position of the water meniscus. A second (search) pin P in the cork on the retort stand is adjusted vertically until, viewed from above, it coincides with no parallax with the image of O formed by refraction at S. The distance \(h = PO\) is read and recorded. The procedure is repeated for \(l = 15,\ 20,\ 25\) and \(30\ \text{cm}\).
Table of readings
| \(l\) (cm) | \(h\) (cm) |
|---|---|
| 10.0 | 7.50 |
| 15.0 | 11.25 |
| 20.0 | 15.00 |
| 25.0 | 18.75 |
| 30.0 | 22.50 |
Graph of \(h\) against \(l\)
Plotting \(h\) on the vertical axis against \(l\) on the 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, \((l_1, h_1) = (10.0,\ 7.50)\) and \((l_2, h_2) = (30.0,\ 22.50)\):
\[ s = \frac{\Delta h}{\Delta l} = \frac{22.50 - 7.50}{30.0 - 10.0} = \frac{15.00}{20.00} = 0.625 \]Evaluation
\[ K_1 = 1 - s = 1 - 0.625 = 0.375 \]\[ K_2 = \frac{1}{K_1} = \frac{1}{0.375} = 0.67 \]Two precautions
- The lower level of the water meniscus was read with the eye at the same horizontal level as the meniscus to avoid parallax error.
- The zero error of the metre rule was noted and allowed for when measuring \(h\) and \(l\).
(b)(i) Total internal reflection of light
Total internal reflection occurs when light travels from an optically denser medium into a less dense medium and the angle of incidence in the denser medium is greater than the critical angle. Under this condition no light is refracted into the less dense medium; all of it is reflected back into the denser medium, obeying the ordinary laws of reflection.
(b)(ii) Apparent upward displacement of the print
Real depth (thickness of glass) \(D = 6\ \text{cm}\), refractive index \(_a\mu_g = 1.5\).
\[ _a\mu_g = \frac{\text{real depth}}{\text{apparent depth}} \quad\Rightarrow\quad \text{apparent depth} = \frac{6}{1.5} = 4\ \text{cm} \]\[ \text{apparent upward displacement} = \text{real depth} - \text{apparent depth} = 6 - 4 = 2.0\ \text{cm} \]The print appears raised (displaced upward) by \(2.0\ \text{cm}\).
