WAEC SSCE - Physics - 2000 (Practical)
Frage 1 Bericht
- Fix a metre rule on the bench with the graduated face up.
- Place the illuminated object at the zero end of the rule and the screen at the other end as illustrated in the diagram above.
- Measure and record D, the distance between the object and the screen. Evaluate D\(^{2}\).
- Place and move the converging lens between the illuminated object and the screen until a diminished sharp image of the object is formed on the screen. Read and record the position, X\(_{1}\), of the lens. From this position, move the lens towards the object until another sharp image of the object is formed on the screen. Read and record the new position x\(_{2}\), of the lens.
- Evaluate and record L (x\(_{1}\) - x\(_{2}\)), L\(^{2}\)) and (D\(^{2}\) - L\(^{2}\))
- Repeat the procedure for D = 90, 80, 70 and 60 cm. In each case, evaluate, L L\(^{2}\) and (D\(^{2}\) - L\(^{2}\)). Tabulate your readings.
- Plot a graph of D\(^{2}\) - L\(^{2}\) on the vertical axis against D on the horizontal axis.
- Determine the slope, S, of the graph and evaluate K = \(\frac{s}{4}\). State two precautions taken to ensure accurate results.
(b)i. Distinguish between a real image and a virtual image.
Draw a ray diagram to show how a converging lens may be used to form a real diminished image of an object.
(a) Lens displacement (conjugate-foci) method
For each fixed object-to-screen distance \(D\), the lens is moved to the two positions \(x_1\) and \(x_2\) that each give a sharp image on the screen. The separation of these positions is \(L = x_1 - x_2\). The readings and evaluated quantities are tabulated below.
| \(D\) (cm) | \(D^{2}\) (cm\(^{2}\)) | \(x_1\) (cm) | \(x_2\) (cm) | \(L=x_1-x_2\) (cm) | \(L^{2}\) (cm\(^{2}\)) | \(D^{2}-L^{2}\) (cm\(^{2}\)) |
|---|---|---|---|---|---|---|
| 100 | 10000 | 80.50 | 18.30 | 62.20 | 3868.84 | 6131.16 |
| 90 | 8100 | 70.20 | 19.00 | 51.20 | 2621.44 | 5478.56 |
| 80 | 6400 | 59.10 | 20.00 | 39.10 | 1528.81 | 4871.19 |
| 70 | 4900 | 46.70 | 22.00 | 24.70 | 610.09 | 4289.91 |
| 60 | 3600 | 29.00 | 7.60 | 21.40 | 457.96 | 3142.04 |
Sample evaluation for the first reading (\(D = 100\ \text{cm}\)):
\[ D^{2} = 100^{2} = 10000\ \text{cm}^{2}, \quad L = 80.50 - 18.30 = 62.20\ \text{cm} \]\[ L^{2} = 62.20^{2} = 3868.84\ \text{cm}^{2}, \quad D^{2}-L^{2} = 10000 - 3868.84 = 6131.16\ \text{cm}^{2} \]Graph of \(D^{2}-L^{2}\) against \(D\)
Plotting \(\left(D^{2}-L^{2}\right)\) on the vertical axis against \(D\) on the horizontal axis gives a straight line through the origin:
Slope and value of K
Reading the slope from the line of best fit using a large triangle, taking two clear points on the line \((D = 100\ \text{cm},\ D^{2}-L^{2} = 6000\ \text{cm}^{2})\) and \((D = 50\ \text{cm},\ D^{2}-L^{2} = 3000\ \text{cm}^{2})\):
\[ S = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6000 - 3000}{100 - 50} = \frac{3000}{50} = 60\ \text{cm} \]\[ K = \frac{S}{4} = \frac{60}{4} = 15\ \text{cm} \](This \(K\) is the focal length of the converging lens, since the displacement method gives \(f = \dfrac{D^{2}-L^{2}}{4D}\), so a graph of \(D^{2}-L^{2}\) against \(D\) has slope \(4f\) and \(K = \tfrac{S}{4} = f\).)
Two precautions
- I avoided parallax error by reading the position of the lens on the metre rule with the line of sight directly above the mark.
- I ensured that a sharp, well-focused image was formed on the screen before taking each reading of the lens position.
(b)(i) Real image versus virtual image
| Real image | Virtual image |
|---|---|
| Formed by the actual intersection of refracted rays. | Formed where the refracted rays only appear to meet when produced backwards. |
| Can be caught (focused) on a screen. | Cannot be caught on a screen. |
| Inverted for a single converging lens. | Upright. |
(b)(ii) Ray diagram: converging lens forming a real, diminished image
With the object placed beyond \(2F\), the converging lens forms a real, inverted and diminished image between \(F\) and \(2F\) on the far side of the lens.
Antwortdetails
(a) Lens displacement (conjugate-foci) method
For each fixed object-to-screen distance \(D\), the lens is moved to the two positions \(x_1\) and \(x_2\) that each give a sharp image on the screen. The separation of these positions is \(L = x_1 - x_2\). The readings and evaluated quantities are tabulated below.
| \(D\) (cm) | \(D^{2}\) (cm\(^{2}\)) | \(x_1\) (cm) | \(x_2\) (cm) | \(L=x_1-x_2\) (cm) | \(L^{2}\) (cm\(^{2}\)) | \(D^{2}-L^{2}\) (cm\(^{2}\)) |
|---|---|---|---|---|---|---|
| 100 | 10000 | 80.50 | 18.30 | 62.20 | 3868.84 | 6131.16 |
| 90 | 8100 | 70.20 | 19.00 | 51.20 | 2621.44 | 5478.56 |
| 80 | 6400 | 59.10 | 20.00 | 39.10 | 1528.81 | 4871.19 |
| 70 | 4900 | 46.70 | 22.00 | 24.70 | 610.09 | 4289.91 |
| 60 | 3600 | 29.00 | 7.60 | 21.40 | 457.96 | 3142.04 |
Sample evaluation for the first reading (\(D = 100\ \text{cm}\)):
\[ D^{2} = 100^{2} = 10000\ \text{cm}^{2}, \quad L = 80.50 - 18.30 = 62.20\ \text{cm} \]\[ L^{2} = 62.20^{2} = 3868.84\ \text{cm}^{2}, \quad D^{2}-L^{2} = 10000 - 3868.84 = 6131.16\ \text{cm}^{2} \]Graph of \(D^{2}-L^{2}\) against \(D\)
Plotting \(\left(D^{2}-L^{2}\right)\) on the vertical axis against \(D\) on the horizontal axis gives a straight line through the origin:
Slope and value of K
Reading the slope from the line of best fit using a large triangle, taking two clear points on the line \((D = 100\ \text{cm},\ D^{2}-L^{2} = 6000\ \text{cm}^{2})\) and \((D = 50\ \text{cm},\ D^{2}-L^{2} = 3000\ \text{cm}^{2})\):
\[ S = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6000 - 3000}{100 - 50} = \frac{3000}{50} = 60\ \text{cm} \]\[ K = \frac{S}{4} = \frac{60}{4} = 15\ \text{cm} \](This \(K\) is the focal length of the converging lens, since the displacement method gives \(f = \dfrac{D^{2}-L^{2}}{4D}\), so a graph of \(D^{2}-L^{2}\) against \(D\) has slope \(4f\) and \(K = \tfrac{S}{4} = f\).)
Two precautions
- I avoided parallax error by reading the position of the lens on the metre rule with the line of sight directly above the mark.
- I ensured that a sharp, well-focused image was formed on the screen before taking each reading of the lens position.
(b)(i) Real image versus virtual image
| Real image | Virtual image |
|---|---|
| Formed by the actual intersection of refracted rays. | Formed where the refracted rays only appear to meet when produced backwards. |
| Can be caught (focused) on a screen. | Cannot be caught on a screen. |
| Inverted for a single converging lens. | Upright. |
(b)(ii) Ray diagram: converging lens forming a real, diminished image
With the object placed beyond \(2F\), the converging lens forms a real, inverted and diminished image between \(F\) and \(2F\) on the far side of the lens.
Frage 2 Bericht
- Using the spring balance provided, determine the weight of object of mass \(M = 5.0\text{g}\). Record this weight as \(W_{1}\).
- Determine the weight of the object when completely immersed in water contained in a beaker as shown in the diagram. Record the weight as \(W_{2}\)
- Determine the weight of the object when it is completely immersed in the liquid labelled 'L'. Record the weight as \(W_{3}\). Evaluate, \(u = (W_{1} - W_{3})\) and \(v = (W_{1} - W_{3})\).
- Repeat the procedure with the objects of masses \(M = 10, 15, 20,\) and \(25\text{g}\). In each case, evaluate \(u = (W_{1} - W_{3})\) and \(v = (W_{1} - W_{3})\) on the vertical axis against \(u = (W_{1} - W_{2}\) on the horizontal axis.
- Determine the slope, \(s\), of the graph. (vii) State two precautions taken to ensure accurate results.
(b)i. A piece of brass of mass \(20.0\text{g}\) is hung on a spring balance from a rigid support and completely immersed in kerosine of density \(8.0 \times 10^{2}\text{ kg m}^{-3}\). Determine the reading on the spring balance. \([g = 10\text{ms}^{-2}]\), density of brass = \(8.0 \times 10^{3}\text{ kg m}^{-3}\) J
ii. State Archimede's principle and the law of floatation.
(a) Determination of the relative density of liquid L by upthrust
The object is hung from the spring balance and its weight in air \(W_1\) is read. It is then fully immersed in water and the reading \(W_2\) taken, and finally fully immersed in liquid L and the reading \(W_3\) taken. The upthrust in water is \(U=(W_1-W_2)\) and the upthrust in L is \(V=(W_1-W_3)\). The procedure is repeated for the five masses.
The diagram of the apparatus is shown below.
Table of readings
| M /g | \(W_1\) /g | \(W_2\) /g | \(W_3\) /g | \(U=(W_1-W_2)\) /g | \(V=(W_1-W_3)\) /g |
|---|---|---|---|---|---|
| 5.0 | 5.00 | 4.00 | 4.20 | 1.00 | 0.80 |
| 10.0 | 10.00 | 8.00 | 8.40 | 2.00 | 1.60 |
| 15.0 | 15.00 | 12.00 | 12.60 | 3.00 | 2.40 |
| 20.0 | 20.00 | 16.00 | 16.80 | 4.00 | 3.20 |
| 25.0 | 25.00 | 20.00 | 21.00 | 5.00 | 4.00 |
Graph of \(V=(W_1-W_3)\) against \(U=(W_1-W_2)\)
Slope of the graph
Taking two points on the line of best fit, \((U_1,V_1)=(1.00,\,0.80)\) and \((U_2,V_2)=(5.00,\,4.00)\):
\[ s=\frac{V_2-V_1}{U_2-U_1}=\frac{4.00-0.80}{5.00-1.00}=\frac{3.20}{4.00}=0.80 \]Since the slope \(s=\dfrac{W_1-W_3}{W_1-W_2}=\dfrac{\text{upthrust in L}}{\text{upthrust in water}}\), the slope is the relative density of liquid L:
\[ \boxed{s=0.80} \](vii) Two precautions
- I ensured the object was completely immersed in the liquid and did not touch the sides or bottom of the beaker.
- I read the spring balance at eye level with the object at rest to avoid parallax error, and I noted and corrected the zero error of the balance.
(b)(i) Reading of the spring balance for the brass
Weight of the brass in air:
\[ W=mg=20\times10^{-3}\times10=0.2\ \text{N} \]Volume of the brass:
\[ V=\frac{m}{\rho_{\text{brass}}}=\frac{20\times10^{-3}}{8.0\times10^{3}}=2.5\times10^{-6}\ \text{m}^3 \]Upthrust = weight of kerosine displaced:
\[ U=\rho_{\text{k}}\,V\,g=8.0\times10^{2}\times2.5\times10^{-6}\times10=0.02\ \text{N} \]Reading on the spring balance = tension in the spring = weight in air \(-\) upthrust:
\[ =0.2-0.02=\boxed{0.18\ \text{N}} \](b)(ii) Statements
Archimedes' principle: When a body is wholly or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces.
Law of flotation: A floating body displaces its own weight of the fluid in which it floats.
Antwortdetails
(a) Determination of the relative density of liquid L by upthrust
The object is hung from the spring balance and its weight in air \(W_1\) is read. It is then fully immersed in water and the reading \(W_2\) taken, and finally fully immersed in liquid L and the reading \(W_3\) taken. The upthrust in water is \(U=(W_1-W_2)\) and the upthrust in L is \(V=(W_1-W_3)\). The procedure is repeated for the five masses.
The diagram of the apparatus is shown below.
Table of readings
| M /g | \(W_1\) /g | \(W_2\) /g | \(W_3\) /g | \(U=(W_1-W_2)\) /g | \(V=(W_1-W_3)\) /g |
|---|---|---|---|---|---|
| 5.0 | 5.00 | 4.00 | 4.20 | 1.00 | 0.80 |
| 10.0 | 10.00 | 8.00 | 8.40 | 2.00 | 1.60 |
| 15.0 | 15.00 | 12.00 | 12.60 | 3.00 | 2.40 |
| 20.0 | 20.00 | 16.00 | 16.80 | 4.00 | 3.20 |
| 25.0 | 25.00 | 20.00 | 21.00 | 5.00 | 4.00 |
Graph of \(V=(W_1-W_3)\) against \(U=(W_1-W_2)\)
Slope of the graph
Taking two points on the line of best fit, \((U_1,V_1)=(1.00,\,0.80)\) and \((U_2,V_2)=(5.00,\,4.00)\):
\[ s=\frac{V_2-V_1}{U_2-U_1}=\frac{4.00-0.80}{5.00-1.00}=\frac{3.20}{4.00}=0.80 \]Since the slope \(s=\dfrac{W_1-W_3}{W_1-W_2}=\dfrac{\text{upthrust in L}}{\text{upthrust in water}}\), the slope is the relative density of liquid L:
\[ \boxed{s=0.80} \](vii) Two precautions
- I ensured the object was completely immersed in the liquid and did not touch the sides or bottom of the beaker.
- I read the spring balance at eye level with the object at rest to avoid parallax error, and I noted and corrected the zero error of the balance.
(b)(i) Reading of the spring balance for the brass
Weight of the brass in air:
\[ W=mg=20\times10^{-3}\times10=0.2\ \text{N} \]Volume of the brass:
\[ V=\frac{m}{\rho_{\text{brass}}}=\frac{20\times10^{-3}}{8.0\times10^{3}}=2.5\times10^{-6}\ \text{m}^3 \]Upthrust = weight of kerosine displaced:
\[ U=\rho_{\text{k}}\,V\,g=8.0\times10^{2}\times2.5\times10^{-6}\times10=0.02\ \text{N} \]Reading on the spring balance = tension in the spring = weight in air \(-\) upthrust:
\[ =0.2-0.02=\boxed{0.18\ \text{N}} \](b)(ii) Statements
Archimedes' principle: When a body is wholly or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces.
Law of flotation: A floating body displaces its own weight of the fluid in which it floats.
Frage 3 Bericht
- Measure and record the e.m.f. of the accumulator provided.
- Connect a circuit as shown in the diagram. S is a standard resistor and R is a resistance box.
- With R = 0\(\Omega\), close the key K. Read and record the ammeter reading I. Evaluate 1\(^{-1}\).
- Repeat the procedure for R=1,2, 3, 4, and 5\(\Omega\). Tabulate your readings.
- Plot a graph of R on the vertical axis and 1\(^{-1}\) on the horizontal axis, starting both axis from the origin (0,0).
- Determine the slope s of the graph and find the intercept C on the vertical axis.
- State two precautions taken to ensure accurate results.
(b)i. State two advantages of a lead-acid accumulator over a Leclanche cell.
ii. A parallel combination of 3\(\Omega\) and 4\(\Omega\) resistors is connected in series with a resistor of 4\(\Omega\) and a battery of negligible internal resistance. Calculate the effective resistance in the circuit.
(a) E.m.f. of the accumulator and the R against \(I^{-1}\) experiment
E.m.f. of the accumulator, \(E = 1.5\ \text{V}\).
Table of readings
| \(R\ (\Omega)\) | \(I\ (\text{A})\) | \(I^{-1}\ (\text{A}^{-1})\) |
|---|---|---|
| 0 | 0.78 | 1.28 |
| 1 | 0.50 | 2.00 |
| 2 | 0.38 | 2.63 |
| 3 | 0.30 | 3.33 |
| 4 | 0.25 | 4.00 |
| 5 | 0.22 | 4.55 |
Graph of \(R\) against \(I^{-1}\)
Slope of the graph
Taking two points on the line of best fit, \((I^{-1}_1, R_1) = (2.00\ \text{A}^{-1},\ 1.1\ \Omega)\) and \((I^{-1}_2, R_2) = (4.00\ \text{A}^{-1},\ 4.1\ \Omega)\):
\[ s = \frac{R_2 - R_1}{I^{-1}_2 - I^{-1}_1} = \frac{4.1 - 1.1}{4.00 - 2.00} = \frac{3.0}{2.00} = 1.5\ \text{V}. \]Intercept on the vertical axis
Producing the line of best fit back to \(I^{-1} = 0\), it cuts the vertical (\(R\)) axis at
\[ C = -1.9\ \Omega. \]Interpretation. For this circuit the current is \(I = \dfrac{E}{R + S + r}\), which rearranges to \(R = E\,(I^{-1}) - (S + r)\). Comparing with \(R = s\,(I^{-1}) + C\): the slope \(s = E = 1.5\ \text{V}\) (equal to the measured e.m.f.), and the intercept \(C = -(S + r) = -1.9\ \Omega\), giving \(S + r = 1.9\ \Omega\).
Two precautions
- I made all connections clean and tight, and opened the key immediately after taking each reading so as not to run down the accumulator.
- I read the ammeter with my eye directly in front of the pointer to avoid errors due to parallax, and I noted and corrected the zero error of the ammeter.
(b)(i) Two advantages of a lead-acid accumulator over a Leclanche cell
- The lead-acid accumulator can be recharged and used again, whereas the Leclanche cell cannot be recharged.
- The lead-acid accumulator has a very low internal resistance, so it can supply a large current for a long time, which the Leclanche cell cannot.
(b)(ii) Effective resistance of the network
The \(3\ \Omega\) and \(4\ \Omega\) resistors are in parallel:
\[ \frac{1}{R_p} = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12}, \qquad R_p = \frac{12}{7} = 1.71\ \Omega. \]This parallel section is in series with the \(4\ \Omega\) resistor (the battery has negligible internal resistance):
\[ R_{\text{eff}} = R_p + 4 = 1.71 + 4 = 5.71\ \Omega. \]The effective resistance in the circuit \(= 5.71\ \Omega\).
Antwortdetails
(a) E.m.f. of the accumulator and the R against \(I^{-1}\) experiment
E.m.f. of the accumulator, \(E = 1.5\ \text{V}\).
Table of readings
| \(R\ (\Omega)\) | \(I\ (\text{A})\) | \(I^{-1}\ (\text{A}^{-1})\) |
|---|---|---|
| 0 | 0.78 | 1.28 |
| 1 | 0.50 | 2.00 |
| 2 | 0.38 | 2.63 |
| 3 | 0.30 | 3.33 |
| 4 | 0.25 | 4.00 |
| 5 | 0.22 | 4.55 |
Graph of \(R\) against \(I^{-1}\)
Slope of the graph
Taking two points on the line of best fit, \((I^{-1}_1, R_1) = (2.00\ \text{A}^{-1},\ 1.1\ \Omega)\) and \((I^{-1}_2, R_2) = (4.00\ \text{A}^{-1},\ 4.1\ \Omega)\):
\[ s = \frac{R_2 - R_1}{I^{-1}_2 - I^{-1}_1} = \frac{4.1 - 1.1}{4.00 - 2.00} = \frac{3.0}{2.00} = 1.5\ \text{V}. \]Intercept on the vertical axis
Producing the line of best fit back to \(I^{-1} = 0\), it cuts the vertical (\(R\)) axis at
\[ C = -1.9\ \Omega. \]Interpretation. For this circuit the current is \(I = \dfrac{E}{R + S + r}\), which rearranges to \(R = E\,(I^{-1}) - (S + r)\). Comparing with \(R = s\,(I^{-1}) + C\): the slope \(s = E = 1.5\ \text{V}\) (equal to the measured e.m.f.), and the intercept \(C = -(S + r) = -1.9\ \Omega\), giving \(S + r = 1.9\ \Omega\).
Two precautions
- I made all connections clean and tight, and opened the key immediately after taking each reading so as not to run down the accumulator.
- I read the ammeter with my eye directly in front of the pointer to avoid errors due to parallax, and I noted and corrected the zero error of the ammeter.
(b)(i) Two advantages of a lead-acid accumulator over a Leclanche cell
- The lead-acid accumulator can be recharged and used again, whereas the Leclanche cell cannot be recharged.
- The lead-acid accumulator has a very low internal resistance, so it can supply a large current for a long time, which the Leclanche cell cannot.
(b)(ii) Effective resistance of the network
The \(3\ \Omega\) and \(4\ \Omega\) resistors are in parallel:
\[ \frac{1}{R_p} = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12}, \qquad R_p = \frac{12}{7} = 1.71\ \Omega. \]This parallel section is in series with the \(4\ \Omega\) resistor (the battery has negligible internal resistance):
\[ R_{\text{eff}} = R_p + 4 = 1.71 + 4 = 5.71\ \Omega. \]The effective resistance in the circuit \(= 5.71\ \Omega\).
