WAEC SSCE - Physics - 2004 (Practical)

The diagram shows the apparatus: a ray box carrying an illuminated cross-wire object, a concave mirror facing it, and a small screen between them. The distance from the ray box (object) to the mirror is b and the distance from the screen (image) to the mirror is a.

(a) The practical

For each object distance \(b\) (20.0, 25.0, 30.0, 35.0, 40.0 cm) the screen is moved until the cross-wire image is sharp, and \(a\) is read off. The mirror equation is

Writing \(l=\dfrac{1}{a}\) and rearranging,

A graph of \(l=\dfrac{1}{a}\) (vertical) against \(\dfrac{1}{b}\) (horizontal) is a straight line of slope \(S=-1\) whose intercept on the vertical axis equals \(\dfrac{1}{f_o}\). Hence the focal length is obtained from that intercept, \(f_o=\dfrac{1}{\text{intercept}}\), and \(S^{-1}=-1\).

Sample table (illustrative)

Two precautions:

• The image on the screen was made as sharp as possible before each reading to avoid parallax and focusing error.
• The distances \(a\) and \(b\) were measured horizontally from the pole of the mirror, with the ray box, screen and mirror kept on the same straight line (aligned axis).

(b)(i) Object 10 cm in front of a concave mirror, f = 15 cm

Here \(u=10\text{ cm}\), \(f=15\text{ cm}\) (object inside the focal point).

Magnification:

The negative \(v\) means the image is behind the mirror. Characteristics of the image: it is virtual, erect (upright), magnified (three times the object size), and formed 30 cm behind the mirror.

(b)(ii) How \(f_o\) was obtained in (a): A graph of \(l=\tfrac{1}{a}\) against \(\tfrac{1}{b}\) was plotted; the intercept on the vertical axis is \(\tfrac{1}{f_o}\), so \(f_o\) is the reciprocal of that intercept.

(a) Measurement and tabulation

The emf of the cell was measured with the voltmeter when the circuit was open:

\[ V_o = 3.0\ \text{V} \]

The circuit was connected as shown, with the ammeter in series and the voltmeter connected across the resistor/load. The key was closed briefly and the rheostat adjusted to obtain the stated current values. The corresponding voltmeter readings were recorded.

For the first reading:

\[ I=0.20\ \text{A} \]

\[ I^{-1}=\frac{1}{0.20}=5.00\ \text{A}^{-1} \]

If \(V=0.19\ \text{V}\), then:

\[ V^{-1}=\frac{1}{0.19}=5.26\ \text{V}^{-1} \]

Table of readings

S/N	\(I\) (A)	\(I^{-1}\) (A-1)	\(V\) (V)	\(V^{-1}\) (V-1)
1	0.20	5.00	0.19	5.26
2	0.25	4.00	0.23	4.35
3	0.30	3.33	0.28	3.57
4	0.35	2.86	0.33	3.03
5	0.40	2.50	0.38 Lies die Unterrichtsnotiz zu Current Electricity (JAMB) Antwortdetails (a) Measurement and tabulation The emf of the cell was measured with the voltmeter when the circuit was open: \[ V_o = 3.0\ \text{V} \] The circuit was connected as shown, with the ammeter in series and the voltmeter connected across the resistor/load. The key was closed briefly and the rheostat adjusted to obtain the stated current values. The corresponding voltmeter readings were recorded. For the first reading: \[ I=0.20\ \text{A} \] \[ I^{-1}=\frac{1}{0.20}=5.00\ \text{A}^{-1} \] If \(V=0.19\ \text{V}\), then: \[ V^{-1}=\frac{1}{0.19}=5.26\ \text{V}^{-1} \] Table of readings S/N \(I\) (A) \(I^{-1}\) (A-1) \(V\) (V) \(V^{-1}\) (V-1) 1 0.20 5.00 0.19 5.26 2 0.25 4.00 0.23 4.35 3 0.30 3.33 0.28 3.57 4 0.35 2.86 0.33 3.03 5 0.40 2.50 0.38 Lies die Unterrichtsnotiz zu Current Electricity (JAMB) Current Electricity Antwort anzeigen Frage 3 Bericht Using the diagram above as a guide, carry out the following instructions. Place the meter rule provided on the knife edge and adjust the position until it balances horizontally. Read and record the balance point, G. Keep the knife edge at this point throughout the experiment. Suspend a mass Q= 50.0g at a point P 30cm from the 0cm end of the rule. On the other side of G, suspend the mass M =30g. Adjust its position until the rule settles down horizontally as shown in the diagram above. Read and record the position R of M. Record the distance, d, between G and R. Also read and record the distance, a, between P and G.  Repeat the procedure for four other values of M = 40, 50, 60, and 70 with Q kept in the same position. Evaluate d\(^{-1}\) in each case. Tabulate your readings. Plot a graph of M on the vertical axis against d\(^{-1}\) on the horizontal axis. Determine the slope,s, of the graph. Evaluate k = \(\frac{s}{Q}\) State two precautions taken to ensure accurate results. (b)i. Explain the moment of a force about a point ii. State the conditions necessary for a body to be in equilibrium when acted upon by a number of parallel end forces Antwort Principle of the experiment When the metre rule rests horizontally on the knife edge at the balance point \(G\), it is in equilibrium under the two suspended weights. Taking moments about \(G\), the anticlockwise moment of \(Q\) equals the clockwise moment of \(M\): \[ Q \times a = M \times d \] where \(a\) is the fixed distance \(PG\) and \(d\) is the distance \(GR\). Rearranging for \(M\): \[ M = (Q\,a)\,d^{-1} \] So a graph of \(M\) (vertical axis) against \(d^{-1}\) (horizontal axis) is a straight line through the origin whose slope is \(s = Q\,a\). Recorded readings Balance point: \(G = 49.50\ \text{cm}\). Mass \(Q = 50.0\ \text{g}\) suspended at \(P = 30.0\ \text{cm}\), so the fixed distance \[ a = PG = 49.50 - 30.00 = 19.50\ \text{cm}. \] For each mass \(M\) the position \(R\) is read, and \(d = R - G\), then \(d^{-1}\) is evaluated. S/N M (g) R (cm) d = R − G (cm) d−1 (cm−1) d−1 ×10−2 (cm−1) a (cm) 1 30.0 82.00 32.50 0.0308 3.08 19.50 2 40.0 73.88 24.38 0.0410 4.10 19.50 3 50.0 69.00 19.50 0.0513 5.13 19.50 4 60.0 65.75 16.25 0.0615 6.15 19.50 5 70.0 63.43 13.93 0.0718 7.18 19.50 Graph of M against d−1 Slope of the graph Taking two widely separated points on the line of best fit, \((d^{-1}_1,\,M_1) = (0.0308\ \text{cm}^{-1},\ 30.0\ \text{g})\) and \((d^{-1}_2,\,M_2) = (0.0718\ \text{cm}^{-1},\ 70.0\ \text{g})\): \[ s = \frac{\Delta M}{\Delta (d^{-1})} = \frac{70.0 - 30.0}{0.0718 - 0.0308} = \frac{40.0}{0.0410} = 975\ \text{g cm}. \] Evaluation of k \[ k = \frac{s}{Q} = \frac{975}{50.0} = 19.5\ \text{g}. \] This equals the fixed distance \(a = PG = 19.50\ \text{cm}\), as expected since \(s = Q\,a\) and therefore \(k = s/Q = a\). Two precautions The scale readings were taken with the eye placed vertically above the mark to avoid the error of parallax. The rule was allowed to come completely to rest in the horizontal position before each reading, and the knife edge was kept fixed at \(G\) throughout the experiment. (b)(i) Moment of a force about a point The moment of a force about a point is the product of the force and the perpendicular distance from that point (the pivot) to the line of action of the force. Its SI unit is the newton metre (N m). (b)(ii) Conditions for equilibrium under parallel forces The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force. The algebraic sum of the moments of the forces about any point is zero, i.e. the total clockwise moment equals the total anticlockwise moment about that point. Lies die Unterrichtsnotiz zu Equilibrium Of Forces (JAMB) Antwortdetails Principle of the experiment When the metre rule rests horizontally on the knife edge at the balance point \(G\), it is in equilibrium under the two suspended weights. Taking moments about \(G\), the anticlockwise moment of \(Q\) equals the clockwise moment of \(M\): \[ Q \times a = M \times d \] where \(a\) is the fixed distance \(PG\) and \(d\) is the distance \(GR\). Rearranging for \(M\): \[ M = (Q\,a)\,d^{-1} \] So a graph of \(M\) (vertical axis) against \(d^{-1}\) (horizontal axis) is a straight line through the origin whose slope is \(s = Q\,a\). Recorded readings Balance point: \(G = 49.50\ \text{cm}\). Mass \(Q = 50.0\ \text{g}\) suspended at \(P = 30.0\ \text{cm}\), so the fixed distance \[ a = PG = 49.50 - 30.00 = 19.50\ \text{cm}. \] For each mass \(M\) the position \(R\) is read, and \(d = R - G\), then \(d^{-1}\) is evaluated. S/N M (g) R (cm) d = R − G (cm) d−1 (cm−1) d−1 ×10−2 (cm−1) a (cm) 1 30.0 82.00 32.50 0.0308 3.08 19.50 2 40.0 73.88 24.38 0.0410 4.10 19.50 3 50.0 69.00 19.50 0.0513 5.13 19.50 4 60.0 65.75 16.25 0.0615 6.15 19.50 5 70.0 63.43 13.93 0.0718 7.18 19.50 Graph of M against d−1 Slope of the graph Taking two widely separated points on the line of best fit, \((d^{-1}_1,\,M_1) = (0.0308\ \text{cm}^{-1},\ 30.0\ \text{g})\) and \((d^{-1}_2,\,M_2) = (0.0718\ \text{cm}^{-1},\ 70.0\ \text{g})\): \[ s = \frac{\Delta M}{\Delta (d^{-1})} = \frac{70.0 - 30.0}{0.0718 - 0.0308} = \frac{40.0}{0.0410} = 975\ \text{g cm}. \] Evaluation of k \[ k = \frac{s}{Q} = \frac{975}{50.0} = 19.5\ \text{g}. \] This equals the fixed distance \(a = PG = 19.50\ \text{cm}\), as expected since \(s = Q\,a\) and therefore \(k = s/Q = a\). Two precautions The scale readings were taken with the eye placed vertically above the mark to avoid the error of parallax. The rule was allowed to come completely to rest in the horizontal position before each reading, and the knife edge was kept fixed at \(G\) throughout the experiment. (b)(i) Moment of a force about a point The moment of a force about a point is the product of the force and the perpendicular distance from that point (the pivot) to the line of action of the force. Its SI unit is the newton metre (N m). (b)(ii) Conditions for equilibrium under parallel forces The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force. The algebraic sum of the moments of the forces about any point is zero, i.e. the total clockwise moment equals the total anticlockwise moment about that point. Lies die Unterrichtsnotiz zu Equilibrium Of Forces (JAMB) Equilibrium Of Forces Antwort anzeigen Vorherige Nächste Frageliste 1 2 3 In Ordnung In Ordnung Bist du sicher? Möchten Sie mit dieser Aktion fortfahren?