WAEC SSCE - Physics - 2013 (Practical)

Test of Practical Knowledge (moments: mass of sand by balancing a metre rule)

Observations

• Mass of the uniform metre rule, \(M = 130\,\text{g}\)
• Mass of the empty measuring cylinder, \(m_1 = 20\,\text{g}\)
• Balance point of the bare rule, \(G = 50.4\,\text{cm}\)

The cylinder is hung at the 2 cm mark of the rule and the knife-edge is moved until the rule balances at position \(K\). Then \(e = K - 2\) is the distance from the knife-edge to the load, and \(f = G - K\) is the distance from the knife-edge to the rule's centre of gravity. Taking moments about the knife-edge:

\[ (m_1 + m_2)\,e = M\,f \quad\Rightarrow\quad m_2 = \frac{M f}{e} - m_1 \]

For example, at \(V = 2\,\text{cm}^3\): \(K = 43.5\), \(e = 41.5\), \(f = 6.9\), so \(m_2 = \dfrac{130 \times 6.9}{41.5} - 20 = 21.61 - 20 = 1.61\,\text{g}\).

Table of readings

\(V\)/cm\(^3\)	\(K\)/cm	\(e = K-2\)/cm	\(f = G-K\)/cm	\(m_2 = \frac{Mf}{e}-m_1\)/g
2	43.5	41.5	6.9	1.61
4	42.7	40.7	7.7	3.44
6	41.9	39.9	8.5	6.37
8	40.1	39.1	9.3	9.42
10	40.4	38.4	10.0	12.18

Graph of \(m_2\) against \(V\)

Plotting \(m_2\) (vertical axis) against \(V\) (horizontal axis) and drawing the best straight line through the points:

Slope of the graph

Reading two well-separated points off the line of best fit, \((V_1, m_2{}_1) = (4.4,\,4.0)\) and \((V_2, m_2{}_2) = (10.4,\,12.8)\):

\[ s = \frac{\Delta m_2}{\Delta V} = \frac{12.8 - 4.0}{10.4 - 4.4} = \frac{8.8}{6.0} = 1.47\ \text{g cm}^{-3} \]

The slope is the mass per unit volume (density) of the sand.

Two precautions

• Draught was avoided and the knife-edge was firmly clamped so that the rule balanced steadily and horizontally before each reading.
• Parallax error was avoided by reading the balance point on the rule and the sand volume in the cylinder with the eye placed directly above the scale.

(b)(i) Mass of 7.5 cm\(^3\) of sand

Reading the best-fit line at \(V = 7.5\,\text{cm}^3\) (shown by the dashed guide lines on the graph) gives

\[ m_2 = 8.4\,\text{g} \]

(b)(ii) Thickness of the gold coin

Mass \(= 102.0\,\text{g}\), cross-sectional area \(= 10.0\,\text{cm}^2\), density \(= 19.3\,\text{g cm}^{-3}\).

\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{102.0}{19.3} = 5.28\,\text{cm}^3 \]\[ \text{thickness} = \frac{\text{Volume}}{\text{Area}} = \frac{5.28}{10.0} = 0.528\,\text{cm} \]

Test of practical knowledge: current through a length of constantan wire

The circuit

The accumulator \(E\), key \(K\), ammeter \(A\), standard resistor \(R = 2\,\Omega\) and the length \(d\) of constantan wire (tapped by the crocodile clip) are joined in series as shown.

Readings recorded before the main experiment

• e.m.f. of the accumulator, \(E = 2.0\ \text{V}\)
• ammeter reading with the clip off the wire, \(I_0 = 0.90\ \text{A}\)

Table of readings

\(d/\text{cm}\)	\(I/\text{A}\)	\(d^{-1}/\text{cm}^{-1}\)
90.0	0.21	0.0111
80.0	0.23	0.0125
70.0	0.25	0.0143
60.0	0.29	0.0167
50.0	0.34	0.0200

Graph of \(I\) against \(d^{-1}\)

Slope of the graph

Two points are taken on the line of best fit:

\((d^{-1}_1, I_1) = (0.0060\ \text{cm}^{-1},\ 0.13\ \text{A})\) and \((d^{-1}_2, I_2) = (0.0240\ \text{cm}^{-1},\ 0.40\ \text{A})\).

\[ S = \frac{\Delta I}{\Delta (d^{-1})} = \frac{0.40 - 0.13}{0.0240 - 0.0060} = \frac{0.27}{0.0180} = 15\ \text{A}\,\text{cm} \]

Intercept on the vertical axis

Where the line of best fit cuts the \(I\)-axis (at \(d^{-1} = 0\)):

\[ c = 0.04\ \text{A} \]

Evaluation of \(k\)

\[ k = \frac{c}{S} = \frac{0.04}{15} = 2.67\times 10^{-3}\ \text{cm}^{-1} \]

Current when \(d = 55\ \text{cm}\)

\(d^{-1} = \dfrac{1}{55} = 0.0182\ \text{cm}^{-1}\). Reading up from \(0.0182\ \text{cm}^{-1}\) to the line of best fit and across to the \(I\)-axis:

\[ I = 0.31\ \text{A} \]

Two precautions

• The key was opened immediately after each reading so that the wire did not heat up (which would change its resistance) and the accumulator did not run down.
• Clean, tight terminal connections were used and the zero error of the ammeter was noted and corrected before each reading.

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(b)(i) Potential difference between two points

The potential difference between two points in an electric circuit is the work done (energy converted from electrical form to other forms) 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 wire depends

• The length of the wire (resistance increases with length).
• The cross-sectional area of the wire (resistance decreases as the area increases).

(It also depends on the material/resistivity of the wire and on its temperature.)

Test of Practical Knowledge: Refraction through a Rectangular Glass Prism

The outline ABCD is traced, the normal NMP drawn with \(|AM| = |DP| = 2.0\,\text{cm}\), and for each angle of incidence \(i\) the emergent ray is located with pins P₃ and P₄. The angle \(\theta\) between the emergent ray and face AB is measured, and \(\cos\theta\) and \(\sin i\) are computed.

Table of Readings

\(i\,/\,^{\circ}\)	\(\theta\,/\,^{\circ}\)	\(\cos\theta\)	\(\sin i\)
5.0	83.0	0.137	0.087
10.0	74.0	0.276	0.174
15.0	65.0	0.423	0.259
20.0	56.0	0.559	0.342
25.0	48.0	0.669	0.423

Graph of \(\cos\theta\) against \(\sin i\)

The points are plotted with \(\cos\theta\) on the vertical axis and \(\sin i\) on the horizontal axis, and a straight line of best fit is drawn through them.

Determination of the Slope

Two widely separated points are read from the line of best fit:

\(A = (\sin i,\ \cos\theta) = (0.150,\ 0.250)\) and \(B = (0.450,\ 0.720)\).

\[ s = \frac{\Delta(\cos\theta)}{\Delta(\sin i)} = \frac{0.720 - 0.250}{0.450 - 0.150} = \frac{0.470}{0.300} = 1.567 \]

Slope, \(s = 1.57\).

Two Precautions

1. The pins were fixed erect (vertical) and spaced about 4 cm apart so that the ray directions were accurately located.
2. Parallax error was avoided by reading the protractor and metre rule with the eye directly above the scale, and sharp, neat traces were made with a well-sharpened pencil.

(b)(i) Laws of Refraction of Light

1. First law: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
2. Second law (Snell's law): The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media, \(\dfrac{\sin i}{\sin r} = n\), or \(n_1 \sin i = n_2 \sin r\), where the symbols have their usual meanings.

(b)(ii) Meaning of "the refractive index of a material is 1.65"

It means that the ratio of the speed of light in vacuum (air) to the speed of light in the material is 1.65, i.e. \(n = \dfrac{c}{v} = 1.65\). Equivalently, for light passing from air into the material, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is 1.65, so light travels 1.65 times more slowly in the material and bends towards the normal on entering it.