WAEC SSCE - Physics - 2002 (Practical)

1.  Measure and record the length XY of the resistance wire provided.
2. Connect the circuit shown in the diagram.
3. With R= O\(\Omega\), close the key, K. Read and record the current 1\(_{o}\) and the voltage drop V\(_{o}\) across the resistance wire.
4. Setting R = 1\(\Omega\). close the key. Read and record the current, I, and the corresponding voltage drop, V across the wire.
5. Repeat the procedure for five other values of R= 5, 10, 20, 40, and 60\(\Omega\). Tabulate your readings.
6. Plot a graph of V on the vertical axis against 1 on the horizontal axis.
7. Determine the slope of the graph
8. State two precautions taken to ensure accurate results.

(b)i. Mention and state the law on which the experiment in (a) is based.

ii. A piece of resistance vire of diameter 0.2 mm and resistance m has a resistivity of 8.8 x 10\(^{-7}\)\(\Omega\)m, calculate the length of the Wire. [\(\pi\) =\(\frac{22}{7}\)]

(a) Readings and graph

Length of resistance wire, XY = 100 cm.

Resistance, R (Ω)	Current, I (A)	Potential difference, V (V)
0	0.15	2.65
1	0.15	2.65
5	0.15	2.65
10	0.10	1.85
20	0.08	1.40
40	0.06	0.90
60	0.04	0.70

Graph of V against I:

Using two widely separated points on the line of best fit, \[ (I_1,V_1)=(0.025\,\text{A},0.50\,\text{V}), \qquad (I_2,V_2)=(0.100\,\text{A},1.75\,\text{V}) \] \[ \text{Slope}=\frac{\Delta V}{\Delta I}=\frac{1.75-0.50}{0.100-0.025}=\frac{1.25}{0.075}=16.7\ \Omega. \] Thus, the resistance of the wire is approximately 16.7 Ω.

Precautions:

1. Connections were clean, tight and correctly made before taking readings.
2. The key was opened immediately after each reading to prevent heating of the resistance wire.

(b)(i)

The experiment is based on Ohm's law. Ohm's law states that the current through a metallic conductor is directly proportional to the potential difference across its ends, provided that temperature and other physical conditions remain constant. Thus,

\[\frac{V}{I}=\text{constant}.\]

(b)(ii)

Diameter of wire, \(d=0.2\,\text{mm}=2.0\times10^{-4}\,\text{m}\).

\[A=\frac{\pi d^2}{4}=\frac{\frac{22}{7}(2.0\times10^{-4})^2}{4}=3.14\times10^{-8}\,\text{m}^2.\]

Using \(R=7\,\Omega\), \(\rho=8.8\times10^{-7}\,\Omega\text{m}\), and \(R=\frac{\rho L}{A}\),

\[L=\frac{RA}{\rho}=\frac{7\times3.14\times10^{-8}}{8.8\times10^{-7}}=0.25\,\text{m}.\]

Length of the wire = \(0.25\,\text{m}\).

Soma maelezo ya somo kwenye Current Electricity (JAMB)

Using the diagram as a guide, carry out the following instructions:

1. Fix a plain sheet of paper on the drawing board.
2. Place the rectangular glass prism on the paper and trace its outline, ABCD. Remove the prism.
3. Draw a normal NMP to meet AB and DC at M and P respectively such that \(|\textbf{AM}| = |\textbf{DP}| = 2.0\text{cm}\).
4. Trace the ray PQ with two pins, P\(_{1}\), and P\(_{2}\), at P and Q respectively such that angle MPQ = i = 5º.
5. Replace the prism on its outline. Trace the emergent ray with two other pins P\(_{3}\) and P\(_{4}\) such that they lie in a straight line with P\(_{2}\) and the image of P\(_{1}\) viewed through the glass prism.
6. Measure and record \(\theta\), the angle between the emergent ray and the face AB of the glass prism.
7. Evaluate \(\cos \theta\) and \(\sin i\).
8. Repeat the procedure for four other values of i= 10°, 15°, 20°, and 25°. Tabulate your readings.
9. Plot a graph of \(\cos \theta\) on the vertical axis against \(\sin i\) on the horizontal axis.
10. Determine the slope of the graph.
11. State two precautions taken to ensure accurate results. Attach your traces to your answer booklet

(b)i. State the laws of refraction of light.

ii. Explain what is meant by the statement the refractive index of a material is 1.65.

(a) Refraction through a rectangular glass prism

The outline ABCD of the rectangular glass prism is traced on the paper. The normal NMP is drawn to meet AB at M and DC at P, with \(|AM|=|DP|=2.0\,\text{cm}\). The incident ray PQ is fixed with pins \(P_1\) and \(P_2\) at the chosen angle \(i=\angle MPQ\). With the prism replaced, the emergent ray is traced with pins \(P_3\) and \(P_4\) set in line with \(P_2\) and the image of \(P_1\) seen through the glass. The angle \(\theta\) between the emergent ray and face AB is measured, and \(\cos\theta\) and \(\sin i\) are evaluated. The set-up is shown below.

Table of readings

\(i\,/\,^{\circ}\)	\(\theta\,/\,^{\circ}\)	\(\cos\theta\)	\(\sin i\)
5	81	0.156	0.087
10	74	0.276	0.174
15	69	0.358	0.259
20	61	0.484	0.342
25	51	0.629	0.423

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

Slope of the graph

The slope is read from two widely separated points on the line of best fit, using the first and last tabulated points \((\sin i,\ \cos\theta)=(0.087,\ 0.156)\) and \((0.423,\ 0.629)\):

\[ \text{slope}=\frac{\Delta(\cos\theta)}{\Delta(\sin i)}=\frac{0.629-0.156}{0.423-0.087}=\frac{0.473}{0.336} \]\[ \boxed{\text{slope}\approx 1.41} \]

Two precautions

1. Parallax error was avoided when reading the angles and when sighting the pins.
2. The optical pins were kept straight (not bent) and fixed vertically, well spaced apart, so that the incident and emergent rays could be aligned accurately.

(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, \[ \frac{\sin i}{\sin r}=\text{constant}. \]

(b)(ii) Meaning of a refractive index of 1.65

It means that the ratio of the speed of light in a vacuum to the speed of light in the material is equal to 1.65:

\[ n=\frac{\text{speed of light in vacuum}}{\text{speed of light in the material}}=1.65. \]

Equivalently, light travels 1.65 times faster in a vacuum than inside the material, and for a ray passing from air into the material \(\dfrac{\sin i}{\sin r}=1.65\).

Soma maelezo ya somo kwenye Electrostatics (JAMB)