WAEC SSCE - Physics - 2005 (Practical)

You are provided with a potentiometer XY, a voltmeter, V, a standard resistor R, an accumulator, E a plug key, K, a jockey, and connecting wires.

1.  Connect a circuit as shown in the diagram above.
2. Close the key and use the jockey to make contact with the potentiometer with XY at a point N such that l = XN= 15cm.
3. Read and record the value of the potential difference V on the voltmeter.
4. Evaluate l\(^{-1}\) and V\(^{-1}\).
5. Repeat the procedure for five other values of I= 25, 35, 45, 55, and 65cm respectively. Read and record the value of V and evaluate V\(^{-1}\) and l\(^{-1}\) in each case. Tabulate your readings.
6. Plot a graph V\(^{-1}\) on the vertical axis against l\(^{-1}\)on the horizontal axis starting both axes from the origin (0,0).
7. Determine the slope, s, of the graph.
8. Evaluate k = \(\frac{1}{s}\)
9. State two precautions taken to ensure accurate re- results.

(b)i. State four factors on which the resistance of a wire depends.

ii. A resistance Wire of length 100cm is connected in a circuit. If the resistance per unit length of the wire is 0.02 \(\Omega\)cm\(^{-1}\), how much heat would be produced in the wire if a voltmeter connected across its ends indicates 1.5V while the current runs for 1 minute?

(a) Potentiometer experiment

Circuit: The accumulator \(E\), key \(K\) and standard resistor \(R\) are joined in series with the potentiometer wire \(XY\). The voltmeter \(V\) is connected between \(X\) and the jockey, so it reads the potential difference across the length \(l = XN\).

With the key closed, the jockey is pressed at \(N\) so that \(l = XN = 15.0\,\text{cm}\) and the voltmeter reading \(V\) is recorded. The procedure is repeated for \(l = 25.0, 35.0, 45.0, 55.0\) and \(65.0\,\text{cm}\). For each length \(l^{-1}\) and \(V^{-1}\) are evaluated.

Table of readings

S/N	\(l\,/\,\text{cm}\)	\(V\,/\,\text{volt}\)	\(l^{-1}\,/\,\text{cm}^{-1}\)	\(V^{-1}\,/\,\text{volt}^{-1}\)
1	15.0	0.60	0.0667	1.667
2	25.0	0.80	0.0400	1.250
3	35.0	1.10	0.0286	0.909
4	45.0	1.30	0.0222	0.769
5	55.0	1.60	0.0182	0.625
6	65.0	1.80	0.0154	0.556

Worked evaluations (row 1): \(l^{-1}=\dfrac{1}{15.0}=0.0667\,\text{cm}^{-1}\), \(V^{-1}=\dfrac{1}{0.60}=1.667\,\text{volt}^{-1}\); (row 6): \(l^{-1}=\dfrac{1}{65.0}=0.0154\,\text{cm}^{-1}\), \(V^{-1}=\dfrac{1}{1.80}=0.556\,\text{volt}^{-1}\).

Graph of \(V^{-1}\) against \(l^{-1}\)

Slope of the graph

Using two points on the line of best fit, \((l^{-1}=0.0154,\ V^{-1}=0.603)\) and \((l^{-1}=0.0667,\ V^{-1}=1.724)\):

\[ s=\frac{\Delta V^{-1}}{\Delta l^{-1}}=\frac{1.724-0.603}{0.0667-0.0154}=\frac{1.121}{0.0513}=21.85\ \text{cm}\,\text{volt}^{-1}. \]

Evaluate \(k=\tfrac{1}{s}\)

\[ k=\frac{1}{s}=\frac{1}{21.85}=0.046\ \text{volt}\,\text{cm}^{-1}. \]

Two precautions

1. I ensured that all connecting wires and terminals were clean and firmly tightened to avoid loose contacts.
2. I read the voltmeter with my eye directly in line with the pointer to avoid parallax error.

(b)(i) Four factors on which the resistance of a wire depends

1. The length of the wire \((R\propto l)\).
2. The cross-sectional area of the wire \((R\propto \tfrac{1}{A})\).
3. The nature (material / resistivity) of the wire.
4. The temperature of the wire.

(b)(ii) Heat produced in the wire

Resistance of the wire:

\[ R = (\text{resistance per unit length})\times(\text{length}) = 0.02\times100 = 2\,\Omega. \]

Given \(V=1.5\,\text{V}\) and \(t=1\,\text{minute}=60\,\text{s}\), the heat produced is

\[ H=\frac{V^{2}t}{R}=\frac{(1.5)^{2}\times60}{2}=\frac{2.25\times60}{2}=\frac{135}{2}=67.5\,\text{J}. \]

Check (using current): \(I=\dfrac{V}{R}=\dfrac{1.5}{2}=0.75\,\text{A}\), so \(H=I^{2}Rt=(0.75)^{2}\times2\times60=0.5625\times120=67.5\,\text{J}.\)

Karanta bayanin darasi akan Current Electricity (JAMB)

You have been provided with a retort stand, clamp and boss, a set of masses, a spiral spring, stopwatch, split cork, and other necessary apparatus. Using the diagram above as a guide, carry out the following instructions;

1. Suspend the spiral spring vertically as shown in the diagram.
2. Suspend a mass hanger on the free end of the spiral spring and add a mass, m 50.0g to the hanger.
3. Pull the hanger gently downwards and release to set it into vertical 0scillations.
4. Determine the time, t, for 20 complete oscillations.
5. Evaluate the period, t, of the oscillation. Also, evaluate T\(^{2}\).
6. Repeat the procedure for four other values of m= 70, 90, 110, and 130g. In each case, determine t and evaluate T and T\(^{2}\). Tabulate your readings.
7. Plot a graph of T\(^{2}\) on the vertical axis against m on the horizontal axis.
8. Determine the slope, s, of the graph and the intercept, l, on the vertical axis.
9. Evaluate k = 4\(\frac{\pi ^{2}}{s}\), Take t = \(\frac{22}{7}\)
10. State two precautions taken to ensure accurate results.

(b)i. Define Young modulus and force constant.

ii. A force of magnitude 500N is applied to the free end of a spiral spring of force constant 1.0 x 10\(^{4}\) Nm\(^{-1}\). Calculate the energy stored in the stretched spring.

Observation and calculations

For each load, the time for 20 complete oscillations was measured twice. The mean time is \(t\), hence

\[T=\frac{t}{20},\qquad T^2=\left(\frac{t}{20}\right)^2.\]

Mass, \(m\) (g)	\(t_1\) (s)	\(t_2\) (s)	Mean \(t\) for 20 oscillations (s)	Period, \(T=t/20\) (s)	\(T^2\) (s2)	\(T^2\times10^3\) (s2)
50.0	5.80	5.80	5.80	0.290	0.084	84
70.0	6.70	6.90	6.80	0.340	0.116	116
90.0	7.80	8.00	7.90	0.395	0.156	156
110.0	8.50	8.70	8.60	0.430	0.185	185
130.0	9.20	9.20	9.20	0.460	0.212	212

The graph is plotted with \(T^2\times10^3\) on the vertical axis and mass on the horizontal axis.

Gradient, intercept and force constant

Using two widely separated points on the straight line, \((m,T^2\times10^3)=(6,10)\) and \((140,234)\),

\[s=\frac{(234-10)\times10^{-3}}{140-6}=1.672\times10^{-3}\ \text{s}^2\text{g}^{-1}=1.672\ \text{s}^2\text{kg}^{-1}.\]

The vertical intercept, \(l\), is approximately \(0\ \text{s}^2\).

Since \(T^2=\dfrac{4\pi^2}{k}m\),

\[k=\frac{4\pi^2}{s}=\frac{4\left(\frac{22}{7}\right)^2}{1.672}=23.6\ \text{N m}^{-1}.\]

Precautions

1. The load was made to oscillate vertically only; conical or sideways oscillations were avoided.
2. Small oscillations were used, and the time for 20 oscillations was measured and repeated before taking the mean.

(b)(i) Definitions

Young modulus is the ratio of tensile stress to tensile strain, provided the elastic limit is not exceeded:

\[E=\frac{F/A}{e/L}=\frac{FL}{Ae}.\]

Force constant is the force required to produce unit extension of a spring:

\[k=\frac{F}{e}.\]

(b)(ii) Energy stored in the spring

\[e=\frac{F}{k}=\frac{500}{1.0\times10^4}=0.050\ \text{m}.\]

\[W=\frac{1}{2}Fe=\frac{1}{2}\times500\times0.050=12.5\ \text{J}.\]

Karanta bayanin darasi akan Elasticity (JAMB)

using the diagram above as a guide:

1. Trace the outline ABC of the equilateral triangular glass prism provided.
2. Remove the prism. Draw a line MN such that it makes an angle \(i = 5^\circ\) with the normal at N on side AB of the outline.
3. Fix two pins at \(P_{1}\) and \(P_{2}\) on MN. Replace the prism on its outline.
4. Looking through the face BC of the prism, fix one pin at \(P_{3}\) and another at \(P_{4}\) Such that they are in a straight line with the images of the pins at \(P_{1}\) and \(P_{2}\).
5. Remove the prism and the pins. Draw a line to join \(P_{4}\) and \(P_{3}\). Produce line \(P_{4}P_{3}\) to meet the line BC of the outline at CQ and line MN produced at P.
6. Draw a normal to BC at Q. Measure and record the angles \(\theta\) and e. Evaluate \(\phi = i + e\).
7. Repeat the procedure, using a different outline in each case, for four other values of \(i = 100\), \(159\), \(20\), and \(25\) respectively. Evaluate \(\phi = i + e\) in each case. Tabulate your readings.
8. Plot a graph of \(\theta\) on the vertical axis against \(\phi\) on the horizontal axis starting both axes from the origin \((0,0)\).
9. Determine the slope of the graph and the intercept on the vertical axis.
10. State two precautions taken to ensure accurate results.

(b)i. Explain what is meant by the statement: the refractive index of glass is 1.5.

ii. Calculate the critical angle of a medium of refractive index 1.65 when light passes from the medium to air.

Principle. A ray of light strikes face AB of the equilateral glass prism at an angle of incidence \(i\) to the normal, is refracted through the glass and emerges from face BC at an angle of emergence \(e\). For each setting the angle \(\theta\) at Q and the emergence angle \(e\) are measured, and \(\phi = i + e\) is evaluated. A graph of \(\theta\) against \(\phi\) is a straight line whose slope and vertical intercept are then read off.

Ray-tracing set-up.

Trace the outline ABC, draw MN at the required angle \(i\) to the normal at N on AB and fix pins P\(_1\), P\(_2\). Replace the prism, and looking through face BC fix pins P\(_3\), P\(_4\) in line with the images of P\(_1\) and P\(_2\). Join and produce P\(_4\)P\(_3\) to meet BC at Q, draw the normal at Q, then measure \(\theta\) and \(e\).

Table of readings.

S/N	\(i\) (°)	\(e\) (°)	\(\theta\) (°)	\(\phi=i+e\) (°)
1	5.0	5.0	130.0	10.0
2	10.0	10.0	140.0	20.0
3	15.0	15.0	150.0	30.0
4	20.0	20.0	160.0	40.0
5	25.0	25.0	170.0	50.0

Graph of \(\theta\) against \(\phi\).

Slope and intercept. Taking two widely separated points on the line of best fit, \((\phi_1,\theta_1)=(0,120)\) and \((\phi_2,\theta_2)=(50,170)\):

\[ \text{slope} = \frac{\Delta\theta}{\Delta\phi} = \frac{170-120}{50-0} = \frac{50}{50} = 1.0 \]

The line cuts the vertical axis at \(\phi = 0\), giving an intercept \(\theta_{\text{intercept}} = 120^\circ\).

Two precautions.

1. The object pins and image pins were fixed erect and viewed with one eye so that they and the images lay in one straight line, avoiding parallax error.
2. A sharp pencil was used for tracing the outline and marking pin positions, and the protractor was read squarely to obtain accurate angles.

(b)(i) The statement that the refractive index of glass is 1.5 means that the speed (or wavelength) of light in air is 1.5 times its speed (or wavelength) in the glass:

\[ n = \frac{\text{speed of light in air}}{\text{speed of light in glass}} = \frac{3}{2} = 1.5 \]

Equivalently, for a ray passing from air into the glass, \(\dfrac{\sin i}{\sin r} = 1.5\).

(b)(ii) Critical angle. At the critical angle \(C\) the ray inside the glass just grazes the surface, so

\[ {}_{a}n_{g} = \frac{1}{\sin C} \;\Rightarrow\; \sin C = \frac{1}{n} = \frac{1}{1.65} = 0.6061 \]\[ C = \sin^{-1}(0.6061) = 37.3^\circ \]

Karanta bayanin darasi akan Electrostatics (JAMB)