WAEC SSCE - Physics - 2019 (Practical)

Question 1 Report

You are provided with a stopwatch, a meter rule, a split cork, retort stand and clamp, a pendulum bob, a piece of thread, and other necessary apparatus.

i. Place the retort stand on a laboratory stool. Clamp the split cork.

ii. Suspend the pendulum bob from the split cork such that the point of support P of the bob is at height \(H = 100\text{cm}\) above the floor Q. The bob should not touch the floor and H should be kept constant throughout the experiment.

iii. Adjust the length of the thread such that the center A of the bob is at a height \(y = \text{AQ} = 20\text{cm}\) from the floor.

iv. Displace the bob such that it oscillates in a horizontal plane.

v. Take the time t for 20 complete oscillations.

vi. Determine the period T of oscillation and evaluate T

vii. Repeat the procedure for four other values of \(y = 30\text{cm}, 40\text{cm}, 50\text{cm},\) and \(60\text{cm}\). In each case, determine T and T.

viii. Tabulate the results.

ix. Plot a graph of T on the vertical axis and y on the horizontal axis, starting both axes from the origin (0,0).

x. Determine the slope, s, of the graph and the intercept c on the vertical axis.

xi. If in this experiment SR= c, calculate R.

x. State two precautions taken to ensure accurate results.

(b) i. The bob of a simple pendulum is displaced a small distance from the equilibrium position and then released to perform simple harmonic motion Identify where its:
\((\propto)\) kinetic energy is maximum
\((\beta)\) acceleration is maximum

ii. An object of weight 120N vibrates with a period of 4.0s when hung from a spring. Calculate the force per unit length of the spring. [\(g = 10\text{ms}^{-2}\), \(\pi = 3.142\)]

Answer

Answer Details

(a) Pendulum experiment

The length of the pendulum is \(l=H-y\), where \(H=100\,\text{cm}\). For each setting, the time for 20 oscillations is measured and

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

S/N	\(y\) (cm)	Time, \(t\), for 20 oscillations (s)	Period, \(T=t/20\) (s)	\(T^2\) (s²)
1	20	50.50	2.525	6.376
2	30	47.50	2.375	5.641
3	40	43.50	2.175	4.731
4	50	40.00	2.000	4.000
5	60	35.80	1.790	3.204

Graph of \(T^2\) against \(y\):

The graph has a negative gradient. From the line of best fit, using two well-separated points on the line, \((20\,\text{cm},6.39\,\text{s}^2)\) and \((60\,\text{cm},3.20\,\text{s}^2)\):

\[\text{gradient}=\frac{3.20-6.39}{60-20}=-0.0798\,\text{s}^2\text{cm}^{-1}.\]

Hence, the magnitude of the slope is \(S=0.080\,\text{s}^2\text{cm}^{-1}\). The intercept on the \(T^2\)-axis is \(c\approx 8.00\,\text{s}^2\).

Since \(SR=c\),

\[R=\frac{c}{S}=\frac{8.00}{0.080}=100\,\text{cm}.\]

Precautions:

The bob was given a small angular displacement and released gently.
The support was kept rigid and the bob was allowed to oscillate freely without draughts or touching the floor.

(b)

(i) The kinetic energy is maximum at the equilibrium position. The acceleration is maximum at either extreme position, where the displacement is maximum.

(ii) \(W=120\,\text{N}\), so

\[m=\frac{W}{g}=\frac{120}{10}=12\,\text{kg}.\]

For a mass-spring system,

\[T=2\pi\sqrt{\frac{m}{k}},\qquad k=\frac{4\pi^2m}{T^2}.\]

\[k=\frac{4(3.142)^2(12)}{(4.0)^2}=29.6\,\text{N m}^{-1}.\]

The force per unit length (spring constant) is \(\boxed{29.6\,\text{N m}^{-1}}\).

Read lesson note on Simple Harmonic Motion (WAEC)

Simple Harmonic Motion

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Question 2 Report

You are provided with a converging lens and holder, a screen, a ray box containing an illuminated object pin, and a meter rule.

i. Place the lens in its holder such that it is facing a distant object seen through a well-lit laboratory window. Move the screen to and fro until a sharp image of the distant object is formed on it. Measure the distance, \(f_{0}\), between the screen and the lens.

ii. Clamp the meter rule securely to the table. Place the illuminated object pin at the end R of the meter rule.

iii. Place the lens at a position P such that \(X = RP =\) 20cm.

iv. Move the screen to a position Q to receive a sharp image of the object. Measure the distance \(Y = PQ\).

v. Evaluate \(Z = (X+Y)\)

vi. Repeat the procedure for five other values of x = 25cm. 3Ocm, 35cm, 40cm and 45cm. In each case, record X,Y and evaluate Z.

vii. Tabulate the results.

viii. Plot a graph with Z on the vertical axis and X on the horizontal axis. Draw a smooth curve through the points.

ix. Determine from your graph the minimum value of \(Z=Z_{0}\) and its corresponding distance

x. Evaluate \(W = ½ \left(\frac{Z_0}{4} + \frac{X_0}{2}\right)\)

xi. State two precautions taken to ensure accurate results.

(b) i. Draw a ray diagram to show how a Convex lens forms an image of magnification less than one.

ii. Name two pairs of features in the human eye and a lens camera that performs similar functions.

Answer

(a) Determination of the focal length of a converging lens

On focusing the distant object sharply on the screen, the focal length obtained was:

\(f_0 = 13.0\text{ cm}\).

The readings obtained with the illuminated object pin were as follows.

S/N	\(X=RP\) (cm)	\(Y=PQ\) (cm)	\(Z=X+Y\) (cm)
1	20	60	80
2	25	38	63
3	30	30	60
4	35	26	61
5	40	24	64
6	45	21	66

Graph of \(Z\) against \(X\):

From the lowest point of the curve,

\(Z_0=60\text{ cm}\), and \(X_0=30\text{ cm}\).

\[W=\frac{1}{2}\left(\frac{Z_0}{4}+\frac{X_0}{2}\right)=\frac{1}{2}\left(\frac{60}{4}+\frac{30}{2}\right)=\frac{1}{2}(15+15)=15\text{ cm}.\]

Precautions

The object pin, optical centre of the lens and centre of the screen were kept at the same height and on one straight line.
The screen was moved until the image was sharply focused, and metre-rule readings were taken with the eye vertically above the mark to avoid parallax.

(b)(i) Ray diagram for a diminished image formed by a convex lens

The object is placed beyond \(2F_1\). The image formed is real, inverted and diminished, between \(F_2\) and \(2F_2\).

Convex lens with the object beyond 2F₁ and a real, inverted, diminished image between F₂ and 2F₂.

(b)(ii) Corresponding parts of the human eye and lens camera

Human eye	Lens camera
Retina	Film or image sensor
Eye lens	Camera lens
Iris	Diaphragm

Read lesson note on Reflection Of Light At Plane And Curved Surfaces (JAMB)

Read lesson note on Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Answer Details

(a) Determination of the focal length of a converging lens

On focusing the distant object sharply on the screen, the focal length obtained was:

\(f_0 = 13.0\text{ cm}\).

The readings obtained with the illuminated object pin were as follows.

S/N	\(X=RP\) (cm)	\(Y=PQ\) (cm)	\(Z=X+Y\) (cm)
1	20	60	80
2	25	38	63
3	30	30	60
4	35	26	61
5	40	24	64
6	45	21	66

Graph of \(Z\) against \(X\):

From the lowest point of the curve,

\(Z_0=60\text{ cm}\), and \(X_0=30\text{ cm}\).

\[W=\frac{1}{2}\left(\frac{Z_0}{4}+\frac{X_0}{2}\right)=\frac{1}{2}\left(\frac{60}{4}+\frac{30}{2}\right)=\frac{1}{2}(15+15)=15\text{ cm}.\]

Precautions

The object pin, optical centre of the lens and centre of the screen were kept at the same height and on one straight line.
The screen was moved until the image was sharply focused, and metre-rule readings were taken with the eye vertically above the mark to avoid parallax.

(b)(i) Ray diagram for a diminished image formed by a convex lens

The object is placed beyond \(2F_1\). The image formed is real, inverted and diminished, between \(F_2\) and \(2F_2\).

(b)(ii) Corresponding parts of the human eye and lens camera

Human eye	Lens camera
Retina	Film or image sensor
Eye lens	Camera lens
Iris	Diaphragm

Read lesson note on Reflection Of Light At Plane And Curved Surfaces (JAMB)

Read lesson note on Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Reflection Of Light At Plane And Curved Surfaces

Properties Of Waves: Reflection, Refraction, Diffraction

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Question 3 Report

You are provided with a battery of e.m.f. E, a key K, a voltmeter, a standard resistor R₀ = \(2\Omega\), a resistance box R, and some connecting wires.

i. Measure and record the e.m.f. E of the battery.

i. Set up a circuit as shown in the diagram above with the key open.

iii. Set the resistance on the resistance box to R 22.

iv. Close the key, read and record the potential difference V on the voltmeter.

v. Evaluate \(V^{-1}\)

vi. Repeat the procedures for five other values of R = \(5\Omega\) \(10\Omega\), \(12\Omega\), \(15\Omega\) and \(20\Omega\). In each case, record V and evaluate V\(^{-1}\)

vii. Tabulate the results.

viii. Plot a graph with R on the vertical axis and \(V^{-1}\) l on the horizontal axis, starting both axes from the origin (0,0).

ix. Determine the slope, s, of the graph and the intercept c on the vertical axis.

x. Calculate \(\propto\) and \(\beta\) from the equations s = R₀ \(\propto\) and c= - (R₀+B).

xi. State two precautions taken to obtain accurate results.

Answer

Determination of the e.m.f. and internal resistance of a battery

The e.m.f. of the battery, measured with the key open, is:

\(E=2.00\ \text{V}\)

With \(R_0=2\ \Omega\), the resistance box was set successively to \(2, 5, 10, 12, 15\) and \(20\ \Omega\). The voltmeter was connected across the standard resistor \(R_0\).

Table of readings

S/N	\(R\ (\Omega)\)	\(V\ (\text{V})\)	\(V^{-1}\ (\text{V}^{-1})\)
1	2	0.800	1.250
2	5	0.500	2.000
3	10	0.308	3.250
4	12	0.267	3.750
5	15	0.222	4.500
6	20	0.174	5.750

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

The graph is a straight line. Using two well-separated points on the line, \((V^{-1}=1.25\ \text{V}^{-1},\ R=2.0\ \Omega)\) and \((V^{-1}=5.75\ \text{V}^{-1},\ R=20.0\ \Omega)\):

\[s=\frac{\Delta R}{\Delta(V^{-1})}=\frac{20.0-2.0}{5.75-1.25}=\frac{18.0}{4.50}=4.0\ \Omega\,\text{V}.\]

From \(R=sV^{-1}+c\),

\[c=R-sV^{-1}=2.0-(4.0\times1.25)=-3.0\ \Omega.\]

Hence,

\[\alpha=\frac{s}{R_0}=\frac{4.0}{2.0}=2.0\ \text{V},\]

\[\beta=-c-R_0=-(-3.0)-2.0=1.0\ \Omega.\]

Therefore, \(\alpha\), the e.m.f. of the battery, is \(2.0\ \text{V}\), while \(\beta\), the internal resistance of the battery, is \(1.0\ \Omega\).

Precautions

The key was closed only while taking a reading, to prevent heating and polarization of the battery.
All connections were clean and tight, and the voltmeter was read at eye level to avoid parallax error.

Read lesson note on Capacitors (JAMB)

Read lesson note on Simple A.C Circuits (JAMB)

Answer Details

Determination of the e.m.f. and internal resistance of a battery

The e.m.f. of the battery, measured with the key open, is:

\(E=2.00\ \text{V}\)

With \(R_0=2\ \Omega\), the resistance box was set successively to \(2, 5, 10, 12, 15\) and \(20\ \Omega\). The voltmeter was connected across the standard resistor \(R_0\).

Table of readings

S/N	\(R\ (\Omega)\)	\(V\ (\text{V})\)	\(V^{-1}\ (\text{V}^{-1})\)
1	2	0.800	1.250
2	5	0.500	2.000
3	10	0.308	3.250
4	12	0.267	3.750
5	15	0.222	4.500
6	20	0.174	5.750

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

The graph is a straight line. Using two well-separated points on the line, \((V^{-1}=1.25\ \text{V}^{-1},\ R=2.0\ \Omega)\) and \((V^{-1}=5.75\ \text{V}^{-1},\ R=20.0\ \Omega)\):

\[s=\frac{\Delta R}{\Delta(V^{-1})}=\frac{20.0-2.0}{5.75-1.25}=\frac{18.0}{4.50}=4.0\ \Omega\,\text{V}.\]

From \(R=sV^{-1}+c\),

\[c=R-sV^{-1}=2.0-(4.0\times1.25)=-3.0\ \Omega.\]

Hence,

\[\alpha=\frac{s}{R_0}=\frac{4.0}{2.0}=2.0\ \text{V},\]

\[\beta=-c-R_0=-(-3.0)-2.0=1.0\ \Omega.\]

Therefore, \(\alpha\), the e.m.f. of the battery, is \(2.0\ \text{V}\), while \(\beta\), the internal resistance of the battery, is \(1.0\ \Omega\).

Precautions

The key was closed only while taking a reading, to prevent heating and polarization of the battery.
All connections were clean and tight, and the voltmeter was read at eye level to avoid parallax error.

Read lesson note on Capacitors (JAMB)

Read lesson note on Simple A.C Circuits (JAMB)

Capacitors

Simple A.C Circuits

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