WAEC SSCE - Physics - 2025 (Practical)

Question 1 Report

1a. (i) Using the spring balance provided, determine the weight of an object of mass M = 5.0 g. Record this weight as W\(_1\).

(ii) Determine the weight of the object when completely immersed in water contained in a beaker as shown in the diagram. Record the weight as W\(_2\).

(iii) Determine the weight of the object when it is completely immersed in the liquid labelled "L". Record the weight as W\(_3\). Evaluate u = (W\(_1\) - W\(_2\)) and v = (W\(_1\) - W\(_3\)).

(iv) Repeat the procedure with the objects of masses M = 10, 15, 20, and 25 g. In each case, evaluate v = (W\(_1\) - W\(_3\)) on the vertical axis against u = (W\(_1\) - W\(_2\)) on the horizontal axis.

(v) Determine the slope, s, of the graph.

(vi) State two precautions taken to ensure accurate results.[21 marks]

bi. A piece of brass of mass 20.0 g is hung on a spring balance from a rigid support and completely immersed in kerosene of density 8.0 × 10\(^2\) kgm\(^{-3}\). Determine the reading on the spring balance. [g = 10 ms\(^{-2}\), density of brass = 8.0 × 10\(^3\) kgm\(^{-3}\)]

(Provide your answer with unit e.g 123.123 m)

bii. Archimedes' Principle and Law of Floatation [2 marks]

Answer

Answer Details

1a. Tables of values

M(g)	W\(_1\)(g)	W\(_2\)(g)	W\(_3\)(g)	U = (W\(_1\) - W\(_2\)(g)	V = (W\(_1\) - W\(_3\)(g)
5.0	5.00	4.20	4.70	0.80	0.30
10.0	10.00	7.90	8.60	2.10	1.40
15.0	15.00	12.75	13.50	2.25	1.50
20.0	17.10	17.10	16.40	2.90	3.60
25.0	22.10	22.10	21.40	2.90	3.60

S = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{3 - 0.8}{2.8 - 1.4}\) = \(\frac{11}{7}\) = 1.57.

Precaution

(i) I carefully avoided parallax error while reading the scale of the spring balance to ensure accurate measurements.

(ii) I checked and accounted for any zero error on the spring balance before taking measurements.

(iii) I handled the liquid carefully to prevent any splashing, which could lead to loss of material or inaccurate readings.

(iv) I ensured that the object did not touch the bottom or sides of the beaker to avoid interference with the force readings.

(v) I thoroughly cleaned the object (mass) before dipping it into the liquid to prevent contamination or alteration of the liquid's properties.

bii) Law of Floatation: The law of floatation states that an object will float in a fluid if its weight is equal to the weight of the fluid it displaces. For a floating object, the weight of the object is balanced by the upward buoyant force.

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Question 2 Report

2a.

(i) Fix a metre rule on the bench with the graduated face up.

(ii) Place the illuminated object at the zero end of the rule and the screen at the other end as illustrated in the diagram above.

(iii) Measure and record D, the distance between the object and the screen. Evaluate D\(^2\)

(iv) Place and move the converging lens between the illuminated object and the screen until a diminished, sharp image of the object is formed on the screen. Read and record the position, x\(_1\), of the lens. From this position, move the lens towards the object until another sharp image of the object is formed on the screen. Read and record the position, x\(_2\), of the lens.

(v) Evaluate and record L = (x\(_1\) - x\(_2\)), L\(^2\) and (D\(^2\) - L\(^2\)).

(vi) Repeat the procedure for D = 90, 80, 70, and 60 cm. In each case, evaluate L, L\(^2\), and (D\(^2\) - L\(^2\)). Tabulate your readings.

(vii) Plot a graph of D\(^2\) - L\(^2\) on the vertical axis against D on the horizontal axis.

(viii) Determine the slope, s, of the graph and evaluate K = \(\frac{\text{s}}{4}\). State two precautions taken to ensure accurate results.[21 marks]

bi. Distinction Between Real and Virtual Images [2 marks]

ii. Draw a ray diagram to show how a converging lens may be used to form a real diminished image of an object [2 marks]

Answer

Answer Details

2a. Table of observations

D(cm)	D\(^2\)(cm\^2\)	X\(_1\)(cm)	L = (X\(_1 - X_2\))	L\(^2\)(cm\(^2\))
100	10000	80.50	62.20	3868.8
90	8100	70.20	51.20	2621.4
80	6400	59.10	39.10	1528.8
70	4900	46.70	24.70	610.09
60	3600	29.00	21.40	457.96

S = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{6250 - 3250}{100 - 60}\) = \(\frac{3000}{40}\) = 75

K = \(\frac{\text{s}}{4}\) = \(\frac{75}{4}\) = 18.75

Precautions:

(i) I carefully avoided parallax error by ensuring my eyes were directly in line with the markings on the metre rule while taking measurements to improve accuracy.

(ii) I adjusted the setup and focused precisely to ensure a sharp and well-defined image was formed on the screen, minimizing errors due to blurriness or misalignment.

(iii) I made sure that the experimental apparatus was stable and free from vibrations to avoid any disturbance during measurements.

(iv) I ensured that the metre rule was properly calibrated and positioned perpendicular to the setup to prevent measurement inaccuracies due to angle errors.

bi. A real image is formed when light rays converge and meet at a point, allowing it to be projected onto a screen. It is typically inverted relative to the object, as seen in concave mirrors or convex lenses when the object is beyond the focal point.

On the other hand, a virtual image is formed when light rays appear to diverge from a point but do not actually meet. This type of image cannot be projected onto a screen and is always upright, as seen in plane mirrors or convex mirrors.

bii. See diagram above.

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Question 3 Report

3a. (i) Measure and record the e.m.f of the accumulator provided.

(ii) Connect the circuit as shown in the diagram. S is a standard resistor, and R is a resistance box.

(iii) With R = 0 Ω, close the key K. Read and record the ammeter reading I. Evaluate I\(^{-1}\).

(iv) Repeat the procedure for R = 1, 2, 3, 4, and 5 Ω. Tabulate your readings.

(v) Plot a graph of R on the vertical axis and I\(^{-1}\) on the horizontal axis, starting both axes from the origin (0,0).

(vi) Determine the slope s of the graph and find the intercept c on the vertical axis.

(vii) State two precautions taken to ensure accurate results. [21 marks]

bi. State two advantages of a lead-acid accumulator over a Leclanche cell. [2 marks]

ii. A parallel combination of 3 Ω and 4 Ω resistors is connected in series with a resistor of 4 Ω and a battery of negligible internal resistance. Calculate the effective resistance in the circuit. [2 marks]

Answer

3a. The e.m.f of the accumulator = 1.5V

Tables of values

R(\(\Omega\))	I(A)	I\(^{-1}\)((A\(^{-1}\))
0	0.78	1.28
1	0.50	2.00
2	0.38	2.63
3	0.30	3.33
4	0.25	4.00
5	0.22	4.55

S = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{4 - 2}{4 - 1}\) = \(\frac{2}{3}\) = 0.67

The intercept on the vertical axis = 1.32\(\Omega\)

Precautions:

(i) I ensured all electrical connections were securely tightened to prevent loose contacts, which could lead to fluctuating or inaccurate readings.

(ii) I avoided parallax error by positioning my eyes directly in line with the scale when reading the ammeter or voltmeter for precise measurements.

(iii) I checked and corrected for any zero error on the ammeter and voltmeter before taking readings to improve accuracy.

(iv) I removed the key from the circuit when not taking measurements to avoid unnecessary heating and potential damage to components.

(v) I took multiple readings at each data point and averaged the values to minimize random errors and ensure more reliable results.

(vi) I ensured the apparatus was properly calibrated and regularly checked for consistency throughout the experiment.

bi. 1. Rechargeability: A lead-acid accumulator is rechargeable, making it more suitable for long-term use, whereas a Leclanché cell is primarily a primary (non-rechargeable) cell.

2. Higher Current Supply: Lead-acid accumulators can deliver higher currents, making them ideal for applications requiring significant power, like in vehicles, while Leclanché cells are suitable for low-current devices.

3. Longer Lifespan: Due to its rechargeable nature, the lead-acid accumulator has a longer operational life compared to a Leclanché cell, which has a limited lifespan and must be replaced after depletion.

4. Stable Voltage: Lead-acid accumulators provide a relatively stable output voltage during discharge, while the voltage of a Leclanché cell drops significantly over time.

bii. Since 3 Ω and 4 Ω are connected in parallel,

\(\frac{1}{\text{R}}\) = \(\frac{1}{3}\) + \(\frac{1}{4}\)

\(\frac{1}{\text{R}}\) = \(\frac{4 + 3}{12}\)

\(\frac{1}{\text{R}}\) = \(\frac{7}{12}\)

R = \(\frac{12}{7}\) = 1.71 Ω

Now, 1.71 Ω is connected in series with 4 Ω.

∴ The effective resistance in the circuit = 1.71 Ω + 4 Ω = 5.71 Ω

Answer Details

3a. The e.m.f of the accumulator = 1.5V

Tables of values

R(\(\Omega\))	I(A)	I\(^{-1}\)((A\(^{-1}\))
0	0.78	1.28
1	0.50	2.00
2	0.38	2.63
3	0.30	3.33
4	0.25	4.00
5	0.22	4.55