WAEC SSCE - Physics - 2025 (Practical)

Detalhes da Resposta

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

Tables of values

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

Detalhes da Resposta

2a. Table of observations

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.

Detalhes da Resposta

1a. Tables of values

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.