JAMB Physics 2010 Past Questions & Answers

This question is about relative velocity. When two objects are moving in the same direction, the velocity of one object as seen by an observer in the other is found by subtracting their speeds. Therefore, the velocity of the faster car relative to the slower car is 130 kmh⁻¹ minus 100 kmh⁻¹, which gives 30 kmh⁻¹.

Common mistake: Students often mistakenly add the speeds together, which is only correct when the cars are moving in opposite directions, leading them to choose 230 kmh⁻¹.

When we ignore air resistance, the only force acting on any falling object is gravity. The acceleration caused by gravity (g) is constant for all objects near the Earth's surface, regardless of their mass or shape. Since both the stone and the feather start from the same height and have the same downward acceleration, they must take the same amount of time to reach the ground.

Common mistake: Students often incorrectly choose D because their everyday experience shows that a heavy stone falls faster than a light feather, but this is due to air resistance, which the question specifically says to ignore.

To find the acceleration, we use the formula that connects initial velocity (u), final velocity (v), time (t), and acceleration (a). The formula is a = (v - u) / t. In this question, the initial velocity u is 25 ms⁻¹, the final velocity v is 45 ms⁻¹, and the time t is 10s. Substituting these values, we get a = $(45 - 25)$ / 10, which simplifies to 20 / 10, giving an acceleration of 2 ms⁻².

Common mistake: Students might mistakenly calculate only the change in velocity $(45 - 25 = 20)$ and forget to divide by the time, leading them to incorrectly choose 20 ms⁻².

Friction is a force that opposes motion between two surfaces that are in contact. By making the surface of a conveyor belt rough, the force of friction between the belt and the load is increased. This increased friction is necessary to prevent the load from slipping backwards as the belt moves, ensuring it is transported effectively.

Common mistake: Students may mistakenly choose option D, confusing the surface texture (roughness) with the overall material strength of the belt, but the roughness is specifically to increase friction, not to make the belt stronger.

First, calculate the Mechanical Advantage (M.A.) which is Load divided by Effort, so M.A. = 800 N / 400 N = 2. The efficiency of a machine is then found using the formula: Efficiency = (Mechanical Advantage / Velocity Ratio) x 100%. Using the given Velocity Ratio of 6, the efficiency is $(2 / 6)$ x 100%, which equals 33.33%. This value is closest to 33.30%.

Common mistake: Students often mistakenly invert the Mechanical Advantage formula, calculating Effort/Load $(400/800 = 0.5)$ and then incorrectly select 50.00% as the efficiency.

Strain energy is the potential energy stored in a stretched material, calculated with the formula: Energy = ½ × Force × extension. First, find the extension (e) by subtracting the original length from the final length: 30.5 cm - 30 cm = 0.5 cm. It is important to convert this extension to the standard unit, metres, by dividing by 100 (0.5 cm = 0.005 m). Now, substitute the values into the formula: Energy = ½ × 300 N × 0.005 m = 0.75 J.

Common mistake: Students often forget to convert the extension from centimetres (0.5 cm) to metres (0.005 m), leading them to incorrectly calculate ½ × 300 × 0.5 = 75 J, which is option C.

This question applies the principle of pressure in fluids, specifically Pascal's principle used in hydraulic presses. The relationship between pressure (P), force (F), and area (A) is given by the formula P = F/A. To find the force, we rearrange the formula to F = P × A. Given the pressure is 100 Pa and the area is 3 m², the force is 100 × 3 = 300 N.

Common mistake: A common mistake is to confuse the value for pressure with the value for force, leading to the incorrect selection of 100 N without considering the effect of the area.

Relative density requires measuring both mass and volume. A beam balance is used to measure the mass of the powder. A density bottle (or pycnometer) is a special flask with a precise volume, specifically designed to accurately determine the volume of solids, especially powders and liquids, by the method of displacement. This combination allows for the accurate calculation of the powder's density and, subsequently, its relative density.

Common mistake: Students might mistakenly choose an Eureka can because it is also used to find the volume of irregular solids, but a density bottle provides much higher accuracy and is the correct instrument for powders.

A hydrometer is an instrument designed to measure the relative density (also called specific gravity) of liquids. It operates on Archimedes' principle of flotation; the denser the liquid, the higher the hydrometer floats. The stem of the hydrometer is calibrated with a scale that allows a direct reading of the liquid's relative density compared to water.

Common mistake: Students often confuse relative density with density and may incorrectly select option B, but a hydrometer is specifically calibrated to measure relative density, which is a ratio and has no units.

Alcohol is preferred over mercury for measuring very cold temperatures because it has a much lower freezing point. Alcohol freezes at about -115°C, while mercury freezes at about -39°C. This means an alcohol thermometer can still work in extremely cold environments where a mercury thermometer would have already frozen and become useless, which is a special advantage.

Common mistake: Students might mistakenly choose 'low boiling point' because it is a well-known property of alcohol, but this is actually a disadvantage as it limits the upper range of temperatures the thermometer can measure.

The problem states that there is no loss of oil, which means the mass of the oil remains constant. We know that mass equals density times volume (m = ρV), so the initial mass must equal the final mass. First, calculate the constant mass using the final values: m = 0.4 kgm⁻³ × 20 m³ = 8 kg. To find the initial volume, rearrange the formula to V₁ = m / ρ₁ and use the initial density: V₁ = 8 kg / 0.8 kgm⁻³, which gives 10 m³.

Common mistake: A common mistake is to correctly calculate the constant mass of the oil as 8 kg but then forget the final step, incorrectly choosing the answer 8 m³ which matches the value for mass.

Heat can be transferred in three ways: conduction, convection, and radiation. Radiation is the transfer of heat through electromagnetic waves, which does not require a medium. All objects with a temperature above absolute zero radiate heat, primarily in the form of invisible infrared rays.

Common mistake: Students might mistakenly choose 'light energy' because very hot objects like fire also produce visible light, but they forget that the main form of radiated heat for most hot bodies is infrared radiation.

This happens because of the special property of water where its solid form (ice) is less dense than its liquid form. This means that a certain mass of ice takes up more space (volume) than the same mass of water. When the ice in the container melts, it becomes water that occupies a smaller volume, causing the overall level to drop.

Common mistake: Students often mistakenly choose that the level remains unchanged, confusing this scenario with the different case of an ice cube floating in a glass of water.

Dew is formed through a process called condensation, which is a change of state from gas to liquid. During the cool early morning hours, the surface of the grass loses heat and becomes cold. When the water vapour (a gas) in the air comes into contact with the cold grass, it cools down and condenses into the small liquid water droplets we call dew.

Common mistake: Students often confuse dew with mist or fog, but mist and fog are tiny water droplets suspended in the air, while dew forms directly on a surface.

A vapour is saturated when the rate at which liquid molecules escape to become vapour (evaporation) is equal to the rate at which vapour molecules return to the liquid (condensation). This state of balance is called a dynamic equilibrium. At this point, the space above the liquid is holding the maximum possible amount of vapour for that specific temperature.

Common mistake: Students might mistakenly choose option D, confusing the condition for saturation with the condition for boiling, which occurs when saturated vapour pressure equals the external atmospheric pressure.

To solve this problem, we use the ideal gas law formula, which is PV = nRT. First, you must convert the temperature from Celsius to Kelvin by adding 273, so 27°C becomes 300 K. Then, rearrange the formula to find the pressure (P = nRT/V) and substitute the given values: P = $(1 × 8.3 × 300)$ / 10⁻². This calculation gives a pressure of 249,000 Nm⁻², which is written in standard form as 2.49 × 10⁵ Nm⁻².

Common mistake: A common mistake is forgetting to convert the temperature from Celsius to Kelvin before calculating, which leads to an incorrect calculation and choosing option B.

The rate at which an object radiates heat energy is determined by its temperature, its surface area, and the nature of its surface (e.g., whether it is black and dull or shiny and polished). According to the Stefan-Boltzmann law, these are the primary factors affecting thermal radiation. The density of the object, which is its mass per unit volume, does not directly influence the rate at which it radiates heat from its surface.

Common mistake: Students might mistakenly choose 'nature of surface' if they forget that black, dull surfaces are better radiators of heat than shiny, polished surfaces.

The relationship between wave velocity (v), frequency (f), and wavelength (λ) is given by the formula v = fλ. First, we find the frequency by rearranging the formula to be f = v / λ, which gives f = 420 / 42 = 10 Hz. The period (T) is the inverse of the frequency (T = 1/f). Therefore, the period is T = 1 / 10 = 0.1 seconds.

Common mistake: A common mistake is to correctly calculate the frequency as 10 Hz but then get confused about how to find the period, possibly making a decimal error to arrive at 1.0 s instead of 0.1 s.

Sound is a mechanical wave, which means it requires a medium (like the particles of air, water, or a solid) to travel. The sound wave makes these particles vibrate, passing the energy along. When air is pumped out of the bottle, the medium is removed, so there are no particles left to transmit the sound vibrations from the bell to the outside.

Common mistake: Students sometimes confuse sound waves with light waves and incorrectly choose option D, forgetting that only electromagnetic waves (like light) can travel through a vacuum.

Reverberation is the persistence of sound in an area after the sound source has stopped. During a thunderstorm, the loud sound of thunder bounces off multiple surfaces like clouds, hills, and buildings. These reflected sounds, or echoes, reach your ear at different times, making it seem like the thunder is rumbling for a long time.

Common mistake: Students might mistakenly choose refraction, thinking the sound is bending through different air layers, but the key phenomenon described is the persistence of sound due to multiple reflections (echoes), which is reverberation.

In a camera, the film is the light-sensitive material at the back where the lens focuses an image to be recorded. Similarly, in the human eye, the retina is the light-sensitive layer of tissue at the back of the eyeball. The eye's lens focuses an image onto the retina, which then sends signals to the brain. Therefore, the film of a camera and the retina of the eye perform the same function of capturing the focused image.

Common mistake: Students might mistakenly choose the iris, confusing the part that captures the image (retina/film) with the part that controls the amount of light entering (iris/diaphragm).

First, find the image distance (v) using the mirror formula: 1/f = 1/u + 1/v. Plugging in the values gives $1/5$ = $1/15$ + 1/v, and solving for v gives an image distance of 7.5 cm. Next, use the magnification formula, M = Image height / Object height = v/u. This means Image height / 4 cm = 7.5 cm / 15 cm, which gives an image height of 2 cm.

Common mistake: Students might mistakenly choose 4 cm, assuming the image size is the same as the object size, which only occurs when the object is placed at the center of curvature (2f).

The relationship between refractive index, real depth, and apparent depth is given by the formula: Refractive Index = Real Depth / Apparent Depth. In this question, the real depth is 10 cm and the refractive index of the ice is 1.50. To find the apparent depth, we rearrange the formula to Apparent Depth = Real Depth / Refractive Index. Therefore, the apparent depth is 10 cm divided by 1.50, which equals 6.67 cm.

Common mistake: Students often make the mistake of multiplying the real depth by the refractive index instead of dividing, which would incorrectly suggest the object appears deeper than it really is.

Short-sightedness, also known as myopia, occurs when the eye's lens focuses light from distant objects in front of the retina, causing them to appear blurry. A concave lens is a diverging lens, meaning it spreads light rays out. Placing a concave lens before the eye diverges the incoming light rays, effectively moving the focal point backward so that it lands correctly on the retina, thus correcting the vision.

Common mistake: Students often confuse the correction for short-sightedness with that for long-sightedness, which is corrected using a convex lens.

White light is actually a mixture of different colours (like in a rainbow), each with its own wavelength. When this light enters a glass prism, it refracts or bends. Because the refractive index of glass is slightly different for each colour, each colour travels at a slightly different speed inside the prism. This difference in speed causes each colour to bend at a different angle, separating them and creating a visible spectrum.

Common mistake: Students might mistakenly choose option A, thinking the colours are somehow stored inside the glass, but the colours are actually components of the white light itself which the prism only separates.