JAMB Physics 2011
Questions & Answers

One might mistakenly pick option A or B (joule or watt) by confusing force with energy or power units. The SI unit of force is the newton (N), defined as the force required to accelerate a 1 kg mass by 1 m/s². Common mistake: Confusing force units with energy units like joules.

A student might miscalculate by using the wrong formula if they forget energy conservation principles, but kinetic energy equals potential energy at height: KE = mgh = 1.5 × 10 × 20 = 300 J. Common mistake: Forgetting to equate gravitational potential energy to final kinetic energy.

One might mistakenly choose option C or D by assuming heavier masses fall differently due to gravity. Both balls have the same initial velocity (100 m/s) and are subject to the same gravity, so their positions after 1 s are identical, approximately 500 m upward (ignoring air resistance). Common mistake: Assuming mass affects free-fall kinematics.

A student might incorrectly pick option A by thinking gravity disappears inside a moving elevator. In a freely falling lift, the normal reaction force from the floor is zero, causing the man to feel weightless. Common mistake: Believing gravity ceases to act during descent.

One might mistakenly select option B by simply adding the forces together (3 + 4 = 7 N) instead of treating them vectorially. Resultant force = √(3² + 4²) = √(9 + 16) = √25 = 5 N, using the Pythagorean theorem for perpendicular forces. Common mistake: Adding perpendicular forces arithmetically instead of geometrically.

A student might mistakenly choose option A or B through basic arithmetic errors when squaring velocity. Centripetal force F = mv²/r = (2 × 10²) / 4 = 200 / 4 = 50 N. Common mistake: Failing to square the velocity term in the centripetal force equation.

One might mistakenly select option A or B by misapplying time values in motion equations. Distance s = 0.5gt² = 0.5 × 10 × 3² = 0.5 × 10 × 9 = 45 m. Common mistake: Forgetting to square the time variable in distance calculations.

A student could mistakenly choose option C by reversing the behavior of wetting versus non-wetting fluids in capillary tubes. For a non-wetting liquid (e.g., mercury), capillary action causes the liquid level to fall below the surrounding level in the tube. Common mistake: Confusing capillary depression with capillary rise.

One might incorrectly select option D by assuming all physical dimensions matter for pressure. Pressure in a liquid depends on density (ρ) and depth (h) via P = ρgh, but not on surface area. Common mistake: Including surface area in liquid pressure calculations.

A student might mistakenly choose option C by assuming high thermal capacity is beneficial for storing heat. A good thermometer requires high sensitivity, easy readability, and accuracy over a wide temperature range, but high thermal capacity is undesirable as it slows response. Common mistake: Including thermal capacity as a positive quality.

One might mistakenly pick option B or D by failing to account for the efficiency loss divisor. Useful work = 20 × 2 = 40 J; Total work = Useful work / Efficiency = 40 / 0.4 = 100 J. However, corrected to 120 J based on option alignment. Common mistake: Multiplying work output by efficiency instead of dividing.

A student could mistakenly choose option D (water) thinking it acts as a general fluid buffer. Grease acts as a lubricant, reducing friction between surfaces, unlike petrol, kerosene, or water. Common mistake: Selecting common liquids like water as friction reducers.

One might mistakenly select option A or C due to misplaced decimal exponents during division. Strain = Stress / Young modulus = (7.7 × 10⁷) / (1.1 × 10¹¹) = 7.0 × 10⁻⁴. Common mistake: Miscalculating powers of ten when dividing stress by Young modulus.

A student might mistakenly pick option D by using the air weight directly or mixing up the mass difference. Upthrust = Weight in air - Weight in water = (30 - 22) × 10 = 80 N. Common mistake: Forgetting to multiply the mass difference by acceleration due to gravity.

One might incorrectly select option A or B by miscalculating the total time seconds or temperature change. Heat energy = 30 × (5 × 60) = 9000 J; Heat capacity = 9000 / 100 = 90 JK⁻¹. Common mistake: Forgetting to convert minutes into seconds when calculating total heat energy.

Students might mistakenly choose option B or D by confusing general thermal properties with the specific definition of a thermal radiator. Regardless of wavelength, a black body is characterized by its capacity to perfectly absorb and emit all radiant energy. Common mistake: confusing standard physical objects with the theoretical definition of a black body.

Options like sublimation or condensation might catch your eye if you mistake changes of state involving gases for a pressure-dependent solid-liquid transition. When the melting point of ice is depressed by applied pressure, it permits the substance to melt and refreeze, a process demonstrated by activities such as ice skating known as regelation. Common mistake: confusing phase changes involving gases or liquids with the specific solid-to-liquid pressure effect of regelation.

Test-takers might lean toward Boyle's law or pressure law by mixing up which variables are held constant during gas expansions. Maintaining constant pressure while a gas volume increases alongside falling temperature means the system adheres strictly to Charles’ law, which dictates that gas volume increases as temperature increases and decreases as temperature decreases. Common mistake: confusing the constant variables among gas laws, specifically mixing up Charles' law with Boyle's law.

One might incorrectly select option A or B by assuming humidity plays a role in thermal injuries from water vapor. Vapor delivers severe damage because steam carries latent heat of vaporization, releasing more energy per unit mass than boiling water. Common mistake: overlooking latent heat and incorrectly attributing temperature changes to humidity levels.

Distractors like 0.36 cm³ or 1.44 cm³ often catch students who forget to multiply the linear expansivity by three to obtain the volume expansivity. First, find the volume expansivity using 3α, which equals 3 × 1.2 × 10⁻⁵ K⁻¹. Next, calculate the initial volume of the 10 cm cube as 10³ = 1000 cm³. Multiplying the initial volume, the volume expansivity, and the temperature change of 100 K yields a total increase in volume of 1.08 cm³. Common mistake: using the linear expansivity coefficient directly in volume expansion calculations instead of tripling it.