JAMB Physics 2011 Past Questions & Answers

The weight a person feels is the normal reaction force from the ground pushing up on them. In a descending lift, this upward force is reduced. When the lift accelerates downwards at the same rate as the acceleration due to gravity (g), both the man and the lift are in a state of free fall. Because they are falling together, the floor of the lift does not push up on the man, so the reaction force is zero, leading to the sensation of weightlessness.

Common mistake: Students often choose option D, wrongly believing that feeling 'weightless' means there is no gravity, but gravity is very much present and is the force causing the free fall.

In the absence of air resistance, all objects moving under the influence of gravity experience the same constant downward acceleration (g), regardless of their mass. Since both balls were thrown with the same initial velocity, they will travel together and have the same position at any given time. Therefore, after one second, they will be at the same height from the point of throw.

Common mistake: The most common mistake is to incorrectly assume that the heavier object will be affected more by gravity and will therefore be at a different position from the lighter object.

To find the distance covered by a freely falling object, we use the equation of motion s = ut + ½gt². Since the object falls from rest, its initial velocity (u) is 0. Using the given time (t) of 3s and the standard value for acceleration due to gravity (g) as 10 m/s², the calculation is s = $(0 * 3)$ + ½ * 10 * 3². This simplifies to s = ½ * 10 * 9, which equals 45 m.

Common mistake: Students often forget the '½' in the formula s = ½gt², mistakenly calculating gt² which gives 90 m (Option A).

To find the centripetal force, we use the formula F = mv²/r, where 'm' is the mass, 'v' is the velocity, and 'r' is the radius of the circle. In this question, the mass (m) is 2 kg, the velocity (v) is 10 m/s, and the radius (r) is 4 m. Substituting these values into the formula gives F = $(2 × 10²)$ / 4, which simplifies to F = $(2 × 100)$ / 4, and finally calculates to 50 N.

Common mistake: Students might mistakenly choose 25 N because they correctly calculate the centripetal acceleration $(v²/r = 10²/4 = 25)$ but forget to multiply it by the mass to find the force.

According to the principle of conservation of energy, the potential energy (P.E.) the hammer has at the top of the roof is completely converted to kinetic energy (K.E.) just before it hits the ground. The formula for potential energy is P.E. = mass × g × height. Using the values given, P.E. = 1.5 kg × 10 ms⁻² × 20.0 m = 300 J. Therefore, the kinetic energy of the hammer just before hitting the ground is also 300 J.

Common mistake: Students might mistakenly apply the '$1/2$' from the kinetic energy formula to the potential energy calculation $(calculating 1/2 * mgh)$, leading to the incorrect answer of 150 J.

This question describes the phenomenon of capillarity, which is related to surface tension and pressure in fluids. When a liquid does not wet the glass, it means the cohesive forces (attraction between the liquid's own molecules) are stronger than the adhesive forces (attraction between the liquid and the glass). This strong cohesion causes the liquid surface to curve downwards (a convex meniscus), leading to a higher pressure inside the liquid just under the surface, which pushes the liquid column down in the tube.

Common mistake: Students often confuse the conditions for capillary rise and fall, and might incorrectly select option D, wrongly associating a non-wetting liquid with a rise in the liquid level.

According to Archimedes' principle, the upthrust on an object is the difference between its weight in air and its apparent weight in the liquid. First, calculate the weight in air: Mass × g = 30 kg × 10 m/s² = 300 N. Next, calculate the apparent weight in water: 22 kg × 10 m/s² = 220 N. The upthrust is therefore the difference, which is 300 N - 220 N = 80 N.

Common mistake: Students often mistakenly choose 220 N, which is the object's apparent weight in water, not the upthrust or the loss in weight.

To find the increase in volume, we use the formula for cubical expansion: ΔV = γV₁Δθ. First, we must find the cubical expansivity (γ) from the given linear expansivity (α) using the relation γ = 3α, which gives γ = 3 × 2.0 × 10⁻⁵ = 6.0 × 10⁻⁵ K⁻¹. The change in temperature (Δθ) is 90°C - 30°C = 60°C. Therefore, the increase in volume is $(6.0 × 10⁻⁵)$ × 40 × 60 = 0.144 cm³.

Common mistake: The most common mistake is forgetting to convert the linear expansivity (α) to cubical expansivity (γ) by multiplying by 3 before using the volume expansion formula.

To solve this problem, you use the formula for Young's modulus, which is defined as the ratio of tensile stress to tensile strain. The formula is Young's Modulus = Tensile Stress / Tensile Strain. To find the strain, you rearrange the formula to get Strain = Tensile Stress / Young's Modulus. Substituting the given values, Strain = $(7.7 × 10⁷ Nm⁻²)$ / $(1.1 × 10¹¹ Nm⁻²)$, which calculates to 7.0 × 10⁻⁴.

Common mistake: A common mistake is to incorrectly subtract the powers of ten, calculating 10⁷ / 10¹¹ to get 10⁻³ instead of 10⁻⁴, which leads to the wrong answer B.

A good thermometer must be able to detect small changes in temperature, which is called high sensitivity. It should also be easy for anyone to read the value, and it must remain accurate whether it's measuring very cold or very hot things. A thermometer should have a *low* thermal capacity so that it doesn't take much heat from the body it is measuring, ensuring the reading is correct and fast.

Common mistake: Students often mistakenly choose option B because they assume 'high' is always a good quality, but a thermometer needs a *low* thermal capacity to avoid altering the temperature of the object being measured.