21 Post-UTME Physics 2011 questions with verified answers and worked explanations. Here are the first 10 — sign up free to practise the full paper.
A body of mass m slides down an inclined plane with a constant velocity. If the angle of the incline is Ø, the coefficient of kinetic friction between the body and the plane is
When a body slides down an incline at a constant velocity, it means the forces are balanced. The force pulling the body down the slope (mg sin Ø) is equal to the kinetic frictional force (f) acting up the slope. Also, the normal reaction (R) is equal to the component of weight perpendicular to the slope (mg cos Ø). Since friction f = μR, we can say mg sin Ø = μ * mg cos Ø. Dividing both sides by mg cos Ø gives μ = sin Ø / cos Ø, which is equal to tan Ø.
Common mistake: Students often confuse the sine and cosine components of weight, which can lead them to incorrectly choose cot Ø if they mix up the numerator and denominator in the final step.
the velocity of a 500 kg car moving long a straight road, changes from 12 m/s to 20 m/s in 5 sec. calculate the average force moving the car.
To find the force, we use Newton's second law, which states that Force = mass × acceleration (F=ma). First, we must calculate the acceleration (a) from the change in velocity: a = (final velocity - initial velocity) / time. Plugging in the values, a = / 5 = 1.6 m/s². Now, we can find the force: F = 500 kg × 1.6 m/s² = 800 N.
Common mistake: Students might mistakenly calculate the force using only the final velocity (F = m * v / t), leading to 500 * 20 / 5 = 2000N, which is incorrect as force depends on acceleration (the change in velocity).
When a 2kg body is at a height 5m above the floor, its velocity is 4m/s. What is its total energy at this height? (acceleration due to gravity = 10m/s2)
The total energy of the body is the sum of its Potential Energy (P.E.) and Kinetic Energy (K.E.). First, calculate the Potential Energy using the formula P.E. = mgh, which gives 2kg x 10m/s² x 5m = 100J. Next, calculate the Kinetic Energy using the formula K.E. = ½mv², which is ½ x 2kg x (4m/s)² = 16J. Finally, add both energies together to get the total energy: 100J + 16J = 116J.
Common mistake: The most common mistake is calculating only the potential energy (100J) or only the kinetic energy (16J), and forgetting that total energy is the sum of both.
The density of sea water is 1030kg/m3. What is the pressure at a depth of 80 m below sea surface? Atmospheric pressure is 1.013 x 105 pa and acceleration due to gravity is 10 m/s2
The total pressure at a depth in a fluid is the sum of the atmospheric pressure on the surface and the pressure due to the fluid column (liquid pressure). First, calculate the liquid pressure using the formula P = hρg, which gives P = 80m * 1030kg/m³ * 10m/s² = 824,000 Pa or 8.24 x 10⁵ Pa. Then, add the atmospheric pressure: Total Pressure = (1.013 x 10⁵ Pa) + (8.24 x 10⁵ Pa) = 9.253 x 10⁵ Pa, which is approximately 9.25 x 10⁵ Pa.
Common mistake: Students often calculate only the liquid pressure (8.24 x 10⁵ Pa) and forget to add the atmospheric pressure, leading them to incorrectly choose option B.
6000 J of heat is delivered to 10g of dry ice at O0C. What is the final temperature if the container has a heat capacity of 20J/K? (specific heat of water = 4200j/kg.k, latent heat of fusion of ice = 3.33 x 105J/kg)
First, calculate the heat needed to melt the 10g (0.01kg) of ice at 0°C using the formula for latent heat, Q₁ = mL. This gives 0.01kg × 3.33 × 10⁵ J/kg = 3330 J. The heat remaining to raise the temperature is 6000 J - 3330 J = 2670 J. This remaining heat warms up both the now-liquid water and the container, so we use the formula Q₂ = (mc + C)ΔT, where C is the container's heat capacity. Solving for the temperature change, ΔT = 2670 / = 2670 / 62, which gives a final temperature of about 43°C.
Common mistake: A common mistake is forgetting to include the heat absorbed by the container, which leads to an incorrect calculation where the final temperature appears to be 63.6°C (Option B).
A sample of radioactive substance, whose half-life is 16 days, registers 32 decays per second. How long will it take for the rate of decay to reduce to 2 decays per second?
The initial decay rate is 32 decays per second. After each half-life of 16 days, this rate is halved. After the first half-life, the rate is 16; after the second, it is 8; after the third, it is 4; and after the fourth, it becomes 2 decays per second. Therefore, it takes 4 half-lives for the rate to reduce to 2, so the total time is 4 multiplied by 16 days, which equals 64 days.
Common mistake: Students often miscalculate the number of half-lives required, sometimes stopping one step early at 4 decays/s (48 days) or going one step too far to 1 decay/s (80 days).
When white light passes through a triangular prism, the emerging rays of light arranged in order of decreasing angle of deviation are
When white light passes through a prism, it splits into its constituent colours in a process called dispersion. The amount of bending, or the angle of deviation, depends on the colour; violet light bends the most, and red light bends the least. The order of colours from the highest deviation to the lowest is Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR). The question asks for the order of decreasing angle of deviation, so the correct sequence from the options is blue, green, yellow, orange.
Common mistake: Students often mistakenly choose the order of increasing angle of deviation (Red, Orange, Yellow, Green), forgetting that red light deviates the least while violet and blue light deviate the most.
The electric field between two parallel plates is E. A particle of mass m and carrying charge q is released at a point half the distance between the plates. The velocity of the particle t seconds after its release is
The force (F) on a particle with charge (q) in an electric field (E) is given by F = qE. According to Newton's second law, this force also equals mass (m) times acceleration (a), so F = ma. By equating these, we get ma = qE, which means the acceleration is a = qE/m. Since the particle is released from rest, its initial velocity is zero, and we can find the final velocity (v) using the equation v = u + at, which becomes v = 0 + (qE/m) * t, simplifying to v = qEt/m.
Common mistake: Students might mistakenly use the equation for displacement (s = ut + ½at²) instead of velocity, leading them to choose an answer involving t².
An airplane increases its speed 36 km/h to 360 km/h in 20.0 s. How far does it travel while accelerating.
First, convert the speeds from km/h to m/s by multiplying by . The initial speed (u) is 36 km/h which is 10 m/s, and the final speed (v) is 360 km/h which is 100 m/s. Use the equation of motion for distance (s) which is s = (u + v)/2 * t. Substituting the values, we get s = * 20, which simplifies to s = * 20 = 55 * 20 = 1100 metres. Finally, convert the distance back to kilometres by dividing by 1000, which gives 1.1 km.
Common mistake: Students often forget to convert the speeds from km/h to the standard unit of m/s before using them in the equations of motion, leading to an incorrect answer.
An object is said to be in simple harmonic motion (SHM) if
Simple Harmonic Motion is a special type of periodic motion where the restoring force is directly proportional to the object's displacement from its equilibrium position. According to Newton's second law (F=ma), this means the acceleration is also directly proportional to the displacement. The motion is always directed back towards the equilibrium position, which is why the acceleration is opposite to the displacement, making option A the correct definition.
Common mistake: Students might mistakenly choose option B, confusing the direct proportionality in the SHM definition with an inverse relationship.
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