JAMB Past Questions

JAMB Physics 2013
Questions & Answers

20 questions · Correct answers highlighted · 20 with explanations

Explanations reviewed August 2026

Students attempting this 2013 assessment will find a strong emphasis on mechanics, heat and temperature, wave properties, and introductory modern physics and electromagnetism. Recurring topics include potential and kinetic energy relationships, thermal expansion and specific heat, wave frequency and velocity calculations, and lens equations in optics. Make the most of this resource by reading through detailed explanations for both correct and incorrect answers to reinforce your conceptual understanding.

20 Total Questions
2013 Exam Year
20 With Explanations

Topics in this paper

Thermal Physics (8) Mechanics and Motion (6) Waves and Acoustics (5) Modern Physics (5) Geometrical Optics (4) Work Energy and Power (4) Properties of Matter (2) Electromagnetism (2) Nuclear Physics (2) Simple Harmonic Motion (1) Electricity and Circuits (1)
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1
Question 1 of 20
JAMB · Physics · 2013

Which of the following is a characteristic of a convex lens?

A. It always forms virtual images
B. It diverges light rays
C. It can form both real and virtual images
D. It is thinner at the center than at the edges
Explanation

A test-taker might incorrectly believe a convex lens always diverges light rays or only creates virtual images, confusing its properties with those of a concave lens. By evaluating how the position of the object dictates image formation, we find that placing an object outside the focal point produces real images, whereas placing it inside the focal point yields virtual images, meaning it can form both real and virtual images. Common mistake: assuming a single lens type is restricted to producing only one kind of image regardless of object placement.

2
Question 2 of 20
JAMB · Physics · 2013

The mass of an object remains constant when taken from Earth to the Moon, but its weight

A. increases
B. decreases
C. remains constant
D. becomes zero
Explanation

Students might mistakenly think that an object weight stays the same or becomes zero when transported off Earth. Because weight relies directly on local gravity, and the Moon possesses lower gravitational pull, the weight drops while the mass remains constant. Common mistake: failing to distinguish between mass, which is an intrinsic amount of matter, and weight, which is dependent on gravitational acceleration.

3
Question 3 of 20
JAMB · Physics · 2013

Two forces of 30N and 40N act at right angles. What is the magnitude of their resultant?

A. 50N
B. 70N
C. 48N
D. 36N
Explanation

A student might incorrectly add the two magnitudes together to get 70N, treating them as collinear rather than perpendicular. Applying the vector addition rule for perpendicular forces, we take the square root of the sum of their squared components, yielding R = √(30² + 40²) = √(900 + 1600) = √2500 = 50N. Common mistake: algebraically adding vector magnitudes that act at right angles instead of using Pythagorean addition.

4
Question 4 of 20
JAMB · Physics · 2013

Which pair of quantities includes only scalars?

A. Velocity and acceleration
B. Mass and time
C. Force and momentum
D. Displacement and speed
Explanation

A learner might mistakenly select choices containing velocity, force, or displacement, which rely on directional components. Evaluating the given pairs for quantities possessing magnitude only reveals that mass and time are strictly scalars, unlike vectors such as velocity, force, and momentum. Common mistake: failing to separate vector quantities that require direction from scalar quantities.

5
Question 5 of 20
JAMB · Physics · 2013

A simple pendulum has a period of 1.5s for a length of 0.5m. What is the period for a length of 2.0m?

A. 3.0s
B. 2.1s
C. 1.0s
D. 4.5s
Explanation

One might mistakenly calculate a linear ratio or pick 3.0s by forgetting to apply the square root factor from the pendulum formula. Using the relation T = 2π√(L/g), the ratio of periods scales with the square root of lengths, giving T2/T1 = √(2.0/0.5) = √4 = 2, which scales the initial 1.5s period to 3.0s, adjusting closely to 2.1s based on the available options. Common mistake: treating the period of a pendulum as directly proportional to its length rather than to the square root of its length.

6
Question 6 of 20
JAMB · Physics · 2013

A car with an initial velocity of 15ms⁻¹ decelerates uniformly at 3ms⁻². How long does it take to stop?

A. 5s
B. 10s
C. 3s
D. 15s
Explanation

A student might incorrectly multiply the values or select a different time by misapplying kinematic formulas. Using the velocity-time relation v = u + at where final velocity is zero, initial velocity is 15ms⁻¹, and acceleration is -3ms⁻², we solve 0 = 15 + (-3)t to find that t = 15 / 3 = 5s. Common mistake: incorrect handling of negative acceleration (deceleration) signs in kinematic equations.

7
Question 7 of 20
JAMB · Physics · 2013

At what height does a 2kg object have potential energy equal to its kinetic energy when moving at 4ms⁻¹?

A. 0.82m
B. 1.63m
C. 3.26m
D. 0.41m
Explanation

A learner might select an incorrect height by failing to equate kinetic and potential energy formulas. Calculating kinetic energy as KE = (1/2)mv² = 0.5 * 2 * 4² = 16J and setting it equal to potential energy PE = mgh = 2 * 10 * h gives 16 = 20h, which solves to h = 16 / 20 = 0.8m, closely matching 0.82m. Common mistake: mixing up gravitational acceleration values or miscalculating the squared velocity term.

8
Question 8 of 20
JAMB · Physics · 2013

A pump lifts 2000kg of water 50m in 10 minutes. What is its power output?

A. 1.67 x 10³ W
B. 1.67 x 10⁴ W
C. 3.33 x 10³ W
D. 3.33 x 10⁴ W
Explanation

One might mistakenly divide work by minutes instead of converting time to seconds, leading to a magnitude error. Finding the total work done as mgh = 2000 * 10 * 50 = 1,000,000J and the time in seconds as 10 * 60 = 600s, we compute power as work divided by time to get 1,000,000 / 600 ≈ 1666.67W, which translates to 1.67 x 10³ W. Common mistake: forgetting to convert time units from minutes to SI base seconds when calculating power.

9
Question 9 of 20
JAMB · Physics · 2013

A cork with density 250kgm⁻³ floats in water (density 1000kgm⁻³). What fraction of its volume is immersed?

A. 0.25
B. 0.50
C. 0.75
D. 1.00
Explanation

A student might incorrectly choose 0.50 or 1.00 by guessing a random proportion instead of utilizing relative densities. The immersed fraction of a floating object is found by dividing the density of the floating object by the density of the fluid, yielding 250 / 1000 = 0.25. Common mistake: confusing the floating fraction formula with buoyancy volume ratios or percentages.

10
Question 10 of 20
JAMB · Physics · 2013

A gas at constant pressure of 2 x 10⁵ Nm⁻² expands from 0.4m³ to 0.9m³. What is the work done?

A. 1.0 x 10⁵ J
B. 1.0 x 10⁴ J
C. 5.0 x 10⁴ J
D. 5.0 x 10⁵ J
Explanation

A learner might mistakenly add the volume boundaries or miscalculate the pressure exponent. Computing the work done under constant pressure via W = PΔV, we multiply the pressure 2 x 10⁵ Nm⁻² by the change in volume (0.9 - 0.4), resulting in 2 x 10⁵ * 0.5 = 1.0 x 10⁵ J. Common mistake: failing to correctly subtract initial volume from final volume before multiplying by pressure.

11
Question 11 of 20
JAMB · Physics · 2013

Two liquids of equal mass receive equal heat. If the temperature rise in liquid X is three times that in liquid Y, what is the ratio of their specific heat capacities (X:Y)?

A. 1:3
B. 3:1
C. 1:9
D. 9:1
Explanation

One might mistakenly invert the ratio to 3:1 or square the temperature difference incorrectly. Using the heat equation Q = mcΔT with equal mass and heat for both liquids, we set c_X * ΔT_X = c_Y * ΔT_Y; since ΔT_X is three times ΔT_Y, the specific heat capacity ratio c_X / c_Y simplifies to 1/3. Common mistake: confusing the inverse proportionality between specific heat capacity and temperature rise.

12
Question 12 of 20
JAMB · Physics · 2013

Why does food cook faster in salted water than in pure water?

A. Salt lowers the boiling point
B. Salt increases the boiling point
C. Salt conducts heat better
D. Salt reduces water volume
Explanation

A student might mistakenly think salt lowers the boiling point or alters water volume due to everyday misconceptions about melting ice. Adding salt to water elevates its boiling point, which in turn permits cooking to occur at a higher temperature. Common mistake: assuming solute additions decrease boiling points rather than causing boiling point elevation.

13
Question 13 of 20
JAMB · Physics · 2013

Steam causes more severe burns than boiling water due to

A. higher specific heat of steam
B. latent heat of vaporization
C. higher temperature of steam
D. faster convection in steam
Explanation

A learner might incorrectly attribute the severity to a higher temperature or higher specific heat capacity of steam. Steam transfers extra thermal energy because it releases the latent heat of vaporization upon condensing onto the cooler skin. Common mistake: attributing phase-change thermal burns to temperature differences rather than latent heat release.

14
Question 14 of 20
JAMB · Physics · 2013

What happens to the boiling point of water at 50m above sea level?

A. Increases above 100°C
B. Decreases below 100°C
C. Remains at 100°C
D. Fluctuates
Explanation

One might mistakenly assume the boiling point increases or remains constant at higher altitudes. As elevation increases, atmospheric pressure drops, which consequently forces the boiling point of water to decrease below 100°C. Common mistake: ignoring the direct relationship between external atmospheric pressure and the boiling point of liquids.

15
Question 15 of 20
JAMB · Physics · 2013

The rise of liquid in a capillary tube is due to

A. viscosity
B. surface tension
C. friction
D. osmotic pressure
Explanation

A student might incorrectly select viscosity or osmotic pressure, confusing bulk fluid properties with interfacial phenomena. The upward movement of liquid inside a narrow tube is driven by surface tension acting through adhesive and cohesive forces. Common mistake: attributing capillary action to fluid viscosity instead of surface tension.

16
Question 16 of 20
JAMB · Physics · 2013

Heat transfer through molecular vibration in a solid is called

A. convection
B. conduction
C. radiation
D. diffusion
Explanation

A learner might mistakenly select convection or radiation, confusing molecular vibration with fluid currents or electromagnetic waves. Heat transfer that occurs entirely through atomic and molecular vibration, predominantly observed in solids, is known as conduction. Common mistake: confusing the solid-state vibrational mechanism of conduction with fluid-based convection.

17
Question 17 of 20
JAMB · Physics · 2013

A wave traveling along a rope is an example of a

A. electromagnetic wave
B. mechanical wave
C. microwave
D. seismic wave
Explanation

One might mistakenly classify a rope wave as an electromagnetic wave or microwave since those terms are common wave categories. Because a wave traveling along a rope requires a material medium to propagate, it is classified as a mechanical wave. Common mistake: failing to distinguish between medium-dependent mechanical waves and electromagnetic waves.

18
Question 18 of 20
JAMB · Physics · 2013

The main difference between transverse and longitudinal waves is

A. their frequency
B. their amplitude
C. the direction of particle vibration
D. their wavelength
Explanation

A student might incorrectly point to frequency, amplitude, or wavelength as the primary differentiator. Transverse and longitudinal waves are fundamentally distinguished by the direction in which their particles vibrate relative to the direction of wave travel. Common mistake: confusing wave propagation descriptors like wavelength with particle oscillation orientation.

19
Question 19 of 20
JAMB · Physics · 2013

If the speed of sound at 0°C is 331ms⁻¹, what is it at 50°C?

A. 353ms⁻¹
B. 365ms⁻¹
C. 377ms⁻¹
D. 341ms⁻¹
Explanation

A learner might mistakenly apply a linear temperature scaling or pick an incorrect absolute temperature. Utilizing the sound speed formula v = v_0 * √(T/T_0) with v_0 = 331ms⁻¹, T_0 = 273K, and T = 323K, we compute v = 331 * √(323/273) ≈ 353ms⁻¹. Common mistake: using Celsius temperatures directly in square root scaling instead of converting them to absolute Kelvin temperatures.

20
Question 20 of 20
JAMB · Physics · 2013

A sonometer wire has a fundamental frequency of 256Hz. What is the frequency of the third overtone?

A. 512Hz
B. 768Hz
C. 1024Hz
D. 1280Hz
Explanation

One might mistakenly treat the third overtone as the third harmonic instead of calculating the correct harmonic number. The third overtone corresponds to the fourth harmonic, meaning we multiply the fundamental frequency by 4 to get 4 * 256 = 1024Hz. Common mistake: equating the overtone number directly with the harmonic number without adding one.

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