JAMB Physics 2003

25 reviewed questions with answers and explanations.

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Question 4

Which property of water helps prevent droplets passing through the small gaps of a tightly woven silk umbrella until the inside is touched?

  1. Osmotic pressure
  2. Capillarity
  3. Surface tension
  4. Viscosity
Answer and explanation

C: Surface tension

Water can form curved films across the small spaces in a fabric. Surface tension helps those films resist passage of water; touching the inside can disturb the film and promote wetting through the fabric.

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Question 10

A uniform 90 cm lever is pivoted at its centre. A 30 N load hangs 15 cm from its left end. What downward force at the right end maintains horizontal equilibrium?

  1. 20 N
  2. 30 N
  3. 60 N
  4. 15 N
Answer and explanation

A: 20 N

The fulcrum is 45 cm from either end. The load arm is 45 −15 =30 cm, while the effort arm is 45 cm. Balancing moments gives F ×45 =30 ×30, so F =20 N.

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Question 15

An aluminium bar has Young’s modulus 7.0 ×10¹⁰ Pa and density 2.7 ×10³ kg/m³. Estimate its longitudinal bar-wave speed using v =√(Y/ρ), to two significant figures.

  1. 3.6 ×10³ m/s
  2. 5.1 ×10³ m/s
  3. 2.8 ×10³ m/s
  4. 4.2 ×10³ m/s
Answer and explanation

B: 5.1 ×10³ m/s

For longitudinal waves in a slender elastic bar, v =√(Y/ρ). Here v =√(7.0 ×10¹⁰/2.7 ×10³) ≈5092 m/s, or 5.1 ×10³ m/s to two significant figures.

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Question 16

Thermal equilibrium between two objects exists when

  1. The heat capacities of both objects are the same
  2. One objects loses heat continuously to the other
  3. The temperatures of both objects are equal
  4. The quantity of heat in both objects is the same.
Answer and explanation

C: The temperatures of both objects are equal

Objects in thermal equilibrium have equal temperatures and no net heat transfer between them. Their heat capacities, masses and internal energies can still differ.

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Question 17

At twice the distance from an isotropic point source of sound in a lossless medium, what factor multiplies the original sound intensity?

  1. 2.00
  2. 0.25
  3. 4.00
  4. 0.50
Answer and explanation

B: 0.25

For a point source radiating uniformly without absorption, intensity is power divided by 4πr². Doubling distance makes the area four times larger, so intensity becomes one quarter of its original value.

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Question 18

In the ideal model of two sufficiently large parallel plane mirrors with reflecting faces toward one another, how many images can repeated reflection form of an object between them?

  1. Four
  2. Two
  3. Eight
  4. Infinitely many
Answer and explanation

D: Infinitely many

In the ideal model, two parallel facing mirrors reflect the object and each other’s images repeatedly. There is no final reflection order, so the construction gives infinitely many images.

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Question 19

A 2000 W heater warms a 5 kg metal object initially at 10°C. Its temperature rises 30°C in 10 minutes. Assuming all supplied heat warms the object, find its total heat capacity.

  1. 1.2 ×10⁴ J/K
  2. 6.0 ×10⁴ J/K
  3. 8.0 ×10³ J/K
  4. 4.0 ×10⁴ J/K
Answer and explanation

D: 4.0 ×10⁴ J/K

The heater supplies Q =Pt =2000 ×600 =1.2 ×10⁶ J. Heat capacity of the whole object is C =Q/ΔT =1.2 ×10⁶/30 =4.0 ×10⁴ J/K. Mass is needed for specific heat capacity, not total heat capacity.

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Question 20

An object is 10 cm from a convex lens and forms a real image on a screen 25 cm from the lens. Find the magnitude of its linear magnification.

  1. 2.5
  2. 1.5
  3. 0.4
  4. 15.0
Answer and explanation

A: 2.5

The magnitude of linear magnification is image distance divided by object distance. Thus |m| =25/10 =2.5;the real image is inverted but 2.5 times as tall.

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Question 22

A vessel wall conducts 1.2 ×10⁶ J in 1 s with a uniform temperature gradient 30 K/m and thermal conductivity 400 W/(m·K). Find the area normal to heat flow.

  1. 1.0 ×10³ m²
  2. 1.0 ×10² m²
  3. 9.0 ×10⁴ m²
  4. 9.0 ×10² m²
Answer and explanation

B: 1.0 ×10² m²

Fourier’s conduction law gives heat rate kA times the temperature-gradient magnitude. Therefore A =(1.2 ×10⁶/1)/(400 ×30) =100 m².

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Question 23

A saturation-pressure table gives 17.50 mmHg at 20°C. Actual water-vapour pressure is 10 mmHg at 20°C. Find relative humidity to the nearest whole percent.

  1. 57%
  2. 17.5%
  3. 10%
  4. 170%
Answer and explanation

A: 57%

Relative humidity is actual vapour pressure divided by saturation vapour pressure at the same temperature, then multiplied by 100. Thus RH =100 ×10/17.50 ≈57.14%, or 57% to the nearest whole percent.

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Question 24

In the paraxial model, where must an object be placed in front of a concave mirror to produce an image at infinity?

  1. At centre of curvature
  2. Between the principal focus and the centre of curvature
  3. At the pole of the mirror
  4. At the principal focus
Answer and explanation

D: At the principal focus

In the paraxial model, a point at the principal focus of a concave mirror reflects into parallel rays. The image distance is therefore infinite.

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Question 25

Neglecting end correction, a pipe of lengthl is closed at one end and open at the other. If sound speed isv, find its fundamental frequency.

  1. v/(2 l)
  2. 2 v/l
  3. v/(5 l)
  4. v/(4 l)
Answer and explanation

D: v/(4 l)

The fundamental mode of a pipe closed at one end has a displacement node at the closed end and an antinode at the open end. Its length is one quarter wavelength, so f =v/(4 l).

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Question 29

Two 2 Ω resistors in parallel are connected in series with a third 2 Ω resistor. Each resistor has a maximum allowed dissipation 18 W. Find the maximum total dissipation without exceeding any resistor’s limit.

  1. 9 W
  2. 27 W
  3. 5 W
  4. 18 W
Answer and explanation

B: 27 W

The single series 2 Ω resistor carries the full currentI, while each parallel 2 Ω branch carriesI/2. Its 18 W limit givesI² ×2 =18, orI =3 A. The branches each dissipate 4.5 W, totaling 27 W.

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Question 31

Which listed pair spans the widest interval in the usual visible spectrum ordering?

  1. Green and yellow
  2. Indigo and violet
  3. Orange and red
  4. Blue and red
Answer and explanation

D: Blue and red

Red and blue occupy well-separated regions of the visible spectrum. Each other pair listed consists of neighbouring colours, so blue and red span the widest spectral interval among these choices.

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Question 34

Which traditional primary cell was commonly used for brief intermittent operation of electric bells?

  1. Nickel-iron accumulator
  2. Lead-acid accumulator
  3. Daniell cell
  4. Leclanché cell
Answer and explanation

D: Leclanché cell

The traditional Leclanché cell was widely used for intermittent electric-bell circuits. Rest periods allow it to recover from polarization, making brief intermittent operation suitable.

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Question 39

When a semiconductor p–n junction diode is forward biased normally, its depletion layer

  1. Narrows
  2. Remains constant
  3. Widens then narrows
  4. Widens
Answer and explanation

A: Narrows

Forward bias opposes the junction’s built-in electric field and lowers its potential barrier. The depletion region narrows, allowing greater carrier injection across the junction.

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Question 40

A nucleus has less mass than the sum of its separated constituent protons and neutrons. What is the energy equivalent of this mass defect called?

  1. Stability
  2. Lost energy
  3. Work function
  4. Binding energy
Answer and explanation

D: Binding energy

A bound nucleus has less rest mass than its separated nucleons. The mass defect multiplied byc² is its binding energy, the energy required to separate it completely into those nucleons.

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Question 41

An electron drops from an excited level to the ground state, emitting light of frequency 8.0 ×10¹⁴ Hz. Find the emitted photon energy using h =6.6 ×10⁻³⁴ J·s.

  1. 5.28 ×10¹⁹ J
  2. 8.25 ×10¹⁹ J
  3. 5.28 ×10⁻¹⁹ J
  4. 8.25 ×10⁻¹⁹ J
Answer and explanation

C: 5.28 ×10⁻¹⁹ J

An emitted photon has energy E =hf. With h =6.6 ×10⁻³⁴ J·s and f =8.0 ×10¹⁴ Hz, E =5.28 ×10⁻¹⁹ J. This equals the electron’s energy-level decrease.

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Question 42

Fission converts 0.01% of a 1.0 g sample’s mass into released energy. Find that energy using c =3.0 ×10⁸ m/s.

  1. 9.0 ×10¹⁰ J
  2. 6.3 ×10¹¹ J
  3. 9.0 ×10¹¹ J
  4. 9.0 ×10⁹ J
Answer and explanation

D: 9.0 ×10⁹ J

A mass loss of 0.01% of 1.0 g is 10⁻⁴ ×10⁻³ =10⁻⁷ kg. Hence E =Δmc² =10⁻⁷(3.0 ×10⁸)² =9.0 ×10⁹ J.

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Question 44

A sinusoidal current with peak 10 A passes through a 12 Ω resistor. Find its average power dissipation over a complete cycle.

  1. 120 W
  2. 20 W
  3. 600 W
  4. 1200 W
Answer and explanation

C: 600 W

For sinusoidal current of peak 10 A, the mean square current is 10²/2 =50 A². Average resistor power is I_RMS²R =50 ×12 =600 W.

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Question 45

Over an ordinary temperature range near room temperature, which statement usually distinguishes a pure metal from an intrinsic semiconductor?

  1. Metal resistance increases with temperature, while intrinsic semiconductor resistance decreases
  2. Metals are always harder than semiconductors
  3. Metals have forbidden band gaps but semiconductors do not
  4. Metal resistance decreases with temperature, while intrinsic semiconductor resistance increases
Answer and explanation

A: Metal resistance increases with temperature, while intrinsic semiconductor resistance decreases

For ordinary metals near room temperature, heating increases lattice scattering and resistance. In an intrinsic semiconductor, heating creates more charge carriers, so resistance generally decreases over that range.

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Question 47

An ammeter reads 1.20 A steadily while 0.990 g of copper is deposited in 40 minutes. The electrochemical equivalent of copper is 3.3 × 10⁻⁴ g C⁻¹. What correction must be added to the ammeter reading?

  1. 0.05 A
  2. 0.06 A
  3. 0.03 A
  4. 0.04 A
Answer and explanation

A: 0.05 A

Faraday’s law gives Q = m/z = 0.990/(3.3 × 10⁻⁴) = 3000 C. The deposition time is 2400 s, so the actual current is 1.25 A. Add 1.25 − 1.20 = 0.05 A to the meter reading.

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Question 48

The maximum kinetic energy of photoelectrons emitted from a metal surface is 0.34 eV. The work function is 1.83 eV. Find the magnitude of the stopping potential.

  1. 1.09 V
  2. 2.17 V
  3. 0.34 V
  4. 1.49 V
Answer and explanation

C: 0.34 V

The stopping potential satisfies eV_s = K_max. An electron with kinetic energy 0.34 eV is stopped by a potential difference of 0.34 V. The work function helps determine the incident photon energy, but is not added to the stopping potential.

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Question 49

At which of the listed angles between velocity v and a nonzero magnetic field B is the force on a moving charge half its maximum magnitude?

  1. 0°
  2. 90°
  3. 45°
  4. 30°
Answer and explanation

D: 30°

The magnetic force magnitude is F = |q|vB sin θ. Half the maximum force requires sin θ = 1/2. Of the listed angles, 30° satisfies this condition; 150° would also do so if it were offered.

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Question 50

The background-subtracted count rate of a radioactive sample is 800 counts per minute. Its half-life is 4 days. With unchanged detection conditions, what will the sample count rate be 16 days later?

  1. 50 counts/min
  2. 25 counts/min
  3. 200 counts/min
  4. 100 counts/min
Answer and explanation

A: 50 counts/min

Sixteen days contains 16/4 = 4 half-lives. The count rate due to the sample therefore falls to 800 × (1/2)⁴ = 50 counts per minute, with the same detector arrangement.

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