JAMB Physics 2007 past questions and answers

47 questions. Try each one, then open the answer to check yourself and read the explanation.

Question 1

Refraction of Light

A glass plate 0.9 cm thick has a refractive index of 1.50. How long does it take for a pulse of light to pass through the plate? [c = 3.0 × 108 ms−1]

  1. A.3.0 × 10−11s
  2. B.4.5 × 10−11s
  3. C.3.0 × 10−10s
  4. D.4.5 × 10−13s
Show the answer and explanation

Answer: B (4.5 × 10−11s)

Explanation

Speed of light in glass = c/n = 3.0 × 108 ÷ 1.5 = 2.0 × 108 m s−1. Time = distance ÷ speed = 0.009 m ÷ 2.0 × 108 = 4.5 × 10−11 s.

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Question 2

Sound

The fundamental frequency of a plucked wire under a tension of 400 N is 250 Hz. When the frequency is changed to 500 Hz at constant length, the tension is

  1. A.40 N.
  2. B.160 N.
  3. C.400 N.
  4. D.1600 N.
Show the answer and explanation

Answer: D (1600 N.)

Explanation

For a stretched wire f ∝ √T. Doubling the frequency (250 Hz to 500 Hz) needs four times the tension: 4 × 400 N = 1600 N.

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Question 3

Dispersion of Light and Colours

The production of pure spectrum could easily be achieved using a

  1. A.triangular prism only.
  2. B.triangular prism with two concave lenses.
  3. C.triangular prism with two convex lenses.
  4. D.glass prism with a pin.
Show the answer and explanation

Answer: C (triangular prism with two convex lenses.)

Explanation

A pure spectrum needs a parallel beam entering the prism (first convex lens) and the separated colours focused sharply on a screen (second convex lens).

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Question 5

Light and Reflection

The instrument used by designers to obtain different colour patterns is called

  1. A.episcope.
  2. B.periscope.
  3. C.kaleidoscope.
  4. D.sextant.
Show the answer and explanation

Answer: C (kaleidoscope.)

Explanation

A kaleidoscope uses inclined plane mirrors to form many images of coloured pieces, giving symmetrical colour patterns.

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Question 6

Lenses and Optical Instruments

When an object is placed between the principal focus and the optical centre of a convex lens, it could be used as a

  1. A.reflecting lens.
  2. B.compound microscope.
  3. C.projector.
  4. D.simple microscope.
Show the answer and explanation

Answer: D (simple microscope.)

Explanation

With the object inside the focal length, a convex lens forms an upright, magnified, virtual image: this is a magnifying glass (simple microscope).

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Question 7

Sound

In a large telecommunications auditorium, perforated absorbent materials are used to line the ceiling so as to

  1. A.reduce the reverberation of sound in the hall.
  2. B.reduce the height of the ceiling from the floor.
  3. C.increase the reverberation of sound in the hall.
  4. D.increase the amount of echo in the hall.
Show the answer and explanation

Answer: A (reduce the reverberation of sound in the hall.)

Explanation

Absorbent materials soak up sound instead of reflecting it, so repeated reflections (reverberation) die out quickly and speech stays clear.

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Question 8

Waves

The phenomenon of light bending round an obstacle is

  1. A.reflection.
  2. B.polarization.
  3. C.refraction.
  4. D.diffraction.
Show the answer and explanation

Answer: D (diffraction.)

Explanation

Diffraction is the spreading or bending of waves round the edges of an obstacle or through a gap.

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Question 9

Modern Physics

The energy E of a photon and its wavelength are related by Eλ=X. The numerical value of X is [h=6.63 × 10−34 Js, c=3 × 108 ms−1]

  1. A.1.99 × 10−27.
  2. B.6.60 × 10−26.
  3. C.1.99 × 10−25.
  4. D.6.60 × 10−28.
Show the answer and explanation

Answer: C (1.99 × 10−25.)

Explanation

E = hc/λ, so Eλ = hc = 6.63 × 10−34 × 3 × 108 = 1.99 × 10−25 J m.

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Question 10

Radioactivity and Nuclear Physics

A radioactive substance has a half-life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?

  1. A.1/16
  2. B.1/8
  3. C.1/4
  4. D.1/32.
Show the answer and explanation

Answer: A (1/16)

Explanation

80 days ÷ 20 days = 4 half-lives. The fraction left is (½)4 = 1/16.

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Question 11

Electronics

Silicon doped with aluminium and germanium doped with arsenic become

  1. A.n- type semiconductors.
  2. B.p- type semiconductors.
  3. C.n- and p- types respectively.
  4. D.p- and n- types respectively.
Show the answer and explanation

Answer: D (p- and n- types respectively.)

Explanation

Aluminium has three valence electrons and creates holes, making p-type silicon. Arsenic has five and gives spare electrons, making n-type germanium.

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Question 12

Modern Physics

A photon of wavelength 6.0 × 10−7m behaves like a particle of a certain mass. The value of that mass is [h=6.63 × 10−34 Js, c=3 × 108 ms−1]

  1. A.1.1 × 10−35 kg.
  2. B.2.2 × 10−27 kg.
  3. C.3.5 × 10−36 kg.
  4. D.2.2 × 10−35 kg.
Show the answer and explanation

Answer: C (3.5 × 10−36 kg.)

Explanation

For a photon, mc2 = hc/λ, so m = h/(λc) = 6.63 × 10−34 ÷ (6.0 × 10−7 × 3 × 108) ≈ 3.7 × 10−36 kg, closest to 3.5 × 10−36 kg.

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Question 13

Radioactivity and Nuclear Physics

During the nuclear reactions described by 23592W → 23593X → 23191Y, the particles emitted are respectively

  1. A.β and β.
  2. B.β and α.
  3. C.α and β.
  4. D.α and α.
Show the answer and explanation

Answer: B (β and α.)

Explanation

W → X: mass number unchanged, atomic number up by 1, so a β particle is emitted. X → Y: mass number down by 4 and atomic number down by 2, so an α particle is emitted.

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Question 14

Electronics

The bond that forms a semiconductor is

  1. A.ionic.
  2. B.metallic.
  3. C.covalent.
  4. D.electrovalent.
Show the answer and explanation

Answer: C (covalent.)

Explanation

In silicon and germanium crystals each atom shares its four valence electrons with four neighbours, forming covalent bonds.

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Question 15

Alternating Current Circuits

The instrument that measures both a.c. and d.c. is

  1. A.an inverter.
  2. B.a current balance.
  3. C.a moving coil ammeter.
  4. D.a moving iron ammeter.
Show the answer and explanation

Answer: D (a moving iron ammeter.)

Explanation

In a moving-iron meter both irons are magnetised the same way whichever way the current flows, so it deflects for both a.c. and d.c. A moving-coil meter works for d.c. only.

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Question 16

Electromagnetic Induction

Lenz's law is a law of the conservation of

  1. A.energy.
  2. B.momentum.
  3. C.electric current.
  4. D.electric charge.
Show the answer and explanation

Answer: A (energy.)

Explanation

The induced current opposes the change causing it, so work must be done to produce it. This is the conservation of energy.

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Question 17

Current Electricity

Three cells each of e.m.f. 1.5 V and internal resistance 2.5Ω are connected as shown in the diagram above. Find the net e.m.f. and the internal resistance.

Diagram for this question
  1. A.4.5V, 7.50 Ω.
  2. B.4.5V, 0.83 Ω.
  3. C.1.5V, 7.50 Ω.
  4. D.1.5V, 0.83 Ω.
Show the answer and explanation

Answer: D (1.5V, 0.83 Ω.)

Explanation

Identical cells in parallel give the e.m.f. of one cell, 1.5 V. Their internal resistances are in parallel: 2.5 ÷ 3 ≈ 0.83 Ω.

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Question 18

Alternating Current Circuits

A 120 V, 60 W lamp is to be operated on 220 V a.c. supply mains. Calculate the value of non-inductive resistance that would be required to ensure that the lamp is run on correct value.

  1. A.500Ω.
  2. B.300Ω.
  3. C.200Ω.
  4. D.100Ω.
Show the answer and explanation

Answer: C (200Ω.)

Explanation

Lamp current I = P/V = 60 ÷ 120 = 0.5 A. The resistor in series must take the other 220 − 120 = 100 V, so R = 100 ÷ 0.5 = 200 Ω.

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Question 19

Magnetism

In the diagram above, if the magnetic field points into the paper and the force on a current-carrying conductor points upwards, what is the direction of the current?

Diagram for this question
  1. A.Downwards
  2. B.Right
  3. C.Left
  4. D.Upwards
Show the answer and explanation

Answer: B (Right)

Explanation

Fleming's left-hand rule: point the first finger into the paper (field) and the thumb up (force); the second finger then points to the right, which is the current direction.

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Question 20

Electromagnetic Induction

The instantaneous value of the induced e.m.f. as a function of time is ε=ε0 sin ωt where ε0 is the peak value of the e.m.f. The instantaneous value of the e.m.f., one quarter of the period, is

  1. A.ε0
  2. B.ε0/2
  3. C.ε0/4
  4. D.0.
Show the answer and explanation

Answer: A (ε0)

Explanation

ω = 2π/T, so at t = T/4 the angle ωt = π/2. ε = ε0 sin 90° = ε0.

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Question 22

Current Electricity

A conductor has a diagram of 1.00 mm and length 2.00 m. If the resistance of the material 0.1Ω, its resistivity is

  1. A.3.93 × 10−8 Ωm.
  2. B.3.93 × 10−6 Ωm.
  3. C.2.55 × 102 Ωm.
  4. D.2.55 × 105 Ωm.
Show the answer and explanation

Answer: A (3.93 × 10−8 Ωm.)

Explanation

Cross-section A = πd2/4 = π(1.0 × 10−3)2/4 = 7.85 × 10−7 m2. ρ = RA/L = 0.1 × 7.85 × 10−7 ÷ 2 = 3.93 × 10−8 Ω m.

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Question 23

Electrostatics

A charge 50 μC has an electric field strength of 360 NC−1 at a certain point. The electric field strength due to another charge 120 μC kept at the same distance apart and in the same medium is

  1. A.18 NC−1
  2. B.144 NC−1
  3. C.150 NC−1
  4. D.864 NC−1
Show the answer and explanation

Answer: D (864 NC−1)

Explanation

At the same distance, E = kq/r2 is proportional to q. E = 360 × 120/50 = 864 N C−1.

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Question 24

Magnetism

Two long parallel wires X and Y carry currents 3A and 5A respectively. If the force experienced per unit length by X is 5 × 10−5 N, the force per unit length experienced by wire Y is

  1. A.5 × 10−4 N.
  2. B.3 × 10−6 N.
  3. C.3 × 10−5 N.
  4. D.5 × 10−5 N.
Show the answer and explanation

Answer: D (5 × 10−5 N.)

Explanation

The force per unit length between two wires is μ0IXIY/2πr, the same expression for both wires (Newton's third law). So Y feels the same 5 × 10−5 N per metre.

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Question 25

Alternating Current Circuits

From the diagram above, the inductive reactance and the resistance R are respectively

Diagram for this question
  1. A.10Ω and 50Ω.
  2. B.20Ω and 50Ω.
  3. C.25Ω and 51Ω.
  4. D.50Ω and 45Ω.
Show the answer and explanation

Answer: A (10Ω and 50Ω.)

Explanation

XL = 2πfL = 2π × (50/π) × 0.1 = 10 Ω. The impedance is 75 ÷ 1.5 = 50 Ω, so R = √(502 − 102) ≈ 49 Ω, about 50 Ω.

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Question 26

Electrical Energy and Power

A 40 kW electric cable is used to transmit electricity through a resistor of resistance 2.0 Ω at 800 V. The power loss as internal energy is

  1. A.5.0 × 102 W.
  2. B.4.0 × 103 W.
  3. C.5.0 × 103 W.
  4. D.4.0 × 102 W.
Show the answer and explanation

Answer: C (5.0 × 103 W.)

Explanation

Current in the cable I = P/V = 40 000 ÷ 800 = 50 A. Power lost as heat = I2R = 2500 × 2.0 = 5000 W = 5.0 × 103 W.

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Question 27

Electrostatics

If two charged plates are maintained at a potential difference of 3 kV, the work done in taking a charge of 600 μC across the field is

  1. A.9.0 J.
  2. B.0.8 J.
  3. C.18.0 J.
  4. D.1.8 J.
Show the answer and explanation

Answer: D (1.8 J.)

Explanation

W = QV = 600 × 10−6 × 3000 = 1.8 J.

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Question 28

Thermal Expansion

The ratio of the coefficient of linear expansion of two metals α1/α2 is 3:4. If, when heated through the same temperature change, the ratio of the increase in lengths of the two metals, e1/e2 is 1:2, the ratio of the original lengths is l1/l2

  1. A.2/3
  2. B.8/3
  3. C.3/8
  4. D.3/2
Show the answer and explanation

Answer: A (2/3)

Explanation

Expansion e = αlΔθ with the same Δθ, so e1/e2 = (α1/α2)(l1/l2). 1/2 = (3/4)(l1/l2), so l1/l2 = 2/3.

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Question 29

Structure of Matter and Kinetic Theory

I. Mass II. Density III. Temperature IV. Nature of substance.
Which of the above affect diffusion?

  1. A.I, II and IV only
  2. B.II, III and IV only
  3. C.I, II, III and IV.
  4. D.I and II only.
Show the answer and explanation

Answer: C (I, II, III and IV.)

Explanation

Lighter, less dense molecules diffuse faster (Graham's law), molecules move faster at higher temperature, and diffusion is quicker in gases than in liquids. So mass, density, temperature and the nature of the substance all matter.

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Question 30

Elasticity

A string of length 4 m is extended by 0.02 m when a load of 0.4 kg is suspended at its end. What will be the length of the string when the applied force is 15 N? [g=10ms−2]

  1. A.5.05 m
  2. B.6.08 m
  3. C.4.05 m
  4. D.4.08 m
Show the answer and explanation

Answer: D (4.08 m)

Explanation

k = F/e = (0.4 × 10) ÷ 0.02 = 200 N m−1. A 15 N force stretches it 15 ÷ 200 = 0.075 m, so the length is 4 + 0.075 ≈ 4.08 m.

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Question 31

Temperature and Thermometers

I. Use a liquid with a high melting point.
II. Use a liquid of high volume expansivity.
III. Use a capillary tube of large diameter.
Which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?

  1. A.II only
  2. B.II and III only
  3. C.I only
  4. D.I and III only
Show the answer and explanation

Answer: A (II only)

Explanation

A thermometer is more sensitive when the liquid expands a lot per degree (high expansivity) and the bore is narrow. A wide tube or a high melting point does not help.

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Question 32

Heat Transfer

The blade of a hoe feels colder to touch in the morning than the wooden handle because the

  1. A.handle contains stored energy in form of heat
  2. B.blade is placed at a lower temperature than the handle
  3. C.handle is a better conductor of heat than the blade
  4. D.blade is a better conductor of heat than the handle
Show the answer and explanation

Answer: D (blade is a better conductor of heat than the handle)

Explanation

Both are at the same temperature, but the metal blade conducts heat away from your hand much faster than wood, so it feels colder.

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Question 34

Density and Upthrust

A piece of iron weighs 250 N in air and 200 N in a liquid of density 1000 kgm−3. The volume of the iron is [g=10ms−2]

  1. A.2.0 × 10−3 m3.
  2. B.2.5 × 10−3 m3.
  3. C.4.5 × 10−3 m3.
  4. D.5.0 × 10−3 m3.
Show the answer and explanation

Answer: D (5.0 × 10−3 m3.)

Explanation

Upthrust = 250 − 200 = 50 N = ρVg, so V = 50 ÷ (1000 × 10) = 5.0 × 10−3 m3.

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Question 35

Change of State and Vapours

I. Increase the melting point of the liquid.
II. Increase the boiling point of the liquid.
III. Decrease the melting point of the liquid.
IV. Decrease the boiling point of the liquid.
Which of the statements above about the effects of increase in pressure in a liquid are correct?

  1. A.II and III only
  2. B.I and II only
  3. C.III and IV only
  4. D.I and III only
Show the answer and explanation

Answer: A (II and III only)

Explanation

Higher pressure makes a liquid boil at a higher temperature (pressure cooker). For water, higher pressure lowers the melting point (regelation). So II and III are correct.

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Question 36

Thermal Expansion

A blacksmith heated a metal whose cubic expansivity is 6.3 × 10−6 K−1. The real expansivity is

  1. A.6.3 × 10−6 K−1.
  2. B.4.2 × 10−6 K−1.
  3. C.2.1 × 10−6 K−1.
  4. D.2.0 × 10−6 K−1.
Show the answer and explanation

Answer: A (6.3 × 10−6 K−1.)

Explanation

For a solid, the cubic expansivity is its real expansivity (the real/apparent distinction applies to liquids in containers), so the real expansivity is 6.3 × 10−6 K−1.

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Question 37

Structure of Matter and Kinetic Theory

The differences observed in solids, liquids and gases may be accounted for by

  1. A.their relative masses.
  2. B.their melting points.
  3. C.the different molecules in each of them.
  4. D.the spacing and forces acting between the molecules.
Show the answer and explanation

Answer: D (the spacing and forces acting between the molecules.)

Explanation

By kinetic theory, solids, liquids and gases differ in how far apart their molecules are and how strongly they attract each other.

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Question 38

Pressure

A reservoir 500 m deep is filled with a fluid of density 850 kgm−3. If the atmospheric pressure is 1.05 × 105 Nm−2, the pressure at the bottom of the reservoir is [g=10ms−2]

  1. A.4.36 × 106 Nm−2.
  2. B.4.25 × 106 Nm−2.
  3. C.4.72 × 106 Nm−2.
  4. D.4.28 × 106 Nm−2.
Show the answer and explanation

Answer: A (4.36 × 106 Nm−2.)

Explanation

Liquid pressure = hρg = 500 × 850 × 10 = 4.25 × 106 N m−2. Adding atmospheric pressure: 4.25 × 106 + 0.105 × 106 ≈ 4.36 × 106 N m−2.

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Question 39

Temperature and Thermometers

A clinical thermometer is different from other mercury in glass thermometers owing to

  1. A.its long stem.
  2. B.the constriction on its stem.
  3. C.its wide range of temperatures.
  4. D.the grade of mercury used in it.
Show the answer and explanation

Answer: B (the constriction on its stem.)

Explanation

The constriction (narrow kink) stops the mercury flowing back into the bulb, so the reading stays after the thermometer leaves the patient.

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Question 40

Quantity of Heat

2 kg of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. What is the increase in temperature of the water? [Specific heat capacity of water = 4200 Jkg−1 K−1]

  1. A.25°C
  2. B.15°C
  3. C.10°C
  4. D.30°C
Show the answer and explanation

Answer: C (10°C)

Explanation

Heat supplied = IVt = 3.5 × 200 × 120 = 84 000 J. Δθ = 84 000 ÷ (2 × 4200) = 10°C.

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Question 41

Work, Energy and Power

A boy drags a bag of rice along a smooth horizontal floor with a force of 2 N applied at an angle of 60° to the floor. The work done after a distance of 3 m is

  1. A.6 J.
  2. B.5 J.
  3. C.4 J.
  4. D.3 J.
Show the answer and explanation

Answer: D (3 J.)

Explanation

Only the horizontal component of the force does work: W = F cos 60° × s = 2 × 0.5 × 3 = 3 J.

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Question 42

Scalars and Vectors

The figure above shows 4 forces 3 N, 10N, 3√3 N and 6 N acting on a particle P. The resultant of the four forces is

Diagram for this question
  1. A.10√3 N.
  2. B.10 N.
  3. C.5√3 N.
  4. D.5 N.
Show the answer and explanation

Answer: B (10 N.)

Explanation

Horizontal: 3 + 10 cos 30° − 6 cos 60° = 3 + 8.66 − 3 = 8.66 N. Vertical: 3√3 + 10 sin 30° − 6 sin 60° = 5.20 + 5 − 5.20 = 5 N. Resultant = √(8.662 + 52) = √100 = 10 N.

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Question 43

Gravitational Field

Two spheres of masses 5.0 kg and 10.0 kg are 0.3 m apart. Calculate the force of attraction between them. [G=6.67 × 10−11 Nm2 kg−2]

  1. A.3.71 × 10−8 N.
  2. B.3.57 × 10−2 N.
  3. C.4.00 × 102 N.
  4. D.3.50 × 10−10 N.
Show the answer and explanation

Answer: A (3.71 × 10−8 N.)

Explanation

F = Gm1m2/r2 = 6.67 × 10−11 × 5 × 10 ÷ 0.09 = 3.71 × 10−8 N.

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Question 44

Motion

A car of mass 1500 kg goes round a circular curve of radius 50 m at a speed of 40 ms−1. The magnitude of the centripetal force on the car is

  1. A.1.2 × 103 N.
  2. B.1.2 × 103 N.
  3. C.4.8 × 103 N.
  4. D.4.8 × 104 N.
Show the answer and explanation

Answer: D (4.8 × 104 N.)

Explanation

F = mv2/r = 1500 × 402 ÷ 50 = 48 000 N = 4.8 × 104 N.

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Question 45

Equilibrium of Forces

The unit of moment of a couple can be expressed in

  1. A.Nm
  2. B.Nm−1
  3. C.Nm2.
  4. D.Nm−2.
Show the answer and explanation

Answer: A (Nm)

Explanation

The moment of a couple is one force × the perpendicular distance between the forces, so its unit is newton × metre (N m).

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Question 46

Simple Machines

A machine has a velocity ratio of 4. If it requires 800 N to overcome a load of 1600 N, what is the efficiency of the machine?

  1. A.2%
  2. B.40%
  3. C.50%
  4. D.60%
Show the answer and explanation

Answer: C (50%)

Explanation

MA = load ÷ effort = 1600 ÷ 800 = 2. Efficiency = MA/VR × 100% = 2/4 × 100% = 50%.

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Question 47

Equilibrium of Forces

In the diagram above, a rod 50 cm long of uniform cross-section is suspended horizontally on a fulcrum, F, by the action of two forces. What is the weight of the rod?

Diagram for this question
  1. A.120N
  2. B.200 N
  3. C.40 N
  4. D.80 N.
Show the answer and explanation

Answer: D (80 N.)

Explanation

The rod's weight W acts at its middle (25 cm), 10 cm right of F. Moments about F: 200 × 10 = 40 × 30 + W × 10, so 2000 = 1200 + 10W and W = 80 N.

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Question 48

Motion

A gramophone record takes 5s to reach its constant angular velocity of 4π rads−1 from rest. Find its constant angular acceleration.

  1. A.20.0π rads−2.
  2. B.1.3π rads−2.
  3. C.0.8π rads−2.
  4. D.0.4π rads−2.
Show the answer and explanation

Answer: C (0.8π rads−2.)

Explanation

Angular acceleration = change in angular velocity ÷ time = 4π ÷ 5 = 0.8π rad s−2.

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Question 49

Simple Machines

A particle of weight 120 N is placed on a plane inclined at an angle 30° to the horizontal. If the plane has an efficiency of 60%, what is the force required to push the weight uniformly up the plane?

  1. A.210 N
  2. B.175 N
  3. C.100 N
  4. D.50 N.
Show the answer and explanation

Answer: C (100 N)

Explanation

VR = 1/sin 30° = 2. MA = efficiency × VR = 0.6 × 2 = 1.2. Effort = load ÷ MA = 120 ÷ 1.2 = 100 N.

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Question 50

Simple Harmonic Motion

From the diagram above, the bob of the pendulum has the fastest speed at

Diagram for this question
  1. A.W.
  2. B.Z.
  3. C.Y.
  4. D.X.
Show the answer and explanation

Answer: D (X.)

Explanation

The bob moves fastest at the lowest point of its swing, X, where all its potential energy has become kinetic energy. It is momentarily at rest at W and Z.

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