JAMB Physics 1993

25 reviewed questions with answers and explanations.

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Question 1

Which of the following quantities has the same unit as the watt?

  1. Force x time
  2. Force x distance
  3. Force x acceleration
  4. Force x velocity
Answer and explanation

D: Force x velocity

A watt is a joule per second. Force multiplied by velocity has units N·m/s =J/s =W, so it has the unit of power.

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Question 4

Two forces of 7 N and 3 N act at right angles. If θ is the angle between the resultant and the 7 N force, which relation holds?

  1. cos θ =3/7
  2. sin θ =3/7
  3. tan θ =3/7
  4. cot θ =3/7
Answer and explanation

C: tan θ =3/7

The resultant has components 7 N along the larger force and 3 N perpendicular to it. Therefore tan θ =opposite/adjacent =3/7.

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Question 5

An aeroplane lands at 180 km/h and slows uniformly to rest in 30 s. What distance does it cover before stopping?

  1. 360 m
  2. 540 m
  3. 750 m
  4. 957 m
Answer and explanation

C: 750 m

The landing speed is 180/3.6 =50 m/s. Uniform deceleration gives average speed(50 +0)/2 =25 m/s, so distance is 25 ×30 =750 m.

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Question 8

Starting with water stored at height, which sequence describes the main energy conversions when a hydroelectric station powers a lamp?

  1. Electrical → mechanical → potential → light
  2. Gravitational potential → mechanical → electrical → light
  3. Mechanical → sound → electrical → light
  4. Kinetic → mechanical → electrical → light
Answer and explanation

B: Gravitational potential → mechanical → electrical → light

Water stored above a turbine has gravitational potential energy. Flowing water turns the turbine, which drives a generator. Electrical energy from the generator is then converted to light by a lamp.

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Question 13

An object of mass 400 g and density 600 kg/m³ hangs at rest from a string with half its volume immersed in paraffin of density 900 kg/m³. Find the string tension. Use g =10 m/s².

  1. 1.0 N
  2. 3.0 N
  3. 4.0 N
  4. 5.0 N
Answer and explanation

A: 1.0 N

The object weighs 0.4 ×10 =4 N. Half its volume displaces paraffin of mass 0.5 ×0.4 ×900/600 =0.3 kg, giving 3 N upthrust. The string tension is 4 −3 =1 N.

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Question 14

With its reference-junction temperature fixed, a thermocouple measures temperature through changes in its

  1. e.m.f. changes with temperature
  2. Resistance changes with temperature
  3. Volume changes with temperature
  4. Pressure changes with resistance.
Answer and explanation

A: e.m.f. changes with temperature

With the reference junction held at a known temperature, a thermocouple’s emf changes with the measuring-junction temperature. Calibration relates that voltage to temperature.

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Question 16

A fixed mass of ideal gas has its volume halved and absolute temperature doubled. What happens to its pressure?

  1. Remains constant
  2. Increases by a factor of 4
  3. Increases by a factor of 3
  4. Decreases by a factor of 4
Answer and explanation

B: Increases by a factor of 4

For a fixed amount of ideal gas, P is proportional toT/V. Doubling absolute temperature while halving volume multiplies pressure by 2/(1/2) =4.

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Question 17

Mix 200 g of water at 90 °C with 100 g of water at 30 °C. Neglect heat loss and the container’s heat capacity. What is the final temperature?

  1. 50 °C
  2. 60 °C
  3. 70 °C
  4. 80 °C
Answer and explanation

C: 70 °C

Ignoring heat loss, heat lost by hot water equals heat gained by cool water:200(90 −T) =100(T −30). Solving gives 300 T =21000, so T =70 °C.

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Question 19

A solid melts at 80 °C. At that temperature,10⁵ J melts 10 g of the solid. Find its specific latent heat of fusion.

  1. 1.00 ×10³ J/kg
  2. 1.25 ×10⁵ J/kg
  3. 1.00 ×10⁷ J/kg
  4. 8.00 ×10⁸ J/kg
Answer and explanation

C: 1.00 ×10⁷ J/kg

At the melting point, heat goes into the phase change. The mass is 10 g =0.010 kg, so specific latent heat is Q/m =10⁵/0.010 =1.00 ×10⁷ J/kg.

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Question 21

A stationary wave has wavelength 60 cm. Find the distance between consecutive antinodes.

  1. 15 cm
  2. 30 cm
  3. 60 cm
  4. 120 cm
Answer and explanation

B: 30 cm

Consecutive antinodes of a stationary wave are half a wavelength apart. Their separation is therefore 60/2 =30 cm.

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Question 22

Which of the following waves can propagate through a vacuum?

  1. High velocity sound waves
  2. Ultrasonic waves
  3. Acoustic waves
  4. Infra-red waves
Answer and explanation

D: Infra-red waves

Infrared radiation is electromagnetic and can travel through vacuum. Sound, ultrasound and other acoustic waves need a material medium for their pressure disturbances.

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Question 23

A radar pulse returns to its transmitting antenna 4 ×10⁻³ s after reflection from an aircraft. How far away is the aircraft? Use c =3 ×10⁸ m/s.

  1. 6.0 ×10² km
  2. 1.2 ×10³ km
  3. 3.0 ×10³ km
  4. 6.0 ×10⁵ km
Answer and explanation

A: 6.0 ×10² km

The measured time includes the outward and return journeys. One-way distance is ct/2 =(3 ×10⁸)(4 ×10⁻³)/2 =6 ×10⁵ m =6 ×10² km.

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Question 26

A concave mirror of radius r forms a real image. Let u and v be the positive object and image distances. Which expression gives the magnitude of linear magnification?

  1. v/r −1
  2. 2 v/r −1
  3. u/r −1
  4. 2 u/r −1
Answer and explanation

B: 2 v/r −1

For a real image, magnification magnitude is v/u. Since 1/u +1/v =2/r, multiplying byv gives v/u +1 =2 v/r. Therefore magnification magnitude is 2 v/r −1.

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Question 27

A thin camera lens has focal length 20 cm. An object is 100 cm from the lens. What lens-to-film distance gives a sharp image?

  1. 17 cm
  2. 20 cm
  3. 25 cm
  4. 100 cm
Answer and explanation

C: 25 cm

Using 1/f =1/u +1/v gives 1/v =1/20 −1/100 =1/25. The film must therefore be 25 cm from the lens to receive the sharp real image.

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Question 30

The property of the eye known as its power of accommodation is controlled by the

  1. Pupil
  2. Vitreous humour
  3. Iris
  4. Ciliary muscles
Answer and explanation

D: Ciliary muscles

Ciliary muscles change tension in the supporting fibres of the lens. This changes lens curvature and optical power, allowing the eye to focus on objects at different distances.

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Question 34

Which of the following correctly explain(s) why soft iron is preferred to steel in electromagnets? I Soft iron is more readily magnetized than steel. II Soft iron is more readily demagnetized than steel. III Soft iron retains magnetism more than steel.

  1. I only
  2. II and III only
  3. I and II only
  4. I, II and III.
Answer and explanation

C: I and II only

Soft iron magnetizes readily when current flows and loses most of that magnetism when current stops. These properties make the electromagnet easy to switch on and off. Steel retains more magnetism.

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Question 37

The terminal voltage of a battery is 4.0 V when supplying a current of 2.0 A, and 2.0 V when supplying a current of 3.0 A. The internal resistance of the battery is

  1. 0.5 Ω
  2. 1.0 Ω
  3. 2.0 Ω
  4. 4.0 Ω
Answer and explanation

C: 2.0 Ω

Terminal voltage is V =E −Ir. The voltage falls by 4 −2 =2 V when current rises by 3 −2 =1 A. Thus r =2/1 =2 Ω; both measurements then give emf 8 V.

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Question 38

A 12 V ideal battery supplies a 1 Ω resistor in series with two parallel branches. Each branch contains two 2 Ω resistors in series. What is the total current supplied by the battery?

  1. 4.00 A
  2. 1.30 A
  3. 0.80 A
  4. 0.75 A
Answer and explanation

A: 4.00 A

Each parallel branch has two 2 Ω resistors in series, giving 4 Ω per branch. Their parallel equivalent is 2 Ω. Adding the 1 Ω series resistor gives 3 Ω, so battery current is 12/3 =4 A.

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Question 41

A 24 V ideal battery is connected across two parallel branches. One branch contains 5 μF and 15 μF capacitors in series; the other contains a 20 μF capacitor. All are initially uncharged. After charging, what is the voltage across the 5 μF capacitor?

  1. 3 V
  2. 6 V
  3. 12 V
  4. 18 V
Answer and explanation

D: 18 V

The 5 μF and 15 μF capacitors are in series across 24 V. Their voltage drops are inversely proportional to capacitance, so the 5 μF capacitor gets 15/(5 +15) ×24 =18 V.

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Question 42

A resistive instrument rated 40 W has resistance 90 Ω. What operating voltage gives its rated power?

  1. 60 V
  2. 150 V
  3. 225 V
  4. 3 600 V
Answer and explanation

A: 60 V

For a resistive instrument, P =V²/R. Therefore V =√(PR) =√(40 ×90) =√3600 =60 V.

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Question 44

For a fixed transmitted power, the primary aim of high-voltage transmission is to

  1. Minimize electrical energy losses due to heat production
  2. Increase the rate of energy transfer by using high voltage
  3. Increase the current in the wires.
  4. Generate electricity at high current and low voltage.
Answer and explanation

A: Minimize electrical energy losses due to heat production

For the same transmitted power, raising voltage reduces current. Since transmission-wire heating is I²R, the smaller current reduces wasted energy.

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Question 45

Find the inductive reactance of a 30.0 mH inductor at 1.30 ×10³ Hz, to one decimal place.

  1. 39.0 Ω
  2. 122.5 Ω
  3. 245.0 Ω
  4. 39000.0 Ω
Answer and explanation

C: 245.0 Ω

Inductive reactance is Xᴸ =2πfL. Converting 30.0 mH to 0.0300 H gives 2π ×1300 ×0.0300 =245.04… Ω, or 245.0 Ω to one decimal place.

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Question 46

Which statements correctly describe cathode rays? I. They consist of small negatively charged particles. II. They can be deflected magnetically but not electrically. III. They consist of fast neutrons deflected by an electric field.

  1. I only
  2. III only
  3. I and II only
  4. II and III only
Answer and explanation

A: I only

Cathode rays are streams of electrons, which carry negative charge. An electric field can deflect them, and a magnetic field can deflect them when they have a velocity component perpendicular to it.

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Question 49

A nucleus has atomic number 88 and mass number 226. After two beta-minus decays and one alpha decay, which notation represents the final nucleus?

  1. ²²²₈₂Z
  2. ²²²₈₈Z
  3. ²²⁶₈₆Z
  4. ²²⁶₈₀Z
Answer and explanation

B: ²²²₈₈Z

Each beta-minus decay raises atomic number by 1 without changing mass number. An alpha decay lowers mass number by 4 and atomic number by 2. The final numbers are A =226 −4 =222 and Z =88 +2 −2 =88.

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Question 50

For a fixed metal surface in the ordinary photoelectric effect, the maximum kinetic energy of emitted photoelectrons depends on the

  1. Intensity of incident radiation
  2. Duration of illumination
  3. Temperature of the radiation source alone
  4. Frequency of incident radiation
Answer and explanation

D: Frequency of incident radiation

For a fixed metal surface, maximum photoelectron kinetic energy is Kmax =hf −φ. Increasing incident frequency above threshold raises the photon energy and therefore the maximum electron energy.

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