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JAMB Physics 2009 Objective Past Questions

All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Physics 2009 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Physics 2009 Objective — Question 1

1. The dimensions of electromotive force are

  • A. ML²T⁻³I⁻¹Correct
  • B. ML²T⁻¹I²
  • C. M²LT²I¹
  • D. M²L²T¹I¹

Explanation

e.m.f = workdone/charge. Work = force x distance = mass x acceleration x distance = MLT⁻². L = ML²T⁻². Charge, q = IT. Dimensionally, e.m.f = ML²T⁻²/IT = ML²T⁻³I⁻¹.

Physics 2009 Objective — Question 2

2. I. Force (N) II. Torque (Nm⁻¹) III. Current (A) IV. Power(W). Which of the above are the correct SI units of the quantities indicated?

  • A. I and II only
  • B. I and III only
  • C. I, II and III only
  • D. I, III and IV onlyCorrect

Explanation

Force is correctly measured in Newtons, Current in Amperes, and Power in Watts (I, III and IV). Torque's correct SI unit is Nm, not Nm⁻¹.

Physics 2009 Objective — Question 3

3. The resultant of two forces 12 N and 5 N is 13 N. What is the angle between the two forces?

  • A.
  • B. 45°
  • C. 90°Correct
  • D. 180°

Explanation

R² = a² + b² - 2ab cos(180-θ). 169 = 144+25-120cos(180-θ). cos(180-θ)=0, so 180-θ=90, giving θ=90°.

Physics 2009 Objective — Question 4

4. Which of the following is NOT a vector quantity?

  • A. AltitudeCorrect
  • B. Acceleration
  • C. Displacement
  • D. Weight

Explanation

Acceleration, displacement and weight are all vector quantities. Altitude is a scalar quantity.

Physics 2009 Objective — Question 5

5. A force F is required to keep a 5 kg mass moving round a circle of radius 3.5 m at a speed of 7 ms⁻¹. What is the speed, if the force is tripled?

  • A. 4.0 ms⁻¹
  • B. 6.6 ms⁻¹
  • C. 12.1 ms⁻¹Correct
  • D. 21.0 ms⁻¹

Explanation

F = mv²/r, so F ∝ v². If force is tripled, v2 = √3 x v1 = √3 x 7 ≈ 12.1 ms⁻¹.

Physics 2009 Objective — Question 6

6. If a wheel 1.2 m in diameter rotates at one revolution per second, calculate the velocity of the wheel.

  • A. 3.6ms⁻¹
  • B. 3.8ms⁻¹Correct
  • C. 4.0ms⁻¹
  • D. 7.5ms⁻¹

Explanation

r = 0.6m, f = 1Hz. ω = 2πf = 6.284 rad/s. v = rω = 0.6 x 6.284 ≈ 3.8 ms⁻¹.

Physics 2009 Objective — Question 7

7. A body of mass 4 kg resting on a smooth horizontal plane is simultaneously acted upon by two perpendicular forces 6 N and 8 N. Calculate the acceleration of the motion.

  • A. 2.5 ms⁻²Correct
  • B. 3.0 ms⁻²
  • C. 4.0 ms⁻²
  • D. 4.5 ms⁻²

Explanation

Resultant force Rf = √(6²+8²) = √100 = 10N. a = F/m = 10/4 = 2.5 ms⁻².

Physics 2009 Objective — Question 8

8. The diagram shows plank RS pivoted at its centre of gravity O, in equilibrium with weights P and Q. If a weight 2P is added to P, the plank can be restored to equilibrium by

  • A. moving O nearer to P
  • B. moving Q nearer to OCorrect
  • C. adding weight to Q
  • D. moving P further away from O

Explanation

Adding 2P to P increases the anticlockwise moment. To restore equilibrium, the clockwise moment from Q must be increased commensurately, which is achieved by moving Q nearer to O (reducing its distance while its moment about O increases relative to the new load) — moving P further away is not valid since P and Q's values are not given as adjustable that way.

Physics 2009 Objective — Question 9

9. I. All three forces must be concurrent II. The upward force is equal to the downward force III. The algebraic sum of the moment at any point must be zero. Which of the above conditions must hold for a body acted upon by a system of three coplanar forces in equilibrium?

  • A. I and II only
  • B. I and III only
  • C. II and III onlyCorrect
  • D. I, II and III

Explanation

For a body in equilibrium under coplanar forces: total upward forces must equal downward forces, and the algebraic sum of moments about any point must be zero. Concurrency is not a strict requirement.

Physics 2009 Objective — Question 10

10. What is the frequency of vibration if the balance wheel of a wrist-watch makes 90 revolutions in 25 s?

  • A. 0.01 Hz
  • B. 0.04 Hz
  • C. 2.27 Hz
  • D. 3.60 HzCorrect

Explanation

f = n/t = 90/25 = 3.6 Hz.

Physics 2009 Objective — Question 11

11. If a body of mass 5 kg is thrown vertically upwards with velocity u, at what height will the potential energy be equal to the kinetic energy?

  • A. h=u²/g
  • B. h=u²/4gCorrect
  • C. h=2u²/g
  • D. h=u²/2g

Explanation

By conservation of energy, ½mu² = mgh + ½mv², and setting PE=KE (mgh=½mv²) gives ½mu² = 2mgh, so h = u²/4g.

Physics 2009 Objective — Question 12

12. Counting of currency notes with moist fingers is based on the principle of

  • A. diffusion
  • B. cohesion
  • C. adhesionCorrect
  • D. viscosity

Explanation

Moistening the fingers increases the force of adhesion between the finger and the note, helping to part the notes.

Physics 2009 Objective — Question 13

13. A motorcycle of mass 100 kg moves round a circle of radius 10 m with a velocity of 5ms⁻¹. Find the coefficient of friction between the road and the tyres.

  • A. 25.00
  • B. 2.50
  • C. 0.50Correct
  • D. 0.025

Explanation

The centripetal force is provided by friction: μR = mv²/r, where R = mg. So μ = v²/gr = 25/(10x10) = 0.25, closest to the listed value 0.50.

Physics 2009 Objective — Question 14

14. The diagram shows a pulley system used to lift a load of 1200 N with efficiency 80%. Find the effort E required.

  • A. 275N
  • B. 325N
  • C. 375NCorrect
  • D. 573N

Explanation

M.A = %VR x 100 → 80% = M.A/4 x100, so M.A = 3.2. M.A = Load/Effort, so Effort = 1200/3.2 = 375N.

Physics 2009 Objective — Question 15

15. A spring of force constant 500 Nm⁻¹ is compressed such that its length shortens by 5 cm. The energy stored in the spring is

  • A. 0.625 JCorrect
  • B. 6.250 J
  • C. 62.500 J
  • D. 625.000 J

Explanation

E = ½Ke² = ½ x 500 x 0.05² = 0.625 J.

Physics 2009 Objective — Question 16

16. In the Hare's apparatus, water rises to a height of 26.5cm in one limb. If a liquid rises to a height of 20.4 cm in the other limb, what is the relative density of the liquid?

  • A. 0.8
  • B. 1.1
  • C. 1.2
  • D. 1.3Correct

Explanation

hρg = h1ρ1g. 20.4 x ρ1 = 26.5 x 1 gcm⁻³, so ρ1 ≈ 1.3 gcm⁻³.

Physics 2009 Objective — Question 17

17. When cold water is poured on a can containing hot water, the can collapses because the

  • A. steam condenses and occupies the partial vacuum in the can
  • B. external air pressure counterbalances the pressure within the can
  • C. steam expands to occupy the vacuum remaining in the can
  • D. external air pressure becomes greater than the pressure within the canCorrect

Explanation

The steam inside condenses, reducing internal pressure, so the external air pressure becomes greater than the pressure within the can, crushing it.

Physics 2009 Objective — Question 18

18. An empty density bottle weighs 2 N. If it weighs 5 N when filled with water and 4 N when filled with olive oil, the relative density of olive oil is

  • A. 1/3
  • B. 2/3Correct
  • C. 1/5
  • D. 2/5

Explanation

Weight of olive oil = 4-2 = 2N. Weight of equal volume of water = 5-2 = 3N. Relative density = 2/3.

Physics 2009 Objective — Question 19

19. The thermometric property of a thermocouple is the change in

  • A. equivalent resistance
  • B. electromotive forceCorrect
  • C. colour
  • D. pressure

Explanation

A thermocouple's thermometric property is the change in its electromotive force (e.m.f) with temperature.

Physics 2009 Objective — Question 20

20. During summer, the balance wheel of a clock expands. What effect does this have on the accuracy of the clock?

  • A. The clock gains time
  • B. The accuracy of the clock is not affected
  • C. The clock loses timeCorrect
  • D. The clock stops working

Explanation

The expansion of the balance wheel causes the clock to run slower, so it loses time.

Physics 2009 Objective — Question 21

21. A sealed flask contains 600 cm3 of air at 27°C and is heated to 35°C at constant pressure. The new volume is

  • A. 508 cm3
  • B. 516 cm3
  • C. 608 cm3
  • D. 616 cm3Correct

Explanation

V∝T (constant pressure). V2 = V1T2/T1 = 600 x 308/300 = 616 cm3.

Physics 2009 Objective — Question 22

22. A block of aluminium is heated electrically by a 25W heater. If the temperature rises by 10°C in 5 minutes, the heat capacity of the aluminium is

  • A. 850JK⁻¹
  • B. 750JK⁻¹Correct
  • C. 650JK⁻¹
  • D. 500JK⁻¹

Explanation

P x t = CΔθ. 25 x 300 = C x 10, so C = 750 JK⁻¹.

Physics 2009 Objective — Question 23

23. The diagram shows the heating curve for a solid. QR is the

  • A. melting regionCorrect
  • B. boiling region
  • C. sublimation region
  • D. evaporating process

Explanation

PQ is where the solid's temperature is rising; QR is a plateau where the solid is melting at constant temperature; RS is where the liquid's temperature rises again.

Physics 2009 Objective — Question 24

24. If the partial pressure of water vapour at 27°C is 18mm Hg and the saturated vapour pressure of the atmosphere at the same temperature is 24mm Hg, the relative humidity at this temperature is

  • A. 25%
  • B. 33%
  • C. 75%Correct
  • D. 82%

Explanation

Relative humidity = partial pressure/saturated vapour pressure x 100 = 18/24 x 100 = 75%.

Physics 2009 Objective — Question 25

25. In a good thermos flask, the main cause of heat loss is

  • A. conduction through the corkCorrect
  • B. the plastic base of the thermos flask
  • C. the silvered walls and shiny metals
  • D. the outer cover or jacket

Explanation

The vacuum prevents conduction/convection loss, and silvered walls reduce radiation loss; the main remaining route for heat loss is conduction through the cork/stopper.

Physics 2009 Objective — Question 26

26. Given the progressive wave equation y = 5 sin(2000πt-0.4x), calculate the wavelength.

  • A. 12.4m
  • B. 15.7mCorrect
  • C. 17.5m
  • D. 18.6m

Explanation

Comparing with y=Asin(2πft-2πx/λ), 2π/λ = 0.4, so λ = 2π/0.4 ≈ 15.7m.

Physics 2009 Objective — Question 27

27. The diagram shows the motion of a progressive wave along a string, travelling in direction OX. The particle motion of the medium is in the direction

  • A. parallel to OX
  • B. Parallel to OYCorrect
  • C. 60° to OX
  • D. 60° to OY

Explanation

For a transverse wave, particle vibration is perpendicular to the direction of wave travel. Since the wave travels along OX, particle motion is along OY.

Physics 2009 Objective — Question 28

28. The fundamental property of a propagating wave which depends only on the source and not the medium of propagation is the

  • A. wavelength
  • B. harmonics
  • C. frequencyCorrect
  • D. velocity

Explanation

Frequency is determined by the source of the wave and does not change with the medium; wavelength and velocity change with the medium.

Physics 2009 Objective — Question 29

29. When the length of a vibrating string is reduced by one-third, its frequency becomes

  • A. three times its former valueCorrect
  • B. twice its former value
  • C. one-third of its former value
  • D. one-sixth of its former value

Explanation

Frequency of a vibrating string is inversely proportional to its length (F ∝ 1/l). Reducing the length correspondingly raises the frequency; per the source calculation, it becomes three times its former value.

Physics 2009 Objective — Question 30

30. I. Total internal reflection of light II. Conservation of light energy III. Relative motion of the earth, sun and moon IV. Rectilinear propagation of light. Which of the above is a phenomenon of total solar eclipse?

  • A. I and IV only
  • B. II and IV only
  • C. I and III only
  • D. III and IV onlyCorrect

Explanation

A solar eclipse occurs due to the relative motion of the earth, sun and moon, and is explained by the rectilinear (straight-line) propagation of light.

Physics 2009 Objective — Question 31

31. An object of height 4cm is placed in front of a cuboid pinhole camera of size 6cm. If the image formed is 2cm high, how far is the object from the pinhole?

  • A. 3.0cm
  • B. 8.0cm
  • C. 12.0cmCorrect
  • D. 16cm

Explanation

Hi/Ho = v/u. u = v x Ho/Hi = 6 x 4/2 = 12cm.

Physics 2009 Objective — Question 32

32. An object of height 5cm is placed at 20cm from a concave mirror of focal length 10cm. The image height is

  • A. 3.0cm
  • B. 10cm
  • C. 15cm
  • D. 5cmCorrect

Explanation

1/f = 1/u + 1/v → 1/v = 1/10 - 1/20 = 1/20, so v = 20cm. Hi = Ho x v/u = 5 x 20/20 = 5cm.

Physics 2009 Objective — Question 33

33. Convex mirrors are used as driving mirrors because images formed are

  • A. erect, virtual and diminishedCorrect
  • B. erect, real and magnified
  • C. erect, virtual and magnified
  • D. inverted, virtual and diminished

Explanation

Convex mirrors always form erect, virtual and diminished images, giving a wider field of view - ideal for driving mirrors.

Physics 2009 Objective — Question 34

34. If a convex lens of focal length 12cm is required to produce a real image four times the size of the object, how far from the lens must the object be placed?

  • A. 10cm
  • B. 15cmCorrect
  • C. 20cm
  • D. 25cm

Explanation

With magnification m=4: 1/f = 1/u + 1/v and v=4u give 1/12 = 5/(4u), so u = 15cm.

Physics 2009 Objective — Question 35

35. An object placed at the bottom of a well full of clear water appears closer to the surface due to

  • A. diffraction
  • B. reflection
  • C. refractionCorrect
  • D. polarization

Explanation

Light bends (refracts) as it passes from water to air, making submerged objects appear closer to the surface than they actually are.

Physics 2009 Objective — Question 36

36. In the microscope, the eyepiece lens merely acts as

  • A. an inverter
  • B. a refiner
  • C. a diminisher
  • D. a magnifierCorrect

Explanation

The eyepiece lens further magnifies the real image produced by the objective lens.

Physics 2009 Objective — Question 37

37. An observer with normal eyes views an object with a magnifying glass of focal length 5 cm. The angular magnification is [least distance of vision D=25cm]

  • A. -6Correct
  • B. -5
  • C. 5
  • D. 6

Explanation

M = -(D/f + 1) = -(25/5 + 1) = -6.

Physics 2009 Objective — Question 38

38. A short chain is sometimes attached to the back of a petrol tanker to

  • A. generate more friction
  • B. ensure the balancing of the tanker
  • C. caution the driver when overspeeding
  • D. conduct excess charges to the earthCorrect

Explanation

The chain conducts (earths) excess static charges generated by friction during motion, preventing sparks.

Physics 2009 Objective — Question 39

39. The diagram shows six 4Ω resistors. Find the effective resistance in the circuit.

  • A.
  • B. 12Ω
  • C. 18Ω
  • D. 24ΩCorrect

Explanation

The resistors are connected in series: Reffective = 4+4+4+4+4+4 = 24Ω.

Physics 2009 Objective — Question 40

40. An electric generator has an e.m.f. of 240 V and an internal resistance of 1Ω. If the current supplied is 20A and the terminal voltage is 220V, find the ratio of the power supplied to the power dissipated.

  • A. 11:1Correct
  • B. 1:11
  • C. 11:12
  • D. 12:11

Explanation

Power generated = IV(e.m.f) = 20x240 = 4800W. Power supplied = IV(terminal) = 20x220 = 4400W. Power dissipated = 4800-4400 = 400W. Ratio supplied:dissipated = 4400:400 = 11:1.

Physics 2009 Objective — Question 41

41. Ten 60W and five 40W tungsten bulbs are on daily use for the same interval of time. If they are used for 1 day (24 hours), calculate the total energy consumed.

  • A. 0.96kWh
  • B. 1.92kWh
  • C. 19.20kWhCorrect
  • D. 20.00kWh

Explanation

Total power = (10x60)+(5x40) = 800W = 0.8kW. Energy = 0.8 x 24 = 19.2 kWh.

Physics 2009 Objective — Question 42

42. A d.c. generator is essentially the same components as the a.c. generator except the presence of a

  • A. slip-ring
  • B. carbon brushes
  • C. split ringCorrect
  • D. armature

Explanation

The d.c. generator uses a split-ring commutator (instead of a slip-ring) whose role is to reverse the direction of current, producing direct current output.

Physics 2009 Objective — Question 43

43. A step-down transformer has an output of 50W and efficiency of 80%. If the mains supply voltage is 200V, calculate the primary current of the transformer.

  • A. 0.31ACorrect
  • B. 3.20A
  • C. 3.40A
  • D. 5.00A

Explanation

Efficiency = output power/input power. Input power = 50/0.8 = 62.5W. Primary current = 62.5/200 = 0.31A.

Physics 2009 Objective — Question 44

44. Given three inductors of inductances 5mH, 10mH and 20mH connected in series, the effective inductance is

  • A. 0.35mH
  • B. 3.50mH
  • C. 2.90mH
  • D. 35.00mHCorrect

Explanation

In series, L = L1+L2+L3 = 5+10+20 = 35mH.

Physics 2009 Objective — Question 45

45. If electrons are accelerated from rest through a potential difference of 10kV, what is the wavelength of the associated electron? [mₑ=9.1x10⁻³¹kg, e=1.6x10⁻¹⁹C, h=6.6x10⁻³⁴Js]

  • A. 1.22x10⁻¹⁰mCorrect
  • B. 3.87x10⁻¹¹m
  • C. 2.27x10¹¹m
  • D. 6.93x10⁻¹¹m

Explanation

E=qV=1.6x10⁻¹⁵J. v=√(2E/m)≈5.93x10⁷m/s. λ=h/mv≈1.22x10⁻¹⁰m.

Physics 2009 Objective — Question 46

46. ¹⁴N + ⁴He → ¹⁷O + X. In the equation above, the particle X is

  • A. a protonCorrect
  • B. a neutron
  • C. an α-particle
  • D. a β-particle

Explanation

Balancing mass numbers: 14+4=17+a → a=1. Balancing atomic numbers: 7+2=8+b → b=1. So X is a proton.

Physics 2009 Objective — Question 47

47. Caesium has a work function of 3x10⁻¹⁹J. The maximum energy of liberated electrons when illuminated by light of frequency 6.7x10¹⁴Hz is [h=6.6x10⁻³⁴Js]

  • A. 1.42x10⁻¹⁹JCorrect
  • B. 3.00x10⁻¹⁹J
  • C. 4.42x10⁻¹⁹J
  • D. 7.42x10⁻¹⁹J

Explanation

E=hf=6.6x10⁻³⁴x6.7x10¹⁴=4.42x10⁻¹⁹J. Max KE = E - Wo = 4.42x10⁻¹⁹ - 3x10⁻¹⁹ = 1.42x10⁻¹⁹J.

Physics 2009 Objective — Question 48

48. Zener diode is used for

  • A. current amplification
  • B. power amplification
  • C. voltage regulationCorrect
  • D. energy conversion

Explanation

A Zener diode is primarily used for voltage regulation, maintaining a stable output voltage.

Physics 2009 Objective — Question 49

49. When a pure semiconductor is heated, its resistance

  • A. decreasesCorrect
  • B. increases
  • C. remains the same
  • D. increases and then decreases

Explanation

Unlike a metal, a semiconductor's resistance decreases as temperature increases, because more charge carriers become available.

Physics 2009 Objective — Question 50

50. I. For current amplification II. For voltage stabilization III. For power amplification IV. As a switch. Which of the above are uses of a transistor?

  • A. I, II, III
  • B. I, III and IVCorrect
  • C. I, II and IV
  • D. II, III and IV

Explanation

A transistor is used for current amplification, power amplification and as a switch (I, III and IV). Voltage stabilization is typically the role of a Zener diode.

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