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.
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.
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.
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.
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.
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.
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.
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.
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.
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%.
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.
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.
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.
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.
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?
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.
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]
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.
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.
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.
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.
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]
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.
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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