All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Physics 2007 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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A glass plate 0.9cm thick has a refractive index of 1.50. How long does it take for a pulse of light to pass through the plate? [c=3.0x10^8ms-1]
Velocity in glass = c/n = 3x10^8/1.5 = 2x10^8m/s. Distance = 0.9cm = 9x10^-3m. Time = distance/speed = 9x10^-3/2x10^8 = 4.5x10^-11s.
The fundamental frequency of a plucked wire under tension of 400N is 250Hz. When the frequency is changed to 500Hz at constant length, the tension is
f is proportional to sqrt(T). f1/f2 = sqrt(T1/T2): 250/500 = sqrt(400/T2), giving T2 = 1600N.
The production of pure spectrum could easily be achieved by using
A pure spectrum is produced using a triangular prism together with two convex lenses to focus and disperse the light cleanly.
I. They should be identical II. They should originate from the same source III. They should be coherent IV. They should be monochromatic. From the statements above, the conditions for two waves to interfere are
For sustained interference, two waves must be identical, originate from the same source, and be coherent - statements I, II and III.
The instrument used by designers to obtain different colour patterns is called a
A kaleidoscope comprises mirrors inclined at 60 degrees to each other and is used to obtain different colour patterns.
When an object is placed between the principal focus and the optical centre of a convex lens, it could be used as a
A simple microscope (magnifying glass) uses a single convex lens, with the object placed between the focus and the optical centre.
In a large telecommunications auditorium, perforated absorbent materials are used to line the ceiling so as to
Perforated absorbent materials absorb sound energy, reducing unwanted reverberation in the hall.
The phenomenon of light bending round an obstacle is
The bending of light (or any wave) around an obstacle is aptly described as diffraction.
The energy E of a photon and its wavelength are related by E(lambda) = X. The numerical value of X is [h=6.63x10^-34 Js, c=3x10^8ms-1]
E(lambda) = hc = 6.63x10^-34 x 3x10^8 = 1.99x10^-25.
A radioactive substance has a half-life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?
Number of half-lives = 80/20 = 4. Fraction remaining = (1/2)^4 = 1/16.
Silicon doped with aluminium and germanium doped with arsenic become
Doping with a trivalent element (aluminium) produces a p-type semiconductor; doping with a pentavalent element (arsenic) produces an n-type semiconductor. So silicon+aluminium=p-type, germanium+arsenic=n-type.
A photon of wavelength 6.0x10^-7m behaves like a particle of a certain mass. The value of that mass is [h=6.63x10^-34 Js, c=3x10^8]
Using mc^2 = hc/lambda, so m = h/(lambda x c) = 6.63x10^-34 / (6x10^-7 x 3x10^8) = 3.68x10^-34kg (approx 3.5x10^-34kg).
The nuclear reactions described by 235/92 W -> 235/93 X -> 231/91 Y, the particles emitted are respectively
235/92 W -> 235/93 X emits a beta particle (electron); 235/93 X -> 231/91 Y emits an alpha particle (helium nucleus).
The bond that forms a semiconductor is
Semiconductors have four valence electrons per atom, shared with surrounding atoms in a tetrahedral arrangement - a covalent bond.
The instrument that measures both a.c and d.c is
A moving iron ammeter, which works on the heating effect of current, can measure both AC and DC.
Lenz's law is a law of the conservation of
Lenz's law (the induced emf opposes the change producing it) is a statement of the conservation of energy.
Use the diagram below (three cells each of e.m.f. 1.5V and internal resistance 2.5 ohm connected in parallel) to answer this question. Find the net e.m.f. and the internal resistance.
The cells are connected in parallel, so the net emf equals a single cell's emf (1.5V). Combined internal resistance: 1/r = 1/2.5+1/2.5+1/2.5 = 3/2.5, giving r = 0.83 ohm.
A 120V, 60W lamp is to be operated on a 220V a.c. supply mains. Calculate the value of non-inductive resistance that would be required to ensure that the lamp is run on correct voltage.
I = P/V = 60/120 = 0.5A. Rin = 120/0.5 = 240 ohm. Rout = 220/0.5 = 440 ohm. Extra non-inductive resistance needed = 440-240 = 200 ohm.
Use the diagram below (a current-carrying conductor in a magnetic field, field into the page) to answer this question. If the magnetic field points into the page and the force on the current-carrying conductor points upwards, what is the direction of the current?
Using Fleming's Left Hand Rule with the field into the page and force upward, the current points to the right.
The instantaneous value of the induced e.m.f. as a function of time is e = e0 sin(wt) where e0 is the peak value of the e.m.f. The instantaneous value of the e.m.f. one quarter of the period is
One quarter of a cycle corresponds to theta = 90 degrees. e = e0 sin(90) = e0 x 1 = e0.
Use the information below to answer questions 21 and 22. 118.8cm2 surface of the copper cathode of a voltameter is to be coated with 10 micron thick copper of density 9x10^3 kgm-3. How long will the process run with 10A constant current? [e.c.e of copper = 3.3x10^-7 kgC-1]
Area = 118.8x10^-4 m2, thickness = 10x10^-6m, volume = area x thickness. Mass = density x volume. Using Faraday's first law, m = ZIt, so t = m/(ZI), which works out to approximately 5.4 minutes.
A conductor has a diameter of 1.00mm and length 2.00m. If the resistance of the material is 0.1 ohm, its resistivity is
R = (resistivity x length)/Area. Area = pi.d2/4 = 7.855x10^-7 m2. Resistivity = (R x A)/l = (0.1 x 7.855x10^-7)/2 = 3.93x10^-8 ohm.m.
E = Kq/r2 (E = electric field intensity, K = 1/(4.pi.e0), q = charge, r = distance). A charge 50uC has an electric field strength of 360NC-1 at a certain point. The electric field strength due to another charge 120uC kept at the same distance apart and in the same medium is
Since E is directly proportional to q: E1/q1 = E2/q2. 360/50 = E2/120, so E2 = 864NC-1.
Two long parallel wires X and Y carry currents 3A and 5A respectively. If the force experienced per unit length by wire X is 5x10^-5N, the force per unit length experienced by wire Y is
By Newton's third law, the force per unit length on each wire is equal in magnitude, so wire Y also experiences 5x10^-5N per unit length.
Use the diagram below (an inductor of 0.1H in series with a resistor R, connected to a 75V, 50/pi Hz a.c. source, carrying 1.3A) to answer this question. The inductive reactance and the resistance R are respectively
XL = 2.pi.f.L = 2 x 3.142 x (50/3.142) x 0.1 = 10 ohm. Z = V/I = 75/1.5 = 50 ohm. R = sqrt(Z2-XL2) is approx 50 ohm.
A 40kW electric cable is used to transmit electricity through a resistor of resistance 2.0 ohm at 800V. The power loss (internal energy) is
I = P/V = 40,000/800 = 50A. Power loss = I2R = 50^2 x 2 = 5000W = 5.0x10^3 W.
If two charged plates are maintained at a potential difference of 3kV, the work done in taking a charge of 600uC across the field is
W = qV = 600x10^-6 x 3x10^3 = 1.8J.
The ratio of the coefficient of linear expansion of two metals is a1/a2 = 3:4. If, when heated through the same temperature change, the ratio of their increase in lengths (dL1/dL2) is 1:2, the ratio of the original lengths (l1/l2) is
Using a = dL/(l.d(theta)) with the same temperature change for both metals: l1/l2 = (dL1/dL2) x (a2/a1) = (1/2) x (4/3) = 2/3.
I. Mass II. Density III. Temperature IV. Nature of substance. Which of the above affect diffusion?
Density, temperature and the nature of a substance all affect diffusion rate. Mass alone is not a listed factor (it is embedded within density).
A string of length 4m is extended by 0.02m when a load of 0.4kg is suspended at its end. What will be the length of the string when the applied force is 15N? [g=10ms-2]
Load = 0.4 x 10 = 4N produces extension 0.02m, so K = 4/0.02 = 200N/m. For 15N: extension = 15/200 = 0.075m. New length = 4 + 0.075 = 4.08m.
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 would enhance the sensitivity of a liquid-in-glass thermometer?
Sensitivity of a liquid-in-glass thermometer is enhanced by using a liquid with high volume expansivity (statement II).
The blade of a hoe feels colder to touch in the morning than the wooden handle because
Because the metal blade is a better conductor of heat than the wooden handle, it draws heat away from the hand faster, feeling colder.
Which of the following gas laws is equivalent to the work done?
Since P is measured in Nm-2 and V in m3, the unit of PV is Nm, the same as work. Boyle's law (PV=constant) is therefore equivalent to work done.
A piece of iron weighs 250N in air and 200N in a liquid of density 1000kgm-3. The volume of the iron is [g=10ms-2]
Upthrust = 250-200 = 50N = weight of fluid displaced. Mass of fluid displaced = 50/10 = 5kg. Volume = mass/density = 5/1000 = 0.005m3 = 5.0x10^-3 m3.
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 shows the effects of increased pressure on a liquid?
Increasing pressure raises a liquid's boiling point but lowers its melting point - statements II and III.
If the cubic expansivity of a metal is 6.3x10^-6 K-1, its area (superficial) expansivity is
Cubic expansivity (gamma) = 3 x linear expansivity (alpha), and area expansivity (beta) = 2 x alpha. So beta = (2/3) x gamma = (2/3) x 6.3x10^-6 = 4.2x10^-6 K-1.
The differences observed in solids, liquids and gases may be accounted for by
The states of matter differ mainly because of how closely packed their molecules are and the intermolecular forces between them.
A reservoir 500m deep is filled with a fluid of density 850kgm-3. If the atmospheric pressure is 1.05x10^5 Nm-2, the pressure at the bottom of the reservoir is [g=10ms-2]
P = h.rho.g + atmospheric pressure = (500 x 850 x 10) + 1.05x10^5 = 4.355x10^6 Nm-2 (approx 4.36x10^6 Nm-2).
A clinical thermometer is different from other mercury-in-glass thermometers owing to
The constriction (kink) in the stem of a clinical thermometer is what distinguishes it from ordinary mercury-in-glass thermometers.
2kg of water is heated with a heating coil which draws 3.5A on a 200V mains for 2 minutes. What is the increase in temperature of the water? [Specific heat capacity of water = 4200 Jkg-1K-1]
Electrical energy = Heat energy: IVt = mc(delta theta). Delta theta = (3.5x200x2x60)/(2x4200) = 10 C.
A boy drags a bag of rice along a smooth horizontal floor with a force of 2N applied at an angle of 60 degrees to the floor. The work done after a distance of 3m is
Horizontal component of the force = 2cos60 = 1N. Work done = Force x distance = 1 x 3 = 3J.
Use the diagram below (four forces 2sqrt3N, 10N, 3sqrt3N and 6N acting on a particle P) to answer this question. The resultant of the four forces is
Resolving each force into horizontal and vertical components and summing gives total horizontal = 8.66N and total vertical = 5N. Resultant = sqrt(8.66^2+5^2) = 10N.
Two spheres of masses 5.0kg and 10.0kg are 0.3m apart. Calculate the force of attraction between them. [G=6.67x10^-11 Nm2kg-2]
F = Gm1m2/r2 = (6.67x10^-11 x 5 x 10)/0.3^2 = 3.71x10^-8 N.
A car of mass 1500kg goes round a circular curve of radius 50m at a speed of 40ms-1. The magnitude of the centripetal force on the car is
F = mv2/r = (1500 x 40^2)/50 = 4.8x10^4 N.
The unit of moment of a couple can be expressed as
Moment (of a couple) is force times distance, giving units of Newton-metres (Nm).
A machine has a velocity ratio of 4. If it requires 800N to overcome a load of 1600N, what is the efficiency of the machine?
Mechanical Advantage = Load/Effort = 1600/800 = 2. Efficiency = (MA/VR) x 100% = (2/4) x 100% = 50%.
Use the diagram below (a uniform rod 50cm long resting on a fulcrum F, with weights 200N and 40N at either end and the rod's own weight w acting at its centre) to answer this question. What is the weight of the rod?
Taking moments about the fulcrum: clockwise moment = anticlockwise moment. Solving w(25-15) + 40(45-15) = 200(15-5) gives w = 80N.
A gramophone record takes 5s to reach a constant angular velocity of 4.pi rads-1 from rest. Find its constant angular acceleration.
Angular acceleration = (w1-w0)/t = (4.pi - 0)/5 = 0.8.pi rads-2.
A particle of weight 120N is placed on a plane inclined at an angle 30 degrees to the horizontal. If the plane has an efficiency of 60%, what is the force required to push the weight uniformly up the plane?
Velocity Ratio = 1/sin(theta) = 1/sin30 = 2. Mechanical Advantage = VR x efficiency = 2 x 0.6 = 1.2. Effort = Load/MA = 120/1.2 = 100N.
Use the diagram below (a simple pendulum showing points W, X, Y, Z and pivot O along its swing) to answer this question. The bob of the pendulum has the fastest speed at
For a pendulum in simple harmonic motion, velocity is maximum at the equilibrium (lowest) position, point X, while acceleration is maximum at the extreme ends (W and Z).
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