JAMB Physics 2008
Questions & Answers

Students could mistakenly pick another height option if they miscalculate the time of flight or confuse the initial launch equations for the two bodies. By setting up the position equations for both the upward-thrown ball and the falling ball, you equate their heights to find the meeting time of four-thirds of a second, which yields a height of nine meters.

Option A or B might be chosen through an incorrect trigonometric setup or by ignoring one of the equilibrium axes. Resolving the forces into horizontal and vertical components allows you to balance the upward vertical tensions against the downward weight of thirty newtons while equating the horizontal components, ultimately solving for thirty newtons.

A student might incorrectly use formulas meant for series spring combinations, leading to a mismatched oscillation period. Parallel springs add their constants directly to give a combined value of three hundred newtons per meter, which can then be applied to the standard period formula to find zero point two five seconds.

Option C or D might be selected if a student forgets to account for the opposite initial direction or misapplies momentum conservation. By applying conservation of momentum alongside the relative velocity equation for elastic collisions, you determine that the two-kilogram mass reverses direction with a velocity of negative two meters per second.

A test-taker might omit the frictional component or the vector resolution of the incline and select a smaller numerical value. Calculating the normal force using the cosine component and the friction using the coefficient yields the necessary opposing forces that, when combined with the down-slope weight component, demand a minimum force of four point two four newtons to push the block upward.

Options A or B might be mistakenly chosen by directly scaling percentages without factoring in the object's constant density. Since the density of the floating object remains unchanged, you equate the product of the first fraction and density to the product of the second fraction and twelve hundred, solving for fifteen hundred kilograms per cubic meter.

A student could choose option A or C by misapplying the natural logarithm or incorrectly evaluating the initial gas state. Using the ideal gas law product for initial pressure and volume to find the constant term, and multiplying it by the natural log of the volume ratio, results in three point four six times ten to the fourth joules.

Option A or B might be chosen if a student forgets to account for the latent heat required to undergo phase change before temperature equilibrium begins. Calculating the heat needed to melt the ice and comparing it with the heat lost by the warm water leaves a remaining thermal energy pool that brings the final mixture temperature to twenty degrees Celsius.

A student might mistakenly calculate efficiency using work directly without converting it to total heat input, or use incorrect temperature ratios. Dividing the work output by the forty percent efficiency gives the total heat input, which then allows you to find the cold reservoir temperature of three hundred sixty kelvin via Carnot efficiency.

Option A, C, or D could be selected if angular frequency and linear wave parameters are incorrectly substituted. Using the maximum acceleration formula containing amplitude and angular frequency, you isolate the wave speed to arrive at twenty-five meters per second.

A student might mistakenly use the fundamental frequency formula without multiplying by the harmonic number, selecting two hundred hertz instead. Finding the fundamental frequency from the wave speed and twice the string length, then scaling it by a factor of four for the fourth harmonic, gives eight hundred hertz.

Options A or B might be chosen if the grating spacing is calculated with incorrect powers of ten or if the diffraction order is omitted. Converting the lines per millimeter into a grating spacing and applying the diffraction grating formula with an order of two yields a sine value of zero point six, corresponding to an angle of thirty-six point nine degrees.

A student might mistakenly multiply the focal lengths or invert the ratio, leading to a fraction or incorrect scaling factor. Angular magnification for this optical setup is found by dividing the focal length of the objective lens by the focal length of the eyepiece, yielding twenty.

Option A or C might be selected if the charge, mass, or magnetic field values are misplaced during calculation. Setting the magnetic force equal to the centripetal force and solving for the radius of the circular trajectory gives zero point two six meters.

Students might mistakenly choose option B (12 V) by incorrectly assuming the voltage splits evenly across unequal capacitors. To find the correct answer, begin by determining the equivalent capacitance, C_eq, which evaluates to (10 × 20) / (10 + 20) = 20/3 μF. Next, compute the total charge stored in the series network using Q = C_eq × V, resulting in (20/3) × 10^-6 × 24 = 160 × 10^-6 C. Because charge remains identical on all components connected in series, the potential difference across the specific 10 μF capacitor is found by dividing this charge by its capacitance, yielding (160 × 10^-6) / (10 × 10^-6) = 16 V. Common mistake: assuming that series capacitors divide the source voltage equally regardless of their individual capacitance values.

Students might pick option A (1 Ω) by mistakenly dividing the internal resistance instead of applying the condition for maximum energy transfer. The principle governing this setup dictates that maximum power is delivered when the load resistance equals the internal resistance. Following the maximum power transfer theorem, the required value for the load resistor directly matches the internal resistance given as 2 Ω. Common mistake: confusing maximum power transfer conditions with impedance matching or minimum resistance pathways.

Students could choose option B (22.5 A) by incorrectly multiplying rather than dividing by the transformer efficiency. To solve this, first relate the output power to the input power using the efficiency formula η = P_out / P_in, which rearranges to 0.9 = 4500 / P_in, meaning P_in = 4500 / 0.9 = 5000 W. With the total input power established, calculate the primary current by dividing P_in by the primary voltage V_p, resulting in 5000 / 200 = 25 A. Common mistake: multiplying the output power by the efficiency instead of dividing.

Students may select option B (4 hours) by incorrectly dividing the total time by an improper integer factor instead of determining the number of half-life cycles. Begin by setting up the activity ratio 200/800, which simplifies to 1/4. Equating this fraction to (1/2)^n reveals that 1/4 = (1/2)^2, meaning the sample has undergone exactly n = 2 half-lives. Dividing the total elapsed time of 12 hours by these 2 half-lives gives 12 / 2 = 6 hours. Common mistake: dividing the time interval by the ratio of the activities instead of the number of half-life cycles.

Students might accidentally select option B (90) by forgetting that beta emission alters the proton count of the nucleus. In beta-minus decay, a neutron converts to a proton, which increases the atomic number by 1. Adding this change to the original atomic number yields 90 + 1 = 91. Common mistake: assuming that beta decay leaves the atomic number unchanged like gamma decay.

Students might choose option B (4.0 x 10^5 ms^-1) by omitting the work function subtraction and treating the photon energy directly as kinetic energy. First, calculate the maximum kinetic energy by subtracting the work function from the photon energy: KE = E - φ = 4.0 × 10^-19 - 2.5 × 10^-19 = 1.5 × 10^-19 J. Next, isolate velocity from the kinetic energy equation KE = (1/2)mv^2 by rewriting it as v = √(2 × KE / m). Substituting the given mass m_e = 9.1 × 10^-31 kg yields √(2 × 1.5 × 10^-19 / 9.1 × 10^-31), which evaluates to approximately 5.0 × 10^5 ms^-1. Common mistake: failing to subtract the work function from the incoming photon energy.