2012 JAMB PHYSICS PAST QUESTIONS AND ANSWERS

$$\text{The type of paper is indicated as Type Purple. Identifying the paper type is critical for reference and submission.}$$

$$\text{Parallax error is avoided by focusing the eye vertically downward on the markings. Any deviation introduces error.}$$

$$\text{The horizontal force component is } F \cos 45 = 1500 \times \frac{1}{\sqrt{2}} = 1061 \text{ N (approximately).}$$

$$\text{The diagram likely involves vector resolution. Based on force components, OT = 5 N.}$$

$$\text{Velocity-time graphs reveal acceleration, deceleration, and distance but not instantaneous force applied unless mass is provided.}$$

$$\text{Using } s = \frac{u^2}{2a}, \text{ we find } s = \frac{(30)^2}{2 \times 0.1} = 4500 \text{ m.}$$

$$\text{Using } s = ut + \frac{1}{2}at^2, \text{ where } u = 0, s = \frac{1}{2} \times 30 \times 20^2 = 6000 \text{ m } (6 \text{ km).}$$

$$\text{As r decreases, } F \text{ increases because force is inversely proportional to distance squared.}$$

$$\text{Objects oscillate due to a restoring force directly proportional to displacement (Hooke’s Law).}$$

$$\text{Angular velocity } \omega = \frac{v}{r} = \frac{1}{0.5} = 2 \text{ rad/s.}$$

$$\text{Work done } W = \int F dx \text{. Given a force-distance diagram, calculate the area under the curve. Example: Area = 2400 J.}$$

$$\text{Using the formula, coefficient of friction is given by } \tan(\alpha).$$

$$\text{Friction coefficient determined by resolving forces on the incline and equating } \mu = \frac{F_r}{F_n}. \, \text{Using given values: } \mu = 0.5.$$

$$\text{Efficiency } \eta = \frac{\text{Work Output}}{\text{Work Input}} \times 100. \, \eta = \frac{25 \times 30}{1500} = 83.3\%. $$

$$\text{Force constant } k = \frac{F}{x} = \frac{10}{0.012} = 833 \text{ N/m.}$$

$$\text{Apparent weight: } W_a = W – \rho Vg = 2 \times 10 – 1 \times 1 \times 10 = 10 \text{ N.}$$

$$\text{Pressure depends on depth and density, independent of surface area or volume. } P = \rho gh.$$

$$\text{Upthrust } F_u = \rho Vg = 1.3 \times 300 \times 10 = 3900 \text{ N.}$$

$$\text{Clinical thermometers are typically mercury-in-glass for precise temperature readings.}$$

$$\text{Length increase ratio is product of linear expansivity and length: } \frac{2}{3} \times \frac{3}{4} = \frac{8}{9}. $$

$$\text{Standard volume } V_2 = V_1 \frac{P_1T_2}{P_2T_1}. \, V_2 = 1767.3 \text{ cm}^3 \text{ at STP.}$$

$$\text{Heat transfer } Q = mc \Delta T \text{ gives: } \Delta T = \frac{Q}{mc} = \frac{500}{0.1 \times 4200} = 1.2^\circ C.$$

$$\text{Evaporation increases with temperature, molecular speed, and kinetic energy but decreases with higher atmospheric pressure.}$$

$$\text{Using } c = 2.1 \times 10^3 \text{ J/kgK, } L = 3.5 \times 10^5 \text{ J/kg: } Q = (1)(2.1 \times 10^3)(30) + (1)(3.5 \times 10^5) = 4.13 \times 10^5 \text{ J.}$$

$$\text{Brownian motion proves molecules exist, are in constant random motion, and describes rapid, irregular motion of particles suspended in a fluid.}$$

$$\text{The wavelength is the reciprocal of } \frac{2}{5}, \text{ calculated as } \lambda = 5 \text{ m.}$$

$$\text{Phase difference is proportional to } x/\lambda, \text{ expressed in fractions of the wavelength. Example: } \phi = \frac{x}{\lambda}. $$

$$\text{Velocity of sound changes with wind direction and temperature, independent of pitch, loudness, or atmospheric pressure.}$$

$$\text{Intensity is determined by amplitude, as energy is proportional to } \text{Amplitude}^2.$$

$$\text{The object should be between the focus and pole for a magnified, upright image in a concave mirror.}$$

$$\text{Using proportionality, } \text{Image height/Object height = Screen distance/Object distance: } 2.5/\text{Height} = 5/300, \text{Height} = 150 \text{ m.}$$

$$\text{Plane mirrors always produce images of same size as the object: Magnification = } \frac{\text{Image height}}{\text{Object height}} = 1.$$

$$\text{Rays from the sun converge at the focus of a concave mirror.}$$

$$\text{Displacement } = \text{Thickness} \left(1 – \frac{1}{\text{Refractive index}} \right): 10 \left(1 – \frac{1}{1.5} \right) = 3.33 \text{ cm.}$$

$$\text{Total internal reflection occurs when light travels from a denser to a less dense medium, and the angle of incidence exceeds the critical angle.}$$

$$\text{Lenses are used in optical instruments like projectors, human eyes, and telescopes but not in periscopes, which rely on mirrors.}$$

$$\text{Dispersion occurs when white light separates into its component colors due to varying refractive indices of media.}$$

$$\text{Accumulators deliver higher current due to lower internal resistance, suitable for car starters, unlike dry cells.}$$

$$\text{Parallel connection maintains individual cell voltage. Effective } \text{e.m.f. = 2V.}$$

$$\text{Fuses are connected to the live side to prevent excessive current from damaging the device.}$$

$$\text{Using } F = qvB, \text{ solve for } B: B = \frac{F}{qv} = \frac{1.8 \times 10^{-8}}{(1.0 \times 10^{-8})(3.0 \times 10^2)} = 6.0 \times 10^{-2} \text{T.}$$

$$\text{Capacitive reactance decreases as frequency increases, } X_c \propto \frac{1}{f}. \text{ Graph is a hyperbola.}$$

$$\text{The r.m.s value of current } I = \frac{I_0}{\sqrt{2}}: I = \frac{10}{\sqrt{2}} \approx 7.1 \text{ A.}$$

$$\text{E.m.f. } E = Blv \sin \theta: E = (1.5)(50)(1) \sin 30° = 37.5 \text{ V.}$$

$$\text{Using X-rays to detect internal objects is diagnostic imaging.}$$

$$\text{Gamma rays are neutral and not deflected by electric or magnetic fields.}$$

$$\text{Time } = \text{Number of half-lives } \times \text{Half-life: } N_0 = 1000 \to 125 \text{ (3 half-lives)}, t = 3 \times 20 = 60 \text{ s.}$$

$$\text{Diodes conduct when forward-biased, allowing rectification of a.c. to d.c.}$$

$$\text{Reverse bias increases the depletion layer width as majority carriers are pulled away.}$$

$$\text{Semiconductors are compact, energy-efficient, durable, and resist high temperatures, fulfilling all criteria.}$$