2012 JAMB PHYSICS PAST QUESTIONS AND ANSWERS

The type of paper is indicated as Type Purple. Identifying the paper type is critical for reference and submission.

Parallax error is avoided by focusing the eye vertically downward on the markings. Any deviation introduces error.

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

The diagram likely involves vector resolution. Based on force components, OT = 5 N.

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

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

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

As r decreases, F increases because force is inversely proportional to distance squared.

Objects oscillate due to a restoring force directly proportional to displacement (Hooke’s Law).

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

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

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

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

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

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

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

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

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

Clinical thermometers are typically mercury-in-glass for precise temperature readings.

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

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

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

Evaporation increases with temperature, molecular speed, and kinetic energy but decreases with higher atmospheric pressure.

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

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

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

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

Velocity of sound changes with wind direction and temperature, independent of pitch, loudness, or atmospheric pressure.

Intensity is determined by amplitude, as energy is proportional to Amplitude$^2$.

The object should be between the focus and pole for a magnified, upright image in a concave mirror.

Using proportionality, Image height/Object height = Screen distance/Object distance: 2.5/Height = 5/300, Height = 150 m.

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

Rays from the sun converge at the focus of a concave mirror.

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

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

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

Dispersion occurs when white light separates into its component colors due to varying refractive indices of media.

Accumulators deliver higher current due to lower internal resistance, suitable for car starters, unlike dry cells.

Parallel connection maintains individual cell voltage. Effective e.m.f. = 2V.

Fuses are connected to the live side to prevent excessive current from damaging the device.

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

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

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

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

Using X-rays to detect internal objects is diagnostic imaging.

Gamma rays are neutral and not deflected by electric or magnetic fields.

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

Diodes conduct when forward-biased, allowing rectification of a.c. to d.c.

Reverse bias increases the depletion layer width as majority carriers are pulled away.

Semiconductors are compact, energy-efficient, durable, and resist high temperatures, fulfilling all criteria.