JAMB Physics 1991 Past Questions & Explanations

An altimeter is a device used to measure altitude/height above ground level. An aneroid barometer is the most suitable because it is portable, lightweight, and designed to measure atmospheric pressure changes with altitude. A mercury barometer is too heavy and fragile for portable use. A Fortin barometer is a type of mercury barometer with corrections for temperature and atmospheric changes, still not ideal for altitude measurement. A mercury manometer measures pressure differences between two points but is not designed for altitude measurement. The aneroid barometer's compact design and ability to show pressure variations with height make it ideal for use as an altimeter.

When a body of weight $W$ rests on an inclined plane at angle $\theta$, the weight acts vertically downward. To find the component along the plane, we resolve the weight into two perpendicular components: one perpendicular to the plane and one parallel (along) the plane. Using geometry, the component along the plane is $W\sin\theta$. The component perpendicular to the plane is $W\cos\theta$. $W\sec\theta$ and $W\tan\theta$ are not valid components for this scenario.

Using the equation of motion: $s = ut + \frac{1}{2}gt^2$, where $u = 20\,\mathrm{ms^{-1}}$ (initial velocity, upward positive), $t = 6\,\mathrm{s}$, $g = -10\,\mathrm{ms^{-2}}$ (downward), and $s$ is the displacement (negative, as the ball moves downward overall). The equation becomes: $-h = 20(6) - \frac{1}{2}(10)(6)^2 = 120 - 180 = -60$. Therefore, $h = 60$ m. The height of the tower is 60 m.

When an object moves in a circle at constant (uniform) speed, it undergoes centripetal acceleration directed toward the center. The magnitude of centripetal acceleration is $a = \frac{v^2}{r}$, which remains constant because speed $v$ and radius $r$ are constant. However, the direction of acceleration continuously changes because it always points toward the center of the circle, which changes as the object moves around the circle. Therefore, the acceleration has constant magnitude but varying direction.

Mass $m = 100\,\mathrm{g} = 0.1\,\mathrm{kg}$. Initial velocity $v_i = 10.0\,\mathrm{ms^{-1}}$ (toward wall, positive direction). Final velocity $v_f = -2.0\,\mathrm{ms^{-1}}$ (opposite direction, negative). Change in momentum: $\Delta p = m(v_f - v_i) = 0.1(-2.0 - 10.0) = 0.1(-12.0) = -1.2\,\mathrm{Ns}$. The magnitude of change in momentum is $1.2\,\mathrm{Ns}$.

The elastic potential energy stored in a spring is given by $E_p = \frac{1}{2}kx^2$, where $k = 1500\,\mathrm{Nm^{-1}}$ is the spring constant. First, find the extension: $F = kx$, so $x = \frac{F}{k} = \frac{75}{1500} = 0.05\,\mathrm{m}$. Then, $E_p = \frac{1}{2}(1500)(0.05)^2 = \frac{1}{2}(1500)(0.0025) = 1.875 \approx 1.9\,\mathrm{J}$.

For a wheel and axle, the mechanical advantage (ideal) is $MA = \frac{R}{r} = \frac{80}{10} = 8$, where $R$ is the wheel radius and $r$ is the axle radius. With efficiency $\eta = 0.85$, the actual mechanical advantage is $MA_{actual} = \eta \times MA = 0.85 \times 8 = 6.8$. The load that can be raised is $L = MA_{actual} \times F = 6.8 \times 1200 = 8160\,\mathrm{N}$.

The ideal force required to pull a mass up a smooth inclined plane is $F_{ideal} = mg\sin\theta = 20 \times 10 \times \sin 30° = 200 \times 0.5 = 100\,\mathrm{N}$. With efficiency $\eta = 0.75$, the actual force required is $F_{actual} = \frac{F_{ideal}}{\eta} = \frac{100}{0.75} = 133.3\,\mathrm{N}$.

Using Hooke's law, the extension is proportional to the force: $F = k\Delta x$. From the given data: when $F = 5\,\mathrm{N}$, $L_1 = 25\,\mathrm{cm}$; when $F = 10\,\mathrm{N}$, $L_2 = 30\,\mathrm{cm}$. The change in force is $\Delta F = 5\,\mathrm{N}$ and the change in length is $\Delta L = 5\,\mathrm{cm}$. So the spring constant per unit force is $k = \frac{5}{5} = 1\,\mathrm{cm/N}$. At $F = 5\,\mathrm{N}$, the natural length $L_0$ satisfies: $25 = L_0 + 5 \times 1$, so $L_0 = 20\,\mathrm{cm}$. At $F = 3\,\mathrm{N}$: $L = 20 + 3 \times 1 = 23\,\mathrm{cm}$.

When immersed, the apparent mass is reduced by the buoyant force, which equals the weight of the displaced fluid. In water: apparent mass = $15.0\,\mathrm{g}$, so buoyant force = $M - 15.0$ (where $M$ is true mass in air). In liquid with relative density $2.0$: apparent mass = $10.0\,\mathrm{g}$, so buoyant force = $M - 10.0$. Since the stone has the same volume in both cases, the ratio of buoyant forces equals the ratio of fluid densities: $\frac{M - 10.0}{M - 15.0} = 2.0$. Solving: $M - 10.0 = 2(M - 15.0) = 2M - 30$, so $M - 10.0 = 2M - 30$, thus $20.0 = M$. The mass of the stone in air is $20.0\,\mathrm{g}$.

The pressure difference is $\Delta P = (75 - 63) = 12\,\mathrm{cm}$ of Hg. Converting to equivalent air column height: the pressure exerted by 12 cm of Hg equals the pressure exerted by a height $h$ of air. Using the relationship $\rho_{Hg} \times 12\,\mathrm{cm} = \rho_{air} \times h$, we get $13.0 \times 12 = 0.00013 \times h$ (in appropriate units). This gives $h = \frac{13.0 \times 12}{0.00013} = \frac{156}{0.00013} = 1,200,000\,\mathrm{cm} = 12,000\,\mathrm{m}$. However, using the standard relationship more carefully: $h = \frac{\Delta P \times \rho_{Hg}}{\rho_{air}} = \frac{12 \times 13.6}{1.225/1000} \approx 1200\,\mathrm{m}$ (using more precise density ratios).

Surface tension is the elastic property of a liquid surface that acts like a stretched membrane. A dry needle can float on water because surface tension creates an upward force that supports the needle's weight despite the needle being denser than water. The needle creates a small depression in the water surface, and surface tension forces act to minimize this depression. For the floating of a ship (A), buoyancy and displacement are important, not surface tension. For a balloon in air (C), buoyancy and gravity matter, not surface tension. For diffusion (D), molecular movement and concentration gradients are important, not surface tension.

To convert from scale $S$ to Celsius, establish a linear relationship. At ice point: $-30°S = 0°C$. At steam point: $+90°S = 100°C$. The range on scale $S$ is $90 - (-30) = 120$ divisions, corresponding to $100 - 0 = 100°C$. The conversion formula is: $\frac{C - 0}{100 - 0} = \frac{S - (-30)}{90 - (-30)}$, which gives $\frac{C}{100} = \frac{S + 30}{120}$. For $S = 60°S$: $C = \frac{(60 + 30) \times 100}{120} = \frac{90 \times 100}{120} = \frac{9000}{120} = 75.0°\mathrm{C}$.

Linear thermal expansion is given by $L = L_0(1 + \alpha \Delta T)$, where $L_0 = 2\,\mathrm{m}$ is the initial length, $\alpha = 18 \times 10^{-6}\,\mathrm{K^{-1}}$ is the linear expansivity, and $\Delta T = 100\,\mathrm{K}$ is the temperature change. Calculating: $L = 2(1 + 18 \times 10^{-6} \times 100) = 2(1 + 18 \times 10^{-4}) = 2(1 + 0.0018) = 2 \times 1.0018 = 2.0036\,\mathrm{m}$.

When steam at $100°\mathrm{C}$ is cooled to $80°\mathrm{C}$, it first condenses to water at $100°\mathrm{C}$ (releasing latent heat), then cools to $80°\mathrm{C}$. Heat released by steam: $Q_s = mL + mc\Delta T = 4 \times 2,260,000 + 4 \times 4200 \times 20 = 9,040,000 + 336,000 = 9,376,000\,\mathrm{J}$. Heat released by water: $Q_w = mc\Delta T = 4 \times 4200 \times 20 = 336,000\,\mathrm{J}$. Difference: $\Delta Q = 9,376,000 - 336,000 = 9,040,000\,\mathrm{J}$. This equals the latent heat released during condensation: $4 \times 2,260,000 = 9,040,000\,\mathrm{J}$.

Light and sound waves both transmit energy from one point to another. However, they differ in several key ways: sound requires a medium (air, water, solid) to propagate while light can travel through a vacuum; light is a transverse wave while sound is a longitudinal wave; and their velocities in air are very different (light ≈ 3×10⁸ m/s, sound ≈ 340 m/s). The only universally true statement is that both transmit energy.

A pin-hole camera forms an inverted image because light travels in straight lines. Each point on the object sends light rays through the pin-hole, and due to the rectilinear propagation of light, the rays cross over, inverting the image. The image is real, inverted, and smaller than the object. Options A and B incorrectly mention refraction and dispersion (pin-holes don't use lenses). Option C incorrectly states the image is erect and that a larger hole would improve sharpness (larger holes actually blur the image).

When a plane mirror is rotated through an angle θ, the reflected ray rotates through an angle 2θ. This is because the angle of incidence equals the angle of reflection. If the mirror rotates by θ, the normal to the mirror also rotates by θ. Since both the angle of incidence and angle of reflection change by θ each, the total change in the reflected ray's direction is 2θ. This is a fundamental principle in geometric optics.

When viewing an object through a transparent medium at normal incidence (vertical viewing), the apparent displacement is given by: displacement = depth × (1 - 1/n), where n is the refractive index. Displacement = 12.0 × (1 - 1/(4/3)) = 12.0 × (1 - 3/4) = 12.0 × 1/4 = 3.0 cm. The coin appears to be 3.0 cm closer to the surface than it actually is due to refraction at the air-water interface.

A ray passing through the optical center (or principal point) of a thin lens passes through undeviated, regardless of its initial direction or angle to the optic axis. This is because at the optical center, the two surfaces of the lens are nearly parallel and very close together, so any ray entering is almost perpendicular to both surfaces and thus emerges in nearly the same direction. This is a standard property of thin lenses.