JAMB Physics 1990 Past Questions & Explanations

A fundamental unit is a unit of measurement that cannot be expressed in terms of other units and forms the basis of a system of units. Newton (N) is a derived unit of force ($ext{kg·m·s}^{-2}$). Watt (W) is a derived unit of power ($ext{kg·m}^2 ext{·s}^{-3}$). Joule (J) is a derived unit of energy ($ext{kg·m}^2 ext{·s}^{-2}$). Second (s) is one of the seven fundamental SI units, representing time. Therefore, the answer is D.

Convert initial speed: v_0 = 90 ext{ kmh}^{-1} = 90 imes rac{1}{3.6} = 25 ext{ ms}^{-1}. Final velocity: $v = 0$ (brought to rest). Time: $t = 10 ext{ s}$. Using the kinematic equation s = rac{v_0 + v}{2} imes t = rac{25 + 0}{2} imes 10 = 12.5 imes 10 = 125 ext{ m}. Alternatively, a = rac{v - v_0}{t} = rac{0 - 25}{10} = -2.5 ext{ ms}^{-2}, then s = v_0 t + rac{1}{2}at^2 = 25(10) + rac{1}{2}(-2.5)(100) = 250 - 125 = 125 ext{ m}.

By conservation of momentum, the initial momentum is zero (particle at rest). After splitting: $0 = m_1v_1 + m_2v_2$. Let $m_1 = m$, $v_1 = v$, $m_2 = M - m$ (remaining mass), and $v_2$ = velocity of second particle (to find). Then: $0 = mv + (M-m)v_2$. Solving for $v_2$: $(M-m)v_2 = -mv$, so $v_2 = -\frac{mv}{M-m}$. The negative sign indicates opposite direction. The magnitude of velocity is $\frac{mv}{M-m}$, which matches option D.

For an object in equilibrium under parallel forces: (i) is TRUE — this is the principle of moments; the net torque must be zero, so clockwise moments equal anticlockwise moments. (ii) is TRUE — this is the condition of translational equilibrium; the net force must be zero in all directions. (iii) is FALSE — resolved components along the x-axis do NOT need to equal those along the y-axis; they must only individually sum to zero along their respective axes. Therefore, statements i and ii are correct.

Power is defined as $P = Fv$, where $F$ is the force applied and $v$ is the velocity. To keep a vehicle moving at constant speed, the engine force must equal the resistance force. Rearranging the power equation: $F = \frac{P}{v}$. This is the force provided by the engine. Option B ($\frac{1}{2}v$) has incorrect dimensions. Option C ($Pv$) would give dimensions of power times velocity. Option D ($\frac{P}{v^2}$) is also dimensionally incorrect.

Work is defined as $W = F \cdot d \cdot \cos\theta$, where $F$ is force, $d$ is displacement, and $\theta$ is the angle between force and displacement. Although the stone is held at height $h$, it is not moving ($d = 0$). Since there is no displacement, work done is zero. The time duration (50 s) is irrelevant; static holding of an object does not constitute work in the physics sense. While gravitational potential energy is stored, the actual work done on the stone during this static holding period is zero.

At the angle of repose (limiting friction), the body is about to slide. The forces along the plane are: component of weight down the plane = $mg\sin\theta$, and maximum static friction up the plane = $\mu_s N = \mu_s mg\cos\theta$. At equilibrium: $mg\sin\theta = \mu_s mg\cos\theta$. Therefore: $\mu_s = \tan\theta = \tan(30°) = \frac{1}{\sqrt{3}} \approx 0.577 \approx 0.58$. Option C is correct.

The velocity ratio (or mechanical advantage for an ideal machine) of an inclined plane is the ratio of the distance moved along the plane to the vertical height gained. If the angle is $\theta$, then for a load moved distance $d$ along the plane, the vertical height gained is $d\sin\theta$. Velocity ratio = $\frac{d}{d\sin\theta} = \frac{1}{\sin\theta}$. For $\theta = 30°$: Velocity ratio = $\frac{1}{\sin(30°)} = \frac{1}{0.5} = 2$. Option A is correct.

The pressure exerted by a liquid column is $P = \rho gh$, where $\rho$ is density, $g$ is acceleration due to gravity, and $h$ is the height of the column. Rearranging: $h = \frac{P}{\rho g} = \frac{102000}{2600 \times 10} = \frac{102000}{26000} = 3.923 \approx 3.92 \text{ m}$. Option C is correct.

(i) is TRUE — land and sea breezes result from differential heating and cooling, causing air to rise and fall in circular patterns, which are natural convection currents. (ii) is TRUE — convection occurs in fluids (liquids and gases) where particles can move freely. Solids cannot undergo convection because atoms are fixed in a lattice structure. (iii) is FALSE — a thermos flask's vacuum prevents heat loss due to both convection and conduction; radiation can still occur but is minimized by reflective surfaces on the walls. Therefore, statements i and ii are correct, making A the answer.

The frequency of light remains constant when entering a different medium. Velocity of light in water: $v = \frac{c}{n} = \frac{3 \times 10^8}{\frac{4}{3}} = 3 \times 10^8 \times \frac{3}{4} = 2.25 \times 10^8 \text{ ms}^{-1}$. Wavelength in water: $\lambda = \frac{v}{f} = \frac{2.25 \times 10^8}{5 \times 10^{14}} = \frac{2.25}{5} \times 10^{-6} = 0.45 \times 10^{-6} = 4.5 \times 10^{-7} \text{ m}$. Option A is correct.

When a wave enters a medium with a different velocity, the frequency of the wave remains constant (frequency is determined by the source). Using the wave equation $v = f\lambda$, if velocity decreases while frequency stays the same, the wavelength must decrease. Option C correctly states that frequency is unaltered while wavelength is decreased. This occurs because $\lambda_{medium} = \frac{v_{medium}}{f}$, and since $v_{medium} < v_{air}$ while $f$ remains constant, $\lambda_{medium} < \lambda_{air}$.

Shadows form when an object blocks light, which is possible because light travels in straight lines. Eclipses occur when one celestial body blocks light from another (e.g., Moon blocking sunlight during a solar eclipse), also due to the straight-line nature of light propagation. Refraction (A) is the bending of light at interfaces. Diffraction (C) is the bending of light around obstacles, which would blur shadow boundaries, not create them. Reflection (D) is the bouncing of light off surfaces. Rectilinear propagation (straight-line travel) is the fundamental property that enables both shadows and eclipses to occur. Option B is correct.

Using the magnification formula for mirrors: $m = \frac{h_i}{h_o} = \frac{-v}{u}$, where $h_i$ is image height, $h_o$ is object height, $v$ is image distance, and $u$ is object distance. Given: $h_o = 3$ cm, $u = 10$ cm, $v = 40$ cm. Magnification: $m = \frac{-40}{10} = -4$ (negative indicates inverted image). Height of image: $h_i = m \times h_o = -4 \times 3 = -12$ cm. The magnitude of the image height is 12 cm. Option D is correct.

When light from an object enters a plane mirror system, the main bright image is formed by direct reflection at the mirror surface. The fainter image results from light undergoing refraction at the glass surface of the mirror before reflecting internally, then refracting again as it exits. This combination of reflection (for the main image) and refraction (for the secondary faint image) explains the observation of two images with different intensities. Options A and B alone cannot explain both images. Option C (diffraction and interference) is not the primary optical phenomenon occurring in this simple mirror scenario.

For a converging lens to produce an erect (upright) and magnified image, the object must be placed within the focal length (between F and the lens). Using the lens magnification formula $m = \frac{v}{u} = 2$ (given magnification is 2×), and the lens equation $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$. Since the image is erect and magnified, $v = -2u$ (virtual image). Substituting into the lens equation: $\frac{1}{20} = \frac{1}{u} + \frac{1}{-2u} = \frac{1}{2u}$, so $u = 10$ cm. This places the object between the lens and focal point, producing the required erect, magnified virtual image.

Short-sightedness (myopia) is a refractive error where the eye's cornea and lens are too powerful, or the eyeball is too long. Consequently, parallel light rays from distant objects are converged (focused) before reaching the retina—i.e., in front of the retina. This creates a blurred image on the retina. Option B is incorrect because converging lenses worsen the condition. Option C is wrong because myopia involves over-convergence, not divergence. Option D is too specific and doesn't accurately describe the optical defect.

Magenta is a color that does not exist in the visible spectrum as a single wavelength; it is a combination of red and blue light. When an object appears magenta, it absorbs the complementary color to magenta. The complement of magenta (red + blue) is green (the remaining color in the visible spectrum). Therefore, a magenta object absorbs green light and reflects red and blue light. Option A is incorrect because magenta reflects both red and blue. Options C and D incorrectly suggest red is absorbed, which contradicts the magenta appearance.