Physics 2019 Jamb Examination

For a body to be in mechanical equilibrium, two conditions must be met. The First Condition is translational equilibrium, which states that the vector sum of all external forces acting on the body must be zero (\( \sum F = 0 \)). The Second Condition is rotational equilibrium, where the sum of all torques (moments) must be zero.

Unstable equilibrium occurs when a slight displacement causes the center of gravity (CG) to lower, creating a moment that tips the object further away from its original position. Typically, a cone resting on its apex (pointy end) is the classic example of unstable equilibrium (II). A cone on its side is in neutral equilibrium (III), and one on its base is stable (I).

Because an irregular stone doesn't have standard dimensions (length, width, radius), we use the displacement method. The stone is submerged in a liquid within a measuring cylinder; the increase in the liquid level (or the volume of water overflow in a Eureka can) equals the volume of the stone.

The color of an opaque object is determined by the wavelengths of light it reflects. When white sunlight (which contains all colors) hits a blue flower, the pigments in the flower absorb most colors (red, orange, yellow, green, violet) and reflect blue light back to our eyes.

1. Find the amplitude (\( A \)): The distance between the two extreme points is 20m, so the amplitude (distance from center to extreme) is \( A = 10 \; \text{m} \).
2. Identify variables: Angular speed \( \omega = 5 \; \text{rad/s} \) and displacement from center \( x = 6 \; \text{m} \).
3. Use the SHM velocity formula:\[ v = \omega \sqrt{A^2 - x^2} \]\[ v = 5 \sqrt{10^2 - 6^2} = 5 \sqrt{100 - 36} \]\[ v = 5 \sqrt{64} = 5 \times 8 = 40 \; \text{m/s} \]

A uniform meter rule has its center of gravity at the 50cm mark. Using the Principle of Moments about the pivot (90cm mark):
Clockwise Moment = Counter-clockwise Moment\[ (0.2 \; \text{kg}) \times (100 \; \text{cm} - 90 \; \text{cm}) = M \times (90 \; \text{cm} - 50 \; \text{cm}) \]\[ 0.2 \times 10 = M \times 40 \]\[ 2 = 40M \implies M = \frac{2}{40} = 0.05 \; \text{kg} \]

In physics, systems naturally evolve toward their state of lowest energy to achieve stability. A state of stable equilibrium corresponds to a position where the potential energy of the object is at a local minimum.

The change in pressure (\( \Delta P \)) for a height change (\( h \)) is \( \Delta P = \rho_{air} g h \). We can convert this change into equivalent height of mercury (\( h_m \)):\[ \rho_m g h_m = \rho_{air} g h \implies h_m = \frac{\rho_{air} \times h}{\rho_m} \]Given \( h = 20 \; \text{km} = 2,000,000 \; \text{cm} \):\[ h_m = \frac{0.00013 \times 2,000,000}{13.6} = \frac{260}{13.6} \approx 19.12 \; \text{cmHg} = 191.2 \; \text{mmHg} \]Pressure at height = Ground Pressure - \( \Delta P \):\[ 760 \; \text{mmHg} - 191.2 \; \text{mmHg} \approx 568.8 \; \text{mmHg} \approx 570 \; \text{mmHg} \]

When an object in unstable equilibrium is slightly disturbed, its center of gravity moves downward. Since Potential Energy is proportional to height (\( P.E. = mgh \)), a lowering of the center of gravity means the potential energy decreases as it falls away from the equilibrium point.

1. Find angular velocity (\( \omega \)): \( \omega = 2\pi f \). With \( f = 2 \; \text{rev/s} \), \( \omega = 4\pi \; \text{rad/s} \).
2. Calculate centripetal force (\( F_c \)):\[ F_c = m\omega^2r \]\[ F_c = 0.5 \times (4\pi)^2 \times 2 \]\[ F_c = 1 \times 16\pi^2 = 16 \times (3.14159)^2 \approx 16 \times 9.87 = 157.9 \; \text{N} \]Rounding to the nearest whole number gives 158N.