2023 - JAMB Physics Past Questions and Answers - page 2

The absolute pressure ($P_{\text{abs}}$) in the tank can be calculated using the equation $P_{\text{abs}}=P_{\text{atm}}+\rho gh$, where:

- $P_{\text{atm}}$ is the atmospheric pressure (given as $101,325\text{Pa}$),
- $\rho$ is the density of mercury ($13,600{\text{kg/m}}^3$),
- $g$ is the acceleration due to gravity ($9.8{\text{m/s}}^2$),
- $h$ is the height difference of the mercury column ($0.25\text{m}$).

Substituting the values:

$$\begin{gathered} P_{\text{abs}}=101,325+13,600\times 9.8\times 0.25 \\ {P}_{\text{abs}}=101,325+33,320 \\ {P}_{\text{abs}}=134,645\text{Pa} \end{gathered}$$

Therefore, the correct answer is: $134,645\text{Pa}$.

The Bohr model of the atom, proposed by Niels Bohr in 1913, describes electrons as orbiting the nucleus in specific energy levels or shells. According to this model, electrons occupy quantized orbits, and each orbit is associated with a specific energy level. Electrons can jump from one energy level to another by absorbing or emitting energy in discrete packets or quanta. The Bohr model was an improvement over the earlier Rutherford model and provided a better explanation for the spectral lines of hydrogen.

The horizontal range (R) of a projectile launched at an angle can be calculated using the formula:

$$R=\frac{v^2\sin (2\theta )}{g}$$

where:
- $v$ is the initial speed of the projectile,
- $\theta$ is the launch angle,
- $g$ is the acceleration due to gravity.

In this case, $v=75\text{m/s}$, $\theta ={22}^{\circ }$, and $g=10{\text{m/s}}^2$.

$$R=\frac{(75{)}^2\sin (2\times {22}^{\circ })}{10}$$

Calculating this expression yields approximately 391 m.

Surface tension is a measure of the force that tends to pull adjacent parts of a liquid's surface together, minimizing the surface area. Mercury has a relatively high surface tension compared to the other liquids listed, making it the correct answer in this case.

The charge ($Q$) on each plate of the capacitor can be calculated using the formula $Q=CV$, where:

- $C$ is the capacitance,
- $V$ is the voltage.

The capacitance ($C$) for a parallel plate capacitor with air as the dielectric is given by $C=\frac{{\epsilon }_{0}A}{d}$, where:

- ${\epsilon }_0$ is the vacuum permittivity ($8.85\times {10}^{-12}\text{F/m}$),
- $A$ is the area of the plates ($0.8{\text{m}}^2$),
- $d$ is the separation between the plates ($0.02\text{m}$).

Let's substitute the values:

$$C=\frac{(8.85\times {10}^{-12}\text{F/m})(0.8{\text{m}}^2)}{0.02\text{m}}=3.54\times {10}^{-10}\text{F}$$

Now, calculate the charge ($Q$):

$$Q=CV=(3.54\times {10}^{-10}\text{F})(120\text{V})=4.25\times {10}^{-8}\text{C}$$

Convert the charge to nanocoulombs:

$$4.25\times {10}^{-8}\text{C}=42.5\text{nC}$$

Therefore, the correct answer is: $42.5\text{nC}$.

The formation of a rainbow involves refraction, dispersion, and total internal reflection. Here's a more accurate summary:

When sunlight shines on a water droplet, the light undergoes refraction as it enters the droplet. The different colours of light are dispersed due to the varying wavelengths. Finally, total internal reflection occurs inside the droplet, leading to the formation of a rainbow.

The rate at which energy is transferred by conduction from the surface to the bottom of the swimming pool ($\frac{q}{t}$) can be calculated using the formula:

$$\frac{q}{t}=K\cdot A\cdot \frac{\mathrm{\Delta }\theta }{L}$$

where:
- $K$ is the thermal conductivity of water ($0.6071\text{W/m}\cdot \text{K}$),
- $A$ is the surface area of the pool ($620{\text{m}}^2$),
- $\mathrm{\Delta }\theta$ is the temperature difference (${25}^{\circ }\text{C}-{15}^{\circ }\text{C}={10}^{\circ }\text{C}$),
- $L$ is the depth of the pool ($1.5\text{m}$).

Substitute the values:

$$\frac{q}{t}=0.6071\times 620\times \frac{10}{1.5}=2509.35\text{W}$$

Convert the result to kilowatts ($\text{kW}$):

$$\frac{2509.35\text{W}}{1000}=2.50935\text{kW}$$

Approximately, the rate of energy transfer is $2.5\text{kW}$, so the correct option is: $2.5\text{kW}$.

The relationship between the number of turns on the primary ($N_1$) and secondary ($N_2$) windings of a transformer is given by the formula:

$$\frac{N_1}{N_2}=\frac{V_1}{V_2}$$

where:
- $N_1$ is the number of turns on the primary winding,
- $N_2$ is the number of turns on the secondary winding,
- $V_1$ is the voltage on the primary side,
- $V_2$ is the voltage on the secondary side.

Given values:
$$V_1=2.2\text{kV}=2200\text{V}$$
$$V_2=110\text{V}$$
$$N_2=25$$

Now, rearrange the formula to solve for $N_1$:

$$N_1=N_2\times \frac{V_1}{V_2}$$

Substitute the values:

$$N_1=25\times \frac{2200}{110}$$

$$N_1=25\times 20$$

$$N_1=500$$

The number of turns on the primary winding is 500 when the secondary winding has 25 turns, and the transformer is used on a 2.2 kV line to deliver 110 V.

Insulators are materials that do not conduct electricity well. Rubber is known for its high electrical resistance, making it a good insulator. On the other hand, silver, copper, and water are conductors of electricity.

Diffraction is the bending of waves around the corners of an obstacle or through an aperture.