2025 - JAMB Physics Past Questions and Answers - page 9

For equilibrium, the sum of clockwise moments about the pivot must equal the sum of anticlockwise moments. The weight of the metre rule acts at its centre, which is the 50 cm mark. The distance between the pivot (10 cm) and the centre is \[50 - 10 = 40 \text{ cm}\].
The string is attached at 85 cm, so its distance from the pivot is \[85 - 10 = 75 \text{ cm}\].
The moment due to the weight is \[100 \times 40\]. The moment due to the tension is \[T \times 75\].
Equating the moments gives
\[100 \times 40 = T \times 75\]
\[4000 = 75T\]
\[T = \frac{4000}{75}\]
\[T = 53.33 \text{ N}\]
The tension in the string is 53.33 N.

Boyle’s law describes the relationship between pressure and volume for a fixed amount of gas at constant temperature. It states that pressure is inversely proportional to volume. This means that when volume increases, pressure decreases in such a way that their product remains constant.
This relationship can be written as
\[P \propto \frac{1}{V}\]
or
\[PV = \text{constant}\]
This shows clearly that pressure and volume are inversely proportional.

When temperature increases, the volume of both the material and the cavity inside it increases. The coefficient of volume expansion is related to the linear coefficient by
\[\gamma = 3\alpha\]
Substituting the value of α gives
\[\gamma = 3 \times (24 \times 10^{-6})\]
\[\gamma = 72 \times 10^{-6} \text{ K}^{-1}\]
The temperature change is
\[\Delta T = 340 - 34 = 306 \text{ K}\]
The change in volume is calculated using
\[\Delta V = V_0 \gamma \Delta T\]
\[\Delta V = 3.48 \times (72 \times 10^{-6}) \times 306\]
\[\Delta V = 0.076896\]
\[\Delta V = 0.076896 \times 3.48\]
\[\Delta V = 0.352 \text{ cm}^3\]
The new volume becomes
\[V = 3.48 + 0.352\]
\[V = 3.832 \text{ cm}^3\]
The new bore volume is approximately 3.83 cm\(^3\).

In a hydraulic press, pressure is transmitted equally throughout the fluid. Pressure is defined as force divided by area, so
\[\frac{F_1}{A_1} = \frac{F_2}{A_2}\]
The area of a circular piston is given by
\[A = \pi r^2\]
Substituting this into the pressure equation gives
\[\frac{F_1}{\pi r_1^2} = \frac{F_2}{\pi r_2^2}\]
The π cancels out, leaving
\[\frac{F_1}{r_1^2} = \frac{F_2}{r_2^2}\]
Rearranging gives
\[\frac{r_2^2}{r_1^2} = \frac{F_2}{F_1}\]
\[\frac{r_2^2}{r_1^2} = \frac{3200}{200}\]
\[\frac{r_2^2}{r_1^2} = 16\]
Taking the square root gives
\[\frac{r_2}{r_1} = 4\]
Therefore
\[r_1 : r_2 = 1 : 4\]

Stress is defined as the force per unit area. The force acting on the wire is the weight of the mass, given by
\[F = mg\]
\[F = 1.5 \times 10\]
\[F = 15 \text{ N}\]
The radius must be converted into metres:
\[0.3 \text{ cm} = 0.003 \text{ m}\]
The cross-sectional area of the wire is
\[A = \pi r^2\]
\[A = \pi (0.003)^2\]
\[A = \pi \times 9 \times 10^{-6}\]
\[A = 2.827 \times 10^{-5} \text{ m}^2\]
The stress is therefore
\[\text{Stress} = \frac{F}{A}\]
\[\text{Stress} = \frac{15}{2.827 \times 10^{-5}}\]
\[\text{Stress} = 5.3 \times 10^5 \text{ N m}^{-2}\]
The stress in the wire is \(5.3 \times 10^5 \text{ N m}^{-2}\).

An erect image formed by a concave mirror is always virtual and occurs when the object is placed between the mirror and the focal point. Since the focal length is 20 cm, the object must be positioned closer than this distance. This is also supported by the magnification being positive and greater than one, which only happens when the object lies between the pole and focus. Therefore, the object distance must be less than 20 cm.

The magnetic force on a charged particle moving in a magnetic field is given by
\[F = qvB \sin \theta\]
where θ is the angle between the velocity and the magnetic field. When the electron moves parallel to the magnetic field, the angle θ equals 0°. Since
\[\sin 0^\circ = 0\]
the force becomes
\[F = qvB \times 0\]
\[F = 0\]
Therefore, the magnetic force acting on the electron is 0.0 N.

Magnetic field lines that form concentric circles around a straight central line are characteristic of the magnetic field created by a straight current-carrying conductor. This circular pattern occurs because electric current flowing through a straight wire produces a magnetic field that wraps around it. Therefore, the source of the magnetic field is a current-carrying straight conductor.

The refractive index of a material relative to air can be found using the relationship
\[n = \frac{1}{\sin C}\]
where C is the critical angle. Substituting the given value gives
\[n = \frac{1}{\sin 45^\circ}\]
Since
\[\sin 45^\circ = \frac{1}{\sqrt{2}}\]
we obtain
\[n = \frac{1}{\frac{1}{\sqrt{2}}}\]
\[n = \sqrt{2}\]
The refractive index of the glass is \(\sqrt{2}\).

A neutral point is a location where the magnetic field produced by a magnet is exactly balanced by another magnetic field, such as the Earth's magnetic field. At this point, the two fields cancel each other completely, resulting in zero net magnetic field. Because the total magnetic field strength is zero, no magnetic influence can be observed, such as compass deflection. This point is therefore called the neutral point.