2025 - JAMB Physics Past Questions and Answers - page 8

The energy stored in a capacitor is given by the expression
\[E = \frac{1}{2}CV^2\]
where \(C\) is the capacitance and \(V\) is the applied voltage.
When capacitors are connected in series, the total capacitance becomes smaller than the capacitance of any individual capacitor.
Since the stored energy depends directly on the capacitance, a decrease in total capacitance results in a decrease in the total energy that can be stored.
Therefore, the reduction in stored energy occurs because of the decrease in overall capacitance.

For an alternating current, the root mean square value is related to the peak current by the relation
\[I_{\text{rms}} = \frac{I_0}{\sqrt{2}}\]
where \(I_0\) is the maximum or peak current.
Substituting the given peak current,
\[I_{\text{rms}} = \frac{7}{\sqrt{2}}\]
The value of \(\sqrt{2}\) is approximately 1.414, so
\[I_{\text{rms}} = \frac{7}{1.414}\]
\[I_{\text{rms}} \approx 4.95 \, \text{A}\]
This value is approximately equal to 5 A.
Therefore, the r.m.s current is 5 A.

Particles suspended in fluids such as liquids and gases move in a random and irregular manner due to constant collisions with fast-moving molecules of the fluid.
This motion does not follow a fixed path and is caused by the continuous impact of surrounding molecules.
This irregular and random motion of particles in a fluid is known as Brownian motion.
Therefore, the correct term for this movement is Brownian motion.

The relationship between volume expansion and temperature change is given by
\[\Delta V = V \gamma \theta\]
where \(\gamma\) is the coefficient of volume expansion.
The coefficient of volume expansion is related to the linear expansivity by
\[\gamma = 3\alpha\]
Substituting the given linear expansivity,
\[\gamma = 3 \times 2.0 \times 10^{-5} = 6.0 \times 10^{-5} \, K^{-1}\]
Substituting the known values into the volume expansion equation,
\[0.0018 = 1 \times (6.0 \times 10^{-5}) \times \theta\]
Rearranging gives
\[\theta = \frac{0.0018}{6.0 \times 10^{-5}}\]
\[\theta = 30^\circ C\]
Therefore, the temperature change is 30°C.

The net force acting on the object is related to its mass and acceleration by Newton’s second law
\[F_{\text{net}} = ma\]
Substituting the values,
\[F_{\text{net}} = 40 \times 0.3 = 12 \, \text{N}\]
The applied force is 18 N, so the frictional force must be the difference between the applied force and the net force
\[F_{\text{friction}} = 18 - 12 = 6 \, \text{N}\]
The coefficient of friction is given by
\[\mu = \frac{F_{\text{friction}}}{N}\]
where the normal reaction \(N = mg\).
Substituting the values,
\[N = 40 \times 10 = 400 \, \text{N}\]
Therefore,
\[\mu = \frac{6}{400}\]
\[\mu = 0.015\]
Hence, the coefficient of friction is 0.015.

Gas molecules move randomly and collide with the walls of their container.
During each collision, momentum is transferred from the molecules to the walls.
Force is defined as the rate of change of momentum, and pressure is defined as force per unit area.
Therefore, pressure represents the rate at which momentum is transferred to the walls per second per unit area.
Hence, pressure equals the rate of change of momentum imparted to the walls per second per unit area.

A thermometer responds quickly when its sensing element can detect temperature changes rapidly.
Resistance thermometers use electrical resistance, which changes almost instantly with temperature.
They also have small sensing elements with low thermal inertia, allowing faster response.
Liquid-in-glass thermometers respond more slowly because liquid expansion takes more time.
Therefore, the resistance thermometer responds fastest to temperature changes.

When two quantities of water at different temperatures are mixed, the heat lost by the hot water equals the heat gained by the cold water.
The heat gained by the cold water is given by
\[Q = mc\Delta T\]
For the cold water,
\[Q = 8 \times 4200 \times (60 - 25)\]
\[Q = 8 \times 4200 \times 35\]
For the hot water cooling from 100°C to 60°C,
\[Q = m \times 4200 \times (100 - 60)\]
\[Q = m \times 4200 \times 40\]
Equating the heat gained and heat lost,
\[m \times 4200 \times 40 = 8 \times 4200 \times 35\]
Canceling 4200 from both sides gives
\[m \times 40 = 8 \times 35\]
\[m = \frac{280}{40}\]
\[m = 7 \, \text{kg}\]
Therefore, the required mass of boiling water is 7 kg.

Specific gravity is defined as the ratio of the density of a substance to the density of water.
This relationship is written as
\[\text{Specific gravity} = \frac{\rho}{\rho_w}\]
Rearranging to find the density of the liquid gives
\[\rho = \text{Specific gravity} \times \rho_w\]
Substituting the given values,
\[\rho = 0.76 \times 1000\]
\[\rho = 760 \, \text{kg/m}^3\]
Therefore, the density of the liquid is \(760 \, \text{kg/m}^3\).

When two surfaces are in contact, irregularities on their surfaces interact and resist motion. This resistance acts parallel to the surface and opposes sliding or attempted sliding motion. This tangential resisting force is known as friction. Therefore, the force described is frictional force.