Physics 2013 Jamb Examination

The relationship between the electromotive force (emf), current, external resistance, and internal resistance is given by the formula:\[ E = I(R + r) \]Where:
\( E = 1.5 \text{ V} \) (emf of the cell)
\( I = 0.3 \text{ A} \) (current)
\( R = 1 \Omega \) (external resistance)
\( r = \text{?} \) (internal resistance)
Substituting the values into the equation:\[ 1.5 = 0.3(1 + r) \]Divide both sides by 0.3:\[ \frac{1.5}{0.3} = 1 + r \]\[ 5 = 1 + r \]\[ r = 5 - 1 = 4 \Omega \]The internal resistance of the cell is 4.0Ω.

Ohm's law states that the current passing through a conductor is directly proportional to the potential difference across it, provided physical conditions like temperature remain constant. Substances that follow this linear relationship are called Ohmic conductors. Metals (metallic conductors) at constant temperature are the classic example of Ohmic conductors.
In contrast, electrolytes and diodes are non-Ohmic because their current-voltage relationship is not linear (the resistance changes with voltage). Glass is an insulator and does not conduct electricity under normal conditions to obey Ohm's law in a practical sense.

First, calculate the total power consumed by all the bulbs:\[ \text{Total Power} = (10 \times 40\text{W}) + (5 \times 100\text{W}) \]\[ \text{Total Power} = 400\text{W} + 500\text{W} = 900\text{W} \]Convert the power to kilowatts (kW) since energy units are in kWh:\[ 900\text{W} = 0.9\text{kW} \]Now, calculate the energy consumed in 10 hours:\[ \text{Energy} = \text{Power} \times \text{Time} = 0.9\text{kW} \times 10\text{h} = 9\text{kWh} \]One 'unit' of electricity is equal to 1 kWh. At a cost of 5N per unit:\[ \text{Total Cost} = 9 \text{ units} \times 5\text{N/unit} = 45\text{N} \]

The power rating of an electrical device is related to the voltage and current by the formula:\[ P = IV \]Where:
\( P = 2000 \text{ W} \)
\( V = 250 \text{ V} \)
To find the maximum current (\(I\)):\[ I = \frac{P}{V} \]\[ I = \frac{2000}{250} \]\[ I = 8 \text{ A} \]Therefore, the maximum current the device can safely take is 8A.

This is analogous to an object falling in a gravitational field. When a charge moves in the direction of the electric force (from high potential to low potential for a positive charge), the electric field does work on the charge. This work results in a decrease in the electric potential energy of the charge. According to the law of conservation of energy, this lost potential energy is converted into kinetic energy, causing the charge to speed up. Thus, the charge loses potential energy and gains kinetic energy.

A galvanometer is a very sensitive instrument that can only handle small currents and has a low resistance. To measure potential difference (voltage) across a component without drawing significant current from the circuit, a voltmeter must have a very high total resistance. This is achieved by connecting a high resistance, known as a 'multiplier', in series with the galvanometer. This ensures that the bulk of the potential difference is dropped across the multiplier, protecting the galvanometer while allowing it to provide a calibrated reading of the total voltage.

Faraday's law of electromagnetic induction states that the magnitude of the induced electromotive force (emf) in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit. While Lenz's law explains the direction of the induced emf (that it opposes the change that created it), Faraday's law is the fundamental principle that defines and explains the existence of induced emfs.

According to Faraday's first law of electrolysis, the mass (\( m \)) of a substance deposited is directly proportional to the quantity of electricity (\( Q \)), where \( Q = It \).\[ m \propto It \implies \frac{m_1}{I_1 t_1} = \frac{m_2}{I_2 t_2} \]Given:
\( m_1 = 1.8 \text{ g} \)
\( I_1 = 2.5 \text{ A} \)
\( t_1 = 3 \text{ hours} \)
\( I_2 = 4 \text{ A} \)
\( t_2 = 4.8 \text{ hours} \)
Substituting into the ratio:\[ \frac{1.8}{2.5 \times 3} = \frac{m_2}{4 \times 4.8} \]\[ \frac{1.8}{7.5} = \frac{m_2}{19.2} \]\[ m_2 = \frac{1.8 \times 19.2}{7.5} \]\[ m_2 = \frac{34.56}{7.5} = 4.608 \text{ g} \]Approximately 4.6g.

The energy levels of an atom are generally given by the formula:\[ E_n = \frac{E_1}{n^2} \]Where \( E_1 \) is the ground state energy (\( n = 1 \)) and \( n \) is the principal quantum number of the energy level.
Given:
\( E_1 = -24.8 \text{ eV} \)
For the third level (\( n = 3 \)):\[ E_3 = \frac{-24.8}{3^2} \]\[ E_3 = \frac{-24.8}{9} \]\[ E_3 \approx -2.755 \text{ eV} \]The energy of the third level is -2.75eV.

In the photoelectric effect, the 'number of photo-electrons ejected per second' corresponds to the photoelectric current. According to the laws of photo-emission, the number of electrons ejected per unit time is directly proportional to the intensity of the incident light, provided the frequency is above the threshold frequency.
The frequency of the beam determines the maximum kinetic energy of the ejected electrons, but not their quantity. The work function and threshold frequency are properties of the metal that determine the minimum energy required to eject an electron.