Physics 2004 Jamb Examination

The energy stored in an inductor can be calculated using the formula:

\[ E = \dfrac{1}{2} LI^2 \]

Where:
- \( E \) is the energy stored,
- \( L \) is the inductance of the coil, and
- \( I \) is the current flowing through the coil.

Given:
- \( I = 2 \) A
- \( \text{Back emf} = 3 \) V
- \( \text{emf} = 12 \) V
- \( t = 0.4 \) s

First, we need to find the inductance of the coil. We know that the back emf is given by:

\[ \text{Back emf} = \text{emf} - L \dfrac{\Delta I}{\Delta t} \]

Substitute the values to find \( L \):

\[ 3 = 12 - L \dfrac{2 - 0}{0.4} \]

\[ 3 = 12 - 5L \]

\[ 5L = 9 \]

\[ L = \dfrac{9}{5} = 1.8 \]

Now, we can find the energy stored:

\[ E = \dfrac{1}{2} \times 1.8 \times 2^2 \]

\[ E = \dfrac{1}{2} \times 1.8 \times 4 \]

\[ E = 3.6 \times 2 \]

\[ E = 7.2 \] J

So, the stored energy in the loop that can be utilized is 7.2 J (Option A).

To find the number of electrons transported per second, we can use the formula:

\[ \text{Charge} = \text{Current} \times \text{Time} \]

Given:
- Current = 16 mA = \( 16 \times 10^{-3} \) A
- Time = 1 s

First, let's convert the current to amperes:

\[ 16 \, \text{mA} = 16 \times 10^{-3} \, \text{A} \]

Now, we can find the charge:

\[ \text{Charge} = 16 \times 10^{-3} \times 1 \]

\[ \text{Charge} = 16 \times 10^{-3} \, \text{C} \]

Now, we know that the charge of one electron is \( 1.6 \times 10^{-19} \) C.

To find the number of electrons, we can divide the total charge by the charge of one electron:

\[ \text{Number of electrons} = \dfrac{\text{Charge}}{\text{Charge of one electron}} \]

\[ \text{Number of electrons} = \dfrac{16 \times 10^{-3}}{1.6 \times 10^{-19}} \]

\[ \text{Number of electrons} = \dfrac{16 \times 10^{-3}}{1.6 \times 10^{-19}} \times \dfrac{1}{10^{-19}} \]

\[ \text{Number of electrons} = \dfrac{16}{1.6} \times 10^{16} \]

\[ \text{Number of electrons} = 10 \times 10^{16} \]

\[ \text{Number of electrons} = 10^{17} \]

So, the number of electrons transported per second is \( 10^{17} \), which is equivalent to \( 1.00 \times 10^{17} \), matching option B.

The difference between x-rays and gamma rays is primarily in their origin and how they are produced:

A. X-rays arise from energy changes in the electronic structure of atoms, typically through interactions between high-speed electrons and matter, while gamma rays originate from the nucleus of an atom during certain nuclear processes like radioactive decay.

So, the correct answer is A. X-rays arise from energy changes in the electronic structure of atoms while gamma rays come from the nucleus.

To find the ratio of the electric power dissipated in the \(6\Omega\) and \(3\Omega\) resistors, we can use the formula for power:

\[ P = \dfrac{V^2}{R} \]

Where:
- \( P \) is the power dissipated,
- \( V \) is the voltage across the resistor, and
- \( R \) is the resistance.

Given that the resistors are in parallel and connected to a \(12V\) battery:

For the \(6\Omega\) resistor:\[ P_1 = \dfrac{(12V)^2}{6\Omega} = \dfrac{144V^2}{6\Omega} = 24W \]

For the \(3\Omega\) resistor:\[ P_2 = \dfrac{(12V)^2}{3\Omega} = \dfrac{144V^2}{3\Omega} = 48W \]

Now, to find the ratio, \(P_1:P_2\):\[ \text{Ratio} = \dfrac{P_1}{P_2} = \dfrac{24W}{48W} = \dfrac{1}{2} \]

So, the ratio of the electric power dissipated in the \(6\Omega\) and \(3\Omega\) resistors respectively is \(1:2\), which corresponds to option B.

To protect a material from the influence of an external magnetic field, it should be kept in a material that can easily redirect or absorb magnetic field lines. The material that can do this effectively is soft iron.

So, the correct option is D. Soft iron ring. Soft iron is often used in magnetic shielding because it can easily become magnetized and demagnetized, effectively redirecting the magnetic field lines away from the material it is protecting.

An electrolyte is a substance that conducts electricity when dissolved in water or molten form by producing ions. Among the options provided, the substance that can act as an electrolyte is:

A. Grape juice

Grape juice contains dissolved ions from various compounds, such as acids and salts, which can conduct electricity when dissolved in water. Therefore, grape juice can be considered an electrolyte.

So, the correct option is A. Grape juice.

Electrical appliances in homes are typically earthed (grounded) primarily for safety reasons, so that:

B. A person touching the appliances is safe from electric shock.

Earthing provides a low-resistance path for current to flow in the event of a fault, ensuring that excess current flows to the ground instead of through a person touching the appliance. This helps prevent electric shocks and reduces the risk of injury or damage due to electrical faults.

So, the correct answer is B. A person touching the appliances is safe from electric shock.

To find the lost voltage of the cell, we can use Ohm's law:

\( V_{lost} = I \times R_{internal} \)

Given:
- Current, \( I = 4\,A \)
- Internal resistance, \( R_{internal} = 0.5\,\Omega \)

\( V_{lost} = 4 \times 0.5 \)

\( V_{lost} = 2\,V \)

So, the lost voltage of the cell is 2 V, which corresponds to option D.

When capacitors are arranged in parallel, their total capacitance is the sum of their individual capacitances.

To achieve the minimum total capacitance, we should arrange the capacitors in parallel in such a way that the smallest capacitance is utilized.

Given capacitors:
- \(C_1 = 0.3 \, \mu F\)
- \(C_2 = 0.5 \, \mu F\)
- \(C_3 = 0.2 \, \mu F\)

To achieve the minimum capacitance, we should only use the smallest capacitance, \(C_3 = 0.2 \, \mu F\).

So, the joint capacitance when arranged to give minimum capacitance is \(0.2 \, \mu F\), which corresponds to option C.

To find the weight of the man, we can use the formula:

\[ \text{Pressure} = \dfrac{\text{Force}}{\text{Area}} \]

Given that the pressure exerted by the man is \(2.6 \times 10^3 \, \text{Nm}^{-3}\) and the area of his feet in contact with the ground is \(4 \times 10^{-2} \, \text{m}^3\), we can rearrange the formula to solve for force:

\[ \text{Force} = \text{Pressure} \times \text{Area} \]

\[ \text{Force} = (2.6 \times 10^3 \, \text{Nm}^{-3}) \times (4 \times 10^{-2} \, \text{m}^3) \]

\[ \text{Force} = (2.6 \times 4) \times (10^3 \times 10^{-2}) \, \text{N} \]

\[ \text{Force} = 10.4 \times 10^1 \, \text{N} \]

\[ \text{Force} = 104 \, \text{N} \]

Therefore, the weight of the man is 104 Newtons.