Step 1: Understanding the Concept:

The fundamental principle of charge quantization states that the total charge (Q) on any body is an integral multiple of the basic unit of charge, which is the charge of a single electron (e). A negative charge on a body indicates an excess of electrons, meaning the body has gained electrons. A positive charge indicates a deficit of electrons, meaning it has lost electrons.


Step 2: Key Formula or Approach:

The formula for quantization of charge is: \[ Q = n \times e \]
where:
\( Q \) is the total charge on the body.
\( n \) is the number of electrons gained or lost.
\( e \) is the magnitude of the charge on one electron, which is \( 1.6 \times 10^{-19} \) C.

We need to find \( n \). We can rearrange the formula as: \[ n = \frac{Q}{e} \]

Step 3: Detailed Explanation:

Given:

Total charge \( Q = -1 \) µC \( = -1 \times 10^{-6} \) C.

The negative sign indicates a gain of electrons.

Charge of one electron \( e = 1.6 \times 10^{-19} \) C (we use the magnitude for the calculation of n).

Now, we calculate the number of electrons (\(n\)): \[ n = \frac{|Q|}{e} = \frac{1 \times 10^{-6} C}{1.6 \times 10^{-19} C} \] \[ n = \frac{1}{1.6} \times 10^{-6 - (-19)} \] \[ n = 0.625 \times 10^{13} \]
To express this in standard scientific notation, we can write: \[ n = 6.25 \times 10^{12} \]

Step 4: Final Answer:

Since the charge is negative, the body has gained electrons. The number of electrons gained is \( 6.25 \times 10^{12} \). Therefore, the correct option is (B).
Quick Tip: Remember that a negative charge always implies a \textbf{gain} of electrons (excess electrons), while a positive charge implies a \textbf{loss} of electrons (deficiency of electrons). This can help you immediately eliminate two of the four options.

Step 1: Understanding the Concept:

The resistance of most conducting materials changes with temperature. For many materials, over a certain temperature range, this relationship is approximately linear. The temperature coefficient of resistance (\(\alpha\)) is a measure of how much the resistance of a material changes per degree Celsius (or Kelvin) of temperature change.


Step 2: Key Formula or Approach:

The relationship between resistance and temperature is given by: \[ R_2 = R_1 [1 + \alpha(T_2 - T_1)] \]
where:
\( R_1 \) is the resistance at the initial temperature \( T_1 \).
\( R_2 \) is the resistance at the final temperature \( T_2 \).
\( \alpha \) is the temperature coefficient of resistance.

To find \( \alpha \), we can rearrange the formula: \[ R_2 - R_1 = R_1 \alpha (T_2 - T_1) \] \[ \alpha = \frac{R_2 - R_1}{R_1(T_2 - T_1)} \]

Step 3: Detailed Explanation:

Given:

Initial resistance \( R_1 = 50 \) Ω.

Initial temperature \( T_1 = 20 \) °C.

Final resistance \( R_2 = 51 \) Ω.

Final temperature \( T_2 = 120 \) °C.

Change in resistance \( \Delta R = R_2 - R_1 = 51 \, \Omega - 50 \, \Omega = 1 \, \Omega \).

Change in temperature \( \Delta T = T_2 - T_1 = 120 \, ^\circC - 20 \, ^\circC = 100 \, ^\circC \).

Now, substitute these values into the formula for \( \alpha \): \[ \alpha = \frac{\Delta R}{R_1 \Delta T} = \frac{1}{50 \times 100} \] \[ \alpha = \frac{1}{5000} \] \[ \alpha = 0.0002 \, ^\circC^{-1} \]
In scientific notation, this is: \[ \alpha = 2 \times 10^{-4} \, ^\circC^{-1} \]

Step 4: Final Answer:

The temperature coefficient of resistance of the material is \( 2 \times 10^{-4} \, ^\circC^{-1} \). Therefore, the correct option is (A).
Quick Tip: Always double-check the formula for \( \alpha \). A common mistake is to divide by \( R_2 \) instead of \( R_1 \), or to forget one of the terms in the denominator. The coefficient relates the change in resistance to the \textbf{original} resistance.

Step 1: Understanding the Concept:

The force experienced by a charged particle moving in a magnetic field is known as the Lorentz force. This force depends on the charge of the particle, its velocity, and the strength of the magnetic field. The force is always perpendicular to both the velocity of the particle and the magnetic field.


Step 2: Key Formula or Approach:

The magnetic force \(\vec{F}\) on a charge \(q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is given by the vector product: \[ \vec{F} = q(\vec{v} \times \vec{B}) \]
The magnitude of this force is given by: \[ F = qvB\sin(\theta) \]
where \( \theta \) is the angle between the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\). Note that the mass of the particle does not appear in this formula.


Step 3: Detailed Explanation:

Case 1 (First Particle):

Mass = \( m \)

Charge = \( q \)

Velocity = \(\vec{V}\)

Magnetic Field = \(\vec{B}\)

The particle moves at a right angle to the magnetic field, so \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \).

The force experienced by this particle is \(\vec{F}_1\). Its magnitude is: \[ F_1 = qVB\sin(90^\circ) = qVB \]
The problem states that this force is \(\vec{F}\), so \( F = qVB \).


Case 2 (Second Particle):

Mass = \( 2m \) (This information is irrelevant for calculating the force).

Charge = \( q' = 2q \)

Velocity = \( \vec{v}' = \frac{\vec{V}}{2} \)

Magnetic Field = \(\vec{B}\) (same field)

The direction is the same, so it also enters at a right angle to the field, \( \theta = 90^\circ \).

The force experienced by this second particle is \(\vec{F}_2\). Its magnitude is: \[ F_2 = q'v'B\sin(90^\circ) = (2q)\left(\frac{V}{2}\right)B(1) \] \[ F_2 = \frac{2}{2} (qVB) = qVB \]

Step 4: Final Answer:

Comparing the magnitudes of the forces, we find that \( F_2 = qVB \) and we already established that \( F = qVB \). Therefore, \( F_2 = F \). The second particle experiences the same force as the first particle. The correct option is (C).
Quick Tip: In problems involving Lorentz force, be aware of distractors. Here, the mass of the particle is given for both cases, but it does not affect the magnetic force. Focus only on the parameters in the formula \( F = qvB\sin(\theta) \).

Step 1: Understanding the Concept:

A current-carrying wire placed in a magnetic field experiences a magnetic force. This force is the result of the collective Lorentz force acting on the individual charge carriers (electrons) moving within the wire. The magnitude of this force depends on the current, the length of the wire in the field, the magnetic field strength, and the angle between the wire and the field.


Step 2: Key Formula or Approach:

The formula for the magnetic force \(\vec{F}\) on a straight wire of length \(\vec{L}\) carrying a current \(I\) in a uniform magnetic field \(\vec{B}\) is: \[ \vec{F} = I(\vec{L} \times \vec{B}) \]
The magnitude of the force is: \[ F = ILB\sin(\theta) \]
where \( \theta \) is the angle between the direction of the current (length vector \(\vec{L}\)) and the magnetic field vector \(\vec{B}\).


Step 3: Detailed Explanation:

Given:

Length of the wire \( L = 5.0 \) cm. We must convert this to SI units (meters): \( L = 0.05 \) m.

Current \( I = 1.0 \) A.

Magnetic field strength \( B = 0.3 \) T.

The magnetic field is applied perpendicular to the length of the wire, so the angle \( \theta = 90^\circ \). We know that \( \sin(90^\circ) = 1 \).

Now, we can substitute these values into the formula: \[ F = (1.0 \, A) \times (0.05 \, m) \times (0.3 \, T) \times \sin(90^\circ) \] \[ F = 1.0 \times 0.05 \times 0.3 \times 1 \] \[ F = 0.05 \times 0.3 \] \[ F = 0.015 \, N \]

Step 4: Final Answer:

The magnitude of the force acting on the wire is 0.015 N. Therefore, the correct option is (D).
Quick Tip: Always convert all given quantities to their base SI units before performing calculations. In this case, converting centimeters to meters is a crucial first step. Forgetting this conversion is a common source of error.

Step 1: Understanding the Concept:

A transformer is a device that transfers electrical energy from one circuit to another through electromagnetic induction. A step-up transformer is one that increases the voltage from the primary coil to the secondary coil (V\textsubscript{s > V\textsubscript{p). For an ideal transformer, the power remains constant (P\textsubscript{in = P\textsubscript{out). Since power P = VI, if the voltage is stepped up, the current must be stepped down to keep the power constant.


Step 2: Detailed Explanation:

The primary application of step-up transformers is in long-distance power transmission. Electrical power is generated at power stations at a relatively moderate voltage. To transmit this power over hundreds of kilometers to cities and towns, significant energy would be lost in the transmission wires due to heating.

The power loss (\(P_{loss}\)) in the transmission lines is given by Joule's law of heating: \[ P_{loss} = I^2 R \]
where \(I\) is the current flowing through the wires and \(R\) is the resistance of the wires.

To minimize this power loss, the current (\(I\)) must be kept as low as possible.

The power transmitted (\(P_{trans}\)) is given by \(P_{trans} = VI\). This means for a given amount of power to be transmitted, \(I = P_{trans}/V\).

Substituting this into the power loss equation: \[ P_{loss} = \left(\frac{P_{trans}}{V}\right)^2 R \]
This equation shows that the power loss is inversely proportional to the square of the transmission voltage (\(P_{loss} \propto 1/V^2\)).

Therefore, by using a step-up transformer at the power station to increase the voltage to a very high value (e.g., hundreds of kilovolts), the current is significantly reduced, which in turn drastically minimizes the power lost as heat in the transmission lines over long distances.


Step 3: Final Answer:

A step-up transformer increases voltage to decrease current, thereby minimizing power loss (\(I^2 R\)) during transmission over long distances. Option (D) correctly describes this function.
Quick Tip: Remember the core principle of power transmission: "High Voltage, Low Current, Low Loss." Step-up transformers are for sending power out over long distances, and step-down transformers are used locally to reduce the voltage to safe levels for consumers.

Step 1: Understanding the Concept:

This problem is based on Faraday's Law of Electromagnetic Induction. The law states that whenever the magnetic flux linked with a closed loop or circuit changes, an electromotive force (emf) is induced in it. The magnitude of this induced emf is equal to the rate of change of magnetic flux.


Step 2: Key Formula or Approach:

According to Faraday's Law, the induced emf (\(\mathcal{E}\)) is given by: \[ \mathcal{E} = - \frac{d\Phi_B}{dt} \]
The negative sign (Lenz's Law) indicates the direction of the induced emf, but since the question asks for the magnitude, we can use the formula: \[ |\mathcal{E}| = \left| \frac{\Delta \Phi_B}{\Delta t} \right| \]
where:
\( \Delta \Phi_B \) is the change in magnetic flux (\( \Phi_{final} - \Phi_{initial} \)).
\( \Delta t \) is the time interval over which the change occurs.


Step 3: Detailed Explanation:

Given:

Initial magnetic flux \( \Phi_{initial} = 3.6 \) Wb.

Final magnetic flux \( \Phi_{final} = 1.6 \) Wb.

Time interval \( \Delta t = 0.4 \) s.

First, calculate the change in magnetic flux: \[ \Delta \Phi_B = \Phi_{final} - \Phi_{initial} = 1.6 \, Wb - 3.6 \, Wb = -2.0 \, Wb \]
Now, calculate the magnitude of the induced emf: \[ |\mathcal{E}| = \left| \frac{\Delta \Phi_B}{\Delta t} \right| = \left| \frac{-2.0 \, Wb}{0.4 \, s} \right| \] \[ |\mathcal{E}| = \frac{2.0}{0.4} \, V \]
To simplify the division, we can multiply the numerator and denominator by 10: \[ |\mathcal{E}| = \frac{20}{4} \, V = 5 \, V \]

Step 4: Final Answer:

The magnitude of the induced emf in the loop is 5 V. Therefore, the correct option is (A).
Quick Tip: When a question asks for the "magnitude" of the induced emf, you don't need to worry about the negative sign from Lenz's Law in your final answer. Just focus on calculating the rate of change of the flux.

Step 1: Understanding the Concept:

This question tests the knowledge of the practical applications of different types of electromagnetic radiation. Water purifiers often use a method of disinfection that does not involve chemicals. This is achieved using a specific type of electromagnetic wave that is effective at destroying microorganisms.


Step 2: Detailed Explanation:

Let's analyze the options:

(A) Ultraviolet (UV) rays: UV radiation, particularly in the UV-C range (wavelengths of 200-280 nm), has a germicidal effect. It works by penetrating the cells of microorganisms like bacteria, viruses, and protozoa and damaging their DNA or RNA. This damage disrupts their ability to reproduce and function, effectively killing or inactivating them. This is a widely used and effective method for water purification.

(B) Infrared (IR) waves: IR radiation is primarily perceived as heat. It does not have the energy per photon required to destroy the DNA of germs and is not used for disinfection in water purifiers.

(C) Visible rays: This is the light that human eyes can detect. It does not have a germicidal effect.

(D) \(\gamma\)-rays (Gamma rays): Gamma rays are highly energetic and are a form of ionizing radiation. They are extremely effective at killing all forms of life and are used for sterilizing medical equipment. However, they are not used in domestic water purifiers due to the significant safety hazards and the complexity of the required equipment.


Step 3: Final Answer:

Based on the analysis, Ultraviolet (UV) rays are the standard electromagnetic radiation used for disinfecting water and killing germs in household water purifiers. Therefore, option (A) is correct.
Quick Tip: Associate different parts of the EM spectrum with common applications. For example: Radio waves (communication), Microwaves (cooking), Infrared (heat/remote controls), Visible (sight), Ultraviolet (sunburn/disinfection), X-rays (medical imaging), Gamma rays (sterilization/cancer therapy).

Step 1: Understanding the Concept:

When two or more thin lenses are placed in contact, they can be treated as a single equivalent lens. The power (and thus the focal length) of this equivalent lens can be calculated by summing the powers of the individual lenses. When an object is placed at infinity, the image is formed at the principal focus of the lens system.


Step 2: Key Formula or Approach:

The formula for the equivalent focal length (\(F\)) of two thin lenses with focal lengths \(f_1\) and \(f_2\) placed in contact is: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
Also, for any lens, when the object is at infinity (\(u = \infty\)), the image is formed at the focal point (\(v = F\)).


Step 3: Detailed Explanation:

Given:

Focal length of the first lens, \(f_1 = +40\) cm (a converging/convex lens).

Focal length of the second lens, \(f_2 = -20\) cm (a diverging/concave lens).

Object distance, \(u = \infty\).


First, we calculate the equivalent focal length \(F\) of the combination: \[ \frac{1}{F} = \frac{1}{+40} + \frac{1}{-20} \] \[ \frac{1}{F} = \frac{1}{40} - \frac{1}{20} \]
To subtract the fractions, we find a common denominator, which is 40. \[ \frac{1}{F} = \frac{1}{40} - \frac{2}{40} \] \[ \frac{1}{F} = \frac{1 - 2}{40} = -\frac{1}{40} \]
Therefore, the equivalent focal length is: \[ F = -40 \, cm \]
The negative sign indicates that the combination of lenses acts as a diverging lens.


Now, we determine the image position. Since the object is placed at infinity, the image will be formed at the principal focus of the equivalent lens. \[ v = F = -40 \, cm \]
The negative sign for the image distance, by sign convention, means the image is formed on the same side as the object, which is "in front of the combination". The image is virtual.


Step 4: Final Answer:

The image is formed at a distance of 40 cm in front of the lens combination. Therefore, the correct option is (B).
Quick Tip: A negative equivalent focal length means the lens combination behaves as a diverging lens. Remember that for an object at infinity, a diverging lens forms a virtual image at its focal point on the same side as the incident light.

Step 1: Understanding the Concept:

The de-Broglie hypothesis states that all matter exhibits wave-like properties. The wavelength associated with a particle, known as its de-Broglie wavelength, is inversely proportional to its momentum.


Step 2: Key Formula or Approach:

The de-Broglie wavelength (\(\lambda\)) of a particle is given by the formula: \[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

Since momentum \(p = mv\), where \(m\) is the mass and \(v\) is the velocity of the particle, the formula can also be written as: \[ \lambda = \frac{h}{mv} \]

Step 3: Detailed Explanation:

We are given an electron and a proton moving with the same velocity, \(v\).

Let \(m_e\) be the mass of the electron and \(m_p\) be the mass of the proton.

Let \(\lambda_e\) be the de-Broglie wavelength of the electron and \(\lambda_p\) be the de-Broglie wavelength of the proton.


Using the de-Broglie formula for the electron: \[ \lambda_e = \frac{h}{m_e v} \quad \cdots (1) \]
Using the de-Broglie formula for the proton: \[ \lambda_p = \frac{h}{m_p v} \quad \cdots (2) \]
We need to find the ratio \((\lambda_e/\lambda_p)\). We can do this by dividing equation (1) by equation (2): \[ \frac{\lambda_e}{\lambda_p} = \frac{\left(\frac{h}{m_e v}\right)}{\left(\frac{h}{m_p v}\right)} \] \[ \frac{\lambda_e}{\lambda_p} = \frac{h}{m_e v} \times \frac{m_p v}{h} \]
The terms \(h\) and \(v\) cancel out from the numerator and the denominator: \[ \frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} \]

Step 4: Final Answer:

The ratio of the de-Broglie wavelengths is equal to the inverse ratio of their masses. Therefore, \((\lambda_e/\lambda_p) = m_p/m_e\). The correct option is (B).
Quick Tip: For particles with the same velocity, the de-Broglie wavelength is inversely proportional to mass (\(\lambda \propto 1/m\)). Since a proton is much heavier than an electron (\(m_p > m_e\)), the electron will have a much longer wavelength (\(\lambda_e > \lambda_p\)). This helps confirm that the ratio \(\lambda_e/\lambda_p\) should be greater than 1.

Step 1: Understanding the Concept:

According to Bohr's model, when an electron in an atom transitions from a higher energy level (\(n_i\)) to a lower energy level (\(n_f\)), it emits a photon. The energy of this photon is equal to the energy difference between the two levels (\(\Delta E = E_{n_i} - E_{n_f}\)). The frequency (\(f\)) of the emitted photon is directly proportional to its energy, according to the Planck-Einstein relation, \(E = hf\), where \(h\) is Planck's constant. Therefore, the transition with the greatest energy difference will emit a photon with the greatest frequency.


Step 2: Key Formula or Approach:

The energy of an electron in the n-th orbit of a hydrogen atom is given by: \[ E_n = -\frac{13.6}{n^2} \, eV \]
The energy difference for a transition from \(n_i\) to \(n_f\) is: \[ \Delta E = E_{n_i} - E_{n_f} = -13.6 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \, eV \]
Greatest frequency corresponds to the greatest energy difference \( \Delta E \). We need to compare the \( \Delta E \) for each given transition.


Step 3: Detailed Explanation:

The energy levels in a hydrogen atom are not equally spaced; the gap between consecutive levels decreases as \(n\) increases. The largest energy gaps are found for transitions to the ground state (\(n=1\)). Let's calculate the energy difference for each option:

(A) n=3 to n=1: \[ \Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \left( 1 - \frac{1}{9} \right) = 13.6 \left( \frac{8}{9} \right) \approx 12.09 \, eV \]
(B) n=3 to n=2: \[ \Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right) \approx 1.89 \, eV \]
(C) n=4 to n=3: \[ \Delta E = 13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 \left( \frac{7}{144} \right) \approx 0.66 \, eV \]
(D) n=4 to n=2: \[ \Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{3}{16} \right) \approx 2.55 \, eV \]

Step 4: Final Answer:

Comparing the calculated energy differences, the transition from n=3 to n=1 has the largest value (12.09 eV). Therefore, this transition will emit a photon with the greatest energy and thus the greatest frequency. The correct option is (A).
Quick Tip: You can often solve this type of question without calculation by visualizing the energy level diagram. The energy levels get closer together as 'n' increases. Therefore, the largest energy drop (and highest frequency photon) will always be for the transition that covers the largest "vertical distance" on the diagram, which corresponds to the largest change in \(1/n^2\), particularly for transitions ending at the n=1 ground state.

Step 1: Understanding the Concept:

A general-purpose p-n junction diode is a semiconductor device that allows current to flow primarily in one direction. Its behavior is described by its I-V (current-voltage) characteristic curve, which has distinct regions of operation: forward bias, reverse bias, and breakdown.


Step 2: Detailed Explanation:

Let's analyze the operating regions of a diode:


Forward Bias Region: When the p-type side is connected to a higher potential than the n-type side, the diode is forward-biased.

Before cut-in voltage: For small forward voltages (less than the cut-in voltage, which is \(\approx 0.3\)V for Germanium and \(\approx 0.7\)V for Silicon), the depletion region is still significant, and only a very small current flows. The diode has very high resistance and is effectively "off".
Beyond cut-in voltage: Once the forward voltage exceeds the cut-in voltage (also called the knee voltage or threshold voltage), the depletion region becomes very narrow, the potential barrier is overcome, and the diode's resistance drops sharply. A large current starts to flow through the diode. This is the normal "on" state or conducting region for a diode.

Reverse Bias Region: When the n-type side is connected to a higher potential than the p-type side, the diode is reverse-biased. The depletion region widens, and the resistance becomes extremely high. Only a very small leakage current, called the reverse saturation current, flows. The diode is effectively "off". "Beyond the reverse saturation current region" typically refers to the breakdown region, where a large reverse current flows, usually damaging a general-purpose diode. "Before the reverse saturation current region" is not a standard term.


Step 3: Final Answer:

The primary function of a diode is to conduct current when it is forward-biased. This significant conduction happens only when the applied voltage is greater than the cut-in voltage. Therefore, a general-purpose diode is normally used in the region beyond the cut-in voltage. The correct option is (A).
Quick Tip: Think of a diode as a one-way electronic valve. The "cut-in voltage" is the minimum pressure (voltage) required to open the valve in the forward direction. Before this pressure, it's closed; after this pressure, it's open and allows flow (current).

Step 1: Understanding the Concept:

Doping is the process of intentionally introducing impurities into an intrinsic (pure) semiconductor to alter its electrical properties.


n-type semiconductor: Formed by doping a semiconductor (like Silicon, Group 14) with a pentavalent impurity (5 valence electrons, Group 15). The fifth electron becomes a free charge carrier, making electrons the majority carriers.
p-type semiconductor: Formed by doping a semiconductor with a trivalent impurity (3 valence electrons, Group 13). The impurity atom creates a deficiency of one electron, known as a "hole," which acts as a positive charge carrier. Holes become the majority carriers.


Step 2: Detailed Explanation:

The host semiconductor is Silicon (Si), which is a tetravalent element (4 valence electrons) from Group 14 of the periodic table. To produce a p-type semiconductor, we need to dope it with a trivalent impurity (an element from Group 13).

Let's analyze the given options based on their group in the periodic table:

(A) Phosphorus (P): It is in Group 15. It is a pentavalent impurity. Doping Si with P creates an n-type semiconductor.

(B) Arsenic (As): It is in Group 15. It is a pentavalent impurity. Doping Si with As creates an n-type semiconductor.

(C) Antimony (Sb): It is in Group 15. It is a pentavalent impurity. Doping Si with Sb creates an n-type semiconductor.

(D) Boron (B): It is in Group 13. It is a trivalent impurity. When Boron is added to Silicon, each Boron atom replaces a Silicon atom but can only form three covalent bonds, leaving one electron deficiency or "hole". This creates a p-type semiconductor.


Step 3: Final Answer:

Boron is the trivalent impurity among the options that will produce a p-type semiconductor when doped in silicon. Therefore, the correct option is (D).
Quick Tip: Use a mnemonic to remember the dopants: \textbf{P}entavalent dopants like \textbf{P}hosphorus give \textbf{P}ositive free charges (this is a common error, it gives Negative charges!). A better mnemonic is: \textbf{n-type} has \textbf{n}egative charge carriers (electrons), made from Group 15 (e.g., Phosphorus, Arsenic). \textbf{p-type} has \textbf{p}ositive charge carriers (holes), made from Group 13 (e.g., \textbf{B}oron, \textbf{A}luminum, \textbf{G}allium - think "BAG").

Step 1: Analyzing the Assertion (A):

The assertion states that a continuous change of magnetic flux is required to maintain an induced emf. This is a direct consequence of Faraday's Law of Electromagnetic Induction, which is mathematically expressed as \( \mathcal{E} = -d\Phi_B/dt \). The induced emf (\(\mathcal{E}\)) is proportional to the \textit{rate of change of magnetic flux (\(d\Phi_B/dt\)). If the flux stops changing (\(d\Phi_B/dt = 0\)), the induced emf becomes zero. Therefore, to \textit{maintain an emf, the flux must change \textit{continuously. The Assertion (A) is true.


Step 2: Analyzing the Reason (R):

The reason describes how an induced current creates a magnetic field that opposes the very change in flux that produced it. This is a precise statement of Lenz's Law. Lenz's Law explains the direction of the induced current and is a manifestation of the conservation of energy. The Reason (R) is also true.


Step 3: Evaluating the Connection between Assertion and Reason:

Now we must determine if the Reason (R) is the correct explanation for the Assertion (A).

Assertion (A) is about the \textit{condition for the existence of an induced emf (a changing flux). It is explained by Faraday's Law.

Reason (R) is about the \textit{direction of the induced current/emf (opposition to the change). It is explained by Lenz's Law.

While both statements are correct and related to the same phenomenon (electromagnetic induction), Lenz's Law (Reason) does not explain \textit{why a changing flux is necessary to produce an emf. It only explains the direction of the effect once it is produced. The fundamental explanation for the Assertion is Faraday's Law itself. Therefore, the Reason is not the correct explanation for the Assertion.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true statements, but Reason (R) is not the correct explanation of Assertion (A). Therefore, the correct option is (B).
Quick Tip: For Assertion-Reason questions, follow a two-step check. First, verify if A and R are individually true. Second, ask "Why is A true?" and see if R is the answer. In this case, the answer to "Why is a continuous change of flux needed?" is "Because the emf is the rate of change of flux (Faraday's Law)," not "Because the induced current opposes the change (Lenz's Law)."

Step 1: Analyzing the Assertion (A):

The assertion states that at resonance in a series LCR circuit, the impedance (Z) is purely resistive (Z = R). In a series LCR circuit, resonance is the condition where the net reactance is zero. This leads to the minimum possible impedance and maximum current. At this point, the circuit behaves as a purely resistive circuit. Hence, Assertion (A) is true.


Step 2: Analyzing the Reason (R):

The reason provides the general formula for the impedance of a series LCR circuit: \(Z = \sqrt{R^2 + (X_C - X_L)^2}\). It also correctly states the condition for resonance: inductive reactance equals capacitive reactance (\(X_L = X_C\)). Hence, Reason (R) is also true.


Step 3: Evaluating the Connection between Assertion and Reason:

To check if the reason explains the assertion, we can substitute the resonance condition from the reason into the impedance formula also given in the reason.

Given \(X_L = X_C\), we have \(X_C - X_L = 0\).

Substituting this into the impedance formula: \[ Z = \sqrt{R^2 + (0)^2} = \sqrt{R^2} = R \]
This derivation shows that the reason directly and correctly explains why the impedance at resonance is equal to the resistance.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) provides the correct mathematical explanation for Assertion (A). Therefore, the correct option is (A).
Quick Tip: Remember that resonance in a series LCR circuit is a special condition where \(X_L = X_C\). This makes the term \((X_C - X_L)\) zero, simplifying the impedance to its minimum value, which is just the resistance R. The circuit becomes non-reactive.

Step 1: Analyzing the Assertion (A):

The formula for fringe width (\(\beta\)) in YDSE is \(\beta = \frac{\lambda D}{d}\), where \(d\) is the separation between the two coherent sources (slits). The assertion states that an interference pattern is observed when the sources are "infinitely close," which means \(d \rightarrow 0\).

If \(d \rightarrow 0\), the fringe width \(\beta \rightarrow \infty\).

An infinitely large fringe width means that the first bright or dark fringe would be at infinity. In practice, the entire screen would be illuminated almost uniformly, and distinct fringes would not be visible. Therefore, an interference pattern is not observed. Assertion (A) is false.


Step 2: Analyzing the Reason (R):

The reason states that the fringe width in YDSE does not depend on the separation between the two slits. As seen from the formula \(\beta = \frac{\lambda D}{d}\), the fringe width (\(\beta\)) is inversely proportional to the slit separation (\(d\)). This means it very much depends on the separation. A smaller separation leads to a wider fringe pattern. Therefore, Reason (R) is false.


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are false statements. Therefore, the correct option is (D).
Quick Tip: Always remember the fringe width formula \(\beta = \frac{\lambda D}{d}\) and the relationship between the variables. To get a clear, measurable interference pattern, you need a small but non-zero slit separation (\(d\)) and a large screen distance (\(D\)).

Step 1: Analyzing the Assertion (A):

The assertion states that a photon behaves as a particle. This is a fundamental concept of the quantum theory of light. Phenomena like the photoelectric effect and Compton scattering can only be explained by considering light as being composed of discrete packets of energy called photons, which exhibit particle-like properties (e.g., they can collide with electrons). Thus, Assertion (A) is true.


Step 2: Analyzing the Reason (R):

The reason states that a photon possesses both energy and momentum. According to the Planck-Einstein relation, a photon has energy \(E = hf\), and according to de Broglie's relation, it has momentum \(p = h/\lambda\). The possession of definite, quantized energy and momentum are key characteristics of a particle. Thus, Reason (R) is also true.


Step 3: Evaluating the Connection between Assertion and Reason:

The particle-like behavior of a photon is defined by its properties. The fact that it carries a specific amount of energy and momentum and can transfer these in collisions is precisely why we describe it as a particle. The reason provides the core physical properties that justify the assertion. Therefore, Reason (R) is the correct explanation for Assertion (A).


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). Therefore, the correct option is (A).
Quick Tip: Light exhibits wave-particle duality. Phenomena like interference and diffraction show its wave nature. Phenomena like the photoelectric effect show its particle nature. A photon is the "quantum" or particle of light.

Step 1: Understanding the Concept:

Drift velocity (\(v_d\)) is the average velocity attained by charged particles (like electrons) in a material due to an electric field. This velocity is related to the electric field (E), the charge of the electron (e), the mass of the electron (m), and the relaxation time (\(\tau\)). The relaxation time is the average time between successive collisions of an electron with the ions in the conductor.


Step 2: Key Formula or Approach:

The formula for drift velocity is: \[ v_d = \frac{eE}{m}\tau \]
The electric field (E) inside a uniform wire of length L connected to a potential difference V is given by: \[ E = \frac{V}{L} \]
By combining these, we can find the relaxation time \(\tau\). \[ \tau = \frac{v_d m}{eE} = \frac{v_d m L}{eV} \]

Step 3: Detailed Explanation:

First, list the given values and convert them to SI units:

Length of the wire, \(L = 20.0\) cm \( = 0.20 \) m.

Voltage of the battery, \(V = 4\) V.

Drift velocity, \(v_d = 0.8\) mm/s \( = 0.8 \times 10^{-3} \) m/s.


We also need the standard values for the charge and mass of an electron:

Charge of an electron, \(e = 1.6 \times 10^{-19}\) C.

Mass of an electron, \(m = 9.1 \times 10^{-31}\) kg.


Now, calculate the electric field \(E\): \[ E = \frac{V}{L} = \frac{4 \, V}{0.20 \, m} = 20 \, V/m \]

Next, rearrange the drift velocity formula to solve for the relaxation time \(\tau\): \[ \tau = \frac{v_d m}{eE} \]
Substitute the known values into this equation: \[ \tau = \frac{(0.8 \times 10^{-3} \, m/s) \times (9.1 \times 10^{-31} \, kg)}{(1.6 \times 10^{-19} \, C) \times (20 \, V/m)} \] \[ \tau = \frac{7.28 \times 10^{-34}}{32 \times 10^{-19}} \] \[ \tau = \frac{7.28}{32} \times 10^{-34 - (-19)} \] \[ \tau = 0.2275 \times 10^{-15} \, s \]
Expressing this in standard scientific notation: \[ \tau \approx 2.28 \times 10^{-16} \, s \]

Step 4: Final Answer:

The relaxation time of the electrons is approximately \(2.28 \times 10^{-16}\) seconds.
Quick Tip: A common mistake is forgetting to convert units to the SI system (cm to m, mm/s to m/s). Always perform this conversion before substituting values into formulas to ensure the final answer has the correct units and magnitude.

Step 1: Understanding the Concept:

This problem involves using the mirror formula for a concave mirror. We need to apply the Cartesian sign convention correctly. For a concave mirror, the focal length is negative. Since the image formed is real, it will be on the same side as the object, and its distance (v) will also be negative. The object distance (u) is always negative for a real object.


Step 2: Key Formula or Approach:

The mirror formula is: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance.

Magnification is also relevant: \(m = -\frac{v}{u}\). Since the image is real, it must be inverted, so \(m\) is negative.


Step 3: Detailed Explanation:

Given information with sign convention:

Focal length, \(f = -10\) cm (concave mirror).

The image is real.

The image distance is twice the object distance: \(|v| = 2|u|\).


Since the object is real, \(u\) is negative. Let \(u = -x\).

Since the image is real (formed by a single concave mirror), it is on the same side of the mirror as the object, so \(v\) is also negative.

Therefore, \(v = -2x\).


Now, substitute these into the mirror formula: \[ \frac{1}{-10} = \frac{1}{-2x} + \frac{1}{-x} \] \[ -\frac{1}{10} = -\frac{1}{2x} - \frac{2}{2x} \] \[ -\frac{1}{10} = -\frac{3}{2x} \] \[ \frac{1}{10} = \frac{3}{2x} \]
Cross-multiplying gives: \[ 2x = 30 \implies x = 15 \, cm \]

So, the object distance is \(u = -x = -15\) cm.

The image distance is \(v = -2x = -30\) cm.


The question asks for the distance between the object and the image. Both are on the principal axis in front of the mirror. \[ Distance = |v - u| = |-30 \, cm - (-15 \, cm)| \] \[ Distance = |-30 + 15| = |-15| = 15 \, cm \]

Step 4: Final Answer:

The distance between the object and the image is 15 cm.
Quick Tip: For concave mirrors, a real image is always formed on the same side as the object. The distance between them is simply the difference in their positions from the pole, \(|v - u|\). Careful application of sign convention is key to solving mirror and lens problems correctly.

Step 1: Understanding the Concept:

This problem requires us to first find the focal length of the lens from its given power. Then, using the lens formula, we can determine the position of the image. The sign of the image distance will tell us the nature of the image (real or virtual).


Step 2: Key Formula or Approach:

The relationship between power (\(P\)) and focal length (\(f\)) of a lens is: \[ P = \frac{1}{f} \quad (where f is in meters) \]
The thin lens formula is: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance.


Step 3: Detailed Explanation:

Given information with sign convention:

Power of the lens, \(P = +4.0\) D. The positive sign indicates it is a converging (convex) lens.

Object distance, \(u = -1.50\) m (object is placed in front of the lens).


First, calculate the focal length (\(f\)): \[ f = \frac{1}{P} = \frac{1}{+4.0} = +0.25 \, m \]
The focal length is \(f = +25\) cm.


Now, use the lens formula to find the image distance (\(v\)): \[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substitute the values (in meters): \[ \frac{1}{v} = \frac{1}{0.25} + \frac{1}{-1.50} \] \[ \frac{1}{v} = 4 - \frac{1}{1.5} \] \[ \frac{1}{v} = 4 - \frac{2}{3} \] \[ \frac{1}{v} = \frac{12 - 2}{3} = \frac{10}{3} \]
Solving for \(v\): \[ v = +\frac{3}{10} = +0.30 \, m \] \[ v = +30 \, cm \]

Step 4: Final Answer:

Position: The image is formed at a distance of 0.30 m (or 30 cm) from the lens.

Nature: Since the image distance \(v\) is positive, the image is formed on the opposite side of the lens from the object. This means the image is real. For a single lens, real images are always inverted.

Therefore, the image is real, inverted, and located 30 cm from the lens on the side opposite to the object.
Quick Tip: Remember the sign convention for power and focal length: Positive power means a converging (convex) lens with a positive focal length. Negative power means a diverging (concave) lens with a negative focal length. This initial check helps in verifying the final answer.

Step 1: Understanding the Concept:

This problem deals with single-slit diffraction. When light passes through a narrow slit, it spreads out, creating a diffraction pattern of bright and dark fringes on a screen. The central bright fringe (central maximum) is the widest and brightest. Its width is defined by the positions of the first dark fringes (minima) on either side of it. The angular width is the angle subtended by the central maximum at the slit.


Step 2: Key Formula or Approach:

The condition for the first minimum in a single-slit diffraction pattern is given by: \[ a \sin\theta = \lambda \]
where \(a\) is the slit width, \(\lambda\) is the wavelength of light, and \(\theta\) is the angular position of the first minimum.

The central maximum extends from \(-\theta\) to \(+\theta\). Therefore, the total angular width of the central maximum is \(2\theta\).

For small angles (which is usually the case in these experiments), we can approximate \(\sin\theta \approx \theta\) (where \(\theta\) is in radians). So, \(a\theta \approx \lambda\), or \(\theta \approx \lambda/a\).

The angular width is \(W_\theta = 2\theta = \frac{2\lambda}{a}\).


Step 3: Detailed Explanation:

First, list the given values and convert them to SI units:

Wavelength, \(\lambda = 700\) nm \( = 700 \times 10^{-9} \) m \( = 7 \times 10^{-7} \) m.

Slit width, \(a = 0.2\) mm \( = 0.2 \times 10^{-3} \) m \( = 2 \times 10^{-4} \) m.

The screen distance \(D=1\) m is not needed for calculating the angular width.


Now, calculate the angular width of the central maximum using the formula: \[ W_\theta = \frac{2\lambda}{a} \]
Substitute the values: \[ W_\theta = \frac{2 \times (7 \times 10^{-7} \, m)}{2 \times 10^{-4} \, m} \] \[ W_\theta = \frac{14 \times 10^{-7}}{2 \times 10^{-4}} \] \[ W_\theta = 7 \times 10^{-7 - (-4)} \] \[ W_\theta = 7 \times 10^{-3} \, radians \]

Step 4: Final Answer:

The angular width of the central maximum is \(7 \times 10^{-3}\) radians.
Quick Tip: Distinguish between angular width (\(2\theta\)) and linear width (\(2x\)). Angular width (\(2\lambda/a\)) is independent of the screen distance \(D\). Linear width (\(2x = 2D\theta = 2D\lambda/a\)) depends on \(D\). Read the question carefully to see which one is asked.

Step 1: Understanding the Concept:

For an electron in a hydrogen atom, the total energy (\(E_{total}\)), kinetic energy (\(KE\)), and potential energy (\(PE\)) are interrelated. These relationships can be derived from Bohr's model or the Virial theorem for an inverse-square force law (like the electrostatic force). The key is to remember the specific relationships between these three energy quantities.


Step 2: Key Formula or Approach:

The standard relationships between the energies for an electron in any orbit 'n' of a hydrogen-like atom are:

Kinetic Energy (\(KE\)) = - Total Energy (\(E_{total}\))
Potential Energy (\(PE\)) = 2 \(\times\) Total Energy (\(E_{total}\))
Potential Energy (\(PE\)) = -2 \(\times\) Kinetic Energy (\(KE\))

We are given \(E_{total} = -13.6\) eV for the ground state (n=1).


Step 3: Detailed Explanation:

(a) The potential energy of the electron.

Using the relationship \(PE = 2 \times E_{total}\): \[ PE = 2 \times (-13.6 \, eV) \] \[ PE = -27.2 \, eV \]

(b) The kinetic energy of the electron.

Using the relationship \(KE = - E_{total}\): \[ KE = -(-13.6 \, eV) \] \[ KE = +13.6 \, eV \]
Note that kinetic energy must always be a positive value.


(c) The ratio of the magnitude of the total energy to the kinetic energy of the electron.

First, find the magnitude of the total energy: \[ |E_{total}| = |-13.6 \, eV| = 13.6 \, eV \]
The kinetic energy is \(KE = 13.6\) eV.

Now, calculate the ratio: \[ Ratio = \frac{|E_{total}|}{KE} = \frac{13.6 \, eV}{13.6 \, eV} = 1 \]

Step 4: Final Answer:

(a) The potential energy is -27.2 eV.

(b) The kinetic energy is +13.6 eV.

(c) The ratio of the magnitude of the total energy to the kinetic energy is 1.
Quick Tip: A good way to remember the energy relations: The Total energy is negative, indicating a bound system. Kinetic energy is always positive. The Potential energy is negative and has twice the magnitude of the total energy. So, if \(E_{total} = -X\), then \(KE = +X\) and \(PE = -2X\).

Step 1: Understanding the Concept:

Dynamic resistance (or AC resistance) of a p-n junction diode is the resistance offered by the diode to a changing current. It is defined as the ratio of a small change in voltage across the diode to the corresponding small change in current through it, for a specific operating point on its V-I characteristic curve.


Step 2: Key Formula or Approach:

The formula for dynamic resistance (\(r_d\)) is: \[ r_d = \frac{\Delta V}{\Delta I} \]
where \(\Delta V\) is the change in voltage and \(\Delta I\) is the corresponding change in current.


Step 3: Detailed Explanation:

Given:

Initial voltage, \(V_1 = 0.90\) V.

Final voltage, \(V_2 = 0.95\) V.

Change in current, \(\Delta I = 0.005\) A.


First, calculate the change in voltage (\(\Delta V\)): \[ \Delta V = V_2 - V_1 = 0.95 \, V - 0.90 \, V = 0.05 \, V \]

Now, use the formula for dynamic resistance: \[ r_d = \frac{\Delta V}{\Delta I} = \frac{0.05 \, V}{0.005 \, A} \] \[ r_d = \frac{50}{5} \, \Omega = 10 \, \Omega \]

Step 4: Final Answer:

The dynamic resistance of the diode is 10 Ω.
Quick Tip: Dynamic resistance is different from static resistance (\(R = V/I\)). Dynamic resistance is relevant for AC signals superimposed on a DC bias, while static resistance is for DC operation. The slope of the V-I graph at a point gives \(1/r_d\).

Step 1: Understanding the Concept:

Every real voltage source, like a battery or cell, has an electromotive force (emf) and an internal resistance. The emf is the ideal potential difference it can provide, while the terminal voltage is the actual potential difference available at its terminals, which is affected by the internal resistance and the current flow.


Step 2: Detailed Explanation:

1. Electromotive Force (emf):

The emf (\(\mathcal{E}\)) of a cell is defined as the maximum potential difference between its two terminals when the cell is in an open circuit (i.e., when no current is drawn from it). Alternatively, it can be defined as the work done by the cell's non-electrostatic forces in moving a unit positive charge from the negative terminal to the positive terminal inside the cell.


2. Terminal Voltage:

The terminal voltage (\(V\)) of a cell is the potential difference between its two terminals when the cell is in a closed circuit (i.e., when current is flowing).

When the cell is discharging (providing current \(I\) to an external circuit), the terminal voltage is given by \(V = \mathcal{E} - Ir\), where \(r\) is the internal resistance of the cell. The term \(Ir\) represents the potential drop across the internal resistance.


Can Terminal Voltage Exceed EMF?

Yes, the terminal voltage of a cell can exceed its emf.

Explanation: This occurs when the cell is being charged. During charging, an external power source drives a current (\(I\)) into the cell, from its positive terminal to its negative terminal. In this case, the external source has to overcome both the cell's emf and the potential drop across its internal resistance. The equation for the terminal voltage becomes: \[ V = \mathcal{E} + Ir \]
Since \(I\) and \(r\) are positive, the term \(Ir\) is positive, making the terminal voltage \(V\) greater than the emf \(\mathcal{E}\).


Step 3: Final Answer:

The terminal voltage of a cell can exceed its emf during the process of charging.
Quick Tip: Remember the conditions: Discharging: \(V = \mathcal{E} - Ir \implies V < \mathcal{E}\) Charging: \(V = \mathcal{E} + Ir \implies V > \mathcal{E}\) Open circuit (\(I=0\)): \(V = \mathcal{E}\)

Step 1: Understanding the Concept:

Current density is a microscopic quantity that describes the flow of charge at a point inside a conductor. It relates the macroscopic current to the cross-sectional area. The equation \(\vec{j} = \sigma\vec{E}\) is the microscopic or vector form of Ohm's law, which relates the current density at a point to the electric field at that point through the material's conductivity.


Step 2: Detailed Explanation:

1. Current Density:

Current density (\(\vec{j}\)) at a point within a conductor is defined as the electric current flowing per unit area of cross-section, taken perpendicular to the direction of current flow. It is a vector quantity whose direction is the same as the direction of motion of positive charge carriers (or opposite to the direction of motion of electrons). \[ j = \frac{I}{A} \]

2. SI Unit:

The SI unit of current density is amperes per square meter (\(A/m^2\)).


3. Derivation of \(\vec{j} = \sigma\vec{E}\):

We start with the expression for drift velocity (\(v_d\)) of an electron in a conductor under the influence of an external electric field (\(E\)): \[ v_d = \left(\frac{e\tau}{m}\right)E \]
where \(e\) is the magnitude of the electron charge, \(m\) is its mass, and \(\tau\) is the average relaxation time.

The total current (\(I\)) flowing through a conductor with cross-sectional area \(A\) and free electron density \(n\) is given by: \[ I = n A e v_d \]
Substitute the expression for \(v_d\) into the current equation: \[ I = n A e \left(\frac{e\tau}{m}\right)E = \frac{n e^2 A \tau}{m} E \]
Now, we define current density \(j = I/A\): \[ j = \frac{I}{A} = \frac{n e^2 \tau}{m} E \]
The term \(\frac{n e^2 \tau}{m}\) depends only on the material properties and temperature. This term is called the electrical conductivity of the material and is denoted by \(\sigma\). \[ \sigma = \frac{n e^2 \tau}{m} \]
Substituting \(\sigma\) back into the equation for \(j\), we get: \[ j = \sigma E \]
Since the direction of conventional current density (\(\vec{j}\)) is the same as the direction of the electric field (\(\vec{E}\)), we can write this in vector form: \[ \vec{j} = \sigma \vec{E} \]
This is the microscopic form of Ohm's law.
Quick Tip: To remember the derivation, follow the chain of logic: Electric field (\(E\)) causes acceleration, which leads to drift velocity (\(v_d\)). Drift velocity leads to current (\(I\)). Current per area is current density (\(j\)). This connects \(j\) back to \(E\).

Step 1: Understanding the Concept:

Electrostatic fields are produced by static charges, while magnetic fields are produced by moving charges (currents). Both are fundamental fields in electromagnetism, and they share some properties while differing in others.


Step 2: Detailed Explanation:

Similarity:

Both the electrostatic field and the magnetic field obey the principle of superposition. This means that the net field at any point due to a collection of sources (charges for electrostatic field, currents for magnetic field) is the vector sum of the fields produced by each individual source.

(Other possible similarities: Both are vector fields, both exert forces on charges, and both fields' strength decreases with distance from the source.)


Difference:

The fundamental difference lies in their sources and the nature of their field lines.

Electrostatic field lines originate from positive charges and terminate on negative charges. They do not form closed loops. This is because isolated electric charges (monopoles) exist.
Magnetic field lines always form continuous, closed loops. They do not have a starting or ending point. This is a consequence of the fact that isolated magnetic poles (magnetic monopoles) have never been observed to exist. Quick Tip: A key way to remember the difference is through Gauss's Laws. Gauss's Law for electricity (\(\oint \vec{E} \cdot d\vec{A} = q_{enc}/\epsilon_0\)) shows that charges are sources. Gauss's Law for magnetism (\(\oint \vec{B} \cdot d\vec{A} = 0\)) shows that there are no magnetic monopoles (no sources), hence the closed loops.

Step 1: Understanding the Concept:

A current-carrying circular loop or coil produces a magnetic field at its center. The strength of this field depends on the current, the radius of the coil, and the number of turns. We need to find the radii for both cases using the given length of the wire, \(l\).


Step 2: Key Formula or Approach:

The magnetic field (\(B\)) at the center of a circular coil with \(n\) turns, radius \(r\), carrying current \(I\) is given by: \[ B = \frac{\mu_0 n I}{2r} \]

Step 3: Detailed Explanation:

Let the current in both cases be \(I\).


Case 1: Single circular loop

Number of turns, \(n_1 = 1\).

The length of the wire is equal to the circumference of the loop. \[ l = 2\pi r_1 \implies r_1 = \frac{l}{2\pi} \]
The magnetic field at the center is \(B_1\): \[ B_1 = \frac{\mu_0 n_1 I}{2r_1} = \frac{\mu_0 (1) I}{2(l/2\pi)} = \frac{\mu_0 \pi I}{l} \]

Case 2: Circular coil of two turns

Number of turns, \(n_2 = 2\).

The length of the wire is equal to twice the circumference of the coil. \[ l = n_2 \times (2\pi r_2) = 2 \times (2\pi r_2) = 4\pi r_2 \implies r_2 = \frac{l}{4\pi} \]
The magnetic field at the center is \(B_2\): \[ B_2 = \frac{\mu_0 n_2 I}{2r_2} = \frac{\mu_0 (2) I}{2(l/4\pi)} = \frac{2\mu_0 I}{l/2\pi} = \frac{4\mu_0 \pi I}{l} \]

Ratio of the magnetic fields

We need to find the ratio \(B_1 / B_2\). \[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 \pi I}{l}}{\frac{4\mu_0 \pi I}{l}} = \frac{1}{4} \]

Step 4: Final Answer:

The ratio of the magnetic fields at their centers (\(B_1 : B_2\)) is 1 : 4.
Quick Tip: For a fixed length of wire \(l\) bent into a coil of \(n\) turns, the radius is \(r = l/(2\pi n)\). The magnetic field is \(B = \frac{\mu_0 n I}{2r} = \frac{\mu_0 n I}{2(l/2\pi n)} = \frac{\mu_0 \pi n^2 I}{l}\). This shows that \(B \propto n^2\). Therefore, the ratio of fields for \(n_1=1\) and \(n_2=2\) will be \(1^2 : 2^2\) or 1:4.

Step 1: Understanding the Concept:

Total internal reflection (TIR) is an optical phenomenon where a light ray traveling in a denser medium is completely reflected back into the same medium when it strikes the interface with a rarer medium at a large enough angle.


Step 2: Detailed Explanation:

There are two necessary conditions for total internal reflection to occur:

Medium of Propagation: The light ray must be traveling from an optically denser medium to an optically rarer medium. (For example, from water to air, or from glass to water).
Angle of Incidence: The angle of incidence (\(i\)) in the denser medium must be greater than the critical angle (\(i_c\)) for the pair of media. The critical angle is the specific angle of incidence for which the angle of refraction is 90°.

So, the two conditions are:

Light must travel from denser to rarer medium.
Angle of incidence \(i > i_c\). Quick Tip: Remember the acronym "D-R-I-G" - Denser to Rarer, Incident angle Greater than critical angle. This can help you recall the two essential conditions for TIR quickly.

Step 1: Understanding the Concept:

When an object in a denser medium is viewed from a rarer medium, its image appears to be at a shallower depth than its actual depth. The difference between the real depth and the apparent depth is known as the "normal shift". This shift is the height by which the image appears to be raised.


Step 2: Key Formula or Approach:

The apparent depth (\(d_{app}\)) is related to the real depth (\(d_{real}\)) and the refractive index (\(\mu\)) of the denser medium by: \[ d_{app} = \frac{d_{real}}{\mu} \]
The normal shift (\(\Delta x\)), which is the height the image is raised, is given by: \[ \Delta x = d_{real} - d_{app} = d_{real} - \frac{d_{real}}{\mu} = d_{real} \left(1 - \frac{1}{\mu}\right) \]

Step 3: Detailed Explanation:

Given:

Real depth (thickness of the glass slab), \(d_{real} = 3.0\) cm.

Refractive index of the glass, \(\mu = 1.5\).


We need to find the normal shift (\(\Delta x\)). Using the formula: \[ \Delta x = d_{real} \left(1 - \frac{1}{\mu}\right) \]
Substitute the given values: \[ \Delta x = 3.0 \, cm \left(1 - \frac{1}{1.5}\right) \]
Since \(1.5 = 3/2\), \(1/1.5 = 2/3\). \[ \Delta x = 3.0 \left(1 - \frac{2}{3}\right) \] \[ \Delta x = 3.0 \left(\frac{3-2}{3}\right) = 3.0 \left(\frac{1}{3}\right) \] \[ \Delta x = 1.0 \, cm \]

Step 4: Final Answer:

The height through which the image of the dot is raised is 1.0 cm.
Quick Tip: For quick calculations, remember that for glass with \(\mu=1.5=3/2\), the normal shift is always one-third of the real thickness (\(\Delta x = t/3\)). For water with \(\mu=1.33 \approx 4/3\), the shift is about one-fourth of the real depth (\(\Delta x \approx t/4\)).

Step 1: Understanding the Concept:

Reactance is the opposition offered by an inductor or a capacitor to the flow of alternating current (AC). Unlike resistance, reactance is frequency-dependent.


Step 2: Detailed Explanation:

1. Reactance of a Capacitor (Capacitive Reactance, \(X_C\)):

The capacitive reactance is given by the formula: \[ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \]
where \(f\) is the frequency of the AC source and \(C\) is the capacitance.

From the formula, we can see that \(X_C\) is inversely proportional to the frequency \(f\).

At low frequencies (\(f \to 0\), i.e., DC), \(X_C \to \infty\). This means a capacitor offers very high opposition to low-frequency AC and completely blocks DC current.
At high frequencies (\(f \to \infty\)), \(X_C \to 0\). This means a capacitor offers very low opposition to high-frequency AC, essentially acting as a short circuit.


2. Reactance of an Inductor (Inductive Reactance, \(X_L\)):

The inductive reactance is given by the formula: \[ X_L = \omega L = 2\pi f L \]
where \(f\) is the frequency of the AC source and \(L\) is the inductance.

From the formula, we can see that \(X_L\) is directly proportional to the frequency \(f\).

At low frequencies (\(f \to 0\), i.e., DC), \(X_L \to 0\). This means an inductor offers very low opposition to low-frequency AC and acts like a simple connecting wire for DC current.
At high frequencies (\(f \to \infty\)), \(X_L \to \infty\). This means an inductor offers very high opposition to high-frequency AC, effectively blocking it. Quick Tip: A simple way to remember: Inductors oppose change, so they oppose high-frequency (rapidly changing) currents more (\(X_L \propto f\)). Capacitors store charge and get "filled up" by slow currents, so they block low-frequency currents more (\(X_C \propto 1/f\)).

Step 1: Understanding the Concept:

In a purely inductive AC circuit, the current is limited by the inductive reactance (\(X_L\)). We can use the AC equivalent of Ohm's law to find the peak current, where reactance plays the role of resistance.


Step 2: Key Formula or Approach:

The peak current (\(I_0\)) is given by: \[ I_0 = \frac{V_0}{X_L} \]
where \(V_0\) is the peak voltage and \(X_L\) is the inductive reactance.

The inductive reactance is calculated as: \[ X_L = 2\pi f L \]

Step 3: Detailed Explanation:

Given:

Self-inductance, \(L = 0.5\) H.

Peak voltage, \(V_0 = 314\) V.

Frequency, \(f = 50\) Hz.

We can use the approximation \(\pi \approx 3.14\).


First, calculate the inductive reactance (\(X_L\)): \[ X_L = 2\pi f L \] \[ X_L = 2 \times 3.14 \times 50 \, Hz \times 0.5 \, H \] \[ X_L = 2 \times 3.14 \times 25 = 50 \times 3.14 = 157 \, \Omega \]

Now, calculate the peak value of the current (\(I_0\)): \[ I_0 = \frac{V_0}{X_L} \]
Notice that the peak voltage \(V_0 = 314\) V is exactly \(100 \times 3.14\) V. \[ I_0 = \frac{314 \, V}{157 \, \Omega} \] \[ I_0 = 2 \, A \]

Step 4: Final Answer:

The peak value of the current in the circuit is 2 A.
Quick Tip: In exam problems, look for numbers that are multiples of \(\pi\) (like 314, 628, 157). Recognizing that \(314 \approx 100\pi\) and \(157 \approx 50\pi\) can simplify calculations significantly, especially when dealing with formulas involving \(2\pi f\).

Step 1: Understanding the Concept:

Electromagnetic (EM) waves are disturbances in electric and magnetic fields that propagate through space, carrying energy. They are described by Maxwell's equations and have several distinct properties.


Step 2: Detailed Explanation:

Two important characteristics of electromagnetic waves are:

Transverse Nature: EM waves are transverse in nature. This means that the oscillating electric field vector (\(\vec{E}\)) and the oscillating magnetic field vector (\(\vec{B}\)) are both perpendicular to each other and also perpendicular to the direction of wave propagation.
Propagation in Vacuum: EM waves do not require any material medium for their propagation. They can travel through a vacuum at a constant speed, known as the speed of light, \(c\), which is approximately \(3 \times 10^8\) m/s. This speed is a universal constant, given by the relation \(c = 1/\sqrt{\mu_0 \epsilon_0}\). Quick Tip: Other key characteristics you can mention include: EM waves are produced by accelerating charges, they carry energy and momentum, and the electric and magnetic fields oscillate in the same phase.

Step 1: Understanding the Concept:

Displacement current was introduced by James Clerk Maxwell to make Ampere's circuital law consistent for time-varying fields. It is not a current due to the flow of charge but rather a current equivalent that arises from a changing electric field.


Step 2: Detailed Explanation:

Displacement Current (\(I_d\)):

Displacement current is a quantity that arises from a time-varying electric field or electric flux. It is defined by the expression: \[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \(\epsilon_0\) is the permittivity of free space and \(\frac{d\Phi_E}{dt}\) is the rate of change of electric flux. A changing electric field produces a magnetic field in the same way that a conduction current (flow of charges) does. The total current in the modified Ampere's law is the sum of conduction current (\(I_c\)) and displacement current (\(I_d\)).


Setup of Displacement Current during Capacitor Charging:

Consider a parallel plate capacitor being charged by a battery.

When the circuit is closed, a conduction current (\(I_c\)) starts flowing in the connecting wires as charges move from the battery to the capacitor plates.
As charge accumulates on the capacitor plates, a time-varying electric field (\(\vec{E}(t)\)) is established in the region between the plates.
This changing electric field leads to a changing electric flux (\(\Phi_E(t)\)) through any surface placed between the plates.
According to Maxwell's theory, this rate of change of electric flux (\(d\Phi_E/dt\)) gives rise to the displacement current (\(I_d\)) in the gap between the plates.
It can be shown that the magnitude of the displacement current between the plates is exactly equal to the magnitude of the conduction current in the wires (\(I_c = I_d\)). This ensures that Kirchhoff's current law is valid at any point and that current is continuous throughout the circuit.

In summary, the conduction current in the wires is converted into displacement current in the capacitor gap.
Quick Tip: Think of displacement current as the "missing piece" that makes Ampere's law work everywhere, especially in places where there is no charge flow, like the vacuum between capacitor plates. It ensures the continuity of current in a circuit.

Step 1: Understanding the Concept:

Bohr's model of the atom introduced quantized energy levels and angular momentum to explain the stability of atoms and their line spectra. De Broglie later provided a physical interpretation for Bohr's quantization rule by proposing that electrons behave as waves, forming standing waves in their orbits.


Step 2: Detailed Explanation:

Postulates of Bohr Model of Hydrogen Atom:


Stationary Orbits: An electron revolves around the nucleus in certain specific circular orbits called stationary orbits. While in these orbits, the electron does not radiate energy, contrary to classical electromagnetic theory.
Quantization of Angular Momentum: The angular momentum (\(L\)) of an electron in a stationary orbit is an integral multiple of \(h/(2\pi)\), where \(h\) is Planck's constant.
\[ L = m_e v_n r_n = \frac{nh}{2\pi} \]
where \(n = 1, 2, 3, \dots\) is the principal quantum number.
Frequency Condition: An atom emits a photon of radiation when an electron makes a transition from a higher energy stationary orbit (\(E_i\)) to a lower energy stationary orbit (\(E_f\)). The frequency (\(f\)) of the emitted photon is given by:
\[ hf = E_i - E_f \]


Derivation using Bohr's Second Postulate:

According to Bohr's second postulate, the angular momentum of an electron in the n^th orbit is given by: \[ m v r = \frac{nh}{2\pi} \]
where \(m\) is the mass of the electron, \(v\) is its speed, and \(r\) is the radius of the orbit.


We can rearrange this equation as: \[ 2\pi r = n \left(\frac{h}{mv}\right) \quad \cdots (1) \]

Now, according to the de Broglie hypothesis, a particle of mass \(m\) moving with speed \(v\) has a wavelength (\(\lambda\)) associated with it, given by: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
where \(p = mv\) is the momentum of the particle.


Substitute this expression for the de Broglie wavelength (\(\lambda\)) into equation (1): \[ 2\pi r = n \lambda \]

The term \(2\pi r\) represents the circumference of the n^th orbit.

Therefore, we have shown that: \[ \textbf{Circumference of n^th orbit} = n \times (\textbf{de Broglie wavelength}) \]

This result implies that an allowed electron orbit is one in which an integral number of electron de Broglie wavelengths fit exactly, forming a stationary or standing wave pattern.
Quick Tip: This derivation provides a beautiful link between Bohr's early quantum idea (quantized angular momentum) and de Broglie's later wave-particle duality concept. The condition \(2\pi r = n\lambda\) is the condition for constructive interference of the electron wave with itself, leading to a stable standing wave pattern.

Step 1: Understanding the Concept:

Resistivity (also known as specific electrical resistance) is an intrinsic property of a material that quantifies how strongly it resists the flow of electric current. Unlike resistance, which depends on the object's dimensions, resistivity is a property of the material itself.


Step 2: Key Formula or Approach:

The resistance (\(R\)) of a uniform conductor is directly proportional to its length (\(L\)) and inversely proportional to its cross-sectional area (\(A\)). The constant of proportionality is the resistivity (\(\rho\)). \[ R = \rho \frac{L}{A} \]
From this, resistivity can be defined as \(\rho = \frac{RA}{L}\).


Step 3: Detailed Explanation:

Definition: Resistivity of a material is defined as the resistance of a conductor of that material having a unit length and a unit cross-sectional area.

Alternatively, it can be defined from the microscopic form of Ohm's law, \(\vec{E} = \rho \vec{j}\), as the ratio of the magnitude of the electric field (\(E\)) to the magnitude of the current density (\(j\)) at a point inside the material. \[ \rho = \frac{E}{j} \]
Its SI unit is the ohm-meter (\(\Omega \cdot m\)).
Quick Tip: To easily remember the definition, consider a cube of the material with side length 1 meter. The resistivity is simply the resistance measured between any two opposite faces of this cube.

Step 1: Understanding the Concept:

The resistivity of a material is determined by two main factors: the number density of free charge carriers (\(n\)) and the average time between collisions of these carriers (relaxation time, \(\tau\)). The formula is \(\rho = \frac{m}{ne^2\tau}\). Temperature affects both \(n\) and \(\tau\) differently in metals and semiconductors.


Step 2: Detailed Explanation:

(i) Resistivity of a Metal:

Effect: The resistivity of a metal increases with a rise in temperature.

Justification: In metals, the number density of free electrons (\(n\)) is very large and remains almost constant with temperature. However, as the temperature increases, the metal ions (the lattice) vibrate with greater amplitude about their mean positions. This increases the frequency of collisions between the free electrons and the vibrating ions. As a result, the average time between collisions, the relaxation time (\(\tau\)), decreases.

Since resistivity is inversely proportional to the relaxation time (\(\rho \propto 1/\tau\)), a decrease in \(\tau\) leads to an increase in \(\rho\).


(ii) Resistivity of a Semiconductor:

Effect: The resistivity of a semiconductor decreases with a rise in temperature.

Justification: In semiconductors, an increase in temperature provides sufficient thermal energy to break more covalent bonds. This process creates more free electrons and holes, which act as charge carriers. Therefore, the number density of charge carriers (\(n\)) increases significantly with temperature.

While the relaxation time (\(\tau\)) also decreases due to more frequent collisions, the increase in carrier density (\(n\)) is the dominant effect.

Since resistivity is inversely proportional to the carrier density (\(\rho \propto 1/n\)), the exponential increase in \(n\) causes a sharp decrease in \(\rho\), even though \(\tau\) is also decreasing.
Quick Tip: Think of it this way: In metals, the number of cars (electrons) on the highway is fixed; increasing temperature just adds more obstacles (vibrating ions), slowing traffic down (higher resistivity). In semiconductors, increasing temperature is like opening new on-ramps, adding many more cars (carriers) to the highway, which drastically increases the overall flow (lower resistivity) despite the obstacles.

Step 1: Understanding the Concept:

A charged particle in a uniform electric field experiences a constant electric force. According to Newton's second law, this constant force produces a constant acceleration. The magnitude and direction of this acceleration depend on the particle's charge and mass.


Step 2: Key Formula or Approach:

The electric force on a particle with charge \(q\) in an electric field \(\vec{E}\) is \(\vec{F} = q\vec{E}\).

The acceleration of the particle is given by Newton's second law, \(\vec{a} = \frac{\vec{F}}{m} = \frac{q\vec{E}}{m}\).


Step 3: Detailed Explanation:

Let \(\vec{E}\) be the uniform electric field between the capacitor plates.


For the electron:

Charge \(q_e = -e\), mass = \(m_e\).

Force on the electron: \(\vec{F}_e = (-e)\vec{E}\). The direction of the force is opposite to the direction of the electric field.

Acceleration of the electron: \(\vec{a}_e = \frac{-e\vec{E}}{m_e}\). The magnitude is \(a_e = \frac{eE}{m_e}\).


For the proton:

Charge \(q_p = +e\), mass = \(m_p\).

Force on the proton: \(\vec{F}_p = (+e)\vec{E}\). The direction of the force is the same as the direction of the electric field.

Acceleration of the proton: \(\vec{a}_p = \frac{+e\vec{E}}{m_p}\). The magnitude is \(a_p = \frac{eE}{m_p}\).


Comparison:

Direction: Since the forces are in opposite directions (one with \(\vec{E}\), one against \(\vec{E}\)), the accelerations are also in opposite directions.

Magnitude: We know that the mass of a proton (\(m_p\)) is much greater than the mass of an electron (\(m_p \approx 1836 \, m_e\)). Since the magnitude of acceleration is inversely proportional to mass (\(a \propto 1/m\)) and \(m_p > m_e\), it follows that the magnitude of the proton's acceleration is lesser than the magnitude of the electron's acceleration (\(a_p < a_e\)).


Step 4: Final Answer:

The acceleration of the proton is lesser in magnitude and opposite in direction compared to that of the electron. Therefore, option (D) is correct.
Quick Tip: The electric force on the proton and electron has the same magnitude because their charges have the same magnitude. However, due to its much larger mass (inertia), the proton is "harder to accelerate" and thus has a smaller acceleration.

Step 1: Understanding the Concept:

This question tests the fundamental properties of a uniform electric field, such as the one found between the plates of a parallel-plate capacitor, and its relationship with electric potential and the motion of charges within it.


Step 2: Detailed Explanation:

Let's analyze each statement:

(A) The electric field between the plates is directed from the positively charged plate to the negatively charged plate.

This is the standard convention for the direction of an electric field. Field lines originate from positive charges and terminate on negative charges. So, this statement is correct.


(B) The magnitude of electric field is the same at all points between the plates.

For a parallel-plate capacitor, the electric field in the region between the plates is uniform (constant in both magnitude and direction), assuming we can neglect the fringing effects near the edges. So, this statement is correct.


(C) The acceleration of the electrons will be constant between the plates.

The acceleration of the electron is given by \(\vec{a} = \frac{q\vec{E}}{m} = \frac{-e\vec{E}}{m_e}\). Since the charge (\(-e\)), mass (\(m_e\)), and electric field (\(\vec{E}\)) are all constant, the acceleration of the electron is also constant. So, this statement is correct.


(D) The potential increases in the direction of the electric field between the plates.

The relationship between electric field (\(E\)) and electric potential (\(V\)) is given by \(E = -dV/dr\). The negative sign indicates that the electric field points in the direction of the steepest decrease in electric potential. In other words, electric potential decreases as we move in the direction of the electric field. The positive plate is at a higher potential than the negative plate. Therefore, this statement is incorrect.


Step 3: Final Answer:

The question asks for the incorrect statement. Based on the analysis, statement (D) is incorrect.
Quick Tip: Think of the electric field like a gravitational field pointing downhill. As you move downhill (in the direction of the field), your gravitational potential energy decreases. Similarly, as you move in the direction of the electric field, the electric potential decreases.

Step 1: Understanding the Concept:

This problem applies the Work-Energy Theorem. When the electron is released from rest, the electric field between the capacitor plates does work on it. This work is converted entirely into the electron's kinetic energy, causing it to accelerate and gain speed.


Step 2: Key Formula or Approach:

Work done by the electric field on a charge \(q\) moving through a potential difference \(V\) is \(W = qV\).

The change in kinetic energy (\(\Delta KE\)) is given by \(KE_{final} - KE_{initial} = \frac{1}{2}mv^2 - 0\).

By the Work-Energy Theorem, Work Done = Change in Kinetic Energy.


Step 3: Detailed Explanation:

The electron has charge \(q = e\) (in magnitude) and is accelerated through a potential difference \(V\).

The work done on the electron by the electric field is: \[ W = e \times V \]
The electron starts from rest, so its initial kinetic energy is zero. Its final kinetic energy when it has reached a speed \(v\) is: \[ KE = \frac{1}{2}mv^2 \]
According to the Work-Energy Theorem: \[ W = \Delta KE \] \[ eV = \frac{1}{2}mv^2 \]
We need to solve for the maximum speed, \(v\). Rearranging the equation: \[ v^2 = \frac{2eV}{m} \] \[ v = \sqrt{\frac{2eV}{m}} \]

Step 4: Final Answer:

The maximum speed gained by the electron is \(\sqrt{\frac{2eV}{m}}\). Therefore, option (C) is correct.
Quick Tip: This is a classic and very important result in electromagnetism. Memorizing the formula \(v = \sqrt{\frac{2qV}{m}}\) for a particle of charge \(q\) and mass \(m\) accelerated from rest through a potential \(V\) can save time in exams.

Step 1: Understanding the Concept:

For a uniform electric field, such as the one between the plates of a parallel-plate capacitor, the potential difference (voltage) is directly proportional to the electric field strength and the distance between the points (or plates).


Step 2: Key Formula or Approach:

The relationship between uniform electric field \(E\), potential difference \(V\), and the distance \(d\) between the plates is given by: \[ V = E \times d \]

Step 3: Detailed Explanation:

Given:

Electric field, \(E = 200\) V/m.

Distance between plates, \(d = 5\) mm.

First, we must convert the distance to SI units (meters): \[ d = 5 \, mm = 5 \times 10^{-3} \, m \]
Now, we can calculate the potential difference \(V\): \[ V = E \times d = (200 \, V/m) \times (5 \times 10^{-3} \, m) \] \[ V = 1000 \times 10^{-3} \, V \] \[ V = 1.0 \, V \]

Step 4: Final Answer:

The potential difference between the two plates is 1.0 V. Therefore, option (D) is correct.
Quick Tip: Always ensure that all quantities are in their base SI units before performing calculations. A common mistake is to forget to convert millimeters to meters, which would lead to an incorrect answer by a factor of 1000.

Step 1: Understanding the Concept:

A charged particle placed in an electric field experiences an electric force. The magnitude of this force is the product of the magnitude of the charge and the strength of the electric field.


Step 2: Key Formula or Approach:

The electric force \(F\) on a charge \(q\) in an electric field \(E\) is given by: \[ F = qE \]

Step 3: Detailed Explanation:

Given:

The electric field is the same as in the previous part, \(E = 200\) V/m.

The particle is a proton, so its charge is the elementary charge, \(q = e = 1.6 \times 10^{-19}\) C.

Now, we calculate the force \(F\) acting on the proton: \[ F = qE = (1.6 \times 10^{-19} \, C) \times (200 \, V/m) \] \[ F = 320 \times 10^{-19} \, N \]
To write this in standard scientific notation: \[ F = 3.2 \times 10^2 \times 10^{-19} \, N = 3.2 \times 10^{-17} \, N \]

Step 4: Final Answer:

The force acting on the proton is \(3.2 \times 10^{-17}\) N. Therefore, option (C) is correct.
Quick Tip: For a uniform field between capacitor plates, the force on a charged particle is constant and independent of its position between the plates (ignoring edge effects).

Step 1: Understanding the Concept:

This is a definitional question based on the quantum theory of light. Max Planck proposed that energy is quantized, and Albert Einstein extended this idea to light, postulating that light itself consists of discrete packets of energy.


Step 2: Detailed Explanation:

The case-study paragraph itself begins with "A photon is a quantum of energy." A quantum (plural: quanta) is the minimum amount of any physical entity involved in an interaction. In the context of electromagnetic radiation, this discrete packet or quantum of energy is called a photon.

Neutrons are subatomic particles found in the nucleus. Molecules are groups of atoms bonded together. Quarks are elementary particles that make up protons and neutrons. None of these is the quantum of electromagnetic energy.


Step 3: Final Answer:

The packet of electromagnetic energy is called a photon. Therefore, option (A) is correct.
Quick Tip: Often, in case-study questions, the answer to a definitional question can be found directly within the provided text. Read the passage carefully first.

Step 1: Understanding the Concept:

A photon, despite having no rest mass, possesses momentum. The momentum of a photon is related to its energy and the speed of light.


Step 2: Key Formula or Approach:

The energy of a photon is given by \(E = hf\) and its momentum is given by the de Broglie relation \(p = h/\lambda\). The relationship between frequency \(f\) and wavelength \(\lambda\) for light is \(c = f\lambda\).

Combining these, we can derive a direct relationship between energy and momentum: \[ E = hf = h\left(\frac{c}{\lambda}\right) = \left(\frac{h}{\lambda}\right)c = pc \]
Therefore, the momentum \(p\) can be calculated as: \[ p = \frac{E}{c} \]

Step 3: Detailed Explanation:

Given:

Energy of the photon, \(E = 3.3 \times 10^{-19}\) J.

The speed of light in vacuum, \(c = 3 \times 10^8\) m/s.

Using the formula \(p = E/c\): \[ p = \frac{3.3 \times 10^{-19} \, J}{3 \times 10^8 \, m/s} \] \[ p = 1.1 \times 10^{-19 - 8} \, kg.m/s \] \[ p = 1.1 \times 10^{-27} \, kg.m/s \]

Step 4: Final Answer:

The momentum of the photon is \(1.1 \times 10^{-27}\) kg.m/s. Therefore, option (C) is correct.
Quick Tip: The relation \(E = pc\) is a cornerstone of relativistic physics and is extremely useful for problems involving photons. Remembering this simple formula can make calculations much faster than working with frequency and wavelength individually.

Step 1: Understanding the Concept:

The total power emitted by a light source is the total energy emitted per second. Since the light consists of discrete photons, the total power is the product of the number of photons emitted per second and the energy of a single photon.


Step 2: Key Formula or Approach:

1. Energy of a single photon: \(E = hf\), where \(h\) is Planck's constant and \(f\) is the frequency.

2. Total power: \(P = nE\), where \(n\) is the number of photons emitted per second.

Combining these, we get \(P = n(hf)\), which can be rearranged to find \(n\): \[ n = \frac{P}{hf} \]

Step 3: Detailed Explanation:

Given:

Frequency, \(f = 5.0 \times 10^{14}\) Hz.

Power, \(P = 3.315\) mW = \(3.315 \times 10^{-3}\) W.

Planck's constant, \(h \approx 6.63 \times 10^{-34}\) J·s.


First, calculate the energy of one photon: \[ E = hf = (6.63 \times 10^{-34} \, J·s) \times (5.0 \times 10^{14} \, Hz) \] \[ E = 33.15 \times 10^{-20} \, J = 3.315 \times 10^{-19} \, J \]

Next, calculate the number of photons emitted per second, \(n\): \[ n = \frac{P}{E} = \frac{3.315 \times 10^{-3} \, W}{3.315 \times 10^{-19} \, J} \] \[ n = 1 \times 10^{-3 - (-19)} \, s^{-1} \] \[ n = 1 \times 10^{16} \, photons/second \]

Step 4: Final Answer:

The number of photons emitted per second is \(1.0 \times 10^{16}\). Therefore, option (C) is correct.
Quick Tip: Look for convenient numbers in the problem. Here, \(3.315\) is half of \(6.63\), which is a common value for Planck's constant (often approximated). The frequency \(5.0 \times 10^{14}\) makes the product \(hf\) easy to relate to the given power. Recognizing these patterns can speed up your calculation.

Step 1: Understanding the Concept:

This problem uses Einstein's photoelectric equation, which is based on the conservation of energy. The energy of an incident photon is used to overcome the metal's work function (the minimum energy to free an electron), and any remaining energy is converted into the kinetic energy of the emitted photoelectron.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation: \[ KE_{max} = E_{photon} - \phi_0 \]
where \(E_{photon} = hf\) and \(\phi_0\) is the work function. So, \[ KE_{max} = hf - \phi_0 \]
We need to work with consistent units, so we will convert the work function from electron-volts (eV) to Joules (J).


Step 3: Detailed Explanation:

Given:

Work function, \(\phi_0 = 2.14\) eV.

Frequency of incident light, \(f = 6 \times 10^{14}\) Hz.

Constants: \(h = 6.63 \times 10^{-34}\) J·s and \(1\) eV = \(1.6 \times 10^{-19}\) J.


First, calculate the energy of the incident photon in Joules: \[ E_{photon} = hf = (6.63 \times 10^{-34} \, J·s) \times (6 \times 10^{14} \, Hz) \] \[ E_{photon} = 39.78 \times 10^{-20} \, J = 3.978 \times 10^{-19} \, J \]

Next, convert the work function to Joules: \[ \phi_0 = 2.14 \, eV \times (1.6 \times 10^{-19} \, J/eV) \] \[ \phi_0 = 3.424 \times 10^{-19} \, J \]

Now, calculate the maximum kinetic energy: \[ KE_{max} = E_{photon} - \phi_0 = (3.978 \times 10^{-19} \, J) - (3.424 \times 10^{-19} \, J) \] \[ KE_{max} = 0.554 \times 10^{-19} \, J \]
This is approximately \(0.55 \times 10^{-19}\) J.


Step 4: Final Answer:

The maximum kinetic energy of the emitted electrons is \(0.55 \times 10^{-19}\) J. Therefore, option (D) is correct.
Quick Tip: When dealing with photoelectric effect problems, you can calculate photon energy in eV using the formula \(E(eV) = \frac{hf}{e} = \frac{(6.63 \times 10^{-34})f}{1.6 \times 10^{-19}} \approx (4.14 \times 10^{-15})f\). This allows you to work directly in eV, which can be faster if the work function is also given in eV.

Step 1: Understanding the Concept:

The work function (\(\phi_0\)) is the minimum energy required to remove an electron from a metal surface. The threshold frequency (\(f_0\)) is the minimum frequency of incident light that can cause photoemission. These two quantities are directly related by Planck's equation.


Step 2: Key Formula or Approach:

The relationship between work function and threshold frequency is: \[ \phi_0 = hf_0 \]
We can rearrange this to find the threshold frequency: \[ f_0 = \frac{\phi_0}{h} \]
Again, we must use consistent units (Joules for energy).


Step 3: Detailed Explanation:

Given:

Work function, \(\phi_0 = 2.21\) eV.

Constants: \(h = 6.63 \times 10^{-34}\) J·s and \(1\) eV = \(1.6 \times 10^{-19}\) J.


First, convert the work function to Joules: \[ \phi_0 = 2.21 \, eV \times (1.6 \times 10^{-19} \, J/eV) \] \[ \phi_0 = 3.536 \times 10^{-19} \, J \]

Now, calculate the threshold frequency: \[ f_0 = \frac{\phi_0}{h} = \frac{3.536 \times 10^{-19} \, J}{6.63 \times 10^{-34} \, J·s} \] \[ f_0 \approx 0.5333 \times 10^{15} \, Hz \]
Expressing this in standard scientific notation: \[ f_0 \approx 5.3 \times 10^{14} \, Hz \]

Step 4: Final Answer:

The threshold frequency for the metal is \(5.3 \times 10^{14}\) Hz. Therefore, option (C) is correct.
Quick Tip: A useful shortcut for energy conversions: The product \(hc \approx 1240\) eV·nm. If you are given the threshold wavelength (\(\lambda_0\)), you can find the work function in eV directly using \(\phi_0(eV) = 1240/\lambda_0(nm)\). Conversely, you could find \(\lambda_0\) and then convert to frequency.

Step 1: Understanding the Concept:

The total electrostatic potential energy of a system of charges in an external electric field is the sum of two components: (1) the potential energy of the charges due to their position in the external field, and (2) the mutual potential energy of the charges due to their interaction with each other. This total energy represents the total work done in assembling the system of charges from infinity.


Step 2: Key Formula or Approach:

The process of building the system step-by-step is used to find the total work done, which equals the potential energy.

Work done to bring charge \(q_1\) from infinity to \(\vec{r}_1\) is \(W_1\).
Work done to bring charge \(q_2\) from infinity to \(\vec{r}_2\) in the presence of the external field and charge \(q_1\) is \(W_2\).
Total potential energy \(U = W_1 + W_2\).

The potential of an external field \(\vec{E}\) at a point \(\vec{r}\) is denoted by \(V(\vec{r})\). The potential at \(\vec{r}_2\) due to a charge \(q_1\) at \(\vec{r}_1\) is \(\frac{1}{4\pi\epsilon_0} \frac{q_1}{|\vec{r}_2 - \vec{r}_1|}\).


Step 3: Detailed Explanation:

Let \(V(\vec{r})\) be the potential due to the external electric field \(\vec{E}\).


1. Bringing charge \(q_1\):

The work done in bringing the charge \(q_1\) from infinity to the point \(\vec{r}_1\) against the external field is: \[ W_1 = q_1 V(\vec{r}_1) \]

2. Bringing charge \(q_2\):

When we bring the charge \(q_2\) from infinity to the point \(\vec{r}_2\), work has to be done against two fields:

(i) The external field \(\vec{E}\), for which the potential at \(\vec{r}_2\) is \(V(\vec{r}_2)\). The work done is \(q_2 V(\vec{r}_2)\).
(ii) The field produced by the charge \(q_1\) already at \(\vec{r}_1\). The potential at \(\vec{r}_2\) due to \(q_1\) is \(V_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1}{|\vec{r}_2 - \vec{r}_1|}\). The work done is \(q_2 V_{12}\).

So, the total work done in bringing \(q_2\) is: \[ W_2 = q_2 V(\vec{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|} \]

3. Total Potential Energy:

The total electrostatic potential energy (\(U\)) of the system is the sum of the work done in assembling the charges: \[ U = W_1 + W_2 \] \[ U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|} \]
This expression represents the sum of the potential energies of the individual charges in the external field plus their mutual interaction potential energy. Let \(r_{12} = |\vec{r}_2 - \vec{r}_1|\). \[ U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}} \] Quick Tip: Remember to account for all sources of potential. For a system of charges in an external field, you must add the "self-energy" of each charge in the field (\(qV\)) and the "interaction energy" for every pair of charges (\(kq_1q_2/r\)).

Step 1: Understanding the Concept:

This problem applies Coulomb's Law, which describes the force between two point charges. The force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.


Step 2: Key Formula or Approach:

Coulomb's Law is given by: \[ F = k \frac{q_1 q_2}{r^2} \]
where \(k = \frac{1}{4\pi\epsilon_0}\). We will analyze how changes in \(r\) and \(q\) affect \(F\).


Step 3: Detailed Explanation:

The initial force is \(F_{initial} = 1.5 \times 10^{-4}\) N.


(i) The distance between the charges is doubled.

Let the initial distance be \(r\). The new distance is \(r_{new} = 2r\).

The new force, \(F_{new}\), will be: \[ F_{new} = k \frac{q_1 q_2}{(r_{new})^2} = k \frac{q_1 q_2}{(2r)^2} = k \frac{q_1 q_2}{4r^2} \] \[ F_{new} = \frac{1}{4} \left(k \frac{q_1 q_2}{r^2}\right) = \frac{1}{4} F_{initial} \]
Calculating the new force: \[ F_{new} = \frac{1}{4} \times (1.5 \times 10^{-4} \, N) = 0.375 \times 10^{-4} \, N \]
So, the force is reduced to one-fourth of its original value.


(ii) The magnitude of each charge is doubled but the distance remains the same.

Let the initial charges be \(q_1\) and \(q_2\). The new charges are \(q'_{1} = 2q_1\) and \(q'_{2} = 2q_2\). The distance \(r\) remains the same.

The new force, \(F'_{new}\), will be: \[ F'_{new} = k \frac{(q'_{1})(q'_{2})}{r^2} = k \frac{(2q_1)(2q_2)}{r^2} = 4 \left(k \frac{q_1 q_2}{r^2}\right) \] \[ F'_{new} = 4 F_{initial} \]
Calculating the new force: \[ F'_{new} = 4 \times (1.5 \times 10^{-4} \, N) = 6.0 \times 10^{-4} \, N \]
So, the force becomes four times its original value.


Step 4: Final Answer:

(i) When the distance is doubled, the force becomes \(0.375 \times 10^{-4}\) N.

(ii) When each charge is doubled, the force becomes \(6.0 \times 10^{-4}\) N.
Quick Tip: Remember the dependencies in Coulomb's Law: \(F \propto q_1q_2\) and \(F \propto 1/r^2\). This allows you to quickly determine the effect of changing these parameters without re-calculating everything from scratch. Just apply the scaling factors.

Step 1: Understanding the Concept:

Both electrostatic and gravitational forces are fundamental forces of nature that act over a distance. They both follow an inverse-square law, but they differ in their strength, the property of matter they act upon, and their nature (attractive/repulsive).


Step 2: Detailed Explanation:

Here is a point-by-point differentiation:

\begin{tabular{|p{4cm|p{4cm|
\hline
Electrostatic Force & Gravitational Force

\hline
Acts between electric charges. & Acts between masses.

\hline
Can be either attractive (between opposite charges) or repulsive (between like charges). & Is always attractive.

\hline
It is a very strong force. For two protons, it is about \(10^{36}\) times stronger than the gravitational force between them. & It is the weakest of the four fundamental forces.

\hline
Depends on the medium between the charges. The force is maximum in a vacuum. & Is independent of the medium between the masses.

\hline
Can be shielded by placing the charges inside a conductive enclosure. & Cannot be shielded.

\hline
\end{tabular Quick Tip: Remember the key differences: Gravity is always attractive and acts on mass, while the electrostatic force can be attractive or repulsive and acts on charge. The most striking difference is their relative strength; the electrostatic force is immensely stronger.

Step 1: Understanding the Concept:

The particle remains stationary, which means it is in equilibrium. This implies that the net force acting on it is zero. The two forces acting on the particle are the gravitational force (acting downwards) and the electrostatic force (acting upwards). For equilibrium, these two forces must be equal in magnitude and opposite in direction.


Step 2: Key Formula or Approach:

1. Gravitational force: \(F_g = mg\), directed downwards.

2. Electric field between plates: \(E = V/d\).

3. Electrostatic force: \(F_e = qE\).

4. Equilibrium condition: \(F_e = F_g \implies qE = mg\).


Step 3: Detailed Explanation:

Given:

Potential difference, \(V = 480\) V.

Distance between plates, \(d = 3.0\) cm \( = 0.03\) m.

Mass of the particle, \(m = 4 \times 10^{-11}\) kg.

Acceleration due to gravity, \(g \approx 9.8\) m/s\textsuperscript{2 (or 10 m/s\textsuperscript{2 for simplicity if allowed). Let's use \(g=9.8\) m/s\textsuperscript{2.


The upper plate is at a positive potential, and the lower plate is at a negative potential. Therefore, the electric field \(\vec{E}\) is directed from the upper plate to the lower plate (i.e., downwards).


(i) Nature of the charge on the particle

The gravitational force \(F_g = mg\) acts downwards.

For the particle to be stationary, the electrostatic force \(F_e\) must be directed upwards to balance the gravitational force.

Since the electric field \(\vec{E}\) is downwards, and the electrostatic force \(\vec{F}_e\) on the particle is upwards (opposite to \(\vec{E}\)), the charge \(q\) on the particle must be negative. (\(\vec{F}_e = q\vec{E}\); if \(\vec{F}_e\) and \(\vec{E}\) are in opposite directions, \(q\) is negative).


(ii) The magnitude of the charge on the particle

First, calculate the electric field strength: \[ E = \frac{V}{d} = \frac{480 \, V}{0.03 \, m} = \frac{48000}{3} \, V/m = 16000 \, V/m = 1.6 \times 10^4 \, V/m \]
Now, apply the equilibrium condition: \[ |F_e| = |F_g| \] \[ |q|E = mg \]
Solve for the magnitude of the charge, \(|q|\): \[ |q| = \frac{mg}{E} = \frac{(4 \times 10^{-11} \, kg) \times (9.8 \, m/s^2)}{1.6 \times 10^4 \, V/m} \] \[ |q| = \frac{39.2 \times 10^{-11}}{1.6 \times 10^4} = 24.5 \times 10^{-15} \, C \] \[ |q| = 2.45 \times 10^{-14} \, C \]

Step 4: Final Answer:

(i) The nature of the charge is negative.

(ii) The magnitude of the charge is \(2.45 \times 10^{-14}\) C.
Quick Tip: This is a classic Millikan oil drop experiment-type problem. The key is always to balance the forces. Draw a free-body diagram to visualize the forces: gravity always acts down, and the electric force direction depends on the charge and field direction.

Step 1: Understanding the Concept:

This question explores the magnetic field produced by a long, straight, current-carrying conductor, based on Oersted's discovery and the Biot-Savart Law.


Step 2: Detailed Explanation:

(i) Right-Hand Thumb Rule:

If you imagine holding the current-carrying conductor in your right hand such that your thumb points in the direction of the conventional current, then the direction in which your fingers curl around the conductor gives the direction of the magnetic field lines. The magnetic field lines are concentric circles centered on the wire.


(ii) Variation with Distance:

For a long, straight conductor, the magnitude of the magnetic field (\(B\)) at a perpendicular distance (\(r\)) from the conductor is inversely proportional to the distance. \[ B \propto \frac{1}{r} \]

(iii) Factors Affecting the Magnetic Field:

The formula for the magnetic field due to a long, straight wire is \(B = \frac{\mu_0 \mu_r I}{2\pi r}\). Based on this formula, the factors (other than the distance \(r\)) that affect the magnitude of the magnetic field are:

Current (\(I\)): The magnetic field is directly proportional to the magnitude of the current flowing through the conductor. A larger current produces a stronger magnetic field.
Magnetic Permeability of the Medium (\(\mu = \mu_0 \mu_r\)): The magnetic field depends on the nature of the medium surrounding the conductor. It is proportional to the magnetic permeability of the medium. Quick Tip: Remember the key differences in how fields decrease with distance: The electric field from a point charge and the magnetic field from a long straight wire both decrease as \(1/r\). But the electric field from a point charge is \(E \propto 1/r^2\), not \(1/r\). The magnetic field from a long straight wire is \(B \propto 1/r\).

Step 1: Understanding the Concept:

A magnetic dipole (like a bar magnet or a current loop) placed in a uniform external magnetic field experiences a torque that tries to align its magnetic dipole moment with the external field. The potential energy of the dipole depends on its orientation with respect to the field, and the conditions of stable and unstable equilibrium correspond to minima and maxima of this potential energy.


Step 2: Key Formula or Approach:

The torque (\(\vec{\tau}\)) on a magnetic dipole with moment \(\vec{m}\) in a magnetic field \(\vec{B}\) is \(\vec{\tau} = \vec{m} \times \vec{B}\). The magnitude is \(\tau = mB\sin\theta\).

The potential energy (\(U\)) of the dipole is \(U = -\vec{m} \cdot \vec{B} = -mB\cos\theta\).

Equilibrium occurs when the net torque is zero (\(\tau = 0\)). This happens when \(\sin\theta = 0\), which means \(\theta = 0^\circ\) or \(\theta = 180^\circ\).


Step 3: Detailed Explanation:

(i) Stable Equilibrium:


Condition: Stable equilibrium is achieved when the potential energy of the dipole is at its minimum.
Orientation: The potential energy \(U = -mB\cos\theta\) is minimum when \(\cos\theta\) is maximum, i.e., \(\cos\theta = +1\). This occurs when the angle \(\theta\) between the magnetic dipole moment \(\vec{m}\) and the magnetic field \(\vec{B}\) is 0°.
Description: In this orientation, the magnetic dipole moment \(\vec{m}\) is parallel to the magnetic field \(\vec{B}\). The net torque on the dipole is zero (\(\tau = mB\sin(0^\circ) = 0\)). If the dipole is slightly displaced from this position, the resulting torque will act to restore it back to the \(\theta=0^\circ\) orientation. The potential energy is \(U_{min} = -mB\).


(ii) Unstable Equilibrium:


Condition: Unstable equilibrium is achieved when the potential energy of the dipole is at its maximum.
Orientation: The potential energy \(U = -mB\cos\theta\) is maximum when \(\cos\theta\) is minimum, i.e., \(\cos\theta = -1\). This occurs when the angle \(\theta\) between the magnetic dipole moment \(\vec{m}\) and the magnetic field \(\vec{B}\) is 180°.
Description: In this orientation, the magnetic dipole moment \(\vec{m}\) is anti-parallel to the magnetic field \(\vec{B}\). The net torque is also zero (\(\tau = mB\sin(180^\circ) = 0\)). However, if the dipole is slightly displaced from this position, the resulting torque will act to flip it completely around, moving it away from the equilibrium position and towards the stable equilibrium position. The potential energy is \(U_{max} = +mB\). Quick Tip: Think of a compass needle in the Earth's magnetic field. It naturally aligns with its north pole pointing north (\(\theta=0^\circ\)). This is stable equilibrium. If you force it to point south (\(\theta=180^\circ\)), it is technically in equilibrium (no torque), but any slight nudge will cause it to flip back to the stable north-pointing position.

Step 1: Understanding the Concept:

The motion of a charged particle in a uniform magnetic field is governed by the Lorentz force, \(\vec{F} = q(\vec{v} \times \vec{B})\). When the velocity is perpendicular to the field, this force provides the necessary centripetal force for circular motion. The radius of this circle depends on mass, velocity, charge, and magnetic field strength. We will analyze how changing these parameters affects the path.


Step 2: Key Formula or Approach:

For circular motion, the magnetic force provides the centripetal force: \[ qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB} \]
where \(r\) is the radius of the circular path.


Step 3: Detailed Explanation:

(i) The strength of magnetic field is decreased.

The radius is inversely proportional to the magnetic field strength (\(r \propto 1/B\)). If \(B\) is decreased, the radius \(r\) of the circular path will increase. The particle will follow a circular path of a larger radius.


(ii) The velocity of the particle is decreased.

The radius is directly proportional to the velocity (\(r \propto v\)). If \(v\) is decreased, the radius \(r\) of the circular path will decrease. The particle will follow a circular path of a smaller radius.


(iii) The velocity has components perpendicular and parallel to \(\vec{B}\).

Let \(\vec{v} = \vec{v}_{\perp} + \vec{v}_{\parallel}\). The perpendicular component, \(\vec{v}_{\perp}\), will cause the particle to move in a circle (radius \(r = mv_{\perp}/qB\)). The parallel component, \(\vec{v}_{\parallel}\), is unaffected by the magnetic field (\(\vec{F} = q(\vec{v}_{\parallel} \times \vec{B}) = 0\)), so it will cause the particle to move with constant velocity along the direction of the magnetic field. The combination of these two motions results in a helical (spiral) path, where the axis of the helix is parallel to the magnetic field.


(iv) An electric field balances the magnetic force.

If an electric field \(\vec{E}\) is applied such that the electrostatic force \(\vec{F}_e = q\vec{E}\) balances the magnetic force \(\vec{F}_m = q(\vec{v} \times \vec{B})\), the net force on the particle becomes zero (\(\vec{F}_{net} = \vec{F}_e + \vec{F}_m = 0\)). According to Newton's first law, a particle with zero net force will continue to move with a constant velocity. Therefore, the particle will move in a straight line with constant velocity \(\vec{v}\). This is the principle of a velocity selector.


(v) The magnetic field is suddenly removed.

If the magnetic field is removed (\(B=0\)), the magnetic force on the particle becomes zero. With no force acting on it, the particle will continue to move in a straight line with the velocity it had at the instant the field was removed. This straight-line path will be tangent to the circular path at the point where the field was switched off.
Quick Tip: The formula \(r = mv/qB\) is central to understanding the motion of charges in B-fields. Remember that \(mv\) is momentum (\(p\)), so \(r=p/qB\). This helps in quickly analyzing the effect of changing any of these four parameters on the radius of the trajectory.

Step 1: Understanding the Concept:

When a rectangular current-carrying loop is placed in a uniform magnetic field, the magnetic forces on its sides can produce a turning effect, or torque. This torque tends to rotate the loop to align its magnetic dipole moment with the external magnetic field.


Step 2: Key Formula or Approach:

We will analyze the magnetic force, \(\vec{F} = I(\vec{L} \times \vec{B})\), on each of the four sides of the rectangular loop. Then, we will calculate the net torque produced by these forces.


Step 3: Detailed Explanation:

Consider a rectangular loop PQRS with length \(l\) (sides PQ and RS) and breadth \(b\) (sides QR and SP). It carries a steady current \(I\) and is placed in a uniform magnetic field \(\vec{B}\). Let \(\hat{n}\) be the unit vector normal to the plane of the loop, and let the angle between \(\hat{n}\) and \(\vec{B}\) be \(\theta\).


Forces on the sides QR and SP (breadth \(b\)):

Let the current in side SP be directed along the positive y-axis and in QR along the negative y-axis. The forces on these sides are: \[ \vec{F}_{SP} = I(\vec{b} \times \vec{B}) \quad and \quad \vec{F}_{QR} = I(-\vec{b} \times \vec{B}) \]
These two forces, \(\vec{F}_{SP}\) and \(\vec{F}_{QR}\), are equal in magnitude and opposite in direction. They are also collinear (acting along the axis of rotation of the loop). Therefore, they cancel each other out and produce no net force and no net torque.


Forces on the sides PQ and RS (length \(l\)):

The force on side PQ is \(\vec{F}_{PQ}\). Its magnitude is \(F_{PQ} = IlB\sin(90^\circ) = IlB\), as the side is perpendicular to the field. Its direction is perpendicular to the plane containing the side and the magnetic field.
The force on side RS is \(\vec{F}_{RS}\). Its magnitude is also \(F_{RS} = IlB\). Its direction is opposite to that of \(\vec{F}_{PQ}\).

These two forces, \(\vec{F}_{PQ}\) and \(\vec{F}_{RS}\), are equal, parallel, and opposite. They form a couple and exert a torque on the loop.


Calculation of Torque:

The torque is given by the product of the magnitude of one of the forces and the perpendicular distance (lever arm) between them. \[ \tau = Force \times Lever Arm \]
Looking at the loop from the side, the lever arm is the perpendicular distance between the lines of action of \(\vec{F}_{PQ}\) and \(\vec{F}_{RS}\), which is \(b\sin\theta\).
\[ \tau = (IlB) \times (b\sin\theta) \] \[ \tau = I(lb)B\sin\theta \]
Since the area of the loop is \(A = lb\), the expression becomes: \[ \tau = IAB\sin\theta \]
If the coil has \(N\) turns, the torque is \(\tau = NIAB\sin\theta\).


The quantity \(NIA\) is defined as the magnitude of the magnetic dipole moment, \(\vec{m}\). So, \(\tau = mB\sin\theta\).

In vector form, the torque is given by the cross product: \[ \vec{\tau} = \vec{m} \times \vec{B} \] Quick Tip: The final expression \(\vec{\tau} = \vec{m} \times \vec{B}\) is analogous to the torque on an electric dipole in an electric field, \(\vec{\tau} = \vec{p} \times \vec{E}\). Remembering this analogy can help you recall the formula and its behavior.

Step 1: Understanding the Concept:

This problem involves the interference of light in a Young's double-slit experiment (YDSE). We need to use the formula for the position of dark fringes (minima) on the screen to determine the wavelength of the light.


Step 2: Key Formula or Approach:

The position (\(y_n\)) of the n-th dark fringe (minimum) from the central maximum in a YDSE is given by: \[ y_n = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d} \]
where \(\lambda\) is the wavelength of light, \(D\) is the distance from the slits to the screen, and \(d\) is the separation between the slits. The distance between two minima is the difference in their positions.


Step 3: Detailed Explanation:

Given:

Slit separation, \(d = 0.15\) mm \( = 0.15 \times 10^{-3}\) m.

Screen distance, \(D = 3.0\) m.

Distance between the first and eighth minima = 8.0 cm \( = 0.08\) m.


Position of the first minimum (for n = 1): \[ y_1 = \left(1 - \frac{1}{2}\right) \frac{\lambda D}{d} = \frac{1}{2} \frac{\lambda D}{d} \]
Position of the eighth minimum (for n = 8): \[ y_8 = \left(8 - \frac{1}{2}\right) \frac{\lambda D}{d} = \frac{15}{2} \frac{\lambda D}{d} \]

The distance between these two minima is: \[ \Delta y = y_8 - y_1 = \frac{15}{2} \frac{\lambda D}{d} - \frac{1}{2} \frac{\lambda D}{d} = \frac{14}{2} \frac{\lambda D}{d} = 7 \frac{\lambda D}{d} \]
We are given that this distance is 8.0 cm. \[ 7 \frac{\lambda D}{d} = 0.08 \, m \]
Now, we solve for the wavelength \(\lambda\): \[ \lambda = \frac{0.08 \times d}{7 \times D} \]
Substitute the given values: \[ \lambda = \frac{0.08 \, m \times (0.15 \times 10^{-3} \, m)}{7 \times 3.0 \, m} \] \[ \lambda = \frac{0.012 \times 10^{-3}}{21} = \frac{12 \times 10^{-6}}{21} \] \[ \lambda \approx 0.5714 \times 10^{-6} \, m \]
This can be expressed in nanometers (1 nm = \(10^{-9}\) m): \[ \lambda \approx 571.4 \times 10^{-9} \, m = 571.4 \, nm \]

Step 4: Final Answer:

The wavelength of the light used is approximately 571.4 nm.
Quick Tip: The distance between any \(N\) consecutive bright or dark fringes is simply \((N-1)\beta\), where \(\beta = \lambda D/d\) is the fringe width. In this case, the distance between the 1st and 8th minima is the distance covering 7 full fringes, so \(7\beta = 8.0\) cm. This is a quicker way to set up the problem.

Step 1: Understanding the Concept:

When two coherent waves superpose, the resultant amplitude (and hence intensity) at a point depends on the phase difference between the waves at that point. We need to find the resultant amplitude using the principle of superposition and then relate it to intensity to find the maximum and minimum possible intensities.


Step 2: Key Formula or Approach:

1. Principle of superposition to find resultant amplitude \(R\).

2. Relation between intensity and amplitude: \(I \propto A^2\).

3. Conditions for constructive (\(I_{max}\)) and destructive (\(I_{min}\)) interference.


Step 3: Detailed Explanation:

Let the two coherent waves be represented by: \[ y_1 = A_1 \sin(\omega t) \] \[ y_2 = A_2 \sin(\omega t + \phi) \]
where \(\phi\) is the phase difference between them.

According to the principle of superposition, the resultant displacement is \(y = y_1 + y_2\). The amplitude \(R\) of the resultant wave is given by: \[ R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} \]
The intensity of a wave is proportional to the square of its amplitude, i.e., \(I \propto A^2\). So, the resultant intensity \(I_R\) is given by: \[ I_R \propto R^2 \implies I_R = k(A_1^2 + A_2^2 + 2A_1A_2\cos\phi) \]
Since \(I_1 = kA_1^2\) and \(I_2 = kA_2^2\), we can write \(\sqrt{I_1I_2} = kA_1A_2\). The equation becomes: \[ I_R = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi \]

Maximum Intensity (\(I_{max}\)):

Intensity will be maximum when interference is constructive, which occurs when \(\cos\phi = +1\). \[ I_{max} = I_1 + I_2 + 2\sqrt{I_1I_2} = (\sqrt{I_1} + \sqrt{I_2})^2 \]
In terms of amplitude, the resultant amplitude is \(R_{max} = A_1 + A_2\), so: \[ I_{max} \propto (A_1 + A_2)^2 \]

Minimum Intensity (\(I_{min}\)):

Intensity will be minimum when interference is destructive, which occurs when \(\cos\phi = -1\). \[ I_{min} = I_1 + I_2 - 2\sqrt{I_1I_2} = (\sqrt{I_1} - \sqrt{I_2})^2 \]
In terms of amplitude, the resultant amplitude is \(R_{min} = |A_1 - A_2|\), so: \[ I_{min} \propto (A_1 - A_2)^2 \]

Ratio of Minimum to Maximum Intensity:

The ratio is: \[ \frac{I_{min}}{I_{max}} = \frac{k(A_1 - A_2)^2}{k(A_1 + A_2)^2} \] \[ \frac{I_{min}}{I_{max}} = \left( \frac{A_1 - A_2}{A_1 + A_2} \right)^2 \]

Step 4: Final Answer:

The derived expression for the ratio of minimum to maximum intensities is \(\left( \frac{A_1 - A_2}{A_1 + A_2} \right)^2\).
Quick Tip: This ratio is often used to define the "visibility" or "contrast" of interference fringes. High contrast (good visibility) occurs when \(I_{min}\) is close to zero, which happens when the amplitudes \(A_1\) and \(A_2\) are nearly equal.

Step 1: Understanding the Concept:

This problem involves the refraction of light through a prism under specific conditions. We need to apply the geometry of the prism and Snell's law at the refracting surfaces to find the desired relationship.


Step 2: Key Formula or Approach:

1. Snell's Law at the first surface (AB): \(n_1 \sin i = n_2 \sin r_1\).

2. Prism angle relation: \(A = r_1 + r_2\).

3. Condition for normal emergence from the second surface (AC).


Step 3: Detailed Explanation:

Given:

Angle of incidence, \(i = A/2\).

Angle of prism = \(A\).

Refractive index of prism material = \(n\).

The medium outside is air, with refractive index \(n_1 = 1\).


The ray emerges normally from the face AC. This means the angle of emergence is zero (\(e = 0\)). When a ray emerges normally, it strikes the surface at an angle of 90°, so the angle of refraction at the second face is also zero (\(r_2 = 0\)).


Using the prism angle relation: \[ A = r_1 + r_2 \]
Since \(r_2 = 0\), we get: \[ A = r_1 \]

Now, apply Snell's Law at the first face, AB: \[ n_{air} \sin(i) = n_{prism} \sin(r_1) \] \[ 1 \times \sin(i) = n \times \sin(r_1) \]
Substitute the known values \(i = A/2\) and \(r_1 = A\): \[ \sin\left(\frac{A}{2}\right) = n \sin(A) \]
We can use the trigonometric identity for the double angle, \(\sin(A) = 2\sin(A/2)\cos(A/2)\): \[ \sin\left(\frac{A}{2}\right) = n \left[ 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right) \right] \]
Assuming the prism angle A is not zero, \(\sin(A/2) \neq 0\), so we can divide both sides by \(\sin(A/2)\): \[ 1 = 2n \cos\left(\frac{A}{2}\right) \]

Step 4: Final Answer:

The relation between A and n is \(1 = 2n \cos(A/2)\). This can also be written as \(n = \frac{1}{2\cos(A/2)}\).
Quick Tip: The conditions "emerges normally" (\(e=0, r_2=0\)) or "grazing incidence/emergence" (\(i=90^\circ\) or \(e=90^\circ\)) are common in prism problems and significantly simplify the geometry. Always start by applying these special conditions.

Step 1: Understanding the Concept:

The focal length of a lens depends not only on the curvature of its surfaces and its own refractive index but also on the refractive index of the surrounding medium. The Lens Maker's formula relates these quantities.


Step 2: Key Formula or Approach:

The Lens Maker's formula is: \[ \frac{1}{f} = \left(\frac{n_{lens}}{n_{medium}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
We will apply this formula for the lens in air and then in the liquid and take their ratio to find the new focal length.


Step 3: Detailed Explanation:

Let \(n_g\) be the refractive index of the glass lens and \(n_m\) be the refractive index of the surrounding medium.

Given:

Focal length in air, \(f_{air} = +20\) cm (convex).

Refractive index of the lens, \(n_g = 1.5\).

Refractive index of the liquid, \(n_l = 1.25\).


Case 1: Lens in air

The medium is air, so \(n_m = n_{air} = 1\). \[ \frac{1}{f_{air}} = \left(\frac{n_g}{n_{air}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] \[ \frac{1}{20} = \left(\frac{1.5}{1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (0.5) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \cdots (1) \]

Case 2: Lens in liquid

The medium is liquid, so \(n_m = n_l = 1.25\). Let the new focal length be \(f_{liquid}\). \[ \frac{1}{f_{liquid}} = \left(\frac{n_g}{n_l} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] \[ \frac{1}{f_{liquid}} = \left(\frac{1.5}{1.25} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{1.50 - 1.25}{1.25}\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] \[ \frac{1}{f_{liquid}} = \left(\frac{0.25}{1.25}\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{1}{5}\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \cdots (2) \]

Now, divide equation (1) by equation (2): \[ \frac{1/20}{1/f_{liquid}} = \frac{0.5 \times (\frac{1}{R_1} - \frac{1}{R_2})}{\frac{1}{5} \times (\frac{1}{R_1} - \frac{1}{R_2})} \] \[ \frac{f_{liquid}}{20} = \frac{0.5}{1/5} = \frac{1/2}{1/5} = \frac{5}{2} = 2.5 \] \[ f_{liquid} = 20 \times 2.5 = 50 \, cm \]

Step 4: Final Answer:

The focal length of the lens when immersed in the liquid is 50 cm.
Quick Tip: When a converging lens (\(n_g\)) is immersed in a medium (\(n_m\)), its focal length increases if \(n_m < n_g\). The lens remains converging. If \(n_m = n_g\), the lens becomes invisible and its focal length becomes infinite. If \(n_m > n_g\), the converging lens will behave as a diverging lens (its focal length becomes negative).