Step 1: Understanding the Concept:

When a body acquires a positive charge, it means it has lost electrons.

Electrons have a definite mass. Therefore, the loss of electrons results in a decrease in the total mass of the body.

The charge on a body is quantized, meaning it is an integral multiple of the elementary charge of an electron (\(e\)).


Step 2: Key Formula or Approach:

1. Use the formula for quantization of charge to find the number of electrons lost:
\[ q = ne \]
where \(q\) is the total charge acquired, \(n\) is the number of electrons transferred, and \(e\) is the charge of a single electron (\(1.6 \times 10^{-19}\) C).

2. Calculate the total mass change (\(\Delta m\)) by multiplying the number of electrons (\(n\)) by the mass of a single electron (\(m_e\)).
\[ \Delta m = n \times m_e \]
where \(m_e \approx 9.1 \times 10^{-31}\) kg.


Step 3: Detailed Explanation:

Given data:

Charge acquired by the body, \(q = 8.0 \times 10^{-12}\) C.

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

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


First, we calculate the number of electrons (\(n\)) lost by the body.

From the quantization of charge, \(q = ne\), we have:
\[ n = \frac{q}{e} = \frac{8.0 \times 10^{-12} C}{1.6 \times 10^{-19} C} \] \[ n = \frac{8.0}{1.6} \times 10^{-12 - (-19)} \] \[ n = 5 \times 10^{7} \]
So, the body has lost \(5 \times 10^{7}\) electrons.


Next, we calculate the total decrease in mass due to the loss of these electrons.
\[ \Delta m = n \times m_e \] \[ \Delta m = (5 \times 10^{7}) \times (9.1 \times 10^{-31} kg) \] \[ \Delta m = (5 \times 9.1) \times 10^{7 - 31} kg \] \[ \Delta m = 45.5 \times 10^{-24} kg \]
To express this in standard scientific notation, we can write it as:
\[ \Delta m = 4.55 \times 10^{-23} kg \]

Step 4: Final Answer:

Since the body acquired a positive charge, it lost electrons, and its mass decreased.

The decrease in mass is \(4.55 \times 10^{-23}\) kg.

This corresponds to option (C).
Quick Tip: Remember that a positive charge on a body implies a deficit of electrons, leading to a decrease in mass. A negative charge implies an excess of electrons, leading to an increase in mass. The change in mass is often very small but not negligible in problems like this.

Step 1: Understanding the Concept:

Current density (\(j\)) is the amount of current flowing per unit area. When the current density is not uniform, the total current (\(I\)) must be found by integrating the current density over the given area. The relationship is \(I = \int j \cdot dA\).

For a cylindrical conductor with current density varying with the radial distance \(r\), we consider a small elemental ring of radius \(r\) and thickness \(dr\) as our area element \(dA\).


Step 2: Key Formula or Approach:

1. The current \(dI\) through an infinitesimally thin ring of radius \(r\) and thickness \(dr\) is given by:
\[ dI = j \cdot dA \]
The area of this elemental ring is \(dA = 2\pi r \, dr\).

2. Substitute the given current density \(j = \alpha r\) and the elemental area \(dA\) into the equation for \(dI\).
\[ dI = (\alpha r)(2\pi r \, dr) = 2\pi \alpha r^2 \, dr \]
3. Integrate \(dI\) from \(r=0\) to \(r = R/2\) to find the total current \(I\) flowing through that portion of the conductor.
\[ I = \int_{0}^{R/2} dI = \int_{0}^{R/2} 2\pi \alpha r^2 \, dr \]

Step 3: Detailed Explanation:

Given data:

Current density, \(j = \alpha r\).

We need to find the current from \(r=0\) to \(r = R/2\).


Let's perform the integration:
\[ I = \int_{0}^{R/2} 2\pi \alpha r^2 \, dr \]
Since \(2\pi\alpha\) are constants, we can take them out of the integral.
\[ I = 2\pi \alpha \int_{0}^{R/2} r^2 \, dr \]
The integral of \(r^2\) is \(\frac{r^3}{3}\).
\[ I = 2\pi \alpha \left[ \frac{r^3}{3} \right]_{0}^{R/2} \]
Now, we apply the limits of integration.
\[ I = 2\pi \alpha \left( \frac{(R/2)^3}{3} - \frac{(0)^3}{3} \right) \] \[ I = 2\pi \alpha \left( \frac{R^3/8}{3} - 0 \right) \] \[ I = 2\pi \alpha \left( \frac{R^3}{24} \right) \] \[ I = \frac{\pi \alpha R^3}{12} \]

Step 4: Final Answer:

The total current is \(I = \frac{\pi \alpha}{12} R^3\).

Since \(\pi\), \(\alpha\), and 12 are constants, the current \(I\) is directly proportional to \(R^3\).
\[ I \propto R^3 \]
This corresponds to option (C).
Quick Tip: For problems involving non-uniform distributions (like charge density, mass density, or current density) over an area or volume, the standard approach is always to define a small differential element, express the quantity for that element, and then integrate over the specified limits to find the total quantity.

Step 1: Understanding the Concept:

A charged particle moving in a magnetic field experiences a magnetic force, known as the Lorentz force. This force is always perpendicular to both the velocity of the particle (\(\vec{v}\)) and the magnetic field (\(\vec{B}\)).

When the initial velocity of the particle is perpendicular to the magnetic field, the magnetic force provides the necessary centripetal force for the particle to move in a circular path. The plane of this circle is perpendicular to the direction of 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 Lorentz force equation:
\[ \vec{F} = q(\vec{v} \times \vec{B}) \]
The direction of this force can be determined using the Right-Hand Palm Rule or Fleming's Left-Hand Rule (for positive charge).


Step 3: Detailed Explanation:

From the given figure:

1. The charge of the particle is \(+q\).

2. The magnetic field \(\vec{B}\) is uniform and directed into the plane of the paper. In the given coordinate system, this corresponds to the negative Z-direction (\(-\hat{k}\)). So, \(\vec{B} = -B\hat{k}\).

3. The particle enters the field at a point on the positive X-axis with an initial velocity \(\vec{v}\) directed along the negative Y-direction (\(-\hat{j}\)). So, \(\vec{v} = -v\hat{j}\).


Now, let's find the direction of the magnetic force \(\vec{F}\) at the point of entry.

Using the cross product:
\[ \vec{F} = q(\vec{v} \times \vec{B}) = q((-v\hat{j}) \times (-B\hat{k})) \] \[ \vec{F} = qvB (\hat{j} \times \hat{k}) \]
We know that \(\hat{j} \times \hat{k} = \hat{i}\).
\[ \vec{F} = qvB \hat{i} \]
The force is directed along the positive X-axis.


Alternatively, using Fleming's Left-Hand Rule:

- Point the forefinger in the direction of the magnetic field (\(\vec{B}\)), which is into the page.

- Point the middle finger in the direction of the velocity of the positive charge (\(\vec{v}\)), which is downwards.

- The thumb points in the direction of the force (\(\vec{F}\)), which is to the right (positive X-direction).


The force \(\vec{F}\) (\(+\hat{i}\)) is perpendicular to the velocity \(\vec{v}\) (\(-\hat{j}\)). This perpendicular force acts as a centripetal force, causing the particle to follow a circular path.

The plane of motion contains both the velocity vector and the force vector. Since \(\vec{v}\) is in the Y-direction and \(\vec{F}\) is in the X-direction, the motion occurs in the XY plane.

Since the magnetic field is confined to the region \(y>0\), the particle will trace a part of a circle. As it completes half a circle, it will exit the magnetic field region. Thus, the path described is a semicircular path in the XY plane.


Step 4: Final Answer:

The particle's velocity is perpendicular to the magnetic field, resulting in circular motion. The force and velocity vectors lie in the XY-plane, so the circular path is in the XY plane. Assuming the field exists only for \(y>0\), the particle describes a semicircle.

This corresponds to option (B).
Quick Tip: To quickly determine the path of a charged particle in a magnetic field: 1. Check if \(\vec{v}\) is parallel or anti-parallel to \(\vec{B}\). If so, the force is zero, and the path is a straight line. 2. Check if \(\vec{v}\) is perpendicular to \(\vec{B}\). If so, the path is a circle in the plane perpendicular to \(\vec{B}\). 3. If \(\vec{v}\) has components both parallel and perpendicular to \(\vec{B}\), the path is a helix. Always use the right-hand rule for positive charges and the left-hand rule for negative charges to find the direction of the force.

Step 1: Understanding the Concept:

The torque (\(\tau\)) experienced by a bar magnet with magnetic moment \(\vec{M}\) placed in a uniform magnetic field \(\vec{B}\) is given by the cross product \(\vec{\tau} = \vec{M} \times \vec{B}\).

The magnitude of the torque is \(\tau = MB \sin\theta\), where \(\theta\) is the angle between the magnetic moment \(\vec{M}\) and the magnetic field \(\vec{B}\).


Step 2: Key Formula or Approach:

1. Determine the initial torque (\(\tau_{initial}\)) when the magnet is at right angles to the field.

2. Determine the final angle (\(\theta_{final}\)) for which the torque (\(\tau_{final}\)) is half of the initial torque.

3. Calculate the angle of rotation, which is the difference between the initial and final angles.
\[ \tau = MB \sin\theta \]

Step 3: Detailed Explanation:

Initial Condition:

The bar magnet is initially at right angles to the magnetic field.

So, the initial angle is \(\theta_{initial} = 90^{\circ}\).

The initial torque is:
\[ \tau_{initial} = MB \sin(90^{\circ}) = MB(1) = MB \]

Final Condition:

The magnet is rotated until the torque becomes one-half of its initial value.
\[ \tau_{final} = \frac{1}{2} \tau_{initial} = \frac{1}{2} MB \]
Let the final angle be \(\theta_{final}\). The final torque is also given by:
\[ \tau_{final} = MB \sin(\theta_{final}) \]
Equating the two expressions for \(\tau_{final}\):
\[ MB \sin(\theta_{final}) = \frac{1}{2} MB \] \[ \sin(\theta_{final}) = \frac{1}{2} \]
This implies that \(\theta_{final} = 30^{\circ}\) or \(\theta_{final} = 150^{\circ}\). Since the magnet is rotated from 90\(^{\circ}\), the smaller change in angle corresponds to \(\theta_{final} = 30^{\circ}\).


Angle of Rotation:

The angle through which the magnet is rotated is the difference between the initial and final angles.
\[ Angle of rotation = \theta_{initial} - \theta_{final} \] \[ Angle of rotation = 90^{\circ} - 30^{\circ} = 60^{\circ} \]

Step 4: Final Answer:

The angle through which the bar magnet is rotated is 60\(^{\circ}\).

This corresponds to option (C).
Quick Tip: The torque on a magnetic dipole is maximum when it is perpendicular to the magnetic field (\(\theta = 90^{\circ}\)) and zero when it is parallel or anti-parallel (\(\theta = 0^{\circ}\) or \(\theta = 180^{\circ}\)). Always carefully read whether the question asks for the final angle or the angle of rotation.

Step 1: Understanding the Concept:

Magnetic materials are classified based on their response to an external magnetic field.

Paramagnetic materials are weakly attracted by an external magnetic field. They have permanent magnetic dipoles that are randomly oriented, but they align partially in the direction of the applied field. Examples include Aluminium, Calcium, and salts of transition metals like Copper Chloride.

Diamagnetic materials are weakly repelled by an external magnetic field. They do not have permanent magnetic dipoles. When placed in a magnetic field, a dipole moment is induced in a direction opposite to the applied field. Examples include water, Sodium Chloride (NaCl), copper, and bismuth.

Ferromagnetic materials are strongly attracted by an external magnetic field. They can be permanently magnetized. Examples include iron, cobalt, and nickel.


Step 2: Detailed Explanation:

We need to identify the material that is not paramagnetic from the given options.

(A) Aluminium (Al): It is a classic example of a paramagnetic material. Its atoms have unpaired electrons, leading to a net magnetic moment.

(B) Sodium Chloride (NaCl): This is an ionic compound. Both Na\(^{+}\) and Cl\(^{-}\) ions have completely filled electron shells, meaning there are no unpaired electrons. Materials with no unpaired electrons are typically diamagnetic.

(C) Calcium (Ca): It is a paramagnetic material. Although an isolated Ca atom has a filled 4s orbital, in a metallic lattice, the electronic structure allows for paramagnetism.

(D) Copper Chloride (CuCl\(_{2}\)): This is a salt of a transition metal (copper). The Cu\(^{2+}\) ion has unpaired electrons in its d-orbital, making the compound paramagnetic.


Step 3: Final Answer:

Based on the analysis, Aluminium, Calcium, and Copper Chloride are paramagnetic, whereas Sodium Chloride is diamagnetic. Therefore, Sodium Chloride is the material that is not paramagnetic.

This corresponds to option (B).
Quick Tip: A simple rule of thumb for identifying magnetic properties: materials with unpaired electrons in their atomic or molecular orbitals are generally paramagnetic or ferromagnetic. Materials where all electrons are paired are diamagnetic. Ionic compounds formed from elements in groups 1 and 17 (like NaCl) are often diamagnetic.

Step 1: Understanding the Concept:

In an AC (alternating current) circuit, the current and voltage vary sinusoidally with time. Standard AC measuring instruments, like an ammeter or voltmeter, are designed to read the Root Mean Square (RMS) value of the current or voltage, not the instantaneous or peak value. The RMS value represents the effective DC equivalent for producing the same amount of heat in a resistor.


Step 2: Key Formula or Approach:

The relationship between the RMS value of current (\(I_{rms}\)) and the maximum or peak value of current (\(I_{max}\) or \(I_0\)) for a sinusoidal AC is given by:
\[ I_{rms} = \frac{I_{max}}{\sqrt{2}} \]
We are given the ammeter reading, which is \(I_{rms}\), and we need to find \(I_{max}\).


Step 3: Detailed Explanation:

Given data:

The reading of the ammeter, \(I_{rms} = 10\) A.


We need to find the maximum value of the current, \(I_{max}\).

Rearranging the formula from Step 2:
\[ I_{max} = I_{rms} \times \sqrt{2} \]
Substituting the given value:
\[ I_{max} = 10 \times \sqrt{2} \] \[ I_{max} = 10\sqrt{2} A \]

Step 4: Final Answer:

The maximum value of the current at any instant in the circuit is \(10\sqrt{2}\) A.

This corresponds to option (A).
Quick Tip: Always remember that AC voltmeters and ammeters measure RMS values unless specified otherwise. To convert from RMS to peak (maximum) value, multiply by \(\sqrt{2}\). To convert from peak to RMS, divide by \(\sqrt{2}\). (\(\sqrt{2} \approx 1.414\)).

Step 1: Understanding the Concept:

In an electromagnetic (EM) wave propagating through a vacuum or free space, the electric field (\(\vec{E}\)) and magnetic field (\(\vec{B}\)) are mutually perpendicular and also perpendicular to the direction of wave propagation. The ratio of the magnitudes of the electric field and magnetic field at any instant is constant and equal to the speed of light in vacuum (\(c\)). This also applies to their amplitudes (\(E_0\) and \(B_0\)).


Step 2: Key Formula or Approach:

The relationship between the amplitude of the electric field (\(E_0\)) and the amplitude of the magnetic field (\(B_0\)) in an EM wave in free space is given by:
\[ c = \frac{E_0}{B_0} \]
where \(c\) is the speed of light in free space, approximately \(3 \times 10^8\) m/s.


Step 3: Detailed Explanation:

Given data:

Amplitude of the electric field, \(E_0 = 1000\) Vm\(^{-1}\).

Speed of light in free space, \(c = 3 \times 10^8\) m/s.


We need to find the amplitude of the magnetic field, \(B_0\).

Rearranging the formula from Step 2:
\[ B_0 = \frac{E_0}{c} \]
Substituting the given values:
\[ B_0 = \frac{1000 Vm^{-1}}{3 \times 10^8 m/s} \] \[ B_0 = \frac{10^3}{3 \times 10^8} T \] \[ B_0 = \frac{1}{3} \times 10^{3-8} T \] \[ B_0 = \frac{1}{3} \times 10^{-5} T \]
Converting the fraction to a decimal:
\[ B_0 \approx 0.333 \times 10^{-5} T \]
To express this in standard scientific notation, we can write it as:
\[ B_0 = 3.33 \times 10^{-6} T \]

Step 4: Final Answer:

The amplitude of the magnetic field in the electromagnetic wave is \(3.33 \times 10^{-6}\) T.

This corresponds to option (D).
Quick Tip: A common mistake is to confuse the formula and write \(c = B_0 / E_0\). Remember that the electric field values (in V/m) are numerically much larger than the magnetic field values (in Tesla) for an EM wave in vacuum. So, you must divide the larger number (\(E_0\)) by the very large speed of light (\(c\)) to get the smaller number (\(B_0\)).

Step 1: Understanding the Concept:

The linear magnification (\(m\)) produced by a spherical mirror provides information about the nature, size, and orientation of the image relative to the object.

The sign of the magnification indicates the nature of the image:

- If \(m\) is positive, the image is virtual and erect.

- If \(m\) is negative, the image is real and inverted.

The magnitude of the magnification indicates the size of the image relative to the object:

- If \(|m| > 1\), the image is magnified (enlarged).

- If \(|m| < 1\), the image is diminished (smaller).

- If \(|m| = 1\), the image is the same size as the object.


Step 2: Detailed Explanation:

Given data:

Magnification, \(m = -2.0\).


Analysis of the given magnification:

1. Sign: The magnification is negative (\(m = -2.0\)). This implies that the image is real and inverted.

2. Magnitude: The magnitude of the magnification is \(|m| = |-2.0| = 2.0\). Since \(|m| > 1\), the image is magnified.


Determining the type of mirror:

Now let's consider the properties of spherical mirrors:

- A convex mirror always forms a virtual, erect, and diminished image, regardless of the object's position. For a convex mirror, the magnification is always in the range \(0 < m < +1\).

- A concave mirror can form different types of images depending on the object's position. It can form a real, inverted, and magnified image. This occurs when the object is placed between the center of curvature (C) and the principal focus (F).


Since the image formed is real, inverted, and magnified, the mirror must be a concave mirror.


Step 3: Final Answer:

Combining the findings:

- The mirror is concave.

- The image is real (and inverted).

This matches the description in option (B).
Quick Tip: Memorize the sign conventions for magnification and the types of images formed by different mirrors. \textbf{Concave Mirror:} Can form real \& inverted OR virtual \& erect images. Forms magnified real images. \textbf{Convex Mirror:} Always forms virtual, erect, and diminished images. \textbf{Sign Convention:} Real image \(\implies\) \(m\) is negative. Virtual image \(\implies\) \(m\) is positive.

Step 1: Understanding the Concept:

This question tests the understanding of the wave-particle duality, a fundamental concept in quantum mechanics. This principle states that all matter and radiation exhibit both wave-like and particle-like properties.

- Light (Photons): Light behaves as a wave in phenomena like interference, diffraction, and polarization. It behaves as a particle (photon) in phenomena like the photoelectric effect and Compton scattering.

- Matter (Electrons): Electrons behave as particles (having mass and charge). The de Broglie hypothesis proposed that they also have a wave-like nature. This was experimentally confirmed by the Davisson-Germer experiment, which showed that a beam of electrons can be diffracted by a crystal lattice.


Step 2: Detailed Explanation:

Let's analyze each statement:

(A) Photons of light show diffraction whereas electrons do not show diffraction.

This statement is incorrect. Light (photons) certainly shows diffraction, which is a wave property. However, electrons also exhibit wave-like properties and show diffraction, as proven by the Davisson-Germer experiment.


(B) Electrons have momentum whereas photons do not have momentum.

This statement is incorrect. Electrons, being particles with mass (\(m\)) and velocity (\(v\)), have momentum (\(p = mv\)). Photons, although massless, also carry momentum, which is given by \(p = E/c = h/\lambda\), where \(E\) is energy, \(c\) is the speed of light, \(h\) is Planck's constant, and \(\lambda\) is the wavelength.


(C) Photons of light and electrons both exhibit dual nature.

This statement is correct. Both light (photons) and matter (electrons) exhibit wave-particle duality. They can behave as waves under certain circumstances (e.g., diffraction) and as particles under others (e.g., collisions).


(D) All electromagnetic radiations do not have photons.

This statement is incorrect. According to the quantum theory of light, all electromagnetic radiation is quantized into discrete packets of energy called photons. The energy of each photon is proportional to the frequency of the radiation.


Step 3: Final Answer:

Based on the analysis, the only correct statement is that both photons and electrons exhibit dual nature.

This corresponds to option (C).
Quick Tip: The concept of wave-particle duality is universal. It applies to everything, from photons to electrons to larger objects. However, the wave nature (de Broglie wavelength, \(\lambda = h/p\)) is only significant and observable for particles with very small momentum, like subatomic particles.

Step 1: Understanding the Concept:

The intensity (\(I\)) of a light beam is defined as the power per unit area. Power is the total energy transferred per unit time. The total energy of the beam is the sum of the energies of all its photons. The energy of a single photon is determined by its frequency (\(f\)) or wavelength (\(\lambda\)).


Step 2: Key Formula or Approach:

1. Energy of a single photon: \(E_{photon} = hf = \frac{hc}{\lambda}\), where \(h\) is Planck's constant.

2. Intensity of the light beam: \(I = \frac{Power}{Area} = \frac{Total Energy}{Area \times time}\).

3. If \(N\) is the number of photons striking an area \(A\) in time \(t\), the intensity is \(I = \frac{N \times E_{photon}}{A \times t}\).

The number of photons per unit area per unit time is \(n = \frac{N}{A \times t}\). So, \(I = n \times E_{photon}\).


Step 3: Detailed Explanation:

First, let's compare the properties of red and blue light. In the visible spectrum, blue light has a shorter wavelength and higher frequency than red light.
\[ \lambda_{blue} < \lambda_{red} \quad and \quad f_{blue} > f_{red} \]
This means the energy of a single blue photon is greater than the energy of a single red photon.
\[ E_{blue} = hf_{blue} \quad and \quad E_{red} = hf_{red} \]
Since \(f_{blue} > f_{red}\), it follows that \(E_{blue} > E_{red}\).


Now, we are given that the intensities are equal:
\[ I_{red} = I_{blue} \]
Let \(n_{red}\) and \(n_{blue}\) be the number of photons per unit area per unit time for the red and blue beams, respectively.

Using the formula \(I = n \times E_{photon}\), we have:
\[ n_{red} \times E_{red} = n_{blue} \times E_{blue} \]
We can rearrange this to find the ratio of the number of photons:
\[ \frac{n_{red}}{n_{blue}} = \frac{E_{blue}}{E_{red}} \]
Since we know that \(E_{blue} > E_{red}\), the ratio \(\frac{E_{blue}}{E_{red}}\) must be greater than 1.
\[ \frac{n_{red}}{n_{blue}} > 1 \implies n_{red} > n_{blue} \]
This means that for the intensities to be equal, the red beam must have more photons per unit area per unit time than the blue beam, to compensate for the lower energy of each red photon.


Let's evaluate the options:

(A) The blue beam has more number of photons than the red beam. (Incorrect)

(B) The red beam has more number of photons than the blue beam. (Correct)

(C) Wavelength of red light is lesser than wavelength of blue light. (Incorrect)

(D) The blue light beam has lesser energy per photon than that in the red light beam. (Incorrect)


Step 4: Final Answer:

To achieve the same intensity as the blue light beam, the red light beam, which consists of lower-energy photons, must have a greater number of photons.

This corresponds to option (B).
Quick Tip: Think of intensity as the total "energy punch" delivered per second. If you are using weaker "punches" (lower-energy photons like red light), you need more of them to deliver the same total punch as a beam with stronger "punches" (higher-energy photons like blue light).

Step 1: Understanding the Concept:

Materials are classified based on their ability to conduct electricity, which depends on the availability of free charge carriers (usually electrons).

- Conductors: Materials that allow electric current to flow easily. They have a large number of free electrons. Metals are excellent conductors.

- Insulators: Materials that resist the flow of electric current. They have very few free electrons.

- Semiconductors: Materials with electrical conductivity between that of conductors and insulators. Their conductivity is sensitive to temperature and impurities.


Step 2: Detailed Explanation:

Let's analyze the materials given in the options:

(A) Sn (Tin): Tin is a metal (specifically, a post-transition metal). Like other metals, it has a crystal lattice structure with a "sea" of delocalized electrons that are free to move. This makes it a good electrical conductor at room temperature.


(B) Mica: Mica is a group of silicate minerals. It is known for its excellent dielectric strength and is widely used as an electrical insulator in capacitors and other electronic components.


(C) Si (Silicon): Silicon is the most common element used in the semiconductor industry. In its pure form at room temperature, it has a much lower conductivity than metals. Its conductivity increases with temperature or by adding impurities (doping). It is not considered a conductor in the same class as metals.


(D) C (Carbon): Carbon exists in different allotropes with vastly different electrical properties. Diamond is an excellent electrical insulator. Graphite, another allotrope, is a conductor due to its layered structure with delocalized pi electrons. Since the allotrope is not specified, and Tin (Sn) is unambiguously a metallic conductor, Sn is the best answer.


Step 3: Final Answer:

Among the given options, Tin (Sn) is a metal and therefore the best example of an electrical conductor at room temperature.

This corresponds to option (A).
Quick Tip: In physics questions, elements are often used to represent classes of materials. Remember the general classification: Metals (like Sn, Cu, Al) are conductors. Non-metals (like S, P) and compounds like Mica are insulators. Metalloids (like Si, Ge) are semiconductors.

Step 1: Understanding the Concept:

A current-carrying conductor produces a magnetic field in the space around it. The pattern of this magnetic field can be visualized using magnetic field lines. For a long, straight conductor, the magnetic field lines are concentric circles. The plane of these circles is perpendicular to the length of the wire, and the center of the circles is the wire itself.


Step 2: Key Formula or Approach:

The direction of the magnetic field lines can be determined by the Right-Hand Thumb Rule (or Right-Hand Grip Rule).

Rule: If you imagine holding the current-carrying wire in your right hand such that your thumb points in the direction of the current, the direction in which your fingers curl gives the direction of the magnetic field lines.


Step 3: Detailed Explanation:

1. Shape of the field lines: For a long straight wire, the magnetic field lines are always concentric circles. This eliminates options (A), (B), and (D) which describe straight lines or helixes.

2. Orientation of the field lines: The wire is held vertically. The magnetic field lines form in planes that are perpendicular to the wire. A plane perpendicular to a vertical line is a horizontal plane. Therefore, the concentric circles are horizontal.

3. Applying the Right-Hand Thumb Rule: The current is in the upward direction. If you point your right thumb upwards along the wire, your fingers curl in an anti-clockwise direction when viewed from above. This confirms the circular path of the field lines.


Combining these points, the magnetic field lines are concentric horizontal circles around the wire.


Step 4: Final Answer:

The shape of the magnetic field lines produced by a long, straight vertical wire with an upward current is concentric horizontal circles.

This corresponds to option (C).
Quick Tip: Always visualize the geometry described. "Vertical wire" means the plane of the magnetic field circles must be "horizontal". The Right-Hand Thumb Rule is essential for determining the direction of the magnetic field for straight wires, while the Right-Hand Grip Rule is used for circular loops and solenoids.

Step 1: Understanding the Concept:

This question deals with the electrical neutrality of doped semiconductors. An n-type semiconductor is formed by adding pentavalent impurity atoms (donors) to an intrinsic semiconductor. It is crucial to understand that the term 'n-type' refers to the majority charge carriers (negative electrons), not the net charge of the material itself.


Step 2: Detailed Explanation:

Analysis of Assertion (A):

An n-type semiconductor is, as a whole, electrically neutral. It is created by doping a neutral intrinsic semiconductor (like silicon) with neutral pentavalent impurity atoms (like phosphorus). Since all the starting components are neutral, the final product must also be electrically neutral. The number of mobile negative charges (electrons) is balanced by the number of fixed positive charges (donor ions) and the mobile positive charges (holes). So, the Assertion (A) is true.


Analysis of Reason (R):

When a pentavalent impurity atom (e.g., Phosphorus) replaces a silicon atom in the crystal lattice, four of its valence electrons form covalent bonds with the neighboring silicon atoms. The fifth valence electron is loosely bound and is easily donated to the conduction band, becoming a free electron. The impurity atom, having lost one electron, becomes a positively charged ion (a donor ion) that is fixed in the lattice. The process starts with a neutral impurity atom and a neutral semiconductor, and results in a free electron and a positive ion. This description is accurate. So, the Reason (R) is true.


Relating Reason and Assertion:

The Reason explains exactly why the Assertion is true. It describes that for every free electron (negative charge carrier) created, a fixed positive donor ion is also created in the lattice. The process starts with neutral atoms. The creation of a free electron-positive ion pair from a neutral atom does not change the overall charge of the crystal. Therefore, the semiconductor remains electrically neutral. The Reason provides a complete and correct explanation for the Assertion.


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains why an n-type semiconductor is electrically neutral.

This corresponds to option (A).
Quick Tip: Do not confuse the type of majority charge carrier with the net charge of the semiconductor. 'n-type' means negative carriers (electrons) are in majority, and 'p-type' means positive carriers (holes) are in majority. Both n-type and p-type semiconductors are electrically neutral as a whole.

Step 1: Understanding the Concept:

This question is about the behavior of a series LCR circuit at the condition of resonance. Resonance occurs when the inductive reactance and capacitive reactance are equal, leading to minimum impedance and maximum current.


Step 2: Key Formula or Approach:

The total impedance (\(Z\)) of a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \(R\) is resistance, \(X_L = \omega L\) is inductive reactance, and \(X_C = 1/(\omega C)\) is capacitive reactance.


Step 3: Detailed Explanation:

Analysis of Assertion (A):

At resonance, the frequency of the AC source is such that the inductive reactance equals the capacitive reactance (\(X_L = X_C\)).

Substituting this condition into the impedance formula:
\[ Z = \sqrt{R^2 + (X_L - X_L)^2} = \sqrt{R^2 + 0} = R \]
Since the impedance \(Z\) becomes equal to the resistance \(R\), the net reactance of the circuit is zero. This means the circuit behaves as if it contains only the resistor. In a purely resistive circuit, the voltage and current are in phase. Therefore, the Assertion (A) is true.


Analysis of Reason (R):

The condition for resonance is \(X_L = X_C\). Substituting the expressions for reactance:
\[ \omega L = \frac{1}{\omega C} \]
Rearranging for the angular frequency \(\omega\):
\[ \omega^2 = \frac{1}{LC} \] \[ \omega = \frac{1}{\sqrt{LC}} \]
This is the correct formula for the resonant angular frequency. So, the Reason (R) is true.


Relating Reason and Assertion:

The Reason states the fundamental condition (\(X_L = X_C\)) that defines resonance. It is this exact condition that causes the reactive part of the impedance \((X_L - X_C)\) to become zero, which in turn causes the circuit to behave purely resistively, as stated in the Assertion. Therefore, the Reason (R) is the correct explanation for the Assertion (A).


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, and the Reason provides the correct physical and mathematical basis for the Assertion.

This corresponds to option (A).
Quick Tip: At resonance in a series LCR circuit, remember these key points: \(X_L = X_C\) Impedance \(Z\) is minimum (\(Z = R\)) Current is maximum (\(I_{max} = V/R\)) Voltage and current are in phase (power factor is 1)

Step 1: Understanding the Concept:

This question differentiates between the phenomena of interference and diffraction. Interference is the superposition of waves from two or more coherent sources, while diffraction is the bending of waves as they pass around an obstacle or through an aperture.


Step 2: Detailed Explanation:

Analysis of Assertion (A):

The double-slit experiment shows an interference pattern because light waves from two slits superpose. If one of the slits is closed, there is no longer a second wave to interfere with the first. The setup becomes a single-slit experiment. Light passing through the single open slit will spread out due to diffraction. This results in a single-slit diffraction pattern on the screen, which consists of a wide central maximum with smaller, less intense secondary maxima on either side. So, the Assertion (A) is true.


Analysis of Reason (R):

The phenomenon of interference is fundamentally about the superposition of waves. To observe a stable interference pattern, there must be at least two coherent waves (waves with a constant phase difference) that overlap in space. A single wave cannot interfere with itself to produce an interference pattern (though it can diffract). Thus, the statement that interference requires at least two waves is correct. So, the Reason (R) is true.


Relating Reason and Assertion:

The Reason explains why the interference pattern disappears when one slit is closed. However, the Assertion is about the appearance of a diffraction pattern. The appearance of the diffraction pattern is a consequence of the wave nature of light passing through the single remaining opening. The Reason does not explain why diffraction occurs, it only explains the condition necessary for interference. Therefore, while both statements are true, the Reason is not the correct explanation for the Assertion.


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are individually true statements from wave optics. However, the Reason explains the absence of interference, not the presence of diffraction.

This corresponds to option (B).
Quick Tip: In a standard Young's Double Slit Experiment (YDSE), you observe a pattern that is actually the interference pattern modulated by the diffraction pattern of a single slit. When you close one slit, the interference disappears, and only the single-slit diffraction pattern remains.

Step 1: Understanding the Concept:

This question relates to the photoelectric effect, specifically the energy distribution of the emitted electrons (photoelectrons). Einstein's photoelectric equation describes the energy conservation in this process.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation:
\[ K_{max} = h\nu - \phi \]
where \(K_{max}\) is the maximum kinetic energy of the emitted photoelectron, \(h\nu\) is the energy of the incident photon, and \(\phi\) is the work function of the metal.


Step 3: Detailed Explanation:

Analysis of Assertion (A):

When monochromatic light (all photons have energy \(h\nu\)) falls on a metal, a photon transfers its entire energy to a single electron. An electron at the very surface of the metal needs the minimum energy to escape, which is the work function \(\phi\). Such an electron will be emitted with the maximum possible kinetic energy, \(K_{max} = h\nu - \phi\). However, an electron from deeper inside the metal will lose some energy in collisions with other atoms on its way to the surface. It will therefore be emitted with a kinetic energy \(K\) that is less than \(K_{max}\). The energy loss can vary, so the emitted electrons have a distribution of kinetic energies ranging from 0 up to \(K_{max}\). Consequently, their speeds also range from zero to a certain maximum value \(v_{max}\). Thus, the Assertion (A) is true.


Analysis of Reason (R):

The work function (\(\phi\)) is defined as the minimum energy required to remove an electron from the surface of a material. It is a characteristic property of the material and has a specific, definite value for each metal. For example, the work function of caesium is different from that of zinc. So, the Reason (R) is true.


Relating Reason and Assertion:

The Reason (that the work function is definite) explains why there is a well-defined maximum kinetic energy (\(K_{max}\)). The value of \(K_{max}\) is fixed because \(h\nu\) and \(\phi\) are both fixed. However, the Reason does not explain why the electrons have speeds ranging from zero up to this maximum. The range of energies is due to energy losses from collisions inside the metal, a concept not mentioned in the Reason. Therefore, the Reason is true but provides only a partial explanation for the phenomenon described in the Assertion. It explains the "maximum value" part but not the "ranging from zero" part.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true statements. However, the Reason does not fully explain the Assertion.

This corresponds to option (B).
Quick Tip: Remember that Einstein's equation gives the maximum kinetic energy. The actual kinetic energy of a photoelectron can be anything from 0 to \(K_{max}\). The existence of a range of energies is experimental proof that photoelectrons come from different depths within the metal.

N/A

Step 1: Understanding the Concept:

As before, the brightness depends on the power dissipated (\(P=I^2R\)). We need to analyze the circuit with the key open and compare the currents through bulbs S and Q.


Step 2: Circuit Analysis when Key K is Opened:

1. When key K is opened, the branch containing bulb P is an open circuit, so no current flows through P. Bulb P goes off.

2. Bulbs S and Q are now connected in series with the battery.

3. The total equivalent resistance of the circuit is:
\[ R'_{total} = R_S + R_Q = R + R = 2R \]

Step 3: Detailed Explanation:

Current through bulbs S and Q:

- The total current flowing from the battery is:
\[ I' = \frac{V}{R'_{total}} = \frac{V}{2R} \]
- In a series circuit, the same current flows through all components. Therefore, the current through bulb S is the same as the current through bulb Q.
\[ I'_S = I'_Q = I' = \frac{V}{2R} \]
Comparing Brightness:

- Since the bulbs are identical (same R) and the current flowing through them is the same, the power dissipated by each bulb is also the same.
\[ P'_S = (I'_S)^2 R = \left(\frac{V}{2R}\right)^2 R = \frac{V^2}{4R} \] \[ P'_Q = (I'_Q)^2 R = \left(\frac{V}{2R}\right)^2 R = \frac{V^2}{4R} \] \[ P'_S = P'_Q \]

Step 4: Final Answer:

Since bulbs S and Q are in series, they carry the same current. As they are identical, they dissipate the same amount of power and therefore have equal brightness.




\begin{quicktipbox
In complex circuits, always simplify series and parallel combinations first to find the total current. Then, work backwards to find the current in each branch. Remember that brightness is proportional to power (\(P = I^2R = V^2/R\)), and for identical components, comparing either current or voltage is sufficient.
\end{quicktipbox Quick Tip: In complex circuits, always simplify series and parallel combinations first to find the total current. Then, work backwards to find the current in each branch. Remember that brightness is proportional to power (\(P = I^2R = V^2/R\)), and for identical components, comparing either current or voltage is sufficient.

Step 1: Understanding the Concept:

The given circuit diagram shows three parallel branches connected across a voltage source. The voltage source is the leftmost branch, consisting of two 10 V cells connected in series. The voltage provided by this source is applied across the other two parallel branches. We can use Ohm's law to find the current in the branch containing the bulb.


Step 2: Key Formula or Approach:

1. Determine the total voltage provided by the cells in series. For cells in series, the total EMF is the sum of individual EMFs: \(V_{total} = V_1 + V_2\).

2. Recognize that the voltage across parallel branches is the same.

3. Apply Ohm's law, \(I = \frac{V}{R}\), to the branch containing the bulb to find the current flowing through it.


Step 3: Detailed Explanation:

Calculating the Source Voltage:

The left branch contains two 10 V cells connected in series. The total EMF provided by this combination is:
\[ V = 10 V + 10 V = 20 V \]
Analyzing the Parallel Branches:

The circuit has three branches connected in parallel:

- Branch 1 (left): The 20 V source.

- Branch 2 (middle): The bulb B with resistance \(R_B = 10 \, \Omega\).

- Branch 3 (right): A 20 \(\Omega\) resistor and a 10 \(\Omega\) resistor in series.

Since these branches are in parallel, the voltage across each of them is the same, which is equal to the source voltage, 20 V.

Calculating the Current through the Bulb:

We are interested in the current flowing through the bulb in the middle branch. Using Ohm's law:
\[ I_{bulb} = \frac{V}{R_B} \]
Substituting the values:
\[ I_{bulb} = \frac{20 V}{10 \, \Omega} = 2 A \]

Step 4: Final Answer:

The current that flows through the bulb B is 2 A.




\begin{quicktipbox
When analyzing complex-looking circuits, try to identify parallel and series connections. Redrawing the circuit can often simplify it. In this case, recognizing that the three main vertical sections are parallel branches is the key to a quick solution.
\end{quicktipbox Quick Tip: When analyzing complex-looking circuits, try to identify parallel and series connections. Redrawing the circuit can often simplify it. In this case, recognizing that the three main vertical sections are parallel branches is the key to a quick solution.

Step 1: Understanding the Concept:

In single-slit diffraction, the condition for secondary maxima is given by the path difference between the waves from the edges of the slit. This condition relates the slit width (\(a\)), the angle of diffraction (\(\theta\)), the order of the maximum (\(n\)), and the wavelength of light (\(\lambda\)).


Step 2: Key Formula or Approach:

The condition for the \(n^{th}\) secondary maximum in a single-slit diffraction pattern is:
\[ a \sin\theta = \left(n + \frac{1}{2}\right)\lambda, \quad for n = 1, 2, 3, \ldots \]
For the first secondary maximum, we set \(n=1\).
\[ a \sin\theta = \frac{3}{2}\lambda \]

Step 3: Detailed Explanation:

Given data:

- Slit width, \(a = 0.55 mm = 0.55 \times 10^{-3} m\).

- Wavelength of light, \(\lambda = 550 nm = 550 \times 10^{-9} m\).

Calculation:

Using the formula for the first secondary maximum:
\[ \sin\theta = \frac{3\lambda}{2a} \]
Substitute the given values into the equation:
\[ \sin\theta = \frac{3 \times (550 \times 10^{-9} m)}{2 \times (0.55 \times 10^{-3} m)} \] \[ \sin\theta = \frac{1650 \times 10^{-9}}{1.1 \times 10^{-3}} \] \[ \sin\theta = 1500 \times 10^{-6} = 1.5 \times 10^{-3} \]
Since the value of \(\sin\theta\) is very small, we can use the small angle approximation, \(\sin\theta \approx \theta\), where \(\theta\) is in radians.
\[ \theta \approx 1.5 \times 10^{-3} radians \]
The question asks for the angle in degrees. We convert radians to degrees:
\[ \theta_{degrees} = \theta_{radians} \times \frac{180^{\circ}}{\pi} \] \[ \theta_{degrees} = (1.5 \times 10^{-3}) \times \frac{180^{\circ}}{3.14159} \] \[ \theta_{degrees} \approx 0.08594^{\circ} \]

Step 4: Final Answer:

Rounding to two significant figures, the angle of diffraction for the first secondary maximum is approximately 0.086\(^{\circ}\).




\begin{quicktipbox
Be careful not to confuse the conditions for maxima and minima in diffraction and interference. For single-slit diffraction, minima are at \(a\sin\theta = n\lambda\) and maxima are approximately at \(a\sin\theta = (n + \frac{1}{2})\lambda\). Also, always ensure all units are consistent (e.g., convert mm and nm to meters) before calculation.
\end{quicktipbox Quick Tip: Be careful not to confuse the conditions for maxima and minima in diffraction and interference. For single-slit diffraction, minima are at \(a\sin\theta = n\lambda\) and maxima are approximately at \(a\sin\theta = (n + \frac{1}{2})\lambda\). Also, always ensure all units are consistent (e.g., convert mm and nm to meters) before calculation.

Step 1: Understanding the Concept:

This problem requires the use of the Lens Maker's formula, which relates the focal length of a lens to its refractive index and the radii of curvature of its two surfaces. An equiconvex lens is a convex lens where both surfaces have the same radius of curvature.


Step 2: Key Formula or Approach:

The Lens Maker's formula is:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \(f\) is the focal length, \(n\) is the refractive index of the lens material, \(R_1\) is the radius of curvature of the first surface, and \(R_2\) is the radius of curvature of the second surface.

For an equiconvex lens, using the sign convention:

- \(R_1 = +R\) (surface facing the incident light is convex)

- \(R_2 = -R\) (second surface is concave from the perspective of light exiting it)


Step 3: Detailed Explanation:

Given data:

- Refractive index, \(n = 1.55\).

- Focal length, \(f = 15.0 cm\).

Applying the formula for an equiconvex lens:

Substitute \(R_1 = R\) and \(R_2 = -R\) into the Lens Maker's formula:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) \] \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} + \frac{1}{R} \right) \] \[ \frac{1}{f} = (n - 1) \frac{2}{R} \]
Now, we rearrange the formula to solve for the radius of curvature, R:
\[ R = 2f(n - 1) \]
Substitute the given numerical values:
\[ R = 2 \times (15.0 cm) \times (1.55 - 1) \] \[ R = 30.0 \times (0.55) \] \[ R = 16.5 cm \]

Step 4: Final Answer:

The radius of curvature of each surface of the equiconvex lens is 16.5 cm.




\begin{quicktipbox
Properly applying the sign convention is crucial when using the Lens Maker's formula. For a biconvex lens, \(R_1\) is positive and \(R_2\) is negative. For a biconcave lens, \(R_1\) is negative and \(R_2\) is positive. For an equiconvex/equiconcave lens, \(|R_1| = |R_2|\).
\end{quicktipbox Quick Tip: Properly applying the sign convention is crucial when using the Lens Maker's formula. For a biconvex lens, \(R_1\) is positive and \(R_2\) is negative. For a biconcave lens, \(R_1\) is negative and \(R_2\) is positive. For an equiconvex/equiconcave lens, \(|R_1| = |R_2|\).

Step 1: Understanding the Concept:

An \(\alpha\)-particle is the nucleus of a helium atom (\(^{4}_{2}He\)). A neutral helium atom consists of a nucleus (2 protons, 2 neutrons) and 2 electrons orbiting it. Therefore, the mass of an \(\alpha\)-particle is the mass of a neutral helium atom minus the mass of its two electrons. The mass of the carbon atom is extra information not required for this calculation if the mass of an electron in 'u' is known.


Step 2: Key Formula or Approach:
\[ Mass of \alpha-particle = Mass of He atom - 2 \times (Mass of an electron) \]
We need the mass of an electron (\(m_e\)) in atomic mass units (u). The standard value is \(m_e \approx 0.00054858\) u.


Step 3: Detailed Explanation:

Given data:

- Mass of a normal helium atom, \(m_{He} = 4.002603\) u.

Known constant:

- Mass of an electron, \(m_e = 0.00054858\) u.

Calculation:

Using the formula from Step 2:
\[ m_{\alpha} = m_{He} - 2 \times m_e \]
Substitute the values:
\[ m_{\alpha} = 4.002603 u - 2 \times (0.00054858 u) \] \[ m_{\alpha} = 4.002603 u - 0.00109716 u \] \[ m_{\alpha} = 4.00150584 u \]
The given mass of the helium atom has six decimal places. It is appropriate to round our result to the same precision.
\[ m_{\alpha} \approx 4.001506 u \]

Step 4: Final Answer:

The mass of an \(\alpha\)-particle is 4.001506 u.




\begin{quicktipbox
In nuclear physics problems, remember that the mass of a nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons (mass defect). Also, the mass of a nucleus is the atomic mass minus the mass of all its electrons. The binding energy of electrons is usually negligible compared to nuclear binding energies.
\end{quicktipbox Quick Tip: In nuclear physics problems, remember that the mass of a nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons (mass defect). Also, the mass of a nucleus is the atomic mass minus the mass of all its electrons. The binding energy of electrons is usually negligible compared to nuclear binding energies.

N/A

Step 1: Understanding the Concept:

In a semiconductor at thermal equilibrium, the product of the concentration of electrons (\(n_e\)) and the concentration of holes (\(n_h\)) is constant and equal to the square of the intrinsic carrier concentration (\(n_i\)). This is known as the law of mass action.


Step 2: Key Formula or Approach:

The law of mass action is given by:
\[ n_e \cdot n_h = n_i^2 \]
We can rearrange this formula to find the electron concentration, \(n_e\).


Step 3: Detailed Explanation:

Given data from the problem:

- Intrinsic carrier concentration, \(n_i = 5 \times 10^{8} m^{-3}\).

- Hole concentration in the extrinsic semiconductor, \(n_h = 8 \times 10^{12} m^{-3}\).

Calculation:

We need to find the electron concentration, \(n_e\).
\[ n_e = \frac{n_i^2}{n_h} \]
Substitute the given values:
\[ n_e = \frac{(5 \times 10^{8} m^{-3})^2}{8 \times 10^{12} m^{-3}} \] \[ n_e = \frac{25 \times 10^{16}}{8 \times 10^{12}} m^{-3} \] \[ n_e = 3.125 \times 10^{(16-12)} m^{-3} \] \[ n_e = 3.125 \times 10^{4} m^{-3} \]

Step 4: Final Answer:

The electron concentration in the extrinsic semiconductor is \(3.125 \times 10^{4} m^{-3}\). Notice that as expected for a p-type semiconductor, the minority carrier concentration (\(n_e\)) is very small compared to the majority carrier concentration (\(n_h\)).
Quick Tip: The law of mass action (\(n_e n_h = n_i^2\)) is a fundamental relationship in semiconductor physics. It shows that if you increase the concentration of one type of carrier (e.g., holes in p-type), the concentration of the other type (electrons) must decrease to keep the product constant at a given temperature.

N/A

Step 1: Understanding the Concept:

Electric current \(I\) is the rate of flow of charge \(Q\), i.e., \(I = dQ/dt\). Therefore, the total charge that passes through a cross-section in a given time interval is the integral of the current over that interval, \(Q = \int I \, dt\). For a current-time (I-t) graph, this corresponds to the area under the curve.


Step 2: Detailed Explanation:

Calculating Charge Q\(_{1}\):

The charge \(Q_1\) is the area under the I-t graph from \(t=1\) s to \(t=2\) s.

This area is a rectangle.

- Height (Current) = 2.0 A

- Width (Time interval) = \(2 s - 1 s = 1 s\)
\[ Q_1 = Area of rectangle = Height \times Width \] \[ Q_1 = 2.0 A \times 1 s = 2.0 C \]

Calculating Charge Q\(_{2}\):

The charge \(Q_2\) is the area under the I-t graph from \(t=4\) s to \(t=6\) s.

This area is a triangle.

- Base (Time interval) = \(6 s - 4 s = 2 s\)

- Height (Maximum Current) = 1.5 A
\[ Q_2 = Area of triangle = \frac{1}{2} \times Base \times Height \] \[ Q_2 = \frac{1}{2} \times 2 s \times 1.5 A = 1.5 C \]

Comparing the Charges:

We have \(Q_1 = 2.0\) C and \(Q_2 = 1.5\) C.

Clearly, \(Q_1 > Q_2\).

The ratio of the charges is:
\[ \frac{Q_1}{Q_2} = \frac{2.0}{1.5} = \frac{20}{15} = \frac{4}{3} \]

Step 3: Final Answer:

The charge passed during the first interval is \(Q_1 = 2.0\) C, and during the second interval is \(Q_2 = 1.5\) C. Thus, \(Q_1\) is greater than \(Q_2\).




\begin{quicktipbox
For any graph-based problem in physics, first identify the physical meaning of the slope and the area under the graph. For an I-t graph, the slope (\(dI/dt\)) represents the rate of change of current, and the area (\(\int I dt\)) represents the total charge passed.
\end{quicktipbox Quick Tip: For any graph-based problem in physics, first identify the physical meaning of the slope and the area under the graph. For an I-t graph, the slope (\(dI/dt\)) represents the rate of change of current, and the area (\(\int I dt\)) represents the total charge passed.

N/A

N/A

Step 1: Understanding the Concept:

The EMF (E) and internal resistance (r) are constant characteristics of a given battery. We can write the circuit equation for two different external resistances and then solve the system of two equations to find the internal resistance r.


Step 2: Setting up the Equations:

The general relationship between EMF, current, and resistances is \(E = I(R+r)\).

For the first case, when the external resistance is \(R_1\), the current is \(I_1\). So, we have:
\[ E = I_1(R_1 + r) \quad \cdots(1) \]
For the second case, when the external resistance is \(R_2\), the current is \(I_2\). So, we have:
\[ E = I_2(R_2 + r) \quad \cdots(2) \]

Step 3: Solving for Internal Resistance (r):

Since the EMF, E, is the same in both equations, we can equate the right-hand sides of (1) and (2):
\[ I_1(R_1 + r) = I_2(R_2 + r) \]
Expand the terms:
\[ I_1 R_1 + I_1 r = I_2 R_2 + I_2 r \]
Now, we collect all terms containing r on one side and all other terms on the other side.
\[ I_1 r - I_2 r = I_2 R_2 - I_1 R_1 \]
Factor out r from the left side:
\[ r(I_1 - I_2) = I_2 R_2 - I_1 R_1 \]
Finally, divide by \((I_1 - I_2)\) to isolate r:
\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]
This is the expression for the internal resistance in terms of the given quantities.




\begin{quicktipbox
A battery's terminal voltage is equal to its EMF only in an open circuit (\(I=0\)). When the battery supplies current, the terminal voltage is always less than the EMF (\(V = E - Ir\)). The maximum possible current (short-circuit current) and maximum power transfer are important concepts related to internal resistance.
\end{quicktipbox Quick Tip: A battery's terminal voltage is equal to its EMF only in an open circuit (\(I=0\)). When the battery supplies current, the terminal voltage is always less than the EMF (\(V = E - Ir\)). The maximum possible current (short-circuit current) and maximum power transfer are important concepts related to internal resistance.

Step 1: Understanding the Concept:

The Biot-Savart law is an equation in electromagnetism that describes the magnetic field generated by a constant electric current. It relates the magnetic field to the magnitude, direction, length, and proximity of the electric current.


Step 2: Detailed Explanation:

According to the Biot-Savart law, the magnetic field \(d\vec{B}\) at a point P due to a small current element \(I d\vec{l}\) of a current-carrying conductor is:

1. Directly proportional to the current \(I\).

2. Directly proportional to the length of the element \(dl\).

3. Directly proportional to the sine of the angle between the element \(d\vec{l}\) and the position vector \(\vec{r}\) from the element to the point P.

4. Inversely proportional to the square of the distance \(r\) from the element to the point P.


In vector form, this is expressed as:
\[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3} \]
where:

- \(d\vec{B}\) is the differential magnetic field vector.

- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7}\) T\(\cdot\)m/A).

- \(I\) is the current in the wire.

- \(d\vec{l}\) is the vector representing the current element.

- \(\vec{r}\) is the position vector from the current element to the point P.

- \(r\) is the magnitude of the position vector \(\vec{r}\).



\hrule


% Solution for 23(b)
23. (b)

% Correct Answer
Correct Answer: The magnitude of the resultant magnetic field is 2.0 \(\times\) 10\(^{-8}\) T, and the direction is into the plane (-z direction).



% Solution
Solution:


Step 1: Understanding the Concept:

The net magnetic field at a point due to multiple current sources is the vector sum of the magnetic fields produced by each source individually (Principle of Superposition). The magnetic field due to a long straight wire is calculated, and the directions are determined using the right-hand thumb rule.


Step 2: Key Formula or Approach:

The magnitude of the magnetic field (\(B\)) at a perpendicular distance \(d\) from an infinitely long straight wire carrying current \(I\) is given by:
\[ B = \frac{\mu_0 I}{2\pi d} \]

Step 3: Detailed Explanation:

Given data:

- Current in each wire, \(I = 2.0\) A.

- Coordinates of point P = (4 m, 5 m).


Magnetic Field due to wire along xx' axis (Wire 1):

- The current flows along the +x axis.

- The perpendicular distance of point P(4, 5) from the x-axis is \(d_1 = 5\) m.

- Magnitude of the magnetic field:
\[ B_1 = \frac{\mu_0 I}{2\pi d_1} = \frac{(4\pi \times 10^{-7} T\cdotm/A) \times 2.0 A}{2\pi \times 5 m} = \frac{4 \times 10^{-7}}{5} T = 0.8 \times 10^{-7} T \]
- Direction: Using the right-hand thumb rule, if the thumb points in the +x direction, the fingers curl out of the page at point P. So, the direction is along the +z axis (\(\hat{k}\)).
\[ \vec{B}_1 = 0.8 \times 10^{-7} \hat{k} T \]

Magnetic Field due to wire along yy' axis (Wire 2):

- The current flows along the +y axis.

- The perpendicular distance of point P(4, 5) from the y-axis is \(d_2 = 4\) m.

- Magnitude of the magnetic field:
\[ B_2 = \frac{\mu_0 I}{2\pi d_2} = \frac{(4\pi \times 10^{-7} T\cdotm/A) \times 2.0 A}{2\pi \times 4 m} = \frac{4 \times 10^{-7}}{4} T = 1.0 \times 10^{-7} T \]
- Direction: Using the right-hand thumb rule, if the thumb points in the +y direction, the fingers curl into the page at point P. So, the direction is along the -z axis (\(-\hat{k}\)).
\[ \vec{B}_2 = -1.0 \times 10^{-7} \hat{k} T \]

Resultant Magnetic Field:

The resultant magnetic field \(\vec{B}_{net}\) is the vector sum of \(\vec{B}_1\) and \(\vec{B}_2\).
\[ \vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = (0.8 \times 10^{-7} \hat{k}) + (-1.0 \times 10^{-7} \hat{k}) = -0.2 \times 10^{-7} \hat{k} T \] \[ \vec{B}_{net} = -2.0 \times 10^{-8} \hat{k} T \]

Step 4: Final Answer:

- The magnitude of the resultant magnetic field is \(|\vec{B}_{net}| = 2.0 \times 10^{-8}\) T.

- The direction is along the negative z-axis, which is into the plane of the paper.




\begin{quicktipbox
When dealing with magnetic fields from multiple wires, calculate the contribution (both magnitude and direction) from each wire separately. Then, add them as vectors. The right-hand thumb rule is your most essential tool for finding directions.
\end{quicktipbox Quick Tip: When dealing with magnetic fields from multiple wires, calculate the contribution (both magnitude and direction) from each wire separately. Then, add them as vectors. The right-hand thumb rule is your most essential tool for finding directions.

Step 1: Understanding the Concept:

This problem is based on the principles of electromagnetic induction, specifically Faraday's Law and Lenz's Law. Lenz's law states that the direction of the induced current in a conductor by a changing magnetic flux is such that the magnetic field created by the induced current opposes the change in the initial magnetic flux.


Step 2: Initial Magnetic Field Direction:

First, we determine the direction of the magnetic field produced by the primary coil (Coil 2). The current in Coil 2 flows from the positive terminal of the battery (at N) towards the negative terminal (at M) through the resistor R\(_2\). Using the right-hand grip rule for the solenoid, if we curl our fingers in the direction of the current, our thumb points towards the left. Thus, Coil 2 produces a steady magnetic field pointing to the left. This field creates a magnetic flux through Coil 1, also directed to the left.


24. (a) Coil '2' is moving towards coil '1'.

Justification: When Coil 2 moves towards Coil 1, the magnetic flux directed to the left through Coil 1 increases. According to Lenz's law, the induced current in Coil 1 must oppose this increase. To do this, it will generate its own magnetic field directed to the right.

Direction: Using the right-hand grip rule for Coil 1, to create a magnetic field to the right, the induced current must flow from point P to O through the resistor R\(_1\).


24. (b) Coil '2' is moving away from coil '1'.

Justification: When Coil 2 moves away from Coil 1, the magnetic flux directed to the left through Coil 1 decreases. According to Lenz's law, the induced current in Coil 1 must oppose this decrease (i.e., it must try to reinforce the original field). To do this, it will generate its own magnetic field directed to the left.

Direction: Using the right-hand grip rule for Coil 1, to create a magnetic field to the left, the induced current must flow from point O to P through the resistor R\(_1\).


24. (c) The resistance R\(_2\) is increased.

Justification: The current in Coil 2 is given by \(I_2 = E/R_2\). When the resistance R\(_2\) is increased, the current \(I_2\) decreases. This causes the magnetic field produced by Coil 2 (which points to the left) to decrease in strength. Consequently, the magnetic flux directed to the left through Coil 1 decreases. This is the same change in flux as in case (b). The induced current in Coil 1 will oppose this decrease by creating its own magnetic field in the same direction as the original field, i.e., to the left.

Direction: As in case (b), to create a magnetic field to the left, the induced current in Coil 1 must flow from point O to P through the resistor R\(_1\).




\begin{quicktipbox
Lenz's Law is about opposing the change in flux, not the flux itself. If flux is increasing, the induced field is opposite. If flux is decreasing, the induced field is in the same direction. Remember: "The induced effect is one of opposition to the cause."
\end{quicktipbox Quick Tip: Lenz's Law is about opposing the \textit{change in flux, not the flux itself. If flux is increasing, the induced field is opposite. If flux is decreasing, the induced field is in the same direction. Remember: "The induced effect is one of opposition to the cause."

Three characteristics of electromagnetic (EM) waves are:


Transverse Nature: The electric field vector (\(\vec{E}\)) and the magnetic field vector (\(\vec{B}\)) are mutually perpendicular to each other and also perpendicular to the direction of propagation of the wave.
Speed: EM waves travel in a vacuum with a constant speed, the speed of light, \(c \approx 3 \times 10^8\) m/s.
No Material Medium Required: EM waves do not require any material medium for their propagation and can travel through a vacuum.

Other possible characteristics include:

They are produced by accelerated charges.
They carry energy and momentum which are transported from one point to another.
The ratio of the magnitudes of the electric and magnetic fields is constant and equal to the speed of light, \(E/B = c\).



\hrule


% Solution for 25(b)
25. (b)

% Correct Answer
Correct Answer: Displacement current exists between the capacitor plates. It is caused by the time-varying electric flux (\(I_D = \epsilon_0 \frac{d\Phi_E}{dt}\)) and ensures current continuity.



% Solution
Solution:

Step 1: Understanding the Concept:

Displacement current is a concept introduced by James Clerk Maxwell. It is not a current due to the flow of charge but is associated with a time-varying electric field. It was needed to make Ampere's circuital law consistent for time-varying fields and to explain the propagation of electromagnetic waves.


Step 2: Explanation:

Where it exists:

During the charging of a capacitor, the displacement current exists in the space or dielectric medium between the two plates of the capacitor. In the connecting wires, there is a normal conduction current due to the flow of electrons.


How it exists:

1. When a capacitor is being charged, the amount of charge \(Q\) on its plates increases with time.

2. The electric field \(E\) between the plates is proportional to the charge (\(E = Q/(\epsilon_0 A)\)). As \(Q\) changes with time, the electric field \(E\) also changes with time.

3. This time-varying electric field creates a changing electric flux (\(\Phi_E = E \cdot A\)) through any surface between the plates.

4. Maxwell proposed that this changing electric flux is equivalent to a current, which he named the displacement current (\(I_D\)). The magnitude of this current is given by: \[ I_D = \epsilon_0 \frac{d\Phi_E}{dt} \]
This displacement current produces a magnetic field in the region between the plates, just as a conduction current would. It ensures that the principle of continuity of current is upheld throughout the circuit; the conduction current in the wires is exactly equal to the displacement current in the gap between the plates (\(I_C = I_D\)).




\begin{quicktipbox
Remember that displacement current is a consequence of a changing electric field, not moving charges. It completes the circuit "conceptually" across the capacitor gap, allowing Ampere's law to be applied universally.
\end{quicktipbox Quick Tip: Remember that displacement current is a consequence of a changing electric field, not moving charges. It completes the circuit "conceptually" across the capacitor gap, allowing Ampere's law to be applied universally.

Step 1: Understanding the Concept:

When light travels from one medium to another, its frequency remains constant, but its speed and wavelength change. The refractive index (\(n\)) of a medium is defined as the ratio of the speed of light in vacuum (\(c\)) to the speed of light in the medium (\(v\)), i.e., \(n = c/v\). The fringe width in a Young's Double Slit Experiment (YDSE) depends directly on the wavelength of light in the medium.

Let \(\lambda_{air}\) be the wavelength in air/vacuum. In a medium of refractive index \(n\), the wavelength becomes \(\lambda_{medium} = \lambda_{air}/n\).


26. (a) Effect on Speed, frequency and wavelength

Given: The new liquid has a refractive index \(n\) which is greater than the refractive index of water (\(n_{water}\)). So, \(n > n_{water}\).


Frequency: The frequency of light is a characteristic of the source. It does not change when the light enters a different medium. Therefore, the frequency will remain unchanged.

Speed: The speed of light in a medium is given by \(v = c/n\). Since \(n > n_{water}\), we have \(c/n < c/n_{water}\). This means the speed of light in the new liquid will be less than the speed in water. The speed will decrease.

Wavelength: The wavelength in the medium is \(\lambda_{medium} = \lambda_{air}/n\). Since \(n > n_{water}\), we have \(\lambda_{air}/n < \lambda_{air}/n_{water}\). This means the wavelength of light in the new liquid will be shorter than the wavelength in water. The wavelength will decrease.



26. (b) Effect on fringe width, shape and position of central maximum


Fringe width (\(\beta\)): The fringe width in YDSE is given by the formula \(\beta = \frac{\lambda D}{d}\), where \(\lambda\) is the wavelength of light in the medium. As established in part (a), the wavelength \(\lambda\) decreases when water is replaced by the new liquid. Since \(\beta \propto \lambda\), the fringe width will decrease. The fringes will become closer to each other.

Shape of interference fringes: The shape of the fringes depends on the geometry of the source slits. For two straight parallel slits, the fringes are hyperbolic in shape, which are approximately straight parallel bands near the center of the screen. Changing the medium does not alter the geometry of the experiment. Therefore, the shape of the fringes will remain unchanged.

Shift in the position of central maximum: The central maximum is the point on the screen where the path difference from the two slits is zero. As the entire setup is immersed in the new liquid, the optical paths from both slits to the geometric center of the screen are still equal (\(n \times S_1P = n \times S_2P\)). No additional path difference is introduced. Therefore, the position of the central maximum will not shift.





\begin{quicktipbox
A key takeaway for YDSE in a medium: frequency is constant, while wavelength \(\lambda\) and fringe width \(\beta\) are inversely proportional to the refractive index \(n\). If a thin transparent sheet is placed in front of only one slit, the entire pattern, including the central maximum, will shift. But if the entire medium is changed, there is no shift.
\end{quicktipbox Quick Tip: A key takeaway for YDSE in a medium: frequency is constant, while wavelength \(\lambda\) and fringe width \(\beta\) are inversely proportional to the refractive index \(n\). If a thin transparent sheet is placed in front of only one slit, the entire pattern, including the central maximum, will shift. But if the entire medium is changed, there is no shift.

Step 1: Einstein's Photoelectric Equation:

The foundation for explaining all these phenomena is Einstein's photoelectric equation, which expresses the conservation of energy in the photoelectric effect. When a photon of energy \(h\nu\) strikes an electron in a metal, a part of its energy is used to overcome the work function (\(\phi_0\)) of the metal, and the rest is converted into the kinetic energy of the photoelectron. The equation is:
\[ K_{max} = h\nu - \phi_0 \]
where \(K_{max}\) is the maximum kinetic energy of the emitted electron, \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi_0\) is the work function of the metal.


27. (a) Existence of Threshold Frequency

Explanation: For photoelectric emission to occur, the emitted electron must have a positive kinetic energy, i.e., \(K_{max} > 0\).

From Einstein's equation, this means:
\[ h\nu - \phi_0 > 0 \implies h\nu > \phi_0 \]
This implies that the energy of the incident photon (\(h\nu\)) must be greater than the work function (\(\phi_0\)).

If we define the threshold frequency (\(\nu_0\)) as the minimum frequency required to cause emission, then at this frequency, the photon energy is just equal to the work function:
\[ h\nu_0 = \phi_0 \implies \nu_0 = \frac{\phi_0}{h} \]
Therefore, for any emission to happen, the condition is \(\nu > \nu_0\). If the frequency of the incident light is less than this certain minimum value \(\nu_0\) (the threshold frequency), no matter how intense the light is, photoelectric emission will not occur because no single photon has enough energy to liberate an electron.


27. (b) Effect of Frequency and Intensity on K\(_{max}\)

Explanation:

Effect of Frequency: Einstein's equation, \(K_{max} = h\nu - \phi_0\), shows a direct linear relationship between the maximum kinetic energy (\(K_{max}\)) and the frequency of incident light (\(\nu\)). As the frequency \(\nu\) increases (and is above the threshold frequency), the energy of each photon increases, and thus the maximum kinetic energy of the emitted photoelectrons increases.

Effect of Intensity: The intensity of light is related to the number of photons incident per unit area per unit time. Increasing the intensity means increasing the number of photons, but it does not change the energy of each individual photon (\(h\nu\)). According to the one-photon, one-electron interaction model, each electron absorbs a single photon. Therefore, increasing the number of photons will increase the number of photoelectrons emitted (the photoelectric current), but it will not change the maximum kinetic energy with which any single electron is emitted. Thus, \(K_{max\) is independent of the intensity.


27. (c) The V\(_0\) vs \(\nu\) Graph

Explanation: The cut-off voltage or stopping potential (\(V_0\)) is the minimum negative potential applied to the collector plate that is sufficient to stop even the most energetic photoelectrons from reaching it. This means the work done by the stopping potential on the electron is equal to its maximum kinetic energy:
\[ K_{max} = e V_0 \]
where \(e\) is the charge of an electron.

Substituting this into Einstein's photoelectric equation:
\[ e V_0 = h\nu - \phi_0 \]
Rearranging this equation to express \(V_0\) as a function of \(\nu\):
\[ V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \]
This equation is in the form of a straight line, \(y = mx + c\), where:

- The y-variable is the cut-off voltage \(V_0\).

- The x-variable is the frequency \(\nu\).

- The slope of the line is \(m = \frac{h}{e}\).

- The y-intercept is \(c = -\frac{\phi_0}{e}\).

Since \(h\) (Planck's constant) and \(e\) (electron charge) are universal constants, the slope of the \(V_0\) versus \(\nu\) graph is a constant value \(\frac{h}{e}\) for any metal surface.



\begin{quicktipbox
Einstein's photoelectric equation is a cornerstone of quantum mechanics. Memorize it and understand how it explains each experimental observation: threshold frequency (energy conservation), \(K_{max}\) vs. intensity (particle nature), and the linear \(V_0\)-\(\nu\) relationship.
\end{quicktipbox Quick Tip: Einstein's photoelectric equation is a cornerstone of quantum mechanics. Memorize it and understand how it explains each experimental observation: threshold frequency (energy conservation), \(K_{max}\) vs. intensity (particle nature), and the linear \(V_0\)-\(\nu\) relationship.

Step 1: Understanding Doping:

Extrinsic semiconductors are created by doping an intrinsic semiconductor. The type of dopant determines which charge carrier becomes more numerous.


n-type semiconductor: Formed by doping with pentavalent impurities (e.g., Phosphorus in Silicon). These "donor" atoms provide extra free electrons.
p-type semiconductor: Formed by doping with trivalent impurities (e.g., Boron in Silicon). These "acceptor" atoms create vacancies or holes.

Step 2: Identifying Carriers:


In an n-type semiconductor: The concentration of free electrons (\(n_e\)) is much higher than the concentration of holes (\(n_h\)). Therefore:

Majority charge carriers are electrons.
Minority charge carriers are holes.

In a p-type semiconductor: The concentration of holes (\(n_h\)) is much higher than the concentration of free electrons (\(n_e\)). Therefore:

Majority charge carriers are holes.
Minority charge carriers are electrons.




\hrule


% Solution for 28(b)
28. (b)

% Correct Answer
Correct Answer: Diffusion current is due to the movement of majority carriers across the junction due to a concentration gradient. Drift current is due to the movement of minority carriers under the influence of the depletion region's electric field.



% Solution
Solution:

Step 1: Formation of a p-n Junction:

When a p-type semiconductor is brought into contact with an n-type semiconductor, a p-n junction is formed. Due to the difference in concentration of charge carriers on the two sides, two processes begin simultaneously: diffusion and drift.


Step 2: Formation of Diffusion Current:


There is a high concentration of holes on the p-side and a low concentration of holes on the n-side.
There is a high concentration of electrons on the n-side and a low concentration of electrons on the p-side.
Due to this concentration gradient, majority carriers start to move across the junction. Holes diffuse from the p-side to the n-side, and electrons diffuse from the n-side to the p-side.
This net flow of charge carriers due to the concentration difference constitutes the diffusion current. The direction of conventional diffusion current is from the p-side to the n-side (in the direction of hole movement).


Step 3: Formation of Drift Current:


As electrons diffuse from n to p and holes diffuse from p to n, they leave behind uncovered positive ions (donors) on the n-side and uncovered negative ions (acceptors) on the p-side near the junction.
This layer of fixed positive and negative ions creates a region devoid of mobile charge carriers, known as the depletion region or space-charge region.
Within this depletion region, an internal electric field is established, pointing from the positive ions on the n-side to the negative ions on the p-side (i.e., from n to p).
This electric field exerts a force on the minority charge carriers. It pushes minority electrons from the p-side towards the n-side and minority holes from the n-side towards the p-side.
This movement of minority carriers under the influence of the internal electric field constitutes the drift current. The direction of conventional drift current is from the n-side to the p-side, opposite to the diffusion current.

In an unbiased p-n junction at equilibrium, the diffusion current and the drift current are equal in magnitude and opposite in direction, resulting in a net current of zero across the junction.




\begin{quicktipbox
A helpful analogy for diffusion vs. drift:

Diffusion: Like a drop of ink spreading out in water due to concentration differences. It's a natural tendency.
Drift: Like leaves being carried along by a river's current (the electric field). It's a forced movement.

In equilibrium, these two opposing flows balance each other perfectly.
\end{quicktipbox Quick Tip: A helpful analogy for diffusion vs. drift: \textbf{Diffusion:} Like a drop of ink spreading out in water due to concentration differences. It's a natural tendency. \textbf{Drift:} Like leaves being carried along by a river's current (the electric field). It's a forced movement. In equilibrium, these two opposing flows balance each other perfectly.

Step 1: Understanding the Concept:

The given arrangement of three plates can be viewed as a combination of two separate capacitors. We need to identify these capacitors and determine if they are connected in series or parallel to find the equivalent capacitance of the system.


Step 2: Key Formula or Approach:

1. The capacitance of a parallel plate capacitor with plate area A, separation d, and filled with a dielectric of constant K is \(C = \frac{K \epsilon_0 A}{d}\).

2. The equivalent capacitance for capacitors in parallel is \(C_{eq} = C_1 + C_2 + \ldots\).


Step 3: Detailed Explanation:

The system forms two capacitors:

- Capacitor C\(_1\) is formed by plates P\(_1\) and P\(_2\).
- Capacitor C\(_2\) is formed by plates P\(_2\) and P\(_3\).
The plate area is \(A = L^2\) and the separation is \(d\). Both are filled with a dielectric K.

The capacitance of each is: \[ C_1 = C_2 = \frac{K \epsilon_0 L^2}{d} \]
Now, let's look at the connections:

- Plate P\(_2\) is connected to point A.
- Plates P\(_1\) and P\(_3\) are connected to point B.
This means C\(_1\) is connected between A and B, and C\(_2\) is also connected between A and B. Therefore, the two capacitors are in a parallel combination.

The equivalent capacitance of the system is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 = \frac{K \epsilon_0 L^2}{d} + \frac{K \epsilon_0 L^2}{d} \] \[ C_{eq} = \frac{2 K \epsilon_0 L^2}{d} \]

Step 4: Final Answer:

The total capacitance of the system is the sum of the two parallel capacitors, which is \(\frac{2\epsilon_0 KL^2}{d}\).

This corresponds to option (C).
Quick Tip: To analyze multi-plate capacitor arrangements, always identify pairs of adjacent plates as individual capacitors. Then, trace the wiring to see if they are connected in series (end-to-end) or parallel (across the same two points).

Step 1: Understanding the Concept:

The charge on a capacitor plate is given by the product of its capacitance and the potential difference across it (\(Q = CV\)). We need to find the charge on plate P\(_1\), which is one of the plates of capacitor C\(_1\).


Step 2: Key Formula or Approach:

1. Identify the capacitor to which plate P\(_1\) belongs (C\(_1\)).

2. Determine the potential difference across C\(_1\).

3. Use the formula \(Q = CV\) to calculate the charge.


Step 3: Detailed Explanation:

Plate P\(_1\) forms capacitor C\(_1\) with plate P\(_2\). As determined in the previous question, C\(_1\) is connected directly between points A and B.

The potential difference across capacitor C\(_1\) is therefore equal to the potential difference between A and B, which is V.

The capacitance of C\(_1\) is:
\[ C_1 = \frac{K \epsilon_0 L^2}{d} \]
The magnitude of the charge on the plates of capacitor C\(_1\) is:
\[ Q_1 = C_1 V = \left( \frac{K \epsilon_0 L^2}{d} \right) V = \frac{\epsilon_0 V K L^2}{d} \]
Point A (connected to P\(_2\)) is at a positive potential with respect to point B (connected to P\(_1\)). This means the inner surface of plate P\(_2\) will have a charge of \(+Q_1\) and the inner surface of plate P\(_1\) will have a charge of \(-Q_1\).

The question asks for the charge on plate P\(_1\). The options are given as magnitudes, so we consider the magnitude of the charge.


Step 4: Final Answer:

The magnitude of the charge on plate P\(_1\) is \(Q_1 = \frac{\epsilon_0 V K L^2}{d}\).

This corresponds to option (B).
Quick Tip: For capacitors in parallel, the voltage across each capacitor is the same. For capacitors in series, the charge on each capacitor is the same. Use this fundamental rule to quickly find the charge or voltage for any component in a combination.

Step 1: Understanding the Concept:

For a parallel plate capacitor, there is a uniform electric field in the region between the plates. The magnitude of this field is related to the potential difference across the plates and the distance between them.


Step 2: Key Formula or Approach:

The relationship between the uniform electric field (E), potential difference (V), and plate separation (d) is:
\[ E = \frac{V}{d} \]

Step 3: Detailed Explanation:

The region between P\(_1\) and P\(_2\) corresponds to capacitor C\(_1\).

- The potential difference across the plates P\(_1\) and P\(_2\) is the potential difference between points A and B, which is given as V.

- The separation between the plates P\(_1\) and P\(_2\) is given as d.

Substituting these values into the formula:
\[ E = \frac{V}{d} \]

Step 4: Final Answer:

The electric field in the region between P\(_1\) and P\(_2\) is \(V/d\).

This corresponds to option (A).
Quick Tip: Remember the formula \(E = V/d\) is valid for a uniform electric field, which is a good approximation for the field inside a parallel plate capacitor, away from the edges. This relationship is fundamental in electrostatics.

Step 1: Understanding the Concept:

This question asks us to find the plate separation of an equivalent air-filled capacitor. We need to equate the capacitance of this new capacitor to the equivalent capacitance of the original system that we calculated in part (i).


Step 2: Key Formula or Approach:

1. Capacitance of the original system: \(C_{eq} = \frac{2 K \epsilon_0 L^2}{d}\).

2. Capacitance of an air-filled capacitor (\(K_{air}=1\)) with area \(L^2\) and new separation \(d_{new}\) is \(C_{air} = \frac{\epsilon_0 L^2}{d_{new}}\).

3. Set \(C_{air} = C_{eq}\) and solve for \(d_{new}\).


Step 3: Detailed Explanation:

We are given that the new capacitor must have the same capacitance as the original system.
\[ C_{air} = C_{eq} \]
Substitute the expressions for the capacitances:
\[ \frac{\epsilon_0 L^2}{d_{new}} = \frac{2 K \epsilon_0 L^2}{d} \]
We can cancel the term \(\epsilon_0 L^2\) from both sides of the equation.
\[ \frac{1}{d_{new}} = \frac{2K}{d} \]
Now, we solve for the new separation, \(d_{new}\):
\[ d_{new} = \frac{d}{2K} \]

Step 4: Final Answer:

The separation of the equivalent air capacitor would be \(\frac{d}{2K}\).

This corresponds to option (C).
Quick Tip: Introducing a dielectric of constant K into a capacitor increases its capacitance by a factor of K. This means to get the same capacitance with air (K=1), you would need to change the geometry, for example, by drastically reducing the plate separation.

Step 1: Understanding the Concept:

Initially, the capacitor system is charged by an external source. This process involves separating charges: the source pulls negative charge from one terminal and deposits positive charge on the other. The system of conducting plates itself remains electrically neutral overall. When the terminals are connected, the separated charges are allowed to recombine.


Step 2: Detailed Explanation:

1. Charging Process: When the potential V is applied, the source pulls electrons from plates P\(_1\) and P\(_3\) (connected to B) and deposits them on plate P\(_2\) (connected to A). This is a conceptual view; in reality, electrons move from B to A. This leaves P\(_1\) and P\(_3\) with a net positive charge and P\(_2\) with a net negative charge. As A is at a positive potential relative to B, plate P\(_2\) acquires a positive charge, say \(+Q_{total}\), and plates P\(_1\) and P\(_3\) together acquire a total negative charge of \(-Q_{total}\). The entire three-plate system, if considered isolated, has a net charge of \((+Q_{total}) + (-Q_{total}) = 0\). The battery only separates the existing charges.

2. Removing the Source: After the system is charged, the source is disconnected. The charges \(+Q_{total}\) on P\(_2\) and \(-Q_{total}\) on (P\(_1\)+P\(_3\)) are now isolated and stored.

3. Connecting A and B: A and B are then connected by a conducting wire. This means plate P\(_2\) is now electrically connected to plates P\(_1\) and P\(_3\). This creates a path for the excess positive charge on P\(_2\) to flow and neutralize the excess negative charge on P\(_1\) and P\(_3\).

4. Final State: The charges will flow until the potential is uniform everywhere on the connected conductors (P\(_1\), P\(_2\), and P\(_3\)). Since the total net charge of the entire isolated system was zero initially, after recombination, the net charge on the system will also be Zero.


Step 3: Final Answer:

Connecting the terminals A and B with a wire short-circuits the capacitor system, allowing the stored positive and negative charges to neutralize each other. The final net charge on the system will be zero.

This corresponds to option (D).
Quick Tip: A capacitor does not create charge; it only stores separated charge. The total charge on an isolated capacitor (sum of charges on both plates) is always zero. Short-circuiting a charged capacitor allows the separated charges to recombine, resulting in zero stored charge and zero potential difference.

Step 1: Understanding the Concept:

The given arrangement of three plates can be viewed as a combination of two separate capacitors. We need to identify these capacitors and determine if they are connected in series or parallel to find the equivalent capacitance of the system.


Step 2: Key Formula or Approach:

1. The capacitance of a parallel plate capacitor with plate area A, separation d, and filled with a dielectric of constant K is \(C = \frac{K \epsilon_0 A}{d}\).

2. The equivalent capacitance for capacitors in parallel is \(C_{eq} = C_1 + C_2 + \ldots\).


Step 3: Detailed Explanation:

The system forms two capacitors:

- Capacitor C\(_1\) is formed by plates P\(_1\) and P\(_2\).
- Capacitor C\(_2\) is formed by plates P\(_2\) and P\(_3\).
The plate area is \(A = L^2\) and the separation is \(d\). Both are filled with a dielectric K.

The capacitance of each is: \[ C_1 = C_2 = \frac{K \epsilon_0 L^2}{d} \]
Now, let's look at the connections:

- Plate P\(_2\) is connected to point A.
- Plates P\(_1\) and P\(_3\) are connected to point B.
This means C\(_1\) is connected between A and B, and C\(_2\) is also connected between A and B. Therefore, the two capacitors are in a parallel combination.

The equivalent capacitance of the system is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 = \frac{K \epsilon_0 L^2}{d} + \frac{K \epsilon_0 L^2}{d} \] \[ C_{eq} = \frac{2 K \epsilon_0 L^2}{d} \]

Step 4: Final Answer:

The total capacitance of the system is the sum of the two parallel capacitors, which is \(\frac{2\epsilon_0 KL^2}{d}\).

This corresponds to option (C).
Quick Tip: To analyze multi-plate capacitor arrangements, always identify pairs of adjacent plates as individual capacitors. Then, trace the wiring to see if they are connected in series (end-to-end) or parallel (across the same two points).

Step 1: Understanding the Concept:

The charge on a capacitor plate is given by the product of its capacitance and the potential difference across it (\(Q = CV\)). We need to find the charge on plate P\(_1\), which is one of the plates of capacitor C\(_1\).


Step 2: Key Formula or Approach:

1. Identify the capacitor to which plate P\(_1\) belongs (C\(_1\)).

2. Determine the potential difference across C\(_1\).

3. Use the formula \(Q = CV\) to calculate the charge.


Step 3: Detailed Explanation:

Plate P\(_1\) forms capacitor C\(_1\) with plate P\(_2\). As determined in the previous question, C\(_1\) is connected directly between points A and B.

The potential difference across capacitor C\(_1\) is therefore equal to the potential difference between A and B, which is V.

The capacitance of C\(_1\) is:
\[ C_1 = \frac{K \epsilon_0 L^2}{d} \]
The magnitude of the charge on the plates of capacitor C\(_1\) is:
\[ Q_1 = C_1 V = \left( \frac{K \epsilon_0 L^2}{d} \right) V = \frac{\epsilon_0 V K L^2}{d} \]
Point A (connected to P\(_2\)) is at a positive potential with respect to point B (connected to P\(_1\)). This means the inner surface of plate P\(_2\) will have a charge of \(+Q_1\) and the inner surface of plate P\(_1\) will have a charge of \(-Q_1\).

The question asks for the charge on plate P\(_1\). The options are given as magnitudes, so we consider the magnitude of the charge.


Step 4: Final Answer:

The magnitude of the charge on plate P\(_1\) is \(Q_1 = \frac{\epsilon_0 V K L^2}{d}\).

This corresponds to option (B).
Quick Tip: For capacitors in parallel, the voltage across each capacitor is the same. For capacitors in series, the charge on each capacitor is the same. Use this fundamental rule to quickly find the charge or voltage for any component in a combination.

Step 1: Understanding the Concept:

For a parallel plate capacitor, there is a uniform electric field in the region between the plates. The magnitude of this field is related to the potential difference across the plates and the distance between them.


Step 2: Key Formula or Approach:

The relationship between the uniform electric field (E), potential difference (V), and plate separation (d) is:
\[ E = \frac{V}{d} \]

Step 3: Detailed Explanation:

The region between P\(_1\) and P\(_2\) corresponds to capacitor C\(_1\).

- The potential difference across the plates P\(_1\) and P\(_2\) is the potential difference between points A and B, which is given as V.

- The separation between the plates P\(_1\) and P\(_2\) is given as d.

Substituting these values into the formula:
\[ E = \frac{V}{d} \]

Step 4: Final Answer:

The electric field in the region between P\(_1\) and P\(_2\) is \(V/d\).

This corresponds to option (A).
Quick Tip: Remember the formula \(E = V/d\) is valid for a uniform electric field, which is a good approximation for the field inside a parallel plate capacitor, away from the edges. This relationship is fundamental in electrostatics.

Step 1: Understanding the Concept:

This question asks us to find the plate separation of an equivalent air-filled capacitor. We need to equate the capacitance of this new capacitor to the equivalent capacitance of the original system that we calculated in part (i).


Step 2: Key Formula or Approach:

1. Capacitance of the original system: \(C_{eq} = \frac{2 K \epsilon_0 L^2}{d}\).

2. Capacitance of an air-filled capacitor (\(K_{air}=1\)) with area \(L^2\) and new separation \(d_{new}\) is \(C_{air} = \frac{\epsilon_0 L^2}{d_{new}}\).

3. Set \(C_{air} = C_{eq}\) and solve for \(d_{new}\).


Step 3: Detailed Explanation:

We are given that the new capacitor must have the same capacitance as the original system.
\[ C_{air} = C_{eq} \]
Substitute the expressions for the capacitances:
\[ \frac{\epsilon_0 L^2}{d_{new}} = \frac{2 K \epsilon_0 L^2}{d} \]
We can cancel the term \(\epsilon_0 L^2\) from both sides of the equation.
\[ \frac{1}{d_{new}} = \frac{2K}{d} \]
Now, we solve for the new separation, \(d_{new}\):
\[ d_{new} = \frac{d}{2K} \]

Step 4: Final Answer:

The separation of the equivalent air capacitor would be \(\frac{d}{2K}\).

This corresponds to option (C).
Quick Tip: Introducing a dielectric of constant K into a capacitor increases its capacitance by a factor of K. This means to get the same capacitance with air (K=1), you would need to change the geometry, for example, by drastically reducing the plate separation.

Step 1: Understanding the Concept:

Initially, the capacitor system is charged by an external source. This process involves separating charges: the source pulls negative charge from one terminal and deposits positive charge on the other. The system of conducting plates itself remains electrically neutral overall. When the terminals are connected, the separated charges are allowed to recombine.


Step 2: Detailed Explanation:

1. Charging Process: When the potential V is applied, the source separates charge. Plate P\(_2\) (connected to A) acquires a total positive charge of \(+Q_{total}\), and plates P\(_1\) and P\(_3\) (connected to B) together acquire a total negative charge of \(-Q_{total}\). The entire three-plate system, if considered as an isolated unit, has a net charge of \((+Q_{total}) + (-Q_{total}) = 0\). The battery only moves pre-existing charges from one part of the system to another.

2. Removing the Source: After the system is charged, the source is disconnected. The charges \(+Q_{total}\) on P\(_2\) and \(-Q_{total}\) on (P\(_1\)+P\(_3\)) are now isolated and stored.

3. Connecting A and B: A and B are then connected by a conducting wire. This means plate P\(_2\) is now electrically connected to plates P\(_1\) and P\(_3\). This creates a path for the excess positive charge on P\(_2\) to flow and neutralize the excess negative charge on P\(_1\) and P\(_3\).

4. Final State: The charges will flow until the potential is uniform everywhere on the connected conductors (P\(_1\), P\(_2\), and P\(_3\)). Since the total net charge of the entire isolated system was zero initially, after recombination, the net charge on the system will also be Zero.


Step 3: Final Answer:

Connecting the terminals A and B with a wire short-circuits the capacitor system, allowing the stored positive and negative charges to neutralize each other. The final net charge on the system will be zero.

This corresponds to option (D).
Quick Tip: A capacitor does not create charge; it only stores separated charge. The total charge on an isolated capacitor (sum of charges on both plates) is always zero. Short-circuiting a charged capacitor allows the separated charges to recombine, resulting in zero stored charge and zero potential difference.

Step 1: Understanding the Concept:

According to the problem description, for the electron to maintain a stable circular orbit, the electrostatic force of attraction exerted by the proton on the electron must be equal to the centripetal force required for the circular motion.


Step 2: Key Formula or Approach:

1. Electrostatic Force (Coulomb's Law): \(F_e = K \frac{|q_1 q_2|}{r^2}\). For a proton (\(+e\)) and an electron (\(-e\)), this is \(F_e = \frac{Ke^2}{r^2}\).

2. Centripetal Force: \(F_c = \frac{mv^2}{r}\).

3. Equate the forces: \(F_c = F_e\).


Step 3: Detailed Explanation:

Setting the centripetal force equal to the electrostatic force:
\[ \frac{mv^2}{r} = \frac{Ke^2}{r^2} \]
We need to solve this equation for the speed, v.

Multiply both sides by r:
\[ mv^2 = \frac{Ke^2}{r} \]
Divide both sides by m:
\[ v^2 = \frac{Ke^2}{mr} \]
Take the square root of both sides to find v:
\[ v = \sqrt{\frac{Ke^2}{mr}} \]

Step 4: Final Answer:

The expression for the speed of the electron is \(\sqrt{\frac{Ke^2}{mr}}\).

This corresponds to option (C).
Quick Tip: This force-balance equation is the starting point for many derivations in the Bohr model. Remembering that the electrostatic force provides the centripetal force is key to solving for orbital speed, radius, and energy.

Step 1: Understanding the Concept:

The total energy (E) of the electron in its orbit is the sum of its kinetic energy (KE) and its electrostatic potential energy (PE).


Step 2: Key Formula or Approach:

1. Total Energy: \(E = KE + PE\).

2. Kinetic Energy: \(KE = \frac{1}{2}mv^2\).

3. Potential Energy: \(PE = \frac{Kq_1q_2}{r} = \frac{K(e)(-e)}{r} = -\frac{Ke^2}{r}\).


Step 3: Detailed Explanation:

First, let's find the kinetic energy in terms of r. From the force balance in the previous question, we found:
\[ mv^2 = \frac{Ke^2}{r} \]
The kinetic energy is half of this value:
\[ KE = \frac{1}{2}mv^2 = \frac{1}{2} \left( \frac{Ke^2}{r} \right) = \frac{Ke^2}{2r} \]
The potential energy of the electron-proton system is:
\[ PE = -\frac{Ke^2}{r} \]
Now, we find the total energy by adding KE and PE:
\[ E = KE + PE = \frac{Ke^2}{2r} + \left( -\frac{Ke^2}{r} \right) \] \[ E = \frac{Ke^2}{2r} - \frac{2Ke^2}{2r} = -\frac{Ke^2}{2r} \]

Step 4: Final Answer:

The expression for the total energy of the electron is \(-\frac{Ke^2}{2r}\).

This corresponds to option (B).
Quick Tip: For circular orbits under an inverse-square force like gravity or electrostatics, remember these useful relations (Virial Theorem): Total Energy \(E = -KE\) and Total Energy \(E = \frac{1}{2}PE\). This can save a lot of time in calculations.

Step 1: Understanding the Concept:

When an electron transitions to a lower energy state, a photon is emitted. The energy of this photon is equal to the decrease in the total energy of the electron. The change in any quantity is defined as (Final Value - Initial Value). We also need to find the corresponding change in the electron's kinetic energy.


Step 2: Key Formula or Approach:

1. Energy of an emitted photon: \(E_{photon} = E_{initial} - E_{final} = \frac{hc}{\lambda}\).

2. Change in total energy: \(\Delta E_{total} = E_{final} - E_{initial}\). Therefore, \(\Delta E_{total} = -E_{photon}\).

3. Relationship between total energy and kinetic energy (from part ii): \(E_{total} = -KE\).

4. Change in kinetic energy: \(\Delta KE = KE_{final} - KE_{initial}\). From the relation \(E_{total} = -KE\), we get \(\Delta E_{total} = -\Delta KE\).


Step 3: Detailed Explanation:

Calculate Photon Energy:

First, calculate the energy of the emitted photon in eV. A useful formula for this is:
\[ E (in eV) = \frac{1240 eV\cdotnm}{\lambda (in nm)} \]
Given \(\lambda = 500\) nm:
\[ E_{photon} = \frac{1240}{500} eV = 2.48 eV \]
Calculate Change in Total Energy (\(\Delta E_{total}\)):

Since a photon is emitted, the electron moves to a lower energy state, so the total energy of the electron decreases. The change in total energy is equal to the negative of the emitted photon's energy.
\[ \Delta E_{total} = E_{final} - E_{initial} = -E_{photon} = -2.48 eV \]
Calculate Change in Kinetic Energy (\(\Delta KE\)):

We know that for the Bohr model, \(E_{total} = -KE\). Therefore, a change in total energy is related to the change in kinetic energy by:
\[ \Delta E_{total} = \Delta(-KE) = -(KE_{final} - KE_{initial}) = -\Delta KE \]
So, \(\Delta KE = -\Delta E_{total}\).
\[ \Delta KE = -(-2.48 eV) = +2.48 eV \]
This makes physical sense: as the electron moves to a lower total energy level, it gets closer to the nucleus (smaller r), moves faster, and thus its kinetic energy increases.


Step 4: Final Answer:

The change in total energy is -2.48 eV, and the change in kinetic energy is +2.48 eV.

This corresponds to option (C).
Quick Tip: When a photon is emitted, the atom's total energy decreases, so \(\Delta E_{total}\) is negative. Conversely, when a photon is absorbed, the atom's total energy increases, and \(\Delta E_{total}\) is positive. Always check the signs.

Step 1: Understanding the Concept:

The frequency of revolution (\(f\)) of an electron in a circular orbit is the number of revolutions it completes per second. It is related to the electron's speed (\(v\)) and the radius of its orbit (\(r\)) by the formula \(f = \frac{v}{2\pi r}\). To find how the frequency depends on the principal quantum number (\(n\)), we need to know the dependencies of \(v\) and \(r\) on \(n\) from Bohr's model.


Step 2: Key Formula or Approach:

1. In the Bohr model for the hydrogen atom, the radius of the n\(^{th}\) orbit is proportional to \(n^2\):
\[ r_n \propto n^2 \]
2. The speed of the electron in the n\(^{th}\) orbit is inversely proportional to \(n\):
\[ v_n \propto \frac{1}{n} \]
3. The frequency of revolution is \(f_n = \frac{v_n}{2\pi r_n}\).


Step 3: Detailed Explanation:

Using the relation for frequency, we can find its proportionality with n:
\[ f_n \propto \frac{v_n}{r_n} \]
Now, substitute the proportionalities for \(v_n\) and \(r_n\):
\[ f_n \propto \frac{1/n}{n^2} \] \[ f_n \propto \frac{1}{n \cdot n^2} = \frac{1}{n^3} \]
Thus, the frequency of revolution of the electron is proportional to the inverse cube of the principal quantum number.


Step 4: Final Answer:

The frequency of revolution is proportional to \(1/n^3\).

This corresponds to option (D).
Quick Tip: For the Bohr model of hydrogen, it's useful to memorize the proportionalities of key quantities with the principal quantum number \(n\): Radius \(r \propto n^2\), Velocity \(v \propto 1/n\), Energy \(E \propto 1/n^2\), and Frequency \(f \propto 1/n^3\). These are frequently tested.

Step 1: Understanding the Concept:

In the Bohr model of the hydrogen atom, both the energy and the radius of an electron's orbit are quantized, meaning they can only take on discrete values determined by the principal quantum number, \(n\). We need to identify the initial and final quantum numbers from the given energy levels and then calculate the change in the orbital radius.


Step 2: Key Formula or Approach:

1. 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.

2. The radius of the n\(^{th}\) orbit is given by \(r_n = n^2 \times r_1\), where \(r_1\) is the radius of the ground state orbit (the Bohr radius, approximately 0.53 \AA).


Step 3: Detailed Explanation:

Identify the quantum numbers:

- The final state is the ground state, which corresponds to \(n_{final} = 1\). The energy of the ground state is \(E_1 = -\frac{13.6}{1^2} = -13.6\) eV.

- The initial state has an energy of \(E_{initial} = -3.4\) eV. We can find the initial quantum number \(n_{initial}\) using the energy formula:
\[ E_{initial} = -\frac{13.6}{n_{initial}^2} \] \[ -3.4 = -\frac{13.6}{n_{initial}^2} \] \[ n_{initial}^2 = \frac{-13.6}{-3.4} = 4 \] \[ n_{initial} = 2 \]
So, the transition is from the first excited state (\(n=2\)) to the ground state (\(n=1\)).


Calculate the radii:

- Given the ground state radius, \(r_1 = 0.53\) \AA.

- The radius of the initial orbit (\(n=2\)) is:
\[ r_2 = (2)^2 \times r_1 = 4 \times 0.53 \AA = 2.12 \AA \]
Calculate the change in radius:

The change in the radius of the orbit is the difference between the initial and final radii.
\[ \Delta r = r_{initial} - r_{final} = r_2 - r_1 \] \[ \Delta r = 2.12 \AA - 0.53 \AA = 1.59 \AA \]

Step 4: Final Answer:

The radius of the orbit changes by 1.59 \AA.

This corresponds to option (C).
Quick Tip: The energy levels of hydrogen are --13.6 eV, --3.4 eV, --1.51 eV, etc. It's helpful to recognize that --3.4 eV corresponds to the n=2 level directly, which can speed up problem-solving.

N/A

Step 1: Understanding the Concept:

The work done by an external agent in moving a system of charges from one configuration to another is equal to the change in the electrostatic potential energy of the system. \[ W = \Delta U = U_{final} - U_{initial} \]

Step 2: Key Formula or Approach:

The potential energy of a system of three point charges \(q_1, q_2, q_3\) is: \[ U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right) \]

Step 3: Calculating Initial Potential Energy (\(U_{initial}\)):

Initial Configuration: Charges are at vertices A, B, C of an equilateral triangle with side \(s = 0.2\) m.
- \(q_A = -2 nC = -2 \times 10^{-9} C\)
- \(q_B = -1 nC = -1 \times 10^{-9} C\)
- \(q_C = +5 nC = 5 \times 10^{-9} C\)
- \(r_{AB} = r_{BC} = r_{CA} = s = 0.2\) m \[ U_{initial} = \frac{1}{s} \frac{1}{4\pi\epsilon_0} (q_A q_B + q_B q_C + q_C q_A) \]
Let's calculate the products of charges:
- \(q_A q_B = (-2)(-1) \times 10^{-18} = 2 \times 10^{-18} C^2\)
- \(q_B q_C = (-1)(5) \times 10^{-18} = -5 \times 10^{-18} C^2\)
- \(q_C q_A = (5)(-2) \times 10^{-18} = -10 \times 10^{-18} C^2\)
Sum of products = \((2 - 5 - 10) \times 10^{-18} = -13 \times 10^{-18} C^2\). \[ U_{initial} = \frac{1}{0.2} (9 \times 10^9) (-13 \times 10^{-18}) = 5 \times (9 \times 10^9) \times (-13 \times 10^{-18}) \] \[ U_{initial} = -585 \times 10^{-9} J = -5.85 \times 10^{-7} J \]

Step 4: Calculating Final Potential Energy (\(U_{final}\)):

Final Configuration: Charges \(q_A, q_B, q_C\) are at points A\(_1\), B\(_1\), C\(_1\) respectively. A\(_1\), B\(_1\), C\(_1\) are midpoints of AB, BC, CA. These points form another equilateral triangle with side \(s' = s/2 = 0.1\) m. \[ U_{final} = \frac{1}{s'} \frac{1}{4\pi\epsilon_0} (q_A q_B + q_B q_C + q_C q_A) \]
The sum of products of charges is the same as before. \[ U_{final} = \frac{1}{0.1} (9 \times 10^9) (-13 \times 10^{-18}) = 10 \times (9 \times 10^9) \times (-13 \times 10^{-18}) \] \[ U_{final} = -1170 \times 10^{-9} J = -11.7 \times 10^{-7} J \]

Step 5: Calculating Work Done:
\[ W = U_{final} - U_{initial} = (-11.7 \times 10^{-7}) - (-5.85 \times 10^{-7}) \] \[ W = (-11.7 + 5.85) \times 10^{-7} J \] \[ W = -5.85 \times 10^{-7} J \]



\begin{quicktipbox
Work done in rearranging a system of charges depends only on the initial and final potential energies of the configurations, not on the path taken. Always calculate \(U_{final}\) and \(U_{initial}\) carefully and find the difference. A negative work done means the electric field did positive work and the system moved to a lower potential energy state.
\end{quicktipbox Quick Tip: Work done in rearranging a system of charges depends only on the initial and final potential energies of the configurations, not on the path taken. Always calculate \(U_{final}\) and \(U_{initial}\) carefully and find the difference. A negative work done means the electric field did positive work and the system moved to a lower potential energy state.

N/A

Step 1: Understanding the Concept:

The electric potential at a point due to a system of point charges is the algebraic sum of the potentials due to each individual charge. We are given a condition that the net potential at a specific point is zero.


Step 2: Key Formula or Approach:

The potential V at a distance r from a point charge q is given by \(V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}\).

For a system of charges, the total potential is \(V_{total} = V_1 + V_2 + \ldots\).

The condition given is \(V_{total} = 0\).


Step 3: Detailed Explanation:

Let the given charges and points be:
- Charge \(q_1 = +2 nC = +2 \times 10^{-9}\) C at the origin O(0, 0, 0).
- Unknown charge \(q_2\) at point P\(_1\)(0, 0, -6 m).
- We want the potential to be zero at point P\(_2\)(0, 0, 2 m).
Let's calculate the distances:
- The distance of P\(_2\) from \(q_1\) is \(r_1 = \sqrt{(0-0)^2 + (0-0)^2 + (2-0)^2} = 2\) m.
- The distance of P\(_2\) from \(q_2\) is \(r_2 = \sqrt{(0-0)^2 + (0-0)^2 + (2 - (-6))^2} = \sqrt{8^2} = 8\) m.
The total potential at P\(_2\) is \(V_{P_2} = V_1 + V_2\). We set this to zero. \[ V_{P_2} = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2} = 0 \] \[ \frac{q_1}{r_1} + \frac{q_2}{r_2} = 0 \implies \frac{q_2}{r_2} = -\frac{q_1}{r_1} \]
Now, solve for \(q_2\): \[ q_2 = -q_1 \left( \frac{r_2}{r_1} \right) \]
Substitute the known values: \[ q_2 = -(+2 \times 10^{-9} C) \left( \frac{8 m}{2 m} \right) \] \[ q_2 = -(2 \times 10^{-9} C) \times 4 = -8 \times 10^{-9} C \] \[ q_2 = -8 nC \]

Step 4: Final Answer:

The required charge is negative in type and has a magnitude of 8 nC.




\begin{quicktipbox
Potential is a scalar quantity, so you only need to perform algebraic addition. Be careful with the signs of the charges and the final result. A common mistake is to forget that the potential from the second charge must be equal and opposite to the first.
\end{quicktipbox Quick Tip: Potential is a scalar quantity, so you only need to perform algebraic addition. Be careful with the signs of the charges and the final result. A common mistake is to forget that the potential from the second charge must be equal and opposite to the first.

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Step 1: Understanding the Concept:

As the conducting loop moves through the magnetic field, the magnetic flux linked with it changes. According to Faraday's law of induction, this change in flux induces an EMF (\(\epsilon = -d\Phi_B/dt\)), which drives a current in the loop. This current then interacts with the magnetic field, resulting in a magnetic (Lorentz) force on the loop. We analyze this process in three stages: entering the field, moving within the field, and exiting the field.


Step 2: Detailed Analysis of Motion:

Stage 1: The loop is entering the magnetic field.

- As the loop enters the field, the area inside the field increases, causing the magnetic flux directed into the page to increase.

- Induced Current: According to Lenz's law, the induced current will flow in a direction that opposes this increase. It will create its own magnetic field pointing out of the page. Using the right-hand grip rule, the induced current flows in the anti-clockwise direction.

- Magnetic Force: The leading vertical arm of the loop is now inside the field and carries an upward current. According to the Lorentz force rule (\(\vec{F} = I\vec{L} \times \vec{B}\)), this arm experiences a magnetic force directed to the left, opposing the loop's motion. The forces on the top and bottom segments cancel each other out. To maintain a constant velocity, an external force equal in magnitude and directed to the right must be applied.


Stage 2: The loop is moving completely inside the magnetic field.

- When the entire loop is within the uniform magnetic field, the magnetic flux (\(\Phi_B = B \cdot A\)) linked with it is constant, as both B and the area A inside the field are constant.

- Induced Current: Since the flux is not changing (\(d\Phi_B/dt = 0\)), the induced EMF and the induced current are zero.

- Magnetic Force: The Lorentz forces on the two vertical arms are equal in magnitude (since they carry the same current, which is zero, or if there were a current, they would be opposite) and opposite in direction. The forces on the horizontal arms are also equal and opposite. Therefore, the net magnetic force on the loop is zero. No external force is needed to maintain constant velocity.


Stage 3: The loop is exiting the magnetic field.

- As the loop exits the field, the area inside the field decreases, causing the magnetic flux directed into the page to decrease.

- Induced Current: According to Lenz's law, the induced current will flow in a direction that opposes this decrease, i.e., it will try to reinforce the field. It creates its own magnetic field pointing into the page. Using the right-hand grip rule, the induced current flows in the clockwise direction.

- Magnetic Force: The trailing vertical arm is still inside the field and carries a downward current. This arm experiences a magnetic force directed to the left, again opposing the loop's motion. To maintain a constant velocity, an external force directed to the right must be applied.




\begin{quicktipbox
The key to analyzing motional EMF problems is to track the change in magnetic flux. An induced current only appears when the flux is changing. This happens when the loop enters or leaves a field, or if the field itself is changing in time. When the loop is fully inside a uniform field, the flux is constant, and there is no induced current.
\end{quicktipbox Quick Tip: The key to analyzing motional EMF problems is to track the change in magnetic flux. An induced current only appears when the flux is changing. This happens when the loop enters or leaves a field, or if the field itself is changing in time. When the loop is fully inside a uniform field, the flux is constant, and there is no induced current.

N/A

Let the side of the square loop be \(s = l/2\). The loop moves with constant velocity \(v\). The field region has a width \(l\).

- Time taken for the loop to enter the field: \(t_{enter} = \frac{side}{velocity} = \frac{l/2}{v} = \frac{l}{2v}\).

- Time spent fully inside the field: The loop travels a distance of \(l - l/2 = l/2\) while fully inside. \(t_{inside} = \frac{l/2}{v} = \frac{l}{2v}\).

- Time taken to exit the field: \(t_{exit} = \frac{l/2}{v} = \frac{l}{2v}\).


Plot of Magnetic Flux (\(\phi\)) vs. Time (t):

Let \(t=0\) be the instant the loop starts entering the field.

- For \(0 \le t \le \frac{l}{2v}\) (entering): The area inside the field increases linearly with time (\(A = (l/2) \cdot vt\)). Thus, the flux \(\phi = BA\) increases linearly from 0 to its maximum value, \(\phi_{max} = B \cdot (l/2)^2 = \frac{Bl^2}{4}\).

- For \(\frac{l}{2v} \le t \le \frac{l}{v}\) (fully inside): The entire loop is in the field, so the area is constant. The flux remains constant at its maximum value, \(\phi = \phi_{max} = \frac{Bl^2}{4}\).

- For \(\frac{l}{v} \le t \le \frac{3l}{2v}\) (exiting): The area inside the field decreases linearly with time. The flux decreases linearly from \(\phi_{max}\) back to 0.

The graph is a trapezoid.


Flux (\(\phi\)) vs. Time (t) Graph

(A textual description as a plot cannot be rendered)

- The t-axis is marked at \(t=0, l/(2v), l/v, 3l/(2v)\).
- The \(\phi\)-axis is marked at \(\phi=0\) and \(\phi_{max} = Bl^2/4\).
- The graph is a straight line from (0, 0) to (\(l/(2v), \phi_{max}\)).
- It is a horizontal line from (\(l/(2v), \phi_{max}\)) to (\(l/v, \phi_{max}\)).
- It is a straight line from (\(l/v, \phi_{max}\)) down to (\(3l/(2v), 0\)).


Plot of Induced EMF (E) vs. Time (t):

The induced EMF is given by \(E = -\frac{d\phi}{dt}\). It is the negative of the slope of the \(\phi\)-t graph.

- For \(0 < t < \frac{l}{2v}\) (entering): The slope of the \(\phi\)-t graph is constant and positive: \(\frac{\Delta\phi}{\Delta t} = \frac{Bl^2/4}{l/2v} = \frac{Blv}{2}\). Therefore, the EMF is constant and negative: \(E = -\frac{Blv}{2}\).

- For \(\frac{l}{2v} < t < \frac{l}{v}\) (fully inside): The slope of the \(\phi\)-t graph is zero. Therefore, the EMF is \(E=0\).

- For \(\frac{l}{v} < t < \frac{3l}{2v}\) (exiting): The slope of the \(\phi\)-t graph is constant and negative: \(\frac{\Delta\phi}{\Delta t} = \frac{0 - Bl^2/4}{l/2v} = -\frac{Blv}{2}\). Therefore, the EMF is constant and positive: \(E = -(-\frac{Blv}{2}) = +\frac{Blv}{2}\).

The graph consists of rectangular pulses.


EMF (E) vs. Time (t) Graph

(A textual description as a plot cannot be rendered)

- The t-axis is marked at \(t=0, l/(2v), l/v, 3l/(2v)\).
- The E-axis is marked at \(E=0, -Blv/2, +Blv/2\).
- The graph is a horizontal line at \(E = -Blv/2\) from \(t=0\) to \(t=l/(2v)\).
- It is a horizontal line at \(E=0\) from \(t=l/(2v)\) to \(t=l/v\).
- It is a horizontal line at \(E = +Blv/2\) from \(t=l/v\) to \(t=3l/(2v)\).




\begin{quicktipbox
Remember that EMF is induced only when magnetic flux is changing. The magnitude of the induced EMF is proportional to the rate of change of flux (\(|\epsilon| = |d\phi/dt|\)). A constant rate of change (linear increase/decrease in flux) results in a constant non-zero EMF.
\end{quicktipbox Quick Tip: Remember that EMF is induced only when magnetic flux is changing. The magnitude of the induced EMF is proportional to the rate of change of flux (\(|\epsilon| = |d\phi/dt|\)). A constant rate of change (linear increase/decrease in flux) results in a constant non-zero EMF.

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(I) Identification of Elements X and Y

Step 1: Analyzing Element X:

- It is given that for element X, the voltage leads the current by \(\pi/2\) (or 90\(^{\circ}\)).

- This is the characteristic property of a pure inductor.

- Phasor Diagram for Inductor (X): The current phasor (\(I\)) is taken along the positive x-axis. Since the voltage leads the current by 90\(^{\circ}\), the voltage phasor (\(V_L\)) is drawn along the positive y-axis, 90\(^{\circ}\) anti-clockwise from the current phasor.


(Phasor diagram description: A vector for I is shown along the +x axis. A vector for V\(_L\) is shown along the +y axis, with an anti-clockwise angle of 90\(^{\circ}\) indicated between them.)


Step 2: Analyzing Element Y:

- It is given that for element Y, the voltage lags behind the current by \(\pi/2\) (or 90\(^{\circ}\)).

- This is the characteristic property of a pure capacitor.

- Phasor Diagram for Capacitor (Y): The current phasor (\(I\)) is taken along the positive x-axis. Since the voltage lags the current by 90\(^{\circ}\), the voltage phasor (\(V_C\)) is drawn along the negative y-axis, 90\(^{\circ}\) clockwise from the current phasor.


(Phasor diagram description: A vector for I is shown along the +x axis. A vector for V\(_C\) is shown along the -y axis, with a clockwise angle of 90\(^{\circ}\) indicated between them.)


(II) Resonance in Series LC(R) Circuit

Step 1: Condition for Resonance:

When element X (Inductor L) and element Y (Capacitor C) are connected in series to the AC source, they form a series LC circuit (or an LCR circuit if we consider the inherent resistance R of the circuit).

Resonance is the condition where the net reactance of the circuit is zero, which happens when the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)).
\[ X_L = X_C \]
Step 2: Expression for Resonant Frequency:

We know \(X_L = \omega L\) and \(X_C = \frac{1}{\omega C}\), where \(\omega = 2\pi\nu\) is the angular frequency.

At the resonant angular frequency, \(\omega_0\):
\[ \omega_0 L = \frac{1}{\omega_0 C} \] \[ \omega_0^2 = \frac{1}{LC} \implies \omega_0 = \frac{1}{\sqrt{LC}} \]
The resonant frequency (\(\nu_0\)) is then:
\[ 2\pi\nu_0 = \frac{1}{\sqrt{LC}} \implies \nu_0 = \frac{1}{2\pi\sqrt{LC}} \]
Step 3: Impedance at Resonance:

The total impedance (Z) of a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
At resonance, \(X_L = X_C\), so the reactive part \((X_L - X_C)\) becomes zero.
\[ Z_{resonance} = \sqrt{R^2 + 0} = R \]
The impedance of the circuit at resonance is at its minimum value and is equal to the total ohmic resistance (R) of the circuit.




\begin{quicktipbox
A useful mnemonic for AC circuits is "ELI the ICE man". For an Inductor (L), EMF (E) or Voltage leads Current (I). For a Capacitor (C), Current (I) leads EMF (E) or Voltage.
\end{quicktipbox Quick Tip: A useful mnemonic for AC circuits is "ELI the ICE man". For an Inductor (L), EMF (E) or Voltage leads Current (I). For a Capacitor (C), Current (I) leads EMF (E) or Voltage.

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Step 1: Understanding the Given Information:

- The prism is equilateral, which means the angle of the prism is \(A = 60^{\circ}\).
- The refractive index of the prism material is \(n = \sqrt{3}\).


Step 2: Calculate the Angle of Minimum Deviation (\(\delta_m\)).

We use the prism formula which relates the refractive index, the angle of the prism, and the angle of minimum deviation: \[ n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Substitute the known values: \[ \sqrt{3} = \frac{\sin\left(\frac{60^{\circ} + \delta_m}{2}\right)}{\sin\left(\frac{60^{\circ}}{2}\right)} \] \[ \sqrt{3} = \frac{\sin\left(30^{\circ} + \frac{\delta_m}{2}\right)}{\sin(30^{\circ})} \]
Since \(\sin(30^{\circ}) = 1/2\): \[ \sqrt{3} = \frac{\sin\left(30^{\circ} + \frac{\delta_m}{2}\right)}{1/2} \] \[ \sin\left(30^{\circ} + \frac{\delta_m}{2}\right) = \frac{\sqrt{3}}{2} \]
The angle whose sine is \(\sqrt{3}/2\) is \(60^{\circ}\). Therefore: \[ 30^{\circ} + \frac{\delta_m}{2} = 60^{\circ} \] \[ \frac{\delta_m}{2} = 60^{\circ} - 30^{\circ} = 30^{\circ} \] \[ \delta_m = 60^{\circ} \]
The angle of minimum deviation is 60\(^{\circ}\).


Step 3: Calculate the Angle of Incidence (i).

For the condition of minimum deviation, the angle of incidence (\(i\)) and the angle of emergence (\(e\)) are equal, and they are related to A and \(\delta_m\) by the formula: \[ i = \frac{A + \delta_m}{2} \]
Substitute the values of A and the calculated \(\delta_m\): \[ i = \frac{60^{\circ} + 60^{\circ}}{2} = \frac{120^{\circ}}{2} \] \[ i = 60^{\circ} \]
The angle of incidence for minimum deviation is 60\(^{\circ}\).




\begin{quicktipbox
For an equilateral prism (\(A=60^\circ\)), a quick check is to remember that when the angle of minimum deviation is equal to the angle of the prism (\(\delta_m = A\)), then the refractive index is \(n = 2\cos(A/2)\) and the angle of incidence is \(i=A\). Here \(\delta_m = 60^\circ = A\), so \(i=60^\circ\), which matches our result.
\end{quicktipbox Quick Tip: For an equilateral prism (\(A=60^\circ\)), a quick check is to remember that when the angle of minimum deviation is equal to the angle of the prism (\(\delta_m = A\)), then the refractive index is \(n = 2\cos(A/2)\) and the angle of incidence is \(i=A\). Here \(\delta_m = 60^\circ = A\), so \(i=60^\circ\), which matches our result.

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Step 1: Understanding the Concept:

In an interference pattern, the central maximum is a bright fringe located at the center. The minima (dark fringes) are located between the maxima (bright fringes). The distance from the central maximum to the first minimum is half the fringe width.


Step 2: Key Formula or Approach:

The position of the n\(^{th}\) dark fringe (minimum) from the central maximum is given by: \[ y_{n, dark} = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d} = \left(n - \frac{1}{2}\right) \beta \]
For the first minimum, we use n = 1.


Step 3: Detailed Explanation:

Given data:

- Fringe width, \(\beta = 5 mm\).

Calculation:

For the first minimum, we set n = 1 in the formula: \[ y_{1, dark} = \left(1 - \frac{1}{2}\right) \beta = \frac{1}{2}\beta \]
Substitute the value of \(\beta\): \[ y_{1, dark} = \frac{1}{2} \times 5 mm = 2.5 mm \]

Step 4: Final Answer:

The first minimum will be at a distance of 2.5 mm from the central maximum.




\begin{quicktipbox
Remember that fringe width (\(\beta\)) is the distance between two consecutive bright fringes or two consecutive dark fringes. The distance between a bright fringe and the adjacent dark fringe is always \(\beta/2\).
\end{quicktipbox Quick Tip: Remember that fringe width (\(\beta\)) is the distance between two consecutive bright fringes or two consecutive dark fringes. The distance between a bright fringe and the adjacent dark fringe is always \(\beta/2\).

Step 1: Understanding the Concept:

This question deals with single-slit diffraction. The "distance between first order minima on both sides of the central maximum" is another way of asking for the width of the central maximum. The central maximum is flanked by the first minima on either side.


Step 2: Key Formula or Approach:

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

For small angles, \(\sin\theta \approx \tan\theta = \frac{y}{D}\), where \(y\) is the linear distance of the minimum from the center of the pattern on a screen kept at a distance D.

So, the position of the n\(^{th}\) minimum is \(y_n = \frac{n\lambda D}{a}\).

The width of the central maximum (W) is the distance between the first minima (n=1) on both sides: \(W = 2y_1\).


Step 3: Detailed Explanation:

Given data:

- Wavelength, \(\lambda = 650 nm = 650 \times 10^{-9} m\).
- Slit width, \(a = 0.6 mm = 0.6 \times 10^{-3} m\).
- Distance to the screen, \(D = 60 cm = 0.6 m\).

Calculation:

The width of the central maximum is: \[ W = 2y_1 = \frac{2\lambda D}{a} \]
Substitute the given values: \[ W = \frac{2 \times (650 \times 10^{-9} m) \times (0.6 m)}{0.6 \times 10^{-3} m} \]
The term 0.6 in the numerator and denominator cancels out. \[ W = 2 \times 650 \times 10^{-9 - (-3)} m \] \[ W = 1300 \times 10^{-6} m \] \[ W = 1.3 \times 10^{-3} m = 1.3 mm \]

Step 4: Final Answer:

The distance between the first order minima on both sides of the central maximum is 1.3 mm.




\begin{quicktipbox
Do not confuse the formulas for interference and diffraction. The width of the central maximum in diffraction (\(W = 2\lambda D/a\)) is twice the width of the secondary maxima (\(\approx \lambda D/a\)). In contrast, all bright fringes in an ideal YDSE have the same width.
\end{quicktipbox Quick Tip: Do not confuse the formulas for interference and diffraction. The width of the central maximum in diffraction (\(W = 2\lambda D/a\)) is twice the width of the secondary maxima (\(\approx \lambda D/a\)). In contrast, all bright fringes in an ideal YDSE have the same width.