CBSE Class 12 Physics Set 1 - (55/2/1) Question Paper 2026 with Solution PDF

Concept:
The electrostatic potential energy of a system of point charges is given by the sum of potential energies of all distinct pairs: \[ U = \frac{1}{4\pi\epsilon_0} \sum \frac{q_i q_j}{r_{ij}} \]

Since the charges are placed at vertices of an equilateral triangle, every separation is equal to \(l\).

Step 1: Identify all charge pairs
The three charges are: \[ 2q,\quad -2q,\quad q \]

Pairs:
- \((2q, -2q)\)
- \((2q, q)\)
- \((-2q, q)\)

Step 2: Compute each interaction energy

1. Between \(2q\) and \(-2q\): \[ U_1 = \frac{1}{4\pi\epsilon_0}\cdot \frac{(2q)(-2q)}{l} = \frac{-4q^2}{4\pi\epsilon_0 l} \]

2. Between \(2q\) and \(q\): \[ U_2 = \frac{1}{4\pi\epsilon_0}\cdot \frac{2q\cdot q}{l} = \frac{2q^2}{4\pi\epsilon_0 l} \]

3. Between \(-2q\) and \(q\): \[ U_3 = \frac{1}{4\pi\epsilon_0}\cdot \frac{-2q\cdot q}{l} = \frac{-2q^2}{4\pi\epsilon_0 l} \]

Step 3: Add all contributions \[ U = \frac{1}{4\pi\epsilon_0 l}(-4q^2 + 2q^2 - 2q^2) \]
\[ U = \frac{-4q^2}{4\pi\epsilon_0 l} \]
\[ U = \frac{-q^2}{\pi\epsilon_0 l} \]

Final Answer: \[ \boxed{\frac{-q^2}{\pi\epsilon_0 l}} \] Quick Tip: For multi-charge systems, always compute pairwise energies only—never treat it like a single charge system.

When two conducting spheres are connected, charge redistributes until both reach the same potential.

Step 1: Condition of equilibrium \[ V_1 = V_2 \Rightarrow \frac{q'_1}{r_1} = \frac{q'_2}{r_2} \]

Step 2: Total charge conservation \[ q'_1 + q'_2 = q_1 + q_2 \]

Step 3: Final potential
Common potential: \[ V = \frac{1}{4\pi\epsilon_0}\frac{q_1+q_2}{r_1 + r_2 (incorrect assumption)} \]

This is WRONG because potential depends on individual radii, not sum.

Correct derivation gives: \[ V = \frac{1}{4\pi\epsilon_0}\frac{q_1+q_2}{\frac{r_1 r_2}{r_1 + r_2}} \]

So option (C) is incorrect.

Final Answer: \[ \boxed{(C)} \] Quick Tip: For conducting spheres in contact: always equate potentials, not charges.

Concept:
According to Einstein's photoelectric equation, when a photon of energy \( E \) is incident on a photosensitive metal surface, its energy is entirely absorbed by a single electron near the surface. This energy is utilized in two ways:

Part of the energy is used to liberate the electron from the metal surface. The minimum energy required for this liberation is called the work function of the metal, denoted by \( \phi_0 \).
The remaining part of the photon energy is converted into the maximum kinetic energy (\( K_{max} \)) of the emitted photoelectron.

Mathematically, the energy conservation equation is written as: \[ E = \phi_0 + K_{max} \]
where the energy of the incident photon is given by \( E = h\nu \) (with \( h \) being Planck's constant and \( \nu \) being the frequency of the incident radiation).


Step 1: Expressing the maximum kinetic energy (\( K_{max} \)) in terms of frequency and work function.

By substituting \( E = h\nu \) into Einstein's photoelectric equation, we obtain: \[ h\nu = \phi_0 + K_{max} \]
Rearranging the terms to isolate \( K_{max} \): \[ K_{max} = h\nu - \phi_0 \]
This equation clearly demonstrates that the maximum kinetic energy depends on:

The frequency of the incident radiation (\( \nu \)) - higher frequency photons carry more energy, which directly increases the kinetic energy of the emitted photoelectrons.
The work function of the metal surface (\( \phi_0 \)) - different metals require different amounts of energy to release an electron; a lower work function results in a higher maximum kinetic energy for a given incident frequency.



Step 2: Analyzing the role of intensity.

The intensity of incident radiation \( I \) is proportional to the number of photons striking the photosensitive surface per unit area per unit time. Increasing the intensity increases the number of photoelectrons emitted (photocurrent), but does not alter the energy of individual photons. Consequently, the maximum kinetic energy of individual emitted electrons remains completely independent of the intensity \( I \).

Therefore, \( K_{max} \) depends on both the work function \( \phi_0 \) and the frequency \( \nu \). This matches option (D). Quick Tip: Remember the core distinction in photoelectric effect: - {Energy parameters} (Frequency \( \nu \), wavelength \( \lambda \), work function \( \phi_0 \)) determine the {Kinetic Energy} of the emitted electrons. - {Intensity parameters} (number of photons per second) determine the {number} of emitted electrons (photocurrent).

Concept:
According to Bohr's model of the hydrogen atom, the electron revolves around the nucleus only in certain stable, non-radiating circular paths called stationary orbits. The radius of the \( n \)-th Bohr orbit \( r_n \) is given by the formula: \[ r_n = \frac{\epsilon_0 n^2 h^2}{\pi m e^2} \]
where:

\( \epsilon_0 \) is the permittivity of free space,
\( h \) is Planck's constant,
\( m \) is the mass of the electron,
\( e \) is the elementary charge,
\( n \) is the principal quantum number (\( n = 1, 2, 3, \ldots \)).

From this formula, we can establish that the radius of the \( n \)-th orbit is directly proportional to the square of the principal quantum number: \[ r_n \propto n^2 \quad \Rightarrow \quad r_n = k n^2 \]
where \( k = \frac{\epsilon_0 h^2}{\pi m e^2} \approx 0.529\,Å \) is a constant.


Step 1: Formulating the distance between consecutive orbits.

Let the distance between consecutive orbits be denoted as \( \Delta r \). The consecutive orbits corresponding to a given quantum number \( n \) are the \( (n+1) \)-th orbit and the \( n \)-th orbit.
The distance \( \Delta r \) between these two consecutive orbits is: \[ \Delta r = r_{n+1} - r_n \]
Substituting the relation \( r_n = k n^2 \) into this expression: \[ \Delta r = k (n + 1)^2 - k n^2 \]


Step 2: Simplifying the algebraic expression.

Expanding the squared term inside the equation: \[ \Delta r = k \left( (n^2 + 2n + 1) - n^2 \right) \]
Subtracting the \( n^2 \) terms: \[ \Delta r = k (2n + 1) \]


Step 3: Analyzing the limit for large values of \( n \).

We are interested in the behavior of this distance for very large values of the principal quantum number (\( n \gg 1 \)).
In the term \( (2n + 1) \), as \( n \) becomes extremely large, the constant term \( 1 \) becomes negligible compared to \( 2n \): \[ 2n + 1 \approx 2n \quad (for n \rightarrow \infty) \]
Substituting this approximation back into our expression for \( \Delta r \): \[ \Delta r \approx k (2n) = (2k) n \]
Since \( 2k \) is a constant, we can write: \[ \Delta r \propto n \]
Thus, for large values of \( n \), the distance between consecutive orbits is directly proportional to \( n \). This matches option (A). Quick Tip: To quickly find proportionalities for consecutive differences of polynomial terms: If \( y_n \propto n^p \), then the consecutive difference \( y_{n+1} - y_n \propto n^{p-1} \) for large \( n \). Here, since \( r_n \propto n^2 \), the difference \( \Delta r \propto n^{2-1} = n^1 \).

Concept:
The direction of the magnetic field vector \( \vec{B} \) produced by a current-carrying element is given by the Biot-Savart Law: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \hat{r})}{r^2} \]
where:

\( I \) is the current flowing through the wire,
\( d\vec{l} \) is a small vector element of the length in the direction of the current flow,
\( \hat{r} \) is the unit position vector pointing from the current element to the observation point,
\( \vec{B} \) is the magnetic field vector.

The directional behavior of the magnetic field can be determined by the vector cross product of the current direction unit vector \( \hat{dl} \) and the position unit vector \( \hat{r} \): \[ \hat{B} \propto \hat{dl} \times \hat{r} \]


Step 1: Identify the coordinate unit vectors for the system.

Let the standard Cartesian coordinate unit vectors be:

\( \hat{i} \) along the \( +x \) direction,
\( \hat{j} \) along the \( +y \) direction,
\( \hat{k} \) along the \( +z \) direction.



Step 2: Express the given parameters in vector form.

1. The wire lies along the y-axis and the current is flowing in the negative y-direction. Therefore, the direction of the current element \( d\vec{l} \) is: \[ \hat{dl} = -\hat{j} \]
2. The observation point is located at \( (50\,cm, 0, 0) \), which lies on the positive x-axis. Therefore, the position vector \( \vec{r} \) pointing from the wire (origin/y-axis) to the point is along the positive x-direction: \[ \hat{r} = \hat{i} \]


Step 3: Evaluate the vector cross product to find the direction of \( \vec{B} \).

The direction of the magnetic field \( \hat{B} \) is given by the cross product: \[ \hat{B} = \hat{dl} \times \hat{r} \]
Substitute the unit vectors into this expression: \[ \hat{B} = (-\hat{j}) \times \hat{i} = -(\hat{j} \times \hat{i}) \]
Using the cyclic properties of unit vector cross products: \[ \hat{i} \times \hat{j} = \hat{k} \quad \Rightarrow \quad \hat{j} \times \hat{i} = -\hat{k} \]
Substitute \( \hat{j} \times \hat{i} = -\hat{k} \) back into the direction equation: \[ \hat{B} = -(-\hat{k}) = \hat{k} \]
Since \( \hat{k} \) represents the positive z-axis, the magnetic field points along the positive z-axis. This matches option (A). Quick Tip: Use the Right-Hand Grip Rule: 1. Point your right thumb in the direction of the current (\(-\hat{j}\), downwards). 2. Let your fingers curl around the wire. 3. At the point on the positive x-axis (\(+\hat{i}\), to the right), your fingers point out of the page, which represents the positive z-axis (\(+\hat{k}\)).

Concept:
Magnetic materials are classified into three major categories based on their behavior in an external magnetic field and the value of their magnetic susceptibility (\( \chi_m briefly \)):

Diamagnetic Materials: These materials are weakly repelled by magnetic fields. They have a negative and small magnetic susceptibility (\( -1 < \chi_m < 0 \)). Examples include Copper (Cu), Bismuth (text{Bi), Water, and Silicon.
Paramagnetic Materials: These materials are weakly attracted by magnetic fields. They have a positive and small magnetic susceptibility (\( 0 < \chi_m < \epsilon \)). Examples include Aluminum (text{Al), Sodium, Calcium, and Oxygen.
Ferromagnetic Materials: These materials are strongly attracted by magnetic fields and can be permanently magnetized. They have a positive and very large magnetic susceptibility (\( \chi_m \gg 1 \)). Examples include Nickel (text{Ni), Iron, and Cobalt.



Step 1: Analyzing the given options.

Let us evaluate the magnetic properties of each material listed in the choices:

Copper (Cu): It is a diamagnetic material. Its magnetic susceptibility is negative and small (( chi_{text{Cu} approx -9.6 times 10^{-6} )).
Aluminum (Al): It is a paramagnetic material. Its magnetic susceptibility is positive and small (\( \chi_{Al} \approx 2.2 \times 10^{-5} \)).
Bismuth (Bi): It is a strongly diamagnetic material. Its magnetic susceptibility is negative and small (\( \chi_{Bi} \approx -1.66 \times 10^{-4} \)).
Nickel (Ni): It is a ferromagnetic material. Its magnetic susceptibility is positive and extremely large (\( \chi_{Ni} \sim 10^2 to 10^3 digital \)).



Step 2: Conclusion.

Among the choices, Aluminum (Al) is the only material that possesses a positive and small value of magnetic susceptibility. This matches option (B). Quick Tip: Remember the susceptibility classification: - textbf{Diamagnetic: \( \chi < 0 \) (small, negative) - textbf{Paramagnetic:} \( \chi > 0 \) (small, positive) - textbf{Ferromagnetic:} \( \chi \gg 0 \) (large, positive)

Concept:
A galvanometer can be converted into an ammeter of a larger range by connecting a very low resistance called a shunt (\( S \)) in parallel with the galvanometer coil.
Let:

\( G \) be the resistance of the galvanometer,
\( I_g \) be the current required for full-scale deflection in the galvanometer,
\( I \) be the total current range of the converted ammeter,
\( S \) be the value of the shunt resistance connected in parallel.

Since the galvanometer and the shunt resistor are connected in parallel, the potential difference across both branches must be equal: \[ V_{galvanometer} = V_{shunt} \] \[ I_g \times G = (I - I_g) \times S \]


Step 1: Setting up the values from the problem statement.

We are given:

Galvanometer resistance, \( G = 27\,\Omega \)
Shunt resistance, \( S = 3\,\Omega \)
Maximum range of the ammeter, \( I = 10\,mA = 10 \times 10^{-3}\,A \)

We need to determine the full-scale deflection current of the galvanometer, which is \( I_g \).


Step 2: Solving the parallel potential difference equation for \( I_g \).

Using the formula: \[ I_g \cdot G = (I - I_g) \cdot S \]
Substitute the given values into this equation: \[ I_g \cdot (27) = (10\,mA - I_g) \cdot (3) \]
Divide both sides of the equation by 3: \[ 9 I_g = 10\,mA - I_g \]
Add \( I_g \) to both sides to group the like terms: \[ 9 I_g + I_g = 10\,mA \] \[ 10 I_g = 10\,mA \]
Dividing both sides by 10 gives: \[ I_g = \frac{10\,mA}{10} = 1\,mA \]
Thus, the galvanometer will show full-scale deflection for a current of \( 1\,mA \). This matches option (C). Quick Tip: For converting a galvanometer into an ammeter, use the quick ratio formula: \[ I_g = I \left( \frac{S}{G + S} \right) \] Substituting the values: \( I_g = 10\,mA \left( \frac{3}{27 + 3} \right) = 10\,mA \left( \frac{3}{30} \right) = 1\,mA \).

Concept:
According to the **Principle of Homogeneity of Dimensions**, a physical equation is dimensionally correct and valid only if the dimensions (and consequently the units) of each term on both sides of the equation are identical.
In any mathematical expression involving addition or subtraction: \[ X = Y + Z - W \]
the physical quantities \( X \), \( Y \), \( Z \), and \( W \) must all share the exact same units and dimensions.

Here, the given relation is: \[ \phi = 5 At^2 + Bt - 2C \]
where:

\( \phi \) represents magnetic flux, measured in Webers (\(Wb\)),
\( t \) represents time, measured in seconds (\(s\)).



Step 1: Finding the SI unit of \( A \).

Applying the principle of homogeneity, the unit of the term \( 5 At^2 \) must be equal to the unit of magnetic flux \( \phi \): \[ [Unit of 5 At^2] = [Unit of \phi] \]
Since numerical constants like \( 5 \) are dimensionless and carry no units: \[ [Unit of A] \times [Unit of t]^2 = [Unit of \phi] \]
Substitute the known SI units: \[ [Unit of A] \times s^2 = Wb \]
Solving for the unit of \( A \): \[ [Unit of A] = \frac{Wb}{s^2} = Wb\,s^{-2} \]


Step 2: Finding the SI unit of \( B \).

Similarly, applying the principle of homogeneity to the second term \( Bt \): \[ [Unit of Bt] = [Unit of \phi] \] \[ [Unit of B] \times [Unit of t] = [Unit of \phi] \]
Substitute the known SI units: \[ [Unit of B] \times s = Wb \]
Solving for the unit of \( B \): \[ [Unit of B] = \frac{Wb}{s} = Wb\,s^{-1} \]


Step 3: Conclusion.

The SI units of \( A \) and \( B \) are \( Wb\,s^{-2} \) and \( Wb\,s^{-1} \) respectively. This matches option (C). Quick Tip: To quickly find the units of coefficients in polynomial equations, simply divide the unit of the left-hand side quantity by the corresponding power of the independent variable: - Unit of \( A \) = \(\frac{[Flux]}{[Time]^2} = Wb\,s^{-2}\) - Unit of \( B \) = \(\frac{[Flux]}{[Time]^1} = Wb\,s^{-1}\)

Concept:
Capacitive reactance: \[ X_C = \frac{1}{\omega C} \]

Let \(x = \frac{1}{\omega}\), then: \[ X_C = \frac{x}{C} \]

Thus slope of graph: \[ slope = \frac{1}{C} \]

Step 1: Use given angles
Slope \(m = \tan\theta\)

For \(C_1\): \[ m_1 = \tan 45^\circ = 1 \]

For \(C_2\): \[ m_2 = \tan 30^\circ = \frac{1}{\sqrt{3}} \]

Step 2: Ratio of slopes \[ \frac{m_1}{m_2} = \frac{1}{1/\sqrt{3}} = \sqrt{3} \]

Since: \[ m \propto \frac{1}{C} \Rightarrow \frac{m_1}{m_2} = \frac{C_2}{C_1} \]
\[ \frac{C_1}{C_2} = \frac{1}{\sqrt{3}} \]

Final Answer: \[ \boxed{(D)} \] Quick Tip: In graph-based physics questions, slope interpretation is the key step.

Concept:
This problem is based on **Faraday's Law of Electromagnetic Induction** and **Lenz's Law**.

When a bar magnet falls toward a closed conducting loop or solenoid, the magnetic flux linked with the solenoid increases.
According to Faraday's law, a change in magnetic flux induces an electromotive force (emf) and consequently a current in the closed circuit.
Lenz's law states that the direction of the induced current is always such that it opposes the change in magnetic flux that produced it. Mathematically, this is expressed as:
\[ e = -\frac{d\phi}{dt} \]



Step 1: Analyzing the direction of magnetic flux change.

The bar magnet is oriented vertically with its North (N) pole facing downwards. As it is dropped from above, it falls towards the upper face of the vertically placed closed solenoid.
Because the North pole is getting closer to the top end of the solenoid, the downward-directed magnetic flux passing through the upper turns of the coil increases rapidly with time.


Step 2: Applying Lenz's Law to find the induced magnetic polarity.

To oppose the continuous increase of downward magnetic flux caused by the approaching North pole, the upper face of the solenoid must develop a magnetic polarity that exerts a repulsive force on the falling magnet.
Therefore, the top face of the solenoid must behave as a **North pole** to repel the approaching North pole of the bar magnet and try to slow down its descent.


Step 3: Determining the direction of the induced current.

According to the clock rule of circular currents:

A face behaving as a North pole corresponds to current flowing in the **anticlockwise (counter-clockwise) direction** when viewed by an observer looking directly at that face.
A face behaving as a South pole corresponds to current flowing in the clockwise direction.

Since an observer looking down from the top views the upper face of the solenoid as a North pole, the induced current must flow in the anticlockwise direction. This matches option (A). Quick Tip: Lenz's Law simplified: - Approaching pole \(\rightarrow\) Solenoid creates the **same pole** to repel it. - Receding pole \(\rightarrow\) Solenoid creates the **opposite pole** to attract it. - Since a North pole is approaching the top face, the top face becomes a North pole. Looking from the top, North corresponds to **anticlockwise** current.

Concept:
An electromagnetic (EM) wave consists of time-varying sinusoidal electric and magnetic fields that oscillate perpendicular to each other and also perpendicular to the direction of wave propagation.

If an electromagnetic wave is propagating along the positive x-axis in free space, the equations representing the space and time variations of the electric field vector \( \vec{E} \) (say along the y-axis) and the magnetic field vector \( \vec{B} \) (say along the z-axis) can be written as: \[ E_y = E_0 \sin(kx - \omega t) \] \[ B_z = B_0 \sin(kx - \omega t) \]
where:

\( E_0 \) and \( B_0 \) are the amplitudes (peak values) of the electric and magnetic fields respectively.
\( k = \frac{2\pi}{\lambda} \) is the wave number.
\( \omega = 2\pi\nu \) is the angular frequency.
\( (kx - \omega t) \) represents the phase of the respective field at a position \( x \) and time \( t \).



Step 1: Comparing the phase angles of the fields.

Let us examine the argument inside the sinusoidal functions for both fields:

The phase angle of the electric field oscillation is: \( \theta_E = kx - \omega t \)
The phase angle of the magnetic field oscillation is: \( \theta_B = kx - \omega t \)

Both expressions are exactly identical at any given coordinate \( x \) and at any specific time instant \( t \).


Step 2: Calculating the phase difference.

The phase difference \( \Delta \theta \) between the electric field and magnetic field is defined as the subtraction of their individual phase angles: \[ \Delta \theta = \theta_E - \theta_B \] \[ \Delta \theta = (kx - \omega t) - (kx - \omega t) = 0\,radian \]
This mathematical result signifies that the electric and magnetic fields reach their maximum values (crests) at the exact same spatial points and at the same time instances. Similarly, they cross zero and reach their minimum values simultaneously. They are completely in phase, meaning their phase difference is zero. This matches option (A). Quick Tip: Do not confuse **spatial orientation** with **phase variation**: - The electric and magnetic fields are textbf{space-wise perpendicular} to each other (\( 90^\circ \) or \( \frac{\pi}{2} \) spatial angle). - But they are textbf{time-wise in phase}, meaning their phase difference is textbf{zero}.

Concept:
According to Bohr's third postulate, when an electron makes a transition from a higher energy level with initial principal quantum number \( n_i \) to a lower energy level with final principal quantum number \( n_f \), a single photon is emitted. The wavelength \( \lambda' \) of this emitted photon is given by the Rydberg formula: \[ \frac{1}{\lambda'} = R \cdot Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where:

\( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7\,m^{-1} \)),
\( Z \) is the atomic number of the atom (for hydrogen atom, \( Z = 1 \)),
\( n_i \) is the initial higher orbit level,
\( n_f \) is the final lower orbit level.

For a hydrogen atom (\( Z = 1 \)), the equation simplifies directly to: \[ \frac{1}{\lambda'} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]


Step 1: Analyzing the first transition (\( n = 5 \) to \( n = 1 \)).

We are given that when the electron falls from \( n_i = 5 \) to \( n_f = 1 \), the wavelength of the emitted photon is \( \lambda \).
Let us substitute these integer values into the Rydberg equation: \[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{5^2} \right) \] \[ \frac{1}{\lambda} = R \left( \frac{1}{1} - \frac{1}{25} \right) \]
Taking a common denominator of 25 inside the bracket: \[ \frac{1}{\lambda} = R \left( \frac{25 - 1}{25} \right) = R \left( \frac{24}{25} \right) \]
Isolating the Rydberg constant \( R \) gives us our first structural relationship: \[ R = \frac{25}{24\lambda} \quad \cdots (1) \]


Step 2: Analyzing the second transition (\( n = 5 \) to \( n = 2 \)).

Let the wavelength of the photon emitted during this second transition be denoted as \( \lambda_{5\rightarrow2} \). Here, the electron falls from initial orbit \( n_i = 5 \) to final orbit \( n_f = 2 \).
Substituting these values into the formula: \[ \frac{1}{\lambda_{5\rightarrow2}} = R \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \] \[ \frac{1}{\lambda_{5\rightarrow2}} = R \left( \frac{1}{4} - \frac{1}{25} \right) \]
Finding a common denominator for the fractions inside the parenthesis, which is \( 4 \times 25 = 100 \): \[ \frac{1}{\lambda_{5\rightarrow2}} = R \left( \frac{25 - 4}{100} \right) = R \left( \frac{21}{100} \right) \quad \cdots (2) \]


Step 3: Substituting the value of \( R \) from Equation (1) into Equation (2).

Now, substitute the expression for \( R = \frac{25}{24\lambda} \) obtained in Step 1 directly into Equation (2): \[ \frac{1}{\lambda_{5\rightarrow2}} = \left( \frac{25}{24\lambda} \right) \times \left( \frac{21}{100} \right) \]
We can perform a step-by-step simplification of the numerical fraction:

Simplify \( \frac{25}{100} \): Since \( 100 = 4 \times 25 \), this reduces to \( \frac{1}{4} \).
Simplify \( \frac{21}{24} \): Both numbers are divisible by 3. \( 21 \div 3 = 7 \) and \( 24 \div 3 = 8 \). This reduces to \( \frac{7}{8} \).

Putting these reduced fractions back together: \[ \frac{1}{\lambda_{5\rightarrow2}} = \frac{1}{8\lambda} \times \frac{7}{4} \]
Multiply the terms in the numerator and denominator: \[ \frac{1}{\lambda_{5\rightarrow2}} = \frac{7}{32\lambda} \]


Step 4: Inverting the fraction to find the final wavelength \( \lambda_{5\rightarrow2} \).

Taking the reciprocal of both sides of our simplified equation yields: \[ \lambda_{5\rightarrow2} = \frac{32}{7}\lambda \]
Thus, the wavelength of the photon emitted in the transition from \( n = 5 \) to \( n = 2 \) is equal to \( \frac{32}{7}\lambda \). This matches option (D). Quick Tip: To avoid calculating \( R \) explicitly, simply set up a direct ratio of the two equations: \[ \frac{\lambda_{2}}{\lambda_{1}} = \frac{\left( \frac{1}{1^2} - \frac{1}{5^2} \right)}{\left( \frac{1}{2^2} - \frac{1}{5^2} \right)} = \frac{\frac{24}{25}}{\frac{21}{100}} = \frac{24}{25} \times \frac{100}{21} = \frac{24 \times 4}{21} = \frac{96}{21} = \frac{32}{7} \] Thus, \( \lambda_2 = \frac{32}{7}\lambda_1 \).

Concept:

According to Einstein's photoelectric equation,
\[ h\nu=\phi+K_{\max} \]

or
\[ K_{\max}=h\nu-\phi. \]

The photoelectric effect depends mainly on two parameters:


Intensity of incident light
Frequency of incident light


The intensity of light determines the number of photons incident per second, whereas the frequency determines the energy carried by each photon.



Assertion Analysis:

The given frequency satisfies
\[ \nu>\nu_0, \]

where \(\nu_0\) is the threshold frequency of the photosensitive surface.

Therefore photoemission takes place.

When the intensity of incident light is increased while keeping the frequency unchanged, a larger number of photons strike the metal surface per unit time.

As a result, a greater number of electrons are emitted from the surface every second.

Since photocurrent is directly proportional to the number of photoelectrons reaching the collector,
\[ Photocurrent \propto Number of emitted electrons. \]

Hence increasing intensity increases the photocurrent.

Therefore, the Assertion is TRUE.



Reason Analysis:

The stopping potential \(V_0\) is related to the maximum kinetic energy of emitted photoelectrons by
\[ eV_0=K_{\max}. \]

Using Einstein's equation,
\[ eV_0=h\nu-\phi. \]

This equation clearly shows that the stopping potential depends on the frequency of the incident radiation.

As the frequency increases above the threshold frequency,
\[ h\nu-\phi \]

increases and therefore the stopping potential also increases.

Hence the Reason is also TRUE.



Relationship Between Assertion and Reason:

The Assertion talks about the increase in photocurrent due to increase in intensity.

The Reason talks about increase in stopping potential due to increase in frequency.

These are two different aspects of the photoelectric effect.
\[ Intensity \longrightarrow Photocurrent \]

whereas
\[ Frequency \longrightarrow Kinetic Energy and Stopping Potential. \]

Thus, although both statements are individually true, the Reason does not explain the Assertion.



Final Answer:
\[ \boxed{(B)} \] Quick Tip: Intensity controls the number of emitted electrons and hence photocurrent. Frequency controls the energy of emitted electrons and hence stopping potential.

Concept:

A p-n junction diode contains a depletion region around the junction.

Due to diffusion of charge carriers, a built-in electric field is established across the junction which creates a potential barrier known as the barrier potential.

This barrier potential opposes the further diffusion of majority charge carriers.

The behaviour of the diode depends upon whether it is forward biased or reverse biased.



Assertion Analysis:

In forward biasing,
\[ p-side \rightarrow positive terminal \]

and
\[ n-side \rightarrow negative terminal. \]

The external applied voltage opposes the built-in potential barrier.

As a consequence, the depletion region becomes thinner and the barrier potential decreases.

Because the barrier potential is reduced, majority charge carriers can cross the junction more easily, resulting in a large forward current.

Therefore, the statement
\[ ``barrier potential increases'' \]

is incorrect.

The barrier potential actually decreases during forward biasing.

Hence, the Assertion is FALSE.



Reason Analysis:

The built-in potential of the junction and the externally applied forward-bias voltage act in opposite directions.

The purpose of forward biasing is precisely to oppose and reduce the junction barrier.

Therefore the statement that the applied voltage acts in the same direction as the built-in potential is incorrect.

In reality,
\[ Forward Bias Voltage \]

acts opposite to
\[ Built-in Potential. \]

Hence the Reason is also FALSE.



Relationship Between Assertion and Reason:

Since the Assertion is false and the Reason is also false, the given pair falls under the category:
\[ Assertion False, Reason False. \]



Final Answer:
\[ \boxed{(D)} \] Quick Tip: Forward bias reduces the depletion width and barrier potential, thereby increasing current. Reverse bias increases the depletion width and barrier potential, thereby reducing current.

Concept:

One of the most important consequences of the wave nature of light is the phenomenon of interference.

When two coherent light waves overlap in space, the resultant intensity at a point depends upon the phase difference between the waves.

The principle of superposition states that the resultant displacement at any point is equal to the algebraic sum of the individual displacements produced by the two waves.

Depending on the phase relationship, interference may be constructive or destructive.



Assertion Analysis:

The assertion states that light added to light can produce darkness.

At first sight this statement may appear contradictory because we normally expect the addition of two light beams to produce brighter illumination.

However, according to wave optics, when two coherent light waves meet with a phase difference of

[
pi
]

or an odd multiple of (pi),

their amplitudes cancel each other.

If the amplitudes of the two interfering waves are equal, the resultant amplitude becomes zero.

Since intensity is proportional to the square of amplitude,

[
I propto A^2,
]

the intensity at that point also becomes zero.

As a result, a dark fringe is produced even though light from two sources is present.

Thus, light added to light can indeed produce darkness.

Therefore, the Assertion is TRUE.



Reason Analysis:

The reason states that when two coherent light waves interfere, darkness is produced at positions of destructive interference.

For destructive interference, the path difference between the two waves is

[
(2n+1)frac{lambda{2,
]

where (n=0,1,2,ldots)

or equivalently, the phase difference is

[
(2n+1)pi.
]

Under this condition, the crest of one wave coincides with the trough of the other wave, leading to cancellation of amplitudes.

Consequently, the resultant intensity becomes minimum and dark fringes are formed.

Therefore, the Reason is also TRUE.



Relationship Between Assertion and Reason:

The phenomenon described in the Reason is exactly the physical explanation of the Assertion.

Darkness is obtained because of destructive interference between coherent light waves.

Thus the Reason correctly explains why light added to light can produce darkness.



Final Answer:

[
(A)
] Quick Tip: Interference is redistribution of energy, not destruction of light.

Concept:

The electrical power consumed by a resistor connected across a source of constant voltage (V) is given by

[
P=frac{V^2{R.
]

Thus, for a fixed voltage source, power is inversely proportional to resistance.

When resistors are connected in series, their equivalent resistance increases, causing the current in the circuit to decrease.

As a result, the total power consumed by the combination may become smaller than the power consumed by an individual resistor connected directly across the same source.



Assertion Analysis:

Let the resistances of the two heaters be (R_1) and (R_2).

Since the heaters are rated at powers (P_1) and (P_2) when connected across the same voltage (V),

[
P_1=frac{V^2{R_1
]

and

[
P_2=frac{V^2{R_2.
]

Given

[
P_2>P_1.
]

Therefore,

[
R_2]

When the heaters are connected in series, the equivalent resistance becomes

[
R_s=R_1+R_2.
]

Since

[
R_1+R_2>R_1,
]

the power consumed by the series combination is

[
P_s=frac{V^2{R_1+R_2.
]

Because the denominator is larger,

[
P_s]

Hence, the power consumed by the series combination is less than that consumed by heater (P_1) alone across the same source.

Therefore, the Assertion is TRUE.



Reason Analysis:

The reason claims that power consumed by an electrical device connected across a dc voltage source is directly proportional to resistance.

For a constant voltage source,

[
P=frac{V^2{R.
]

This clearly shows that power is inversely proportional to resistance.

If resistance increases, power decreases.

If resistance decreases, power increases.

Therefore the statement given in the Reason is incorrect.

Hence, the Reason is FALSE.



Relationship Between Assertion and Reason:

The Assertion is true because connecting the heaters in series increases the equivalent resistance, which reduces the total power consumed.

The Reason is false because power is not directly proportional to resistance for a constant voltage source.

Therefore, the Assertion is true but the Reason is false.



Final Answer:

[
(C)
] Quick Tip: For constant voltage, higher resistance means lower power.

Concept:
Semiconductors are materials whose electrical conductivity lies between conductors and insulators. Their conductivity depends strongly on temperature and doping.

Step 1: Intrinsic Semiconductor
An intrinsic semiconductor is a pure semiconductor without any intentional impurity doping.

Step 2: Extrinsic Semiconductor
An extrinsic semiconductor is formed by adding impurities (doping) to intrinsic semiconductors to increase conductivity.

Step 3: Key Differences

(i) Purity
- Intrinsic: Pure material (Si, Ge)
- Extrinsic: Impurity added (donor or acceptor atoms)

(ii) Charge carriers
- Intrinsic: Equal electrons and holes (\(n = p\))
- Extrinsic: Unequal carriers; majority and minority carriers exist

(iii) Conductivity
- Intrinsic: Low conductivity
- Extrinsic: High conductivity due to doping

(iv) Dependence
- Intrinsic: Depends mainly on temperature
- Extrinsic: Depends on temperature and doping concentration

Final Answer:
Intrinsic and extrinsic semiconductors differ mainly in purity and charge carrier concentration. Quick Tip: Doping increases conductivity by many orders of magnitude.

Concept:
De Broglie wavelength: \[ \lambda = \frac{h}{p} \quad,\quad p = \sqrt{2mK} \]

So: \[ \lambda = \frac{h}{\sqrt{2mK}} \]

Step 1: Alpha particle and proton masses \[ m_\alpha = 4m_p \]



Case (i): Same kinetic energy
\[ \lambda \propto \frac{1}{\sqrt{m}} \]

So: \[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p}{m_\alpha}} = \sqrt{\frac{m_p}{4m_p}} = \frac{1}{2} \]
\[ \boxed{\left(\frac{\lambda_\alpha}{\lambda_p}\right)_1 = \frac{1}{2}} \]



Case (ii): Same potential difference

Kinetic energy: \[ K = qV \]

Alpha particle charge = \(2e\), proton charge = \(e\)

So: \[ K_\alpha = 2eV,\quad K_p = eV \]

Now: \[ \lambda = \frac{h}{\sqrt{2mK}} \]
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p K_p}{m_\alpha K_\alpha}} \]

Substitute: \[ = \sqrt{\frac{m_p (eV)}{(4m_p)(2eV)}} \]
\[ = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}} \]
\[ \boxed{\left(\frac{\lambda_\alpha}{\lambda_p}\right)_2 = \frac{1}{2\sqrt{2}}} \]



Final Answer: \[ (i) \frac{1}{2}, \quad (ii) \frac{1}{2\sqrt{2}} \] Quick Tip: Always include both mass and charge when potential difference is given.

Concept:
For minimum deviation in prism: \[ i = e,\quad r = \frac{A}{2} \]

Given:
Equilateral prism: \[ A = 60^\circ \]

Step 1: Given relation \[ i = \frac{3}{4}A = \frac{3}{4} \times 60^\circ = 45^\circ \]

Step 2: Refraction angle
At minimum deviation: \[ r = \frac{A}{2} = 30^\circ \]

Step 3: Snell’s law \[ \mu = \frac{\sin i}{\sin r} \]
\[ \mu = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2} \]

Step 4: Speed of light \[ v = \frac{c}{\mu} = \frac{3 \times 10^8}{\sqrt{2}} \approx 2.12 \times 10^8 \ m/s \]
\[ \boxed{v \approx 2.0 \times 10^8 \ m/s} \] Quick Tip: At minimum deviation, prism behaves symmetrically: \(i = e\).

Concept:
The resistance of a conductor increases with temperature according to: \[ R_T = R_0 \left(1 + \alpha \Delta T \right) \]
where:

\(R_0\) = resistance at initial temperature \(T_0\)
\(R_T\) = resistance at temperature \(T\)
\(\alpha\) = temperature coefficient of resistance
\(\Delta T = T - T_0\)


Since voltage is constant: \[ R = \frac{V}{I} \]



Step 1: Initial resistance
\[ R_0 = \frac{V}{I_1} = \frac{220}{2.9} \]
\[ R_0 \approx 75.86\ \Omega \]



Step 2: Final (steady) resistance
\[ R_T = \frac{V}{I_2} = \frac{220}{2.5} \]
\[ R_T = 88\ \Omega \]



Step 3: Apply temperature dependence of resistance
\[ \frac{R_T}{R_0} = 1 + \alpha (T - T_0) \]

Substitute values: \[ \frac{88}{75.86} = 1 + (1.7 \times 10^{-4})(T - 27) \]



Step 4: Simplify left-hand side
\[ \frac{88}{75.86} \approx 1.160 \]

So: \[ 1.160 = 1 + (1.7 \times 10^{-4})(T - 27) \]



Step 5: Solve for temperature
\[ 0.160 = (1.7 \times 10^{-4})(T - 27) \]
\[ T - 27 = \frac{0.160}{1.7 \times 10^{-4}} \]
\[ T - 27 \approx 941.18 \]
\[ T \approx 968^{\circ}C \]



Final Answer: \[ \boxed{T \approx 9.7 \times 10^2\ ^{\circ}C} \] Quick Tip: For heating elements: - Use \(R = \frac{V}{I}\) - Then apply \(R_T = R_0(1 + \alpha \Delta T)\) - Always compute resistance ratio first for faster solving.

Concept:
For a concave mirror, the mirror formula is: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Magnification is: \[ m = \frac{dv}{du} \approx \frac{f^2}{(u - f)^2} \]
For a small object along principal axis, image length is obtained using longitudinal magnification.



Step 1: Given data
\[ f = -20 cm, \quad u_1 = -25 cm \]

Pencil length = 5 cm, so far end: \[ u_2 = -30 cm \]



Step 2: Find image positions using mirror formula

For near end: \[ \frac{1}{-20} = \frac{1}{v_1} + \frac{1}{-25} \]
\[ \frac{1}{v_1} = -\frac{1}{20} + \frac{1}{25} = \frac{-5 + 4}{100} = -\frac{1}{100} \]
\[ v_1 = -100 cm \]



For far end: \[ u_2 = -30 \]
\[ \frac{1}{-20} = \frac{1}{v_2} + \frac{1}{-30} \]
\[ \frac{1}{v_2} = -\frac{1}{20} + \frac{1}{30} = \frac{-3 + 2}{60} = -\frac{1}{60} \]
\[ v_2 = -60 cm \]



Step 3: Image length
\[ Image length = |v_2 - v_1| \]
\[ = |-60 - (-100)| = 40 cm \]



Final Answer: \[ \boxed{40 \ cm} \] Quick Tip: For extended objects along principal axis: Compute image position of both ends separately using mirror formula.

Concept:
In Young’s double slit experiment, position of bright fringe: \[ y = \frac{n \lambda D}{d} \]

For coincidence of bright fringes: \[ n_1 \lambda_1 = n_2 \lambda_2 \]



Step 1: Given data
\[ \lambda_1 = 500 nm, \quad \lambda_2 = 600 nm \]



Step 2: Condition for coincidence
\[ n_1 \cdot 500 = n_2 \cdot 600 \]

Divide: \[ \frac{n_1}{n_2} = \frac{600}{500} = \frac{6}{5} \]

Smallest integers: \[ n_1 = 6, \quad n_2 = 5 \]



Step 3: Find position of coincidence


Using: \[ y = \frac{n \lambda D}{d} \]

Take \( \lambda_1 \): \[ y = \frac{6 \times 500 \times 10^{-9} \times 1.8}{0.3 \times 10^{-3}} \]



Step 4: Simplification
\[ y = \frac{6 \times 500 \times 1.8}{0.3} \times 10^{-6} \]
\[ = \frac{5400}{0.3} \times 10^{-6} = 18000 \times 10^{-6} \]
\[ y = 1.8 \times 10^{-2} m \]
\[ y = 18 mm \]



Final Answer: \[ \boxed{18 \ mm} \] Quick Tip: For two wavelengths in YDSE: First find LCM condition \(n_1\lambda_1 = n_2\lambda_2\), then substitute in fringe formula.

Concept:
To categorize different nuclear species (nuclides), we analyze their atomic number (\(Z\)), mass number (\(A\)), and neutron number (\(N\)):

Isotopes: Nuclides that have the same atomic number \(Z\) (i.e., the same number of protons) but different mass numbers \(A\) (i.e., different numbers of neutrons).
Isotones: Nuclides that have different atomic numbers \(Z\) and different mass numbers \(A\), but possess the exact same number of neutrons \(N\), where \(N = A - Z\).



Step 1: Analyze the given nuclides for their proton and neutron counts.

Let us write down the values of \(A\) (mass number), \(Z\) (atomic number/proton number), and \(N = A - Z\) (neutron number) for each of the given four nuclides:

For Carbon-12 (\( ^{12}_{6}C \)):
\[ Z_1 = 6, \quad A_1 = 12 \quad \Rightarrow \quad N_1 = A_1 - Z_1 = 12 - 6 = 6 neutrons. \]

For Mercury-198 (\( ^{198}_{80}Hg \)):
\[ Z_2 = 80, \quad A_2 = 198 \quad \Rightarrow \quad N_2 = A_2 - Z_2 = 198 - 80 = 118 neutrons. \]

For Carbon-14 (\( ^{14}_{6}C \)):
\[ Z_3 = 6, \quad A_3 = 14 \quad \Rightarrow \quad N_3 = A_3 - Z_3 = 14 - 6 = 8 neutrons. \]

For Gold-197 (\( ^{197}_{79}Au \)):
\[ Z_4 = 79, \quad A_4 = 197 \quad \Rightarrow \quad N_4 = A_4 - Z_4 = 197 - 79 = 118 neutrons. \]



Step 2: Identify the isotopes.

By definition, isotopes must share the same atomic number \(Z\). Looking at our analysis in Step 1:

\( ^{12}_{6}C \) has \(Z = 6\).
\( ^{14}_{6}C \) has \(Z = 6\).

Since both have the same atomic number (\(Z = 6\)) but different mass numbers (\(12 \neq 14\)), the pair of isotopes is: \[ \left( ^{12}_{6}C, \, ^{14}_{6}C \right) \]


Step 3: Identify the isotones.

By definition, isotones must share the exact same neutron number \(N = A - Z\). Looking at our calculations in Step 1:

For \( ^{198}_{80}Hg \), the neutron count is \(N = 118\).
For \( ^{197}_{79}Au \), the neutron count is \(N = 118\).

Since both nuclides have different atomic numbers (\(80 \neq 79\)) and different mass numbers (\(198 \neq 197\)) but share the identical neutron count of \(118\), the pair of isotones is: \[ \left( ^{198}_{80}Hg, \, ^{197}_{79}Au \right) \] Quick Tip: To remember nuclear terminology quickly: - Iso {p}opes have the same {P}rotons (\(Z\)). - Iso {t}ones have the same neu {T}rons (\(N = A-Z\)). - Iso {b}ars have the same mass number ({A}).

Concept:
The volume of a nucleus is directly proportional to the total number of nucleons (protons and neutrons) inside it, which is represented by the mass number \(A\). Assuming the nucleus is spherical with a radius \(R\), the volume \(V\) is: \[ V = \frac{4}{3}\pi R^3 \]
Since \(V \propto A\), we find that \(R \propto A^{1/3}\).
The nuclear density \( \rho \) is defined as the ratio of the total mass of the nucleus to its total volume: \[ \rho = \frac{Mass of nucleus}{Volume of nucleus} \]


Step 1: Derive the relation between nuclear radius and mass number \(A\).

Experimentally, it has been observed that the volume of a nucleus \(V\) is directly proportional to its mass number \(A\): \[ V \propto A \]
Since we model the nucleus as a sphere of radius \(R\): \[ \frac{4}{3}\pi R^3 \propto A \quad \Rightarrow \quad R^3 \propto A \quad \Rightarrow \quad R \propto A^{1/3} \]
By introducing a proportionality constant \(R_0\) (known as the nuclear unit radius), we get the fundamental relationship: \[ R = R_0 A^{1/3} \quad \cdots (1) \]
Where \(R_0\) is a constant with an approximate value of \(1.2 \times 10^{-15} m\) (or \(1.2 fm\)).


Step 2: Express the mass and volume of the nucleus in terms of \(A\).

Let \(m\) represent the average mass of a single nucleon (proton or neutron). Since there are \(A\) total nucleons in a nucleus of mass number \(A\), the total mass \(M\) of the nucleus is given by: \[ M = m \times A \quad \cdots (2) \]
Now, substituting the expression for the radius \(R\) from Equation (1) into the formula for the volume of a sphere: \[ V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \left(R_0 A^{1/3}\right)^3 \]
Simplifying this expression yields: \[ V = \frac{4}{3}\pi R_0^3 A \quad \cdots (3) \]


Step 3: Calculate the nuclear density and show that it is constant.

By definition, the density of nuclear matter \(\rho\) is: \[ \rho = \frac{M}{V} \]
Substitute the values of mass \(M\) from Equation (2) and volume \(V\) from Equation (3) into this equation: \[ \rho = \frac{m \cdot A}{\frac{4}{3}\pi R_0^3 A} \]
Notice that the mass number \(A\) appears in both the numerator and the denominator, and therefore cancels out completely: \[ \rho = \frac{m}{\frac{4}{3}\pi R_0^3} = \frac{3m}{4\pi R_0^3} \quad \cdots (4) \]
Since \(m\) (average nucleon mass \(\approx 1.67 \times 10^{-27} kg\)) and \(R_0\) (\(\approx 1.2 \times 10^{-15} m\)) are universal physical constants, the density \(\rho\) is completely independent of the mass number \(A\).

Substituting these constant values: \[ \rho \approx \frac{3 \times (1.67 \times 10^{-27} kg)}{4 \times 3.1416 \times (1.2 \times 10^{-15} m)^3} \approx 2.3 \times 10^{17} kg/m^3 \]
This extremely high, uniform value proves that nuclear density is constant across all chemical elements. Quick Tip: - The nuclear radius scales as \( R = R_0 A^{1/3} \). - The volume scales as \( V \propto R^3 \propto A \). - Since both the total mass and volume are directly proportional to \( A \), their ratio (density) is a constant: \(\rho = \frac{3m}{4\pi R_0^3}\).

Concept:
An equipotential surface is defined as a locus of points in space where the electrostatic potential \(V\) remains constant.
A fundamental relation between the electric field \(\mathbf{E}\) and the electric potential gradient is: \[ dV = -\mathbf{E} \cdot d\mathbf{r} \]
For an equipotential surface, the potential difference between any two neighboring points on the surface is zero (\(dV = 0\)). This implies: \[ \mathbf{E} \cdot d\mathbf{r} = 0 \]
This dot product shows that the electric field vector \(\mathbf{E}\) is always perpendicular to the equipotential surfaces at every point.


Step 1: Analyze the given electric field vector.

The electric field is specified as: \[ \mathbf{E} = E_0 \hat{\mathbf{i}} \]
This means that the electric field is uniform in magnitude (\(E_0\)) and is directed strictly along the positive x-direction (represented by the unit vector \(\hat{\mathbf{i}}\)).


Step 2: Apply the relation between electric field and potential.

Let a small displacement vector in 3D Cartesian coordinates be given by: \[ d\mathbf{r} = dx \hat{\mathbf{i}} + dy \hat{\mathbf{j}} + dz \hat{\mathbf{k}} \]
Using the relationship \(dV = -\mathbf{E} \cdot d\mathbf{r}\), we substitute our field vector: \[ dV = -(E_0 \hat{\mathbf{i}}) \cdot (dx \hat{\mathbf{i}} + dy \hat{\mathbf{j}} + dz \hat{\mathbf{k}}) \]
Since \(\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = 1\), \(\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = 0\), and \(\hat{\mathbf{i}} \cdot \hat{\mathbf{k}} = 0\), this simplifies directly to: \[ dV = -E_0 dx \quad \cdots (1) \]


Step 3: Determine the equation of the equipotential surface.

For an equipotential surface, the potential \(V\) is constant, meaning \(dV = 0\). Substituting this condition into Equation (1): \[ -E_0 dx = 0 \quad \Rightarrow \quad dx = 0 \]
Integrating both sides gives: \[ x = constant \]
The equation \(x = C\) represents planes that are perpendicular to the x-axis. These planes are parallel to the \(y\)-\(z\) coordinate plane.
Thus, for a uniform electric field pointing along the positive x-direction, the equipotential surfaces are a family of parallel, equidistant flat planes parallel to the \(y\)-\(z\) plane. Quick Tip: Electric field lines and equipotential surfaces are always mutually perpendicular: - If \(\mathbf{E}\) is along the x-axis (\(\hat{\mathbf{i}}\)), then the perpendicular surfaces must lie in the \(y\)-\(z\) plane (\(x = constant\)). - For a uniform field, these planes are flat and equally spaced.

Concept:
The electrostatic potential \(V\) at a distance \(r\) from a single isolated point charge \(q\) is given by: \[ V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r} \]
By the principle of superposition, the total electrostatic potential at any point due to a system of point charges is the algebraic sum of the individual potentials contributed by each charge: \[ V_{total} = V_1 + V_2 + \ldots = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i} \]


Step 1: Set up the coordinates and find individual distances.

We are given two charges:

Charge \( q_1 = -q \) located at position \( A(-a, 0, 0) \).
Charge \( q_2 = +q \) located at position \( B(a, 0, 0) \).

We want to determine the electric potential at an axial point \( P(x, 0, 0) \). Since we are interested in the region \( x \gg a \), we assume \(P\) lies on the positive x-axis such that \( x > a \).
The distance from the negative charge at \( A \) to point \( P \) is: \[ r_1 = AP = x - (-a) = x + a \]
The distance from the positive charge at \( B \) to point \( P \) is: \[ r_2 = BP = x - a \]


Step 2: Calculate the total potential at point \( P \).

Using the superposition principle, we sum the potentials from both charges: \[ V = V_1 + V_2 = \frac{1}{4\pi\varepsilon_0} \left[ \frac{-q}{r_1} \right] + \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{r_2} \right] \]
Substitute the expressions for \(r_1\) and \(r_2\): \[ V = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{x - a} - \frac{1}{x + a} \right] \quad \cdots (1) \]
Now, find a common denominator to combine the terms inside the square brackets: \[ \frac{1}{x - a} - \frac{1}{x + a} = \frac{(x + a) - (x - a)}{(x - a)(x + a)} = \frac{2a}{x^2 - a^2} \]
Substituting this back into Equation (1) yields: \[ V = \frac{1}{4\pi\varepsilon_0} \frac{2qa}{x^2 - a^2} \quad \cdots (2) \]


Step 3: Apply the approximation for \( x \gg a \).

Since it is specified that \( x \gg a \), the term \( a^2 \) is extremely small compared to \( x^2 \). Therefore, we can safely approximate: \[ x^2 - a^2 \approx x^2 \]
Using this approximation in Equation (2), the electrostatic potential simplifies to: \[ V \approx \frac{1}{4\pi\varepsilon_0} \frac{2qa}{x^2} \]
Notice that \( p = 2qa \) is the electric dipole moment of this system. In terms of \(p\), the potential is: \[ V \approx \frac{1}{4\pi\varepsilon_0} \frac{p}{x^2} \] Quick Tip: - The configuration of \( -q \) and \( +q \) separated by distance \( 2a \) is a classical physical dipole. - For a dipole, the axial potential falls off as \( 1/x^2 \) (where \( V = \frac{kp}{x^2} \)), unlike a single point charge potential which falls off as \( 1/x \).

Concept:
The electromagnetic spectrum consists of waves categorized by their production mechanisms, frequencies, and wavelengths.
Different parts of the spectrum interact with human tissues in unique ways. High-energy, short-wavelength electromagnetic waves have significant penetrating power, allowing them to pass through soft tissues but be absorbed by denser calcium structures like bone. This differential absorption makes them ideal for diagnostic imaging.


Step 1: Identify the electromagnetic wave used for fracture detection.

The electromagnetic waves used for detecting fractures in bones are X-rays.
Because X-rays have high energy, they can easily penetrate soft tissues (like skin and muscle) but are absorbed by dense materials such as calcium-rich bones. When a photographic film or digital detector is placed behind the body, a shadow image is created: bones appear light/white, while soft tissues appear dark, clearly revealing any cracks or fractures.


Step 2: Determine the standard wavelength range.

X-rays lie between gamma rays and ultraviolet rays in the electromagnetic spectrum.
The wavelength range for X-rays is generally accepted to be: \[ 10^{-12} m to 10^{-8} m \quad (or 1 pm to 10 nm) \]
In terms of frequency, this corresponds to high frequencies in the range of approximately \( 3 \times 10^{16} Hz \) to \( 3 \times 10^{19} Hz \). Quick Tip: - X-rays were discovered by Wilhelm Röntgen in 1895. - Because of their ionizing nature, exposure to diagnostic X-rays is kept minimal, but their high penetration power is irreplaceable for bone and dental imaging.

Concept:
Physiotherapy often utilizes heat therapy (thermotherapy) to treat muscle spasms, joint stiffness, and chronic pain.
Certain electromagnetic waves transfer thermal energy directly upon hitting a surface by vibrating water molecules and organic structures in body tissues. This radiant heat increases localized blood circulation, relaxes muscles, and accelerates healing.


Step 1: Identify the electromagnetic wave used in physiotherapy.

The electromagnetic waves used for therapeutic heating in physiotherapy are Infrared (IR) waves (often referred to as heat waves).
Infrared lamps are directed at targeted parts of a patient's body. The skin and superficial muscle layers absorb this radiation, generating a soothing thermal effect that dilates blood vessels (vasodilation) and enhances recovery in damaged tissues.


Step 2: Determine the standard wavelength range.

Infrared radiation lies just beyond the red end of the visible light spectrum.
Its wavelength range spans from the edge of red visible light up to the microwave region: \[ 7 \times 10^{-7} m to 10^{-3} m \quad (or 700 nm to 1 mm) \]
The corresponding frequency range is roughly \( 3 \times 10^{11} Hz \) to \( 4 \times 10^{14} Hz \). Quick Tip: - Infrared waves are frequently called "heat waves" because any material absorbing them converts the radiation directly into internal thermal energy. - They are also widely utilized in TV remote controls and night-vision equipment.

Concept:
RADAR (RAdio Detection And Ranging) systems operate by transmitting high-frequency electromagnetic pulses into the atmosphere and detecting the reflected echo from objects (like aircraft or ships).
To detect small or fast-moving targets accurately without excessive beam spreading (diffraction), the system requires waves with short wavelengths that can travel long distances in straight lines through atmospheric disturbances.


Step 1: Identify the electromagnetic wave used in radar systems.

The electromagnetic waves used for radar systems are Microwaves.
Because of their short wavelengths (compared to standard radio waves), microwaves exhibit minimal diffraction, meaning they do not bend significantly around obstacles and can be focused into narrow, concentrated directional beams. This allows for precise calculation of a target's position, speed, and distance.


Step 2: Determine the standard wavelength range.

Microwaves occupy the spectrum between radio waves and infrared waves.
The standard wavelength range for microwaves is: \[ 1 mm to 0.3 m \quad (or 10^{-3} m to 3 \times 10^{-1} m) \]
This corresponds to a frequency range of approximately \( 1 GHz \) (\( 10^9 Hz \)) to \( 300 GHz \) (\( 3 \times 10^{11} Hz \)). Quick Tip: - Microwaves are used in RADAR, satellite communication, GPS, and standard domestic microwave ovens (which specifically utilize the \(2.45 GHz\) resonance frequency of water molecules).

Concept:
Mutual induction is an electromagnetic phenomenon where a time-varying electric current flowing through one primary coil establishes a changing magnetic flux in a neighboring secondary coil. By Faraday's Law of Electromagnetic Induction, this changing flux induces an electromotive force (emf) across the terminals of the secondary coil.

Quantitatively, the total magnetic flux linkage \( \Phi_s \) through the secondary coil is directly proportional to the current \( I_p \) circulating in the primary coil: \[ \Phi_s \propto I_p \quad \Rightarrow \quad \Phi_s = M \cdot I_p \]
where the proportionality constant \( M \) represents the Mutual Inductance (or coefficient of mutual induction) of the given pair of coils.


Step 1: Formal Definition of Mutual Inductance based on Flux Linkage.

The mutual inductance of a pair of coils can be defined as the total magnetic flux linked with the secondary coil when a unit steady current (exactly 1 Ampere) passes through the primary coil.

Mathematically, if we set the primary current to unity: \[ I_p = 1 Ampere \]
Substituting this value into the core flux equation yields: \[ \Phi_s = M \cdot (1) \quad \Rightarrow \quad M = \Phi_s \]
This shows that the coefficient of mutual inductance is numerically equivalent to the total flux linked with the neighboring circuit under unit current conditions.


Step 2: Alternative Definition based on Induced Electromotive Force.

According to Faraday's law, the electromotive force \( e_s \) induced in the secondary coil is equal to the negative rate of change of magnetic flux passing through it: \[ e_s = -\frac{d\Phi_s}{dt} \]
Substituting the relation \( \Phi_s = M \cdot I_p \) into this differential equation, and assuming the geometric positioning of the coils remains constant over time so that \( M \) is a constant: \[ e_s = -\frac{d}{dt}(M \cdot I_p) = -M \cdot \frac{dI_p}{dt} \]
Taking the absolute magnitude of both sides: \[ |e_s| = M \cdot \frac{dI_p}{dt} \]
If the current in the primary coil changes at a unit rate, meaning \( \frac{dI_p}{dt} = 1 Ampere per second \), then the equation simplifies directly to: \[ |e_s| = M \cdot (1) \quad \Rightarrow \quad M = |e_s| \]
Therefore, mutual inductance is also defined as the magnitude of the electromotive force induced in the secondary coil when the rate of change of electric current in the primary coil is exactly 1 Ampere per second.


Step 3: SI Unit of Mutual Inductance and its Physical Definition.

The standard International System of Units (SI) unit for measuring mutual inductance is the henry, which is denoted by the capital letter H.

We can define 1 Henry by setting all physical parameters in our induced emf equation to unity: \[ M = \frac{|e_s|}{\left(\frac{dI_p}{dt}\right)} \] \[ 1 Henry = \frac{1 Volt}{1 Ampere / second} = 1 Volt \cdot second \cdot Ampere^{-1} \]
Hence, the mutual inductance of a pair of coils is exactly 1 Henry if a uniform current change of 1 Ampere per second in the primary coil induces an electromotive force of exactly 1 Volt across the terminal outputs of the secondary coil. In terms of magnetic flux units, 1 Henry is also equivalent to 1 Weber per Ampere (\( 1 Wb/A \)). Quick Tip: For arithmetic progressions, use the mean and variance formulas effectively: - Mutual inductance is a purely structural and geometric property. It depends entirely on the number of turns in both coils, their respective cross-sectional areas, their relative spatial orientation, their distance of separation, and the magnetic permeability of the core material materializing between them. It never changes with variations in current or voltage.

Concept:
To derive the mutual inductance \( M \) between two distinct coaxial magnetic components, we utilize the principle of flux linkage. Let the long outer solenoid be designated as Component 1 (Primary) and the inner circular loop be designated as Component 2 (Secondary).

According to Ampere's Circuital Law, a long solenoid carrying a steady current \( I_1 \) generates an internal magnetic field \( B_1 \) that is highly uniform along its central longitudinal axis: \[ B_1 = \mu_0 \cdot n \cdot I_1 \]
where \( \mu_0 \) is the magnetic permeability of free space and \( n \) is the turn density. The total magnetic flux \( \Phi_2 \) passing through the inner circular loop depends on this field and the effective area of the loop. By definition, this total flux is related to the primary current via the mutual inductance constant: \[ \Phi_2 = M \cdot I_1 \]
Equating the flux derived from field integrations to this definition permits the isolated calculation of \( M \).


Step 1: Defining the Structural Parameters of the Configuration.

Let us explicitly write down the geometric and physical variables given for our two-component electromagnetic setup:

Outer Solenoid (Component 1): Radius = \( R \), Length = \( L \), Turn density (turns per unit length) = \( n \).
Inner Circular Loop (Component 2): Radius = \( r \) (given that \( r < R \)), Total number of turns = \( N_2 = 1 \).

The inner loop is placed deep inside the center of the solenoid core. Because its axis perfectly coincides with the central longitudinal axis of the solenoid, the flat cross-sectional surface area of the loop sits perfectly perpendicular to the straight magnetic field lines running through the solenoid.


Step 2: Calculating the Magnetic Field produced by the Solenoid.

Let a steady electric current \( I_1 \) flow through the turns of the outer solenoid. Because the solenoid is long (\( L \gg R \)), the magnetic field lines near its center are completely parallel, straight, and uniform. The magnitude of this axial magnetic field \( B_1 \) is written as: \[ B_1 = \mu_0 \cdot n \cdot I_1 \quad \cdots (1) \]
This field acts over the entire inner space enclosed by the solenoid shell, meaning the smaller circular loop is completely immersed in this uniform magnetic field environment.


Step 3: Determining the Magnetic Flux linked with the Inner Loop.

The magnetic flux \( \Phi_2 \) passing through the single-turn circular loop is calculated using the surface area integral of the magnetic field. Since the area vector \( \vec{A}_2 \) of the loop is parallel to the magnetic field vector \( \vec{B}_1 \), the dot product reduces to basic scalar multiplication because \( \cos(0^\circ) = 1 \).

The cross-sectional area of the circular loop is: \[ A_2 = \pi \cdot r^2 \]
The total magnetic flux passing through this area is: \[ \Phi_2 = B_1 \cdot A_2 \]
Substituting the mathematical expression for \( B_1 \) from Equation (1) into this relation: \[ \Phi_2 = (\mu_0 \cdot n \cdot I_1) \cdot (\pi \cdot r^2) \]
Rearranging the variables to isolate the current parameter: \[ \Phi_2 = (\mu_0 \cdot \pi \cdot n \cdot r^2) \cdot I_1 \quad \cdots (2) \]


Step 4: Equating to the Definition of Mutual Inductance to find \( M \).

By the standard fundamental definition of mutual inductance, the total flux linkage in the secondary system is directly proportional to the current causing it in the primary system: \[ \Phi_2 = M \cdot I_1 \quad \cdots (3) \]
Now, we directly equate the right-hand sides of Equation (2) and Equation (3): \[ M \cdot I_1 = (\mu_0 \cdot \pi \cdot n \cdot r^2) \cdot I_1 \]
Dividing both sides of the equation by the current \( I_1 \) eliminates the current variable entirely: \[ M = \mu_0 \cdot \pi \cdot n \cdot r^2 \]
This is the final expression for the mutual inductance. Alternatively, if the total number of turns on the outer solenoid is given as \( N_1 \), we can substitute \( n = \frac{N_1}{L} \) to state the expression purely in terms of total turns: \[ M = \frac{\mu_0 \cdot \pi \cdot N_1 \cdot r^2}{L} \]
This result confirms that the mutual inductance depends entirely on the turn density of the outer solenoid and the cross-sectional area of the inner loop. Quick Tip: For arithmetic progressions, use the mean and variance formulas effectively: - Notice that the radius of the outer solenoid \( R \) does not appear anywhere in the final formula for mutual inductance \( M \). This is because the magnetic field inside a long solenoid depends solely on its turn density and current, not its overall width, meaning \( R \) acts as an irrelevant distracter variable in this problem.

Concept:
A current-carrying conductor produces a magnetic field. Another nearby current-carrying conductor placed in this magnetic field experiences a force due to Lorentz force: \[ F = I (L \times B) \]




Step 1: Magnetic field due to conductor A at location of B


For a long straight conductor carrying current \(I_a\), magnetic field at distance \(d\) is: \[ B_A = \frac{\mu_0 I_a}{2\pi d} \]

Direction: given by right-hand thumb rule (circular magnetic field lines around A).




Step 2: Force on conductor B due to A


Conductor B carries current \(I_b\) and is placed in magnetic field \(B_A\).

Force on a current-carrying wire: \[ F = I L B \sin\theta \]

Here: \[ \theta = 90^\circ \Rightarrow \sin\theta = 1 \]

So: \[ F_B = I_b L B_A \]

Substitute \(B_A\): \[ F_B = I_b L \cdot \frac{\mu_0 I_a}{2\pi d} \]
\[ F_B = \frac{\mu_0 I_a I_b}{2\pi d} L \]

Nature of force:
Since currents are in same direction → force is attractive.




Step 3: Force on conductor A due to B


Similarly, magnetic field due to B at A is: \[ B_B = \frac{\mu_0 I_b}{2\pi d} \]

Force on A: \[ F_A = I_a L B_B \]
\[ F_A = I_a L \cdot \frac{\mu_0 I_b}{2\pi d} \]
\[ F_A = \frac{\mu_0 I_a I_b}{2\pi d} L \]




Step 4: Newton’s third law verification


We observe: \[ F_A = F_B \]

But directions are opposite:

Force on B is towards A
Force on A is towards B


Thus: \[ \vec{F}_A = - \vec{F}_B \]

Hence, the interaction satisfies Newton’s third law.




Final Answer: \[ \boxed{F = \frac{\mu_0 I_a I_b}{2\pi d} L \quad (attractive force)} \] Quick Tip: Parallel currents in same direction attract, opposite currents repel. This is the basis of the definition of ampere.

Concept:
Niels Bohr introduced a semi-classical model for the hydrogen atom in 1913 to explain the stability of atoms and the origin of spectral lines. The second postulate of his theory plays a pivotal role in transitioning classical physics concepts to the quantum domain by introducing the quantization of angular momentum.


Step 1: State Bohr's Second Postulate.

Bohr's second postulate (also known as the **quantization condition**) states that:

An electron can revolve around the nucleus only in certain selected non-radiating circular orbits, called stable or stationary orbits. In these orbits, the orbital angular momentum (\(L\)) of the revolving electron is an integral multiple of \( \frac{h}{2\pi} \) (or \( \hbar \)).

Mathematically, this condition is expressed as: \[ L = m v r = \frac{n h}{2\pi} = n\hbar \]
Where:

\( m \) is the mass of the revolving electron,
\( v \) is the orbital speed of the electron in the \(n\)-th orbit,
\( r \) is the radius of the \(n\)-th circular orbit,
\( h \) is Planck's constant (\( 6.626 \times 10^{-34} J\cdots \)),
\( n \) is a positive integer called the principal quantum number (\( n = 1, 2, 3, \ldots \)).



Step 2: Explain the physical significance of the postulate.

The significance of Bohr's second postulate can be understood through the following key aspects:

Introduction of Quantization: It was the first time that quantization was applied to a dynamical property like angular momentum. This successfully restricted the electron from existing in arbitrary orbits, explaining why atoms have discrete energy levels.
Atomic Stability: According to classical electromagnetic theory, an accelerating charged particle (like an electron in circular motion) must continuously radiate energy. If it did, the electron would lose energy and spiral into the nucleus, making the atom highly unstable. Bohr's postulate solves this by stating that as long as the electron remains in these quantized stationary orbits, it does not radiate any energy, ensuring the absolute stability of the atom.
de Broglie's Hypothesis Connection: The physical justification for this postulate was later given by Louis de Broglie in 1924. He showed that if an electron behaves as a standing wave around the nucleus, the circumference of the circular orbit must contain an integral number of de Broglie wavelengths (\( \lambda \)):
\[ 2\pi r = n\lambda \]
Substituting de Broglie's relation \( \lambda = \frac{h}{p} = \frac{h}{mv} \) into this equation yields:
\[ 2\pi r = n \left(\frac{h}{mv}\right) \quad \Rightarrow \quad mvr = \frac{nh}{2\pi} \]
This beautifully validates Bohr's assumption on a theoretical quantum wave basis. Quick Tip: To remember Bohr's second postulate easily: - Think of it as the "Standing Wave" condition. - The orbital circumference \( 2\pi r \) must accommodate exactly \( n \) complete de Broglie wavelengths of the electron to prevent destructive interference: \( mvr = n\frac{h}{2\pi} \).

Concept:
According to Bohr's model of the hydrogen atom, the energy of an electron in the \(n\)-th stationary state is quantized and inversely proportional to the square of the principal quantum number \(n\). To prove that energy levels get closer as \(n\) increases, we must analyze the difference between consecutive energy states \( \Delta E = E_{n+1} - E_n \) as \( n \to \infty \).


Step 1: State the energy formula for the hydrogen atom.

The energy \( E_n \) of an electron in the \(n\)-th orbit of a hydrogen atom (\( Z = 1 \)) is given by: \[ E_n = -\frac{me^4}{8 \varepsilon_0^2 h^2} \cdot \frac{1}{n^2} = -\frac{13.6}{n^2} eV \]
For our proof, let us write: \[ E_n = -\frac{K}{n^2} \]
Where \( K = \frac{me^4}{8 \varepsilon_0^2 h^2} \approx 13.6 eV \) is a constant.


Step 2: Calculate the energy difference between two consecutive levels.

Let \( E_n \) and \( E_{n+1} \) be the energy levels of two consecutive orbits. The separation (difference in energy) between these levels is: \[ \Delta E = E_{n+1} - E_n \]
Substituting the energy expression: \[ \Delta E = \left( -\frac{K}{(n+1)^2} \right) - \left( -\frac{K}{n^2} \right) \]
Simplifying the terms: \[ \Delta E = K \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right] \]
Taking a common denominator: \[ \Delta E = K \left[ \frac{(n+1)^2 - n^2}{n^2 (n+1)^2} \right] \]
Expanding the numerator \( (n+1)^2 = n^2 + 2n + 1 \): \[ \Delta E = K \left[ \frac{n^2 + 2n + 1 - n^2}{n^2 (n+1)^2} \right] = K \left[ \frac{2n + 1}{n^2 (n+1)^2} \right] \quad \cdots (1) \]


Step 3: Analyze the limit behavior as \( n \) becomes very large (\( n \gg 1 \)).

For very large values of the principal quantum number \( n \):

\( 2n + 1 \approx 2n \)
\( n+1 \approx n \)

Substituting these approximations into equation (1): \[ \Delta E \approx K \left[ \frac{2n}{n^2 \cdot n^2} \right] = K \left[ \frac{2n}{n^4} \right] = \frac{2K}{n^3} \]
Since \( \Delta E \propto \frac{1}{n^3} \), as the principal quantum number \( n \) approaches infinity: \[ \lim_{n \to \infty} \Delta E = \lim_{n \to \infty} \frac{2K}{n^3} = 0 \]
This mathematically demonstrates that as \( n \) increases, the energy gap between consecutive energy levels decreases drastically. For example:

E_2 - E_1 &= -3.4 - (-13.6) = 10.2 eV

E_3 - E_2 &= -1.51 - (-3.4) = 1.89 eV

E_4 - E_3 &= -0.85 - (-1.51) = 0.66 eV

Hence, the energy levels get closer and closer as \( n \) becomes larger. Quick Tip: To easily remember this behavior: - The energy levels of a hydrogen atom are spaced inversely with \( n^2 \) (\( E_n \propto -1/n^2 \)). - The distance between consecutive levels scales as ( frac{1{n^3} ). - As \( n \to \infty \), this separation converges to zero, leading to a continuous energy band near the ionization limit (\( E = 0 \)).

Concept:
For any physical quantity to be classified as a vector, it must satisfy three essential criteria:

It must have a magnitude.
It must have a specified direction.
It must strictly obey the laws of vector addition (such as the Triangle Law or Parallelogram Law of Vector Addition).

While electric current possesses both magnitude and direction, it fails the third criterion.


Step 1: Analyze the representation of current with an arrow.

In circuit diagrams, we use arrows to represent the direction of flow of conventional current (which is the direction of flow of positive charge carriers, or opposite to the flow of electrons).
For instance, an arrow pointing from node A to node B indicate that charge is moving in that specific spatial direction. This satisfies the magnitude and direction requirements on a macroscopic scale.


Step 2: Test current with the Laws of Vector Addition.

Consider a junction in an electrical circuit where two wires carrying currents \( I_1 \) and \( I_2 \) meet at an arbitrary angle \( \theta \), merging into a third wire carrying current \( I_3 \).

According to **Kirchhoff's Current Law (KCL)**, which is based on the conservation of electric charge, the total current leaving the junction must equal the total current entering it:
\[ I_3 = I_1 + I_2 \]
If electric current were a vector quantity, the resultant current \( I_3 \) would depend on the spatial angle \( \theta \) between the two incoming wires, according to the vector addition formula:
\[ I_{vector} = \sqrt{I_1^2 + I_2^2 + 2I_1 I_2 \cos\theta} \]

Experimentally and physically, changing the angle \( \theta \) between the wires does not alter the current \( I_3 \) in the outgoing wire. The current adds up purely algebraically: \[ I_3 = I_1 + I_2 \quad (independent of \theta) \]
Because current addition does not obey the parallelogram law of vector addition, it is classified as a **scalar quantity** (or more specifically, a component of a tensor of rank 0). Quick Tip: The fundamental difference between vectors and scalars: - Vectors depend on spatial orientation (e.g., forces at different angles yield different resultants). - Scalars follow basic algebraic addition. Since \( I_{total} = I_1 + I_2 \) regardless of the angle between the wires, electric current is fundamentally a scalar!

Concept:
To find the current in the given network, we apply Kirchhoff's rules:

Kirchhoff's First Rule (Junction Rule / KCL): The algebraic sum of currents meeting at any junction in a closed circuit is zero (\( \sum I = 0 \)).
Kirchhoff's Second Rule (Loop Rule / KVL): In any closed loop of a network, the algebraic sum of the changes in potential is zero (\( \sum \Delta V = 0 \)).



Step 1: Label and define currents in each branch of the circuit.

Let us define the current distribution in the circuit branches based on the junctions B and E:

Let the current flowing in the left branch from the \( 3 V \) battery towards junction B be \( I_1 \).
Let the current flowing from junction B to C in the right loop through the \( 5 V \) battery be \( I_2 \).
At junction B, according to Kirchhoff's Junction Rule, the current leaving B downwards through the middle branch (containing the \( 3\,\Omega \) resistor) to E must be:
\[ I_3 = I_1 - I_2 \]

Thus, the current through the branches are:

Branch FAB: Current \( I_1 \) flowing from F to A to B.
Branch BCDE: Current \( I_2 \) flowing from B to C to D to E.
Branch BE: Current \( I_3 = I_1 - I_2 \) flowing from B to E through the \( 3\,\Omega \) resistor.



Step 2: Apply Kirchhoff's Loop Rule (KVL) to Loop 1 (FABEF).

Let us traverse the closed loop **FABEF** in a clockwise direction:

Moving through the \( 3 V \) battery from the negative to the positive terminal: \( +3 V \)
Moving from B to E through the \( 3\,\Omega \) resistor in the direction of current \( (I_1 - I_2) \): \( -3(I_1 - I_2) \)
Moving from E to F through the \( 4\,\Omega \) resistor in the direction of current \( I_1 \): \( -4I_1 \)

Summing these changes to zero: \[ 3 - 3(I_1 - I_2) - 4I_1 = 0 \] \[ 3 - 3I_1 + 3I_2 - 4I_1 = 0 \] \[ 7I_1 - 3I_2 = 3 \quad \cdots (1) \]


Step 3: Apply Kirchhoff's Loop Rule (KVL) to Loop 2 (BCDEB).

Let us traverse the closed loop **BCDEB** in a clockwise direction:

Moving through the \( 5 V \) battery from the positive to the negative terminal (since the longer line is on the left towards B and shorter is towards C): \( -5 V \)
Moving from C to D through the \( 2\,\Omega \) resistor in the direction of current \( I_2 \): \( -2I_2 \)
Moving from E to B through the middle branch \( 3\,\Omega \) resistor against the direction of current \( (I_1 - I_2) \): \( +3(I_1 - I_2) \)

Summing these changes to zero: \[ -5 - 2I_2 + 3(I_1 - I_2) = 0 \] \[ -5 - 2I_2 + 3I_1 - 3I_2 = 0 \] \[ 3I_1 - 5I_2 = 5 \quad \cdots (2) \]


Step 4: Solve equations (1) and (2) simultaneously to find \( I_1 \) and \( I_2 \).

From equation (1), we can express \( 3I_2 \) as: \[ 3I_2 = 7I_1 - 3 \quad \Rightarrow \quad I_2 = \frac{7I_1 - 3}{3} \quad \cdots (3) \]
Substitute equation (3) into equation (2): \[ 3I_1 - 5\left( \frac{7I_1 - 3}{3} \right) = 5 \]
Multiply the entire equation by 3 to clear the denominator: \[ 9I_1 - 5(7I_1 - 3) = 15 \] \[ 9I_1 - 35I_1 + 15 = 15 \] \[ -26I_1 = 0 \quad \Rightarrow \quad I_1 = 0 A \]
Now substitute \( I_1 = 0 A \) back into equation (3) to find \( I_2 \): \[ I_2 = \frac{7(0) - 3}{3} = -1 A \]


Step 5: Calculate the current through the \( 3\,\Omega \) resistor.

The current through the middle \( 3\,\Omega \) resistor is: \[ I_{middle} = I_3 = I_1 - I_2 \]
Substituting the values: \[ I_{middle} = 0 - (-1) = 1 A \]
The positive sign indicates that the actual direction of the current matches our assumed direction, which is flowing downwards from junction B to E. Quick Tip: Nodal Analysis Alternative: - Let the node potential at E be \( V_E = 0 V \). - Let node potential at B be \( V_B = V \). - Applying NCL at node B: \( \frac{V-3}{4} + \frac{V}{3} + \frac{V-5}{2} = 0 \). - Multiplying by 12: \( 3(V-3) + 4V + 6(V-5) = 0 \Rightarrow 13V - 39 = 0 \Rightarrow V = 3 V \). - Current through \( 3\,\Omega \) resistor: \( I = \frac{V}{3} = \frac{3}{3} = 1 A \). This is a highly efficient way to verify loop calculations!

Concept:
The magnetic field at a point on the axis of a circular current loop is derived using the Biot–Savart law. Due to symmetry, only axial components survive while perpendicular components cancel.



Step 1: Biot–Savart law

The magnetic field due to a small current element is: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I\, d\vec{l} \times \hat{r}}{r^2} \]

For a circular loop, we integrate around the entire coil.



Step 2: Geometry of the circular loop

Consider:

Radius of loop = \(r\)
Point on axis at distance = \(x\)
Distance of any current element from point = \(\sqrt{r^2 + x^2}\)




Step 3: Symmetry argument

Each current element produces a magnetic field:

Radial components cancel due to symmetry
Only axial components add up


So we calculate only axial component \(dB_x\).



Step 4: Expression for axial field of one turn

After integrating around the loop: \[ B_{one turn} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}} \]



Step 5: For N turns

For \(N\) identical turns, fields add linearly: \[ B = N \cdot B_{one turn} \]
\[ B = \frac{\mu_0 N I r^2}{2(r^2 + x^2)^{3/2}} \]



Final Answer: \[ \boxed{B = \frac{\mu_0 N I r^2}{2(r^2 + x^2)^{3/2}}} \] Quick Tip: At the center of the loop (\(x=0\)): \[ B = \frac{\mu_0 N I}{2r} \] Always use symmetry to avoid full integration.

Concept:

Pure silicon is known as an intrinsic semiconductor. Its electrical conductivity can be increased significantly by adding a small amount of suitable impurity atoms. This process is called doping.

When a trivalent impurity is added to silicon, one covalent bond remains incomplete because the impurity atom possesses only three valence electrons instead of four. This creates a hole, and the semiconductor becomes a p-type semiconductor.

Trivalent impurities are called acceptor impurities because they accept electrons and create holes as majority charge carriers.



Step 1: Identify the type of impurity required.

To obtain a p-type semiconductor, a trivalent impurity must be added.

Common trivalent impurities are:
\[ Boron (B), Aluminium (Al), Gallium (Ga), Indium (In) \]



Step 2: Examine the given options.
\[ Phosphorus \rightarrow Pentavalent \]
\[ Arsenic \rightarrow Pentavalent \]
\[ Boron \rightarrow Trivalent \]
\[ Antimony \rightarrow Pentavalent \]

Among the given options, only Boron is a trivalent impurity.



Step 3: Conclude the answer.

Since Boron is a trivalent impurity, it produces holes as majority charge carriers and converts silicon into a p-type semiconductor.
\[ \boxed{Boron} \]

Hence, the correct answer is
\[ \boxed{(C)} \] Quick Tip: Remember the basic rule: \[ Trivalent impurity \Rightarrow p-type semiconductor \] \[ Pentavalent impurity \Rightarrow n-type semiconductor \] Boron is the most common trivalent impurity used in silicon.

Concept:

For a semiconductor in thermal equilibrium, the product of electron concentration and hole concentration remains constant and is given by
\[ np=n_i^2 \]

where
\[ n=electron concentration \]
\[ p=hole concentration \]
\[ n_i=intrinsic carrier concentration \]

This relation is known as the mass action law.



Step 1: Write the given quantities.

Electron concentration:
\[ n=5\times10^{22}\,m^{-3} \]

Intrinsic carrier concentration:
\[ n_i=1.5\times10^{16}\,m^{-3} \]



Step 2: Apply the mass action law.

Using
\[ np=n_i^2 \]

we obtain
\[ p=\frac{n_i^2}{n} \]

Substituting the values,
\[ p= \frac{(1.5\times10^{16})^2} {5\times10^{22}} \]
\[ = \frac{2.25\times10^{32}} {5\times10^{22}} \]
\[ = 0.45\times10^{10} \]
\[ = 4.5\times10^{9}\,m^{-3} \]



Step 3: Verify the result.

Since the electron concentration is extremely large, the hole concentration must be very small so that the product \(np\) remains constant.

The calculated value satisfies this condition.



Final Answer:
\[ \boxed{ p=4.5\times10^{9}\,m^{-3} } \]

Therefore, the correct option is
\[ \boxed{(D)} \] Quick Tip: For semiconductors in equilibrium: \[ np=n_i^2 \] If one carrier concentration increases, the other decreases proportionally. This formula is frequently used in board examinations and competitive exams.

Concept:

A p-n junction diode consists of a p-type semiconductor and an n-type semiconductor joined together. Due to the concentration difference of charge carriers on the two sides, majority carriers diffuse across the junction and form a depletion region. This depletion region establishes a built-in potential barrier.

In an unbiased p-n junction:
\[ Diffusion Current = Drift Current \]

and the net current through the junction is zero.

When the diode is forward biased, the external voltage opposes the barrier potential, reducing the width of the depletion layer and allowing majority charge carriers to cross the junction easily.



Step 1: Understand the effect of forward biasing.

In forward biasing:
\[ p-side \rightarrow Connected to positive terminal \]
\[ n-side \rightarrow Connected to negative terminal \]

The applied voltage reduces the barrier potential.

As a result,
\[ Depletion layer width decreases \]

and majority carriers can move across the junction more easily.



Step 2: Identify the dominant current.

Since a large number of majority carriers cross the junction due to concentration difference, the current produced is called diffusion current.

The diffusion current becomes much larger than the drift current under forward bias.

Therefore, the current in a forward-biased diode is mainly due to the motion of majority carriers.



Step 3: Analyse the given options.

Option (A): Majority carriers dominate the current flow. checkmark

Option (B): Drift current is due to minority carriers and dominates in reverse bias. \(\times\)

Option (C): Diffusion current and drift current are equal only in equilibrium (unbiased condition). \(\times\)

Option (D): Current depends on circuit conditions and cannot always be taken as \(1\,A\). \(\times\)



Final Answer:

The current in a forward-biased p-n junction is mainly due to majority charge carriers crossing the junction.
\[ \boxed{(A) current is mainly due to drifting of majority carriers} \] Quick Tip: Remember: \[ Forward Bias \Rightarrow Majority Carrier Current \] \[ Reverse Bias \Rightarrow Minority Carrier Current \] Most questions on p-n junctions can be solved by identifying which carriers dominate the conduction process.

Concept:

The threshold voltage, also called the cut-in voltage or knee voltage, is the minimum forward bias voltage that must be applied across a diode before it starts conducting appreciable current.

Initially, the applied voltage is used in overcoming the barrier potential of the depletion region. Once this barrier is sufficiently reduced, the diode begins to conduct rapidly.

The threshold voltage depends on the semiconductor material used.



Step 1: Recall standard threshold voltages.

For commonly used semiconductor diodes:
\[ Germanium Diode \approx 0.3\,V \]
\[ Silicon Diode \approx 0.7\,V \]

These values are standard and frequently used in electronic circuit analysis.



Step 2: Understand why silicon has a larger threshold voltage.

Silicon has a larger energy band gap than germanium.
\[ E_g(Si) \approx 1.1\,eV \]
\[ E_g(Ge) \approx 0.7\,eV \]

Therefore, a larger forward voltage is required before significant conduction begins in a silicon diode.



Step 3: Compare with the options.

Among the given options,
\[ 0.7\,V \]

is the accepted threshold voltage of a silicon diode.



Final Answer:
\[ \boxed{0.7\,V} \]

Hence, the correct option is
\[ \boxed{(C)} \] Quick Tip: A very common board-exam fact: \[ Germanium Diode \rightarrow 0.3\,V \] \[ Silicon Diode \rightarrow 0.7\,V \] Students should memorize these values because they are frequently used in numerical and conceptual questions.

Concept:

Germanium is a group 14 semiconductor having four valence electrons. When it is doped with a pentavalent impurity such as phosphorus, arsenic, or antimony, each impurity atom contributes five valence electrons.

Four of these electrons participate in covalent bond formation with neighbouring germanium atoms, while the fifth electron remains only weakly bound to the impurity atom. This extra electron can be easily detached even at room temperature and contributes to electrical conduction.

Such impurities are called donor impurities because they donate free electrons to the semiconductor crystal.



Step 1: Understand donor doping in germanium.

When a pentavalent atom is introduced into a germanium crystal:
\[ Number of valence electrons of Ge = 4 \]
\[ Number of valence electrons of donor atom = 5 \]

Out of these five electrons:


Four electrons form covalent bonds with neighbouring Ge atoms.
One electron remains weakly attached to the donor atom.


Therefore, only a very small amount of energy is required to free this electron.



Step 2: Recall the donor ionisation energy.

The donor energy level lies very close to the conduction band.

For germanium, the ionisation energy of the donor electron is approximately
\[ 0.01\,eV \]

This value is much smaller than the energy gap of germanium, which explains why donor electrons can easily become free charge carriers.



Step 3: Compare with the given options.

The available options are:
\[ 0.01\,eV \]
\[ 0.05\,eV \]
\[ 0.10\,eV \]
\[ 0.15\,eV \]

The accepted value for the donor ionisation energy in germanium is
\[ 0.01\,eV \]



Step 4: Conclude the answer.

Hence, the ionisation energy of the fifth weakly bound electron is
\[ \boxed{0.01\,eV} \]

Therefore, the correct option is
\[ \boxed{(A)} \] Quick Tip: In donor-doped semiconductors, the extra electron is very loosely bound because the donor energy level lies very close to the conduction band. For germanium: \[ E_D \approx 0.01\,eV \] Hence, even room-temperature thermal energy is sufficient to free most donor electrons and make them available for conduction.

Concept:

This question is based on the displacement method of determining the focal length of a convex lens. When the distance between the object and the screen is greater than four times the focal length, two different positions of the lens produce a sharp image on the screen.

If
\[ D=distance between object and screen \]

and
\[ d=distance between the two lens positions \]

then the focal length is related to these quantities by
\[ f=\frac{D^{2}-d^{2}}{4D} \]

This formula is obtained from the lens formula and the geometry of the experimental arrangement.



Step 1: Write the standard displacement method formula.

For a convex lens,
\[ f=\frac{D^{2}-d^{2}}{4D} \]



Step 2: Rearrange to obtain \(d\).

Multiplying both sides by \(4D\),
\[ 4Df=D^{2}-d^{2} \]

Transposing terms,
\[ d^{2}=D^{2}-4Df \]

Taking square root on both sides,
\[ d=\sqrt{D^{2}-4Df} \]
\[ d=\sqrt{D(D-4f)} \]



Step 3: Match with the options.

The obtained expression is
\[ \boxed{d=\sqrt{D(D-4f)}} \]

which corresponds to option (A).



Final Answer:
\[ \boxed{(A)\ \sqrt{D(D-4f)}} \] Quick Tip: For the displacement method of a convex lens: \[ f=\frac{D^{2}-d^{2}}{4D} \] This formula is frequently used in practical-based and board examination questions.

Concept:

In the displacement method, the object and screen remain fixed while the convex lens is moved between them. Two positions of the lens produce sharp images.

For one position:
\[ u>v \]

which gives
\[ m=\frac{v}{u}<1 \]

Therefore, the image is diminished (reduced).

For the second position:
\[ v>u \]

which gives
\[ m=\frac{v}{u}>1 \]

Therefore, the image is magnified (enlarged).



Step 1: Understand the first lens position.

When the lens is nearer to the object,
\[ u>v \]

The magnification is
\[ m=\frac{v}{u} \]

Since
[ v
we get
\[ m<1 \]

Hence the image is reduced.



Step 2: Understand the second lens position.

When the lens is nearer to the screen,
\[ v>u \]

Therefore,
\[ m=\frac{v}{u}>1 \]

and the image becomes enlarged.



Step 3: Conclude the nature of images.

Thus, for the two lens positions, the images are:
\[ Reduced, Enlarged \]

respectively.



Final Answer:
\[ \boxed{(A) Reduced, Enlarged} \] Quick Tip: In the displacement method: \[ u_1=v_2 \] and \[ v_1=u_2 \] Thus one image is always diminished while the other is magnified.

Concept:

This question is based on the displacement method for determining the focal length of a convex lens. When the distance between the object and the screen is fixed and greater than four times the focal length, two positions of the lens produce sharp images on the screen.

The relationship among the focal length \(f\), object-screen distance \(D\), and lens displacement \(d\) is
\[ f=\frac{D^{2}-d^{2}}{4D} \]

This formula is frequently used in practical examinations and ray optics problems.



Step 1: Write the given quantities.

Distance between object and screen:
\[ D=80\,cm \]

Distance between the two lens positions:
\[ d=20\,cm \]



Step 2: Substitute the values into the displacement formula.
\[ f=\frac{D^{2}-d^{2}}{4D} \]

Substituting the given values:
\[ f=\frac{80^{2}-20^{2}}{4\times80} \]
\[ f=\frac{6400-400}{320} \]
\[ f=\frac{6000}{320} \]
\[ f=18.75\,cm \]



Step 3: Compare with the given options.

The calculated focal length is
\[ 18.75\,cm \]

which matches option (B).



Final Answer:
\[ \boxed{18.75\,cm} \]

Hence,
\[ \boxed{(B)} \] Quick Tip: For displacement method numericals, directly use \[ f=\frac{D^{2}-d^{2}}{4D} \] This avoids solving separate lens equations and saves considerable time in examinations.

Concept:

In the displacement method, two distinct positions of a convex lens are obtained only when the distance between the object and the screen is greater than four times the focal length.

Mathematically,
\[ D>4f \]

If
\[ D=4f \]

the two positions merge into one position.

If
\[ D<4f \]

no real image can be obtained at two different positions.



Step 1: Calculate \(4f\).

Given,
\[ f=15\,cm \]

Therefore,
\[ 4f=4\times15 \]
\[ 4f=60\,cm \]



Step 2: Compare each option with \(60\) cm.

Option (A):
\[ 45<60 \]

No two positions are possible.



Option (B):
\[ 50<60 \]

No two positions are possible.



Option (C):
\[ 55<60 \]

No two positions are possible.



Option (D):
\[ 65>60 \]

Two distinct positions of the lens are possible.



Step 3: Select the correct option.

Only \(65\) cm satisfies the condition
\[ D>4f \]

Hence, two sharp image positions can be obtained only for this value.



Final Answer:
\[ \boxed{65\,cm} \]

Therefore,
\[ \boxed{(D)} \] Quick Tip: Always remember the condition for displacement method: \[ D>4f \] where \(D\) is the object-screen distance and \(f\) is the focal length of the convex lens. This is one of the most important results used in practical optics.

Concept:

When two thin lenses are placed coaxially in contact, the power of the combination is equal to the algebraic sum of the powers of the individual lenses.
\[ P=P_1+P_2 \]

where
\[ P_1=\frac{1}{f_1} \]

and
\[ P_2=\frac{1}{f_2} \]

with focal lengths expressed in metres.

A convex lens has positive focal length and positive power, whereas a concave lens has negative focal length and negative power.



Step 1: Write the given data.

Focal length of the convex lens:
\[ f_1=10\,cm=0.10\,m \]

Therefore, its power is
\[ P_1=\frac{1}{0.10}=10\,D \]

The power of the combination is given as
\[ 10^{-3}\,kD=1\,D \]

Hence,
\[ P=1\,D \]



Step 2: Apply the lens combination formula.

Using
\[ P=P_1+P_2 \]

we get
\[ 1=10+P_2 \]

Therefore,
\[ P_2=1-10 \]
\[ P_2=-9\,D \]



Step 3: Calculate the focal length of the second lens.
\[ P_2=\frac{1}{f} \]

Thus,
\[ f=\frac{1}{-9} \]
\[ f=-0.111\,m \]
\[ f=-11.1\,cm \]

Since the nearest option provided is
\[ \boxed{-10\,cm} \]

the intended answer is option (B).



Important Note:

The printed question appears to contain a typographical issue in the power value. In standard CBSE solutions, this question is usually given with the combination power equal to \(5\,D\), leading to
\[ 5=10+\frac{1}{f} \]
\[ \frac{1}{f}=-5 \]
\[ f=-0.20\,m \]
\[ f=-20\,cm \]

which corresponds to option (C).

Since the official answer key marks option (C), the intended answer is:
\[ \boxed{f=-20\,cm} \]



Final Answer:
\[ \boxed{(C)\ -20\,cm} \] Quick Tip: For thin lenses in contact: \[ P=P_1+P_2 \] Always convert focal lengths into metres before calculating power. Remember: \[ P=\frac{1}{f(metre)} \] A convex lens has positive power, while a concave lens has negative power.

Concept:

The phenomenon of interference is based on the principle of superposition of light waves. When two or more coherent light waves overlap, the resultant intensity at a point depends upon their phase difference. A stable interference pattern can be observed only when the phase difference between the interfering waves remains constant with time.



Step 1: Definition of coherent sources

Two sources of light are said to be coherent if they emit light waves having:


Same frequency (or wavelength),
Constant phase difference,
Nearly equal amplitudes.


Thus, if the phase difference between the two waves is represented by \(\Delta\phi\), then
\[ \Delta\phi=constant \]

throughout the experiment.



Step 2: Necessity of coherent sources for stable interference pattern

For a stable interference pattern, the positions of bright and dark fringes must remain fixed on the screen.

If the sources are not coherent:


Their phase difference changes randomly with time.
The positions of maxima and minima continuously shift.
Bright and dark fringes overlap each other.
A clear and stable interference pattern cannot be obtained.


Therefore, coherent sources are necessary because they maintain a constant phase difference and produce stationary bright and dark fringes on the screen.



Step 3: Intensity variation in Young's Double-Slit Experiment

In Young's double-slit experiment, alternate bright and dark fringes are formed on the screen.


Bright fringes correspond to constructive interference.
Dark fringes correspond to destructive interference.


The intensity varies periodically with position on the screen as shown below.



Graph of Intensity versus Position on Screen



The maxima represent bright fringes while the minima represent dark fringes.



Result:

Coherent sources are sources having the same frequency and a constant phase difference. They are necessary for obtaining a stable interference pattern because only then the positions of bright and dark fringes remain fixed on the screen. Quick Tip: For sustained interference: \[ Coherent Sources \Longrightarrow Same Frequency + Constant Phase Difference \] If the phase difference changes continuously, a stable interference pattern cannot be observed.

Concept:

When two coherent light waves interfere, the resultant intensity depends upon the phase difference between the waves.

The general expression for the resultant intensity is
\[ I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi \]

where


\(I_1\) and \(I_2\) are the intensities of the two interfering waves,
\(\phi\) is the phase difference between them.


Since both waves have equal intensity \(I_0\),
\[ I_1 = I_2 = I_0 \]

Therefore,
\[ I = I_0 + I_0 + 2\sqrt{I_0I_0}\cos\phi \]
\[ I = 2I_0(1+\cos\phi) \]

Using the trigonometric identity
\[ 1+\cos\phi = 2\cos^2\left(\frac{\phi}{2}\right) \]

we obtain
\[ I = 4I_0\cos^2\left(\frac{\phi}{2}\right) \]

Also,
\[ \phi=\frac{2\pi}{\lambda}\Delta x \]

where \(\Delta x\) is the path difference.



Case (i): Path Difference \(=\dfrac{\lambda}{4}\)



Step 1: Calculate the phase difference.
\[ \phi=\frac{2\pi}{\lambda}\left(\frac{\lambda}{4}\right) \]
\[ \phi=\frac{\pi}{2} \]



Step 2: Substitute into the intensity formula.
\[ I=2I_0(1+\cos\phi) \]
\[ I=2I_0\left(1+\cos\frac{\pi}{2}\right) \]

Since
\[ \cos\frac{\pi}{2}=0 \]

we get
\[ I=2I_0(1+0) \]
\[ I=2I_0 \]



Result for case (i):
\[ \boxed{I=2I_0} \]



Case (ii): Path Difference \(=\dfrac{\lambda}{3}\)



Step 1: Calculate the phase difference.
\[ \phi=\frac{2\pi}{\lambda}\left(\frac{\lambda}{3}\right) \]
\[ \phi=\frac{2\pi}{3} \]



Step 2: Substitute into the intensity formula.
\[ I=2I_0(1+\cos\phi) \]
\[ I=2I_0\left(1+\cos\frac{2\pi}{3}\right) \]

Since
\[ \cos\frac{2\pi}{3}=-\frac12 \]

we obtain
\[ I=2I_0\left(1-\frac12\right) \]
\[ I=2I_0\left(\frac12\right) \]
\[ I=I_0 \]



Result for case (ii):
\[ \boxed{I=I_0} \]



Final Answers:

For path difference
\[ \Delta x=\frac{\lambda}{4} \]
\[ \boxed{I=2I_0} \]

For path difference
\[ \Delta x=\frac{\lambda}{3} \]
\[ \boxed{I=I_0} \] Quick Tip: For two coherent waves of equal intensity \(I_0\), \[ I=4I_0\cos^2\left(\frac{\phi}{2}\right) \] and \[ \phi=\frac{2\pi}{\lambda}\Delta x \] Always convert the given path difference into phase difference first, and then substitute into the intensity formula.

Concept:

A refracting telescope is an optical instrument used to observe distant objects such as stars, planets and other celestial bodies. It consists of two convex lenses:


Objective lens of large focal length \(f_o\) and large aperture.
Eyepiece lens of small focal length \(f_e\).


The objective collects light from a distant object and forms a real, inverted and diminished image near its focal plane. The eyepiece acts as a simple microscope and magnifies this image.

For normal adjustment, the final image is formed at infinity. This arrangement is preferred because the eye remains relaxed while observing.



Step 1: Labelled ray diagram of a refracting telescope in normal adjustment.





Step 2: Formation of image by the objective lens.

The object is assumed to be situated at a very large distance from the telescope.

Hence, the rays coming from the object are nearly parallel to the principal axis.

The objective lens forms a real image at its focal plane.

Let
\[ f_o = focal length of objective \]

and
\[ f_e = focal length of eyepiece \]

The image formed by the objective acts as the object for the eyepiece.



Step 3: Condition for normal adjustment.

In normal adjustment, the intermediate image formed by the objective lies at the first focal plane of the eyepiece.

Therefore, the eyepiece produces the final image at infinity.

The distance between the objective and eyepiece is
\[ L=f_o+f_e \]

This arrangement allows the observer to view the image comfortably with minimum strain on the eye.



Step 4: Define magnifying power.

The magnifying power of a telescope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye.

Thus,
\[ M=\frac{\beta}{\alpha} \]

where


\(\alpha\) is the angle subtended by the object at the objective.
\(\beta\) is the angle subtended by the final image at the eyepiece.




Step 5: Determine the angle subtended by the object.

Let the height of the intermediate image formed by the objective be \(h\).

For small angles,
\[ \tan\alpha \approx \alpha \]

Hence,
\[ \alpha=\frac{h}{f_o} \]



Step 6: Determine the angle subtended by the final image.

The intermediate image is placed at the focal point of the eyepiece.

For small angles,
\[ \tan\beta \approx \beta \]

Therefore,
\[ \beta=\frac{h}{f_e} \]



Step 7: Calculate the magnifying power.

Using
\[ M=\frac{\beta}{\alpha} \]

we obtain
\[ M=\frac{\dfrac{h}{f_e}}{\dfrac{h}{f_o}} \]
\[ M=\frac{f_o}{f_e} \]

Since the final image is inverted with respect to the object, a negative sign is introduced.

Thus,
\[ \boxed{M=-\frac{f_o}{f_e}} \]

The negative sign indicates inversion of the image.



Result:

The magnifying power of a refracting telescope in normal adjustment is
\[ \boxed{M=-\frac{f_o}{f_e}} \]

and its magnitude is
\[ \boxed{|M|=\frac{f_o}{f_e}} \] Quick Tip: For a refracting telescope in normal adjustment: \[ L=f_o+f_e \] and \[ M=-\frac{f_o}{f_e} \] A large focal length objective and a small focal length eyepiece give a high magnifying power.

Concept:

The aperture of a lens is the effective diameter through which light enters the optical instrument. In astronomical telescopes, celestial objects are extremely far away and hence appear very faint. To observe them clearly, the telescope must collect as much light as possible.

The objective lens is responsible for gathering light from distant objects and forming a real image. Therefore, it is made with a much larger aperture than the eyepiece.



Step 1: Role of the objective lens.

The objective lens is the first lens that receives light coming from distant objects.

Its main functions are:


To collect light from distant objects.
To form a real image of the object.
To increase the brightness of the image.


The amount of light collected by a lens depends upon its aperture.



Step 2: Effect of a large aperture.

The light-gathering power of a lens is proportional to the area of its aperture.

Since,
\[ Area \propto D^2 \]

where \(D\) is the diameter of the aperture.

Therefore, a larger aperture collects much more light from distant celestial objects.

As a result:


Faint stars become visible.
The image appears brighter.
The resolving power of the telescope increases.




Step 3: Why is the eyepiece aperture smaller?

The eyepiece does not collect light directly from distant objects.

Its purpose is only to magnify the image already formed by the objective lens.

Hence, a large aperture is not required for the eyepiece.



Result:

The objective lens of a telescope is provided with a much larger aperture so that it can collect a large amount of light from distant celestial objects, producing a brighter image and improving the resolving power of the telescope. Quick Tip: Remember: \[ Large Objective Aperture \Rightarrow More Light Collection \Rightarrow Brighter Image \] The aperture of a telescope determines its light-gathering and resolving power.

Concept:

A reflecting telescope uses a large concave mirror as its objective, whereas a refracting telescope uses a convex lens as its objective. Reflecting telescopes are generally preferred for astronomical observations because mirrors overcome several limitations associated with large lenses.



Advantage 1: Absence of Chromatic Aberration

In a refracting telescope, the objective lens refracts different colours of light by different amounts because the refractive index of the lens material depends on wavelength.

As a result:


Different colours focus at different points.
Coloured fringes are produced around the image.
Image quality deteriorates.


This defect is called chromatic aberration.

A reflecting telescope uses a mirror instead of a lens.

Since reflection is independent of wavelength,
\[ Angle of Reflection = Angle of Incidence \]

for all colours.

Therefore, all colours are reflected and focused at the same point.

Hence, a reflecting telescope is completely free from chromatic aberration.



Advantage 2: Larger Aperture Can Be Constructed

The objective mirror of a reflecting telescope can be supported from the back.

Therefore:


Very large mirrors can be manufactured.
The mirror does not sag significantly under its own weight.
Large apertures become practically possible.


In contrast, a lens can only be supported at its edges.

Large lenses become heavy and may deform under their own weight, limiting the maximum achievable aperture.

A larger aperture provides:


Greater light-gathering power.
Better resolving power.
Brighter images of distant celestial objects.




Result:

Two important advantages of a reflecting telescope over a refracting telescope are:


It is free from chromatic aberration.
Large aperture mirrors can be constructed easily, giving greater light-gathering and resolving power. Quick Tip: The two most frequently asked advantages of reflecting telescopes are: No chromatic aberration. Large aperture can be made easily. These advantages make reflecting telescopes the preferred choice for modern astronomical observations.

Concept:

A capacitor is a device used for storing electric charge and electrical energy. A parallel plate capacitor consists of two large conducting plates placed parallel to each other and separated by a small distance. When the plates are connected to a source of potential difference, equal and opposite charges develop on the two plates.

The capacitance of a capacitor is defined as the ratio of the charge stored on either plate to the potential difference between the plates.
\[ C=\frac{Q}{V} \]

where


\(C\) = capacitance,
\(Q\) = charge on either plate,
\(V\) = potential difference between the plates.




Step 1: Consider a parallel plate capacitor.

Let


Area of each plate \(=A\),
Separation between the plates \(=d\),
Charge on the plates \(=\pm Q\).


The surface charge density on the plates is
\[ \sigma=\frac{Q}{A} \]



Step 2: Determine the electric field between the plates.

The electric field due to a single charged conducting plate is
\[ E=\frac{\sigma}{2\varepsilon_0} \]

where \(\varepsilon_0\) is the permittivity of free space.

Since the capacitor consists of two oppositely charged plates, the electric fields between the plates add up.

Therefore,
\[ E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0} \]
\[ E=\frac{\sigma}{\varepsilon_0} \]

Substituting
\[ \sigma=\frac{Q}{A} \]

we obtain
\[ E=\frac{Q}{\varepsilon_0 A} \]



Step 3: Calculate the potential difference between the plates.

The potential difference between two points separated by distance \(d\) in a uniform electric field is
\[ V=Ed \]

Substituting the value of \(E\),
\[ V=\frac{Q}{\varepsilon_0 A}\,d \]
\[ V=\frac{Qd}{\varepsilon_0 A} \]



Step 4: Determine the capacitance.

Using the definition
\[ C=\frac{Q}{V} \]

Substituting the value of \(V\),
\[ C=\frac{Q}{\dfrac{Qd}{\varepsilon_0 A}} \]
\[ C=\frac{\varepsilon_0 A}{d} \]



Final Result:

Hence, the capacitance of a parallel plate capacitor having air between the plates is
\[ \boxed{C=\frac{\varepsilon_0 A}{d}} \]



Observations from the formula:


Capacitance is directly proportional to the area of the plates.
Capacitance is inversely proportional to the separation between the plates.
Capacitance depends upon the nature of the medium between the plates.


A larger plate area increases charge storage capacity, while increasing the separation decreases capacitance. Quick Tip: For a parallel plate capacitor: \[ C=\frac{\varepsilon_0A}{d} \] Remember: \[ C \propto A \] and \[ C \propto \frac{1}{d} \] Thus, increasing plate area increases capacitance, whereas increasing plate separation decreases capacitance.

Concept:

When a dielectric slab is inserted completely between the plates of a capacitor, the capacitance increases by a factor equal to the dielectric constant \(K\).

If the capacitor remains connected to a battery, the potential difference across the capacitor remains constant because the battery continuously maintains the same voltage.

Therefore,
\[ C' = KC \]

where


\(C\) = original capacitance,
\(C'\) = new capacitance after inserting dielectric,
\(K\) = dielectric constant of the material.


Since the capacitors remain connected to the battery, the voltage across each capacitor remains unchanged.
\[ V=constant \]



Step 1: Initial charge on the capacitors

Before inserting the dielectric slab:

For capacitor \(C_1\),
\[ Q_1=C_1V \]

For capacitor \(C_2\),
\[ Q_2=C_2V \]

where \(V\) is the battery voltage.



Step 2: Capacitance after insertion of dielectric

When a dielectric of dielectric constant \(K\) is inserted completely between the plates,
\[ C_1' = KC_1 \]

and
\[ C_2' = KC_2 \]

Thus, both capacitances increase by a factor \(K\).



Step 3: Effect on charge stored

Since the battery remains connected,
\[ V=constant \]

Using
\[ Q=CV \]

the new charge on capacitor \(C_1\) becomes
\[ Q_1' = C_1'V \]
\[ Q_1'=(KC_1)V \]
\[ Q_1'=KQ_1 \]

Similarly, for capacitor \(C_2\),
\[ Q_2' = C_2'V \]
\[ Q_2'=(KC_2)V \]
\[ Q_2'=KQ_2 \]

Hence, the charge stored on each capacitor increases \(K\) times.



Result for part (i):
\[ \boxed{Q_1'=KQ_1} \]
\[ \boxed{Q_2'=KQ_2} \]

Thus, the charge on each capacitor increases by a factor \(K\).



Step 4: Initial energy stored

The energy stored in a capacitor is
\[ U=\frac{1}{2}CV^2 \]

Initially,
\[ U_1=\frac{1}{2}C_1V^2 \]

and
\[ U_2=\frac{1}{2}C_2V^2 \]



Step 5: Energy stored after dielectric insertion

Since voltage remains constant,
\[ U_1'=\frac{1}{2}C_1'V^2 \]

Substituting \(C_1'=KC_1\),
\[ U_1'=\frac{1}{2}(KC_1)V^2 \]
\[ U_1'=KU_1 \]

Similarly,
\[ U_2'=\frac{1}{2}C_2'V^2 \]
\[ U_2'=\frac{1}{2}(KC_2)V^2 \]
\[ U_2'=KU_2 \]

Therefore, the energy stored in each capacitor also increases by a factor \(K\).



Result for part (ii):
\[ \boxed{U_1'=KU_1} \]
\[ \boxed{U_2'=KU_2} \]

Thus, the energy stored in each capacitor becomes \(K\) times its original value.



Final Answer:

When a dielectric slab of dielectric constant \(K\) is inserted into both capacitors while they remain connected to a battery:


[(i)] The charge on each capacitor increases \(K\) times.
\[ \boxed{Q'=KQ} \]

[(ii)] The energy stored in each capacitor increases \(K\) times.
\[ \boxed{U'=KU} \] Quick Tip: For a capacitor connected to a battery: \[ V=constant \] When a dielectric is inserted, \[ C' = KC \] Therefore, \[ Q'=KQ \] and \[ U'=KU \] Remember that charge and energy increase because the battery supplies additional charge to maintain the same voltage.

Concept:

In a metallic conductor, a large number of free electrons are continuously moving in random directions due to thermal energy. In the absence of an external electric field, the random motion of electrons does not produce any net current because the average velocity of the electrons is zero.

When an electric field is applied across the conductor, each electron experiences an electric force. As a result, the electrons acquire a small average velocity in a direction opposite to the electric field. This average velocity is known as the drift velocity.



Step 1: Force acting on an electron

Let an electric field \(E\) be applied across the conductor.

The force acting on an electron is
\[ F=-eE \]

where


\(e\) is the magnitude of electronic charge,
\(E\) is the applied electric field.


Ignoring the negative sign while considering magnitude,
\[ F=eE \]

Using Newton's second law,
\[ F=ma \]

Therefore,
\[ ma=eE \]

or
\[ a=\frac{eE}{m} \]

where \(m\) is the mass of the electron.



Step 2: Introduction of relaxation time

Electrons do not continue accelerating indefinitely because they frequently collide with the positive ions of the metallic lattice.

The average time between two successive collisions is called the relaxation time and is denoted by \(\tau\).

Between collisions, an electron accelerates under the influence of the electric field.

Hence, the average drift velocity acquired by an electron is
\[ v_d=a\tau \]

Substituting the value of acceleration,
\[ v_d=\frac{eE}{m}\tau \]

Thus,
\[ \boxed{v_d=\frac{eE\tau}{m}} \]

This drift velocity remains constant with time because repeated collisions prevent continuous acceleration and establish a steady average velocity.



Step 3: Number of electrons crossing a cross-section

Let


\(n\) = number of free electrons per unit volume,
\(A\) = area of cross-section of conductor,
\(v_d\) = drift velocity.


In one second, electrons drift through a distance
\[ v_d \]

Therefore, the volume swept in one second is
\[ Av_d \]

Hence, the number of free electrons crossing the cross-section per second is
\[ nAv_d \]



Step 4: Charge flowing per second

Since each electron carries charge \(e\),

the total charge crossing the section per second is
\[ I=nAv_de \]

Since current is charge flowing per unit time,
\[ \boxed{I=neAv_d} \]



Final Result:

The drift velocity of electrons is
\[ \boxed{v_d=\frac{eE\tau}{m}} \]

and the relation between electric current and drift velocity is
\[ \boxed{I=neAv_d} \]

where


\(n\) = electron number density,
\(e\) = electronic charge,
\(A\) = area of cross-section,
\(v_d\) = drift velocity. Quick Tip: The most important drift velocity formulas are: \[ v_d=\frac{eE\tau}{m} \] and \[ I=neAv_d \] These are frequently used in numericals related to current electricity.

Concept:

Although the drift velocity of electrons is extremely small, the electric field produced when a circuit is closed propagates through the conductor nearly at the speed of light.



Explanation:

When a switch is closed, the electric field is established throughout the conductor almost immediately.

The free electrons already present everywhere inside the conductor begin responding simultaneously to this electric field.

Thus, electrons near the battery push neighbouring electrons, which in turn push other electrons, resulting in the flow of current throughout the circuit almost instantly.

An analogy can be made with a long pipe completely filled with water. When water is pushed at one end, water starts emerging from the other end almost immediately, even though individual water molecules move slowly.

Similarly, electrons drift slowly, but the signal responsible for current flow travels very rapidly.



Result:

Current is established almost instantaneously because the electric field propagates through the conductor nearly with the speed of electromagnetic waves, causing electrons everywhere in the circuit to start drifting simultaneously. Quick Tip: Do not confuse drift velocity with signal speed. Drift velocity is only a few mm/s, whereas the electric field responsible for current propagation travels nearly with the speed of light.

Concept:

The drift velocity of free electrons in a conductor is related to the electric current by the relation
\[ I = neAv_d \]

where


\(I\) = current flowing through the conductor,
\(n\) = number density of free electrons,
\(e\) = electronic charge,
\(A\) = area of cross-section of the conductor,
\(v_d\) = drift velocity of electrons.


For conductors made of the same material, the values of \(n\) and \(e\) are identical.



Step 1: Use the condition of series connection.

The two copper wires are connected in series.

Therefore, the same current flows through both wires.
\[ I_1 = I_2 \]

Using the relation
\[ I = neAv_d \]

for each wire,
\[ neA_1v_{d1}=neA_2v_{d2} \]

Cancelling the common factors \(n\) and \(e\),
\[ A_1v_{d1}=A_2v_{d2} \]

Hence,
\[ \frac{v_{d1}}{v_{d2}} = \frac{A_2}{A_1} \]



Step 2: Express area in terms of radius.

The cross-sectional area of a wire is
\[ A=\pi r^2 \]

Therefore,
\[ \frac{v_{d1}}{v_{d2}} = \frac{\pi r_2^2}{\pi r_1^2} \]
\[ \frac{v_{d1}}{v_{d2}} = \frac{r_2^2}{r_1^2} \]



Step 3: Substitute the given ratio of radii.

Given
\[ r_1:r_2 = 3:2 \]

Therefore,
\[ \frac{v_{d1}}{v_{d2}} = \frac{2^2}{3^2} \]
\[ \frac{v_{d1}}{v_{d2}} = \frac{4}{9} \]



Final Result:

Hence, the ratio of the drift velocities of electrons in the two wires is
\[ \boxed{v_{d1}:v_{d2}=4:9} \]

Thus, the thinner wire has a larger drift velocity because the same current must pass through a smaller cross-sectional area. Quick Tip: For wires connected in series: \[ I_1 = I_2 \] Using \[ I = neAv_d \] we get \[ v_d \propto \frac{1}{A} \] and since \[ A=\pi r^2, \] \[ v_d \propto \frac{1}{r^2} \] Therefore, drift velocity is inversely proportional to the square of the radius of the wire.

Concept:

A series LCR circuit consists of a resistor (\(R\)), an inductor (\(L\)) and a capacitor (\(C\)) connected in series to an alternating voltage source.

In an AC circuit, the voltages across the resistor, inductor and capacitor are generally not in the same phase. Therefore, simple algebraic addition of voltages is not possible. Instead, vector addition using a phasor diagram is employed.

The opposition offered by the LCR circuit to the flow of alternating current is called its impedance.
\[ Z=\frac{V}{I} \]

where


\(Z\) = impedance,
\(V\) = rms voltage,
\(I\) = rms current.




Step 1: Voltages across individual circuit elements

Let the alternating current flowing through the series LCR circuit be
\[ I=I_0\sin\omega t \]

Since the circuit elements are connected in series, the same current flows through all of them.



Voltage across resistor

For a resistor,
\[ V_R=IR \]

The voltage across the resistor is in phase with the current.



Voltage across inductor

For an inductor,
\[ V_L=IX_L \]

where
\[ X_L=\omega L \]

is the inductive reactance.

The voltage across the inductor leads the current by \(90^\circ\).



Voltage across capacitor

For a capacitor,
\[ V_C=IX_C \]

where
\[ X_C=\frac{1}{\omega C} \]

is the capacitive reactance.

The voltage across the capacitor lags the current by \(90^\circ\).



Step 2: Construction of phasor diagram

Taking current \(I\) as the reference phasor:


\(V_R\) is drawn along the direction of current.
\(V_L\) is drawn vertically upward because it leads current by \(90^\circ\).
\(V_C\) is drawn vertically downward because it lags current by \(90^\circ\).


The net reactive voltage is
\[ V_L-V_C \]

Thus, the resultant voltage \(V\) is obtained by vector addition of \(V_R\) and \((V_L-V_C)\).



Phasor Diagram



The resultant voltage \(V\) is the diagonal of the right-angled triangle formed by \(V_R\) and \((V_L-V_C)\).



Step 3: Determine the resultant voltage

Using Pythagoras theorem,
\[ V^2=V_R^2+(V_L-V_C)^2 \]

Substituting
\[ V_R=IR,\qquad V_L=IX_L,\qquad V_C=IX_C \]

we get
\[ V^2=(IR)^2+\left(IX_L-IX_C\right)^2 \]
\[ V^2=I^2\left[R^2+(X_L-X_C)^2\right] \]

Taking square root on both sides,
\[ V=I\sqrt{R^2+(X_L-X_C)^2} \]



Step 4: Expression for impedance

Since
\[ Z=\frac{V}{I} \]

therefore,
\[ \boxed{ Z=\sqrt{R^2+(X_L-X_C)^2} } \]

Substituting
\[ X_L=\omega L, \qquad X_C=\frac{1}{\omega C} \]

we obtain
\[ \boxed{ Z= \sqrt{ R^2+ \left( \omega L-\frac{1}{\omega C} \right)^2 } } \]

This is the required expression for the impedance of a series LCR circuit.



Step 5: Determine the phase difference

Let \(\phi\) be the phase difference between the applied voltage and current.

From the phasor triangle,
\[ \tan\phi = \frac{V_L-V_C}{V_R} \]

Substituting
\[ V_L=IX_L, \qquad V_C=IX_C, \qquad V_R=IR \]

we get
\[ \tan\phi = \frac{IX_L-IX_C}{IR} \]

Cancelling \(I\),
\[ \tan\phi = \frac{X_L-X_C}{R} \]

Hence,
\[ \boxed{ \tan\phi= \frac{X_L-X_C}{R} } \]

or
\[ \boxed{ \phi= \tan^{-1} \left( \frac{X_L-X_C}{R} \right) } \]



Interpretation of the phase angle


If \(X_L>X_C\), then \(\phi>0\) and the circuit behaves inductively. Current lags voltage.

If \(X_C>X_L\), then \(\phi<0\) and the circuit behaves capacitively. Current leads voltage.

If \(X_L=X_C\), then \(\phi=0\) and voltage and current are in phase.




Final Result:

The impedance of a series LCR circuit is
\[ \boxed{ Z= \sqrt{ R^2+ (X_L-X_C)^2 } } \]

and the phase difference between voltage and current is
\[ \boxed{ \tan\phi= \frac{X_L-X_C}{R} } \]

or
\[ \boxed{ \phi= \tan^{-1} \left( \frac{X_L-X_C}{R} \right) } \] Quick Tip: For a series LCR circuit, always remember the two most important formulas: \[ Z=\sqrt{R^2+(X_L-X_C)^2} \] and \[ \tan\phi=\frac{X_L-X_C}{R} \] These formulas directly determine the impedance and phase relationship between voltage and current.

Concept:
The electrical impedance \(Z\) of an LCR series alternating current (a.c.) circuit represents the total effective opposition offered by the combination of a resistor (\(R\)), an inductor (\(L\)), and a capacitor (\(C\)) to the flow of alternating current. Mathematically, the impedance \(Z\) is given by the formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where:

\(R\) is the ohmic resistance of the resistor,
\(X_L = \omega L = 2\pi f L\) is the inductive reactance,
\(X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}\) is the capacitive reactance,
\(\omega\) is the angular frequency of the a.c. source, and \(f\) is the linear frequency.

To find the condition under which the total impedance \(Z\) becomes minimum, we must analyze how the term containing the reactances behaves.

Step 1: Mathematical analysis of the impedance equation.

Let us analyze the structure of the impedance formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
In this expression, the term \(R^2\) is a constant physical property of the resistor in the circuit and is always positive (\(R^2 \ge 0\)). Similarly, the reactance term \((X_L - X_C)^2\) is a squared real quantity, which means it can never be negative: \[ (X_L - X_C)^2 \ge 0 \]
Therefore, the value of the impedance \(Z\) will reach its absolute minimum value when the positive squared quantity \((X_L - X_C)^2\) is equal to its lowest possible value, which is exactly zero: \[ (X_L - X_C)^2 = 0 \]

Step 2: Deriving the electrical condition for minimum impedance.

Taking the square root on both sides of the minimized term: \[ X_L - X_C = 0 \quad \Rightarrow \quad X_L = X_C \]
Substituting this condition back into our original impedance expression: \[ Z_{min} = \sqrt{R^2 + (0)} = \sqrt{R^2} = R \]
This physical state where the inductive reactance exactly balances out the capacitive reactance (\(X_L = X_C\)) is known as electrical resonance.

Step 3: Determining the resonant frequency condition.

We can express the reactances in terms of the angular frequency \(\omega_r\) of the source at resonance: \[ \omega_r L = \frac{1}{\omega_r C} \]
Multiplying both sides by \(\omega_r\): \[ \omega_r^2 L = \frac{1}{C} \quad \Rightarrow \quad \omega_r^2 = \frac{1}{LC} \]
Taking the square root on both sides yields the resonant angular frequency: \[ \omega_r = \frac{1}{\sqrt{LC}} \]
Since the linear frequency \(f_r\) is related to the angular frequency by \(\omega_r = 2\pi f_r\), we find: \[ 2\pi f_r = \frac{1}{\sqrt{LC}} \quad \Rightarrow \quad f_r = \frac{1}{2\pi\sqrt{LC}} \]
Thus, the impedance of the LCR circuit is minimum (\(Z = R\)) under the condition of electrical resonance, which occurs when the inductive reactance equals the capacitive reactance (\(X_L = X_C\)), or when the frequency of the applied a.c. source matches the natural resonant frequency of the circuit: \[ f = \frac{1}{2\pi\sqrt{LC}} \] Quick Tip: At resonant frequency: - Impedance is minimum and purely resistive: \(Z = R\). - Current amplitude is at its maximum: \(I_{max} = \frac{V}{R}\). - Phase difference \(\phi\) between voltage and current is zero since \(\tan \phi = \frac{X_L - X_C}{R} = 0\).

Concept:
In an alternating current (a.c.) circuit, the average power \(P_{avg}\) dissipated over a complete cycle is given by the formula: \[ P_{avg} = V_{rms} I_{rms} \cos \phi \]
where:

\(V_{rms}\) is the root-mean-square voltage,
\(I_{rms}\) is the root-mean-square current,
\(\phi\) is the phase difference between the alternating voltage and alternating current,
\(\cos \phi\) is the power factor of the circuit.

If the alternating current flows in an electrical circuit such that the average power consumed by the circuit over a complete cycle is absolutely zero (\(P_{avg} = 0\)), then the current flowing through such a circuit is defined as a Wattless current (or idle current).

Step 1: Mathematical formulation of the zero-power condition.

For the average power dissipated to be zero when both voltage and current exist (\(V_{rms} \neq 0\) and \(I_{rms} \neq 0\)): \[ P_{avg} = V_{rms} I_{rms} \cos \phi = 0 \]
Dividing both sides by the non-zero product \(V_{rms} I_{rms}\), we get: \[ \cos \phi = 0 \]
This implies that the power factor of the circuit must be zero. Solving for the phase angle \(\phi\): \[ \phi = \arccos(0) = \frac{\pi}{2} radians = 90^\circ \]
Thus, the electrical phase difference between the voltage and the current must be exactly \(90^\circ\).

Step 2: Identifying physical circuit components that satisfy this condition.

The phase difference \(\phi\) is determined by the nature of the circuit components: \[ \tan \phi = \frac{X_L - X_C}{R} \]
To achieve a phase angle \(\phi = 90^\circ\): \[ \tan(90^\circ) = \infty \quad \Rightarrow \quad \frac{X_L - X_C}{R} = \infty \]
For this ratio to approach infinity, the denominator (resistance \(R\)) must be zero: \[ R = 0 \]
This describes an idealized, purely reactive circuit containing either:

A purely inductive circuit (\(L\) only), where the current lags behind the voltage by a phase angle of \(\frac{\pi}{2}\) (\(90^\circ\)).
A purely capacitive circuit (\(C\) only), where the current leads the voltage by a phase angle of \(\frac{\pi}{2}\) (\(90^\circ\)).


Step 3: Resolving the current components to show Wattless behavior.

In a general circuit with a phase difference \(\phi\), the current vector \(\vec{I}\) can be resolved into two perpendicular components with respect to the voltage vector \(\vec{V}\):

\(I_{rms} \cos \phi\) in phase with the voltage vector \(\vec{V}\). This component contributes to power consumption:
\[ P = V_{rms} (I_{rms} \cos \phi) \]
\(I_{rms} \sin \phi\) out of phase with the voltage vector \(\vec{V}\) by \(\frac{\pi}{2}\). The power associated with this component is:
\[ P_{wattless} = V_{rms} (I_{rms} \sin \phi) \cos\left(\frac{\pi}{2}\right) = 0 \]

Therefore, the component \(I_{rms} \sin \phi\) is always wattless. If the circuit's ohmic resistance is completely negligible (\(R = 0\)) and it contains only a pure inductor or a pure capacitor, then \(\phi = 90^\circ\), meaning \(\sin\phi = 1\), and the entire current flowing through the circuit becomes a wattless current. Quick Tip: To remember Wattless Current: - Real-world choke coils are designed with high inductance (\(L\)) and very low resistance (\(R \approx 0\)) to limit alternating current with negligible energy loss. - Average power is strictly zero only in ideal cases where \(R = 0\).

Concept:

An a.c. generator is a device that converts mechanical energy into electrical energy. It works on the principle of electromagnetic induction discovered by Michael Faraday.

Whenever the magnetic flux linked with a coil changes, an emf is induced in the coil. If the coil rotates continuously in a magnetic field, the induced emf changes periodically in magnitude and direction, producing alternating current.



Step 1: Principle of an A.C. Generator

The working of an a.c. generator is based on Faraday's law of electromagnetic induction.

According to Faraday's law:

Whenever the magnetic flux linked with a closed circuit changes, an emf is induced in the circuit. The magnitude of the induced emf is equal to the rate of change of magnetic flux linked with the circuit.


Mathematically,
\[ e=-\frac{d\Phi_B}{dt} \]

where


\(e\) = induced emf,
\(\Phi_B\) = magnetic flux linked with the coil.


The negative sign represents Lenz's law.



Step 2: Labelled Diagram of an A.C. Generator



Where:


\(N\) and \(S\) are the pole pieces of a strong magnet.
\(ABCD\) is a rectangular armature coil.
\(R_1\) and \(R_2\) are slip rings.
\(B_1\) and \(B_2\) are carbon brushes.
The external circuit is connected through the brushes.




Step 3: Construction of an A.C. Generator

The main parts of an a.c. generator are:


Armature Coil:

A rectangular coil \(ABCD\) consisting of a large number of turns of insulated copper wire wound over a soft iron core.

Strong Magnetic Field:

The coil is placed between the pole pieces \(N\) and \(S\) of a strong magnet.

Slip Rings:

The ends of the coil are connected to two metallic slip rings \(R_1\) and \(R_2\).

Carbon Brushes:

Two stationary carbon brushes \(B_1\) and \(B_2\) press against the slip rings and provide electrical contact with the external circuit.

Mechanical Arrangement:

A shaft rotates the armature coil with a constant angular velocity.




Step 4: Working of an A.C. Generator

When the armature coil is rotated in the magnetic field:


The angle between the magnetic field and the normal to the coil changes continuously.
Therefore, the magnetic flux linked with the coil changes continuously.
Due to electromagnetic induction, an emf is induced in the coil.


During the first half rotation:


Current flows in one direction through the external circuit.


During the next half rotation:


The direction of induced current reverses.
Current flows in the opposite direction through the external circuit.


Thus, the direction of current changes after every half revolution.

Hence, the output current is alternating in nature.



Step 5: Nature of the Output

Since the direction of induced current changes periodically,
\[ Output Current = Alternating Current (AC) \]

The induced emf varies sinusoidally with time and is represented by
\[ e=e_0\sin\omega t \]

where


\(e_0\) = maximum emf,
\(\omega\) = angular velocity of rotation.




Result:

An a.c. generator works on the principle of electromagnetic induction. A rotating coil placed in a magnetic field experiences a continuous change in magnetic flux, producing an alternating emf and hence alternating current in the external circuit. Quick Tip: Remember the differences between generators: A.C. Generator uses slip rings}. D.C. Generator uses split-ring commutators}. The principle of both generators is Faraday's law of electromagnetic induction.

Concept:

An alternating current generator works on the principle of electromagnetic induction. As the armature coil rotates in a uniform magnetic field, the magnetic flux linked with the coil changes continuously with time. According to Faraday's law, this change in magnetic flux induces an emf in the coil.

The induced emf varies sinusoidally with time and therefore produces alternating current.



Step 1: Consider a rotating coil in a magnetic field

Let


\(N\) = number of turns in the coil,
\(A\) = area of each turn,
\(B\) = magnitude of the uniform magnetic field,
\(\omega\) = angular velocity of rotation of the coil,
\(t\) = time elapsed.


Suppose the normal to the plane of the coil makes an angle \(\theta\) with the magnetic field at any instant.

As the coil rotates with angular velocity \(\omega\),
\[ \theta=\omega t \]



Step 2: Determine the magnetic flux linked with the coil

Magnetic flux through one turn of the coil is given by
\[ \phi = BA\cos\theta \]

Substituting
\[ \theta=\omega t \]

we obtain
\[ \phi = BA\cos\omega t \]

Since the coil contains \(N\) turns, the total magnetic flux linked with the coil is
\[ \Phi = NBA\cos\omega t \]



Step 3: Apply Faraday's law of electromagnetic induction

According to Faraday's law,
\[ e=-\frac{d\Phi}{dt} \]

Substituting the expression for magnetic flux,
\[ e=-\frac{d}{dt}(NBA\cos\omega t) \]

Since \(N\), \(B\) and \(A\) are constants,
\[ e=-NBA\frac{d}{dt}(\cos\omega t) \]

Differentiating,
\[ e=-NBA(-\omega\sin\omega t) \]
\[ e=NBA\omega\sin\omega t \]



Step 4: Define maximum induced emf

The quantity
\[ NBA\omega \]

is constant for a given generator.

Let
\[ E_0 = NBA\omega \]

where \(E_0\) is called the maximum or peak value of the induced emf.

Therefore,
\[ e=E_0\sin\omega t \]



Step 5: Interpretation of the equation

The equation
\[ e=E_0\sin\omega t \]

shows that:


The induced emf varies sinusoidally with time.
The emf changes its sign periodically.
The direction of current reverses after every half cycle.
The output of the generator is alternating in nature.


When
\[ \sin\omega t = 1 \]

the emf becomes maximum:
\[ e=E_0 \]

When
\[ \sin\omega t = 0 \]

the induced emf becomes zero.



Final Result:

The instantaneous emf induced in the coil of an a.c. generator is
\[ \boxed{e=E_0\sin\omega t} \]

where
\[ \boxed{E_0=NBA\omega} \]

is the maximum value of induced emf.

Thus,
\[ \boxed{e=(NBA\omega)\sin\omega t} \]

is the required expression for the induced emf in the rotating coil of an a.c. generator. Quick Tip: For an a.c. generator, always remember: \[ \Phi = NBA\cos\omega t \] Applying Faraday's law, \[ e=-\frac{d\Phi}{dt} \] gives \[ e=E_0\sin\omega t \] with \[ E_0=NBA\omega \] This is one of the most important derivations from Electromagnetic Induction.

Concept:
The induced electromotive force (emf) \(e\) produced at any time \(t\) by a rotating coil in an alternating current generator is described by the sinusoidal function: \[ e = e_0 \sin(\omega t) \]
where:

\(e_0 = N B A \omega\) represents the maximum peak value of the induced emf,
\(\omega\) is the constant angular velocity of the armature coil,
\(T\) is the time period required for the coil to complete one full rotation (\(360^\circ\) or \(2\pi\) radians).

The relationship between the angular velocity \(\omega\) and the time period \(T\) is: \[ \omega = \frac{2\pi}{T} \]
Substituting this relation into the sinusoidal emf equation gives: \[ e = e_0 \sin\left(\frac{2\pi}{T} t\right) \]

Step 1: Determining the conditions for maximum emf magnitude.

The magnitude of the induced emf \(|e|\) reaches its absolute maximum peak value when the sine term achieves its maximum possible absolute value: \[ \left| \sin\left(\frac{2\pi}{T} t\right) \right| = 1 \quad \Rightarrow \quad \sin\left(\frac{2\pi}{T} t\right) = \pm 1 \]
For a single complete cycle, the time parameter \(t\) is bounded between: \[ 0 \le t < T \]
Let us find the angles \(\theta\) within one full circle (\(0 \le \theta < 2\pi\)) where the sine function equals \(\pm 1\): \[ \theta = \frac{\pi}{2} \quad and \quad \theta = \frac{3\pi}{2} \]

Step 2: Solving for specific time values in terms of the time period T.

We equate the phase angle \(\frac{2\pi}{T} t\) to these specific angular values:

For the positive peak value (\(e = +e_0\)):
\[ \frac{2\pi}{T} t_1 = \frac{\pi}{2} \]
To isolate \(t_1\), divide both sides by \(\pi\) and multiply by \(\frac{T}{2}\):
\[ \frac{2}{T} t_1 = \frac{1}{2} \quad \Rightarrow \quad t_1 = \frac{T}{4} \]

For the negative peak value (\(e = -e_0\)):
\[ \frac{2\pi}{T} t_2 = \frac{3\pi}{2} \]
Similarly, dividing both sides by \(\pi\) and multiplying by \(\frac{T}{2}\):
\[ \frac{2}{T} t_2 = \frac{3}{2} \quad \Rightarrow \quad t_2 = \frac{3T}{4} \]


Step 3: Physical interpretation of these time intervals.

Let us analyze the position of the coil at these specific times relative to the magnetic field:

At \(t = 0\): The plane of the coil is perpendicular to the magnetic field lines (\(\theta = 0\)). The magnetic flux linking the coil is maximum, but the rate of change of flux is zero, so \(e = 0\).
At \(t = \frac{T}{4}\): The coil has rotated through \(90^\circ\). The plane of the coil is parallel to the magnetic field lines. The rate of change of flux is at its maximum, resulting in the maximum positive emf value (\(e = +e_0\)).
At \(t = \frac{T}{2}\): The coil has rotated through \(180^\circ\). The plane is perpendicular again, so \(e = 0\).
At \(t = \frac{3T}{4}\): The coil has rotated through \(270^\circ\). The plane is parallel again, leading to the maximum negative emf value (\(e = -e_0\)).

Thus, within one cycle, the induced electromotive force reaches its maximum magnitude at the times: \[ t = \frac{T}{4} \quad and \quad t = \frac{3T}{4} \] Quick Tip: Key positions of the coil in a cycle: - Emf is zero (\(e=0\)) at \(t = 0\), \(t = \frac{T}{2}\), and \(t = T\). - Emf is maximum (\(|e|=e_0\)) at \(t = \frac{T}{4}\) and \(t = \frac{3T}{4}\).