The force on charge \(-q\) due to charge \(Q\) is the Coulomb force, directed towards \(Q\). Let's call this the x-direction.

\(F_x = F_{Coulomb} = \frac{1}{4\pi\epsilon_0} \frac{Q q}{r^2}\).


The force on charge \(-q\) due to the external electric field \(\vec{E}\) is perpendicular to the line joining the charges. Let's call this the y-direction.

\(F_y = F_{Electric} = qE\).


The net force on charge \(-q\) is the vector sum of these two forces.


The angle \(\theta\) the net force makes with the line joining \(Q\) and \(-q\) (the x-direction) is given by \(\tan\theta = \frac{F_y}{F_x}\).

\(\tan\theta = \frac{qE}{\frac{Qq}{4\pi\epsilon_0 r^2}}\).

\(\tan\theta = \frac{qE \cdot 4\pi\epsilon_0 r^2}{Qq} = \frac{4\pi\epsilon_0 E r^2}{Q}\).


Therefore, the angle is \(\theta = \tan^{-1} \left( \frac{4\pi\epsilon_0 E r^2}{Q} \right)\).
Quick Tip: When dealing with multiple forces on a charge, always use vector addition. Break down each force into components along perpendicular axes. The direction of the resultant force can then be easily found using trigonometry (\(\tan\theta = F_y/F_x\)).

The formula for resistance is \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area.


Since the material is the same for all wires, \(\rho\) is constant.


For wire A: \(R_A = \rho \frac{2l}{A/2} = 4 \left( \rho \frac{l}{A} \right)\).


For wire B: \(R_B = \rho \frac{l}{A} = 1 \left( \rho \frac{l}{A} \right)\).


For wire C: \(R_C = \rho \frac{l/2}{2A} = \frac{1}{4} \left( \rho \frac{l}{A} \right)\).


Let \(R_0 = \rho \frac{l}{A}\). Then \(R_A = 4R_0\), \(R_B = R_0\), and \(R_C = 0.25 R_0\).


Comparing the values, we find that \(4R_0 > R_0 > 0.25R_0\).


Thus, the correct order is \(R_A > R_B > R_C\).
Quick Tip: Remember that resistance is directly proportional to the length of the conductor and inversely proportional to its cross-sectional area. A long, thin wire will have a much higher resistance than a short, thick wire of the same material.

The Lorentz force is given by the vector formula \(\vec{F} = q(\vec{v} \times \vec{B})\).


The velocity vector is along the y-axis: \(\vec{v} = v\hat{j}\).


The magnetic field vector is along the x-axis: \(\vec{B} = B\hat{i}\).


We need to compute the cross product \(\vec{v} \times \vec{B}\).

\(\vec{v} \times \vec{B} = (v\hat{j}) \times (B\hat{i}) = vB (\hat{j} \times \hat{i})\).


Using the right-hand rule for unit vectors, we know that \(\hat{j} \times \hat{i} = -\hat{k}\).


Substituting this result, we get \(\vec{v} \times \vec{B} = -vB\hat{k}\).


The force is then \(\vec{F} = q(-vB\hat{k}) = -qvB\hat{k}\).


The direction of the force is along the negative z-axis.
Quick Tip: The direction of the Lorentz force is always perpendicular to both the velocity of the charge and the magnetic field. Use the Right-Hand Rule (for positive charges) or Fleming's Left-Hand Rule to quickly determine the direction of the force. For a negative charge, the force direction is opposite to that found by the rule.

The torque on a magnet in a magnetic field is given by \(\tau = MB \sin\theta\).


Initially, the magnet is at right angles, so the initial angle is \(\theta_1 = 90^\circ\).


The initial torque is \(\tau_{initial} = MB \sin(90^\circ) = MB\).


The magnet is rotated until the new torque is half the initial value.

\(\tau_{final} = \frac{1}{2} \tau_{initial} = \frac{1}{2} MB\).


Let the final angle be \(\theta_2\). Then \(\tau_{final} = MB \sin\theta_2\).

\(\frac{1}{2} MB = MB \sin\theta_2\).


This gives \(\sin\theta_2 = \frac{1}{2}\), which means \(\theta_2 = 30^\circ\).


The question asks for the angle of rotation, which is the change in angle.


Angle of rotation = \(\theta_1 - \theta_2 = 90^\circ - 30^\circ = 60^\circ\).
Quick Tip: Pay close attention to what the question asks for. In this case, it's the "angle of rotation" (\(\Delta\theta\)) and not the "final angle" (\(\theta_2\)). A common mistake is to stop after finding the final angle.

Magnetic permeability of a substance is \(\mu\), and for free space it is \(\mu_0\).


The question asks for a substance where \(\mu < \mu_0\).


This condition is characteristic of diamagnetic materials.


Let's analyze the given options based on their magnetic properties.


(A) Sodium is a paramagnetic material, for which \(\mu > \mu_0\).


(B) Iron is a ferromagnetic material, for which \(\mu \gg \mu_0\).


(C) Aluminium is a paramagnetic material, for which \(\mu > \mu_0\).


(D) Copper is a diamagnetic material, for which \(\mu < \mu_0\).


Therefore, copper has a magnetic permeability less than that of free space.
Quick Tip: Memorize the classification of magnetic materials: \textbf{Diamagnetic} (\(\mu < \mu_0\)): Weakly repelled by magnets. Ex: Copper, Gold, Water. \textbf{Paramagnetic} (\(\mu > \mu_0\)): Weakly attracted by magnets. Ex: Aluminium, Sodium. \textbf{Ferromagnetic} (\(\mu \gg \mu_0\)): Strongly attracted by magnets. Ex: Iron, Cobalt, Nickel.

According to Faraday's law of electromagnetic induction, the induced electromotive force (emf) is given by \(\mathcal{E} = -\frac{d\phi}{dt}\).


Given the magnetic flux \(\phi = 5t^2 + 4t - 2\).


First, we differentiate the flux with respect to time to find the induced emf.

\(\frac{d\phi}{dt} = \frac{d}{dt}(5t^2 + 4t - 2) = 10t + 4\).


So, the induced emf is \(\mathcal{E} = -(10t + 4)\).


We need to find the emf at time \(t = 1\) s.

\(\mathcal{E}(t=1) = -(10(1) + 4) = -14\) V.


The magnitude of the induced emf is \(|\mathcal{E}| = 14\) V.


The magnitude of the induced current is given by Ohm's law, \(I = \frac{|\mathcal{E}|}{R}\).


Given the resistance \(R = 14 \, \Omega\).

\(I = \frac{14 V}{14 \, \Omega} = 1.0\) A.
Quick Tip: The negative sign in Faraday's law (\(\mathcal{E} = -d\phi/dt\)) relates to the direction of the induced current (Lenz's Law). When a question asks for the magnitude of the current or emf, you can ignore the negative sign.

For an electromagnetic wave in free space, the amplitudes of the electric field (\(E_0\)) and magnetic field (\(B_0\)) are related by the speed of light (\(c\)).


The relationship is \(E_0 = c B_0\).


Given the amplitude of the magnetic field, \(B_0 = 6.0 \times 10^{-4}\) T.


The speed of light in free space is \(c = 3 \times 10^8\) m/s.


Now, we calculate the amplitude of the electric field.

\(E_0 = (3 \times 10^8 m/s) \times (6.0 \times 10^{-4} T)\).

\(E_0 = 18 \times 10^{(8-4)} V/m\).

\(E_0 = 18 \times 10^4 V/m\).


To express this in standard scientific notation, we write it as \(1.8 \times 10^5 V/m\).
Quick Tip: A simple way to remember the formula relating electric and magnetic field amplitudes in an EM wave is \(E = cB\). The electric field amplitude is a much larger number than the magnetic field amplitude because it's multiplied by the very large value of \(c\).

A steady current in a long straight wire produces a magnetic field around it.


The direction and shape of the magnetic field lines can be determined by the Right-Hand Grip Rule.


According to this rule, if you point the thumb of your right hand in the direction of the current (upward), your fingers will curl around the wire in the direction of the magnetic field lines.


This curling motion of the fingers describes circles.


Since the wire is vertical, the planes of these circles are horizontal.


Therefore, the magnetic field lines are concentric horizontal circles centered on the wire.
Quick Tip: The Right-Hand Grip Rule is fundamental for determining the direction of the magnetic field around a current-carrying wire. Thumb points with the current (\(I\)), fingers curl in the direction of the magnetic field (\(\vec{B}\)).

The magnification produced by a spherical mirror is given as \(m = -2.0\).


Let's analyze the properties based on the value of magnification.


The sign of the magnification tells us about the nature of the image.


A negative sign (\(m < 0\)) indicates that the image is real and inverted.


A positive sign (\(m > 0\)) indicates that the image is virtual and erect.


Since \(m = -2.0\) is negative, the image is real and inverted.


Now let's determine the type of mirror.


Convex mirrors always produce virtual, erect, and diminished images. Their magnification is always positive and less than 1 (\(0 < m < +1\)).


Concave mirrors can produce real, inverted images (when the object is placed beyond the focus) or virtual, erect images (when the object is placed between the pole and the focus).


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


Therefore, the mirror is concave and the image is real.
Quick Tip: Remember the sign conventions for magnification: \(m\) is negative \(\rightarrow\) Real and Inverted image. \(m\) is positive \(\rightarrow\) Virtual and Erect image. Also, remember that convex mirrors can only form virtual and erect images.

Let's evaluate each statement.


(A) This statement is false. Both photons of light (a wave phenomenon) and electrons (as per de Broglie's hypothesis, proven by Davisson-Germer experiment) exhibit diffraction.


(B) This statement is false. Both electrons and photons have momentum. The momentum of an electron is \(p = mv\), and the momentum of a photon is \(p = h/\lambda\).


(C) This statement is true. This is the principle of wave-particle duality. Light behaves as both a wave (showing interference, diffraction) and a particle (photoelectric effect). Electrons, which are particles, also exhibit wave-like properties (diffraction).


(D) This statement is false. According to the quantum theory of radiation, all electromagnetic radiations consist of discrete packets of energy called photons.


Therefore, the only correct statement is (C).
Quick Tip: The concept of wave-particle duality is central to modern physics. It states that all matter and energy exhibit both wave-like and particle-like properties. Light has photons (particles) and waves. Electrons are particles but have an associated de Broglie wavelength.

Intensity (I) of a light beam is defined as the power per unit area, which can be expressed in terms of photons as \(I = \frac{n(hf)}{A}\), where n is the number of photons passing through area A per unit time, h is Planck's constant, and f is the frequency of light.


The energy of a single photon is given by \(E = hf = \frac{hc}{\lambda}\), where \(\lambda\) is the wavelength.


We know that blue light has a shorter wavelength than red light (\(\lambda_{blue} < \lambda_{red}\)), which means it has a higher frequency (\(f_{blue} > f_{red}\)).


Therefore, the energy of a single blue photon is greater than the energy of a single red photon (\(E_{blue} > E_{red}\)).


The problem states that the intensities are equal: \(I_{red} = I_{blue}\).


Let \(N_{red}\) and \(N_{blue}\) be the number of photons per unit area per unit time for red and blue light, respectively.

\(N_{red} \times E_{red} = N_{blue} \times E_{blue}\).


Since \(E_{blue} > E_{red}\), for the equality to hold, it must be that \(N_{red} > N_{blue}\).


This means the number of photons in the red beam is greater than the number of photons in the blue beam to maintain the same intensity.
Quick Tip: For a fixed intensity, light with lower energy per photon (like red light) must have a higher photon flux (more photons per second) compared to light with higher energy per photon (like blue light). Intensity is a measure of total energy flow, not the energy of individual photons.

Let's analyze the electrical properties of each substance at room temperature.


(A) Sn (Tin) is a metal. Metals have a large number of free electrons available for conduction and are excellent electrical conductors at room temperature.


(B) Mica is a silicate mineral that is widely used as an electrical insulator due to its very high resistivity.


(C) Si (Silicon) is a semiconductor. Its conductivity at room temperature is intermediate, much lower than that of metals but higher than that of insulators.


(D) C (Carbon) can exist in various allotropes. Diamond is an excellent insulator. Graphite is a conductor, but tin is a metal and generally considered a better conductor. In the context of this question, Tin is the most unambiguous choice for an electrical conductor.


Therefore, Sn (Tin) is the electrical conductor among the given options.
Quick Tip: Memorize the basic classification of materials: Conductors (e.g., metals like Cu, Ag, Sn), Insulators (e.g., glass, mica, rubber), and Semiconductors (e.g., Si, Ge). This classification is based on their electrical conductivity at room temperature.

Assertion (A) is true. When one of the two slits is closed, the setup becomes a single-slit experiment. Light passing through the single open slit will diffract, producing a characteristic single-slit diffraction pattern on the screen.


Reason (R) is also true. The phenomenon of interference is based on the superposition principle applied to two or more coherent waves. A stable interference pattern requires at least two such waves.


However, the Reason (R) does not explain the Assertion (A). The assertion is about the occurrence of diffraction from a single slit. The reason states the condition for interference. While the reason explains why an interference pattern disappears when one slit is closed, it does not explain why a diffraction pattern appears. The appearance of the diffraction pattern is due to the bending of waves around the edges of the single slit.


Therefore, both statements are independently true, but the reason is not the correct explanation for the assertion.
Quick Tip: Distinguish clearly between interference and diffraction. Interference is the superposition of waves from two or more distinct coherent sources (like two slits). Diffraction is the bending of waves as they pass through a single aperture or around an obstacle. Both phenomena demonstrate the wave nature of light.

Assertion (A) is true. When monochromatic light strikes a metal surface, photoelectrons are emitted. The maximum kinetic energy of these electrons is given by Einstein's photoelectric equation: \(K_{max} = hf - \phi\), where \(hf\) is the energy of the incident photon and \(\phi\) is the work function. Electrons emitted from the very surface of the metal have this maximum kinetic energy. However, electrons from deeper within the metal may lose some energy in collisions before they escape. This results in them being emitted with kinetic energies ranging from zero up to \(K_{max}\).


Reason (R) is also true. The work function (\(\phi\)) is the minimum energy required to remove an electron from the surface of a particular metal. It is a characteristic constant property for each metal.


The Reason (R) is the correct explanation for the Assertion (A). The fact that there is a definite work function (\(\phi\)) means that for a given frequency (\(f\)), there is a well-defined maximum kinetic energy (\(K_{max}\)). The existence of this specific maximum value is the upper limit of the range mentioned in the assertion. The work function is the key factor that determines this maximum value, thus explaining the "certain maximum value" part of the assertion, which defines the range.
Quick Tip: Einstein's photoelectric equation, \(K_{max} = hf - \phi\), is crucial. It explains that the maximum kinetic energy depends on the light's frequency and the metal's work function. The range of energies (from 0 to \(K_{max}\)) is due to energy loss by electrons from below the surface.

Assertion (A) is true. An n-type semiconductor, despite having an abundance of free electrons (negative charge carriers), is electrically neutral as a whole.


Reason (R) is also true. The process of creating an n-type semiconductor involves doping a neutral intrinsic semiconductor (like silicon) with neutral pentavalent impurity atoms (like phosphorus). Each impurity atom donates one electron to the conduction band, becoming a fixed positive ion (a donor ion) in the crystal lattice.


The Reason (R) is the correct explanation for Assertion (A). The crystal started with neutral atoms (silicon and phosphorus). For every free electron created by a dopant atom, a corresponding positive ion is also created, fixed in the lattice. The total positive charge of the atomic nuclei and fixed donor ions exactly balances the total negative charge of all the electrons (both valence and conduction electrons). Therefore, the overall crystal remains electrically neutral.
Quick Tip: Do not confuse the type of majority charge carrier with the overall charge of the semiconductor. In an n-type semiconductor, electrons are the majority carriers, but the material itself is neutral. Similarly, in a p-type semiconductor, holes are the majority carriers, but the material is also neutral.

Assertion (A) is true. The impedance of a series LCR circuit is given by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), where \(X_L\) is the inductive reactance and \(X_C\) is the capacitive reactance. At resonance, the inductive and capacitive reactances cancel each other out (\(X_L = X_C\)). Therefore, the impedance becomes \(Z = \sqrt{R^2 + 0^2} = R\). Since the impedance is equal to the resistance, the circuit behaves as a purely resistive circuit. The voltage and current are in phase.


Reason (R) is also true. The condition for resonance in a series LCR circuit is that the inductive reactance equals the capacitive reactance, \(X_L = X_C\). Substituting the formulas \(X_L = \omega L\) and \(X_C = \frac{1}{\omega C}\), we get \(\omega L = \frac{1}{\omega C}\). Solving for the angular frequency \(\omega\) gives the resonant angular frequency \(\omega_0 = \frac{1}{\sqrt{LC}}\).


The Reason (R) correctly explains the Assertion (A). The condition \(X_L = X_C\) stated in the reason is the fundamental definition of resonance. It is this exact condition that leads to the reactive part of the impedance becoming zero, causing the circuit to behave purely resistively as stated in the assertion.
Quick Tip: At resonance in a series LCR circuit, impedance (Z) is minimum (\(Z=R\)) and the current is maximum (\(I_{max} = V/R\)). The circuit is purely resistive, meaning the phase angle between voltage and current is zero.

Given:

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

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


(a)

(i) After doping, the hole concentration (\(n_h = 8 \times 10^{12}\)) is much greater than the intrinsic carrier concentration (\(n_i = 5 \times 10^8\)).

Since holes are the majority charge carriers, the semiconductor is p-type.

A p-type semiconductor is formed by doping with a trivalent impurity, also known as an acceptor impurity (e.g., Boron, Aluminium).


(ii) As established above, since holes (\(n_h\)) are the majority carriers, the extrinsic semiconductor formed is a p-type semiconductor.


(b)

For an extrinsic semiconductor in thermal equilibrium, the mass-action law holds: \(n_e \cdot n_h = n_i^2\), where \(n_e\) is the electron concentration.


We need to calculate the electron concentration \(n_e\).

\(n_e = \frac{n_i^2}{n_h}\).

\(n_e = \frac{(5 \times 10^8 m^{-3})^2}{8 \times 10^{12} m^{-3}}\).

\(n_e = \frac{25 \times 10^{16}}{8 \times 10^{12}} m^{-3}\).

\(n_e = 3.125 \times 10^{4} m^{-3}\).


The electron concentration in the extrinsic semiconductor is \(3.125 \times 10^4 m^{-3}\).
Quick Tip: The mass-action law, \(n_e \cdot n_h = n_i^2\), is a fundamental relationship in semiconductor physics. It states that the product of electron and hole concentrations is constant at a given temperature, regardless of doping. If doping increases one type of carrier, the other type must decrease to maintain this constant product.

The formula for fringe width (\(\beta\)) in a Young's double-slit experiment is given by:
\(\beta = \frac{\lambda D}{d}\), where \(\lambda\) is the wavelength, \(D\) is the distance to the screen, and \(d\) is the slit separation.


Given:

Change in screen position, \(\Delta D = -30 cm = -0.3 m\) (negative because it moves towards the slits).

Change in fringe width, \(\Delta \beta = -0.09 mm = -0.09 \times 10^{-3} m\) (negative because fringe width decreases as D decreases).

Slit separation, \(d = 2 mm = 2 \times 10^{-3} m\).


Let the initial distance be \(D_1\) and the final distance be \(D_2\). The initial fringe width is \(\beta_1 = \frac{\lambda D_1}{d}\) and the final is \(\beta_2 = \frac{\lambda D_2}{d}\).


The change in fringe width is \(\Delta \beta = \beta_2 - \beta_1\).
\(\Delta \beta = \frac{\lambda D_2}{d} - \frac{\lambda D_1}{d} = \frac{\lambda}{d}(D_2 - D_1)\).


We know that \(D_2 - D_1 = \Delta D = -0.3\) m.


Substituting the given values:
\(-0.09 \times 10^{-3} = \frac{\lambda}{2 \times 10^{-3}}(-0.3)\).


Now, we solve for the wavelength \(\lambda\).
\(\lambda = \frac{(-0.09 \times 10^{-3}) \times (2 \times 10^{-3})}{-0.3}\).
\(\lambda = \frac{0.18 \times 10^{-6}}{0.3}\).
\(\lambda = 0.6 \times 10^{-6} m\).


Converting to nanometers (\(1 nm = 10^{-9} m\)):
\(\lambda = 600 \times 10^{-9} m = 600 nm\).


The wavelength of light used is 600 nm.
Quick Tip: When dealing with changes in YDSE parameters, it is often easier to work with the change formula, \(\Delta\beta = \frac{\lambda \Delta D}{d}\), rather than setting up two separate equations. This simplifies the algebra.

Part 1: Condition for f = R

We use the Lens Maker's Formula: \(\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\).


For a biconvex lens with equal radii of curvature R, using the sign convention:

The first surface has radius \(R_1 = +R\).

The second surface has radius \(R_2 = -R\).


Substituting these into the formula:
\(\frac{1}{f} = (n-1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (n-1) \left( \frac{1}{R} + \frac{1}{R} \right)\).
\(\frac{1}{f} = (n-1) \frac{2}{R}\). (Equation 1)


The question requires the condition for \(f = R\). So we substitute \(f=R\) into Equation 1.
\(\frac{1}{R} = (n-1) \frac{2}{R}\).
\(1 = 2(n-1)\).
\(1 = 2n - 2\).
\(2n = 3\).
\(n = 1.5\).

The condition is that the refractive index of the lens material must be 1.5.


Part 2: New focal length when one surface is plane

If one surface is made plane, the lens becomes a plano-convex lens.

Let the new radii be \(R_1 = R\) and \(R_2 = \infty\).

Let the new focal length be \(f'\).

Using the Lens Maker's Formula again:
\(\frac{1}{f'} = (n-1) \left( \frac{1}{R} - \frac{1}{\infty} \right)\).
\(\frac{1}{f'} = (n-1) \left( \frac{1}{R} - 0 \right) = \frac{n-1}{R}\).


From Equation 1, we can write \((n-1) = \frac{R}{2f}\).

Substituting this into the expression for \(1/f'\):
\(\frac{1}{f'} = \frac{1}{R} \left( \frac{R}{2f} \right) = \frac{1}{2f}\).

This means the new focal length is \(f' = 2f\).


Given the initial condition that \(f = R\), the new focal length is \(f' = 2R\).
Quick Tip: Always be careful with the sign convention for radii of curvature in the Lens Maker's Formula. For a biconvex lens, \(R_1\) is positive and \(R_2\) is negative. For a plano-convex lens, one radius is finite and the other is infinite.

In Bohr's model of the hydrogen atom, the radius of the \(n^{th}\) orbit is given by the formula:
\(r_n = r_0 n^2\), where \(r_0\) is the Bohr radius (a constant).


The electron makes a transition from an initial state \(n_i = 3\) to a final state \(n_f = 2\).


The initial radius of the orbit (\(n_i = 3\)) is:
\(r_i = r_3 = r_0 (3)^2 = 9r_0\).


The final radius of the orbit (\(n_f = 2\)) is:
\(r_f = r_2 = r_0 (2)^2 = 4r_0\).


The change in radius is \(\Delta r = r_f - r_i\).
\(\Delta r = 4r_0 - 9r_0 = -5r_0\).


The percentage change in the radius is calculated with respect to the initial radius:

Percentage Change = \(\frac{\Delta r}{r_i} \times 100%\).

Percentage Change = \(\frac{-5r_0}{9r_0} \times 100%\).

Percentage Change = \(-\frac{5}{9} \times 100%\).

Percentage Change = \(-55.555...% \approx -55.56%\).


The negative sign indicates that the radius has decreased. So, there is a decrease of 55.56% in the radius of the orbit.
Quick Tip: Key relationships in Bohr's model to remember: Radius \(r_n \propto n^2\), and Energy \(E_n \propto -1/n^2\). When an electron transitions to a lower n, its orbit gets smaller and its energy becomes more negative (it is more tightly bound).

Let the resistance of each identical bulb be R and the battery voltage be V. Brightness is proportional to the power dissipated, \(P = I^2 R\).


(i) Key K is closed

Bulbs P and Q are in parallel. Their equivalent resistance is \(R_{PQ} = \frac{R \times R}{R+R} = \frac{R}{2}\).

This combination is in series with bulb S. The total resistance of the circuit is \(R_{total} = R_S + R_{PQ} = R + \frac{R}{2} = \frac{3R}{2}\).

The total current from the battery flows through S, so \(I_S = \frac{V}{R_{total}} = \frac{V}{3R/2} = \frac{2V}{3R}\).

This current \(I_S\) splits equally between the identical bulbs P and Q.
\(I_P = I_Q = \frac{I_S}{2} = \frac{1}{2} \left( \frac{2V}{3R} \right) = \frac{V}{3R}\).

Now, we compare the power (brightness):
\(P_S = I_S^2 R = \left( \frac{2V}{3R} \right)^2 R = \frac{4V^2}{9R}\).
\(P_Q = I_Q^2 R = \left( \frac{V}{3R} \right)^2 R = \frac{V^2}{9R}\).

Since \(P_S = 4P_Q\), bulb S is four times brighter than bulb Q (and bulb P).


(ii) Key K is open

When the key K is open, the branch with bulb P is incomplete, so no current flows through P.

Bulbs S and Q are now in a simple series circuit.

The total resistance of the circuit is \(R_{total} = R_S + R_Q = R + R = 2R\).

The current flowing through the series circuit is the same for both bulbs.
\(I_S = I_Q = \frac{V}{R_{total}} = \frac{V}{2R}\).

Since the current and resistance are the same for both bulbs, their power dissipation is also the same.
\(P_S = P_Q = I_S^2 R = \left( \frac{V}{2R} \right)^2 R = \frac{V^2}{4R}\).

Therefore, bulbs S and Q have equal brightness.
Quick Tip: For circuits with identical bulbs, brightness is determined by the power dissipated (\(P=I^2R\) or \(P=V^2/R\)). Comparing the current through each bulb is often the most direct way to compare their brightness. A bulb in the main branch (like S when closed) will carry more current than bulbs in parallel branches.

We can solve this circuit using Kirchhoff's Laws. Let the top and bottom junctions be A and B respectively.

Let \(I_1\) be the current from the top cell through the 20 \(\Omega\) resistor.

Let \(I_2\) be the current from the bottom cell through the 10 \(\Omega\) resistor.

Let \(I_B\) be the current flowing through the bulb B (resistance \(R_B = 10 \, \Omega\)).


Applying Kirchhoff's Current Law (KCL) at junction A:

The currents \(I_1\) and \(I_2\) enter the junction and combine to form \(I_B\) which leaves the junction.
\(I_1 + I_2 = I_B\). (Equation 1)


Applying Kirchhoff's Voltage Law (KVL) to the top loop (containing the top cell, 20 \(\Omega\) resistor, and the bulb):

Starting from the negative terminal of the top cell and moving clockwise:
\(+10 - I_1(20) - I_B(10) = 0\).
\(10 - 20I_1 - 10I_B = 0\). (Equation 2)


Applying Kirchhoff's Voltage Law (KVL) to the bottom loop (containing the bottom cell, 10 \(\Omega\) resistor, and the bulb):

Starting from the negative terminal of the bottom cell and moving clockwise:
\(+10 - I_2(10) - I_B(10) = 0\).
\(10 - 10I_2 - 10I_B = 0\). (Equation 3)


Now we solve the system of three equations. From Equation 3:
\(10I_2 = 10 - 10I_B \implies I_2 = 1 - I_B\).

From Equation 2:
\(20I_1 = 10 - 10I_B \implies I_1 = \frac{10 - 10I_B}{20} = 0.5 - 0.5I_B\).


Substitute these expressions for \(I_1\) and \(I_2\) into Equation 1:
\((0.5 - 0.5I_B) + (1 - I_B) = I_B\).
\(1.5 - 1.5I_B = I_B\).
\(1.5 = 2.5I_B\).
\(I_B = \frac{1.5}{2.5} = \frac{15}{25} = \frac{3}{5}\).
\(I_B = 0.6\) A.


The current that flows through the bulb is 0.6 A.
Quick Tip: When dealing with circuits containing multiple loops and sources, Kirchhoff's Laws are a reliable method. Always be systematic: 1. Assign current directions. 2. Apply the junction rule (KCL). 3. Apply the loop rule (KVL) for as many independent loops as needed to get a solvable system of equations.

(a) Majority and Minority Charge Carriers

In a p-type semiconductor:

Majority charge carriers are holes.

Minority charge carriers are electrons.


In an n-type semiconductor:

Majority charge carriers are electrons.

Minority charge carriers are holes.


(b) Formation of Diffusion and Drift Currents

Diffusion Current: When a p-n junction is formed, there is a high concentration of holes on the p-side and a high concentration of electrons on the n-side.

Due to this concentration gradient, holes start to diffuse from the p-side to the n-side, and electrons start to diffuse from the n-side to the p-side.

This movement of charge carriers across the junction constitutes the diffusion current.


Drift Current: As diffusion occurs, a layer of fixed positive ions on the n-side and fixed negative ions on the p-side is created near the junction. This is the depletion region.

This layer of ions creates an internal electric field directed from the n-side to the p-side.

This electric field causes the minority charge carriers (electrons from the p-side and holes from the n-side) to be swept across the junction.

This movement of minority carriers due to the electric field constitutes the drift current.

In an unbiased p-n junction at equilibrium, the diffusion current and drift current are equal in magnitude and opposite in direction.
Quick Tip: Remember the cause for each current type: \textbf{Diffusion} \(\rightarrow\) \textbf{D}ifference in concentration. \textbf{Drift} \(\rightarrow\) \textbf{D}riven by an electric field. At equilibrium, these two currents balance each other perfectly.

(a) Conditions for zero magnetic force

The magnetic Lorentz force is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), or in magnitude, \(F = qvB\sin\theta\).

The force F will be zero if:

1. The particle is uncharged (\(q = 0\)).

2. The particle is stationary (\(\vec{v} = 0\)).

3. The particle moves parallel (\(\theta = 0^\circ\)) or anti-parallel (\(\theta = 180^\circ\)) to the direction of the magnetic field, as \(\sin(0^\circ) = \sin(180^\circ) = 0\).


(b) Force on the square loop

The long straight wire produces a magnetic field \(B = \frac{\mu_0 I_1}{2\pi r}\) at a distance r. The direction is into the page at the location of the loop (by Right-Hand Grip Rule).

The force on a wire segment is \(\vec{F} = I_2(\vec{l} \times \vec{B})\).


Force on arm MN: Distance from wire is \(r = l\). Length is \(l\). Current \(I_2\) is upwards. \(B_{MN} = \frac{\mu_0 I_1}{2\pi l}\). Force direction is towards the wire (attractive).
\(F_{MN} = I_2 l B_{MN} = I_2 l \left(\frac{\mu_0 I_1}{2\pi l}\right) = \frac{\mu_0 I_1 I_2}{2\pi}\).


Force on arm KP: Distance from wire is \(r = l+l=2l\). Length is \(l\). Current \(I_2\) is downwards. \(B_{KP} = \frac{\mu_0 I_1}{2\pi (2l)}\). Force direction is away from the wire (repulsive).
\(F_{KP} = I_2 l B_{KP} = I_2 l \left(\frac{\mu_0 I_1}{4\pi l}\right) = \frac{\mu_0 I_1 I_2}{4\pi}\).


Forces on arms NK and PM: The magnetic field is perpendicular to these arms, but for every small element on NK, there is an element on PM with an equal and opposite force. So, the net force on these arms is zero.


Net Force: The net force on the loop is the vector sum of forces on all arms. Since \(F_{MN}\) is attractive and \(F_{KP}\) is repulsive, and \(F_{MN} > F_{KP}\), the net force is attractive.
\(F_{net} = F_{MN} - F_{KP} = \frac{\mu_0 I_1 I_2}{2\pi} - \frac{\mu_0 I_1 I_2}{4\pi}\).
\(F_{net} = \frac{\mu_0 I_1 I_2}{4\pi}\). The direction is towards the long wire.
Quick Tip: To find the net force on a loop near a straight wire, calculate the force on the parallel sides separately. The force is stronger on the side closer to the wire. The net force will be in the direction of the stronger force. Parallel currents attract; anti-parallel currents repel.

(a) Self-inductance of a solenoid

The magnetic field inside a long solenoid is uniform and given by \(B = \mu_0 n I\).

Here, \(n = N/l\) is the number of turns per unit length. So, \(B = \mu_0 \frac{N}{l} I\).

The magnetic flux through a single turn of the solenoid is \(\phi_{turn} = B \cdot A = \left(\mu_0 \frac{N}{l} I\right) A\).

The total magnetic flux linked with the entire solenoid (N turns) is \(\Phi = N \cdot \phi_{turn}\).
\(\Phi = N \left(\mu_0 \frac{N}{l} I A\right) = \frac{\mu_0 N^2 A}{l} I\).

By definition, the self-inductance L is related to the total flux by \(\Phi = LI\).

Comparing the two expressions for \(\Phi\), we get:
\(L = \frac{\mu_0 N^2 A}{l}\).


(b) Comparing self-inductance from the graph

The relationship between the magnitude of induced emf \(|\epsilon|\) and the rate of change of current \(\frac{dI}{dt}\) is given by:
\(|\epsilon| = L \left| \frac{dI}{dt} \right|\).

This equation is of the form \(y = mx\), where \(y = |\epsilon|\) and \(x = \left| \frac{dI}{dt} \right|\).

The slope of the \(|\epsilon|\) vs \(\frac{dI}{dt}\) graph is therefore equal to the self-inductance L.

Slope \(m = L = \frac{|\epsilon|}{dI/dt}\).

From the given graph, the slope of the line for coil '1' is greater than the slope of the line for coil '2'.

Therefore, coil '1' has a greater value of self-inductance.
Quick Tip: The self-inductance L of a device depends only on its geometry (size, shape, number of turns) and the material inside it. For a solenoid, notice that L is proportional to the square of the number of turns (\(N^2\)).

Einstein's photoelectric equation is \(K_{max} = h\nu - \phi_0\), where \(K_{max}\) is the maximum kinetic energy of photoelectrons, \(h\nu\) is the energy of an incident photon, and \(\phi_0\) is the work function of the metal.


(a) Existence of Threshold Frequency

For a photoelectron to be emitted, its kinetic energy must be positive, i.e., \(K_{max} > 0\).

This implies \(h\nu - \phi_0 > 0\), or \(h\nu > \phi_0\).

So, the frequency \(\nu\) must be greater than a certain minimum value, called the threshold frequency (\(\nu_0\)), where \(\nu_0 = \frac{\phi_0}{h}\).

If the incident frequency \(\nu\) is less than \(\nu_0\), no photoelectric emission will occur, regardless of the light's intensity.


(b) Effect of Frequency and Intensity on \(K_{max}\)

The equation \(K_{max} = h\nu - \phi_0\) directly shows that \(K_{max}\) depends linearly on the frequency \(\nu\).

Increasing the frequency \(\nu\) increases the energy of each photon, thus increasing the maximum kinetic energy of the emitted electrons.

Intensity of light corresponds to the number of photons incident per second. Increasing the intensity increases the number of emitted photoelectrons but does not change the energy of any individual photon (\(h\nu\)).

Therefore, \(K_{max}\) is affected by frequency, not intensity.


(c) Cut-off Voltage vs Frequency Graph

The cut-off (or stopping) voltage \(V_0\) is related to \(K_{max}\) by the equation \(K_{max} = eV_0\).

Substituting this into Einstein's equation gives: \(eV_0 = h\nu - \phi_0\).

Rearranging for \(V_0\): \(V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}\).

This is in the form of a straight-line equation, \(y = mx + c\).

A plot of \(V_0\) (y-axis) versus \(\nu\) (x-axis) will be a straight line.

By comparing the equations, the slope of this line is \(m = \frac{h}{e}\).
Quick Tip: Einstein's photoelectric equation is the key to understanding all aspects of the photoelectric effect. Remember that light energy is quantized into photons (\(E=h\nu\)), and one photon interacts with one electron.

(a) Conditions for Interference

For two coherent light waves with wavelength \(\lambda\), the conditions for interference at a point depend on the path difference (\(\Delta x\)) between the waves reaching that point.

(i) Constructive Interference (Bright Fringes): Occurs when the path difference is an integral multiple of the wavelength.
\(\Delta x = n\lambda\), where \(n = 0, 1, 2, 3, ...\).

(ii) Destructive Interference (Dark Fringes): Occurs when the path difference is an odd integral multiple of half the wavelength.
\(\Delta x = (2n+1)\frac{\lambda}{2}\) or \(\Delta x = (n+\frac{1}{2})\lambda\), where \(n = 0, 1, 2, 3, ...\).


Interference from two sodium lamps:

No, these conditions cannot be applied to light from two independent sodium lamps.

Reason: Two independent light sources are not coherent. The atoms in each lamp emit photons independently and randomly. As a result, the phase difference between the waves from the two lamps fluctuates rapidly and randomly. A stable, observable interference pattern requires a constant phase difference between the sources, i.e., they must be coherent.


(b) Effect of changing colour on fringe width

The fringe width (\(\beta\)) in Young's double-slit experiment is given by the formula \(\beta = \frac{\lambda D}{d}\).

Here, \(\lambda\) is the wavelength of light, D is the screen distance, and d is the slit separation.

The wavelength of red light (\(\lambda_{red}\)) is greater than the wavelength of green light (\(\lambda_{green}\)). In the visible spectrum, \(\lambda_{red} \approx 700\) nm and \(\lambda_{green} \approx 550\) nm.

Since the fringe width \(\beta\) is directly proportional to the wavelength \(\lambda\) (\(\beta \propto \lambda\)), and D and d are kept constant:

Replacing green light with red light will increase the fringe width.

The interference fringes will become wider and more spread out on the screen.
Quick Tip: Coherence is the essential condition for observing a stable interference pattern. This is why in experiments like YDSE, light from a single source is passed through two slits, making the two slits act as two coherent sources.

(i) Derivation of Resistivity

The drift velocity of free electrons in a conductor under an electric field E is given by \(v_d = \frac{eE}{m}\tau\), where e is the charge, m is the mass of an electron, and \(\tau\) is the relaxation time.

The current I flowing through the conductor is related to the drift velocity by \(I = n e A v_d\), where n is the number density of free electrons and A is the cross-sectional area.

Substituting the expression for \(v_d\): \(I = n e A \left( \frac{eE}{m}\tau \right) = \frac{n e^2 A \tau}{m} E\).

The current density is \(J = \frac{I}{A} = \frac{n e^2 \tau}{m} E\).

From the microscopic form of Ohm's law, we have \(J = \sigma E\), where \(\sigma\) is the electrical conductivity.

Comparing the two expressions for J, we find the conductivity: \(\sigma = \frac{n e^2 \tau}{m}\).

Resistivity (\(\rho\)) is the reciprocal of conductivity (\(\rho = 1/\sigma\)).

Therefore, the expression for resistivity is \(\rho = \frac{m}{n e^2 \tau}\).


(ii) Comparison of Charges

The charge Q that passes through a cross-section is given by the area under the current-time (I-t) graph.

The first time interval is from t=0 to t=3 s. The current is constant at I = 2.0 A.
\(Q_1 = Area_1 = Area of rectangle = I \times \Delta t = (2.0 A) \times (3 s) = 6.0 C\).

The second time interval is from t=3 s to t=6 s. The graph is a straight line, forming a trapezoid with the t-axis.

The area of the trapezoid is given by \(\frac{1}{2} \times (sum of parallel sides) \times (height)\).
\(Q_2 = Area_2 = \frac{1}{2} \times (I(t=3) + I(t=6)) \times (6-3)\).

From the graph, \(I(t=3) = 2.0\) A and \(I(t=6) = 1.0\) A.
\(Q_2 = \frac{1}{2} \times (2.0 + 1.0) A \times (3 s) = \frac{1}{2} \times 3.0 \times 3 = 4.5 C\).

Comparing the two charges, \(Q_1 = 6.0\) C and \(Q_2 = 4.5\) C. Therefore, \(Q_1 > Q_2\).
Quick Tip: Remember that charge (\(Q\)) is the integral of current over time (\(Q = \int I dt\)). For simple I-t graphs, this integral is equivalent to calculating the geometric area under the curve.

(I) Current and Maximum Current

In the circuit, the total resistance is the sum of the external resistance R and the internal resistance r.
\(R_{total} = R + r\).

According to Ohm's law, the current I in the circuit is given by:
\(I = \frac{Total EMF}{Total Resistance} = \frac{E}{R+r}\).

The current I will be maximum when the denominator (\(R+r\)) is minimum. Since r is constant, this occurs when the external resistance R is minimum, i.e., \(R=0\) (short circuit).
\(I_{max} = \frac{E}{0+r} = \frac{E}{r}\).

This is the maximum possible current the battery can supply.


(II) Terminal Voltage and Maximum Terminal Voltage

The terminal voltage V across the battery is the potential difference across its terminals. It is also equal to the potential difference across the external resistance R.
\(V = I \cdot R\).

Substituting the expression for I: \(V = \left(\frac{E}{R+r}\right)R = \frac{ER}{R+r}\).

Alternatively, using the loop rule for the battery, \(V = E - Ir\).

To find the maximum possible value of V, we consider the equation \(V = E - Ir\). V is maximum when the term \(Ir\) is minimum.

The minimum possible current is \(I=0\), which occurs when the external resistance is infinite (\(R \to \infty\), an open circuit).

When \(R \to \infty\), \(I \to 0\).

The terminal voltage becomes \(V = E - (0)r = E\).

Thus, the maximum possible value of the terminal voltage is the emf E of the battery itself.
Quick Tip: Remember the key difference: EMF (E) is the work done per unit charge by the battery source. Terminal Voltage (V) is the potential difference available at the terminals for the external circuit. They are equal only in an open circuit (\(I=0\)). Otherwise, \(V = E - Ir\).

(a) Characteristics of Electromagnetic (EM) Waves

1. Transverse Nature: EM waves are transverse in nature. The electric field (\(\vec{E}\)) and magnetic field (\(\vec{B}\)) vectors oscillate perpendicular to each other and also perpendicular to the direction of wave propagation.

2. Speed in Vacuum: All EM waves travel through vacuum with the same constant speed, the speed of light, \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \approx 3 \times 10^8\) m/s.

3. No Medium Required: EM waves do not require any material medium for their propagation. They can travel through empty space (vacuum).

(Other possible characteristics: They carry energy and momentum. The ratio of electric to magnetic field magnitudes is constant, \(E/B=c\).)


(b) Displacement Current in a Capacitor

How it arises: During the charging of a parallel-plate capacitor, the charge on the plates (\(q\)) increases with time. This causes the electric field (\(E = q/(\epsilon_0 A)\)) between the plates to increase. Consequently, the electric flux (\(\Phi_E = E \cdot A\)) through the space between the plates changes with time.

According to Maxwell's theory, a changing electric flux produces a magnetic field in the same way that a conduction current does. This source of magnetic field is termed the displacement current (\(I_D\)).
\(I_D = \epsilon_0 \frac{d\Phi_E}{dt}\).


Where it exists: The displacement current exists in the region between the capacitor plates.

In the connecting wires, there is a conduction current (\(I_C = dq/dt\)) due to the flow of electrons. Between the plates, there is no flow of charge, so the conduction current is zero.

The displacement current \(I_D\) in the gap is exactly equal to the conduction current \(I_C\) in the wires, thus ensuring that the total current is continuous throughout the entire circuit.
Quick Tip: Maxwell's great contribution was realizing that a changing electric field can create a magnetic field. This concept, embodied in the displacement current, unified electricity and magnetism and predicted the existence of electromagnetic waves.

The electrostatic force of attraction between the proton and electron provides the necessary centripetal force for the circular motion.


Electrostatic Force, \(F_e = K \frac{(e)(e)}{r^2} = \frac{Ke^2}{r^2}\).


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


Equating the two forces: \(F_e = F_c\).

\(\frac{Ke^2}{r^2} = \frac{mv^2}{r}\).


Solving for \(v^2\): \(v^2 = \frac{Ke^2}{r^2} \cdot \frac{r}{m} = \frac{Ke^2}{mr}\).


Taking the square root to find the speed v:

\(v = \sqrt{\frac{Ke^2}{mr}}\).
Quick Tip: The fundamental principle in analyzing atomic orbits is equating the electrostatic force with the centripetal force. This relationship is the starting point for deriving expressions for velocity, energy, and radius in the Bohr model.

The total energy (E) is the sum of the kinetic energy (KE) and the potential energy (PE).

\(E = KE + PE\).


From the result in part (i), we found \(mv^2 = \frac{Ke^2}{r}\).


Kinetic energy is \(KE = \frac{1}{2}mv^2 = \frac{1}{2}\left(\frac{Ke^2}{r}\right) = \frac{Ke^2}{2r}\).


The electrostatic potential energy of the electron-proton system is \(PE = -K \frac{e^2}{r}\).


Total energy is \(E = \frac{Ke^2}{2r} + \left(-\frac{Ke^2}{r}\right) = \frac{Ke^2 - 2Ke^2}{2r}\).

\(E = -\frac{Ke^2}{2r}\).
Quick Tip: In the Bohr model for hydrogen, remember the relationship between the different forms of energy: Total Energy (E) = - Kinetic Energy (KE) = Potential Energy (PE) / 2. This is a very useful shortcut.

When a photon is emitted, the total energy of the atom decreases. The change in total energy (\(\Delta E_{total}\)) is equal to the negative of the emitted photon's energy.


Energy of the emitted photon, \(E_{photon} = \frac{hc}{\lambda}\).


Using the shortcut formula for energy in eV: \(E_{photon} = \frac{1240 eV\cdotnm}{\lambda (nm)}\).

\(E_{photon} = \frac{1240}{500} eV = 2.48 eV\).


Change in total energy of the electron: \(\Delta E_{total} = E_{final} - E_{initial} = -E_{photon} = -2.48 eV\).


From the relationship \(E = -KE\), we can write \(KE = -E\).


The change in kinetic energy is \(\Delta KE = KE_{final} - KE_{initial} = (-E_{final}) - (-E_{initial}) = -(E_{final} - E_{initial})\).

\(\Delta KE = -(\Delta E_{total})\).

\(\Delta KE = -(-2.48 eV) = +2.48 eV\).


So the changes are -2.48 eV and 2.48 eV respectively.
Quick Tip: When an electron in an atom moves to a lower energy level (more negative total energy), it becomes more tightly bound, its speed increases, and thus its kinetic energy increases.

The frequency of revolution (f) is the reciprocal of the time period (T), and is given by \(f = \frac{v}{2\pi r}\).


In Bohr's model, the velocity of the electron in the n\(^{th}\) orbit is \(v_n \propto \frac{1}{n}\).


The radius of the n\(^{th}\) orbit is \(r_n \propto n^2\).


Substituting these proportionalities into the frequency formula:

\(f_n \propto \frac{v_n}{r_n} \propto \frac{1/n}{n^2}\).

\(f_n \propto \frac{1}{n^3}\).


Thus, the frequency of revolution is proportional to \(1/n^3\).
Quick Tip: It's helpful to memorize the key proportionalities in the Bohr model: Radius \(r_n \propto n^2\), Velocity \(v_n \propto 1/n\), and Energy \(E_n \propto -1/n^2\). From these, other quantities like frequency (\(f \propto v/r\)) can be quickly derived.

The energy of the n\(^{th}\) state in a hydrogen atom is given by \(E_n = \frac{-13.6}{n^2}\) eV.


The initial state has energy \(E_i = -3.4\) eV. Let's find the initial quantum number \(n_i\).

\(-3.4 = \frac{-13.6}{n_i^2} \implies n_i^2 = \frac{-13.6}{-3.4} = 4 \implies n_i = 2\).


The final state is the ground state, for which the quantum number is \(n_f = 1\).


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


Given \(r_1 = 0.53\) Å.


Initial radius (for \(n_i=2\)): \(r_i = (2)^2 \times r_1 = 4 \times 0.53 Å = 2.12 Å\).


Final radius (for \(n_f=1\)): \(r_f = (1)^2 \times r_1 = 1 \times 0.53 Å = 0.53 Å\).


The change in the radius of the orbit is \(\Delta r = |r_f - r_i| = |0.53 - 2.12| Å = |-1.59| Å\).


The radius changes by 1.59 Å.
Quick Tip: The energy levels of hydrogen are quantized as \(E_n = -13.6/n^2\) eV. Memorizing the first few energy levels (E1=-13.6, E2=-3.4, E3=-1.51 eV) can save calculation time in exams.

The given arrangement consists of two capacitors connected in parallel.


The first capacitor (C\(_1\)) is formed by plates P\(_1\) and P\(_2\).

The second capacitor (C\(_2\)) is formed by plates P\(_2\) and P\(_3\).


Plate P\(_2\) is a common plate connected to point A. Plates P\(_1\) and P\(_3\) are connected together to point B. This is a parallel combination.


The capacitance of a single parallel plate capacitor with a dielectric is \(C = \frac{K \epsilon_0 A}{d}\).


Here, the area of each plate is \(A = L^2\).


So, \(C_1 = \frac{K \epsilon_0 L^2}{d}\) and \(C_2 = \frac{K \epsilon_0 L^2}{d}\).


For a parallel combination, the total capacitance is the sum of individual capacitances.

\(C_{total} = C_1 + C_2 = \frac{K \epsilon_0 L^2}{d} + \frac{K \epsilon_0 L^2}{d} = \frac{2K \epsilon_0 L^2}{d}\).
Quick Tip: When plates are stacked and connected alternately, they form a parallel combination of capacitors. The number of capacitors formed is one less than the number of plates.

The charge on plate P\(_1\) is the charge stored in the first capacitor (C\(_1\)) formed by plates P\(_1\) and P\(_2\).


Since the capacitors are in parallel, the potential difference across C\(_1\) is the same as the total potential difference, which is V.


The charge on a capacitor is given by the formula \(Q = CV\).

\(Q_1 = C_1 V\).


We know that \(C_1 = \frac{K \epsilon_0 L^2}{d}\).


Substituting this, we get the magnitude of the charge on plate P\(_1\):

\(Q_1 = \left( \frac{K \epsilon_0 L^2}{d} \right) V = \frac{\epsilon_0 V K L^2}{d}\).

(Since A is at a positive potential and P\(_1\) is connected to B, the charge on P\(_1\) will be negative, but the options refer to the magnitude).
Quick Tip: In a parallel combination of capacitors, the voltage across each capacitor is the same and is equal to the voltage applied across the combination. The total charge is the sum of the charges on individual capacitors.

For a uniform electric field, the relationship between the field strength (E), potential difference (V), and distance (d) is \(E = \frac{V}{d}\).


The region between plates P\(_1\) and P\(_2\) forms a capacitor.


The potential difference across these two plates is given as V.


The distance (separation) between these plates is given as d.


Therefore, the electric field (E) in the region between P\(_1\) and P\(_2\) is:

\(E = \frac{V}{d}\).
Quick Tip: The formula \(E=V/d\) is valid for the magnitude of the uniform electric field between the plates of a parallel plate capacitor. The direction of the field is from the plate at higher potential to the plate at lower potential.

Let the equivalent air capacitor have a plate separation of \(d'\). The dielectric is air, so K=1.


The capacitance of this air capacitor is \(C_{air} = \frac{\epsilon_0 A}{d'} = \frac{\epsilon_0 L^2}{d'}\).


From part (i), the capacitance of the given system is \(C_{system} = \frac{2K \epsilon_0 L^2}{d}\).


We are given that the capacitances are equal: \(C_{air} = C_{system}\).

\(\frac{\epsilon_0 L^2}{d'} = \frac{2K \epsilon_0 L^2}{d}\).


We can cancel the term \(\epsilon_0 L^2\) from both sides.

\(\frac{1}{d'} = \frac{2K}{d}\).


Solving for the new separation \(d'\):

\(d' = \frac{d}{2K}\).
Quick Tip: Inserting a dielectric of constant K into a capacitor increases its capacitance by a factor of K. Alternatively, to get the same capacitance as a dielectric-filled capacitor, an air-filled capacitor would need a much smaller plate separation.

Initially, the capacitor system is charged by a battery. The battery acts like a pump, moving charge from one terminal to the other.


It moves a total positive charge \(Q_{total} = C_{total}V\) to plate P\(_2\) (connected to A) and an equal amount of negative charge to plates P\(_1\) and P\(_3\) (connected to B).


Charge on P\(_2\): \(+Q_{total} = +\frac{2K \epsilon_0 L^2 V}{d}\).


Charge on (P\(_1\) + P\(_3\)): \(-Q_{total} = -\frac{2K \epsilon_0 L^2 V}{d}\).


The net charge of the entire isolated system (all three plates considered together) is the sum of the charges on all plates:

\(Q_{system} = Q_{P_1} + Q_{P_2} + Q_{P_3} = (+Q_{total}) + (-Q_{total}) = 0\).


The capacitor as a whole is electrically neutral.


When the source is removed and points A and B are connected by a wire, the plates are short-circuited. The excess positive charge on P\(_2\) will flow through the wire to neutralize the negative charge on P\(_1\) and P\(_3\).


This process is an internal redistribution of charge. The total charge of the isolated system does not change.


Since the net charge on the system was initially zero, it remains zero after the connection is made.
Quick Tip: A capacitor, when charged by a battery, stores energy by separating charges, but it does not create net charge. The entire device remains electrically neutral. Short-circuiting a charged capacitor simply allows the separated charges to recombine, bringing the potential difference to zero.

Part (i): Focal length of concave lens

Case 1: Convex lens only

Given: Focal length of convex lens, \(f_1 = +10\) cm. Object distance, \(u = -30\) cm.

Using the lens formula: \(\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u}\).
\(\frac{1}{10} = \frac{1}{v_1} - \frac{1}{-30} = \frac{1}{v_1} + \frac{1}{30}\).
\(\frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}\).

The initial image position is \(v_1 = +15\) cm.


Case 2: Combination of lenses

A concave lens is placed in contact. The screen is moved by 45 cm. Since the concave lens is diverging, it will shift the final image further away.

New image distance, \(v_2 = v_1 + 45 cm = 15 + 45 = 60\) cm.

The object distance remains \(u = -30\) cm.

Let F be the equivalent focal length of the combination.
\(\frac{1}{F} = \frac{1}{v_2} - \frac{1}{u} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}\).

The equivalent focal length is \(F = +20\) cm.


For lenses in contact, \(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\), where \(f_2\) is the focal length of the concave lens.
\(\frac{1}{20} = \frac{1}{10} + \frac{1}{f_2}\).
\(\frac{1}{f_2} = \frac{1}{20} - \frac{1}{10} = \frac{1-2}{20} = -\frac{1}{20}\).

Therefore, the focal length of the concave lens is \(f_2 = -20\) cm.


Part (ii): Prism

For an equilateral prism, the angle of the prism is \(A = 60^\circ\).

Given refractive index, \(n = \sqrt{3}\).

Using the prism formula for the angle of minimum deviation (\(\delta_m\)):
\(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\).
\(\sqrt{3} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin(30^\circ)}\).

Since \(\sin(30^\circ) = 1/2\):
\(\sqrt{3} \times \frac{1}{2} = \sin\left(\frac{60^\circ + \delta_m}{2}\right)\).
\(\sin\left(\frac{60^\circ + \delta_m}{2}\right) = \frac{\sqrt{3}}{2}\).

This implies \(\frac{60^\circ + \delta_m}{2} = 60^\circ\).
\(60^\circ + \delta_m = 120^\circ\).
\(\delta_m = 60^\circ\).

For the case of minimum deviation, the angle of incidence (i) is given by:
\(i = \frac{A + \delta_m}{2} = \frac{60^\circ + 60^\circ}{2} = 60^\circ\).
Quick Tip: For combinations of thin lenses in contact, the reciprocal of the equivalent focal length is the algebraic sum of the reciprocals of individual focal lengths. Remember to use the sign convention consistently. For a prism at minimum deviation, the light ray passes symmetrically through it (\(i=e, r_1=r_2\)).

Part (i): Young's Double Slit Experiment

Given: Wavelength \(\lambda = 633 nm = 633 \times 10^{-9}\) m.

Screen distance \(D = 5.0\) m.

Fringe width \(\beta = 5 mm = 5 \times 10^{-3}\) m.


(I) Slit Separation (d)

The formula for fringe width is \(\beta = \frac{\lambda D}{d}\).

Rearranging to find d:
\(d = \frac{\lambda D}{\beta}\).
\(d = \frac{(633 \times 10^{-9} m) \times (5.0 m)}{5 \times 10^{-3} m}\).
\(d = 633 \times 10^{-6} m = 0.633 mm\).


(II) Position of First Minimum

The position of the n\(^{th}\) dark fringe (minimum) from the center is given by \(y_n = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d} = \left(n - \frac{1}{2}\right)\beta\).

For the first minimum, n = 1.
\(y_1 = \left(1 - \frac{1}{2}\right)\beta = \frac{1}{2}\beta\).
\(y_1 = \frac{1}{2} \times (5 mm) = 2.5 mm\).


Part (ii): Single Slit Diffraction

Given: Wavelength \(\lambda = 650 nm = 650 \times 10^{-9}\) m.

Slit width \(a = 0.6 mm = 0.6 \times 10^{-3}\) m.

Screen distance \(D = 60 cm = 0.6\) m.


The position of the first minimum from the center is given by \(y = \frac{\lambda D}{a}\).
\(y = \frac{(650 \times 10^{-9} m) \times (0.6 m)}{0.6 \times 10^{-3} m}\).
\(y = 650 \times 10^{-6} m = 0.65 mm\).

This is the distance from the center to the first minimum on one side.

The distance between the first order minima on both sides is the width of the central maximum, which is \(2y\).

Distance = \(2 \times 0.65 mm = 1.3 mm\).
Quick Tip: Be careful to distinguish between formulas for interference and diffraction. For YDSE, the bright fringe separation (fringe width) is \(\beta = \lambda D/d\). For single-slit diffraction, the width of the central maximum (distance between first minima) is \(W = 2\lambda D/a\).

Part (i): Electric Field of a Dipole

The setup describes an electric dipole with charges at (a, 0) and (-a, 0). The point P(0, y) is on the equatorial line.

The distance from each charge to point P is \(r = \sqrt{a^2 + y^2}\).

The electric field due to +q at (a,0) is \(\vec{E}_+\). The electric field due to -q at (-a,0) is \(\vec{E}_-\).

The magnitudes of these fields are equal: \(E_+ = E_- = E = \frac{kq}{r^2} = \frac{kq}{a^2+y^2}\).

The vertical components (\(y\)-components) of the fields cancel out due to symmetry.

The horizontal components (\(x\)-components) add up. Both point in the negative x-direction.
\(E_{net} = E_+ \cos\theta + E_- \cos\theta = 2E \cos\theta\), where \(\theta\) is the angle with the x-axis.

From the geometry, \(\cos\theta = \frac{a}{r} = \frac{a}{\sqrt{a^2 + y^2}}\).
\(E_{net} = 2 \left(\frac{kq}{a^2+y^2}\right) \left(\frac{a}{\sqrt{a^2 + y^2}}\right) = \frac{2kqa}{(a^2+y^2)^{3/2}}\).

The direction is along the negative x-axis. So, in vector form: \(\vec{E}_{net} = \frac{-2kqa}{(a^2+y^2)^{3/2}}\hat{i}\).

This can be written in terms of dipole moment \(p = q(2a)\) as \(\vec{E}_{net} = \frac{-kp}{(a^2+y^2)^{3/2}}\hat{i}\).


For a far off point, \(y \gg a\). We can neglect \(a^2\) in the denominator.
\(\vec{E}_{net} \approx \frac{-kp}{(y^2)^{3/2}}\hat{i} = \frac{-kp}{y^3}\hat{i}\).


Part (ii): Work Done

Work done is the change in the potential energy of the system: \(W = U_{final} - U_{initial}\).

The potential energy of a system of three charges is \(U = k\left(\frac{q_A q_B}{r_{AB}} + \frac{q_B q_C}{r_{BC}} + \frac{q_C q_A}{r_{CA}}\right)\).

Given: \(q_A = -2 nC\), \(q_B = -1 nC\), \(q_C = +5 nC\).

Initial state: The charges are at the vertices of an equilateral triangle of side \(r_{initial} = 0.2\) m.
\(U_{initial} = \frac{9\times10^9}{0.2} [(-2)(-1) + (-1)(5) + (5)(-2)] \times 10^{-18}\).
\(U_{initial} = 4.5\times10^{10} [2 - 5 - 10] \times 10^{-18} = 4.5\times10^{10} [-13] \times 10^{-18} = -58.5 \times 10^{-8} J\).

Final state: The charges are at the midpoints of the sides. The new triangle formed by the midpoints is also equilateral, with side length \(r_{final} = \frac{0.2}{2} = 0.1\) m.
\(U_{final} = \frac{9\times10^9}{0.1} [(-2)(-1) + (-1)(5) + (5)(-2)] \times 10^{-18}\).
\(U_{final} = 9\times10^{10} [-13] \times 10^{-18} = -117 \times 10^{-8} J\).

Work Done, \(W = U_{final} - U_{initial} = (-117 \times 10^{-8}) - (-58.5 \times 10^{-8}) J\).
\(W = -58.5 \times 10^{-8} J = -5.85 \times 10^{-7} J\).
Quick Tip: The work done in rearranging a system of charges depends only on the initial and final potential energies of the system, not on the path taken. Remember that potential energy is a scalar quantity.

Consistency of Gauss's Law and Coulomb's Law

Let's derive Coulomb's Law from Gauss's Law. Consider a point charge q at the origin.

Gauss's Law states: \(\oint \vec{E} \cdot d\vec{S} = \frac{q_{enc}}{\epsilon_0}\).

To find the field at a distance y, we choose a spherical Gaussian surface of radius y, concentric with the charge.

By symmetry, the electric field \(\vec{E}\) must be radial and have the same magnitude at all points on the surface.

Therefore, \(\vec{E}\) is parallel to \(d\vec{S}\) everywhere on the surface, so \(\vec{E} \cdot d\vec{S} = E dS\).
\(\oint E dS = E \oint dS = E (4\pi y^2)\).

The enclosed charge is \(q_{enc} = q\).

Equating the two, \(E (4\pi y^2) = \frac{q}{\epsilon_0}\), which gives \(E = \frac{q}{4\pi\epsilon_0 y^2}\).

The force on a test charge \(q_0\) placed at this point would be \(F = q_0 E = \frac{1}{4\pi\epsilon_0}\frac{q q_0}{y^2}\), which is Coulomb's Law. Thus, the two are consistent.


Electric Field of a Spherical Shell

Consider a thin spherical shell of radius r with total charge Q uniformly distributed on its surface.

Let's find the field at a point P at a distance y from the center.


(I) Outside the shell (y > r)

We draw a concentric spherical Gaussian surface of radius y.

The total charge enclosed by this surface is the entire charge of the shell, \(q_{enc} = Q\).

From Gauss's Law, \(\oint \vec{E} \cdot d\vec{S} = \frac{Q}{\epsilon_0}\).

By symmetry, \(E (4\pi y^2) = \frac{Q}{\epsilon_0}\).
\(E = \frac{Q}{4\pi\epsilon_0 y^2}\). (The field is the same as if all charge were concentrated at the center).


(II) Inside the shell (y < r)

We draw a concentric spherical Gaussian surface of radius y.

Since all the charge resides on the surface of the shell (at radius r), there is no charge enclosed by this Gaussian surface.
\(q_{enc} = 0\).

From Gauss's Law, \(\oint \vec{E} \cdot d\vec{S} = \frac{0}{\epsilon_0} = 0\).

Since the area \(4\pi y^2\) is not zero, the electric field magnitude must be zero.
\(E = 0\).
Quick Tip: Gauss's Law is extremely powerful for calculating electric fields in situations with high symmetry (spherical, cylindrical, planar). The key is to choose a Gaussian surface on which the electric field magnitude is constant and the angle between \(\vec{E}\) and \(d\vec{S}\) is simple (0 or 90 degrees).

Part (i): Lenz's Law and Conservation of Energy

Lenz's Law Statement: The direction of the induced electromotive force (emf) and hence the induced current in a closed circuit is always such that it opposes the change in magnetic flux that produces it.

Connection to Conservation of Energy: Consider moving the N-pole of a magnet towards a closed loop. As the magnet approaches, the magnetic flux through the loop increases. According to Lenz's law, the induced current will create its own magnetic field to oppose this increase. It does this by forming an N-pole on the face of the loop towards the magnet, which repels the approaching magnet. An external agent must do mechanical work against this repulsive force to move the magnet. This mechanical work done by the external agent is converted into electrical energy in the loop, which then dissipates as heat. If the induced current were in the opposite direction (assisting the change), it would create an S-pole, pulling the magnet in. This would accelerate the magnet, increasing the flux change and the current further, leading to a runaway creation of energy from nothing, which violates the principle of conservation of energy. Thus, Lenz's law is a direct consequence of energy conservation.


Part (ii): Loop in Magnetic Field

Let the region of magnetic field start at x=0 and end at x=l. The loop has side \(s=l/2\).

Time to enter the field: \(t_1 = s/v = l/(2v)\).

Time when fully entered to start of exit: \(t_2 = l/v\).

Time when fully exited: \(t_3 = (l+s)/v = (l+l/2)/v = 3l/(2v)\).


(I) Directions of Induced Current

Entering: As the loop enters, the magnetic flux into the page is increasing. To oppose this, the induced current will create a magnetic field out of the page. By the right-hand rule, the induced current must flow in the counter-clockwise direction.

Leaving: As the loop leaves, the magnetic flux into the page is decreasing. To oppose this, the induced current will create a magnetic field into the page. By the right-hand rule, the induced current must flow in the clockwise direction.


(II) Plots of Flux (\(\phi\)) and EMF (E)

Magnetic Flux \(\phi(t) = B \cdot A_{inside}(t)\). Maximum flux is \(\phi_{max} = B(l/2)^2 = Bl^2/4\).

Induced EMF \(E(t) = -d\phi/dt\). The magnitude of EMF while entering or leaving is \(|E| = Bv(l/2)\).
Quick Tip: When drawing flux and EMF graphs for a loop moving through a magnetic field, remember that EMF is the negative slope of the flux-time graph. A linearly changing flux results in a constant non-zero EMF, and a constant flux results in zero EMF.

Part (i): Peak and RMS Values

Peak Value (\(I_0, V_0\)): This is the maximum value or amplitude that an alternating current or voltage attains during its cycle.

RMS Value (\(I_{rms}, V_{rms}\)): The Root Mean Square value is the effective value of an AC current/voltage. It is defined as the equivalent steady DC value that would produce the same heating effect (power dissipation) in a given resistor over a complete cycle.

Relation: The RMS value is related to the peak value by:
\(I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0\) and \(V_{rms} = \frac{V_0}{\sqrt{2}} \approx 0.707 V_0\).


Part (ii): AC Circuit Elements

(I) Identification and Phasor Diagrams

Element X: "Voltage leads the current by \(\pi/2\)". This is the characteristic behavior of an ideal inductor (L). Mnemonic: ELI (EMF leads Current in an Inductor).


Element Y: "Voltage lags behind the current by \(\pi/2\)". This is the characteristic behavior of an ideal capacitor (C). Mnemonic: ICE (Current leads EMF in a Capacitor).


(II) Resonance in Series L-C Circuit

When the inductor (X) and capacitor (Y) are connected in series with a resistor R (real circuits have some resistance) to the source, it forms a series LCR circuit.

Condition for Resonance: Resonance occurs when the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)), causing the net reactance of the circuit to be zero.
\(X_L = X_C\).

Resonant Frequency:
\(2\pi\nu_0 L = \frac{1}{2\pi\nu_0 C}\).
\((2\pi\nu_0)^2 = \frac{1}{LC}\).
\(2\pi\nu_0 = \frac{1}{\sqrt{LC}}\).

The resonant frequency is \(\nu_0 = \frac{1}{2\pi\sqrt{LC}}\).

Impedance Value: The impedance of a series LCR circuit is \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).

At resonance, \(X_L - X_C = 0\).

Therefore, the impedance at resonance is minimum and equals the resistance of the circuit: \(Z = \sqrt{R^2 + 0} = R\).

(If the components are considered ideal and no resistor is present, R=0 and the impedance would be zero).
Quick Tip: Use the mnemonics "ELI the ICE man" to remember the phase relationships in AC circuits. For an inductor (L), Voltage (E) leads Current (I). For a capacitor (C), Current (I) leads Voltage (E). At resonance in a series LCR circuit, the circuit behaves as if it is purely resistive.