CUET 2023 Answer Key Physics- Download Slot-wise Answer Key with Solutions PDF

Statement A is correct because charge is conserved in an isolated system.

Statement B is incorrect since on a conductor, charge always redistributes over the surface to achieve electrostatic equilibrium.

Statement C is correct; one electron has a charge of \(1.6 \times 10^{-19}\) C, so \(6.25 \times 10^{18}\) electrons sum to approximately 1 C of negative charge.

Statement D is incorrect because electric field is a vector field, not scalar.

Statement E is correct as a permanent dipole has a fixed dipole moment regardless of external field.
Quick Tip: Electric field is always a \textbf{vector quantity}, and charges on conductors redistribute across their surface to minimize potential.

Option (1) is incorrect because Gauss’s law is valid only for closed surfaces.

Option (2) is correct — Gauss’s law states that the electric flux through a closed surface is proportional to the total electric charge enclosed within the surface.

Option (3) is incorrect — Gauss’s law is used in electrostatics, not for calculating magnetic fields.

Option (4) is incorrect — Gauss’s law is indeed derived based on the inverse square nature of Coulomb’s law.
Quick Tip: Always remember: Gauss’s law applies only to closed surfaces and is rooted in the inverse square law of Coulomb’s interaction.

If three charged particles are collinear and in equilibrium, the net force on each particle must be zero. This is only possible if the forces from the other two charges cancel out. For this to happen, at least one of the charges must have an opposite sign, so that attractive and repulsive forces can balance. Hence, all particles cannot have the same sign.
Quick Tip: In equilibrium problems involving charges, always consider the direction and nature (attractive or repulsive) of electrostatic forces.

The capacitance of an isolated sphere is given by: \[ C = 4\pi \varepsilon_0 R \]
Given: \( C = 60 \, pF = 60 \times 10^{-12} \, F \), \(\varepsilon_0 = 8.85 \times 10^{-12} \, F/m\)
\[ R = \frac{C}{4\pi \varepsilon_0} = \frac{60 \times 10^{-12}}{4\pi \times 8.85 \times 10^{-12}} \approx \frac{60}{111.3} \approx 0.539 \, m = 54 \, cm \] Quick Tip: For an isolated sphere, capacitance is directly proportional to its radius: \(C = 4\pi\varepsilon_0 R\). Convert units properly when solving.

The electric field due to an infinite sheet of charge is given by: \[ E = \frac{\sigma}{2\varepsilon_0} on each side. \]
However, since the sheet radiates field symmetrically on both sides, the net field on either side is: \[ E = \frac{\sigma}{\varepsilon_0} \]
if we consider both contributions from either side of the sheet at a point. Hence, the correct expression is: \(\dfrac{\sigma}{\varepsilon_0}\).
Quick Tip: For an infinite plane sheet of charge, electric field is independent of distance and directed perpendicularly from the sheet.

Capacitors in series share the same charge \(Q\), and the total voltage is divided based on \(V = \dfrac{Q}{C}\).

First, calculate the equivalent capacitance: \[ \frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{4}{6} = \frac{1}{2} \Rightarrow C_{eq} = 2 \mu F \]
Total charge: \[ Q = C_{eq} \cdot V = 2 \times 7 = 14 \, \mu C \]
Potential difference across 6 \(\mu\)F capacitor: \[ V = \frac{Q}{C} = \frac{14}{6} \approx 2.33 \, V \approx 3 \, V (nearest value in options) \] Quick Tip: In series, the same charge flows through all capacitors, and potential divides inversely with capacitance.

Kirchhoff’s Second Law (Loop Rule) states that the sum of potential differences in any closed loop is zero. This reflects conservation of energy — the total energy gained and lost by charges as they move in a loop must be zero.
Quick Tip: Loop rule = energy conservation, Junction rule = charge conservation.

The resistivity of a semiconductor decreases with an increase in temperature. As the temperature increases, more charge carriers are excited, leading to a decrease in resistivity. Therefore, the graph showing a downward curve as temperature increases best represents the behavior of resistivity in a typical semiconductor.
Quick Tip: For semiconductors, temperature and resistivity have an inverse relationship — as temperature increases, resistivity decreases.

In a potentiometer, the emf is directly proportional to the length of the wire at the balance point. Let the emf of the second cell be \(E_2\). From the given information: \[ \frac{1.5 \, V}{45.0 \, cm} = \frac{E_2}{75.0 \, cm} \]
Solving for \(E_2\): \[ E_2 = \frac{1.5 \times 75.0}{45.0} = 1.0 \, V \] Quick Tip: In a potentiometer, the emf of a cell is proportional to the length of the wire at the balance point.

Let the resistances of the heater and bulb be \(R_H\) and \(R_B\), respectively.
The power consumed by the heater and bulb in series can be calculated using: \[ P_H = \frac{V^2}{R_H}, \quad P_B = \frac{V^2}{R_B} \]
The total resistance of the series combination is \(R_{total} = R_H + R_B\). The total power supplied is: \[ P_{total} = \frac{V^2}{R_{total}} \]
By substituting the values and solving for individual powers, we get: \[ P_B = 124.8 \, W, P_H = 33.25 \, W \] Quick Tip: For series combinations, the total power is divided between the components according to their resistances.

For the power delivered to the load resistor in each case, we use the power formula: \[ P = \frac{V^2}{R_{total}} \]
If the power is the same for both resistors, we equate the powers for both configurations: \[ \frac{V^2}{R_1 + r} = \frac{V^2}{R_2 + r} \]
Solving this equation gives the internal resistance \(r = \frac{R_1 + R_2}{2}\).
Quick Tip: In power calculation problems, use the equivalent resistance formula for the specific configuration and equate powers to solve for unknowns.

When a bar of soft iron is placed in a magnetic field parallel to it, the magnetic field lines become denser inside the iron and bend to follow the shape of the bar. This results in a more concentrated field with curved lines of force, which are represented in option (2).
Quick Tip: When a ferromagnetic material like soft iron is placed in a magnetic field, it tends to concentrate the magnetic field lines inside it, bending the lines in the process.

To protect the galvanometer from large currents, a high resistance wire is connected in series to limit the amount of current passing through the galvanometer. This ensures that the current is not large enough to cause damage.
Quick Tip: Always use a high resistance wire in series when protecting delicate instruments like galvanometers from high current.

A stationary charge (zero speed) generates only an electric field, not a magnetic field. Moving charges or oscillating charges create both electric and magnetic fields, but a static charge only creates an electric field.
Quick Tip: Remember, only moving charges create magnetic fields. A static charge produces only an electric field.

For a long straight conducting wire carrying a current, the magnetic field around the wire is constant at every point at a fixed distance from the wire. It does not depend on the distance from the wire but is a function of the current and the geometry of the wire. The field lines are circular and centered on the wire.
Quick Tip: The magnetic field around a current-carrying wire is constant at a given distance and follows a circular pattern.

The coercivity is the magnetic field strength required to demagnetize the magnet. To demagnetize the magnet, the solenoid must generate a magnetic field with strength equal to the coercivity. The formula for the magnetic field inside a solenoid is: \[ B = \mu_0 \frac{N}{L} I \]
Where:
- \(B\) is the magnetic field strength,
- \(\mu_0 = 4\pi \times 10^{-7} \, T m/A\) is the permeability of free space,
- \(N = 112\) is the number of turns,
- \(L = 1.6 \, m\) is the length of the solenoid, and
- \(I\) is the current.

We set the magnetic field \(B\) equal to the coercivity: \[ B = 140 \, A/m \]
Solving for the current \(I\): \[ I = \frac{140 \times 1.6}{4\pi \times 10^{-7} \times 112} \approx 2.25 \, A \] Quick Tip: To demagnetize a bar magnet using a solenoid, the magnetic field strength inside the solenoid must match the coercivity of the magnet.

The given circuit involves two current-carrying conductors forming a network with a point \(P\) located near them. The magnetic field at \(P\) due to each segment of the circuit can be calculated using Ampère's law and the Biot-Savart law. The total magnetic field at point \(P\) is a summation of the contributions from each segment. The correct expression for the magnetic field at \(P\) is given by: \[ |B| = \frac{\mu_0 I}{4\pi R} \left(\frac{3}{2} \pi \right) \] Quick Tip: For calculating magnetic fields from current-carrying conductors, use the Biot-Savart law and consider the contributions from each section of the wire.

Eddy currents are circulating currents induced within conductors when they are exposed to changing magnetic fields. They are used in magnetic brakes, speedometers, and induction furnaces, but transformers are not typically based on eddy currents. Transformers operate on the principle of electromagnetic induction, not on the generation of eddy currents.
Quick Tip: Eddy currents are beneficial in applications like magnetic brakes and induction furnaces, but unwanted in transformers, as they lead to energy loss.

The relationship between the rms voltage (\(V_{rms}\)) and the peak voltage (\(V_{peak}\)) in an a.c. circuit is given by: \[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \]
Given: \[ V_{rms} = 100\sqrt{2} \, V \]
So, \[ V_{peak} = V_{rms} \times \sqrt{2} = 100\sqrt{2} \times \sqrt{2} = 200 \, V \] Quick Tip: The peak voltage is \(\sqrt{2}\) times the rms voltage in an a.c. circuit.

When a conducting loop is moved through a magnetic field, the induced emf and the resulting current will generate forces on the segments of the wire. The forces are given by: \[ F = I L B \]
Where \(I\) is the induced current, \(L\) is the length of the wire, and \(B\) is the magnetic field strength.

- \(F_A\) acts on the side entering the magnetic field, with the least magnetic interaction.
- \(F_B\) and \(F_D\) are the forces acting on the loop's sides where the magnetic field is strongest, as they are directly aligned with the field.
- \(F_C\) is the force acting on the side of the wire where the magnetic field is cutting through the loop the most efficiently.

Therefore, the ranking is: \[ F_C > F_B = F_D > F_A \] Quick Tip: When considering the forces on a conducting loop in a magnetic field, the force is strongest where the field is most aligned and the length of the wire interacts with the magnetic flux.

For DC, the inductors act as short circuits (because their reactance is zero) and the capacitors act as open circuits (because their reactance is infinite). Therefore, the impedance for a DC source is simply \( Z = R_1 \).

For high-frequency AC, the inductors act as high-impedance elements, and the capacitors act as low-impedance elements. Therefore, the impedance for a high-frequency AC source is \( Z = R_2 + R_3 \), as the inductors contribute effectively to the impedance while the capacitors' reactance approaches zero.
Quick Tip: For DC, capacitors behave like open circuits, and inductors behave like short circuits. For high-frequency AC, capacitors behave like short circuits, and inductors behave like open circuits.

In AC circuits, the impedance \(Z\) is calculated depending on whether the components are resistors, inductors, or capacitors:
- For a circuit containing a resistor and an inductor in series, the impedance is: \[ Z = \sqrt{R^2 + X_L^2} \]
- For a circuit containing a resistor and a capacitor in series, the impedance is: \[ Z = \sqrt{R^2 + X_C^2} \]
- For a circuit containing only an inductor and a capacitor in series, the impedance is: \[ Z = |X_L - X_C| \]
Thus, the correct answer is A, B, and C only.
Quick Tip: Impedance is calculated by considering the resistive, inductive, and capacitive components of the circuit, following the given formulas based on the arrangement.

For the DC case, the impedance is simply the resistance \(R\), and the current is given by: \[ I_{dc} = \frac{V}{R} \quad \Rightarrow \quad R = \frac{V}{I_{dc}} = \frac{100}{1} = 100 \, \Omega \]

For the AC case, the total impedance \(Z\) is the sum of the resistance \(R\) and the inductive reactance \(X_L\): \[ I_{ac} = \frac{V}{Z} \quad \Rightarrow \quad Z = \frac{V}{I_{ac}} = \frac{100}{0.5} = 200 \, \Omega \]

The impedance \(Z\) for an inductive circuit is given by: \[ Z = \sqrt{R^2 + X_L^2} \]
Substitute the values: \[ 200 = \sqrt{100^2 + X_L^2} \quad \Rightarrow \quad X_L = \sqrt{200^2 - 100^2} = 173.2 \, \Omega \]

The inductive reactance \(X_L\) is related to the inductance \(L\) by: \[ X_L = 2 \pi f L \]
Substitute the frequency \(f = 50 \, Hz\): \[ 173.2 = 2 \pi \times 50 \times L \quad \Rightarrow \quad L = \frac{173.2}{2 \pi \times 50} \approx 0.55 \, H \] Quick Tip: For a solenoid in an AC circuit, the impedance is the sum of the resistance and the inductive reactance, and the inductance can be calculated from the reactance.

The electromagnetic spectrum is divided into different regions based on frequency ranges. Let's match the entries:

- X-rays (A) have frequencies in the range of \( 1 \times 10^{16} \, Hz \) to \( 3 \times 10^{21} \, Hz \), matching with List II entry I.
- Microwaves (B) have frequencies in the range of \( 1 \times 10^{9} \, Hz \) to \( 3 \times 10^{11} \, Hz \), matching with List II entry II.
- Radiowaves (C) have frequencies in the range of \( 5 \times 10^{5} \, Hz \) to \( 5 \times 10^{9} \, Hz \), matching with List II entry IV.
- Gamma rays (D) have frequencies in the range of \( 1 \times 10^{18} \, Hz \) to \( 5 \times 10^{22} \, Hz \), matching with List II entry III.

Thus, the correct answer is: \[ A-I, B-II, C-IV, D-III \] Quick Tip: Different regions of the electromagnetic spectrum are characterized by their frequency ranges. Always match the given ranges to their respective regions for accurate identification.

For an electromagnetic wave traveling in vacuum, the electric field (\(E\)) and magnetic field (\(B\)) are perpendicular to each other and to the direction of wave propagation. The relationship between the electric and magnetic fields is given by the equation: \[ \frac{E_0}{B_0} = \frac{\omega}{k} \]
This relationship holds true for an electromagnetic wave traveling through vacuum.
Quick Tip: For any electromagnetic wave traveling through vacuum, the ratio of the electric field amplitude to the magnetic field amplitude is equal to the ratio of angular frequency to the wave number.

Total internal reflection occurs when light travels from a denser medium to a less dense medium and the angle of incidence is greater than the critical angle. The light undergoes total internal reflection when it travels from a medium with a higher refractive index to one with a lower refractive index.

- In the case of air to water, the light travels from air (refractive index \(n \approx 1\)) to water (refractive index \(n \approx 1.33\)). Total internal reflection can occur for certain angles of incidence.
- For other options, the refractive index of the second medium is either equal or smaller, preventing total internal reflection.
Quick Tip: Total internal reflection occurs when light moves from a denser medium to a less dense one and when the angle of incidence exceeds the critical angle.

The first maximum in a single-slit diffraction pattern occurs when the angle \(\theta\) satisfies the condition: \[ a \sin \theta = m\lambda \quad (m = 1) \]
For the first maximum, this becomes: \[ e \sin \left(\frac{\pi}{3}\right) = \lambda \]
Using \(\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), we get: \[ e \times \frac{\sqrt{3}}{2} = \lambda \quad \Rightarrow \quad e = \frac{\lambda}{\sqrt{3}} \] Quick Tip: In single-slit diffraction, the angle of the first minimum and maximum are related to the slit width and wavelength.

The fringe width in Young's double slit experiment is given by: \[ w = \frac{\lambda D}{d} \]
Where:
- \(\lambda\) is the wavelength of light,
- \(D\) is the distance between the screen and the slits,
- \(d\) is the separation between the slits.

When the wavelength changes and the slit separation is tripled, the new fringe width becomes: \[ w_{new} = \frac{720 \times D}{3d} = 3 \times \frac{640 \times D}{d} = 3 \times 0.8 \, mm = 2.4 \, mm \] Quick Tip: The fringe width in Young's double slit experiment is directly proportional to the wavelength and inversely proportional to the slit separation.

The angular size of the image formed by the telescope is given by: \[ \theta_{image} = \frac{\theta_{object} \times f_{eyepiece}}{f_{objective}} \]
Where:
- \(\theta_{object} = \frac{1}{2^\circ} \),
- \(f_{objective} = 50 \, cm\),
- \(f_{eyepiece} = 2 \, cm\).

Substituting the values, we get: \[ \theta_{image} = \frac{\frac{1}{2^\circ} \times 2}{50} = 1.27^\circ \] Quick Tip: In an astronomical telescope, the angular size of the image is determined by the focal lengths of the objective and eyepiece.

For a spherical lens, the focal length \( f \) is given by the lens formula: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For the air lens: \[ \frac{1}{f_1} = (\mu_{glass} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{10} - \frac{1}{10} \right) \] \[ \frac{1}{f_1} = 0.5 \times \left( \frac{2}{10} \right) = \frac{0.1}{10} \quad \Rightarrow \quad f_1 = 15 \, cm \]

When a liquid with refractive index \( \mu_2 \) is filled, the effective refractive index becomes: \[ \frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For \( \mu_2 = 1.5 \), the focal length is: \[ \frac{1}{f_2} = (1.5 - 1) \left( \frac{2}{10} \right) = \frac{1}{5} \quad \Rightarrow \quad f_2 = 30 \, cm \] Quick Tip: The focal length of a lens is inversely proportional to the radii of curvature and the refractive index difference. When changing the refractive index, the focal length changes significantly.

The magnification produced by a concave mirror is given by the relation: \[ M = \frac{v}{u} = 1 + \frac{f}{u} \]
Where \( v \) is the image distance, \( u \) is the object distance, and \( f \) is the focal length of the mirror.
For magnification \( M = +1.5 \), the object should be placed between \( F \) and \( 2F \) for a concave mirror.
Quick Tip: For a concave mirror, when the object is placed between \( F \) and \( 2F \), the image formed is real, inverted, and magnified, producing a positive magnification.

A distant light source produces approximately plane wavefronts because the rays coming from the source are nearly parallel. For practical purposes, the wavefront can be considered a plane as the source is at an effectively infinite distance.
Quick Tip: When light travels from a distant source, the wavefronts can be approximated as plane waves due to the parallel nature of the rays.

The deBroglie wavelength is given by the equation: \[ \lambda = \frac{h}{\sqrt{2mE}} \]
Since the proton, deuteron, and electron have the same energy, their deBroglie wavelengths depend on their masses. The electron, having the least mass, will have the longest deBroglie wavelength, followed by the proton and deuteron. The \(\alpha\)-particle, having the highest mass, will have the shortest wavelength. Thus, the correct order of wavelengths is \( \lambda_{\infty} < \lambda_d < \lambda_p = \lambda_e \).
Quick Tip: For particles with the same energy, the deBroglie wavelength decreases as the particle's mass increases.

According to the photoelectric effect, the stopping potential \( V_0 \) is related to the intensity of the incident radiation by the equation: \[ V_0 = \frac{h}{e} \left( \nu - \nu_0 \right) \]
At constant frequency, increasing intensity does not increase the energy per photon but increases the number of photons striking the surface, which in turn increases the number of electrons ejected. However, after a certain intensity threshold, increasing intensity does not change the stopping potential significantly, causing the curve to level off. This is why the correct curve is an initial sharp increase followed by leveling off.
Quick Tip: For the photoelectric effect, the stopping potential depends on the frequency of light, not its intensity. Intensity affects the number of ejected electrons, but the stopping potential only increases with frequency.

A - The photoelectric current depends on the intensity of the incident radiation because higher intensity means more photons striking the surface, resulting in more electrons being ejected.

B - The stopping potential is indeed related to the maximum kinetic energy of emitted electrons, but it is independent of intensity, depending only on the frequency of the incident radiation.

C - Photoelectric emission happens due to the absorption of a photon by an electron, which is correct.

D - The photoelectric effect adheres to the law of conservation of energy, as the energy from the photons is converted to the kinetic energy of emitted electrons.
Quick Tip: Intensity affects the number of emitted electrons in the photoelectric effect, but not their kinetic energy, which is determined by the frequency of the incident light.

The energy released during a fusion process is given by the equation: \[ E_{released} = \left( m_1 + m_2 - M \right) c^2 \]
This formula is derived from the mass defect in the fusion process, where the combined mass of the products is less than the sum of the masses of the reactants. The mass difference is converted into energy according to Einstein’s equation \( E = mc^2 \).
Quick Tip: In nuclear fusion, the energy released is directly proportional to the mass defect between the combined masses of the reactants and the product nucleus.

Nuclear force is the fundamental force that binds nucleons (protons and neutrons) together in an atomic nucleus. It is a strong force, effective only over very short distances (on the order of femtometers), and is independent of the charges of the particles involved (i.e., charge independent).
Quick Tip: The nuclear force is short-range, meaning it only operates within the nucleus, and it does not depend on the electric charge of the particles involved.

The energy levels of the hydrogen atom are depicted as follows:
- The energy difference between level D and A (13.6 eV to 0.85 eV) corresponds to the longest wavelength \( \lambda_A \).
- The energy difference between levels C and A (3.4 eV to 0.85 eV) corresponds to \( \lambda_B \).
- The energy difference between B and A (1.5 eV to 0.85 eV) corresponds to \( \lambda_C \).
- The smallest energy difference corresponds to \( \lambda_D \).

Thus, the order of wavelengths from longest to shortest is: \( \lambda_A > \lambda_B > \lambda_C > \lambda_D \).
Quick Tip: In atomic transitions, the larger the energy difference, the shorter the wavelength of the emitted radiation.

- A: \( ^{222}_{86}Rn \to ^{218}_{84}Po \) undergoes alpha decay, emitting a \( \alpha \)-particle, so the correct match is \( A-II \).
- B: \( ^{214}_{83}Bi \to ^{214}_{84}Po \) undergoes beta decay, emitting a \( \beta \)-particle, so the correct match is \( B-I \).
- C: \( ^{234}_{90}Th \to ^{234}_{92}U \) undergoes alpha decay, emitting a \( \alpha \)-particle, so the correct match is \( C-IV \).
- D: \( ^{22}_{11}Na \to ^{22}_{10}Na \) undergoes beta decay, emitting a \( \beta \)-particle, so the correct match is \( D-III \).

Thus, the correct matching is: A-II, B-I, C-IV, D-III.
Quick Tip: Alpha decay results in the emission of a \( \alpha \)-particle, while beta decay results in the emission of a \( \beta \)-particle.

- Statement 1: Pure silicon doped with a trivalent impurity (such as boron) results in a p-type semiconductor, which has more holes as majority carriers. This statement is correct.

- Statement 2: In an n-type semiconductor, the majority carriers are electrons, not holes. This statement is also correct.

- Statement 3: In a p-type semiconductor, the majority carriers are holes, and the minority carriers are electrons. Therefore, this statement is correct.

- Statement 4: The resistivity of an intrinsic semiconductor decreases with an increase in temperature, which is incorrect. In intrinsic semiconductors, resistivity decreases as temperature increases due to an increase in charge carriers, making this statement the incorrect one.
Quick Tip: For semiconductors, remember that resistivity decreases with temperature due to the increased movement of charge carriers in the material.

- The circuit consists of two AND gates. The first AND gate takes inputs \(A\) and \(B\), while the output of the first AND gate is passed to a NOT gate. The NOT gate inverts the output of the AND gate.

- Therefore, the operation performed by this circuit is a NAND operation (the negation of an AND operation). The result of the AND operation is negated by the NOT gate, giving the NAND result.
Quick Tip: Remember that a NAND gate is essentially an AND gate followed by a NOT gate, i.e., the output is the negation of an AND operation.

The circuit shown consists of resistors and a diode. To find the potential difference between points A and B, we first analyze the circuit behavior. Since the diode is forward biased (due to the potential difference from the 24V battery), it will conduct. The resistors \( R_1 \) and \( R_2 \) are in series, with a total resistance of: \[ R_{total} = R_1 + R_2 = 8\,k\Omega + 8\,k\Omega = 16\,k\Omega \]
The total current \( I \) through the circuit can be found using Ohm’s law: \[ I = \frac{V}{R_{total}} = \frac{24\,V}{16\,k\Omega} = 1.5\,mA \]
Now, the potential drop across \( R_2 \) (which is the same as \( R_1 \)) is: \[ V_{R_2} = I \times R_2 = 1.5\,mA \times 8\,k\Omega = 12\,V \]
The remaining potential drop across \( R_1 \) (which corresponds to the potential difference between A and B) is: \[ V_{AB} = V - V_{R_2} = 24\,V - 12\,V = 4\,V \]

Thus, the potential difference between points A and B is 4V.
Quick Tip: To calculate potential differences in circuits with resistors and diodes, remember to use Ohm’s law to find the current and then determine the voltage drops across resistors.

- Zener Diode (A): Zener diodes are typically used as voltage regulators due to their ability to maintain a steady voltage across them when reverse biased. Therefore, the correct match is A-I.

- LED (B): LEDs (Light Emitting Diodes) are primarily used to emit light when current flows through them, and they are commonly used to detect optical signals. Hence, the correct match is B-II.

- Rectifier (C): A rectifier is used to convert AC to DC, a basic operation in power supplies and electronic devices. Thus, C-IV is the correct match.

- Photo diode (D): A photo diode is used to detect optical signals and convert them to electrical signals. Thus, D-III is the correct match.


Thus, the correct matching of List I with List II is: \[ A-I, B-II, C-III, D-IV \] Quick Tip: When matching components to their uses, think about their main functions in circuits. For example, LEDs are for light, Zener diodes for voltage regulation, rectifiers for AC to DC conversion, and photo diodes for optical signal detection.

We use the formula for the transmission distance: \[ d = \sqrt{2 \cdot h \cdot R} \]
Substitute the values: \[ d = \sqrt{2 \cdot 180 \cdot 6.4 \times 10^6}
d = \sqrt{2.304 \times 10^9}
d = 48 \times 10^3 \, m = 48 \, km \]
Hence, the transmission distance is 48 km.
Quick Tip: The transmission distance depends on the height of the tower and the radius of the Earth. Use the formula \(d = \sqrt{2 \cdot h \cdot R}\) for such calculations.

Matching each item from List I with List II:
- A. Modulation: This refers to the retrieval of information from the carrier wave at the receiver, so it matches with I.

- B. Baseband signals: These refer to the frequency range over which an equipment operates, so it matches with III.

- C. Demodulation: This is the process of superimposition of a signal on a high-frequency wave, so it matches with II.

- D. Bandwidth: This represents the band of frequencies representing the original signal, so it matches with I.
Quick Tip: In communication systems, modulation, demodulation, baseband signals, and bandwidth are key concepts used to transfer and retrieve information. Matching them with their definitions helps in understanding how signals are processed.

Let the distance between the charge \(q\) and the point where the potential is zero be \(x\). The distance between the point and the charge \(-3q\) will be \(12 - x\).
The potential at a point due to a point charge is given by the formula: \[ V = \frac{kQ}{r} \]
where \(k\) is the electrostatic constant, \(Q\) is the charge, and \(r\) is the distance from the charge.
For the potential to be zero, the potentials due to both charges must be equal in magnitude and opposite in sign. Hence, we have: \[ \frac{kq}{x} = \frac{k(3q)}{12 - x} \]
Simplifying the equation: \[ \frac{1}{x} = \frac{3}{12 - x} \]
Cross multiplying: \[ 12 - x = 3x \] \[ 12 = 4x \] \[ x = 3 \, cm \]
Thus, the point at which the potential is zero is 4 cm from the charge \(q\).
Quick Tip: When dealing with the potential of point charges, remember that the potential is a scalar quantity. Therefore, to find the point of zero potential, we set the magnitudes of the potentials equal and solve for the distance.

The mobility of electrons depends on the temperature and the structure of the conductor. It is the ability of an electron to move in a material under the influence of an electric field. Mobility decreases with an increase in temperature, because higher temperatures cause more collisions between electrons and atoms, thus decreasing the electron's velocity. However, the mobility does not depend on the potential difference, as it is related to the internal properties of the conductor. Therefore, the correct options are: \[ C: does not depend on potential difference, \quad E: independent to the temperature of conductor \] Quick Tip: Remember, the mobility of electrons is affected by the temperature of the material and its crystal lattice. It is not directly affected by the applied potential difference.

The displacement current is a quantity that appears in Maxwell's equations. It is proportional to the rate of change of the electric flux \(\phi_E\). The general form of the displacement current is: \[ I_d = \mu_0 \epsilon_0 \frac{d\phi_E}{dt} \]
where \(\mu_0\) is the permeability of free space, \(\epsilon_0\) is the permittivity of free space, and \(\frac{d\phi_E}{dt}\) is the time rate of change of the electric flux.

Thus, the correct option is: \[ \mu_0 \epsilon_0 \frac{d\phi_E}{dt} \] Quick Tip: Displacement current is crucial in Maxwell's equations to ensure that the equation is consistent in both electric and magnetic fields, particularly when there is a time-varying electric field.

The given graph shows the relationship between magnification \(m\) and image distance \(v\). From the graph, we observe that the magnification \(m\) is related to \(v\) in a manner that can be interpreted in terms of the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Where \(f\) is the focal length of the lens, \(v\) is the image distance, and \(u\) is the object distance. Using the given relationship between \(m\) and \(v\) from the graph, and analyzing the geometry and algebra behind the graph, we can deduce that the focal length of the lens is given by:
\[ f = \frac{b^2}{ac} \]

Thus, the correct answer is \(\frac{b^2}{ac}\). Quick Tip: To solve questions involving lenses and magnification, remember to use the lens formula and the relationships between magnification and object/image distances.

Based on Bohr's atomic model:

- A. Radius of electron orbit: The radius of an electron’s orbit is directly proportional to \(n^2\). Therefore, the match is A-I.

- B. Angular momentum of electron: The angular momentum is directly proportional to \(n\). Therefore, the match is B-II.

- C. Velocity of electron: The velocity of an electron is inversely proportional to \(n\). Therefore, the match is C-III.

- D. Energy of electron: The energy is inversely proportional to \(n^2\). Therefore, the match is D-IV.
Quick Tip: When studying Bohr’s atomic model, remember that quantities related to the electron’s orbit, such as radius and angular momentum, have specific relationships with the principal quantum number \(n\). The energy of the electron decreases as \(n\) increases.