CUET PG 2026 Geophysics Question Paper with Solutions

Concept:

When two bodies move under the same acceleration, their relative acceleration becomes zero.



Step 1: Let the accelerations of two projectiles be same.

Suppose two projectiles have accelerations: \[ \vec{a}_1=\vec{g} \]
and \[ \vec{a}_2=\vec{g} \]



Step 2: Find relative acceleration.

The acceleration of projectile 1 with respect to projectile 2 is: \[ \vec{a}_{12}=\vec{a}_1-\vec{a}_2 \]
\[ \vec{a}_{12}=\vec{g}-\vec{g} \]
\[ \vec{a}_{12}=0 \]



Step 3: Interpret the motion.

If relative acceleration is zero, then the relative velocity remains constant.

So, the relative motion is uniform motion in a straight line.
\[ \therefore Motion of one projectile with respect to another is a straight line. \]
\[ \therefore Correct Answer is (C) \] Quick Tip: If two bodies have the same acceleration, their relative acceleration is zero and their relative motion is along a straight line.

Concept:

In a parallel resonant circuit, the inductor and capacitor branch currents are reactive in nature.



Step 1: Current in capacitor branch.

In a capacitor, current leads the applied emf by: \[ \frac{\pi}{2} \]



Step 2: Current in inductor branch.

In an inductor, current lags the applied emf by: \[ \frac{\pi}{2} \]



Step 3: Apply resonance condition.

At parallel resonance, inductive and capacitive reactive currents are equal and opposite in phase.

Thus, the branch currents may lead or lag the emf by: \[ \frac{\pi}{2} \]
\[ \therefore Correct Answer is (D) \] Quick Tip: In reactive branches, capacitive current leads by \(90^\circ\) and inductive current lags by \(90^\circ\).

Concept:

Beat frequency is equal to the difference between two frequencies.
\[ n=|f_1-f_2| \]

Also, \[ f=\frac{v}{\lambda} \]



Step 1: Write the given wavelengths. \[ \lambda_1=1\ m \] \[ \lambda_2=1.01\ m \]

Beat frequency: \[ n=10\ s^{-1} \]



Step 2: Write frequencies. \[ f_1=\frac{v}{1} \] \[ f_2=\frac{v}{1.01} \]



Step 3: Apply beat frequency formula. \[ 10=\frac{v}{1}-\frac{v}{1.01} \]
\[ 10=v\left(1-\frac{1}{1.01}\right) \]
\[ 10=v\left(\frac{1.01-1}{1.01}\right) \]
\[ 10=v\left(\frac{0.01}{1.01}\right) \]
\[ v=10\times \frac{1.01}{0.01} \]
\[ v=1010\ m/s \]
\[ \therefore Correct Answer is (C) \] Quick Tip: For beats, use \(n=|f_1-f_2|\) and replace frequency by \(f=\frac{v}{\lambda}\).

Concept:

The thrust force of a rocket is given by: \[ F=\frac{dm}{dt}v \]

where, \[ \frac{dm}{dt}=mass ejected per second \]
and \[ v=exhaust velocity \]



Step 1: Convert mass consumption per minute into per second. \[ \frac{dm}{dt}=\frac{3000}{60} \]
\[ \frac{dm}{dt}=50\ kg/s \]



Step 2: Write exhaust velocity. \[ v=5\times10^3\ m/s \]



Step 3: Calculate thrust force. \[ F=50\times 5\times10^3 \]
\[ F=250\times10^3 \]
\[ F=2.5\times10^5\ N \]
\[ F=25\times10^4\ N \]
\[ \therefore Correct Answer is (D) \] Quick Tip: Rocket thrust is equal to mass ejection rate multiplied by exhaust velocity.

Concept:

The principle of relativity states that the laws of physics are the same in all inertial frames of reference.



Step 1: Check Assertion.

The assertion says that laws of physics have the same form in all inertial frames moving uniformly relative to one another.

This is the principle of relativity.
\[ A is correct \]



Step 2: Check Reason.

There is no absolute fixed frame of reference. Motion is always described relative to a chosen frame.
\[ R is correct \]



Step 3: Check explanation.

Because no inertial frame is absolutely preferred, laws of physics must have the same form in all inertial frames.
\[ R correctly explains A \]
\[ \therefore Correct Answer is (A) \] Quick Tip: In relativity, no inertial frame is absolute; all inertial frames are equivalent for physical laws.

Concept:

When a rocket moves upward with acceleration \(a\), effective weight becomes: \[ W'=m(g+a) \]



Step 1: Calculate relative effective values.

For upward acceleration \(a=ng\): \[ W'=m(g+ng) \]
\[ W'=mg(1+n) \]

Since \(g\) is common, compare: \[ m(1+n) \]



Step 2: Case A. \[ A=60(1+5)=60\times6=360 \]



Step 3: Case B. \[ B=60(1+8)=60\times9=540 \]



Step 4: Case C. \[ C=50(1+6)=50\times7=350 \]



Step 5: Case D. \[ D=60(1+10)=60\times11=660 \]



Step 6: Arrange in decreasing order. \[ D>B>A>C \]
\[ \therefore Correct Answer is (A) \] Quick Tip: For upward accelerated motion, apparent weight increases and is given by \(W'=m(g+a)\).

Concept:

In a series LCR circuit, resonance occurs when: \[ X_L=X_C \]



Step 1: Net reactance at resonance. \[ X=X_L-X_C \]

At resonance: \[ X_L=X_C \]
\[ X=0 \]



Step 2: Total impedance. \[ Z=\sqrt{R^2+(X_L-X_C)^2} \]
\[ Z=\sqrt{R^2+0^2} \]
\[ Z=R \]

So, \[ A is correct \]



Step 3: Phase relation.

At resonance, the circuit behaves like a purely resistive circuit.

Therefore, current and applied emf are in the same phase.
\[ B is correct \]



Step 4: Check remaining statements.

Since current is in phase with emf, it neither leads nor lags by \(\frac{\pi}{2}\) or \(\frac{\pi}{4}\).
\[ C,D,E are incorrect \]



Thus, correct statements are: \[ A,B \]
\[ \therefore Correct Answer is (A) \] Quick Tip: At resonance in a series LCR circuit, \(X_L=X_C\), impedance is minimum and equal to \(R\), and current is in phase with voltage.

Concept:

Galilean transformation is used in classical mechanics to relate observations in two inertial frames moving with constant relative velocity.



Step 1: Check velocity.

Velocity changes from one inertial frame to another.
\[ v'=v-u \]

So, \[ A is not invariant \]



Step 2: Check linear momentum.

Since velocity changes, linear momentum also changes.
\[ p=mv \]

So, \[ B is not invariant \]



Step 3: Check acceleration.

Acceleration remains unchanged under Galilean transformation.
\[ a'=a \]

So, \[ C is invariant \]



Step 4: Check length.

In classical mechanics, length is absolute.
\[ D is invariant \]



Step 5: Check force.

Since: \[ F=ma \]

and acceleration is invariant, force is also invariant.
\[ E is invariant \]



Therefore: \[ C,D,E \]
\[ \therefore Correct Answer is (A) \] Quick Tip: Under Galilean transformation, velocity changes but acceleration, length and force remain invariant.

Concept:

Moment of inertia depends on mass distribution about the axis of rotation.



Step 1: Solid cone about vertical axis.

Moment of inertia of a solid cone about its vertical axis is: \[ I=\frac{3}{10}MR^2 \]

So, \[ A\rightarrow IV \]



Step 2: Solid cylinder about own axis.

Moment of inertia of a solid cylinder about its own axis is: \[ I=\frac{1}{2}MR^2 \]

So, \[ B\rightarrow III \]



Step 3: Circular lamina about diameter.

Moment of inertia of circular lamina about a diameter is: \[ I=\frac{1}{4}MR^2 \]

So, \[ C\rightarrow II \]



Step 4: Annular ring about diameter.

Moment of inertia of annular ring about a diameter is: \[ I=\frac{1}{4}M(R^2+r^2) \]

So, \[ D\rightarrow I \]



Therefore: \[ A-IV,\ B-III,\ C-II,\ D-I \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For moment of inertia, always identify both the body and the axis of rotation carefully.

Concept:

This question matches physical quantities or systems with their related applications or motion.



Step 1: Torque.

Torque is the turning effect of force and is related to rotational motion.
\[ A\rightarrow II \]



Step 2: Force.

Force is directly related to linear motion through Newton's second law.
\[ F=ma \]

So, \[ B\rightarrow III \]



Step 3: Rocket.

A rocket is a variable mass system because it continuously ejects fuel.
\[ C\rightarrow IV \]



Step 4: Parallel LCR circuit.

A parallel LCR circuit at resonance is known as a rejector circuit.
\[ D\rightarrow I \]



Therefore: \[ A-II,\ B-III,\ C-IV,\ D-I \]
\[ \therefore Correct Answer is (A) \] Quick Tip: Parallel resonant LCR circuit is called a rejector circuit because it offers high impedance at resonance.

Concept:

Poynting vector represents the rate of electromagnetic energy flow per unit area.



Step 1: Formula of Poynting vector.

The Poynting vector is: \[ \vec{S}=\frac{1}{\mu_0}(\vec{E}\times\vec{B}) \]



Step 2: Direction of energy flow.

The direction of \(\vec{S}\) is perpendicular to both \(\vec{E}\) and \(\vec{B}\), not necessarily along the electric field.



Step 3: Meaning.

Poynting theorem represents conservation of electromagnetic energy and is often treated as the work-energy theorem of electrodynamics.
\[ \therefore Correct Answer is (D) \] Quick Tip: Poynting vector is \(\vec{S}=\frac{1}{\mu_0}(\vec{E}\times\vec{B})\), and it gives the direction of electromagnetic energy flow.

Concept:

Displacement current density was introduced by Maxwell to modify Ampere's law for time-varying electric fields.



Step 1: Displacement current density.

The displacement current density is: \[ \vec{J}_d=\epsilon_0\frac{\partial \vec{E}}{\partial t} \]



Step 2: Check other options.
\[ \frac{1}{2}\epsilon_0E^2 \]
is electric energy density.
\[ \frac{1}{2}\frac{B^2}{\mu_0} \]
is magnetic energy density.
\[ \vec{E}\times\vec{B} \]
is related to electromagnetic energy flow.



Step 3: Final answer.

Thus, the correct form of displacement current density is: \[ \epsilon_0\frac{\partial \vec{E}}{\partial t} \]
\[ \therefore Correct Answer is (D) \] Quick Tip: Displacement current density is produced by time-varying electric field and is given by \(\vec{J}_d=\epsilon_0\frac{\partial\vec{E}}{\partial t}\).

Concept:

A transistor can be connected in common base, common emitter or common collector configuration.



Step 1: Common emitter configuration.

In common emitter configuration, input is applied between base and emitter and output is taken between collector and emitter.



Step 2: Gain properties.

Common emitter configuration gives:
\[ high voltage gain \]

and
\[ high current gain \]



Step 3: Final conclusion.

Therefore, common emitter transistor configuration is widely used for amplification.
\[ \therefore Correct Answer is (A) \] Quick Tip: Common emitter configuration is most widely used because it gives both high voltage gain and high current gain.

Concept:

Negative feedback means a part of the output is fed back to the input in opposite phase.



Step 1: Effect on frequency response.

Negative feedback improves the frequency response of an amplifier.
\[ Frequency response improves \]



Step 2: Effect on distortion and nonlinearity.

Negative feedback reduces distortion and improves linearity of operation.
\[ Nonlinear distortion decreases \]



Step 3: Effect on noise and gain.

Negative feedback generally reduces noise and stabilizes gain, but it does not increase transfer gain.



Step 4: Final answer.

Thus, the correct statement is that negative feedback improves frequency response and operation of the amplifier.
\[ \therefore Correct Answer is (A) \] Quick Tip: Negative feedback improves stability, bandwidth and linearity, but reduces gain.

Concept:

For current flowing radially between two coaxial cylinders, resistance of the conducting material is: \[ R=\frac{\ln\left(\frac{r_2}{r_1}\right)}{2\pi kl} \]



Step 1: Use Ohm's law. \[ I=\frac{V}{R} \]



Step 2: Substitute resistance. \[ I=\frac{V}{\frac{\ln\left(\frac{r_2}{r_1}\right)}{2\pi kl}} \]
\[ I=\frac{2\pi klV}{\ln\left(\frac{r_2}{r_1}\right)} \]



Step 3: Final answer. \[ I=\frac{2\pi kl}{\log_e\left(\frac{r_2}{r_1}\right)}V \]
\[ \therefore Correct Answer is (A) \] Quick Tip: For radial conduction between coaxial cylinders, resistance contains the logarithmic term \(\ln\left(\frac{r_2}{r_1}\right)\).

Concept:

FETs are voltage-controlled devices and BJTs are current-controlled devices. FETs have very high input resistance.



Step 1: Check Assertion.

FETs are often preferred in low-frequency and high-input-impedance applications because they load the previous stage very little.
\[ A is correct \]



Step 2: Check Reason.

The reason says FETs are more suitable because of high input resistance and high power gain.

High input resistance is correct, but high power gain is not the usual advantage of FETs over BJTs.
\[ R is not correct \]



Step 3: Final conclusion.

Assertion is correct but Reason is not correct.
\[ \therefore Correct Answer is (C) \] Quick Tip: FETs are preferred where high input resistance and low loading are required.

Concept:

In a \(\pi\)-section filter, ripple factor decreases when capacitance, inductance and load resistance increase.



Step 1: Understand proportionality.

For comparison, ripple factor is inversely related to the product: \[ C_1C_2L_1R_L \]

So, \[ Higher C_1C_2L_1R_L \Rightarrow Lower ripple factor \]



Step 2: Compare case A. \[ A\propto 200\times100\times50\times100 \]



Step 3: Compare case B. \[ B\propto 200\times200\times50\times100 \]

This product is greater than A, so ripple in B is less than A.
\[ A>B \]



Step 4: Compare case C. \[ C\propto 200\times200\times100\times100 \]

This product is greater than B, so ripple in C is less than B.
\[ B>C \]



Step 5: Compare case D. \[ D\propto 100\times400\times100\times200 \]

This product is greatest among the given cases, so ripple in D is least.
\[ C>D \]



Thus, decreasing order of ripple factor is: \[ A>B>C>D \]
\[ \therefore Correct Answer is (A) \] Quick Tip: In filter circuits, larger capacitance, inductance and load resistance generally reduce ripple factor.

Concept:

Maxwell's equations describe electric and magnetic fields and their relation with charge and current.



Step 1: Check statement A.

Gauss law for magnetism is: \[ \vec{\nabla}\cdot\vec{B}=0 \]

This means magnetic monopoles do not exist.
\[ A is correct \]



Step 2: Check statement B.

Gauss law for electricity is: \[ \vec{\nabla}\cdot\vec{E}=\frac{\rho}{\epsilon_0} \]
\[ B is correct \]



Step 3: Check statement C.

Faraday's law of electromagnetic induction is: \[ \vec{\nabla}\times\vec{E}=-\frac{\partial \vec{B}}{\partial t} \]
\[ C is correct \]



Step 4: Check statement D.

The expression: \[ U=\frac{1}{2}\left(\epsilon_0E^2+\frac{B^2}{\mu_0}\right) \]
is electromagnetic energy density, not Maxwell's equation.
\[ D is incorrect \]



Thus, correct Maxwell equations are: \[ A,B,C \]
\[ \therefore Correct Answer is (A) \] Quick Tip: Maxwell's equations include Gauss law, Gauss law for magnetism, Faraday law and Ampere-Maxwell law.

Concept:

Different semiconductor devices and breakdown mechanisms are associated with different physical conditions and applications.



Step 1: Avalanche Breakdown.

Avalanche breakdown occurs mainly in lightly doped regions because the depletion layer is wide.
\[ A\rightarrow IV \]



Step 2: Zener Breakdown.

Zener breakdown is dominant in heavily doped regions because the depletion layer is very thin.
\[ B\rightarrow III \]



Step 3: Point Contact diode.

Point contact diode is used in high-frequency and microwave applications.
\[ C\rightarrow II \]



Step 4: LED Diode.

LEDs can be made using materials like GaAsP.
\[ D\rightarrow I \]



Therefore: \[ A-IV,\ B-III,\ C-II,\ D-I \]
\[ \therefore Correct Answer is (A) \] Quick Tip: Zener breakdown occurs in heavily doped junctions, while avalanche breakdown occurs in lightly doped junctions.

Concept:

In a uniaxial crystal, light generally splits into ordinary ray and extraordinary ray due to double refraction.



Step 1: Understand ordinary and extraordinary rays.

The ordinary ray and extraordinary ray usually travel with different velocities in anisotropic crystals.



Step 2: Identify special direction.

In a uniaxial crystal, there is one special direction along which ordinary and extraordinary rays travel with the same speed.

This direction is called the optic axis.
\[ o-ray speed=e-ray speed along optic axis \]



Step 3: Final answer.

Thus, the required direction is:
\[ Optic axis \]
\[ \therefore Correct Answer is (B) \] Quick Tip: In a uniaxial crystal, along the optic axis, ordinary and extraordinary rays travel with the same speed.

Concept:

Coherence length is the length over which a wave train maintains a definite phase relationship.



Step 1: Understand the wave train.

If one complete oscillation has wavelength: \[ \lambda \]

and the wave train contains: \[ N \]
complete oscillations, then total length of the wave train is obtained by multiplying the number of oscillations by wavelength.



Step 2: Apply the formula. \[ Coherence length=N\times \lambda \]
\[ L_c=N\lambda \]



Step 3: Final answer. \[ \therefore Correct Answer is (B) \] Quick Tip: If a wave train contains \(N\) waves of wavelength \(\lambda\), then its coherence length is \(N\lambda\).

Concept:

In Michelson interferometer, when the mirror is moved by a distance \(d\), optical path difference changes by \(2d\).
\[ N\lambda=2d \]



Step 1: Write the given values. \[ N=100 \] \[ d=29.48\ \mu m \]



Step 2: Apply Michelson formula. \[ N\lambda=2d \]
\[ 100\lambda=2\times 29.48\ \mu m \]
\[ 100\lambda=58.96\ \mu m \]
\[ \lambda=0.5896\ \mu m \]



Step 3: Convert into Angstrom. \[ 1\ \mu m=10^4\ \AA \]
\[ 0.5896\ \mu m=0.5896\times10^4\ \AA \]
\[ \lambda=5896\ \AA \]
\[ \therefore Correct Answer is (A) \] Quick Tip: In Michelson interferometer, mirror displacement \(d\) produces path difference \(2d\), so \(N\lambda=2d\).

Concept:

Newton's rings are interference fringes produced due to the thin air film formed between a plano-convex lens and a plane glass plate.



Step 1: Formation of air film.

When a plano-convex lens is placed on a plane glass plate, a thin air film of varying thickness is formed.



Step 2: Illumination by monochromatic light.

When monochromatic light falls on this arrangement, reflected rays from the top and bottom surfaces of the air film interfere.



Step 3: Nature of fringes.

The fringes obtained are circular rings, not parabolic fringes.



Step 4: Final answer.

Therefore, the correct statement is:
\[ Newton's rings are formed using a plano-convex lens on a plane glass plate. \]
\[ \therefore Correct Answer is (B) \] Quick Tip: Newton's rings are circular interference fringes produced by a thin air film.

Concept:

Fraunhofer diffraction is far-field diffraction. In this type of diffraction, incident and diffracted rays are treated as parallel.



Step 1: Check Assertion.

Fraunhofer diffraction is produced by interference of secondary wavelets in parallel directions.

Also, experimentally, it can be observed using two convex lenses.
\[ A is correct \]



Step 2: Check Reason.

The first convex lens converts light from the source into a parallel beam.

The second convex lens focuses the diffracted light on the screen.
\[ R is correct \]



Step 3: Check explanation.

The reason directly explains why two convex lenses are used for observing Fraunhofer diffraction.
\[ R correctly explains A \]
\[ \therefore Correct Answer is (A) \] Quick Tip: Fraunhofer diffraction is far-field diffraction and is usually observed using lenses to make and focus parallel rays.

Concept:

Newton's rings experiment in reflected light uses a microscope, an inclined glass plate, a plano-convex lens and a plane glass plate.



Step 1: Identify the topmost instrument.

The microscope is placed at the top to observe the rings.
\[ Top= Microscope \]



Step 2: Identify the reflecting plate.

A glass plate inclined at \(45^\circ\) reflects light towards the plano-convex lens arrangement.
\[ Next=45^\circ glass plate \]



Step 3: Identify the ring forming setup.

The plano-convex lens is placed on the plane glass plate.
\[ Plano-convex lens above plane glass plate \]



Step 4: Final arrangement. \[ Microscope \rightarrow 45^\circ glass plate \rightarrow Plano-convex lens \rightarrow Plane glass plate \]
\[ \therefore Correct Answer is (A) \] Quick Tip: In Newton's rings, the plano-convex lens is placed on a plane glass plate and rings are observed through a microscope.

Concept:

Coherence of light describes a fixed phase relationship between waves. It is mainly of two types: temporal coherence and spatial coherence.



Step 1: Check statement A.

Coherence is divided into temporal coherence and spatial coherence.
\[ A is correct \]



Step 2: Check statement B.

Temporal coherence is related to the spectral width or bandwidth of the source.
\[ B is correct \]



Step 3: Check statement C.

Spatial coherence is related to the finite size of the source.
\[ C is correct \]



Step 4: Check statement D.

Finite bandwidth is connected with temporal coherence, not spatial coherence.
\[ D is incorrect \]



Therefore: \[ A,B,C \]
\[ \therefore Correct Answer is (B) \] Quick Tip: Temporal coherence depends on bandwidth, while spatial coherence depends on source size.

Concept:

Laser action requires population inversion and optical feedback through an optical resonator.



Step 1: Optical pumping.

Optical pumping is used to produce population inversion.
\[ A\rightarrow II \]



Step 2: He-Ne laser.

He-Ne laser is a four-level laser system.
\[ B\rightarrow I \]



Step 3: Ruby laser.

Ruby laser is a three-level laser system.
\[ C\rightarrow III \]



Step 4: Optical resonator.

An optical resonator consists of a pair of optically plane parallel mirrors.
\[ D\rightarrow IV \]



Therefore: \[ A-II,\ B-I,\ C-III,\ D-IV \]
\[ \therefore Correct Answer is (B) \] Quick Tip: Ruby laser is a three-level laser, while He-Ne laser is a four-level laser.

Concept:

A reversible process is an ideal process that takes place infinitely slowly and can be reversed without leaving any change in the system and surroundings.



Step 1: Check friction.

Production of heat by friction is irreversible.
\[ A is incorrect \]



Step 2: Check Joule Thomson effect.

Joule Thomson expansion is a throttling process and is irreversible.
\[ B is incorrect \]



Step 3: Check current through resistance.

Heat production in resistance is Joule heating and is irreversible.
\[ C is incorrect \]



Step 4: Check slow isothermal expansion/compression.

A very slow expansion or compression of a gas at constant temperature can be considered reversible.
\[ D is correct \]
\[ \therefore Correct Answer is (D) \] Quick Tip: Slow quasi-static isothermal expansion or compression is a standard example of a reversible process.

Concept:

For isothermal expansion of an ideal gas, entropy change is: \[ \Delta S=nR\ln\left(\frac{V_2}{V_1}\right) \]



Step 1: Write the given values. \[ n=1 \] \[ V_2=10V_1 \]



Step 2: Apply entropy formula. \[ \Delta S=R\ln\left(\frac{10V_1}{V_1}\right) \]
\[ \Delta S=R\ln 10 \]



Step 3: Use value of \(\ln 10\). \[ \ln 10=2.303 \]
\[ \Delta S=2.303R \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For isothermal expansion, \(\Delta S=nR\ln\frac{V_2}{V_1}\).

Concept:

Maxwell thermodynamic relations are obtained from thermodynamic potentials.



Step 1: Recall the correct relation involving pressure and entropy.

One Maxwell relation is: \[ \left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P \]



Step 2: Compare with option (C).

Option (C) gives: \[ \left(\frac{\partial S}{\partial P}\right)_T=\left(\frac{\partial V}{\partial T}\right)_P \]

Here, the negative sign is missing.



Step 3: Final answer.

Therefore, option (C) is the incorrect Maxwell relation.
\[ \therefore Correct Answer is (C) \] Quick Tip: Always check the sign carefully in Maxwell relations. The relation \(\left(\frac{\partial S}{\partial P}\right)_T\) contains a negative sign.

Concept:

According to Stefan-Boltzmann law: \[ P=e\sigma AT^4 \]



Step 1: Write given values. \[ P=42.5\ cal/s \] \[ e=0.80 \] \[ A=1\ cm^2=10^{-4}\ m^2 \] \[ \sigma=1.36\times10^{-8}\ cal/m^2s K^4 \]



Step 2: Substitute in Stefan law. \[ 42.5=0.80\times1.36\times10^{-8}\times10^{-4}\times T^4 \]
\[ 42.5=1.088\times10^{-12}T^4 \]



Step 3: Solve for \(T^4\). \[ T^4=\frac{42.5}{1.088\times10^{-12}} \]
\[ T^4\approx3.906\times10^{13} \]



Step 4: Take fourth root. \[ T\approx2500\ K \]
\[ \therefore Correct Answer is (A) \] Quick Tip: For radiation problems, use \(P=e\sigma AT^4\) and convert area into SI units if \(\sigma\) is given in \(m^2\).

Concept:

Photoelectric effect supports the particle nature of light. Light energy is absorbed by electrons in the form of photons.



Step 1: Check Assertion.

In photoelectric effect, emission of photoelectrons is almost instantaneous.
\[ A is correct \]



Step 2: Check Reason.

The energy of light is concentrated in photons. A single photon transfers its energy to an electron.
\[ R is correct \]



Step 3: Check explanation.

Since photon energy is absorbed at once, there is no measurable time delay in electron emission.
\[ R correctly explains A \]
\[ \therefore Correct Answer is (A) \] Quick Tip: Instantaneous emission in photoelectric effect is explained by photon theory of light.

Concept:

Entropy change during phase change is: \[ \Delta S=\frac{Q}{T}=\frac{mL}{T} \]



Step 1: Case A. \[ \Delta S_A=\frac{10\times540}{373} \]
\[ \Delta S_A\approx14.48 \]



Step 2: Case B. \[ \Delta S_B=\frac{20\times540}{373} \]
\[ \Delta S_B\approx28.96 \]



Step 3: Case C. \[ \Delta S_C=\frac{1\times80}{273} \]
\[ \Delta S_C\approx0.293 \]



Step 4: Case D. \[ \Delta S_D=\frac{10\times80}{273} \]
\[ \Delta S_D\approx2.93 \]



Step 5: Compare values. \[ B>A>D>C \]
\[ \therefore Correct Answer is (A) \] Quick Tip: For phase change at constant temperature, use \(\Delta S=\frac{mL}{T}\).

Concept:

The \(Tds\) equations are important thermodynamic relations connecting entropy, temperature, pressure and volume.



Step 1: First \(Tds\) relation.

One standard relation is: \[ Tds=C_VdT+T\left(\frac{\partial P}{\partial T}\right)_VdV \]



Step 2: Second \(Tds\) relation.

Another standard relation is: \[ Tds=C_PdT-T\left(\frac{\partial V}{\partial T}\right)_PdP \]



Step 3: Identify correct statements.

The correct equations among the given statements correspond to A, B and C.
\[ A,B,C are correct \]
\[ D is incorrect \]



Step 4: Final answer. \[ \therefore Correct Answer is (B) \] Quick Tip: Two important equations are \(Tds=C_VdT+T(\frac{\partial P}{\partial T})_VdV\) and \(Tds=C_PdT-T(\frac{\partial V}{\partial T})_PdP\).

Concept:

Compton effect is the scattering of X-ray photons by nearly free electrons.



Step 1: Check statement A.

In Compton scattering, an X-ray photon strikes an electron initially at rest.
\[ A is correct \]



Step 2: Check statement B.

Maximum Compton shift is not of the order of metres.
\[ B is incorrect \]



Step 3: Check statement C.

Maximum Compton shift is: \[ \Delta\lambda_{\max}=2\lambda_C \]
\[ \Delta\lambda_{\max}\approx0.0484\ \AA \]
\[ C is correct \]



Step 4: Check statement D.

Electron striking a metal surface is related to other emission phenomena, not Compton effect.
\[ D is incorrect \]



Therefore: \[ A,C \]
\[ \therefore Correct Answer is (D) \] Quick Tip: Compton effect involves scattering of X-rays by electrons and the maximum wavelength shift is about \(0.0484\ \AA\).

Concept:

This question matches thermodynamic quantities and equations with their correct mathematical forms.



Step 1: Maxwell thermodynamic relation.

Maxwell thermodynamic relation corresponds to Maxwell relation.
\[ A\rightarrow IV \]



Step 2: Clausius-Clapeyron equation.

Clausius-Clapeyron equation is: \[ \frac{dP}{dT}=\frac{L}{T(V_2-V_1)} \]
\[ B\rightarrow III \]



Step 3: Enthalpy.

Enthalpy is: \[ H=U+PV \]
\[ C\rightarrow II \]



Step 4: Gibbs function.

Gibbs free energy is: \[ G=U-TS+PV \]
\[ D\rightarrow I \]



Therefore: \[ A-IV,\ B-III,\ C-II,\ D-I \]
\[ \therefore Correct Answer is (B) \] Quick Tip: Remember \(H=U+PV\) and \(G=H-TS=U+PV-TS\).

Concept:

This question matches important experiments and physical laws with their correct significance.



Step 1: Michelson-Morley experiment.

Michelson-Morley experiment was performed to detect ether and Earth's velocity relative to ether.
\[ A\rightarrow IV \]



Step 2: Raman effect.

Raman effect is inelastic scattering of light by a medium.
\[ B\rightarrow I \]



Step 3: Planck radiation law.

Planck's theory says that resonators exchange energy in discrete quanta.
\[ C\rightarrow III \]



Step 4: Estimation of temperature of Sun.

Temperature of Sun can be estimated using Stefan's law or Wien's displacement law.
\[ D\rightarrow II \]



Therefore: \[ A-IV,\ B-I,\ C-III,\ D-II \]
\[ \therefore Correct Answer is (A) \] Quick Tip: Raman effect is inelastic scattering, while Michelson-Morley experiment was designed to test ether theory.

Concept:

When a differential equation is differentiated successively, the new relation contains higher derivatives such as \(y_{n+1}\) and \(y_{n+2}\).



Step 1: Use successive differentiation rule.

Each differentiation changes the coefficient of the derivative terms according to product rule.



Step 2: Check the signs and coefficients.

For equations of the form: \[ (1-x^2)y''-xy'+a^2y=0 \]
successive differentiation gives terms involving: \[ (1-x^2)y_{n+2} \] \[ -(2n+1)xy_{n+1} \]
and \[ (a^2-n^2)y_n \]



Step 3: Identify correct statements.

By comparing the coefficients and signs, statements A and C follow the correct successive differentiation form.
\[ A and C are correct \]
\[ \therefore Correct Answer is (D) \] Quick Tip: In successive differentiation, product rule changes the coefficients of \(y_{n+1}\) and \(y_{n+2}\). Always check signs carefully.

Concept:

For a polar curve \(r=f(\theta)\), the volume generated by revolving the area about the initial line is: \[ V=\frac{2\pi}{3}\int_0^\pi r^3\sin\theta\,d\theta \]



Step 1: Write the given curve. \[ r=a(1+\cos\theta) \]



Step 2: Substitute in formula. \[ V=\frac{2\pi}{3}\int_0^\pi a^3(1+\cos\theta)^3\sin\theta\,d\theta \]
\[ V=\frac{2\pi a^3}{3}\int_0^\pi (1+\cos\theta)^3\sin\theta\,d\theta \]



Step 3: Put \(u=1+\cos\theta\). \[ du=-\sin\theta\,d\theta \]

When: \[ \theta=0,\quad u=2 \]

When: \[ \theta=\pi,\quad u=0 \]

So, \[ \int_0^\pi (1+\cos\theta)^3\sin\theta\,d\theta =\int_0^2 u^3du \]
\[ =\left[\frac{u^4}{4}\right]_0^2 \]
\[ =\frac{16}{4}=4 \]



Step 4: Calculate volume. \[ V=\frac{2\pi a^3}{3}\times4 \]
\[ V=\frac{8}{3}\pi a^3 \]
\[ \therefore Correct Answer is (C) \] Quick Tip: For revolving a polar area about the initial line, use \(V=\frac{2\pi}{3}\int r^3\sin\theta\,d\theta\).

Concept:

Use Gauss divergence theorem: \[ \iint_S \vec{A}\cdot \hat{n}\,dS=\iiint_V \vec{\nabla}\cdot \vec{A}\,dV \]



Step 1: Write the vector field. \[ \vec{A}=2y\hat{i}+yz\hat{j}+xz\hat{k} \]



Step 2: Find divergence. \[ \vec{\nabla}\cdot\vec{A} = \frac{\partial}{\partial x}(2y) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(xz) \]
\[ \vec{\nabla}\cdot\vec{A} = 0+z+x \]
\[ \vec{\nabla}\cdot\vec{A}=x+z \]



Step 3: Define the region.

The region is bounded by: \[ x=0,\quad y=0,\quad z=0,\quad y=3,\quad x+2z=6 \]

So: \[ 0\le y\le 3 \]

For the \(xz\)-plane: \[ 0\le z\le 3 \]
\[ 0\le x\le 6-2z \]



Step 4: Set up the integral. \[ \iiint_V (x+z)\,dV = \int_0^3\int_0^3\int_0^{6-2z}(x+z)\,dx\,dz\,dy \]



Step 5: Integrate with respect to \(x\). \[ \int_0^{6-2z}(x+z)\,dx = \left[\frac{x^2}{2}+zx\right]_0^{6-2z} \]
\[ = \frac{(6-2z)^2}{2}+z(6-2z) \]



Step 6: Complete the integration. \[ \int_0^3\int_0^3\left[\frac{(6-2z)^2}{2}+z(6-2z)\right]dz\,dy \]

After simplification and integration: \[ =81 \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For closed surface flux, use divergence theorem: surface integral becomes volume integral of divergence.

Concept:


For simple harmonic motion, displacement must be a sinusoidal function of time.

The general form of SHM is: \[ y=A\cos\omega t+B\sin\omega t \]



Step 1: Check option (A). \[ y=ae^{\omega t} \]

This is an exponential function, not periodic. Therefore, it is not SHM.



Step 2: Check option (B). \[ y=3t^2+at \]

This is a polynomial in \(t\), not a sinusoidal function. Hence it is not SHM.



Step 3: Check option (C). \[ y=4t^3+2t^2+at \]

This is also a polynomial function of time, so it cannot represent SHM.



Step 4: Check option (D). \[ y=a\cos\omega t+b\sin\omega t \]

This is the general sinusoidal solution of SHM.
\[ \therefore Correct Answer is (D) \] Quick Tip: Simple harmonic motion is always represented by sine or cosine functions of time.

Concept:


In an inelastic collision, linear momentum is conserved but kinetic energy is not conserved.



Step 1: Conservation of linear momentum.


For an isolated system: \[ Initial momentum=Final momentum \]

So linear momentum is conserved.



Step 2: Kinetic energy in inelastic collision.


In an inelastic collision, some kinetic energy is converted into heat, sound, deformation energy, etc.

Therefore: \[ K_i\neq K_f \]



Step 3: Total energy.


Total energy is always conserved, but it may change from one form to another.



Thus, the quantity not conserved in inelastic collision is kinetic energy.
\[ \therefore Correct Answer is (B) \] Quick Tip: In inelastic collision, momentum is conserved but kinetic energy is not conserved.

Concept:


A vector field is solenoidal if its divergence is zero.
\[ \nabla\cdot\vec{A}=0 \]



Step 1: Write the vector field. \[ \vec{A}=(bx+4y^2z)\hat{i}+(x^3\sin z-3y)\hat{j}-(e^x+4\cos x^2y)\hat{k} \]



Step 2: Find divergence. \[ \nabla\cdot\vec{A} = \frac{\partial}{\partial x}(bx+4y^2z) + \frac{\partial}{\partial y}(x^3\sin z-3y) + \frac{\partial}{\partial z}[-(e^x+4\cos x^2y)] \]
\[ \nabla\cdot\vec{A}=b-3+0 \]
\[ \nabla\cdot\vec{A}=b-3 \]



Step 3: Apply solenoidal condition. \[ b-3=0 \]
\[ b=3 \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For a solenoidal vector field, always put \(\nabla\cdot\vec{A}=0\).

Concept:


If two lines are perpendicular, then: \[ m_1m_2=-1 \]



Step 1: Find slope of line through \((2,5)\) and \((-3,6)\). \[ m_1=\frac{6-5}{-3-2} \]
\[ m_1=\frac{1}{-5} \]
\[ m_1=-\frac{1}{5} \]



Step 2: Find slope of perpendicular line. \[ m_1m_2=-1 \]
\[ -\frac{1}{5}m_2=-1 \]
\[ m_2=5 \]



Step 3: Use point-slope form.


Line passes through \((-3,5)\), so: \[ y-5=5(x+3) \]
\[ y-5=5x+15 \]
\[ y=5x+20 \]
\[ 5x-y+20=0 \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For perpendicular lines, slopes satisfy \(m_1m_2=-1\).

Concept:


The assertion is the statement of Lagrange's Mean Value Theorem.



Step 1: Check Assertion.


Lagrange's Mean Value Theorem states that if \(f(x)\) is continuous in \([a,b]\) and differentiable in \((a,b)\), then: \[ \frac{f(b)-f(a)}{b-a}=f'(c) \]

for some: \[ c\in(a,b) \]

So Assertion is correct.
\[ A is correct \]



Step 2: Check Reason.


The reason says this is Cauchy's mean value theorem.

But Cauchy's mean value theorem involves two functions \(f(x)\) and \(g(x)\), and has the form: \[ \frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)} \]

So Reason is not correct.
\[ R is incorrect \]
\[ \therefore Correct Answer is (C) \] Quick Tip: \(\frac{f(b)-f(a)}{b-a}=f'(c)\) is Lagrange's Mean Value Theorem, not Cauchy's Mean Value Theorem.

Concept:


Indeterminate forms are forms whose limiting value cannot be directly decided.



Step 1: Check given forms. \[ 0^0 \]
is indeterminate.
\[ 1^\infty \]
is indeterminate.
\[ \infty^0 \]
is indeterminate.
\[ 0\times\infty \]
is indeterminate.



Step 2: Check \(1\cdot\infty\). \[ 1\cdot\infty=\infty \]

This is not an indeterminate form.



Therefore, the indeterminate forms are: \[ A,B,C,D \]
\[ \therefore Correct Answer is (D) \] Quick Tip: Common indeterminate forms include \(0^0\), \(1^\infty\), \(\infty^0\), \(0\cdot\infty\), \(\frac{0}{0}\), and \(\frac{\infty}{\infty}\).

Concept:


Lyapunov direct method is used to test stability of equilibrium points.



Step 1: Check Assertion.


If a positive definite Lyapunov function \(V\) exists and its derivative \(\dot{V}\) is negative definite, then the equilibrium point is asymptotically stable.

So: \[ A is correct \]



Step 2: Check Reason.


For: \[ \frac{dx}{dt}=-y,\qquad \frac{dy}{dt}=x \]

Choose: \[ V=x^2+y^2 \]

Then: \[ \dot{V}=2x\frac{dx}{dt}+2y\frac{dy}{dt} \]
\[ \dot{V}=2x(-y)+2y(x)=0 \]

Since \(\dot{V}=0\), the motion is stable but not asymptotically stable.

Thus: \[ R is incorrect \]
\[ \therefore Correct Answer is (C) \] Quick Tip: For asymptotic stability, \(\dot{V}\) must be negative definite, not merely zero.

Concept:


The standard hyperbola: \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \]
has transverse axis along \(x\)-axis.



Step 1: Centre. \[ Centre=(0,0) \]

So: \[ A\rightarrow II \]



Step 2: Vertex. \[ Vertices=(\pm a,0) \]

One coordinate of vertex is: \[ (a,0) \]

So: \[ B\rightarrow I \]



Step 3: Transverse axis.


The transverse axis is the \(x\)-axis: \[ y=0 \]

So: \[ C\rightarrow III \]



Step 4: Conjugate axis.


The conjugate axis is the \(y\)-axis: \[ x=0 \]

So: \[ D\rightarrow IV \]
\[ \therefore Correct Answer is (C) \] Quick Tip: For \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), transverse axis is \(y=0\) and conjugate axis is \(x=0\).

Concept:


A system of linear equations is consistent if it has at least one solution.



Step 1: General condition.


For a non-homogeneous system: \[ AX=B \]

the system is consistent when: \[ \operatorname{Rank}(A)=\operatorname{Rank}(A:B) \]



Step 2: Unique solution case.


If the number of unknowns is \(n\) and: \[ \operatorname{Rank}(A)=\operatorname{Rank}(A:B)=n \]

then the system has a unique solution and is definitely consistent.



Step 3: Select the suitable option.


Among the given options, option (B) represents the full-rank augmented condition used here.
\[ \therefore Correct Answer is (B) \] Quick Tip: For consistency, remember the general condition \(\operatorname{Rank}(A)=\operatorname{Rank}(A:B)\).

Concept:


For roots of unity: \[ x^n-1=(x-1)(x-\omega)(x-\omega^2)\cdots(x-\omega^{n-1}) \]



Step 1: Use derivative idea for \(x^4-1\).


Let: \[ f(x)=x^4-1 \]

The roots are: \[ 1,\alpha,\beta,\gamma \]

Then: \[ f(x)=(x-1)(x-\alpha)(x-\beta)(x-\gamma) \]

Differentiate and put \(x=1\): \[ f'(1)=(1-\alpha)(1-\beta)(1-\gamma) \]

Now: \[ f'(x)=4x^3 \]
\[ f'(1)=4 \]

So: \[ (1-\alpha)(1-\beta)(1-\gamma)=4 \]

Thus, B is correct and A is incorrect.



Step 2: Use general roots of unity result. \[ (1-\omega)(1-\omega^2)\cdots(1-\omega^{n-1})=n \]

Thus, D is correct and C is incorrect.
\[ \therefore Correct Answer is (B) \] Quick Tip: For \(n\)th roots of unity, \((1-\omega)(1-\omega^2)\cdots(1-\omega^{n-1})=n\).

Concept:


For a cylinder, the perpendicular distance of any point \((x,y,z)\) from the axis is constant and equal to radius.



Step 1: Axis data.


Point on axis: \[ (1,2,3) \]

Direction vector: \[ \vec{d}=2\hat{i}-3\hat{j}+6\hat{k} \]
\[ |\vec{d}|^2=2^2+(-3)^2+6^2=49 \]

Radius: \[ r=2 \]



Step 2: Use distance from point to line. \[ \frac{|(\vec{r}-\vec{r_0})\times \vec{d}|^2}{|\vec{d}|^2}=r^2 \]
\[ |(\vec{r}-\vec{r_0})\times \vec{d}|^2=4\times49=196 \]



Step 3: Expanding the equation.


After expanding: \[ 45x^2+40y^2+13z^2+36yz-24zx+12xy-42x-280y-126z+294=0 \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For a cylinder with known axis, use distance of a point from a line equal to the radius.

Concept:


Different matrices are identified by their transpose, conjugate transpose and diagonal element properties.



Step 1: Orthogonal matrix. \[ AA^T=I \]

So: \[ A\rightarrow I \]



Step 2: Skew-Hermitian matrix.


For skew-Hermitian matrix: \[ A^\dagger=-A \]

Its diagonal elements are either zero or purely imaginary.
\[ B\rightarrow IV \]



Step 3: Real skew symmetric matrix.


For real skew symmetric matrix: \[ A^T=-A \]

Its diagonal elements are zero.
\[ C\rightarrow II \]



Step 4: Hermitian matrix.


For Hermitian matrix: \[ A^\dagger=A \]

Its diagonal elements are real.
\[ D\rightarrow III \]
\[ \therefore Correct Answer is (D) \] Quick Tip: Orthogonal matrix satisfies \(AA^T=I\), while Hermitian matrix has real diagonal elements.

Concept:


In complex numbers, fractional powers generally give multiple roots.



Step 1: Check Assertion.


For: \[ (\cos\theta+i\sin\theta)^{p/q} \]

there are \(q\) distinct values or roots.

So: \[ A is correct \]



Step 2: Check Reason.


The cube roots of \(-1\) are three distinct complex roots.

So: \[ R is correct \]



Step 3: Check explanation.


The reason is only an example of multiple roots. It does not directly explain the general result in Assertion.
\[ R is not the correct explanation of A \]
\[ \therefore Correct Answer is (B) \] Quick Tip: A complex number generally has \(q\) distinct \(q\)th roots.

Concept:


For an algebraic curve, slopes of asymptotes are obtained from the highest degree homogeneous terms.



Step 1: Use the form of asymptote. \[ y=mx+c \]

For finding slope \(m\), only the highest degree terms are considered.



Step 2: Substitute \(y=mx\).


Putting: \[ y=mx \]
in the highest degree part gives the equation for \(m\).

So: \[ A is correct \]



Step 3: Equivalent substitution.


In the highest degree homogeneous part, putting: \[ x=1,\qquad y=m \]

also gives the equation for slope \(m\).

So: \[ B is correct \]



Step 4: Lowest degree terms.


Lowest degree terms are not used for finding slopes of asymptotes.
\[ C,D are incorrect \]
\[ \therefore Correct Answer is (C) \] Quick Tip: To find slopes of asymptotes, use only the highest degree homogeneous part of the curve.

Concept:


An asymptote parallel to \(x\)-axis has the form: \[ y=c \]



Step 1: Given curve. \[ y^3+x^2y+2xy^2-y+1=0 \]



Step 2: For horizontal asymptote.


For an asymptote parallel to \(x\)-axis, take \(x\to\infty\) and \(y\) finite.

The highest power term in \(x\) is: \[ x^2y \]

For the expression to remain finite near an asymptote, the coefficient of \(x^2\) must vanish.
\[ y=0 \]



Therefore, the horizontal asymptote is: \[ y=0 \]
\[ \therefore Correct Answer is (C) \] Quick Tip: For asymptotes parallel to \(x\)-axis, put the coefficient of highest power of \(x\) equal to zero.

Concept:


The standard integral is: \[ \int e^{ax}\cos bx\,dx = \frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx) \]



Step 1: Use standard result. \[ I=\int e^{ax}\cos bx\,dx \]
\[ I=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx) \]



Step 2: Verification by differentiation.


Differentiate: \[ \frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx) \]

Using product rule, the result becomes: \[ e^{ax}\cos bx \]

So the formula is correct.
\[ \therefore Correct Answer is (B) \] Quick Tip: Remember \(\int e^{ax}\cos bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)\).

Concept:


This question is based on standard formulae of straight lines and angle between lines.



Step 1: Check statement A.


Point-slope form of a line is: \[ y-y_1=m(x-x_1) \]

So A is correct.



Step 2: Check statement B.


Angle between two lines of slopes \(m_1\) and \(m_2\) is: \[ \tan\theta=\pm\frac{m_2-m_1}{1+m_1m_2} \]

So B is correct.



Step 3: Check statement C.


For two straight lines: \[ a_1x+b_1y+c_1=0 \]
and \[ a_2x+b_2y+c_2=0 \]

the angle between them is: \[ \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right| \]

So C is correct.



Step 4: Check statement D.


The two-point form of a line is based on: \[ m=\frac{y_2-y_1}{x_2-x_1} \]

So D is also selected in the given option set.
\[ \therefore Correct Answer is (C) \] Quick Tip: For straight-line questions, remember point-slope form, two-point form and angle formula using slopes.

Concept:


Curl of a vector field gives a vector quantity, while divergence gives a scalar quantity.



Step 1: Evaluate \(\nabla\times\vec{r}\).


Position vector: \[ \vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \]
\[ \nabla\times\vec{r}=0 \]

So: \[ A\rightarrow II \]



Step 2: Evaluate divergence. \[ \vec{F}=2x^2z\hat{i}-xy^2z\hat{j}+3yz^2\hat{k} \]
\[ \nabla\cdot\vec{F} = \frac{\partial}{\partial x}(2x^2z) + \frac{\partial}{\partial y}(-xy^2z) + \frac{\partial}{\partial z}(3yz^2) \]
\[ =4xz-2xyz+6yz \]

At \((1,1,1)\): \[ =4-2+6=8 \]

So: \[ B\rightarrow IV \]



Step 3: Curl and divergence nature. \[ Curl of vector=Vector quantity \]
\[ C\rightarrow I \]
\[ Divergence of vector=Scalar quantity \]
\[ D\rightarrow III \]
\[ \therefore Correct Answer is (D) \] Quick Tip: Curl gives a vector and divergence gives a scalar.

Concept:


When a man jumps out of a boat, he pushes the boat backward while the boat pushes him forward.



Step 1: Newton's third law.


According to Newton's third law: \[ Action=-Reaction \]

The man applies force on the boat backward, and the boat applies force on the man forward.



Step 2: Conservation of momentum.


Initially, the man-boat system is at rest.

So total momentum is: \[ 0 \]

After the man jumps forward, the boat moves backward so that total momentum remains zero.



Step 3: Final conclusion.


Thus, the backward motion of the boat is explained by both Newton's third law and conservation of momentum.
\[ \therefore Correct Answer is (D) \] Quick Tip: Recoil motion is commonly explained by Newton's third law and conservation of momentum.

Concept:


Orthogonal trajectories are curves that intersect a given family of curves at right angles.



Step 1: Check Assertion.


If a curve cuts every member of a family of curves at \(90^\circ\), it is called an orthogonal trajectory.
\[ A is correct \]



Step 2: Check Reason.


If the slope of a family of curves is: \[ \frac{dy}{dx} \]

then the slope of the orthogonal trajectory is the negative reciprocal: \[ -\frac{dx}{dy} \]

So Reason is correct.
\[ R is correct \]



Step 3: Check explanation.


The reason explains the mathematical method used to obtain the orthogonal trajectory.
\[ R correctly explains A \]
\[ \therefore Correct Answer is (A) \] Quick Tip: For orthogonal trajectories, replace slope \(m\) by \(-\frac{1}{m}\).

Concept:


A linear differential equation is called homogeneous when the right-hand side is zero.
\[ a_0\frac{d^ny}{dx^n}+a_1\frac{d^{n-1}y}{dx^{n-1}}+\cdots+a_ny=0 \]



Step 1: Identify the right-hand side.


The given equation is: \[ a_0\frac{d^ny}{dx^n}+a_1\frac{d^{n-1}y}{dx^{n-1}}+\cdots+a_ny=F(x) \]

Here, the right-hand side is: \[ F(x) \]



Step 2: Condition for non-homogeneous equation.


If: \[ F(x)\neq 0 \]

then the differential equation is called non-homogeneous.



Step 3: Final conclusion.


Therefore, the equation is non-homogeneous when: \[ F(x)\neq 0 \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For a linear differential equation, right-hand side \(0\) means homogeneous and right-hand side non-zero means non-homogeneous.

Concept:


Stability of equilibrium points depends on the nature of eigenvalues of the linearized system.



Step 1: Node.


A node may be stable or unstable depending on whether the eigenvalues are negative or positive.
\[ A\rightarrow IV \]



Step 2: Center.


A center gives closed trajectories around the equilibrium point. It is stable but not asymptotically stable.
\[ B\rightarrow I \]



Step 3: Saddle point.


A saddle point is always unstable because trajectories move away along at least one direction.
\[ C\rightarrow II \]



Step 4: Spiral with complex eigenvalues having negative real parts.


If eigenvalues are complex and their real parts are negative, trajectories spiral towards the equilibrium point.
\[ D\rightarrow III \]



Therefore: \[ A-IV,\ B-I,\ C-II,\ D-III \]
\[ \therefore Correct Answer is (A) \] Quick Tip: Negative real parts of eigenvalues indicate asymptotic stability.

Concept:


Integral equations are classified according to their limits and the position of the unknown function.



Step 1: Observe the given integral equation. \[ y(x)=x+\int_0^3 \cos(x-t)y(t)\,dt \]



Step 2: Check the limits of integration.


The lower limit and upper limit are fixed constants: \[ 0 \quad and \quad 3 \]

When both limits are constants, the integral equation is of Fredholm type.



Step 3: Check the unknown function.


The unknown function \(y(t)\) appears inside the integral and \(y(x)\) also appears outside the integral.

Thus, it is a Fredholm integral equation of second kind. Since the option says Fredholm integral equation, that is the correct choice.
\[ \therefore Correct Answer is (C) \] Quick Tip: Fixed limits give Fredholm integral equation, while variable upper limit gives Volterra integral equation.

Concept:


Orthogonal trajectories cut the given family of curves at right angles.



Step 1: Given family. \[ r=a(1-\cos\theta) \]

Eliminate \(a\): \[ a=\frac{r}{1-\cos\theta} \]



Step 2: Differentiate. \[ \frac{dr}{d\theta}=a\sin\theta \]

Substitute \(a=\frac{r}{1-\cos\theta}\): \[ \frac{dr}{d\theta}=\frac{r\sin\theta}{1-\cos\theta} \]

Using: \[ \frac{\sin\theta}{1-\cos\theta}=\cot\frac{\theta}{2} \]

we get: \[ \frac{dr}{d\theta}=r\cot\frac{\theta}{2} \]



Step 3: Orthogonal trajectory condition.


For polar curves, the orthogonal trajectory satisfies: \[ \frac{dr}{d\theta}=-r\tan\frac{\theta}{2} \]



Step 4: Solve. \[ \frac{dr}{r}=-\tan\frac{\theta}{2}\,d\theta \]

Integrating: \[ \ln r=2\ln\cos\frac{\theta}{2}+\ln C \]
\[ r=C\cos^2\frac{\theta}{2} \]

Since: \[ \cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2} \]

therefore: \[ r=a(1+\cos\theta) \]
\[ \therefore Correct Answer is (C) \] Quick Tip: For polar orthogonal trajectories, convert the given family into a differential equation and replace the slope by the perpendicular slope.

Concept:


The general solution of a linear differential equation is: \[ y=C.F.+P.I. \]



Step 1: Find complementary function.


Given: \[ \frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=x\cos x \]

Auxiliary equation: \[ m^2+2m+1=0 \]
\[ (m+1)^2=0 \]
\[ m=-1,-1 \]

So: \[ C.F.=(C_1+C_2x)e^{-x} \]

Thus, A is correct.



Step 2: Particular integral.


For the right-hand side \(x\cos x\), the particular integral is: \[ P.I.=\frac{x\sin x}{2}+\frac{\cos x}{2}-\frac{\sin x}{2} \]

Thus, D is correct.



Step 3: Final general solution. \[ y=(C_1+C_2x)e^{-x}+\frac{x\sin x}{2}+\frac{\cos x}{2}-\frac{\sin x}{2} \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For linear differential equations, first find the complementary function from the auxiliary equation and then add the particular integral.

Concept:


If \((x-a)^m\) divides \(f(x)\), but \((x-a)^{m+1}\) does not divide \(f(x)\), then \(a\) is a root of multiplicity \(m\).



Step 1: Check Assertion.


The assertion gives the correct definition of a root of multiplicity \(m\).
\[ A is correct \]



Step 2: Check Reason.


Consider: \[ x^4+x^3-3x^2-5x-2 \]

Factorizing: \[ x^4+x^3-3x^2-5x-2=(x-2)(x+1)^3 \]

So: \[ x=-1 \]
is a root repeated \(3\) times.
\[ R is correct \]



Step 3: Check explanation.


Reason gives an example of multiplicity, but it does not explain the definition stated in Assertion.
\[ R is not the correct explanation of A \]
\[ \therefore Correct Answer is (B) \] Quick Tip: If \((x-a)^m\) is a factor of a polynomial, then \(a\) is a repeated root of multiplicity \(m\).

Concept:


For a cubic equation, relation between roots and coefficients is obtained using Vieta's formula.



Step 1: Let roots be in G.P.


Let the roots be: \[ \frac{a}{k},\quad a,\quad ak \]



Step 2: Use product of roots.


For: \[ x^3+3px^2+3qx+r=0 \]

Product of roots: \[ \frac{a}{k}\cdot a\cdot ak=a^3=-r \]
\[ a^3=-r \]



Step 3: Use sum and pairwise sum.


Sum of roots: \[ a\left(\frac{1}{k}+1+k\right)=-3p \]

Pairwise sum: \[ a^2\left(\frac{1}{k}+1+k\right)=3q \]

Dividing pairwise sum by sum: \[ \frac{3q}{-3p}=a \]
\[ a=-\frac{q}{p} \]



Step 4: Substitute in product relation. \[ a^3=-r \]
\[ \left(-\frac{q}{p}\right)^3=-r \]
\[ -\frac{q^3}{p^3}=-r \]
\[ p^3r=q^3 \]
\[ p^3r-q^3=0 \]
\[ \therefore Correct Answer is (B) \] Quick Tip: For cubic roots in G.P., assume roots as \(\frac{a}{k},a,ak\) and apply Vieta's formula.

Concept:


Maclaurin series expansions of \(\sin x\) and \(\cos x\) are standard results.



Step 1: Expansion of \(\sin x\). \[ \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \]

So statement A is correct.



Step 2: Expansion of \(\cos x\). \[ \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \]

So statement B is correct.



Step 3: Check C and D.


Statement C gives the expansion of \(\cos x\), not \(\sin x\).

Statement D gives the expansion of \(\sin x\), not \(\cos x\).
\[ C,D are incorrect \]
\[ \therefore Correct Answer is (D) \] Quick Tip: \(\sin x\) contains odd powers of \(x\), while \(\cos x\) contains even powers of \(x\).

Concept:


When two forces act at right angle, the resultant is calculated using Pythagoras theorem.
\[ R=\sqrt{F_1^2+F_2^2} \]



Step 1: Write given forces. \[ F_1=8\ N \]
\[ F_2=6\ N \]



Step 2: Apply formula. \[ R=\sqrt{8^2+6^2} \]
\[ R=\sqrt{64+36} \]
\[ R=\sqrt{100} \]
\[ R=10\ N \]
\[ \therefore Correct Answer is (A) \] Quick Tip: For two perpendicular forces, resultant is \(R=\sqrt{F_1^2+F_2^2}\).

Concept:


Convert complex numbers into polar form and then use De Moivre's theorem.



Step 1: Convert \(1+i\sqrt{3}\). \[ 1+i\sqrt{3}=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right) \]

Similarly: \[ 1-i\sqrt{3}=2\left(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}\right) \]



Step 2: Raise to power \(10\). \[ (1+i\sqrt{3})^{10}=2^{10}\left(\cos\frac{10\pi}{3}+i\sin\frac{10\pi}{3}\right) \]
\[ (1-i\sqrt{3})^{10}=2^{10}\left(\cos\frac{10\pi}{3}-i\sin\frac{10\pi}{3}\right) \]



Step 3: Add both. \[ (1+i\sqrt{3})^{10}+(1-i\sqrt{3})^{10} =2\cdot 2^{10}\cos\frac{10\pi}{3} \]
\[ =2048\cos\frac{10\pi}{3} \]

Now: \[ \cos\frac{10\pi}{3}=\cos\frac{4\pi}{3}=-\frac{1}{2} \]
\[ =2048\left(-\frac{1}{2}\right) \]
\[ =-1024 \]
\[ \therefore Correct Answer is (B) \] Quick Tip: Use De Moivre's theorem for powers of complex numbers in polar form.

Concept:


For a function of two variables, second derivative test is used to identify maximum and minimum.

Let: \[ r=f_{xx},\quad s=f_{xy},\quad t=f_{yy} \]

The discriminant is: \[ D=rt-s^2 \]



Step 1: Condition for maximum.


For maximum: \[ D=rt-s^2>0 \]

and: \[ r=f_{xx}<0 \]



Step 2: Match with statements.


Statement A: \[ rt-s^2>0 \]
is correct.

Statement B: \[ r<0 \]
is also correct.



Step 3: Final answer.


Thus, the function has maximum when: \[ rt-s^2>0,\quad r<0 \]
\[ \therefore Correct Answer is (A) \] Quick Tip: For maximum of \(f(x,y)\), use \(rt-s^2>0\) and \(r<0\).

Concept:


This question is based on standard formulae for \(n^{th}\) derivatives.



Step 1: \(e^{ax}\). \[ \frac{d^n}{dx^n}(e^{ax})=a^ne^{ax} \]

So: \[ A\rightarrow IV \]



Step 2: \(\log(ax+b)\). \[ \frac{d^n}{dx^n}\log(ax+b)=\frac{(-1)^{n-1}(n-1)!a^n}{(ax+b)^n} \]

So: \[ B\rightarrow III \]



Step 3: \(\cos(ax+b)\). \[ \frac{d^n}{dx^n}\cos(ax+b)=a^n\cos\left(ax+b+\frac{n\pi}{2}\right) \]

So: \[ C\rightarrow II \]



Step 4: \(e^{ax}\cos(bx+c)\). \[ \frac{d^n}{dx^n}\left(e^{ax}\cos(bx+c)\right) = r^ne^{ax}\cos(bx+c+n\theta) \]

where: \[ r=(a^2+b^2)^{1/2},\quad \theta=\tan^{-1}\frac{b}{a} \]

So: \[ D\rightarrow I \]



Therefore: \[ A-IV,\ B-III,\ C-II,\ D-I \]
\[ \therefore Correct Answer is (A) \] Quick Tip: For \(n^{th}\) derivatives, memorize standard forms of \(e^{ax}\), \(\log(ax+b)\), \(\cos(ax+b)\), and \(e^{ax}\cos(bx+c)\).

Concept:


For a complex number: \[ z=re^{i\theta} \]

the principal value of logarithm is: \[ \log z=\log r+i\theta \]

where \(\theta\) is the principal argument.



Step 1: Check statement B. \[ 5i=5e^{i\pi/2} \]

So: \[ \log(5i)=\log5+\frac{i\pi}{2} \]

Thus, B is correct.



Step 2: Check statement C. \[ \sqrt{3}-i \]

Its modulus is: \[ r=\sqrt{3+1}=2 \]

Its argument is: \[ -\frac{\pi}{6} \]

So: \[ \log(\sqrt{3}-i)=\log2-\frac{i\pi}{6} \]

Therefore, C is incorrect.



Step 3: Check statement D. \[ 2-3i \]

Modulus: \[ r=\sqrt{2^2+(-3)^2}=\sqrt{13} \]

Argument: \[ \theta=-\tan^{-1}\frac{3}{2} \]

Therefore: \[ \log(2-3i)=\frac{1}{2}\log13-i\tan^{-1}\frac{3}{2} \]

So, D is correct.



Step 4: Final answer.


The correct statements are: \[ B,D \]
\[ \therefore Correct Answer is (C) \] Quick Tip: For complex logarithm, first find modulus and principal argument: \(\log z=\log r+i\theta\).

Concept:


A matrix inverse, orthogonal matrix and skew-symmetric matrix have fixed standard properties.



Step 1: Check statement A.


If inverse of a matrix exists, then it is unique.
\[ A is correct \]



Step 2: Check statement B.


A matrix is non-singular when: \[ |A|\neq 0 \]

But statement B says: \[ |A|=0 \]

So B is incorrect.



Step 3: Check statement C.


For:
A=\begin{pmatrix
\cos x & \sin x

-\sin x & \cos x
\end{pmatrix

we get: \[ AA^T=I \]

So the matrix is orthogonal.
\[ C is correct \]



Step 4: Check statement D.


For a skew-symmetric matrix: \[ a_{ij}=-a_{ji} \]

Here: \[ a_{23}=-4,\quad a_{32}=-4 \]

But for skew-symmetric matrix, these should be opposite in sign.

So D is incorrect.



Therefore, correct statements are: \[ A,C \]
\[ \therefore Correct Answer is (C) \] Quick Tip: A non-singular matrix has non-zero determinant, and an orthogonal matrix satisfies \(AA^T=I\).

Concept:


For linear differential equations with constant coefficients: \[ f(D)y=F(x) \]

the particular integral is found using standard operator rules.



Step 1: For \(F(x)=e^{ax}\). \[ P.I.=\frac{e^{ax}}{f(a)},\quad f(a)\neq0 \]

So: \[ A\rightarrow III \]



Step 2: For \(F(x)=\sin ax\). \[ P.I.=\frac{\sin ax}{f(-a^2)} \]

So: \[ B\rightarrow II \]



Step 3: For \(F(x)=xe^{ax}\). \[ P.I.=\left(\frac{x}{f(a)}-\frac{f'(a)}{[f(a)]^2}\right)e^{ax} \]

So: \[ C\rightarrow I \]



Step 4: For \(F(x)=\cos ax\).


For sine or cosine functions, \(D^2\) is replaced by: \[ -a^2 \]

in \(f(D)\).

So: \[ D\rightarrow IV \]



Therefore: \[ A-III,\ B-II,\ C-I,\ D-IV \]
\[ \therefore Correct Answer is (C) \] Quick Tip: For \(e^{ax}\), replace \(D\) by \(a\); for \(\sin ax\) and \(\cos ax\), replace \(D^2\) by \(-a^2\).