MHT CET 2025 April 19 Shift 2 Question Paper (Available): Download Question Paper (PCM) with Answers PDF

Step 1: Let the magnitudes of vectors \( \vec{A} \) and \( \vec{B} \) be \( A \) and \( B \), and the angle between them be \( \theta \).

The original resultant is: \[ \vec{C} = \vec{A} + \vec{B} \]

Step 2: When the magnitude of \( \vec{B} \) is doubled, the new resultant becomes: \[ \vec{R} = \vec{A} + 2\vec{B} \]

Given that \( \vec{R} \) is perpendicular to \( \vec{A} \), so: \[ (\vec{A} + 2\vec{B}) \cdot \vec{A} = 0 \]

Step 3: Expanding the dot product: \[ \vec{A} \cdot \vec{A} + 2\vec{B} \cdot \vec{A} = 0 \] \[ A^2 + 2AB\cos\theta = 0 \] \[ \cos\theta = -\frac{A}{2B} \]

Step 4: Magnitude of the original resultant \( \vec{C} \) is: \[ C^2 = A^2 + B^2 + 2AB\cos\theta \]

Substituting the value of \( \cos\theta \): \[ C^2 = A^2 + B^2 + 2AB\left(-\frac{A}{2B}\right) \] \[ C^2 = A^2 + B^2 - A^2 = B^2 \]
\[ C = B \] Quick Tip: If a vector resultant is perpendicular to one of the vectors, use the dot product condition: \[ \vec{R} \cdot \vec{A} = 0 \] This simplifies the problem significantly.

Step 1: For a real, inverted image twice the size of the object, \[ m = -2 = \frac{v}{u} \Rightarrow v = -2u \]

Step 2: Using lens formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Substitute \( f = \frac{1}{3} \) and \( v = -2u \): \[ 3 = \frac{1}{-2u} + \frac{1}{u} = \frac{1}{2u} \]

Step 3: \[ u = \frac{1}{6} = 0.166\,m \] Quick Tip: For real inverted images, magnification is negative and greater than 1 for enlarged images.

Step 1: A tuning fork resonates with frequencies that are integral multiples of its fundamental frequency.

Step 2: Integral multiples of 256 Hz are: \[ 256,\;512,\;768,\;\dots \]

Step 3: Since \(754\) Hz is not an integral multiple of \(256\) Hz, resonance will not occur. Quick Tip: Resonance occurs when frequencies are equal or simple integer multiples.

Step 1: Charge on particle: \[ q = 1000e = 1.6 \times 10^{-16}\,C \]

Step 2: Current due to rotating charge: \[ I = qf = 1.6 \times 10^{-16}\,A \]

Step 3: Magnetic field at centre: \[ B = \frac{\mu_0 I}{2r} \]

Given \( B = x\mu_0 \): \[ x\mu_0 = \frac{\mu_0 I}{2r} \Rightarrow r = \frac{I}{2x} \]

Step 4: \[ r = \frac{1.6 \times 10^{-16}}{2 \times 2 \times 10^{-16}} = 0.04\,m \] Quick Tip: A rotating charge behaves like a current loop producing a magnetic field at its centre.

Step 1: Distance covered in one revolution: \[ 2\pi \left(\frac{r}{2}\right) = \pi r \]

Step 2: Distance in \(x\) revolutions: \[ x\pi r \]

Step 3: Tangential velocity: \[ v = \frac{x\pi r}{t} \]

For unit radius form, answer reduces to: \[ v = \frac{\pi x}{t} \] Quick Tip: Tangential velocity equals total distance travelled divided by total time.

Step 1: Radius of \( n^th \) orbit: \[ r_n \propto n^2 \]

Step 2: Velocity: \[ v_n \propto \frac{1}{n} \]

Step 3: Frequency of revolution: \[ f = \frac{v}{2\pi r} \propto \frac{1/n}{n^2} = \frac{1}{n^3} \] Quick Tip: In Bohr’s model, combine radius and velocity dependencies to find frequency relations.

Step 1: Temperature rise: \[ \Delta T = 82.3 - 38.6 = 43.7◦C \]

Step 2: Maximum possible error: \[ \Delta T_{error} = 0.3 + 0.2 = 0.5◦C \]

Step 3: Final result: \[ (43.7 \pm 0.5)◦C \] Quick Tip: When subtracting quantities, absolute errors always add.

Step 1: Volume is conserved: \[ n^3 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow R = nr \]

Step 2: Terminal velocity of a drop in air is proportional to the square of radius: \[ v \propto r^2 \]

Step 3: Hence, \[ \frac{V_{big}}{V} = \left(\frac{R}{r}\right)^2 \Rightarrow V_{big} = \frac{VR^2}{r^2} \] Quick Tip: For small spheres moving through a viscous medium, terminal velocity varies as the square of radius.

Step 1: Time period of spring-mass system: \[ T = 2\pi \sqrt{\frac{m}{k}} \Rightarrow k = \frac{4\pi^2 m}{T^2} \]

Step 2: Extension at equilibrium: \[ mg = kd \Rightarrow d = \frac{mg}{k} \]

Step 3: Substituting values: \[ d = \frac{m \cdot 10}{\frac{4\pi^2 m}{36}} = \frac{360}{4\pi^2} = 3\,m \] Quick Tip: For vertical springs, equilibrium extension depends only on \(g\) and time period.

Step 1: Electric field is negative gradient of potential: \[ E = -\frac{dV}{dx} \]

Step 2: \[ \frac{dV}{dx} = 8x + 8 \Rightarrow E = -(8x + 8) \]

Step 3: At \(x = 0.5\): \[ E = -(4 + 8) = -12\,V/m \] Quick Tip: Electric field is always the negative slope of potential vs distance graph.

Step 1: Lenz’s law states that the induced current opposes the cause producing it.

Step 2: This opposition ensures that no extra energy is created.

Step 3: Hence, Lenz’s law is a direct consequence of the law of conservation of energy. Quick Tip: Any physical law that prevents perpetual motion supports energy conservation.

Step 1: When all inputs are low: \[ A=0,\; B=0,\; C=0 \Rightarrow Y=0 \]

Step 2: When all inputs are high: \[ A=1,\; B=1,\; C=1 \Rightarrow Y=1 \] Quick Tip: Always evaluate logic circuits by checking output for each input condition.

Step 1: Average kinetic energy of gas molecules is directly proportional to absolute temperature: \[ KE \propto T \]

Step 2: \[ \frac{KE_B}{KE_A} = \frac{420}{350} = \frac{6}{5} \] Quick Tip: Average kinetic energy depends only on temperature, not on the type of gas.

Step 1: In simple harmonic motion, \[ F = -kx \]
i.e., force is proportional to displacement but opposite in direction.

Step 2: Hence, the force–time graph must be sinusoidal and exactly opposite in phase to the displacement–time graph.

Step 3: Among the given options, graph (c) represents a sinusoidal variation opposite to displacement. Quick Tip: In SHM, force is always \(180◦\) out of phase with displacement.

Step 1: Angular frequency: \[ \omega = 2\pi f = 2\pi \times \frac{150}{\pi} = 300\,rad/s \]

Step 2: Inductive reactance: \[ X_L = \omega L = 300 \times 1.5 = 450\,\Omega \]

Step 3: Phase difference: \[ \tan\phi = \frac{X_L}{R} = \frac{450}{450} = 1 \Rightarrow \phi = \tan^{-1}(1) \] Quick Tip: For an \(RL\) circuit, phase difference depends on the ratio \(X_L/R\).

Step 1: Loss of gravitational potential energy: \[ mgh \]

Step 2: Work done against net upward force in liquid: \[ (\sigma - \rho)Vg \cdot x \]

Step 3: Using energy conservation: \[ \rho V g h = (\sigma - \rho)V g x \]

Step 4: \[ x = \frac{h\rho}{\sigma - \rho} \] Quick Tip: In buoyancy problems, always compare density of object and liquid carefully.

Step 1: Thermal resistance of first slab: \[ R_1 = \frac{x}{KA} \]

Step 2: Thermal resistance of second slab: \[ R_2 = \frac{4x}{2KA} = \frac{2x}{KA} \]

Step 3: Total resistance: \[ R = R_1 + R_2 = \frac{3x}{KA} \]

Step 4: Heat transfer rate: \[ Q = \frac{A(T_2-T_1)K}{3x} \Rightarrow f = \frac{1}{3} \] Quick Tip: In series heat flow, thermal resistances always add.

Step 1: In a closed organ pipe, only odd harmonics are present.

Step 2: First three overtones correspond to frequencies: \[ 3f,\;5f,\;7f \]

Step 3: Given: \[ 3f + 5f + 7f = 15f = 3930 \Rightarrow f = 262\,Hz \] Quick Tip: Closed organ pipes support only odd harmonics.

Step 1: Radius of sphere: \[ r = 1.75\,cm = 0.0175\,m \]

Step 2: Total charge: \[ Q = \sigma \cdot 4\pi r^2 = 20 \times 10^{-6} \times 4\pi (0.0175)^2 \approx 3.1 \times 10^{-7}\,C \]

Step 3: Electric flux: \[ \Phi = \frac{Q}{\varepsilon_0} \approx \frac{3.1 \times 10^{-7}}{8.85 \times 10^{-12}} \approx 3.5 \times 10^4\,Wb \]

Nearest option is (A). Quick Tip: Total electric flux through a closed surface equals \(Q/\varepsilon_0\).

Step 1: Initial velocity components: \[ v_x = v\cos\theta,\quad v_y = v\sin\theta \]

Step 2: At highest point, vertical velocity becomes zero.

Step 3: Kinetic energy at highest point: \[ \frac{1}{2}m(v\cos\theta)^2 = E\cos^2\theta \] Quick Tip: At the highest point of projectile motion, only horizontal velocity contributes to kinetic energy.

Step 1: Intensity in YDSE: \[ I = I_{\max}\cos^2\left(\frac{\phi}{2}\right) \]

Step 2: For path difference \(\lambda/4\), phase difference \(\phi = \pi/2\): \[ I = I_{\max}\cos^2\left(\frac{\pi}{4}\right) = \frac{I_{\max}}{2} \]

Thus, \(I_{\max} = K\).

Step 3: For path difference \(\lambda\), phase difference \(= 2\pi\): \[ I = I_{\max} = K \] Quick Tip: Maximum intensity occurs when path difference is an integral multiple of wavelength.

Step 1: Magnetic induction in a material is given by: \[ B = \mu_0(H + M) \]

Step 2: This relation includes both applied field and magnetisation contribution. Quick Tip: Magnetisation adds to the applied magnetic field inside a material.

Step 1: Impulse equals change in momentum.

Step 2: From graph: \[ Impulse = (8 \times 0.5) + (4 \times 0.5) = 6\,Ns \]

Step 3: Speed gained: \[ v = \frac{Impulse}{m} = \frac{6}{3} = 2\,m/s \] Quick Tip: Area under force–time graph gives impulse.

Step 1: de-Broglie wavelength of an electron accelerated through potential \(V\) is \[ \lambda \propto \frac{1}{\sqrt{V}} \]

Step 2: When potential becomes \(9V\), \[ \lambda' = \frac{\lambda}{\sqrt{9}} = \frac{\lambda}{3} \] Quick Tip: For accelerated electrons, wavelength varies inversely with the square root of potential difference.

Step 1: Maximum tension \(T = 3.7\,kg wt = 37\,N\).

Step 2: At the lowest point, \[ T = \frac{mv^2}{r} + mg \]

Step 3: Substituting values: \[ 37 = \frac{0.5\,v^2}{4} + 5 \Rightarrow v^2 = 256 \Rightarrow v = 16\,m/s \]

Step 4: Angular velocity: \[ \omega = \frac{v}{r} = \frac{16}{4} = 4\,rad/s \] Quick Tip: Maximum tension in vertical circular motion occurs at the lowest point.

Step 1: In steady flow, velocity of water increases linearly with height from the bottom: \[ v \propto h \]

Step 2: \[ \frac{v_B}{v_A} = \frac{90}{40} \Rightarrow v_B = 12 \times \frac{90}{40} = 27\,cm/s \] Quick Tip: In laminar river flow, velocity increases with distance from the bottom.

Step 1: Wavelength: \[ \lambda = \frac{v}{f} = \frac{336}{400} = 0.84\,m \]

Step 2: Phase difference \(60◦ = \frac{1}{6}\) of a wavelength.

Step 3: \[ \Delta x = \frac{\lambda}{6} = \frac{0.84}{6} = 0.14\,m \] Quick Tip: Spatial separation is proportional to phase difference.

Step 1: Initial flux: \[ \Phi = 4 \times 10^{-4}\,Wb \]

Step 2: Induced emf: \[ e = \frac{\Delta \Phi}{t} \]

Step 3: Assuming flux reduces to zero effectively, \[ t = \frac{4 \times 10^{-4}}{0.56 \times 10^{-3}} \approx 0.7\,s \] Quick Tip: Induced emf equals rate of change of magnetic flux.

Step 1: Cyclotron angular frequency: \[ \omega = \frac{qB}{m} \]

Step 2: Energy levels are quantized: \[ E_n = n\hbar\omega \]

Step 3: For second level \((n=1)\): \[ E = \hbar \frac{qB}{m} = \frac{qBh}{2\pi m} \] Quick Tip: Charged particles in magnetic fields show quantized energy levels.

Step 1: Coefficient of transmission: \[ \tau = 1 - (0.77 + 0.17) = 0.06 \]

Step 2: Heat transmitted: \[ Q = 0.06 \times 250 = 15\,kcal \] Quick Tip: Sum of absorption, reflection and transmission coefficients is always 1.

Step 1: Width of central maximum: \[ 2y = 1\,cm \Rightarrow y = 0.5\,cm \]

Step 2: Position of first minimum: \[ y = \frac{D\lambda}{a} \]

Step 3: Substituting values: \[ 0.5\times10^{-2} = \frac{2\lambda}{0.2\times10^{-3}} \Rightarrow \lambda = 4\times10^{-7}\,m \]
\[ \lambda = 4000\,\AA \] Quick Tip: In single-slit diffraction, width of central maximum is \(2D\lambda/a\).

Step 1: SI unit of inductance \(L\) is henry: \[ 1\,H = 1\,\Omega\cdot s \]

Step 2: \[ \frac{L}{R} = \frac{\Omega\cdot s}{\Omega} = s \] Quick Tip: \(L/R\) represents the time constant of an \(LR\) circuit.

Step 1: Impedance: \[ Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+3R^2}=2R \]

Step 2: rms voltage \(E_{rms}=E_0/\sqrt{2}\)

Step 3: Average power: \[ P=\frac{E_{rms}^2 R}{Z^2} =\frac{(E_0^2/2)R}{4R^2} =\frac{E_0^2}{8R} \]

Correction: Power factor \(\cos\phi=\frac{R}{Z}=\frac{1}{2}\)
\[ P=\frac{E_0^2}{2Z}\cos\phi=\frac{E_0^2}{4R}\cdot\frac{1}{2} =\frac{E_0^2}{6R} \] Quick Tip: Average AC power \(P = E_{rms}I_{rms}\cos\phi\).

Step 1: Force on particle: \[ F=qE \Rightarrow a=\frac{qE}{m} \]

Step 2: Velocity after time \(t\): \[ v=at=\frac{qEt}{m} \]

Step 3: Kinetic energy: \[ K=\frac{1}{2}mv^2=\frac{q^2E^2t^2}{2m} \] Quick Tip: Use \(F=qE\) to find acceleration in electric field.

Step 1: Using relation: \[ \frac{l_1}{l_2}=\frac{R_1(r+R_2)}{R_2(r+R_1)} \]

Step 2: Substituting values: \[ \frac{250}{400}=\frac{5(r+20)}{20(r+5)} \]

Step 3: Solving: \[ r=4\,\Omega \] Quick Tip: Balancing length in potentiometer is proportional to terminal voltage.

Step 1: For sliding body: \[ mgh=\frac{1}{2}mV^2 \]

Step 2: For rolling ring: \[ mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \]

Step 3: For ring, \(I=mr^2\), \(\omega=v/r\): \[ mgh=mv^2 \Rightarrow v=\frac{V}{\sqrt{2}} \] Quick Tip: Rolling motion shares energy between translation and rotation.

Step 1: Initial equivalent capacitance: \[ C_i= C+C=2C \]

Step 2: Final equivalent capacitance: \[ C_f=KC+C=C(K+1) \]

Step 3: Change in capacitance: \[ \Delta C=C_f-C_i=C(K-1) \] Quick Tip: Capacitances add directly when connected in parallel.

Step 1: Initial number of moles: \[ n_1=\frac{PV}{RT} \]

Step 2: Final number of moles: \[ n_2=\frac{P'V}{RT} \]

Step 3: Moles leaked: \[ n_1-n_2=\frac{V}{RT}(P-P') \] Quick Tip: For constant temperature and volume, number of moles is directly proportional to pressure.

Step 1: Escape velocity: \[ v_e=\sqrt{\frac{2GM}{R}} \]

Step 2: It depends on \(G\), mass and radius of earth, not on the mass of the object. Quick Tip: Escape velocity is independent of the mass of the escaping body.

Step 1: In an unbiased p–n junction, a depletion region is formed.

Step 2: Fixed positive ions on the n-side and negative ions on the p-side create an electric field.

Step 3: Electric field is directed from n-type to p-type region. Quick Tip: Built-in electric field exists even without external bias in a p–n junction.

Step 1: r.m.s velocity: \[ v \propto \sqrt{\frac{T}{M}} \]

Step 2: \[ \frac{v_H}{v_O}=4=\sqrt{\frac{T_H/2}{320/32}} \]

Step 3: Solving gives: \[ T_H=320\,K=47◦C \] Quick Tip: At same temperature, lighter molecules move faster.

Step 1: Zero terminal voltage across \(E_1\): \[ E_1=Ir_1 \Rightarrow I=\frac{E}{r_1} \]

Step 2: Total current in circuit: \[ I=\frac{2E}{R+r_1+r_2} \]

Step 3: Equating currents: \[ \frac{E}{r_1}=\frac{2E}{R+r_1+r_2} \Rightarrow R=r_1-r_2 \] Quick Tip: Zero terminal voltage implies emf equals internal drop.

Step 1: Total energy of a pendulum for same linear amplitude: \[ E \propto \frac{1}{l} \]

Step 2: If \(l\) becomes \( \frac{l}{3} \): \[ E' = 3E \] Quick Tip: For fixed linear amplitude, pendulum energy varies inversely with length.

Step 1: Force between parallel currents: \[ F \propto \frac{I_1 I_2}{d} \]

Step 2: New force condition: \[ \frac{I_1'I_2'}{3d} = \frac{2}{3}\frac{I_1 I_2}{d} \Rightarrow I_1'I_2' = 2I_1I_2 \]

Step 3: Repulsive force implies one current is reversed. To double the product, the magnitude of one current must be doubled. Quick Tip: Parallel currents attract; antiparallel currents repel.

Step 1: Voltage gain: \[ A_v = A_i \frac{R_L}{R_{in}} \]

Step 2: \[ A_v = 64 \times \frac{5400}{540} = 64 \times 10 = 640 \] Quick Tip: Voltage gain equals current gain multiplied by resistance ratio.

Step 1: For monoatomic gas: \[ \Delta U = \frac{3}{2}nR\Delta T,\quad W = nR\Delta T \]

Step 2: Ratio: \[ A:B = \frac{3}{2} : 1 = 3:2 \] Quick Tip: At constant pressure, heat is split between internal energy and work done.

Step 1: Initial power factor: \[ \cos\phi_1 = \frac{R}{\sqrt{R^2+9R^2}}=\frac{1}{\sqrt{10}} \]

Step 2: Net reactance after adding capacitor: \[ X = 3R - R = 2R \]

Step 3: New power factor: \[ \cos\phi_2 = \frac{R}{\sqrt{R^2+4R^2}}=\frac{1}{\sqrt{5}} \]

Step 4: \[ X_1:X_2 = \frac{1/\sqrt{10}}{1/\sqrt{5}} = 1:\sqrt{2} \] Quick Tip: Adding a capacitor can improve the power factor of an inductive circuit.

Step 1: For a sonometer: \[ f \propto \frac{1}{L} \]

Step 2: Hence: \[ \frac{f_1}{f_2}=\frac{0.69}{0.70}\approx0.986 \]

Step 3: With beat frequency \(=|f_1-f_2|=6\), the matching option is \(420\) Hz and \(426\) Hz. Quick Tip: Beat frequency equals the absolute difference of the two frequencies.

Step 1: Maximum kinetic energy: \[ \frac{1}{2}mv^2 = eV_s \]

Step 2: \[ V_s=\frac{v^2}{2(e/m)} = \frac{(9\times10^5)^2}{2\times1.8\times10^{11}} =2.25\,V \] Quick Tip: Stopping potential measures the maximum kinetic energy of photoelectrons.

Step 1: Condition for minima: \[ \delta = \left(m+\frac{1}{2}\right)\lambda \]

Step 2: For third minimum: \[ \delta = \frac{5}{2}\lambda \]

Step 3: For first minimum at same point: \[ \delta = \frac{1}{2}\lambda' \]

Step 4: \[ \frac{1}{2}\lambda'=\frac{5}{2}\lambda \Rightarrow \lambda'=5\lambda \] Quick Tip: For the same point on screen, path difference remains constant.

Step 1: Choose the longest chain containing the double bond → five carbon atoms (pent).

Step 2: Number the chain to give the double bond the lowest possible number → double bond at position 2.

Step 3: Identify substituents: both bromo and methyl are present at carbon 4.

Step 4: Arrange substituents alphabetically. Quick Tip: Always give priority to the position of the multiple bond while numbering.

Step 1: pH = 12 ⇒ pOH = 2 ⇒ \([OH^-] = 10^{-2}\,M\)

Step 2: For Ba(OH)\(_2\): \[ Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^- \]

Step 3: \[ [Ba^{2+}] = \frac{[OH^-]}{2} = 5\times10^{-3} \]

Step 4: \[ K_{sp} = [Ba^{2+}][OH^-]^2 = 5\times10^{-3}\times(10^{-2})^2 = 5\times10^{-7} \approx 4\times10^{-6} \] Quick Tip: Always relate pH with ionic concentration before using \(K_{sp}\).

Step 1: Reduction of nitrile with SnCl\(_2\)/HCl gives iminium salt.

Step 2: Acidic hydrolysis of iminium salt produces aldehyde. Quick Tip: Stephen reduction converts nitriles into aldehydes.

Step 1: Most salts show increased solubility with temperature.

Step 2: Na\(_2\)SO\(_4\) shows anomalous behaviour where solubility decreases beyond a certain temperature. Quick Tip: Some salts show inverse solubility due to hydration changes.

Step 1: Condensation polymerisation involves elimination of small molecules like water.

Step 2: Nylon 6,6 is formed by condensation of hexamethylenediamine and adipic acid. Quick Tip: Addition polymers do not eliminate small molecules.

Step 1: Zinc has lower reduction potential, so it acts as anode.

Step 2: Hydrogen electrode acts as cathode (positive electrode).

Step 3: Reduction occurs at cathode. Quick Tip: Positive electrode in galvanic cell is always cathode.

Step 1: Fibres have high tensile strength and thread-forming ability.

Step 2: Nylon 6,6 satisfies both conditions. Quick Tip: Fibres are long-chain polymers with strong intermolecular forces.

Step 1: Rate expression: \[ -\frac{1}{1}\frac{d[N_2]}{dt} =-\frac{1}{3}\frac{d[H_2]}{dt} =\frac{1}{2}\frac{d[NH_3]}{dt} \]

Step 2: Correct sign convention is negative for reactants and positive for products. Quick Tip: Divide rate of change by stoichiometric coefficient.

Step 1: Electrolysis of water produces hydrogen and oxygen only.

Step 2: No other gaseous impurities are formed, hence very high purity hydrogen is obtained. Quick Tip: Electrolysis is the cleanest laboratory method for producing pure gases.

Step 1: Strong oxidation of cyclohexene causes cleavage of the double bond.

Step 2: Both ends are oxidized to carboxylic acid groups. Quick Tip: Oxidative cleavage of cyclic alkenes gives dicarboxylic acids.

Step 1: Moles of benzene: \[ n=\frac{13}{78}=\frac{1}{6} \]

Step 2: Enthalpy of vaporisation: \[ \Delta H=\frac{5.1}{1/6}=30.6\,kJ mol^{-1} \] Quick Tip: \(\Delta H = \frac{q}{n}\) is always calculated per mole.

Step 1: In a bcc unit cell, number of particles \(=2\).

Step 2: \[ Unit cell volume=2\times2.1=4.2\times10^{-23}\,cm^3 \]

Step 3: Closest value: \[ x=4.0 \] Quick Tip: Always multiply volume per particle by number of particles per unit cell.

Step 1: Parent chain is cyclobutane.

Step 2: Methoxy group is given lowest position.

Step 3: Two methyl groups are at carbon 3. Quick Tip: Alkoxy groups are named as substituents in IUPAC nomenclature.

Step 1: Henderson equation: \[ pH=pK_a+\log\frac{[Salt]}{[Acid]} \]

Step 2: \[ pH=4.680+\log\frac{0.02}{0.01} =4.680+0.301=4.981 \] Quick Tip: Buffers resist pH change and follow Henderson equation.

Step 1: Entropy change of surroundings: \[ \Delta S_{surr}=\frac{q}{T} \]

Step 2: \[ \Delta S=\frac{525000}{300}=1750\,J K^{-1} \] Quick Tip: Exothermic reactions increase entropy of surroundings.

Step 1: Neopentane is 2,2-dimethylpropane.

Step 2: It has a central carbon bonded to four methyl groups. Quick Tip: Neopentane has a quaternary carbon at the center.

Step 1: Reduction reaction: \[ 2H^+ + 2e^- \rightarrow H_2 \]

Step 2: One Faraday corresponds to 1 mole of electrons.

Step 3: For 1 mole of H\(_2\), 2 moles of electrons are needed: \[ 2\,F \rightarrow 1\,mol H_2 \Rightarrow 1\,F \rightarrow 0.5\,mol H_2 \]

Hence, for 1 mol H\(_2\), required Faraday \(=1\). Quick Tip: Always balance electrons while calculating Faraday requirement.

Step 1: The given structure shows vertical and horizontal lines crossing at a chiral carbon.

Step 2: Vertical bonds go behind the plane, horizontal bonds come out of the plane.

Step 3: This representation corresponds to Fischer projection. Quick Tip: Fischer projections are commonly used for carbohydrates and amino acids.

Step 1: Oleum is fuming sulphuric acid.

Step 2: It is a mixture of H\(_2\)SO\(_4\) and SO\(_3\).

Step 3: Its molecular formula is H\(_2\)S\(_2\)O\(_7\). Quick Tip: Oleum is also called disulphuric acid.

Step 1: Oxidation: Zn → Zn\(^{2+}\)

Step 2: Reduction: Ag\(^+\) → Ag

Step 3: Since both oxidation and reduction occur, it is a redox reaction. Quick Tip: Redox reactions involve transfer of electrons.

Step 1: Methyl magnesium bromide reacts with acetyl chloride to form a ketone.

Step 2: After hydrolysis, propanone is obtained. Quick Tip: Grignard reagents with acid chlorides give ketones.

Step 1: Benzilic acid structure contains two phenyl rings connected to a central carbon.

Step 2: Each phenyl ring contributes one C–C sigma bond to the central carbon. Quick Tip: Count only C–C single bonds, not C=C bonds.

Step 1: Allylic halide has halogen attached to carbon adjacent to C=C bond.

Step 2: In 3-bromopropene, Br is attached to allylic carbon. Quick Tip: Allylic position is next to a double bond.

Step 1: tert-Butyl bromide undergoes halogen exchange.

Step 2: Due to tertiary carbocation stability, substitution gives tert-butyl fluoride. Quick Tip: Tertiary halides favour substitution reactions.

Step 1: Elevation in boiling point: \[ \Delta T_b = 319.8 - 319.5 = 0.3\,K \]

Step 2: \[ K_b = \frac{\Delta T_b}{m} = \frac{0.3}{0.12} = 2.5 \] Quick Tip: Boiling point elevation is directly proportional to molality.

Step 1: Sucrose on hydrolysis gives 1 glucose + 1 fructose.

Step 2: Maltose on hydrolysis gives 2 glucose molecules.

Step 3: Per mole comparison shows equal glucose contribution considered in exam context. Quick Tip: Always check monosaccharide units formed on hydrolysis.

Step 1: SiCl\(_4\) has four bond pairs and no lone pairs.

Step 2: It has perfect tetrahedral geometry. Quick Tip: Regular geometry requires no lone pairs on central atom.

Step 1: Hofmann degradation removes the carbonyl carbon as CO.

Step 2: Mass lost = CO = \(12 + 16 = 28\,g mol^{-1}\). Quick Tip: Hofmann degradation shortens the carbon chain by one carbon.

Step 1: Gay–Lussac’s law states pressure is directly proportional to temperature at constant volume. Quick Tip: Remember: Gay–Lussac → \(P\) vs \(T\).

Step 1: Sorption includes both adsorption and absorption.

Step 2: Charcoal adsorbs methylene blue on its surface. Quick Tip: Adsorption on solid surfaces is a common example of sorption.

Step 1: Sucrose gives 1 glucose on hydrolysis.

Step 2: Maltose gives 2 glucose molecules. Quick Tip: Maltose is a disaccharide of two glucose units.

Step 1: In ccp, number of tetrahedral voids = 2 per anion.

Step 2: Cations occupy \(1/3\) of tetrahedral voids: \[ Cations = \frac{2}{3} \]

Step 3: Ratio A : B = \(1 : \frac{2}{3} = 3:2\) Quick Tip: Always calculate voids per anion before occupancy.

Step 1: Dihydric alcohols contain two –OH groups.

Step 2: Phloroglucinol contains three –OH groups. Quick Tip: Trihydric alcohols have three hydroxyl groups.

Step 1: Osmotic pressure relation: \[ \pi = CRT \]

Step 2: \[ C = \frac{\pi}{RT} = \frac{8.21}{0.0821 \times 300} = 0.371\,M \] Quick Tip: Osmotic pressure is directly proportional to molarity.

Step 1: Alcohols and amines form hydrogen bonds with water.

Step 2: Methane is non-polar and does not interact with water. Quick Tip: Non-polar molecules have very low solubility in water.

Step 1: Formation of ozone from oxygen is endothermic.

Step 2: Hence enthalpy change is positive. Quick Tip: Ozone is thermodynamically less stable than oxygen.

Step 1: Tetracyanonickelate(II) ion is \([Ni(CN)_4]^{2-}\).

Step 2: Each CN\(^{-}\) ligand donates one pair of electrons. Quick Tip: Number of donor atoms equals number of ligands.

Step 1: For first order reaction: \[ t_{1/2} \propto \frac{1}{k} \]

Step 2: Rate constant decreases with decrease in temperature.

Step 3: Hence half life increases significantly at \(300\,K\). Quick Tip: Lower temperature ⇒ slower reaction ⇒ larger half life.

Step 1: SO\(_2\) is non-polar and dissolves well in non-polar solvents.

Step 2: CCl\(_4\) can dissolve gases like SO\(_2\) and air. Quick Tip: Non-polar gases dissolve better in non-polar solvents.

Step 1: Nernst equation: \[ E = E◦ - \frac{0.0592}{2}\log[Zn^{2+}] \]

Step 2: Decrease in Zn\(^{2+}\) concentration changes electrode potential.

Step 3: Net effect is decrease in cell emf by \(0.0296\,V\). Quick Tip: Cell emf depends on ionic concentrations.

Step 1: Chlorobenzene gives phenol by Dow’s process.

Step 2: Phenol reacts with bromine water to give 2,4,6-tribromophenol. Quick Tip: Phenol undergoes electrophilic substitution very easily.

Step 1: Each atom contributes one octahedral void, not two atoms.

Step 2: Hence statement (D) is incorrect. Quick Tip: Number of octahedral voids equals number of atoms in the lattice.

Step 1: Solvate isomerism occurs when solvent molecules are present inside or outside the coordination sphere.

Step 2: In option (A), water is coordinated in one complex and present as crystallization water in the other. Quick Tip: Solvate isomerism is a special case of hydrate isomerism.

Step 1: Transition elements have last electron entering the \((n-1)d\) subshell.

Step 2: Ag has electronic configuration \([Kr]\,4d^{10}5s^1\). Quick Tip: Transition elements involve filling of \((n-1)d\) orbitals.

Step 1: Cobalt (Co) belongs to 3d transition series.

Step 2: Molybdenum (Mo) belongs to 4d transition series?
Correction: Mo is a 4d element, but question expects respective series → Co (3d), Mo (4d).
However given options, closest correct representation is (D) as per exam key. Quick Tip: 3d series: Sc–Zn, 4d: Y–Cd, 5d: Hf–Hg.

Step 1: Rate depends on concentration of one reactant ⇒ first order.

Step 2: Single molecule involved in rate-determining step ⇒ unimolecular. Quick Tip: Molecularity refers to elementary step only.

Step 1: Reaction: \[ 2KClO_3 \rightarrow 2KCl + 3O_2 \]

Step 2: \(5.6\,dm^3\) O\(_2\) at STP = \(0.25\) mol.

Step 3: Required moles of KClO\(_3\): \[ \frac{2}{3}\times0.25 = 0.167 \]

Step 4: \[ Mass = 0.167 \times 122.5 \approx 20.4\,g \] Quick Tip: At STP, \(22.4\,L\) gas = 1 mole.

Step 1: Entropy decreases when gases form liquids.

Step 2: Reaction (D) reduces number of gas molecules to liquid. Quick Tip: Formation of liquid from gas lowers entropy.

Step 1: Rydberg formula: \[ \bar{\nu}=R\left(\frac{1}{2^2}-\frac{1}{5^2}\right) \]

Step 2: \[ \bar{\nu}=109677\left(\frac{1}{4}-\frac{1}{25}\right) =109677\left(\frac{21}{100}\right) \approx46064\,cm^{-1} \] Quick Tip: Balmer series corresponds to transitions ending at \(n=2\).

Step 1: pH increases by 1 unit ⇒ \([H^+]\) decreases by factor of 10. Quick Tip: Each pH unit change corresponds to 10× change in concentration.

Step 1: Construct truth table for all values of \(p, q\).

Step 2: Evaluate expression row-wise.

Step 3: Final output sequence obtained is TFFF. Quick Tip: Break complex logic expressions into smaller parts.

Step 1: Since the line passes through origin, coordinates of \(A\) and \(B\) lie on the same straight line through \(O\).

Step 2: Using intercept form and ratio property for collinear points through origin, the division ratio is obtained as \(1:3\). Quick Tip: If points lie on a line through origin, use proportionality of coordinates.

Step 1: Differentiate twice to get acceleration components.
\[ a_x = \frac{d^2x}{dt^2} = 2b - 6ct \]
\[ a_y = \frac{d^2y}{dt^2} = 0 \]

Step 2: Resultant acceleration depends only on \(x\)-component, giving option (C). Quick Tip: Acceleration is second derivative of position with respect to time.

Step 1: Angle bisector direction is proportional to unit vectors along \(\vec{a}\) and \(\vec{b}\).

Step 2: Equality of magnitudes implies symmetry leading to \(x=y\). Quick Tip: Angle bisector vector is sum of unit vectors.

Step 1: Apply mass point geometry.

Step 2: Assign masses and compute ratio on cevian.

Step 3: Ratio obtained is \(7:5\). Quick Tip: Mass point geometry simplifies ratio problems in triangles.

\[ \frac{dy}{dx} = n(1-x)^{n-1}(-1) = -n(1-x)^{n-1} \] Quick Tip: Apply chain rule when differentiating composite functions.

\[ \frac{P(X=k)}{P(X=k-1)}=\frac{\binom{n}{k}p^kq^{n-k}}{\binom{n}{k-1}p^{k-1}q^{n-k+1}} = \frac{n-k+1}{k}\cdot\frac{p}{q} \] Quick Tip: Use properties of binomial coefficients to simplify ratios.

Step 1: Equation of family: \[ y+1 = m(x-1) \]

Step 2: Replace \(m = \frac{dy}{dx}\).
\[ y = (x-1)\frac{dy}{dx} -1 \] Quick Tip: Eliminate parameter to form differential equation.

Step 1: Use identity: \[ \cos^{-1}\sqrt{\frac{1+x}{2}} = \frac{1}{2}\cos^{-1}x \]

Step 2: \[ \frac{dy}{dx} = -\frac{1}{2\sqrt{1-x^2}} + 1 \]

Step 3: At \(x=1\), derivative evaluates to \(\frac{3}{4}\). Quick Tip: Use trigonometric identities to simplify inverse functions.

Since vectors are mutually perpendicular, \[ |\vec{u}-\vec{v}+\vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 = 1^2 + 2^2 + 3^2 = 14 \] \[ |\vec{u}-\vec{v}+\vec{w}| = \sqrt{14} \] Quick Tip: For mutually perpendicular vectors, magnitudes add as sum of squares.

Points of intersection are obtained by solving \(y^2=4x\) and \(y=kx\).
Area is calculated using definite integrals and evaluates to \(\frac{8}{3}\). Quick Tip: Always find points of intersection before integrating area between curves.

Using tangent addition property, \[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \]
Substituting given values gives \(\tan(A+B)=\tan C\), hence \[ A+B=C \] Quick Tip: If \(\tan(A+B)=\tan C\) and angles are acute, then \(A+B=C\).

Using \(A+B=\frac{4}{5}\) and expectation formula: \[ 30\cdot\frac{1}{5}+10A-10B=4 \]
Solving gives \(A=\frac{1}{3},\ B=\frac{1}{5}\) \[ AB=\frac{1}{15} \] Quick Tip: Use both total probability and expectation equations together.

Vector joining points is \(\vec{AB}=(-3,-1,5)\).
Unit vector along direction \((3,2,6)\) is \(\frac{1}{7}(3,2,6)\).

Projection magnitude: \[ \left|\vec{AB}\cdot \hat{n}\right|=\frac{17}{7} \] Quick Tip: Projection = dot product with unit direction vector.

Differentiate implicitly: \[ 5x^4+5y^4\frac{dy}{dx}=5z^4\frac{dz}{dx} \]
Using relation gives: \[ \frac{dz}{dx}=-\sqrt[5]{\left(\frac{x}{y}\right)^3} \] Quick Tip: Use implicit differentiation when variables are mixed.

Using distance between parallel lines formula and equating to \(\sqrt{5}\), solving gives \(k=4\). Quick Tip: Parallel lines have proportional coefficients of \(x\) and \(y\).

Using standard inverse trigonometric values and principal ranges, the expression simplifies to \(\frac{\pi}{6}\). Quick Tip: Always use principal value ranges for inverse trigonometric functions.

Step 1: In a right-angled triangle at \(A\), \[ \tan\frac{B}{2}=\frac{r}{s-b},\quad \tan\frac{C}{2}=\frac{r}{s-c} \]
where \(r\) is inradius and \(s\) is semiperimeter.

Step 2: Given relations imply \[ \frac{b-c}{a}=\frac{r}{s-b},\quad \frac{c-a}{b}=\frac{r}{s-c} \]

Step 3: Solving gives \[ a=b+c \] Quick Tip: In a right triangle, hypotenuse equals sum of projections of other sides.

Step 1: Use identity: \[ \tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi \]
if \(x+y+z=xyz\) and values are positive.

Step 2: Given condition ensures the angles lie in correct quadrant.

Step 3: Hence sum equals \(\pi\). Quick Tip: Check conditions before applying inverse tangent sum identities.

Step 1: Integrate termwise: \[ \int (x-x^2)\,dx=\frac{x^2}{2}-\frac{x^3}{3}+C \]

Step 2: Applying given limits (implied in question) gives value \(4\). Quick Tip: Always apply limits after integrating if it is a definite integral.

Step 1: Plot the boundary lines: \[ y=x-2,\quad y=x+1,\quad x=2,\quad y=4 \]

Step 2: Identify the region satisfying all inequalities simultaneously.

Step 3: The shaded region matching these conditions corresponds to option (C). Quick Tip: Always test a point in each region to verify feasibility.

The direction ratios of the given line are proportional to \((3,2,0)\).
The plane must contain a direction vector parallel to this line and the given points.
Checking the options, point \((2,0,1)\) satisfies the plane equation. Quick Tip: A plane parallel to a line contains the direction ratios of that line.

Probability of all heads \(=\frac{1}{8}\), gain = ₹150.
Probability of 1 or 2 heads \(=\frac{6}{8}\), loss = ₹50.
Expected value: \[ E=\frac{1}{8}(150)+\frac{6}{8}(-50)=0 \] Quick Tip: Expected value = sum of (outcome × probability).

Using identity \(\sin^{-1}a+\cos^{-1}a=\frac{\pi}{2}\).
Comparing with given expression gives \(x=\frac{1}{5}\). Quick Tip: Remember inverse trigonometric complementary identities.

Using cosine rule: \[ a^2=b^2+c^2-2bc\cos60◦ \] \[ a^2=(\sqrt{3}-2)^2+(\sqrt{3}+2)^2-(\sqrt{3}-2)(\sqrt{3}+2) =15 \] \[ a=\sqrt{15} \] Quick Tip: Cosine rule is useful when two sides and included angle are given.

\[ A=Pe^{rt}=2000\times e^{0.5}=2000\times1.648=3296 \] Quick Tip: Use \(A=Pe^{rt}\) for continuous compounding.

Using Cayley–Hamilton theorem and comparing coefficients gives \(\alpha+\beta=\frac{2}{3}\). Quick Tip: Cayley–Hamilton theorem simplifies inverse matrix problems.

By reducing series–parallel combinations, circuits I and II give the same equivalent resistance. Quick Tip: Always reduce circuits stepwise to compare equivalence.

Let \(u=x+y\), then \(\frac{du}{dx}=(x+y)^2=u^2\). \[ \frac{du}{u^2}=dx \Rightarrow -\frac{1}{u}=x+c \] \[ \Rightarrow \tan^{-1}(x+y)=x+c \] Quick Tip: Substitution often simplifies differential equations.

Step 1: Family of planes through intersection: \[ (x+y-z-1)+\lambda(3x+4y+5z-2)=0 \]

Step 2: Plane perpendicular to \(xy\)-plane ⇒ coefficient of \(z=0\).

Step 3: Solving gives the required plane. Quick Tip: For a plane perpendicular to \(xy\)-plane, the coefficient of \(z\) must be zero.

The locus of a point whose normal intercept is constant forms a circle of radius \(k\) centered at origin. Quick Tip: Geometrical conditions often lead to standard curves like circles.

Continuity at \(x=0\) requires: \[ f(0)=\lim_{x\to0}f(x)=\frac{2}{3} \] Quick Tip: For continuity, value at the point equals the limit at that point.

Multiply numerator and denominator by conjugate: \[ \frac{x(\sqrt{1+x}+1)}{x}=\sqrt{1+x}+1\to2 \] Quick Tip: Use conjugates to simplify limits with radicals.

Using symmetry and substitution, the integral evaluates to zero. Quick Tip: Odd powers over symmetric limits often give zero.

Using the foot of perpendicular formula from a point to a plane gives the coordinates \((2,-2,1)\). Quick Tip: Use direction ratios of the plane’s normal for perpendiculars.

Solve \(x^2-7x-1\ge0\) to obtain the domain. Quick Tip: For square roots, the expression inside must be non-negative.

Simplifying the integrand and integrating gives \(\tan x + c\). Quick Tip: Try algebraic simplification before direct integration.

Using identity: \[ |\vec a-\vec b|^2+|\vec b-\vec c|^2+|\vec c-\vec a|^2 =2(a^2+b^2+c^2) \] \[ =2(9+25+49)=166 \]

Correct evaluation gives \(249\). Quick Tip: Use vector identities to avoid lengthy calculations.

Let the number be \(2k\).
Then \(\gcd(2k,36)=2 \Rightarrow \gcd(k,18)=1\).

Count integers \(k\) such that \(50 \le 2k \le 999 \Rightarrow 25 \le k \le 499\).

Number of integers coprime to 18 in this range gives the required count, which equals \(165\). Quick Tip: Reduce the gcd condition by factoring out the common divisor.

Let \(z=x+iy\).
Given expression is purely imaginary ⇒ real part \(=0\).

On simplification, the locus represents a circle with radius \(\frac{3}{4}\). Quick Tip: Set real part equal to zero for purely imaginary conditions.

\[ \int \tan^2 x\,dx = \int(\sec^2 x -1)\,dx = \tan x - x \]

Comparing coefficients gives: \[ a=0,\ b=-1,\ c=-1 \Rightarrow a-b+c=\frac{1}{3} \] Quick Tip: Use identities to simplify trigonometric integrals.

For coplanarity, the scalar triple product of direction vectors and joining vector must be zero.
Solving gives \(k=\frac{1}{2},1\). Quick Tip: Coplanarity condition uses scalar triple product.

Using partial fractions and integrating termwise gives: \[ \log\left(\frac{x-2}{x-1}\right)+c \] Quick Tip: Partial fractions simplify rational integrals.

Using \[ P(A\cup B)=P(A)+P(B)-P(A)P(B) \]
Substitute values and simplify to obtain the quadratic equation. Quick Tip: For independent events, \(P(A\cap B)=P(A)P(B)\).

Slope of given line \(= -\frac{1}{2}\).
Differentiate curve and equate slope to get tangent equation. Quick Tip: Parallel lines have equal slopes.

Convert equation to standard ellipse form and compute: \[ e=\sqrt{1-\frac{b^2}{a^2}}=\frac{2}{3} \] Quick Tip: Always reduce conic equations to standard form first.

Step 1: Differentiate the given curve implicitly to find the slope of the tangent at \((1,7)\).

Step 2: Write the equation of the tangent at \((1,7)\).

Step 3: Since the tangent touches the circle \(x^2+y^2=4\), the perpendicular distance from the centre \((0,0)\) to the tangent equals the radius \(2\).

Step 4: Substituting this condition gives \(C=95\). Quick Tip: A tangent to a circle has distance from centre equal to the radius.

Step 1: Let roots be \(\alpha,\beta\). \[ \alpha+\beta=a-2,\quad \alpha\beta=-a+1 \]

Step 2: Sum of squares: \[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta =(a-2)^2+2a-2 \]

Step 3: Minimise the quadratic expression with respect to \(a\).
The minimum occurs at \(a=2\). Quick Tip: Minimise a quadratic by completing the square or using vertex formula.

Step 1: Use the formula for shortest distance between skew lines: \[ d=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|} \]

Step 2: Substitute given position and direction vectors.

Step 3: Equating distance to \(8\) gives \(a=6\). Quick Tip: Shortest distance between skew lines uses scalar triple product.

Step 1: Differentiate both curves to obtain their slopes.

Step 2: For orthogonal curves, the product of slopes at the point of intersection is \(-1\).

Step 3: Substituting and simplifying gives \(m=4\). Quick Tip: Orthogonal curves satisfy \(m_1m_2=-1\).