BITSAT 2011 Question Paper with Solution PDF: Download Answer Key

Resolve velocity of ball relative to ground by vector addition of car velocity and ball’s velocity relative to car.
Horizontal component \(= 30 + 20\cos60^\circ\), vertical component \(= 20\sin60^\circ\).
Angle of projection is obtained using \(\tan\theta = \frac{v_y}{v_x}\). Quick Tip: Always add velocities vectorially when motion is observed from a moving frame.

Distance in \(n^{th}\) second is \(s_n = u + \frac{a}{2}(2n-1)\).
Using the given condition, velocity after \(t\) seconds is obtained as \(50\ cm/s\). Quick Tip: Use the formula for distance in the \(n^{th}\) second for uniform acceleration problems.

Using vector identity \[ C^2 = A^2 + B^2 + 2AB\cos\theta \]
the angle depends on the sign of \(\cos\theta\), determining whether it is acute, obtuse, or right. Quick Tip: Remember the vector addition formula to relate magnitudes and angles.

Substitute dimensions of all quantities and equate powers of \(L\) to find dimension of \(q\). Quick Tip: Always balance dimensions on both sides of the equation.

Acceleration is inversely proportional to mass.
Time taken varies inversely with acceleration, hence doubles. Quick Tip: If force is constant, increasing mass reduces acceleration.

Find system acceleration using total mass, then calculate tension using \(T = ma\). Quick Tip: Treat connected bodies as one system to find acceleration.

Using energy conservation, compression is proportional to square root of drop height. Quick Tip: Apply conservation of energy for spring compression problems.

Step 1: Torque relation.
\[ \tau = I\alpha \]

Step 2: Substitute values.
\[ I = \frac{31.4}{4} = 7.85 \approx 2.5 \] Quick Tip: Moment of inertia = torque / angular acceleration.

Step 1: Shell theorem.

Spherical shell exerts zero force inside.

Step 2: Only central mass acts.

Effective mass inside = \(M/3\)

Step 3: Change in force.

Difference gives \(\frac{2GMm}{3R^2}\) Quick Tip: Inside a spherical shell, gravitational force is zero.

Step 1: Definition.

Geo-stationary satellite appears fixed relative to Earth.

Step 2: Condition.

Same angular velocity and direction as Earth’s rotation. Quick Tip: Time period of geo-stationary satellite = 24 hours.

Step 1: Extension formula.
\[ \Delta L = \frac{FL}{YA} \]

Step 2: Volume constant.
\(AL = constant\Rightarrow L \propto \frac{1}{A}\)

Step 3: Force relation.
\[ F \propto A^2 \Rightarrow F_2 = 9F \] Quick Tip: For same volume, force ∝ (area)\(^2\).

Step 1: Formula.
\[ Y = \frac{FL}{A\Delta L} \]

Step 2: Substitution.

Gives \(19.6\times10^{11}\) N/m\(^2\). Quick Tip: Convert mm\(^2\) to m\(^2\) carefully.

Step 1: Definition.

Viscosity resists flow between layers. Quick Tip: Higher viscosity means greater internal friction.

Step 1: Stefan’s law.
\[ E \propto T^4 \] Quick Tip: Black body radiation follows Stefan–Boltzmann law.

Step 1: Newton’s law of cooling.

Rate ∝ temperature difference.

Step 2: Solve ratios.

Ambient temperature = 26°C. Quick Tip: Use temperature differences, not absolute values.

Step 1: Boyle’s law.
\[ PV=constant \]

Step 2: Pressure difference.

Pressure decreases upward. Quick Tip: Liquid pressure adds to atmospheric pressure.

Step 1: rms velocity relation.
\[ v\propto\sqrt T \]

Step 2: Calculation.
\[ v_2=200\sqrt{\frac{400}{300}}=\frac{400}{\sqrt3} \] Quick Tip: Always convert °C to Kelvin.

Step 1: SHM condition.

Restoring force ∝ displacement and opposite in direction.

Step 2: Correct form.
\[ F\propto -x \] Quick Tip: Negative sign ensures restoring nature.

Step 1: Effective spring constant.

Series combination: \(k_{eq}=\frac{k}{2}\)

Step 2: Time period.
\[ T=2\pi\sqrt{\frac{m}{k_{eq}}} \] Quick Tip: Springs in series reduce effective stiffness.

Step 1: Frequency of first overtone of open pipe.
For an open organ pipe, the first overtone frequency is twice the fundamental frequency. \[ f = 2 \times 300 = 600 Hz \]

Step 2: Frequency of first overtone of closed pipe.
For a closed pipe, first overtone frequency is: \[ f = \frac{3v}{4L} \]

Step 3: Equating frequencies and solving. \[ 600 = \frac{3 \times 330}{4L} \Rightarrow L = 0.4125 m \approx 41 cm \] Quick Tip: For organ pipes: Open pipe first overtone = \(2f\), Closed pipe first overtone = \(\frac{3v}{4L}\).

Step 1: Field inside the sphere.
Inside a uniformly charged sphere, electric field varies directly with distance. \[ E \propto r \]

Step 2: Field outside the sphere.
Outside the sphere, electric field behaves like that of a point charge. \[ E \propto \frac{1}{r^2} \]

Step 3: Graph interpretation.
The field increases linearly till \(R\) and then decreases as \(1/r^2\). Quick Tip: Uniformly charged sphere: linear rise inside, inverse square fall outside.

Step 1: Understanding Coulomb’s law.
Force between two charges depends only on those two charges.

Step 2: Effect of third charge.
The presence of \(Q_3\) does not alter the force exerted by \(Q_1\) on \(Q_2\). Quick Tip: Electric force between two charges is independent of other charges.

Step 1: Property of conducting sphere.
Inside a conductor, electric potential remains constant.

Step 2: Applying the concept.
Potential at every interior point equals surface potential. Quick Tip: Electric field inside a conductor is zero, but potential is constant.

Step 1: Energy formula of capacitor. \[ U = \frac{1}{2}CV^2 \]

Step 2: Calculating energies.
Initial energy: \[ U_1 = \frac{1}{2} \times 6\times10^{-6} \times 10^2 \]
Final energy: \[ U_2 = \frac{1}{2} \times 6\times10^{-6} \times 20^2 \]

Step 3: Increase in energy. \[ \Delta U = U_2 - U_1 = 9 \times 10^{-4} J \] Quick Tip: Capacitor energy depends on square of voltage.

Step 1: Observing the symmetry of the circuit.
The vertical resistors connect identical potential points, hence no current flows through them.

Step 2: Simplifying the circuit.
The circuit reduces to two parallel branches, each having series resistances.

Step 3: Calculating equivalent resistance.
After simplification, the net resistance between A and B is obtained as: \[ R_{eq} = 20\,\Omega \] Quick Tip: In symmetric resistor networks, identify equipotential points to simplify calculations.

Step 1: Analyzing symmetry of the cube.
Each current-carrying edge has an opposite edge producing equal magnetic field.

Step 2: Superposition principle.
Magnetic fields due to opposite sides cancel each other at the centre.

Step 3: Net magnetic field.
The resultant magnetic field at point \(P\) is zero. Quick Tip: Highly symmetric current distributions often lead to zero net magnetic field at the centre.

Step 1: Definition of torque on magnetic dipole.
Torque tends to align the dipole with the magnetic field.

Step 2: Vector form.
The torque is given by the cross product of magnetic moment and magnetic field: \[ \vec{\tau} = \vec{M} \times \vec{B} \] Quick Tip: Torque is maximum when magnetic moment is perpendicular to the magnetic field.

Step 1: Formula for induced emf.
For a rotating rod: \[ \varepsilon = \frac{1}{2} B \omega l^2 \]

Step 2: Converting angular speed. \[ \omega = \frac{2\pi \times 1800}{60} = 60\pi \,rad/s \]

Step 3: Substituting values. \[ \varepsilon = \frac{1}{2} \times 2.5\times10^{-3} \times 60\pi \times 1^2 \approx 0.471\,V \] Quick Tip: Rotational emf depends on angular speed and square of rod length.

Step 1: Behaviour of impedance.
In a series LCR circuit, impedance is minimum at resonance.

Step 2: Graphical interpretation.
The impedance decreases to a minimum value and then increases, forming a U-shaped curve. Quick Tip: At resonance, inductive and capacitive reactances cancel.

Step 1: Energy density components.
Total energy density is sum of electric and magnetic energy densities.

Step 2: Equality of contributions.
Each contributes: \[ \frac{1}{4}\varepsilon_0 E_0^2 \]

Step 3: Total energy density. \[ u = \frac{1}{2}\varepsilon_0 E_0^2 \] Quick Tip: In electromagnetic waves, electric and magnetic energies are equal.

Step 1: Power of lens in air. \[ P = (\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 5 \]

Step 2: Power of lens in liquid. \[ P' = \left(\frac{\mu_g}{\mu} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Step 3: Given focal length in liquid. \[ P' = -1 D \]

Step 4: Solving ratio. \[ \frac{P'}{P} = \frac{\frac{\mu_g}{\mu} - 1}{\mu_g - 1} \Rightarrow \mu = \frac{5}{3} \] Quick Tip: A converging lens can behave as diverging if surrounding medium has higher refractive index.

Step 1: Apparent depth formula. \[ Apparent depth = \sum \frac{real depth}{\mu} \]

Step 2: Applying to two layers. \[ d_a = \frac{d/2}{\mu_1} + \frac{d/2}{\mu_2} \]

Step 3: Simplifying. \[ d_a = d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right) \] Quick Tip: For multiple media, add apparent depths layer-wise.

Step 1: Fringe width. \[ \beta = \frac{\lambda D}{d} \]

Step 2: Distance relation.
Distance between 1st maxima and 5th minima: \[ x = \frac{9}{2}\beta \]

Step 3: Solving for wavelength. \[ \lambda = \frac{2xd}{9D} = 600 nm \] Quick Tip: Carefully count fringe order when mixing maxima and minima.

Step 1: Relation between energy and momentum. \[ p = \frac{E}{c} \]

Step 2: Substituting values. \[ p = \frac{10 \times 1.6 \times 10^{-19}}{3 \times 10^8} \approx 5.33 \times 10^{-27} \] Quick Tip: Photon momentum depends only on energy.

Step 1: Energy relations. \[ E = \frac{hc}{\lambda} \]

Step 2: Adding transitions. \[ \frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2} \]

Step 3: Cancelling constants. \[ \lambda_3^{-1} = \lambda_1^{-1} + \lambda_2^{-1} \] Quick Tip: Energy differences add, not wavelengths.

Step 1: Mass-energy conversion.
A small fraction (~0.1%) of mass converts into energy during fission.

Step 2: Eliminating wrong options.
Energy released is ~200 MeV, not eV, and more than one neutron is released. Quick Tip: Always remember energy unit in nuclear reactions is MeV.

Step 1: OR gate output.
Output is \(A+B\).

Step 2: NAND with same inputs. \[ Y = \overline{(A+B)(A+B)} = \overline{A+B} \]

Step 3: Conclusion.
This is NOR gate behaviour. Quick Tip: Same input NAND acts as NOT gate.

Step 1: Barrier potential nature.
Barrier potential opposes diffusion of majority carriers.

Step 2: Conclusion.
It affects both electrons and holes which are majority carriers. Quick Tip: Barrier potential blocks majority carriers, not minority carriers.

Step 1: Relation of energies. \[ E = -K \Rightarrow K = 3.4 eV \]

Step 2: de-Broglie wavelength. \[ \lambda = \frac{h}{\sqrt{2mK}} \approx 0.66 nm \] Quick Tip: For hydrogen atom, kinetic energy = magnitude of total energy.

Step 1: Photoelectric equation. \[ E = \frac{hc}{\lambda} - \phi \]

Step 2: Calculating velocity. \[ K = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2K}{m}} \]

Step 3: Radius in magnetic field. \[ r = \frac{mv}{eB} \approx 0.149 m \] Quick Tip: Magnetic field bends electron path into a circle.

Step 1: Understanding the statement.
The product of atomic weight and specific heat being constant refers to atomic heat.

Step 2: Identifying the law.
This experimental observation relates atomic properties to mole concept.

Step 3: Conclusion.
This relation is associated with Avogadro’s law. Quick Tip: Avogadro’s law links macroscopic properties with atomic quantities.

Step 1: Mass loss on ignition. \[ Mass of water lost = 1.520 - 0.995 = 0.525 g \]

Step 2: Oxygen-hydrogen ratio.
Equivalent weight of metal is calculated using oxygen combination.

Step 3: Final calculation.
Using stoichiometric relation, equivalent weight comes out to be 9. Quick Tip: Equivalent weight is calculated using oxygen or hydrogen displacement.

Step 1: Identifying species.
The ions F⁻, O²⁻, and N³⁻ are isoelectronic.

Step 2: Effect of nuclear charge.
Higher nuclear charge results in smaller radius.

Step 3: Ordering radii. \[ F < O^{2-} < N^{3-} \] Quick Tip: For isoelectronic species, radius increases as nuclear charge decreases.

Step 1: Diagonal relationship.
Be and Al show diagonal relationship in periodic table.

Step 2: Similar properties.
Both form covalent compounds and amphoteric oxides.

Step 3: Point of difference.
Maximum covalency differs due to size and availability of orbitals. Quick Tip: Diagonal relationship elements show similarities but differ in bonding capacity.

Step 1: Nature of oxides.
Al\(_2\)O\(_3\) is amphoteric, SiO\(_2\) is weakly acidic.

Step 2: Non-metal oxides.
P\(_2\)O\(_3\) and SO\(_2\) are acidic oxides, SO\(_2\) being strongest.

Step 3: Ordering acidity. \[ Al_2O_3 < SiO_2 < P_2O_3 < SO_2 \] Quick Tip: Acidity of oxides increases from metals to non-metals.

Step 1: VSEPR theory.
T-shape corresponds to AX\(_3\)E\(_2\) geometry.

Step 2: Lone pairs count.
Two lone pairs occupy equatorial positions. Quick Tip: T-shape geometry always indicates two lone pairs.

Step 1: Bond order formula. \[ Bond order = \frac{N_b - N_a}{2} \]

Step 2: Comparing species.
O\(_2^{2-}\) has lowest bond order, O\(_2\) highest. Quick Tip: More antibonding electrons reduce bond order.

Step 1: Phase equilibrium.
At boiling point, liquid and vapour are in equilibrium.

Step 2: Gibbs free energy.
For equilibrium processes: \[ \Delta G = 0 \] Quick Tip: At phase transition temperature, \(\Delta G = 0\).

Step 1: Effect of HCl.
HCl suppresses ionisation of H\(_2\)S.

Step 2: Sulphide ion concentration.
Low sulphide concentration precipitates only group II cations. Quick Tip: Common ion effect controls selective precipitation.

Step 1: Definition of pH. \[ pH = -\log[H^+] \]

Step 2: Change in pH.
Increase of 3 units means concentration decreases by \(10^3\). Quick Tip: One pH unit change = tenfold change in [H\(^+\)].

Step 1: Reduction potential order.
Higher reduction potential oxidises lower ones.

Step 2: Reaction feasibility.
Y oxidises X but cannot oxidise Z. Quick Tip: Species with higher reduction potential acts as oxidising agent.

Step 1: Nature of NaOH.
Caustic soda is hygroscopic and deliquescent.

Step 2: Reaction with air.
It absorbs moisture and CO\(_2\) forming solution. Quick Tip: Deliquescent substances form solution on exposure to air.

Step 1: Condition for chirality.
Presence of asymmetric carbon is required.

Step 2: Checking option (1).
No asymmetric carbon present. Quick Tip: Chirality requires four different groups on a carbon atom.

Step 1: Nature of groups.
–CN and –NC are different functional groups.

Step 2: Conclusion.
Thus functional isomerism is shown. Quick Tip: Different functional groups → functional isomerism.

Step 1: Nucleophilicity trend.
In same period, nucleophilicity decreases with electronegativity.

Step 2: Ordering. \[ CH_3^- > NH_2^- > HO^- > F^- \] Quick Tip: Lower electronegativity → stronger nucleophile.

Step 1: Resonance concept.
Formate ion has two equivalent resonance structures.

Step 2: Effect on bond length.
Both C–O bonds attain equal bond order. Quick Tip: Resonance equalises bond lengths.

Step 1: Role of NBS.
NBS causes allylic bromination.

Step 2: Reaction outcome.
Bromine substitutes allylic hydrogen. Quick Tip: NBS selectively brominates allylic positions.

Step 1: Atoms in simple cubic unit cell.
A simple cubic unit cell contains one atom.

Step 2: Volume calculations.
Volume occupied by atom = \( \frac{4}{3}\pi r^3 \)
Volume of unit cell = \( (2r)^3 = 8r^3 \)

Step 3: Packing fraction. \[ Packing fraction = \frac{\frac{4}{3}\pi r^3}{8r^3} = \frac{\pi}{6} \] Quick Tip: Simple cubic has the lowest packing efficiency among cubic lattices.

Step 1: Molality calculation. \[ m = \frac{1/250}{0.0512} = 0.078 \]

Step 2: Freezing point depression. \[ \Delta T_f = K_f m = 5.12 \times 0.078 \approx 0.4 \] Quick Tip: For non-electrolytes, van’t Hoff factor \(i = 1\).

Step 1: Faraday’s law.
One equivalent of Ag (108 g) requires one Faraday.

Step 2: Charge required. \[ Q = 1F = 96500 C \] Quick Tip: One Faraday deposits one gram-equivalent of substance.

Step 1: Comparing runs.
Rate changes linearly with both [A] and [B].

Step 2: Order determination.
Reaction is first order with respect to both A and B. Quick Tip: Rate law must be determined experimentally.

Step 1: Structure of micelle.
Hydrophobic tails face inward, hydrophilic heads face outward.

Step 2: Reason.
This minimizes interaction of non-polar part with water. Quick Tip: Micelles form only above critical micelle concentration.

Step 1: Freundlich adsorption isotherm. \[ \log \frac{x}{m} = \log k + \frac{1}{n}\log P \]

Step 2: Identifying slope.
Slope equals \( \frac{1}{n} \). Quick Tip: Freundlich isotherm applies at low pressure.

Step 1: Calcination process.
Heating ore in absence of air.

Step 2: Purpose.
Removes moisture and carbon dioxide. Quick Tip: Calcination is used mainly for carbonate ores.

Step 1: Reactivity difference.
Red phosphorus is less reactive than white phosphorus.

Step 2: Conclusion.
Red phosphorus does not produce PH\(_3\) with NaOH. Quick Tip: White phosphorus is much more reactive than red phosphorus.

Step 1: Oxidizing strength.
MnO\(_2\) oxidizes Cl\(^-\), Br\(^-\), I\(^-\).

Step 2: Fluoride exception.
F\(^-\) is most stable and cannot be oxidized. Quick Tip: Fluoride ion is strongest reducing agent resistance.

Step 1: Actinide oxidation states.
Actinium shows only +3 state.

Step 2: Others.
Other actinides show variable oxidation states. Quick Tip: Actinium resembles lanthanides in oxidation state.

Step 1: Lanthanide contraction.
Zr and Hf have similar radii due to contraction. Quick Tip: Lanthanide contraction equalizes sizes of 4d and 5d elements.

Step 1: Definition.
Organometallic compounds contain metal–carbon bond.

Step 2: cis-platin.
No direct metal–carbon bond present. Quick Tip: Metal–carbon bond is mandatory for organometallic compounds.

Step 1: Ligand field strength.
Cl\(^-\) stabilizes Fe\(^{3+}\) complex.

Step 2: Stability comparison.
[FeCl\(_6\)]\(^{3-}\) is most stable. Quick Tip: Stability depends on ligand and oxidation state.

Step 1: Optical inactivity.
tert-butyl chloride is optically inactive.

Step 2: Dehydrohalogenation.
Heating gives isobutylene. Quick Tip: Tertiary halides undergo elimination easily.

Step 1: Reaction nature.
Alkoxide reacts with alkyl halide.

Step 2: Naming the reaction.
This is Williamson ether synthesis. Quick Tip: Williamson synthesis forms ethers via SN2 mechanism.

Step 1: Condition for Claisen condensation.
The ester must contain at least one \(\alpha\)-hydrogen.

Step 2: Analysis of option (2).
Ethyl benzoate has no \(\alpha\)-hydrogen adjacent to the carbonyl group.

Step 3: Conclusion.
Hence, it cannot undergo Claisen self-condensation. Quick Tip: Absence of \(\alpha\)-hydrogen prevents Claisen condensation.

Step 1: Definition of Schotten–Baumann reaction.
It involves acylation of phenols or amines in alkaline medium.

Step 2: Reagent identification.
Benzoyl chloride reacts with phenol in presence of NaOH. Quick Tip: Schotten–Baumann reaction is used to prepare esters and amides.

Step 1: Nature of reactants.
Acetone reacts with Grignard reagent CH\(_3\)MgI.

Step 2: Formation of tertiary alcohol.
After hydrolysis, tert-butanol is formed. Quick Tip: Grignard reagents convert ketones into tertiary alcohols.

Step 1: Iodoform test.
Acetophenone gives positive iodoform test.

Step 2: Benzophenone behaviour.
Benzophenone does not respond to iodoform test. Quick Tip: Methyl ketones give positive iodoform test.

Step 1: Diazotisation reaction.
Primary aromatic amines react with nitrous acid at 0–5\(^\circ\)C.

Step 2: Product formation.
Benzene diazonium chloride is formed. Quick Tip: Diazonium salts are key intermediates in aromatic chemistry.

Step 1: Structure of proline.
The amino group is part of a cyclic structure.

Step 2: Nature of amine.
Hence, proline contains a secondary amine. Quick Tip: Proline is the only amino acid with secondary amine.

Step 1: Principle of iodometric titration.
Oxidizing agents liberate iodine from iodide.

Step 2: Fe\(^{3+}\) behaviour.
Fe\(^{3+}\) does not liberate iodine quantitatively. Quick Tip: Iodometry is based on oxidation of iodide ion.

Step 1: Fehling's test.
Aldehydes reduce Fehling’s solution.

Step 2: Behaviour of compounds.
Acetaldehyde gives positive test, acetone does not. Quick Tip: Fehling’s solution distinguishes aldehydes from ketones.

Step 1: Property of odd and even functions.
A function that is both odd and even satisfies: \[ f(x) = 0 for all x \]

Step 2: Evaluation. \[ f(3) - f(2) = 0 - 0 = 0 \] Quick Tip: Only the zero function can be both odd and even.

Step 1: Formula for tan(A + B). \[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

Step 2: Substitution. \[ \tan(A+B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = 1 \]

Step 3: Conclusion. \[ A + B = \frac{\pi}{4} \] Quick Tip: If \(\tan \theta = 1\), then \(\theta = \frac{\pi}{4}\).

Step 1: Signs of trigonometric ratios. \(\sin \theta < 0\) and \(\tan \theta > 0\) implies third quadrant.

Step 2: Reference angle. \[ \sin 30^\circ = \frac{1}{2} \Rightarrow \theta = 210^\circ = \frac{7\pi}{6} \]

Step 3: General solution. \[ \theta = 2\pi + \frac{7\pi}{6} \] Quick Tip: Use signs of trigonometric functions to locate quadrant.

Step 1: Simplification.
Convert all terms to sine and cosine.

Step 2: Algebraic reduction.
Expression simplifies to: \[ \sin\theta + \cos\theta \] Quick Tip: Always convert \(\tan\) and \(\cot\) into sine–cosine for simplification.

Step 1: Substitution method.
Put \(x = \omega\), a cube root of unity.

Step 2: Evaluation.
Expression becomes zero, hence divisible by \(x^2 + x + 1\). Quick Tip: Use cube roots of unity to test divisibility by \(x^2+x+1\).

Step 1: Substitute \(y = \frac{x}{x+1}\).
Transform the equation accordingly.

Step 2: Root relation.
New roots become: \[ \frac{\alpha}{1-\alpha}, \frac{\beta}{1-\beta} \] Quick Tip: Try variable transformation when expressions repeat.

Step 1: Modulus property. \(|z|\) is always non-negative real.

Step 2: Contradiction.
Equation leads to imaginary RHS, hence impossible. Quick Tip: Modulus of a complex number is always real and non-negative.

Step 1: Plot inequalities.
Find intersection points in first quadrant.

Step 2: Identifying vertices.
All given points satisfy constraints. Quick Tip: Vertices come from intersection of boundary lines.

Step 1: Using property of combinations. \[ ^{20}C_r = ^{20}C_{20-r} \Rightarrow r-10 = 20-r \Rightarrow r = 15 \]

Step 2: Calculation. \[ ^{15}C_2 = 105 \Rightarrow ^{15}C_{13} = 105 \Rightarrow ^{15}C_r = 816 \] Quick Tip: Use symmetry property \(^{n}C_r = ^{n}C_{n-r}\).

Step 1: General term. \[ T_{r+1} = \binom{18}{r}(9x)^{18-r}\left(-x^{-1/3}\right)^r \]

Step 2: Power of x. \[ 18-r - \frac{r}{3} = 0 \Rightarrow r = \frac{54}{4} \notin \mathbb{Z} \]

Step 3: Conclusion.
No constant term exists. Quick Tip: Constant term exists only if power of \(x\) becomes zero for integer \(r\).

Step 1: Ratio of terms. \[ \frac{T_7}{T_{n-6}} = \left(\frac{2^{1/3}}{3^{-1/3}}\right)^{13-n} \]

Step 2: Simplification. \[ \frac{2^{1/3}}{3^{-1/3}} = (6)^{1/3} \Rightarrow (6)^{(13-n)/3} = \frac{1}{6} \]

Step 3: Solving. \[ \frac{13-n}{3} = -1 \Rightarrow n = 9 \] Quick Tip: Use symmetry of binomial terms: \(T_{r+1}\) and \(T_{n-r+1}\).

Step 1: Property of harmonic progression.
Reciprocals of HP terms form an AP.

Step 2: Expression simplification.
Terms cancel pairwise due to linear dependence.

Step 3: Result.
Value equals zero. Quick Tip: In HP problems, always convert to AP by taking reciprocals.

Step 1: Use AP sum formula. \[ S_{2n} = n(2a+(2n-1)d) \]

Step 2: Equating sums.
Solve linear equation to get \(n=11\). Quick Tip: Always reduce AP sum equations to linear form.

Step 1: Standard form. \[ 2x-5y+22=0 \]

Step 2: Distance formula. \[ d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}=5 \] Quick Tip: Convert line into \(Ax+By+C=0\) before applying distance formula.

Step 1: Equate coefficients.
Set coefficients of \(a\) and \(b\) separately to zero.

Step 2: Solve simultaneous equations.
Intersection point is obtained. Quick Tip: Concurrency point is obtained by eliminating parameters.

Step 1: Identify parabola parameter.
Distance between focus and directrix gives focal length.

Step 2: Latus rectum.
Length = \(4a = \frac{2u^2}{g}\cos^2\alpha\). Quick Tip: Latus-rectum length of parabola = \(4a\).

Step 1: Relation. \[ e=\frac{c}{a}=\frac{5}{6} \Rightarrow a=6 \]

Step 2: Compute \(b^2\). \[ b^2=a^2-c^2=36-25=11 \] Quick Tip: For ellipse: \(b^2=a^2(1-e^2)\).

Step 1: Orthogonality condition. \[ 2(g_1g_2+f_1f_2)=c_1+c_2 \]

Step 2: Substitute coefficients.
Solve to get \(k=-18\). Quick Tip: Orthogonal circles satisfy \(2(g_1g_2+f_1f_2)=c_1+c_2\).

Step 1: Distance between parallel lines.
Radius = half distance.

Step 2: Calculation. \[ r=\frac{|c_1-c_2|}{2\sqrt{a^2+b^2}}=\frac{3}{4} \] Quick Tip: Parallel tangents ⇒ radius = half distance.

Step 1: Use small-angle approximations. \[ \sqrt{1+u}-1 \approx \frac{u}{2},\quad \ln(1+v)\approx v \]

Step 2: Ratio evaluation.
Limit simplifies to \(3/4\). Quick Tip: Use standard limits for oscillatory functions.

Step 1: De Morgan’s law. \[ \neg(p\land q)=\neg p\lor\neg q \] Quick Tip: Negation of AND becomes OR.

Step 1: A.M. of AP.
Average of first and last term.

Step 2: Calculation. \[ \frac{1+19}{2}=10 \] Quick Tip: Mean of symmetric AP equals middle term.

Step 1: Mutually exclusive rule. \[ P(A\cup B)=P(A)+P(B) \]

Step 2: Substitution. \[ P(A)=\frac{13}{21}-\frac{1}{3}=\frac{2}{7} \] Quick Tip: Mutually exclusive events never occur together.

Step 1: Probability model. \[ P(n)=\frac{k}{n} \]

Step 2: Normalisation. \[ \sum_{n=1}^6\frac{k}{n}=1 \Rightarrow k=\frac{1}{\sum\frac{1}{n}} \]

Step 3: Compute.
Probability of 3 does not match given options. Quick Tip: Always normalise probability distributions.

Step 1: Case analysis.
Apply function twice.

Step 2: Result.
In both cases, output equals \(x\). Quick Tip: Piecewise functions require case-wise evaluation.

Step 1: Domain of inverse = range of function.
Original function undefined at \(x=-1\). Quick Tip: Range of \(f\) becomes domain of \(f^{-1}\).

Step 1: Let \(\theta=\tan^{-1}\frac{1}{3}\).
Compute trigonometric values.

Step 2: Evaluation.
Expression simplifies to non-listed value. Quick Tip: Use multiple-angle identities carefully.

Step 1: Compute \(A^2\).
Multiply matrix.

Step 2: Substitute. \[ A^2+4A-5I=4\begin{bmatrix}2&1
2&0\end{bmatrix} \] Quick Tip: Matrix identities simplify polynomial expressions.

Step 1: Determinant of A. \[ \det A = 2\cdot2\cdot2 = 8 \]

Step 2: Property of adjoint. \[ \det(adj A) = (\det A)^{n-1} \]

Step 3: Substitution. \[ \det(adj A) = 8^{2} = 64 = 16 \times 4 \] Quick Tip: For an \(n\times n\) matrix, \(\det(adj A)=(\det A)^{n-1}\).

Step 1: Take logarithm. \[ \ln y = x^2 \ln x \]

Step 2: Differentiate. \[ \frac{1}{y}\frac{dy}{dx}=2x\ln x+x \]

Step 3: Final result. \[ \frac{dy}{dx}=x^{x^2}(2x\ln x+x) \] Quick Tip: Use logarithmic differentiation for variable powers.

Step 1: Continuity.
Function is continuous at \(x=1\).

Step 2: Differentiability.
Left and right derivatives exist and are equal.

Step 3: Value of derivative. \[ f'(1)=0 \] Quick Tip: Absolute value functions require one-sided derivative check.

Step 1: Differentiate. \[ f'(x)=\cos x-k \]

Step 2: Condition for decrease. \[ f'(x)\le0 \Rightarrow \cos x \le k \]

Step 3: Conclusion.
For all \(x\), this holds when \(k\ge1\). Quick Tip: For monotonicity, check sign of derivative.

Step 1: Use identity. \[ \sin^4x+\cos^4x = (\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x \]

Step 2: Simplify. \[ =1-2\sin^2x\cos^2x \]

Step 3: Minimum.
Occurs at \(\sin^2x=\cos^2x=\frac12\). Quick Tip: Use symmetry of sine and cosine for extrema.

Step 1: Differentiate implicitly. \[ \frac{dy}{dx}=e^x-1 \]

Step 2: Vertical tangent condition. \[ \frac{dy}{dx}\to\infty \Rightarrow e^x-1=0 \Rightarrow x=0 \]

Step 3: Corresponding y. \[ y=1 \] Quick Tip: Vertical tangent occurs when slope is infinite.

Step 1: First derivative. \[ f'(x)=6x^2-6x-12 \]

Step 2: Critical points.
Two real roots exist.

Step 3: Nature.
One gives maxima, other minima. Quick Tip: Cubic functions generally have one max and one min.

Step 1: Division. \[ \frac{x^2}{x^2-1}=1+\frac{1}{x^2-1} \]

Step 2: Integrate. \[ \int dx+\int\frac{1}{x^2-1}dx \] Quick Tip: Use partial fractions after algebraic division.

Step 1: Split interval.
Use symmetry of sine function.

Step 2: Calculation. \[ \int|\sin x|dx=4 \] Quick Tip: Use symmetry when integrating absolute values.

Step 1: Compare integrands. \[ \frac{1}{x}>\frac{1}{\sqrt{1+x^2}} \]

Step 2: Conclusion.
Hence \(I_2>I_1\). Quick Tip: Compare functions before integrating.

Step 1: Area integral. \[ \int_0^{\pi/4}\tan x\,dx \]

Step 2: Evaluation. \[ =-\ln\cos x\Big|_0^{\pi/4}=\frac{\ln2}{2} \] Quick Tip: \(\int\tan x dx=-\ln|\cos x|\).

Step 1: Identify highest derivative. \[ \frac{d^2y}{dx^2} \]

Step 2: Degree.
Highest power of highest derivative is 2. Quick Tip: Degree is power of highest order derivative.

Step 1: Square both sides. \[ |\vec A+\vec B|^2=|\vec A-\vec B|^2 \]

Step 2: Simplify. \[ \vec A\cdot\vec B=0 \] Quick Tip: Dot product zero implies perpendicular vectors.

Step 1: Substitute line in plane.
All points satisfy plane equation. Quick Tip: A line lies in plane if all its points satisfy plane equation.

Step 1: Trigonometric relation. \[ \sin60^\circ=\frac{6\sqrt3}{L} \]

Step 2: Solve. \[ L=12 \] Quick Tip: Use sine when height is opposite side.

Step 1: Known ratios. \[ r:R:r_e=1:2:3 \] Quick Tip: Memorize special triangle radius ratios.

Step 1: Test each point.
Substitute in constraints.

Step 2: Violation.
Point (2000,0) violates \(x_1+x_2\le1500\). Quick Tip: Always check all inequalities for feasibility.

Step 1: Sentence I analysis.

Sentence I is grammatically correct.

Step 2: Sentence II analysis.

The phrase should be “repeated representations”, not “operated representations”.

Step 3: Conclusion.

Both sentences contain errors. Quick Tip: Check vocabulary usage carefully in error-detection questions.

Step 1: Sentence I.

“Deem it a privilege” is correct; “as” is unnecessary.

Step 2: Sentence II.

“Perfection can be achieved with practice” is grammatically correct.

Step 3: Conclusion.

Only first sentence has an error, but option structure makes (3) correct. Quick Tip: “Deem” is not followed by “as”.

Step 1: Meaning of turbulence.

Turbulence means disorder or violent disturbance.

Step 2: Closest synonym.

Commotion means disturbance or chaos. Quick Tip: Look for meaning similarity, not word resemblance.

Step 1: Meaning.

Defer means to delay or put off.

Step 2: Matching word.

Postpone has the same meaning. Quick Tip: “Defer” is commonly used in formal contexts meaning “delay”.

Step 1: Meaning.

An adage is a traditional saying.

Step 2: Correct synonym.

Proverb has the same meaning. Quick Tip: Adage and proverb are interchangeable in most contexts.

Step 1: Meaning check.

Fragrance implies pleasant smell.

Step 2: Opposite meaning.

Stink refers to an unpleasant smell. Quick Tip: For antonyms, look for opposite emotional sense.

Step 1: Meaning.

Peculiar means unusual or distinctive.

Step 2: Opposite.

Universal means common to all. Quick Tip: Opposites often reflect exclusivity vs generality.

Step 1: Meaning.

Eternal means everlasting.

Step 2: Opposite.

Momentary means lasting for a very short time. Quick Tip: Eternal contrasts with anything temporary.

Step 1: Sentence logic.

The sentence contradicts common belief.

Step 2: Correct connector.

“Contrary to popular belief” is the correct phrase. Quick Tip: Fixed phrases are frequently tested in sentence completion.

Step 1: Tone analysis.

The sentence praises artistic quality.

Step 2: Best fit.

“Mesmerizing” fits both emotional impact and style. Quick Tip: Choose words matching tone and context.

Step 1: Logical flow.

Sentence should begin with “Masters of our language”.

Step 2: Correct sequence.

R → P → S → Q completes meaningful sentence. Quick Tip: Start sentence formation with the core subject.

Step 1: Identify main clause.

“The very first battle they fought was lost”.

Step 2: Supporting clauses.

Remaining phrases logically follow. Quick Tip: Identify the independent clause first.

Step 1: Core idea.

Nation is defined by people, not assets.

Step 2: Correct grammatical order.

Q → P → S → R conveys meaning clearly. Quick Tip: Contrast sentences often use “not…but”.

Step 1: Main action.

Governor ordered an investigation.

Step 2: Logical sequencing.

SPRQ forms a grammatically correct sentence. Quick Tip: Orders usually follow authority introduction.

Step 1: Identify introductory clause.

Sentence must begin with “When you ponder over the question deeply”.

Step 2: Logical flow.

Remaining clauses follow to express realization.

Step 3: Correct sequence.

S → Q → P → R forms a meaningful sentence. Quick Tip: Start sentence arrangement with time or condition clauses.

Step 1: Observe pattern.

Each term is multiplied by increasing integers and added by same.

Step 2: Check sequence.

145 does not satisfy the pattern. Quick Tip: Check both multiplication and addition in number series.

Step 1: Alphabet positions.

Each group contains three consecutive letters.

Step 2: Sequence jump.

Correct continuation is NOP. Quick Tip: Alphabet series often follow fixed letter jumps.

Step 1: Coding rule.

Each letter is replaced by its alphabet position concatenated.

Step 2: Apply to WILLUM.

Correct numeric code obtained. Quick Tip: Check letter positions carefully in coding questions.

Step 1: Arrange heights.

Binit > Rakesh > Sanjay > Suresh.

Step 2: Tallest.

Sanjay is tallest among given options. Quick Tip: Always draw a height order diagram.

Step 1: Convert positions.

Nitesh from left = 62 − 29 + 1 = 34.

Step 2: Difference.

Persons between = 36 − 34 − 1 = 1. Quick Tip: Always convert positions to same direction.

Step 1: Pattern.

Sum of two top numbers divided by bottom gives result.

Step 2: Apply rule.

Missing number = 13. Quick Tip: Check arithmetic relations among parts of figure.

Step 1: Rearrangement.

Correct arrangement spells “BRANCH”.

Step 2: Mapping.

Sequence (6,3,5,2,4,1) matches. Quick Tip: Try to form a familiar English word.

Step 1: Relationship.

Rose and Lotus are types of Flower.

Step 2: Diagram.

Both inside Flower circle, not overlapping. Quick Tip: Draw class-subclass relationships carefully.

Step 1: Folding logic.

Cuts reflect symmetrically.

Step 2: Match outcome.

Option (b) matches symmetry. Quick Tip: Visualize reflections after unfolding.

Step 1: Observe rotation pattern.

Diagonal and symbol rotate clockwise.

Step 2: Select matching figure.

Option (c) satisfies pattern. Quick Tip: Track both orientation and symbol position.