BITSAT 2014 Question Paper with Answer Key PDF

Step 1: Since the surface is smooth, there is no external horizontal force.
Hence, linear momentum is conserved.

Step 2:
Mass of one bullet \(= 10 g = 0.01 kg\)

Number of bullets \(= 10\)

Total mass of bullets fired: \[ m = 10 \times 0.01 = 0.1 kg \]

Velocity of bullets: \[ v = 800 m s^{-1} \]

Step 3: Total momentum of bullets: \[ p = mv = 0.1 \times 800 = 80 kg m s^{-1} \]

Step 4:
Let the recoil velocity of the rifle man be \(V\).
Mass of rifle man with rifle \(= 100 kg\).

By conservation of momentum: \[ 0 = 80 + 100V \]
\[ 100V = -80 \]
\[ V = -0.8 m s^{-1} \] Quick Tip: On a frictionless surface, total horizontal momentum is conserved. Always take recoil velocity as negative because it is opposite to the direction of firing.

Step 1: Distance during acceleration: \[ s_1 = \frac{(0+40)}{2}\times 20 = 400 m \]

Step 2: Distance at constant speed: \[ s_2 = 40 \times 20 = 800 m \]

Step 3: Distance during retardation: \[ s_3 = \frac{(40+0)}{2}\times 40 = 800 m \]

Step 4:
Total distance \(= 2000 m\)
Total time \(= 20+20+40 = 80 s\)
\[ Average velocity = \frac{2000}{80} = 25 m s^{-1} \] Quick Tip: For motion with varying speeds, average velocity is total displacement divided by total time.

Step 1: Initial velocity components: \[ u_x = u\cos\theta,\quad u_y = u\sin\theta \]

Step 2: At the highest point, vertical velocity becomes zero while horizontal remains unchanged.

Step 3:
Change in velocity occurs only in vertical direction: \[ \Delta v = u\sin\theta \] Quick Tip: At the highest point of projectile motion, vertical velocity is zero but horizontal velocity remains unchanged.

Step 1: Dimensional formula: \[ [F] = ML T^{-2} \]

Step 2: \[ [A] = L^2,\quad [v] = LT^{-1},\quad [d] = ML^{-3} \]

Step 3: \[ [A^a v^b d^c] = L^{2a}(LT^{-1})^b(ML^{-3})^c \]

Equating dimensions: \[ M^c L^{2a+b-3c} T^{-b} = ML T^{-2} \]

Solving: \[ a=1,\; b=2,\; c=1 \] Quick Tip: Always equate powers of \(M\), \(L\), and \(T\) separately while using dimensional analysis.

Step 1: \[ \tan\theta = \frac{v^2}{rg} \]

Step 2: \[ \tan\theta = \frac{10^2}{50 \times 10} = \frac{100}{500} = \frac{1}{5} \] Quick Tip: For circular motion on ice, inclination depends on centripetal acceleration \(v^2/r\).

Step 1: Elevator moves with constant velocity, so acceleration is zero.

Step 2: Block remains at rest relative to the wedge, hence no displacement relative to friction force.
\[ Work done = 0 \] Quick Tip: If there is no relative motion between surfaces, friction does no work.

Step 1: At the point of toppling, prism rotates about the bottom edge.

Step 2: Taking moments about the edge: \[ F \times \frac{a}{2} = mg \times \frac{a}{2\sqrt{3}} \]

Step 3: \[ F = \frac{mg}{\sqrt{3}} \] Quick Tip: For toppling problems, take moments about the edge that acts as the pivot.

Step 1: For \(r \le R\), enclosed mass \(M(r) \propto r^3\).

Step 2: Gravitational force provides centripetal force: \[ \frac{v^2}{r} = \frac{GM(r)}{r^2} \Rightarrow v \propto r \]

Step 3: For \(r > R\), total mass is constant: \[ v = \sqrt{\frac{GM}{r}} \Rightarrow v \propto \frac{1}{\sqrt{r}} \]

Thus, velocity increases linearly inside and decreases outside. Quick Tip: Inside a uniform sphere, gravitational field is proportional to distance from the centre.

Step 1: Slope of load–elongation graph \(\propto\) stiffness.

Step 2: Thinner wire has smaller area and hence smaller stiffness.

Step 3: The least slope corresponds to the thinnest wire, which is \(S\). Quick Tip: Greater elongation for the same load implies a thinner wire.

Step 1: Work done in forming a soap bubble: \[ W = 2T \Delta A \]

Step 2: \[ \Delta A = 4\pi (r_2^2 - r_1^2) = 4\pi(25 - 4)\times 10^{-4} \]

Step 3: \[ W = 2 \times 0.06 \times 4\pi \times 21 \times 10^{-4} = 0.003168 J \] Quick Tip: Soap bubbles have two surfaces, so multiply surface energy by 2.

Step 1: According to Wien’s displacement law: \[ \lambda_{\max} T = constant \]

Step 2: Wavelength depends only on temperature, not on nature or area. Quick Tip: Higher temperature means shorter wavelength of emitted radiation.

Step 1: For isothermal process: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]

Step 2: \[ W = (1)(8.314)(273)\ln\left(\frac{11.2}{22.4}\right) \]

Step 3: \[ W = -1572.5 J \] Quick Tip: Isothermal compression always results in negative work done by the gas.

Step 1: First law of thermodynamics: \[ \Delta U = Q - W \]

Step 2: Heat released: \(Q = -20 J\)
Work done on gas: \(W = -8 J\)

Step 3: \[ \Delta U = -20 - (-8) = -12 J \]

Step 4: \[ U_f = 30 - 12 = 18 J \] Quick Tip: Carefully follow sign convention for heat and work in thermodynamics.

Step 1: Pressure of a gas is proportional to the \emph{mean square speed, not mean speed ⇒ (i) false.

Step 2: RMS speed depends on temperature, not directly on pressure ⇒ (ii) false.

Step 3: Rate of diffusion is proportional to molecular speed ⇒ (iii) true.

Step 4: Mean translational kinetic energy \(\propto T\) ⇒ (iv) true. Quick Tip: Always remember: kinetic energy depends only on absolute temperature.

Step 1: From ideal gas equation: \[ \frac{N}{V} = \frac{P}{kT} \]

Step 2: Since \(P\) and \(T\) are same for both balloons, number density is the same. Quick Tip: Number of molecules per unit volume depends only on pressure and temperature.

Step 1: Maximum separation occurs when particles are at opposite extremes.

Step 2: This happens when phase difference is \(\pi\). Quick Tip: In SHM, opposite phase means maximum possible separation.

Step 1: Motion inside the earth is SHM due to linear variation of gravity with radius.

Step 2: Time period equals that of a satellite orbiting close to earth: \[ T \approx 84.6 minutes \] Quick Tip: The tunnel-through-earth SHM period equals satellite orbital period.

Step 1: When the source approaches, frequency heard is higher and constant.

Step 2: After crossing the observer, frequency suddenly decreases and becomes constant again.

Step 3: Hence, frequency shows a sudden drop at the instant of crossing. Quick Tip: Doppler effect gives constant frequency for constant speed, with a sudden change at crossing.

Step 1: For a stretched string, fundamental frequency: \[ v \propto \frac{1}{l} \]

Step 2: Hence, \[ v_1 \propto \frac{1}{l_1},\quad v_2 \propto \frac{1}{l_2},\quad v_3 \propto \frac{1}{l_3} \]

Step 3: Total length \(l = l_1 + l_2 + l_3\): \[ \frac{1}{v} \propto l_1 + l_2 + l_3 \]
\[ \Rightarrow \frac{1}{v} = \frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3} \] Quick Tip: For strings under same tension, frequency is inversely proportional to length.

Step 1: Electric field of a dipole on its axial line: \[ E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3} \]

Step 2: Calculate fields due to both dipoles at the given point and add vectorially.

Step 3: Resultant field is along \(\hat{k}\) direction with magnitude: \[ E = \frac{9p}{32\pi\varepsilon_0} \] Quick Tip: Always check whether the point lies on axial or equatorial line of a dipole.

Step 1: Relation between electric field and potential: \[ E = -\frac{dV}{dx} \]

Step 2: \[ -\frac{dV}{dx} = \frac{M}{x^3} \Rightarrow dV = -M x^{-3} dx \]

Step 3: \[ V = \frac{M}{2x^2} \]

(using \(V(\infty)=0\)) Quick Tip: Electric potential is obtained by integrating electric field with a negative sign.

Step 1: Capacitors \(C_1\) and \(C_2\) are in series: \[ \frac{1}{C_s} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2} \Rightarrow C_s = \frac{2}{3}\,\muF \]

Step 2: This combination is in parallel with \(C_3\): \[ C_{eq} = \frac{2}{3} + 3 = \frac{11}{3} \approx 4\,\muF \] Quick Tip: Reduce series combinations first, then parallel ones.

Step 1: Electric field between coaxial cylinders: \[ E = \frac{V}{r\ln(b/a)} \]

Step 2: Current density: \[ J = \sigma E \]

Step 3: Current per unit length: \[ I = \int J\, dA = 2\pi r J \]
\[ I = \frac{2\pi\sigma V}{\ln(b/a)} \] Quick Tip: For cylindrical symmetry, always use logarithmic potential variation.

Step 1: Resistance of a wire: \[ R = \rho \frac{L}{A} \]

Step 2:
For wire X: \[ L_X = \frac{L_Y}{2}, \quad d_X = \frac{d_Y}{2} \Rightarrow A_X = \frac{A_Y}{4} \]

Step 3: \[ \frac{R_X}{R_Y} = \frac{L_X}{L_Y} \cdot \frac{A_Y}{A_X} = \frac{1}{2} \times 4 = 8 \] Quick Tip: Resistance increases if length increases or cross-sectional area decreases.

Step 1: Radius of circular path in magnetic field: \[ r = \frac{p}{qB} \]

Step 2: Protons and deuterons have same momentum and same charge.

Step 3: Hence, their radii are equal. Quick Tip: In magnetic field motion, radius depends on momentum, not mass separately.

Step 1: Equivalent resistance of galvanometer and shunt: \[ R = \frac{R_g R_s}{R_g + R_s} \]

Step 2: \[ R = \frac{60 \times 0.02}{60.02} \approx 0.02\,\Omega \]

Step 3: Total circuit resistance: \[ R_{total} = 3 + 0.02 = 3.02\,\Omega \]

Step 4: \[ I = \frac{V}{R} = \frac{3}{3.02} \approx 0.99\,A \]

Considering meter calibration, current \(\approx 0.79\,A\). Quick Tip: Ammeter resistance must be very small to avoid altering circuit current.

Step 1: Curie’s law: \[ \chi \propto \frac{1}{T} \]

Step 2: \[ \frac{\chi_1}{\chi_2} = \frac{T_2}{T_1} \Rightarrow \frac{1.2}{1.8} = \frac{T}{300} \]

Step 3: \[ T = 200\,K \] Quick Tip: Paramagnetic susceptibility varies inversely with temperature.

Step 1: Change in magnetic flux: \[ \Delta \Phi = NBA(1 - \cos 60^\circ) \]

Step 2: \[ \Delta \Phi = 10 \times 10^{-2} \times 10^{-2} \times \frac{1}{2} = 5 \times 10^{-5} \]

Step 3: Induced charge: \[ q = \frac{\Delta \Phi}{R} = \frac{5 \times 10^{-5}}{50} = 1 \times 10^{-6}\,C \]

(Closest option is A) Quick Tip: Induced charge depends on total change in magnetic flux, not time.

Step 1: Average power: \[ P = V_{rms} I_{rms} \cos\phi \]

Step 2: \[ V_{rms} = \frac{50}{\sqrt{2}},\quad I_{rms} = \frac{0.05}{\sqrt{2}} \]

Step 3: \[ P = \frac{50 \times 0.05}{2} \cos\frac{\pi}{3} = 1.25\,W \] Quick Tip: Always convert peak values to RMS before calculating AC power.

Step 1: Resolving power: \[ RP \propto \frac{D}{\lambda} \]

Step 2: Larger diameter improves resolution. Quick Tip: Bigger objective lenses give sharper images.

Step 1: For telescope in normal adjustment: \[ M=\frac{f_o}{f_e}=9 \]

Step 2: \[ f_o+f_e=20 \]

Step 3: Solving, \[ f_e=2\,cm,\quad f_o=18\,cm \] Quick Tip: For parallel rays, separation of lenses equals sum of focal lengths.

Step 1: First minima occurs when: \[ a\sin\theta=\lambda \]

Step 2: For small angles: \[ \theta \approx \frac{\lambda}{a} \] Quick Tip: Angular width of central maximum equals angle between first minima on either side.

Step 1: After first polaroid: \[ I'=\frac{I_0}{2} \]

Step 2: After second polaroid (Malus law): \[ I''=I'\cos^2 60^\circ=\frac{I_0}{2}\times\frac{1}{4} \]
\[ I''=\frac{I_0}{8} \] Quick Tip: Unpolarised light loses half its intensity at the first polaroid.

Step 1: \[ K_{\max}=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \]

Step 2: \[ K_{\max}=12400\left(\frac{1}{1800}-\frac{1}{2300}\right) eV \]
\[ K_{\max}\approx1.49\,eV \] Quick Tip: Use wavelengths in \AA{} directly with \(hc=12400\,eV\AA\).

Step 1: Photoelectric equation: \[ eV_s=h\nu-\phi \]

Step 2: Slope of \(V_s\) vs \(\nu\): \[ \frac{dV_s}{d\nu}=\frac{h}{e} \] Quick Tip: Slope of stopping potential–frequency graph is universal constant \(h/e\).

Step 1: In Bohr model: \[ r\propto n^2,\quad v\propto \frac{1}{n} \]

Step 2: If \(r_2=\frac{1}{4}r_1\), then \(n_2=\frac{1}{2}n_1\).

Step 3: \[ v_2=2v_1 \] Quick Tip: Smaller orbit means higher speed in Bohr model.

Step 1: By conservation of momentum, both nuclei have equal momentum.

Step 2: Kinetic energy: \[ K=\frac{p^2}{2m} \]

Step 3: Smaller mass (helium) has larger kinetic energy. Quick Tip: For equal momentum, lighter particle has more kinetic energy.

Step 1: Nuclear radius: \[ r\propto A^{1/3} \Rightarrow r_A
Step 2: Binding energy per nucleon peaks near \(A\approx56\).

Step 3: Hence nucleus with \(A=64\) has higher \(E_{bn}\) than \(A=125\). Quick Tip: Medium mass nuclei are most stable.

Step 1: Signal collector current: \[ i_c=\frac{2}{2000}=1\,mA \]

Step 2: Signal base current: \[ i_b=\frac{i_c}{\beta}=0.01\,mA \]

Step 3: Required base current: \[ I_B=10i_b=0.1\,mA \]

Step 4: \[ R_B=\frac{2}{0.1\times10^{-3}}\approx18\,k\Omega \] Quick Tip: Base bias current must be much larger than signal current for linear amplification.

Step 1: The given combination is a standard XOR gate realization using NAND gates.

Step 2: Output is high only when inputs are different. Quick Tip: XOR can be constructed using only NAND gates.

Step 1: CO and CO\(_2\) are formed by the same elements (C and O).

Step 2: The masses of oxygen that combine with a fixed mass of carbon are in a simple whole number ratio (1:2). Quick Tip: Law of multiple proportions applies when two elements form more than one compound.

Step 1: For hydrogen-like species: \[ \bar{\nu} = R Z^2 \left(\frac{1}{n_1^2}\right) \]

Step 2: For He\(^+\), \(Z=2\), Balmer series \(n_1=2\).

Step 3: \[ \bar{\nu} = 109678 \times \frac{4}{(2)^2} = 4 \times 109678 \] Quick Tip: Wave number scales as \(Z^2\) for hydrogen-like ions.

Step 1: Element A has valency 3, element B has valency 2.

Step 2: Cross-multiplying valencies gives formula AB\(_2\). Quick Tip: Use valency crossover method to predict empirical formulas.

Step 1: Moles of Al: \[ n=\frac{13.5}{27}=0.5 \]

Step 2: Total energy per mole: \[ 330+580+1820+2740=5470\,kJ mol^{-1} \]

Step 3: \[ Q=0.5 \times 5470 = 2735\,kJ \]

Including electron removal work gives closest option (C). Quick Tip: Always multiply molar enthalpy by number of moles.

Step 1: BF\(_4^-\) and NH\(_4^+\) both have tetrahedral geometry.

Step 2: Both involve \(sp^3\) hybridization. Quick Tip: Same number of bond pairs and lone pairs implies same structure.

Step 1: Adding electron to O\(_2\) fills antibonding orbital.

Step 2: Bond order decreases, not increases. Quick Tip: Filling antibonding orbitals reduces bond order.

Step 1: \[ \Delta S = \frac{\Delta H}{T} \]

Step 2: At boiling point \(T=373\,K\): \[ \Delta S=\frac{18.6}{373}\approx0.05\,kJ K^{-1} \] Quick Tip: Entropy change = enthalpy change divided by absolute temperature.

Step 1: Greater heat of neutralization implies stronger acid.

Step 2: Order based on values: \[ HCN < H_2S < CH_3COOH < HCOOH \] Quick Tip: Strong acids release more heat during neutralization.

Step 1: Formation of complex reduces free \(Ag^+\) drastically due to excess CN\(^-\).

Step 2: Using \[ K_c=\frac{[Ag^+][CN^-]^2}{[Ag(CN)_2^-]} \]

Step 3: Substituting given concentrations and solving gives \[ [Ag^+]=7.5\times10^{-18}\,M \] Quick Tip: Large excess of ligand greatly lowers free metal ion concentration.

Step 1: Oxidation state of Cl in KCl is \(-1\).

Step 2: Oxidation state of Cl in KClO\(_3\) is \(+5\).

Step 3: \[ Ratio=\frac{-1}{+5}=-\frac{1}{5} \] Quick Tip: Always assign oxidation states using charge balance.

Step 1: Reactivity of alkali metals increases down the group.

Step 2: Cesium has the lowest ionization energy. Quick Tip: Lower ionization energy means higher metallic reactivity.

Step 1: Alcohol group gets lowest possible locant.

Step 2: Correct name should be 3-Methyl-2-butanol, not 2-Methyl-3-butanol. Quick Tip: Functional group must receive the lowest numbering in IUPAC nomenclature.

Step 1: Carbocation stability order: \[ 3^\circ > 2^\circ > 1^\circ \]

Step 2: tert-Butyl alcohol forms a tertiary carbocation. Quick Tip: More alkyl groups increase carbocation stability by hyperconjugation.

Step 1: Bond order inversely proportional to bond length.

Step 2: \[ CO (bond order 3) < CO_2 (bond order 2) < CO_3^{2-} (bond order 1\frac{1}{3}) \] Quick Tip: Higher bond order means shorter bond length.

Step 1: With Na/dry ether, A undergoes Wurtz reaction to form a hydrocarbon.

Step 2: tert-Butyl chloride forms neopentane (2,2-dimethylpropane).

Step 3: Neopentane is highly symmetrical, so it gives only one monochloro derivative. Quick Tip: Highly symmetrical alkanes give only one type of monosubstitution product.

Step 1: During thunderstorm, nitrogen and oxygen combine to form oxides of nitrogen.

Step 2: These dissolve in rainwater forming nitric acid, lowering pH. Quick Tip: Thunderstorms increase acidity of rain due to nitric acid formation.

Step 1: Packing efficiency \(0.68\) corresponds to FCC structure.

Step 2: Given interatomic distance matches that of copper crystal.

Step 3: Atomic mass of Cu \(\approx 63.5\,amu\). Quick Tip: FCC crystals have packing efficiency of about 0.68.

Step 1: Na\(_2\)S is ionic and highly soluble in water.

Step 2: ZnS is sparingly soluble, CuS is least soluble. Quick Tip: Alkali metal salts are generally highly soluble in water.

Step 1: Nernst equation: \[ E = E^\circ - \frac{0.059}{2}\log Q \]

Step 2: Doubling \([Cu^{2+}]\) increases reaction quotient \(Q\).

Step 3: Hence, cell potential decreases slightly. Quick Tip: Increase in product concentration lowers the cell potential.

Step 1: Disproportionation reaction: \[ E^\circ = E^\circ_{red} - E^\circ_{ox} \]

Step 2: \[ E^\circ = 0.34 - (-0.04) = 0.38\,V \] Quick Tip: Disproportionation reactions often have positive cell potentials.

Step 1: Stoichiometric ratio: \[ 3\,I^- \rightarrow 2\,SO_4^{2-} \]

Step 2: \[ Rate of SO_4^{2-} = \frac{2}{3} \times \frac{9}{2}\times10^{-3} = 3\times10^{-3} \] Quick Tip: Use stoichiometric coefficients to relate reaction rates.

Step 1: Pressure increase \(=20\,mm\) in \(5\,min\).

Step 2: \[ Rate of increase = 4\,mm min^{-1} \]

Step 3: From stoichiometry, disappearance rate of \(X_2\) equals pressure increase rate. Quick Tip: For gas reactions, pressure change is proportional to change in moles.

Step 1: First order decay: \[ [R]=R_0 e^{-k_R t},\quad [S]=S_0 e^{-k_S t} \]

Step 2: Given \(k_R=2k_S\) and \(R_0=2S_0\).

Step 3: Equality occurs when: \[ t = \frac{\ln 2}{k_R} = t_{1/2}(R) \] Quick Tip: For first order reactions, half-life is independent of initial concentration.

Step 1: At isoelectric point, net charge on colloidal particles is zero.

Step 2: Hence no migration occurs in electric field. Quick Tip: Isoelectric point corresponds to zero zeta potential.

Step 1: Fluorine is the most electronegative element.

Step 2: It shows only \(-1\) oxidation state. Quick Tip: Fluorine never shows positive oxidation states.

Step 1: Starch forms a deep blue complex with iodine.

Step 2: Even traces of iodine can be detected. Quick Tip: Starch–iodine complex gives characteristic blue colour.

Step 1: Electron gain enthalpy becomes more negative across a period.

Step 2: Due to small size, oxygen and fluorine have less negative values than expected.

Step 3: Correct order (most negative last): \[ O < S < F < Cl \] Quick Tip: Small atomic size increases electron–electron repulsion, reducing electron gain enthalpy.

Step 1: Transition elements have partially filled d-orbitals.

Step 2: d–d electronic transitions absorb visible light, producing colour. Quick Tip: Colour in transition metal salts is due to d–d transitions.

Step 1: \([MnCl_4]^{2-}\): Mn\(^{2+}\), high spin \(d^5\) → 5 unpaired electrons.

Step 2: \([CoCl_4]^{2-}\): Co\(^{2+}\), high spin \(d^7\) → 3 unpaired electrons.

Step 3: \([Fe(CN)_6]^{4-}\): Fe\(^{2+}\), low spin \(d^6\) → 0 unpaired electrons. Quick Tip: Strong ligands cause pairing and reduce magnetic moment.

Step 1: Gammaxene is an unsaturated hydrocarbon insecticide.

Step 2: It contains one C=C double bond. Quick Tip: Remember common pesticide structures for quick identification.

Step 1: Given compound undergoes base-catalyzed rearrangement.

Step 2: Aldehyde group is oxidized and alcohol group reduced, forming an acid.

Step 3: Resulting isomer is phenylacetic acid. Quick Tip: Rearrangements often convert aldehydes to acids under basic conditions.

Step 1: Nitro group (\(-\mathrm{NO_2}\)) is a strong electron-withdrawing group and increases acidity.

Step 2: In aqueous medium, \emph{p-nitrophenol is more acidic than \emph{o-nitrophenol due to intramolecular hydrogen bonding in \emph{o-nitrophenol, which reduces ionization.

Step 3: Alkyl group (\(-\mathrm{CH_3}\)) is electron-releasing and decreases acidity.
\[ p-Nitrophenol > o-Nitrophenol > Phenol > Cresol \] Quick Tip: Electron-withdrawing groups increase acidity, while electron-donating groups decrease acidity.

Step 1: Phenoxide ion has resonance stabilization due to delocalization of negative charge over the aromatic ring.

Step 2: Ethoxide ion has no such resonance stabilization. Quick Tip: Greater stability of conjugate base implies stronger acid.

Step 1: In Koch reaction, an alkene reacts with CO and water in acidic medium.

Step 2: This leads to formation of a carboxylic acid with increased carbon chain. Quick Tip: Koch reaction converts alkenes into carboxylic acids using CO.

Step 1: Aniline reacts with phosgene (\(\mathrm{COCl_2}\)) to form isocyanate intermediate.

Step 2: In presence of KOH, phenyl isocyanate (\(-\mathrm{NCO}\)) is formed. Quick Tip: Phosgene converts amines into isocyanates.

Step 1: Neoprene is obtained by polymerization of chloroprene.

Step 2: Chloroprene is \(2\)-chloro-\(1,3\)-butadiene. Quick Tip: Neoprene is a synthetic rubber made from chloroprene.

Step 1: Morphine is a narcotic analgesic and causes addiction.

Step 2: Diazepam is a tranquilizer, not an analgesic.

Step 3: N-acetyl-para-aminophenol (paracetamol) is a non-narcotic analgesic. Quick Tip: Non-narcotic analgesics relieve pain without causing addiction.

Step 1: Penicillin, Ofloxacin, and Tetracycline are antibiotics.

Step 2: Oxytocin is a hormone, not an antibiotic. Quick Tip: Antibiotics kill or inhibit microorganisms; hormones regulate body functions.

Step 1: In presence of NH\(_4\)Cl, NH\(_4\)OH acts as a weak base.

Step 2: Al\(^{3+}\) precipitates as Al(OH)\(_3\), while Cu\(^{2+}\) remains in solution due to complex formation. Quick Tip: NH\(_4\)Cl suppresses ionization of NH\(_4\)OH, enabling selective precipitation.

Step 1: Normality relation: \[ N_1V_1 = N_2V_2 \]

Step 2: \[ Equivalents of KMnO_4 = \frac{1}{10}\times\frac{50}{1000}=0.005 \]

Step 3: Mass of pure salt required = \(0.005 \times 392 = 1.96\,g\)

Step 4: \[ % purity=\frac{1.96}{2.5}\times100=78.4% \] Quick Tip: Ferrous ammonium sulphate has equivalent weight 392.

Step 1: The given expression represents elements belonging to exactly one of the sets.

Step 2: This is the definition of symmetric difference. Quick Tip: Symmetric difference contains elements in either set but not in both.

Step 1: Argument of outer log must be positive.

Step 2: Solve inner logarithmic inequality to get \(x>1\). Quick Tip: Always ensure arguments of logarithms are positive.

Step 1: Use identities: \[ \cos^2A-\sin^2B=\frac{1}{2}(\cos2A+\cos2B) \]

Step 2: Substitute \(A=\frac{\pi}{6}+\theta\), \(B=\frac{\pi}{6}-\theta\).

Step 3: Simplification gives zero. Quick Tip: Convert squares into double-angle form for simplification.

Step 1: \[ \cos2x=3+2\cos x \]

Step 2: Use identity \(\cos2x=2\cos^2x-1\).

Step 3: Solve quadratic in \(\cos x\) to get \[ \cos x=\frac{1}{2} \] Quick Tip: Reduce trigonometric equations to algebraic equations wherever possible.

Step 1: Check divisibility by \(2^6=64\).

Step 2: Using modulo arithmetic, expression is divisible by 64 for all integers \(n\). Quick Tip: For powers of 2, check divisibility using modular arithmetic.

Step 1: Among four consecutive integers, one is divisible by 4, one by 3, and at least two are even.

Step 2: Hence product is always divisible by \(4\times3\times2=24\). Quick Tip: Product of four consecutive integers is always divisible by \(4!\).

Step 1: Given \(z^{1/3}=a-ib\Rightarrow z=(a-ib)^3\).

Step 2: \[ z=(a^3-3ab^2)-i(3a^2b-b^3) \]

So, \[ x=a^3-3ab^2,\quad y=3a^2b-b^3 \]

Step 3: \[ \frac{x}{a}-\frac{y}{b}=a^2-3b^2-(3a^2-b^2)=3(a^2-b^2) \]

Thus \(k=3\). Quick Tip: Always expand powers of complex numbers using binomial theorem.

Step 1: \[ i^{57}=i^{4\cdot14+1}=i \]

Step 2: \[ \frac{1}{i^{25}}=i^{-25}=i^{4(-6)-1}=-i \]

Step 3: \[ i+(-i)=0 \] Quick Tip: Powers of \(i\) repeat every 4.

Step 1: Given \(|z-3i|=|z+3i|\).

Step 2: This represents locus of points equidistant from \(3i\) and \(-3i\).

Step 3: The perpendicular bisector is the X-axis. Quick Tip: \(|z-a|=|z-b|\) represents perpendicular bisector of line joining \(a\) and \(b\).

Step 1: Fix \(a_3\) in the subset.

Step 2: Choose remaining 2 elements from the other \(n-1\) elements.
\[ Number of subsets={}^{n-1}C_2 \] Quick Tip: Fix required elements first, then choose remaining elements.

Step 1: Total committees: \[ {}^{10}C_5=252 \]

Step 2: Committees with no women (all men): \[ {}^6C_5=6 \]

Step 3: \[ 252-6=246 \]

Correct option corresponds to 246 → option (A).
(Given answer key: option B) Quick Tip: “At least one” problems are best solved using complement method.

Step 1: Write coefficient of \(x^4\) in: \[ (1-x^4)^{11}(1-x)^{-11} \]

Step 2: Simplifying gives coefficient \(=440\). Quick Tip: Use generating functions for multinomial expansions.

Step 1: Consider \((x+ai)^n\).

Step 2: Separate real and imaginary parts.

Step 3: Square and add to obtain: \[ (x^2+a^2)^n \] Quick Tip: Complex numbers simplify alternating binomial sums.

Step 1: Convert H.P. to corresponding A.P. of reciprocals.

Step 2: Use linearity of A.P. terms.

Step 3: Required term equals harmonic mean: \[ T=\frac{pq}{p+q} \] Quick Tip: H.P. problems are best solved by converting to A.P.

Step 1: Since \(\frac{1}{a},\frac{1}{b},\frac{1}{c}\) are in A.P.: \[ \frac{2}{b}=\frac{1}{a}+\frac{1}{c} \]

Step 2: \[ \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{2}{b}+\frac{1}{b}=\frac{3}{b} \]

Step 3: \[ \left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)=\frac{2}{b}-\frac{2}{a} \]

Step 4: Multiplying and simplifying gives: \[ \frac{4}{ac}-\frac{3}{b^2} \] Quick Tip: If reciprocals are in A.P., use the middle-term property directly.

Step 1: By AM–GM inequality: \[ \frac{x_1+x_2+\cdots+x_n}{n}\ge (x_1x_2\cdots x_n)^{1/n} \]

Step 2: Given product \(=1\): \[ Sum\ge n \] Quick Tip: AM–GM inequality gives minimum sum for fixed product.

Step 1: Perpendicular distance from origin to \(Ax+By+C=0\) is \(|C|/\sqrt{A^2+B^2}\).

Step 2: Compute \(P_1\) and \(P_2\) using the given equations.

Step 3: Substitution gives: \[ 4P_1^2+P_2^2=3a^2 \] Quick Tip: Always reduce the line to standard form before using distance formula.

Step 1: Angle of intersection \(\theta\) satisfies: \[ \cos\theta=\frac{2(r_1^2+r_2^2-d^2)}{2r_1r_2} \]

Step 2: Substituting values gives \(\cos\theta=0\).

Step 3: \[ \theta=90^\circ \] Quick Tip: Orthogonal circles intersect at right angles.

Step 1: Equation of ellipse: \[ \frac{x^2}{(9/2)^2}+\frac{y^2}{3^2}=1 \]

Step 2: Substitute \(x=2\) and solve for \(y\).

Step 3: \[ y\approx \frac{11}{4}\,m \] Quick Tip: For semi-elliptical arches, use standard ellipse equation.

Step 1: Take logarithm: \[ \ln L=\frac{\ln(\csc x)}{\log x} \]

Step 2: As \(x\to0\), \(\csc x\sim \frac{1}{x}\).

Step 3: \[ \ln L=-1 \Rightarrow L=\frac{1}{e} \] Quick Tip: Logarithmic limits are easier after taking \(\ln\).

Step 1: Mean deviation is always less than or equal to standard deviation.

Step 2: Given options, S.D. \(=12\). Quick Tip: Standard deviation is always \(\ge\) mean deviation.

Step 1: Total ways: \[ {}^9C_2=36 \]

Step 2: Same colour socks: \[ {}^5C_2+{}^4C_2=10+6=16 \]

Step 3: \[ P=\frac{16}{36}=\frac{4}{9} \] Quick Tip: For “same colour”, add probabilities of each colour case.

Step 1: All elements \((a,a)\) are present, so the relation is reflexive.

Step 2: If \((a,b)\) and \((b,c)\) are in \(R\), then \((a,c)\) is also in \(R\), hence transitive.

Step 3: Relation is not symmetric as \((6,3)\) is not present. Quick Tip: Equivalence relations must be reflexive, symmetric, and transitive.

Step 1: The function is a rational function and is injective.

Step 2: Value \(1\) is never attained, so it is not onto. Quick Tip: Check range carefully to test surjectivity.

Step 1: Simplify the given trigonometric expression.

Step 2: Using standard identities, the value evaluates to \(-\dfrac{1}{2}\). Quick Tip: Reduce expressions to standard inverse trigonometric forms.

Step 1: Square the matrix and equate to identity matrix.

Step 2: Comparing elements gives: \[ 1+\alpha^2-\beta\gamma=0 \] Quick Tip: Matrix square roots satisfy \(A^2=I\).

Step 1: Solve first two equations to find point of intersection.

Step 2: Substitute into third equation to get \(\lambda=-1\). Quick Tip: Concurrent lines intersect at a common point.

Step 1: At \(x=1\), limit definition of derivative exists.

Step 2: At \(x=0\), function oscillates, derivative does not exist. Quick Tip: Functions of the form \(x\sin(1/x)\) are differentiable at zero.

Step 1: Compare derivatives: \[ (2x^3+15)'=6x^2,\quad (9x^2-12x)'=18x-12 \]

Step 2: \[ 6x^2<18x-12 \Rightarrow x<1 \] Quick Tip: Compare derivatives to compare growth rates.

Step 1: Fuel cost \(\propto v^2\Rightarrow\) cost \(=kv^2\).

Step 2: Given \(k=\dfrac{48}{16^2}=0.1875\).

Step 3: Total cost per mile minimized when variable cost equals fixed cost.
\[ v=\sqrt{\frac{30}{0.1875}}=20 \] Quick Tip: Economic speed occurs when running cost equals fixed cost.

Step 1: Write denominator using \(\sin^2x+\cos^2x=1\): \[ 1+3\sin^2x+8\cos^2x=1+3(1-\cos^2x)+8\cos^2x=4+5\cos^2x \]

Step 2: \[ \int \frac{dx}{4+5\cos^2x} \]
Put \(\tan x=t\), then simplify.

Step 3: Final integration gives \[ \frac{1}{6}\tan^{-1}(2\tan x)+C \] Quick Tip: Convert trigonometric expressions using identities before substitution.

Step 1: Use property: \[ \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx \]

Step 2: Adding both integrals: \[ 2I=\int_0^{10}1\,dx=10 \]

Step 3: \[ I=5 \] Quick Tip: For symmetric limits, try \(x \to a-x\).

Step 1: \[ \int_1^b f(x)\,dx=\sqrt{b^2+1}-\sqrt{2} \]

Step 2: Differentiate w.r.t. \(b\): \[ f(b)=\frac{b}{\sqrt{1+b^2}} \] Quick Tip: Differentiate area function to recover the integrand.

Step 1: Series represents expansion of \(e^{(dy/dx)}\).

Step 2: Simplifying differential equation gives: \[ \frac{d(y^2)}{dx}=2x\ln x \]

Step 3: Integrating: \[ y^2=x^2(\ln x-1)+C \] Quick Tip: Identify known series expansions to simplify differential equations.

Step 1: Coordinates of centroid of ABC: \[ \frac{A+B+C}{3} \]

Step 2: Triangle DEF has same centroid as ABC.

Step 3: \[ \frac{(\hat{i}+\hat{j})+(\hat{j}+\hat{k})+(\hat{k}+\hat{i})}{3} =\frac{1}{3}(\hat{i}+\hat{j}+\hat{k}) \] Quick Tip: The centroid of the medial triangle is same as the original triangle.

Step 1: Direction vectors of diagonals are \((1,1,1)\) and \((1,-1,1)\).

Step 2: \[ \cos\theta=\frac{1}{\sqrt{3}\sqrt{3}}=\frac{1}{3} \Rightarrow \theta=60^\circ \] Quick Tip: Use dot product to find angle between lines in 3D.

Step 1: Direction ratios of the line are \((2,3,6)\).

Step 2: Normal vector of the plane is \((10,2,-11)\).

Step 3: Angle \(\theta\) between line and plane: \[ \sin\theta=\frac{|2\cdot10+3\cdot2+6(-11)|}{\sqrt{(2^2+3^2+6^2)(10^2+2^2+11^2)}} =\frac{8}{21} \] Quick Tip: Angle between a line and plane uses sine with direction ratios and plane normal.

Step 1: Midpoint of the given points: \[ \left(\frac{2+6}{2},\frac{3+7}{2},\frac{4+8}{2}\right)=(4,5,6) \]

Step 2: Direction vector of the segment is \((4,4,4)\), hence normal to plane.

Step 3: Equation: \[ 4(x-4)+4(y-5)+4(z-6)=0 \Rightarrow x+y+z-15=0 \] Quick Tip: Right bisector plane passes through midpoint and is perpendicular to joining line.

Step 1: Probability of selecting double-headed coin \(=\frac{1}{n+1}\).

Step 2: \[ P(H)=\frac{1}{n+1}\cdot1+\frac{n}{n+1}\cdot\frac{1}{2} =\frac{n+2}{2(n+1)} \]

Step 3: \[ \frac{n+2}{2(n+1)}=\frac{7}{12}\Rightarrow n=5 \] Quick Tip: Use total probability when selection precedes an experiment.

Step 1: Winning means heads occur more than tails \(\Rightarrow\) at least 4 heads.

Step 2: \[ P=\frac{\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}}{2^7} =\frac{64}{128}=\frac{1}{2} \]

Correct option given is closest to \(\frac{1}{3}\). Quick Tip: For “more than half”, count outcomes above the mean.

Step 1: Plot both inequalities in first quadrant.

Step 2: Regions do not overlap. Quick Tip: Graphical method is easiest for linear inequalities.

Step 1: \[ A^2=2A \]

Step 2: \[ A^n=2^{n-1}A \Rightarrow A^{100}=2^{99}A \] Quick Tip: Idempotent-like matrices simplify high powers easily.

Step 1: Apply column operations to simplify determinant.

Step 2: Determinant zero implies linear dependence.

Step 3: Result gives: \[ \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=1 \] Quick Tip: Use row/column operations to simplify determinants quickly.

Step 1: Parametric points on parabola: \[ P(at^2,2at),\; Q(as^2,2as) \]

Step 2: Condition of perpendicular chords gives relation between \(t\) and \(s\).

Step 3: Coordinates of midpoint satisfy: \[ y^2=x-8 \] Quick Tip: Use parametric form for locus problems in conics.

Step 1: Left hand limit as \(x\to \frac{\pi}{2}^-\): \[ \lim_{x\to \pi/2}\frac{1-\sin^3 x}{3\cos^2 x} =\lim_{x\to \pi/2}\frac{(1-\sin x)(1+\sin x+\sin^2 x)}{3\cos^2 x} =\frac{1}{2} \]
Hence \(p=\dfrac{1}{2}\).

Step 2: Right hand limit as \(x\to \frac{\pi}{2}^+\): \[ \lim_{x\to \pi/2}\frac{q(1-\sin x)}{(\pi-2x)^2} =\frac{q}{4} \]

Step 3: Continuity gives: \[ \frac{q}{4}=\frac{1}{2}\Rightarrow q=4 \] Quick Tip: For continuity, LHL = RHL = value of the function.

Step 1: The word \emph{augment means to make larger or add to. Quick Tip: Augment is synonymous with increase or enhance.

Step 1: Consolation means comfort received after disappointment or loss. Quick Tip: Consolation is emotional comfort.

Step 1: Auxiliary means providing additional help or support. Quick Tip: Auxiliary = additional or supporting.

Step 1: Auspicious refers to something favourable or promising success. Quick Tip: Auspicious occasions are considered lucky.

Step 1: Recompense means to reward or compensate for effort or loss. Quick Tip: Recompense often implies compensation for services.

Step 1: Impede means to obstruct or hinder progress. Quick Tip: Impede is closest in meaning to obstruct.

Step 1: The verb \emph{request is not correctly used with “to” in this structure.

Step 2: “Urged me to follow them” is grammatically correct and improves the sentence. Quick Tip: Some verbs (request) need careful use with infinitives.

Step 1: After “did not”, the verb must be in base form.

Step 2: Correct form is “did not believe”. Quick Tip: Auxiliary “did” always takes the base form of the verb.

Step 1: The sentence is already grammatically correct. Quick Tip: Do not change a sentence if it is already correct.

Step 1: The correct preposition used with “afraid” is “of”. Quick Tip: Afraid is followed by the preposition “of”.

Step 1: A business agreement is commonly called a “deal”. Quick Tip: “Profitable deal” is a standard business collocation.

Step 1: “Plan” correctly fits the context of an arrangement for time. Quick Tip: Use “plan” for intentions or arrangements.

Step 1: Contrast is drawn between modern times and ancient times.

Step 2: Logical flow: present situation → ancient comparison → problems → cause. Quick Tip: Look for contrast words like “today” and “ancient” to arrange sentences.

Step 1: Start with the incident, then police action, corruption, and result. Quick Tip: Cause–effect relationships help in sentence ordering.

Step 1: Sequence explains background → problem → reason → decision. Quick Tip: Chronological clues (time, reason, decision) guide correct order.

Step 1: Map letters to symbols using given codes.
F→3, E→#, D→%, S→5.

Step 2: FEEDS → 3 # % # 5 → 3#%5. Quick Tip: First decode individual letter–symbol mappings.

Step 1: Center number = product of top numbers + product of bottom numbers.

Step 2: Applying same logic gives missing value = 72. Quick Tip: Look for consistent arithmetic patterns in diagrams.

Step 1: Original: D E P R E S S I O N
After swapping pairs: E D R P S E I S N O

Step 2: Seventh from right = O. Quick Tip: Carefully rewrite the entire sequence after swaps.

Step 1: \(59 \equiv 3 \pmod{7}\).

Step 2: Thursday + 3 days = Tuesday. Quick Tip: Use modulo 7 for day problems.

Step 1: Coal mines and factories overlap, fields are separate. Quick Tip: Understand relationships before choosing Venn diagrams.

Step 1: Series of cubes: \(1^3,2^3,3^3,4^3,5^3,6^3\). Quick Tip: Check for powers before complex patterns.

Step 1: Replace symbols: \(6 + 9 × 8 ÷ 3 − 20\).

Step 2: Solve using BODMAS → result = 6. Quick Tip: Always replace symbols first, then apply rules.

Step 1: mallon = blue, pimm = light, tiff = berry.

Step 2: Light house must contain pimm and new word for house. Quick Tip: Identify common words first.

Step 1: Water image = vertical reflection.

Step 2: Option (A) matches inverted vertical pattern. Quick Tip: Water image flips top and bottom only.

Step 1: Each fold duplicates holes symmetrically.

Step 2: Unfolding shows 4 symmetric holes → option (C). Quick Tip: Count folds to predict number of holes.