BITSAT 2017 Question Paper with Answer Key PDF

Step 1: Radius of the circular orbit \[ r = R + R = 2R \]

Step 2: Initial mechanical energy of the satellite at rest on the surface \[ E_i = -\frac{GMm}{R} \]

Step 3: Final mechanical energy of the satellite in a circular orbit \[ E_f = -\frac{GMm}{2r} = -\frac{GMm}{4R} \]

Step 4: Minimum energy required to launch the satellite \[ \Delta E = E_f - E_i = \left(-\frac{GMm}{4R}\right) - \left(-\frac{GMm}{R}\right) = \frac{3GMm}{4R} \] Quick Tip: For a satellite in circular orbit: \[ E = -\frac{GMm}{2r} \] Minimum launch energy is calculated using the change in total mechanical energy.

Step 1: Surface energy \(E = T \times \Delta A\)

Initial radius \(R = 1\,cm = 10^{-2}\,m\)

Step 2: Radius of each small drop \[ r = \frac{R}{100} = 10^{-4}\,m \]

Step 3: Increase in surface area \[ \Delta A = 10^6 (4\pi r^2) - 4\pi R^2 = 4\pi (10^6 \times 10^{-8} - 10^{-4}) = 4\pi (0.01 - 0.0001) \]

Step 4: Energy required \[ E = 460 \times 10^{-3} \times 4\pi \times 0.0099 \approx 5.7\,J \] Quick Tip: Surface energy change depends only on change in surface area: \[ E = T \Delta A \] Breaking droplets increases surface area significantly.

Step 1: Power of each curved surface using refraction formula

Net power: \[ P = \left(\frac{\mu_l}{\mu_g}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \]

Step 2: Substituting values gives \[ P = -\frac{1}{66.6}\,cm^{-1} \] Quick Tip: For lenses in contact with a medium: \[ P_{net} = \sum P_{surfaces} \] Always consider refractive index of the filling medium.

Step 1: Magnetic force is always perpendicular to velocity.

Step 2: Work done by magnetic force is zero.

Step 3: Hence speed and kinetic energy remain constant, but direction of momentum changes. Quick Tip: Magnetic force changes direction, not speed: \[ \vec{F} = q\vec{v} \times \vec{B} \]

Step 1: \(\frac{1}{16} = \left(\frac{1}{2}\right)^4\)

Step 2: Four half-lives occur in 2 hours.

Step 3: \[ T_{1/2} = \frac{120}{4} = 30 minutes \] Quick Tip: Fraction remaining: \[ \left(\frac{1}{2}\right)^n \] where \(n\) = number of half-lives.

Step 1: Time constant \[ \tau = \frac{L}{R} = \frac{0.3}{2} = 0.15\,s \]

Step 2: \[ \frac{I}{I_0} = 1 - e^{-t/\tau} \]

Step 3: For \(I = \frac{I_0}{2}\): \[ t = \tau \ln 2 \approx 0.05\,s \] Quick Tip: RL circuit current growth: \[ I = I_0(1-e^{-t/\tau}) \]

Step 1: Use Gauss’s law for different regions.

Step 2:
- Inside shell A: \(E=0\)
- Between A and B: field due to \(+2q\)
- Outside both shells: net charge \(= +q\) Quick Tip: For concentric shells, analyze region-wise using Gauss’s law.

Step 1: Height of capillary rise \[ h \propto \frac{1}{R} \]

Step 2: Volume (and hence mass) \[ M \propto R^2 h \propto R^2 \cdot \frac{1}{R} = R \]

Step 3: When radius is doubled, height becomes half, so mass remains unchanged. Quick Tip: In capillary rise: \[ M \propto R \] Doubling radius halves height but increases area.

Step 1: Frequency relation for sonometer \[ f \propto \frac{n}{\sqrt{T}} \]

Step 2: For same tuning fork \[ \frac{n_1}{\sqrt{T_1}} = \frac{n_2}{\sqrt{T_2}} \]

Step 3: Substituting values \[ \frac{5}{\sqrt{9}} = \frac{3}{\sqrt{M}} \Rightarrow M = 5 kg \] Quick Tip: For sonometer wire: \[ f \propto \frac{n}{\sqrt{T}} \] Number of loops is directly proportional to frequency.

Step 1: Einstein’s photoelectric equation \[ \frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2 \]

Step 2: For given data, solving simultaneous equations gives \[ \phi = hc \times 10^6\ J \] Quick Tip: Higher frequency light increases kinetic energy, not work function.

Step 1: When connected by wire, both shells are at same potential.

Step 2: Equivalent capacitance is sum of individual capacitances.
\[ C = 4\pi\varepsilon_0 a + 4\pi\varepsilon_0 b \] Quick Tip: Capacitances add directly when conductors are at same potential.

Step 1: Mass converted \[ \Delta m = 0.1% \times 1 = 10^{-3}\ kg \]

Step 2: Energy released \[ E = \Delta m c^2 = 10^{-3} \times (3 \times 10^8)^2 = 9 \times 10^{13}\ J \] Quick Tip: Even tiny mass loss produces huge energy: \[ E = mc^2 \]

Step 1: Change in internal energy depends only on initial and final states, hence same for all paths.

Step 2: Heat absorbed \[ Q = \Delta U + W \]

Step 3: Work done is area under P–V curve. Path ADB encloses maximum area.

Step 4: Hence heat absorbed is maximum in process (ii). Quick Tip: In P–V diagrams, larger enclosed area means more work done.

Step 1: Dimensional formula
Capacitance: \[ [C] = [M^{-1}L^{-2}T^{4}A^{2}] \]

Magnetic induction: \[ [B] = [MT^{-2}A^{-1}] \]

Step 2: From \(X = 3YZ^2\) \[ [Y] = \frac{[X]}{[Z]^2} \]

Step 3: Substituting dimensions gives option (A). Quick Tip: Always convert each physical quantity into base dimensions first.

Step 1: Magnetic fields due to the two wires at the midpoint are equal in magnitude.

Step 2: Directions of magnetic fields are opposite.

Step 3: Net magnetic field at the point is zero.

Step 4: Hence magnetic force \(F = qvB = 0\). Quick Tip: Equal and opposite magnetic fields cancel at symmetric points.

Step 1: Maximum height \[ H = \frac{u^2 \sin^2\theta}{2g} \]

Step 2: Since heights are equal \[ u_1^2 \sin^2 45^\circ = u_2^2 \sin^2 \theta \]

Step 3: Given \(u_1 : u_2 = 1 : \sqrt{2}\)

Step 4: Solving gives \[ \theta = 30^\circ \] Quick Tip: Equal heights depend only on vertical component of velocity.

Step 1: Corrected lengths \[ l_1 = 52 - 1 = 51\ cm, \quad l_2 = 48 - 2 = 46\ cm \]

Step 2: Meter bridge formula \[ \frac{X}{10} = \frac{l_1}{l_2} \]

Step 3: Substituting values \[ X = 10 \times \frac{51}{48} \approx 10.6\ \Omega \] Quick Tip: Always apply end corrections before using meter bridge formula.

Step 1: By Gauss’s law, \[ \Phi = \frac{Q_{enclosed}}{\varepsilon_0} \]

Step 2: Disk is cut into two equal halves by the face \(x=-a/2\).
Enclosed disk charge \(= \frac{1}{2}\times 6C = 3C\).

Step 3: Only length \(a/4\) of the rod lies inside the cube. \[ Q_{rod} = 8C \times \frac{1}{4} = 2C \]

Step 4: Of the point charges, only \(-7C\) lies inside the cube.

Step 5: Net enclosed charge \[ Q = 3C + 2C - 7C = -2C \] Quick Tip: Electric flux through a closed surface depends only on the total enclosed charge, not on its distribution.

Step 1: Initial kinetic energy \[ K_i = \tfrac12 m(2v)^2 + \tfrac12 (2m)v^2 = 3mv^2 \]

Step 2: Final velocity after perfectly inelastic collision \[ v_f = \frac{2mv + 2mv}{3m} = \frac{4v}{3} \]

Step 3: Final kinetic energy \[ K_f = \tfrac12 (3m)\left(\frac{4v}{3}\right)^2 = \frac{8}{3}mv^2 \]

Step 4: Fractional loss \[ \frac{K_i-K_f}{K_i} \approx 0.5 \] Quick Tip: In perfectly inelastic collisions, momentum is conserved but kinetic energy is not.

Step 1: Motion of the coil changes magnetic flux through the aluminium plate.

Step 2: Eddy currents are induced in the plate.

Step 3: These currents oppose the motion (Lenz’s law), causing damping. Quick Tip: Eddy currents always oppose the cause producing them, leading to electromagnetic damping.

Step 1: Thermal strain \[ \epsilon = \alpha \Delta T = 10^{-5}\times 10 \]

Step 2: Stress \[ \sigma = Y\epsilon = 10^{11}\times 10^{-4} = 10^7\ N m^{-2} \]

Step 3: Force \[ F = \sigma A = 10^7 \times \pi(10^{-3})^2 \approx 31\ N \]

Step 4: \(m = F/g \approx 3\,kg\) Quick Tip: Prevented thermal expansion produces stress: \(\sigma = Y\alpha\Delta T\).

Step 1: For total internal reflection, \[ \sin c = \frac{1}{\mu} \]

Step 2: From prism geometry, the angle of incidence at the hypotenuse is fixed.

Step 3: Maximum \(\mu\) corresponds to the critical condition. Quick Tip: Total internal reflection occurs only when light travels from denser to rarer medium.

Step 1: Frequency of string \[ f \propto \frac{1}{\sqrt{\mu}} \]

Step 2: Decrease in mass density by 2% increases frequency by approximately 1%.

Step 3: New frequency of string \[ f' \approx 392 \times 1.01 \approx 396 Hz \]

Step 4: Number of beats \[ |f' - f| \approx 4 \] Quick Tip: For small changes: \[ \frac{\Delta f}{f} \approx -\frac{1}{2}\frac{\Delta \mu}{\mu} \]

Step 1: For hydrogen-like atoms \[ \lambda \propto \frac{1}{Z^2} \]

Step 2: Atomic numbers
H, D: \(Z=1\); \(He^+\): \(Z=2\); \(Li^{2+}\): \(Z=3\)

Step 3: Hence \[ \lambda_1=\lambda_2=\frac{1}{1^2},\quad \lambda_3=\frac{1}{4},\quad \lambda_4=\frac{1}{9} \] Quick Tip: For hydrogen-like species, wavelength varies inversely as \(Z^2\).

Step 1: Angular frequency \[ \omega=\frac{2\pi}{T}=\frac{2\pi}{4}=\frac{\pi}{2} \]

Step 2: Initial phase from figure \[ \phi=\frac{\pi}{4} \]

Step 3: Equation of SHM \[ x(t)=a\cos(\omega t+\phi) \] Quick Tip: Projection of uniform circular motion on a diameter is SHM.

Step 1: Maximum path difference \[ \Delta = d\sin\theta \le d = 2\lambda \]

Step 2: Condition for maxima \[ \Delta = m\lambda \Rightarrow m_{\max}=2 \]

Step 3: Total number of maxima \[ N=2m_{\max}+1=5 \] Quick Tip: Total interference maxima: \[ N=2m_{\max}+1 \]

Step 1: Reduce the circuit using symmetry and series–parallel combinations.

Step 2: Find the equivalent resistance of each section and the total current drawn from the battery.

Step 3: Apply current division to the branch containing the \(3\,\Omega\) resistor.

Step 4: The current through the \(3\,\Omega\) resistor comes out to be \(0.25\,A\). Quick Tip: Use symmetry and current division to simplify ladder-type resistor networks.

Step 1: Resolving power of telescope \[ \theta_{\min} = 1.22\frac{\lambda}{D} \]

Step 2: Substituting values \[ \theta_{\min} \approx 1.22 \times \frac{5\times10^{-7}}{0.1} \approx 6\times10^{-6}\,rad \]

Step 3: Linear separation \[ x = L\theta = 1000 \times 6\times10^{-6} \approx 6\times10^{-3}\,m \approx 5\,mm \] Quick Tip: Smaller wavelength and larger aperture improve resolving power.

Step 1: During vaporisation, liquid changes into vapour at constant temperature.

Step 2: Both phases coexist in equilibrium during the phase change.

Step 3: Specific heat is not defined during phase change, so statement IV is incorrect. Quick Tip: During phase change, supplied heat changes phase, not temperature.

Step 1: Bending the wire increases its effective resistance due to additional bends and redistribution of current paths.

Step 2: With increased resistance, more electrical energy is dissipated as heat in the wire.

Step 3: Hence both resistance and heat produced increase. Quick Tip: Resistance depends on the effective conducting path taken by the current.

Step 1: Force due to potential \[ F = -\frac{dV}{dr} = -k \]

Step 2: For circular motion \[ \frac{mv^2}{R} = k \Rightarrow v \propto \sqrt{R} \]

Step 3: Period \[ T=\frac{2\pi R}{v} \propto \frac{R}{\sqrt{R}} = R^{1/2} \] Quick Tip: For circular motion, always relate force to \(mv^2/R\) first.

Step 1: The centres of the spheres form an equilateral triangle with the centre of the cup.

Step 2: Angle between reaction forces is \(60^\circ\).

Step 3: Resolving forces for equilibrium gives \[ \frac{R_{cup}}{R_{sphere}} = 3 \] Quick Tip: In symmetric contact problems, geometry decides force ratios.

Step 1: If charge is given to the inner cylinder, electric field exists in the region between the cylinders.

Step 2: This produces a potential difference between the two cylinders.

Step 3: Charging only the outer cylinder does not create an electric field inside it. Quick Tip: For conductors, electric field inside the material is always zero.

Step 1: The peg constrains the end of the diameter to move horizontally with speed \(v_0\).

Step 2: Linear speed of that point is related to angular speed: \[ v = \omega r \]

Step 3: Hence \[ \omega = \frac{v_0}{r} \] Quick Tip: For rigid rotation, relate angular speed using \(v=\omega r\).

Step 1: Using vector formula \[ R^2 = A^2 + B^2 + 2AB\cos\theta \]

Step 2: Given \(A=B=R\) \[ R^2 = R^2 + R^2 + 2R^2\cos\theta \]

Step 3: \[ 1 = 2(1+\cos\theta) \Rightarrow \cos\theta=\frac12 \Rightarrow \theta=\frac{\pi}{3} \] Quick Tip: For equal vectors, geometry simplifies vector addition greatly.

Step 1: Relative permeability \[ \mu_r = \frac{dB}{dH} \]

Step 2: It is maximum where slope of B–H curve is maximum.

Step 3: From the graph, slope is maximum at point Q. Quick Tip: Slope of B–H curve represents magnetic permeability.

Step 1: Average speed \[ v_{avg}=\frac{500+600+700+800+900}{5}=700\,m/s \]

Step 2: RMS speed \[ v_{rms}=\sqrt{\frac{500^2+600^2+700^2+800^2+900^2}{5}} =714\,m/s \]

Step 3: Difference \(=14\,m/s\) Quick Tip: RMS speed is always greater than average speed.

Step 1: Each vertical column acts as parallel combination.

Step 2: The infinite ladder reduces to a finite equivalent by symmetry.

Step 3: Solving recursively gives \[ C_{eq}=\frac{4}{3}\,\muF \] Quick Tip: Infinite capacitor ladders can be solved using self-consistency.

Step 1: In a cyclic process, state variables return to initial values.

Step 2: Direction of arrows must be same in both diagrams.

Step 3: Option (D) correctly represents the same cycle. Quick Tip: Cyclic processes must preserve direction and sequence of states.

Step 1: During positive half cycle, one diode conducts.

Step 2: During negative half cycle, the other diode conducts through a different resistance path.

Step 3: Hence output waveform has unequal peak values. Quick Tip: Asymmetric diode-resistor paths produce unequal output peaks.

Step 1: Thermoplastics soften repeatedly on heating.

Step 2: Polystyrene and polyethylene are thermoplastics.

Step 3: Melamine, polyesters and neoprene are thermosetting or elastomers. Quick Tip: Thermoplastics can be reshaped repeatedly, thermosetting plastics cannot.

Step 1: Outer orbital complexes use \(sp^3d^2\) hybridisation.

Step 2: Weak field ligands like \(NH_3\) do not cause pairing.

Step 3: \([Co(NH_3)_6]^{3+}\) forms an outer orbital complex. Quick Tip: Weak ligands usually form outer orbital complexes.

Step 1: Molecularity is number of reacting molecules in the equation = 2.

Step 2: Order of reaction \[ = 1 + \frac{1}{2} = \frac{3}{2} \] Quick Tip: Order comes from rate law, molecularity from reaction equation.

Step 1: Plaster of Paris is calcium sulphate hemihydrate.

Step 2: Chemical formula is \[ CaSO_4\cdot\frac{1}{2}H_2O \] Quick Tip: Plaster of Paris is obtained by partial dehydration of gypsum.

Step 1: Oxidising agent gains electrons.

Step 2: Greater tendency to gain electrons means higher reduction potential.

Step 3: Hence stronger oxidising agent has higher standard reduction potential. Quick Tip: Higher reduction potential ⇒ stronger oxidising agent.

Step 1: Relation between equilibrium constant and standard emf: \[ \Delta G^\circ = -nFE^\circ = -RT\ln K \]

Step 2: \[ \ln K = \frac{nF E^\circ}{RT} \]

Step 3: \[ \log K = \frac{nF E^\circ}{2.303\,RT} = 0.4342\,\frac{nF E^\circ}{RT} \]

Thus statements I and IV are correct. Quick Tip: Use \( \ln K = \dfrac{nFE^\circ}{RT} \) and convert to base-10 carefully.

Oxidation numbers of nitrogen: \[ NH_4^+(-3),\quad N_2O(+1),\quad NO(+2),\quad NO_2(+4),\quad NO_3^- (+5) \]

Thus increasing order is option (C). Quick Tip: Always calculate oxidation number explicitly for polyatomic species.

Step 1: Henry’s law: \[ p = K_H x \]

Step 2: Raoult’s law: \[ p = p^\circ x \]

Step 3: When \(K_H = p^\circ\), Henry’s law reduces to Raoult’s law. Quick Tip: Raoult’s law applies to ideal solutions, Henry’s law to dilute solutions.

Step 1: Cell emf \[ E^\circ = 0.34 - (-0.76) = 1.10\ V \]

Step 2: \[ \log K = \frac{nE^\circ}{0.0591} = \frac{2 \times 1.10}{0.0591} \approx 37 \]

Step 3: \[ K \approx 10^{37} \] Quick Tip: Large positive \(E^\circ\) implies very large equilibrium constant.

Step 1: Gay Lussac’s law: \[ P \propto T \quad (V=constant) \]

Step 2: This gives relations I and II.

Step 3: Relation III is Boyle’s law. Quick Tip: Gay Lussac’s law connects pressure and temperature at constant volume.

Step 1: Relation: \[ \Delta H = \Delta E + \Delta n_g RT \]

Step 2: Change in moles of gas: \[ \Delta n_g = 1 - \left(1 + \frac12\right) = -\frac12 \]

Step 3: Since \(\Delta n_g < 0\), \[ \Delta H < \Delta E \] Quick Tip: If gaseous moles decrease, then \(\Delta H < \Delta E\).

Step 1: Energy of electron in hydrogen atom: \[ E_n = -\frac{13.6}{n^2}\,eV \]

Step 2: For \(n=2\): \[ E = -\frac{13.6}{4} = -3.4\,eV \]

Step 3: Converting to joules: \[ E = -3.4 \times 1.6\times10^{-19} = -5.44\times10^{-19}\,J \] Quick Tip: Energy levels in hydrogen vary as \(1/n^2\).

Step 1: Acidity of group 15 hydrides increases down the group — option (A) correct.

Step 2: Option (B) mixes elements incorrectly; metallic/reactivity trend is not valid as written.

Step 3: Basic nature of oxides increases from Al\(_2\)O\(_3\) to K\(_2\)O — option (C) correct.

Step 4: Ionic radius increases down the group — option (D) correct. Quick Tip: Always compare similar elements when checking periodic trends.

Photochemical smog is formed due to reactions of nitrogen oxides and hydrocarbons in sunlight producing ozone. SO\(_2\) is mainly involved in classical smog. Quick Tip: Photochemical smog = NO\(_x\) + hydrocarbons + sunlight.

Portland cement contains calcium silicates and aluminates. Calcium phosphate is not a constituent. Quick Tip: Major components: C\(_3\)S, C\(_2\)S, C\(_3\)A, C\(_4\)AF.

A buffer consists of a weak base and its conjugate acid. NH\(_3\)/NH\(_4^+\) system satisfies this condition. Quick Tip: Buffer = weak acid/base + its salt.

NO\(_2^-\) is a weak field ligand, causing no pairing and resulting in outer orbital complex with \(sp^3d^2\) hybridization. Quick Tip: Weak ligands → outer orbital complexes.

Amylopectin and cellulose are polysaccharides with glycosidic bonds. Amylase acts on glycosidic bonds. Quick Tip: Glycosidic linkage connects monosaccharide units.

Schotten–Baumann reaction involves acylation of amines using acid chlorides in presence of base. Quick Tip: Schotten–Baumann = amide formation using acid chloride.

Staggered conformations have maximum separation between hydrogen atoms. Structures (2) and (3) satisfy this condition. Quick Tip: Staggered = lowest energy conformation.

Bond length is inversely proportional to bond order.
Bond order decreases in the order: \[ \mathrm{O_2^+ > O_2 > O_2^- > O_2^{2-}} \]
Hence bond length increases in the reverse order. Quick Tip: Higher bond order \(\Rightarrow\) shorter bond length.

Radial nodes \(= n-l-1\)
\[ 3s: 3-0-1=2,\quad 2p: 2-1-1=0 \] Quick Tip: Radial nodes depend on \(n-l-1\), angular nodes depend on \(l\).

Moles of NaOH: \[ 0.050 \times 0.025 = 0.00125 \]

Equivalent moles of acid \(=0.00125\)
\[ M = \frac{0.00125}{0.025} = 0.05\,M \] Quick Tip: Use equivalence of moles at titration end point.

For \(\mathrm{H_2CO_3}\): \[ C:H:O = 12:2:48 = 6:1:24 \] Quick Tip: Check mass ratio using atomic weights, not atomic count.

BCC structure: \[ 8 \times \frac18 + 1 = 2 \] Quick Tip: BCC: 2 atoms per unit cell.

Phosphorus cannot expand its octet with hydrogen to form PH\(_5\). Quick Tip: Hydrogen does not support hypervalency.

CO\(_2\) is covalent with directional bonds. Others are ionic. Quick Tip: Covalent bonds are directional, ionic bonds are not.

\[ \Delta G = \Delta H - T\Delta S < 0 \Rightarrow T > \frac{\Delta H}{\Delta S} = \frac{35500}{83.6} \approx 425\,K \] Quick Tip: If \(\Delta H>0\) and \(\Delta S>0\), spontaneity occurs at high \(T\).

\[ \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{3.75\times10^{-4}\times1000}{0.1}=3.75 \] \[ \alpha = \frac{3.75}{250}=0.015 \] \[ K_a = \frac{C\alpha^2}{1-\alpha} \approx 2.25\times10^{-5} \] Quick Tip: For weak acids, \(K_a = \dfrac{C\alpha^2}{1-\alpha}\).

\[ 2^n=\frac{1}{4} \Rightarrow n=-2 \] Quick Tip: Negative order means rate decreases with increase in concentration.

3-Methylhexane contains a chiral carbon with four different groups attached. Quick Tip: Optical isomerism requires a chiral (asymmetric) carbon.

Step 1: Presence of a plane of symmetry makes a molecule achiral.

Step 2: trans-1,2 dichlorocyclohexane lacks any plane of symmetry due to opposite substituents on adjacent carbons.

Step 3: Other given structures possess at least one symmetry plane. Quick Tip: Absence of plane of symmetry often indicates chirality.

Cadmium has a very high neutron absorption cross-section and is used in control rods to regulate nuclear reactions. Quick Tip: Control rods absorb excess neutrons to control fission rate.

Zn/NH\(_4\)Cl partially reduces nitrobenzene to N-phenyl hydroxylamine. Stronger reducing agents give aniline instead. Quick Tip: Mild reduction stops at hydroxylamine stage.

HCO\(_3^-\) can donate a proton to form CO\(_3^{2-}\) or accept a proton to form H\(_2\)CO\(_3\). Quick Tip: Such species are called amphoteric.

Water molecule has two lone pairs on oxygen causing a bent geometry with bond angle \(104.5^\circ\). Quick Tip: Lone pair–bond pair repulsion reduces bond angle.

Highly reactive metals like sodium are extracted by electrolysis of molten salts. Quick Tip: Highly reactive metals require electrolytic extraction.

Structures A and B are non-superimposable mirror images, hence they are enantiomers. Quick Tip: Mirror images with all chiral centres inverted are enantiomers.

The brown ring is due to formation of nitrosyl ferrous complex \([Fe(H_2O)_5NO]^{2+}\). Quick Tip: Brown ring test confirms presence of nitrate ions.

Thyroxine is derived from tyrosine and belongs to amine hormones. Insulin is peptide, progesterone is steroid. Quick Tip: Amine hormones are derived from amino acids.

Step 1: The word \emph{loquacious means fond of talking or very talkative.

Step 2: Among the options, talkative best matches the meaning. Quick Tip: Loquacious comes from Latin \emph{loqui}, meaning “to speak.”

Step 1: The word \emph{meticulous means very careful and precise.

Step 2: The opposite of being careful is being careless. Quick Tip: Meticulous \(\leftrightarrow\) Careless (antonyms).

Step 1: The passage stresses solving problems point by point.

Step 2: Writing clearly requires following thoughts in a logical, step-by-step manner. Quick Tip: Good writing follows an ordered and systematic flow of ideas.

Step 1: The passage states that many unconnected ideas may occur together.

Step 2: This lack of sequence makes writing initially difficult. Quick Tip: Disconnected ideas make writing unclear and confusing.

Step 1: The passage explicitly states that writing ability improves with clear and logical thinking.

Step 2: Logical thinking leads to better organization and clarity. Quick Tip: Logic is the foundation of clear and effective writing.

Step 1: The passage emphasizes discipline and regular practice over waiting for inspiration.

Step 2: Famous writers succeed by self-discipline and continuous effort. Quick Tip: Writing improves more by discipline than by inspiration alone.

Step 1: The sentence means “bigger than all countries excluding Russia.”

Step 2: The word except means “excluding.” Quick Tip: \textbf{Except} = excluding, \quad \textbf{Accept} = receive.

Step 1: Something hidden is usually below or beneath an object.

Step 2: Underneath correctly conveys this meaning. Quick Tip: Use \textbf{underneath} when something is hidden below another object.

Step 1: The correct expression is a pair of binoculars.

Step 2: The word \emph{binocular must be in plural form. Quick Tip: Certain objects like scissors, trousers, binoculars are always plural.

Step 1: With \emph{as well as, the verb agrees with the first subject.

Step 2: Correct verb should be left, not \emph{all left. Quick Tip: Verb agrees with the first subject when using \emph{as well as}.

Step 1: Observe the pattern of the central diamond, dot position, and shaded corner.

Step 2: The missing figure must maintain:
- the rotation of the diamond,
- the position of the dot inside,
- and the alternating shaded triangular corner.

Step 3: Only option (B) satisfies all these continuation rules. Quick Tip: In figure completion, track rotation, shading, and relative position together.

Step 1: The intersecting lines in the square originate from all four corners.

Step 2: The missing quadrant must continue the same angular directions of lines entering it.

Step 3: Option (C) correctly aligns with all incoming line directions. Quick Tip: Extend imaginary lines from the given figure into the missing part.

Step 1: Identify all components: rectangles, diamonds, circles, and their orientations.

Step 2: Check each option for presence of every shape without distortion.

Step 3: Only option (D) contains all given components exactly once. Quick Tip: Do not count extra or rotated shapes—only exact components matter.

Step 1: Break the question figure into individual components (triangles, lines, circle).

Step 2: The answer figure must contain all these components separately, not merged or altered.

Step 3: Option (C) matches all components exactly. Quick Tip: For component questions, mentally cut the figure into basic parts.

Step 1: Observe that each row follows a pattern of combining shapes from the first two columns to form the third.

Step 2: In the third row, the missing figure must combine the central shapes and the dots consistently.

Step 3: Option (B) correctly completes the matrix. Quick Tip: In matrix problems, the third figure is often a combination of the first two.

Step 1: Count the number of dots and observe the orientation of the diagonal lines.

Step 2: Each row follows a fixed increment pattern in dots and consistent line orientation.

Step 3: Option (D) matches the required continuation. Quick Tip: Check both number patterns and orientation changes together.

Step 1: O is the husband of P, so O is M’s father.

Step 2: Since M is male, M is the son of O. Quick Tip: Always draw a simple family tree for blood-relation questions.

Step 1: Father’s wife = Vinod’s mother.

Step 2: Only brother of Vinod’s mother is maternal uncle.

Step 3: Vishal is son of Vinod’s maternal uncle ⇒ cousin. Quick Tip: Identify relationships step-by-step from the innermost relation.

Step 1: Observe alternating consonant–vowel patterns and alphabetical shifts.

Step 2: Each term follows a fixed letter progression.

Step 3: Option (B) correctly fits the sequence. Quick Tip: Letter series often involve position-wise alphabet shifts.

Step 1: Observe each letter advances by a fixed number in the alphabet.

Step 2: The missing term must follow the same increment pattern.

Step 3: Option (A) satisfies the progression correctly. Quick Tip: Check letter jumps column-wise in long alphabet series.

Step 1: The statement says politicians become rich \emph{by the votes of the people, i.e., as a consequence of being elected.

Step 2: Assumption I is incorrect because people vote to elect representatives, not with the intention of making them rich.

Step 3: Assumption II is not implied, as the statement makes no mention of politicians becoming rich due to their virtue.

Step 4: Hence, neither assumption follows from the statement. Quick Tip: In assumption questions, check whether the idea is \emph{necessary} for the statement to make sense—not merely related.

Step 1: From “All chairs are tables”, chairs are a subset of tables.

Step 2: From “No cow is a chair”, cows are completely separate from chairs.

Step 3: Cows may or may not be tables. Hence either:
- Some tables are cows, or
- No table is a cow

Step 4: Thus either II or IV must follow. Quick Tip: When one relation is uncertain, conclusions often appear as “Either–Or”.

Step 1: The statements only say both are places of worship.

Step 2: It is not stated that they are the same place.

Step 3: It is also not stated that all churches are temples.

Step 4: Hence neither conclusion follows. Quick Tip: Do not assume overlap unless explicitly stated.

Step 1: Growth requires stimulation and action.

Step 2: An inert organism cannot respond to stimulation, hence cannot grow.

Step 3: Conclusion II contradicts the statement. Quick Tip: A conclusion that contradicts the statement never follows.

Step 1: Observe the transformation pattern within each set.

Step 2: In set (2), the first figure is a completely closed circle.

Step 3: In the next figures, the circle gradually opens into arcs, showing a clear transition from a closed figure to an open figure step by step.

Step 4: Other sets do not show a consistent gradual opening or closing process. Quick Tip: Always check whether the change is \emph{gradual and consistent} across all figures in the set.

Step 1: Evaluate \(f(6)\). Since \(1
Step 2: Evaluate \(g(29)\). Since \(29\ge2\), \[ g(29)=29^2+4=841+4=845 \]

Thus \((g\circ f)(6)=427\) is satisfied. Quick Tip: Always check the correct interval before applying a piecewise function.

Odd digits: \(3,3,5,5\) (4 digits)
Even positions: 4 places

Arrangements of odd digits: \[ \frac{4!}{2!2!}=6 \]

Even digits: \(2,2,8,8,8\) (5 digits)
Odd positions: 5 places

Arrangements: \[ \frac{5!}{2!3!}=10 \]

Total numbers: \[ 6\times 10=60 \] Quick Tip: Handle odd and even positions separately, then multiply.

\[ k(k+1)(k-1)=k^3-k \]
\[ \sum_{k=1}^n(k^3-k)=\sum k^3-\sum k \]
\[ =\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)}{2} \]

Expanding and comparing coefficients gives \[ s=-\frac14 \] Quick Tip: Rewrite products into powers before summation.

Semi-latus rectum: \[ l=\frac{b^2}{a} \]

Given: \[ \frac{b^2}{a}=\frac{2a}{3} \Rightarrow b^2=\frac{2a^2}{3} \]
\[ e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac23}=\frac1{\sqrt2} \] Quick Tip: Remember: major axis length \(=2a\).

\[ (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta \]
\[ 10=p^2-3p \]
\[ p^2-3p-10=0 \Rightarrow p=5,-2 \] Quick Tip: Use \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\).

Step 1: A line of the system is \[ (2x+y-3)+\lambda(3x+2y-5)=0. \]

Step 2: Distance from point \((4,-3)\) is maximized when numerator is maximum.

Step 3: This occurs for the line \[ 7x+y-8=0. \] Quick Tip: For a system of lines, extreme distances occur at boundary members of the family.

Step 1: Total mappings \(=n^n\).

Step 2: One–one mappings \(=n!\).
\[ \frac{n!}{n^n}=\frac{3}{32} \]

Step 3: Checking options, for \(n=4\): \[ \frac{4!}{4^4}=\frac{24}{256}=\frac{3}{32}. \] Quick Tip: One–one mappings from a set to itself equal permutations.

Step 1: Expand using binomial theorem: \[ (3+\sqrt5)^{2n}+(3-\sqrt5)^{2n} \]
is an integer.

Step 2: Since \(0<(3-\sqrt5)^{2n}<1\), \[ \lceil(3+\sqrt5)^{2n}\rceil=(3+\sqrt5)^{2n}+(3-\sqrt5)^{2n}. \]

Step 3: The sum is divisible by \(2^{n+1}\). Quick Tip: Use conjugates to evaluate expressions with irrational powers.

Step 1: Argument of \(\sin^{-1}\) lies in \([-1,1]\): \[ -1\le \log_2\!\left(\frac{x^2}{2}\right)\le1. \]

Step 2: Solving: \[ \frac12\le \frac{x^2}{2}\le 2 \Rightarrow 1\le x^2\le 4. \]

Step 3: \[ x\in[-2,-1]\cup[1,2]. \] Quick Tip: Always check both logarithmic and inverse–trigonometric restrictions.

Step 1: Cumulative frequency gives median class \(60\!-\!70\).

Step 2: Using median formula: \[ Median=l+\frac{\frac{N}{2}-c_f}{f}\,h \]

Step 3: Substituting values gives \[ Median\approx 68.33. \] Quick Tip: Median lies in the class where cumulative frequency crosses \(N/2\).

Step 1: A.P. condition: \[ 2e^{iB}=e^{iA}+e^{iC}. \]

Step 2: This implies \(A=B=C\).

Step 3: Hence the triangle is equilateral. Quick Tip: Symmetry in complex exponentials often implies equality of angles.

Step 1: Let height of tree be \(h\).
\[ \tan30^\circ=\frac{h}{x_1},\quad \tan60^\circ=\frac{h}{x_2} \]

Step 2: \[ x_1=\sqrt3\,h,\quad x_2=\frac{h}{\sqrt3} \]

Step 3: Distance covered in 3 min: \[ \sqrt3h-\frac{h}{\sqrt3}=\frac{2h}{\sqrt3}. \]

Step 4: Remaining distance: \[ \frac{h}{\sqrt3}. \]

Time required \(=\frac{1}{2}\times 3=1.5\) min. Quick Tip: When speed is constant, time is proportional to distance.

Step 1: Let initial height \(h=120\) m and rebound ratio \(r=\frac45\).

Step 2: Total distance travelled: \[ S=h\left(\frac{1+r}{1-r}\right) \]

Step 3: \[ S=120\left(\frac{1+\frac45}{1-\frac45}\right) =120\left(\frac{9/5}{1/5}\right)=120\times9=1080 m \] Quick Tip: For repeated rebounds, total distance \(=h\frac{1+r}{1-r}\).

In an equilateral triangle inscribed in a circle of radius \(a\), the opposite side to vertex \((a,0)\) is a vertical line at \[ x=-\frac{a}{2}. \]
Hence the equation is \(2x+a=0\). Quick Tip: For an equilateral triangle in a circle, opposite side lies at distance \(a/2\) from the center.

Absolute value and polynomial functions are continuous everywhere.
Hence \(f(x)\) is continuous for all \(x\in[-1,1]\). Quick Tip: Absolute value does not affect continuity.

Checking small values: \[ n=1:\; \frac{4}{2}=2=\frac{2!}{1!1!}\;(not <) \] \[ n=2:\; \frac{16}{3}<\frac{24}{4} \]

Thus the inequality holds for \(n\ge2\). Quick Tip: Test boundary values to determine validity range.

For non-trivial solution, determinant of coefficients must be zero.
Solving the determinant condition gives: \[ \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1. \] Quick Tip: Non-zero solution \(\Rightarrow\) determinant \(=0\).

\[ f'(x)=x^x(\ln x+1) \]
\(f'(x)>0\) when \(\ln x+1>0\Rightarrow x>\tfrac1e\).
This interval is not listed. Quick Tip: Check derivative sign for monotonicity.

Extrema occur at \(x=\pm3\).
\[ y(3)=1,\quad y(-3)=-\frac1{11} \]

Hence the range is \(\left[-\frac1{11},1\right]\). Quick Tip: Find extrema using derivative to get range.

As \(x\to\infty\), \[ a_i^{1/x}\approx1+\frac{\ln a_i}{x}. \]

Taking logarithm and limit gives: \[ \ln L=\sum_{i=1}^n \ln a_i \Rightarrow L=a_1a_2\cdots a_n. \] Quick Tip: Use logarithms to evaluate limits of powers.

Using \[ \cot^{-1}a+\cot^{-1}b=\cot^{-1}\!\left(\frac{ab-1}{a+b}\right), \] \[ \cot^{-1}7+\cot^{-1}8=\cot^{-1}\!\left(\frac{56-1}{15}\right)=\cot^{-1}\!\left(\frac{11}{3}\right). \]
Now, \[ \cot^{-1}\!\left(\frac{11}{3}\right)+\cot^{-1}18 =\cot^{-1}\!\left(\frac{66-1}{\frac{11}{3}+18}\right) =\cot^{-1}3. \] Quick Tip: Use the identity for sum of inverse cotangents to simplify stepwise.

Differentiating \[ -\frac{e^x}{1+\sin x} \]
gives \[ \frac{e^x\cos x}{1+\sin x}. \]
Hence the integral equals \(C-\dfrac{e^x}{1+\sin x}\). Quick Tip: Often verify integrals by differentiating the given options.

\(E=\{2,3,5,7\}\), \(F=\{1,2,3\}\) \[ E\cup F=\{1,2,3,5,7\}. \] \[ P(E\cup F)=0.15+0.23+0.12+0.20+0.07=0.77. \] Quick Tip: Union of events means include all outcomes appearing in either set.

\[ \cos x+\cos3x=2\cos2x\cos x=0. \]
So, \[ \cos x=0 \Rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}, \] \[ \cos2x=0 \Rightarrow x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}. \]
Total roots \(=6\). Quick Tip: Convert sums of cosines into products to count roots easily.

Split at \(x=\frac{\pi}{4}\): \[ \int_0^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx =2(\sqrt2-1)=2\sqrt2-2. \] Quick Tip: For absolute value graphs, split the interval where the expression changes sign.

Using series expansions, \[ \log(\cos x)\sim-\frac{x^2}{2},\quad \log(1+x^2)\sim x^2. \]
Hence \(f(x)\sim -\frac{x}{2}\to0\) as \(x\to0\), so \(f\) is continuous.
Derivative at \(0\) also exists. Hence differentiable. Quick Tip: Use series expansions to test continuity and differentiability at 0.

On the boundary \(x+2y=8\): \[ z=3(8-2y)+2y=24-4y\le24. \]
Maximum occurs at \(y=0\Rightarrow z=24\). Quick Tip: For linear programming, maxima occur at boundary points.

Material used \[ M=2\pi rh+\frac54\pi r^2. \]
For fixed volume \(V=\pi r^2h\), \[ M=2\frac{V}{r}+\frac54\pi r^2. \]
Minimizing gives \(h=\frac54 r\Rightarrow \frac{r}{h}=\frac45.\) Quick Tip: Express material in terms of one variable using volume constraint.

Step 1: \[ n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C) \]

Step 2: Since \(n(A\cap C)\) and \(n(A\cap B\cap C)\) are not fixed, multiple values are possible.

Step 3: Valid arrangements give 26, 27, or 28. Quick Tip: If intersections are not fully specified, multiple answers may be possible.

\[ |f(z)|=\frac{|7-z|}{|1-z^2|} \]

Substituting \(z=1+2i\) and simplifying gives \[ |f(z)|=|z|. \] Quick Tip: For complex functions, use modulus properties to simplify.

Using identity: \[ \cos^{-1}\!\left(\frac{1-t^2}{1+t^2}\right)=2\tan^{-1}t, \]
with \(t=\log x\).
\[ f(x)=2\tan^{-1}(\log x) \]
\[ f'(x)=\frac{2}{1+(\log x)^2}\cdot\frac{1}{x} \]

At \(x=e\), \[ f'(e)=\frac{2}{2}\cdot\frac{1}{e}=\frac{1}{e}. \] Quick Tip: Convert inverse trigonometric expressions using standard identities.

Step 1: Divisibility by 3 depends on sum of digits.

Step 2: Statement–2 gives the rule used to count valid numbers in Statement–1.

Step 3: Hence both are true and Statement–2 explains Statement–1. Quick Tip: Check whether Statement–2 logically justifies Statement–1.

Checking option \(y=x-2\), it satisfies tangency conditions for both curves. Quick Tip: Substitute the line in both equations to verify tangency.

Rotation matrices satisfy: \[ R(s)R(t)=R(s+t). \] Quick Tip: Rotation matrices add angles under multiplication.

Integrating by parts, \[ \int x\log\!\left(1+\frac1x\right)dx =\frac{x^2}{2}\log\!\left(1+\frac1x\right)+\cdots \]

Thus \(f(x)=\dfrac12x^2\). Quick Tip: Look at the coefficient of the logarithmic term after integration.

For three unit vectors equally inclined with mutual dot product \(\cos\theta\), \[ |\vec a\cdot(\vec b\times\vec c)|=\sqrt{\det \begin{pmatrix} 1&\cos\theta&\cos\theta
\cos\theta&1&\cos\theta
\cos\theta&\cos\theta&1 \end{pmatrix}} =(1-\cos\theta)\sqrt{1+2\cos\theta}. \] Quick Tip: Use the Gram determinant for \(|\vec a\cdot(\vec b\times\vec c)|\).

Exponents form \(\sum_{k\ge1}\frac{k}{2^{k+1}}=1\).
Hence product \(=2^{\,1}=2\). Quick Tip: Convert products to sums of exponents.

Using \(\frac{{}^{n}C_r}{{}^{\,r+3}C_r}=\frac{n!(3)!}{(n-r)!(r+3)!}\) and summing gives \(\frac{3}{n+3}\).
Thus \(a=n\Rightarrow a-n=0\). Quick Tip: Match the closed form after simplifying binomial ratios.

Expanding after column operations gives \[ \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=1. \] Quick Tip: Use column reductions to factor linear relations.

Using Bayes’ theorem over possible compositions yields probability \(=3/5\). Quick Tip: Condition on observed outcomes and apply Bayes.

Evaluate plane function at endpoints: values \(-2\) and \(+6\).
Hence ratio \(=|{-2}|:|6|=1:3\). Quick Tip: Use section formula via signed distances to the plane.

Using \(x\to\frac{\pi}{2}-x\) and adding, \[ I=\frac12\int_0^{\pi/2}dx=\frac{\pi}{4}. \] Quick Tip: Exploit symmetry with complementary substitutions.

Let \(\vec v=(x,y,z)\). Solve the linear system from dot products to get \((x,y,z)=(1,2,3)\). Quick Tip: Dot products give linear equations in components.

Slopes are \(-b/a\) and \(-a/b\). \[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\frac{|b^2-a^2|}{2ab}. \] Quick Tip: Use slope form for angle between lines.

Direction vectors: \[ \vec{AB}=(k, -1, 3),\quad \vec{BC}=(2, k, -1). \]
Perpendicular \(\Rightarrow \vec{AB}\cdot\vec{BC}=0\): \[ 2k+k(-1)-3=0\Rightarrow k=-1. \] Quick Tip: Perpendicular lines have zero dot product of direction vectors.