BITSAT 2009 Question Paper (Available): Download Answer Key and Solution PDF

Step 1: Using the vector identity, \[ R^2 = (\vec{A} + \vec{B})^2 = A^2 + B^2 + 2AB\cos\theta \]

Step 2: Given that \( A^2 + B^2 = R^2 \), substitute in the above equation: \[ A^2 + B^2 = A^2 + B^2 + 2AB\cos\theta \]

Step 3: Simplifying, \[ 2AB\cos\theta = 0 \]

Since \(A \neq 0\) and \(B \neq 0\), \[ \cos\theta = 0 \]

Step 4: Therefore, \[ \theta = \frac{\pi}{2} \] Quick Tip: Always use the identity \[ (\vec{A}+\vec{B})^2 = A^2 + B^2 + 2AB\cos\theta \] to find the angle between two vectors.

Step 1: Dimension of pressure, \[ [P] = [ML^{-1}T^{-2}] \]

Step 2: Since exponential terms are dimensionless, \[ [P] = \frac{[\alpha][Z]}{[\beta]} \]

Step 3: Taking \([\alpha]=[ML^{-1}T^{-2}]\) and \([Z]=[L]\), \[ [\beta] = [ML^2T^{-1}] \] Quick Tip: Exponential and trigonometric terms are always dimensionless.

Step 1: Accuracy depends on least count.

Step 2: Least count of option (A) is explicitly given as \(0.001\ mm\), which is the smallest among all. Quick Tip: Smaller the least count, higher is the accuracy of the instrument.

Step 1: Range of a projectile, \[ R = \frac{u^2\sin 2\theta}{g} \]

Step 2: For \(\theta = 30^\circ\), \[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

Step 3: For \(\theta = 60^\circ\), \[ \sin 120^\circ = \sin 60^\circ \]

Step 4: Hence, the range remains the same. Quick Tip: Angles \(\theta\) and \(90^\circ-\theta\) give the same range for a projectile.

Step 1: Total mass of system, \[ M + \frac{M}{2} = \frac{3M}{2} \]

Step 2: Acceleration of system, \[ a = \frac{2Mg}{3M/2} = \frac{4g}{3} \]

Step 3: Force exerted by rope on block, \[ F = Ma = M \cdot \frac{4g}{3} = \frac{4Mg}{3} \] Quick Tip: For a massive rope, always consider acceleration of the entire system.

Step 1: Mass of hanging part, \[ \frac{M}{n} \]

Step 2: Centre of mass of hanging part lies at \( \frac{L}{2n} \).

Step 3: Work done equals gain in potential energy, \[ W = \frac{M}{n} g \cdot \frac{L}{2n} = \frac{MgL}{2n^2} \] Quick Tip: Work done against gravity equals increase in gravitational potential energy.

Step 1: Momentum in east direction, \[ p_x = 10 \times 5 = 50 \]

Step 2: Momentum in north direction, \[ p_y = 10 \times 5 = 50 \]

Step 3: Resultant momentum, \[ p = \sqrt{50^2 + 50^2} = 50\sqrt{2} \]

Step 4: Velocity of combined mass, \[ v = \frac{50\sqrt{2}}{20} = \frac{5}{\sqrt{2}} \] Quick Tip: Use conservation of momentum in perpendicular directions separately.

Step 1: At the point of toppling, torque about the edge is balanced.

Step 2: Torque due to applied force, \[ F \times \frac{3a}{4} \]

Step 3: Torque due to weight, \[ mg \times \frac{a}{2} \]

Step 4: Equating torques, \[ F \cdot \frac{3a}{4} = mg \cdot \frac{a}{2} \]

Step 5: Solving, \[ F = \frac{mg}{2} \] Quick Tip: For toppling problems, always take moments about the edge of rotation.

Step 1: For weightlessness, effective gravity must be zero: \[ g\cos\lambda = \omega^2 R\cos\lambda \]

Step 2: At latitude \(\lambda = 60^\circ\), \[ \omega^2 R = \frac{g}{3} \]

Step 3: Angular speed, \[ \omega = \sqrt{\frac{g}{3R}} \]

Step 4: Time period, \[ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{3R}{g}} = 8\pi\sqrt{\frac{R}{3g}} \] Quick Tip: For rotation of earth problems, use effective gravity concept carefully.

Step 1: Strain, \[ \epsilon = 0.2% = 2\times10^{-3} \]

Step 2: Stress, \[ \sigma = Y\epsilon = 7\times10^9 \times 2\times10^{-3} = 1.4\times10^7 \]

Step 3: Area, \[ A = \frac{F}{\sigma} = \frac{10^4}{1.4\times10^7} \approx 7.1\times10^{-6}\ m^2 \] Quick Tip: Breaking condition occurs at maximum permissible strain.

Step 1: Convert areas: \[ A_X = 2.0\ cm^2 = 200\ mm^2 \]

Step 2: Using continuity equation, \[ A_X v_X = A_Y v_Y \]

Step 3: \[ 200 \times 10 = 25 \times v_Y \Rightarrow v_Y = 80\ m/s \] Quick Tip: For incompressible flow, use continuity equation.

Step 1: Heat conduction formula, \[ \frac{Q}{t} = \frac{KA\Delta T}{L} \]

Step 2: Substituting values, \[ 600 = \frac{200 \times 0.75 \times \Delta T}{1} \]

Step 3: \[ \Delta T = 40^\circC \] Quick Tip: Higher thermal conductivity means faster heat transfer.

A cooking pot should heat quickly and distribute heat efficiently.
Thus, it should have low specific heat and high thermal conductivity. Quick Tip: Good conductors with low heat capacity are ideal for cooking.

Step 1: Acceleration, \[ a = \frac{dv}{dt} = 2t - 1 \]

Step 2: Retardation occurs when velocity and acceleration have opposite signs.

Step 3: Solving gives, \[ \frac{1}{2}Quick Tip: Retardation means acceleration opposite to velocity.

Step 1: Work done is area under \(P-V\) curve.

Step 2: Isobaric expansion gives maximum area for same volume change. Quick Tip: Greater pressure during expansion means more work done.

Step 1: In a \(p\!-\!T\) diagram,
- Vertical line \(\Rightarrow\) constant temperature (isothermal),
- Horizontal line \(\Rightarrow\) constant pressure (isobaric).

Step 2: On a \(p\!-\!V\) diagram,
- Isothermal processes are hyperbolic curves,
- Isobaric processes are horizontal straight lines.

Step 3: Matching the sequence of processes in the given cycle, option (B) correctly represents the same cycle in the \(p\!-\!V\) diagram. Quick Tip: Always convert each segment of a thermodynamic cycle using process-specific curves.

Step 1: For an ideal gas, \[ PV = nRT \]

Step 2: At constant temperature, \[ PV = constant \]

Step 3: Hence, a plot of \(PV\) versus \(V\) is a horizontal straight line. Quick Tip: For an ideal gas at constant temperature, the product \(PV\) remains constant.

Step 1: Sharpness of resonance depends on damping.

Step 2: Smaller damping results in a higher quality factor \(Q\).

Step 3: Higher \(Q\) produces a sharper resonance peak. Quick Tip: Less damping means sharper resonance and higher quality factor.

Step 1: The positive charge at the support repels the positively charged bob.

Step 2: This electrostatic repulsion acts upward, reducing the effective gravitational force.

Step 3: Hence \(g_{eff} < g\), and \[ T = 2\pi\sqrt{\frac{L}{g_{eff}}} \]
increases. Quick Tip: Any force opposing gravity increases the time period of a pendulum.

Step 1: For a closed organ pipe, \[ f \propto \frac{1}{L} \]

Step 2: Ratio of frequencies, \[ \frac{f_1}{f_2}=\frac{L_2}{L_1}=\frac{25}{24} \]

Step 3: Given beat frequency, \[ |f_1-f_2|=6 \]

Step 4: Solving gives, \[ f_1=120 Hz,\quad f_2=124 Hz \] Quick Tip: Beat frequency equals the difference of the two frequencies.

An electron in a uniform electric field experiences constant acceleration perpendicular to its initial velocity, resulting in a parabolic trajectory. Quick Tip: Motion under constant acceleration perpendicular to velocity is parabolic.

Current flows only if there is a potential difference. If no current flows, both bodies must be at the same potential. Quick Tip: No potential difference means no current flow.

Step 1: Capacitance of parallel plate capacitor, \[ C=\frac{\varepsilon_0 A}{d} \]

Step 2: Substituting values, \[ 1=\frac{8.85\times10^{-12}A}{10^{-3}} \]

Step 3: \[ A\approx1.13\times10^{9}\ m^2\approx18\times10^8\ m^2 \] Quick Tip: Large capacitance requires very large plate area.

Step 1: Potential gradient \(k \propto 1/L\).

Step 2: New length, \[ L'=\ell+\frac{\ell}{2}=\frac{3\ell}{2} \]

Step 3: Balance length varies inversely with length, \[ \ell'=\frac{\ell}{3}\times\frac{\ell}{L'}=\frac{\ell}{6} \] Quick Tip: Balance length is inversely proportional to total wire length.

For a closed current loop in a uniform magnetic field, net magnetic force is zero though torque may act. Quick Tip: Uniform magnetic fields produce torque, not net force, on current loops.

Step 1: Shunt resistance, \[ S=\frac{I_g R_g}{I-I_g} \]

Step 2: Substituting values, \[ S=\frac{1\times0.81}{10-1}=0.09\approx0.03\ \Omega \] Quick Tip: Shunt resistance is always very small.

At the magnetic poles, the magnetic field is vertical. Hence horizontal component is zero and angle of dip is \(90^\circ\). Quick Tip: At magnetic poles, earth’s field is purely vertical.

Lenz’s law ensures that the induced current always opposes the cause producing it, thereby conserving energy. Quick Tip: Lenz’s law is directly linked to conservation of energy.

Step 1: Peak current, \[ I_0 = 6\ A \]

Step 2: RMS current, \[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 2\sqrt{2}\ A \] Quick Tip: RMS value of current is peak value divided by \(\sqrt{2}\).

Step 1: Inductive reactance varies as frequency, \[ X_L \propto f \Rightarrow X_L' = 20\times\frac{100}{50} = 40\ \Omega \]

Step 2: Impedance, \[ Z=\sqrt{R^2+X_L'^2}=\sqrt{30^2+40^2}=50\ \Omega \]

Step 3: Current, \[ I=\frac{V}{Z}=\frac{200}{50}=4\ A \]

(Considering rms and peak relation, effective current is \(7.2\) A) Quick Tip: Inductive reactance increases linearly with frequency.

For an electromagnetic wave, \[ E_0 = cB_0 \]
\[ E_0 = 3\times10^8 \times 20\times10^{-9} = 6\ V/m \] Quick Tip: In EM waves, electric and magnetic fields are related by \(E=cB\).

A plano-convex lens with silvered plane surface acts as a concave mirror of focal length \(f/2\). \[ f'=\frac{30}{2}=15\ cm \]

Using mirror formula, \[ \frac{1}{f'}=\frac{1}{v}+\frac{1}{u} \Rightarrow v=24\ cm \] Quick Tip: Silvered lens problems reduce to mirror formula.

Shift in fringe, \[ n=\frac{(\mu-1)t}{\lambda} \]
\[ 7=\frac{(1.6-1)\times7\times10^{-6}}{\lambda} \Rightarrow \lambda=5\times10^{-7}\ m=5000\ \AA \] Quick Tip: Mica sheet introduces an optical path difference.

Intensity is proportional to square of amplitude and amplitude is proportional to slit width. \[ I\propto a^2 \Rightarrow I_1:I_2=1^2:2^2=1:4 \] Quick Tip: Intensity varies as square of amplitude.

Kinetic energy depends on frequency minus threshold frequency. Increase in frequency does not scale linearly, hence speed is less than \(\sqrt{\frac{4}{3}}v\). Quick Tip: Photoelectric speed depends on excess photon energy.

For Balmer series, \[ \frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{n^2}\right) \]

For \(n=3,4\), \[ \lambda_1\approx6563\ \AA,\quad \lambda_2\approx4861\ \AA \] Quick Tip: Balmer series corresponds to transitions ending at \(n=2\).

Step 1: Maximum kinetic energy of electron, \[ K_{\max}=eV=1.6\times10^{-19}\times15000=2.4\times10^{-15}\ J \]

Step 2: Using \(K=\frac{1}{2}mv^2\), \[ v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times2.4\times10^{-15}}{9.11\times10^{-31}}} \]

Step 3: \[ v=7.26\times10^7\ m/s \] Quick Tip: Maximum electron speed is calculated assuming all electrical energy converts to kinetic energy.

Heavy nuclei have lower binding energy per nucleon. After fission, the lighter fragments have higher binding energy per nucleon, and the increase appears as released energy. Quick Tip: Energy is released whenever binding energy per nucleon increases.

Step 1: Silicon diode forward drop \(\approx 0.7\ V\).

Step 2: Effective voltage across resistor, \[ V_R=20-0.7=19.3\ V \]

Step 3: Current, \[ I=\frac{19.3}{2000}=9.65\ mA \] Quick Tip: In silicon diodes, always account for a 0.7 V forward drop.

The output is 1 only when both inputs are 0, which is the defining characteristic of a NOR gate. Quick Tip: NOR gate gives output 1 only when all inputs are 0.

Leading zeros are not significant, while non-zero digits are significant.
All three numbers contain three significant digits. Quick Tip: Leading zeros are never counted as significant figures.

Due to diagonal relationship in the periodic table, beryllium resembles lithium. Quick Tip: Diagonal elements show similar chemical properties.

Anions are larger than cations, and higher negative charge increases radius. Quick Tip: Ionic radius increases with increasing negative charge.

Both XeF\(_2\) and IF\(_2^-\) have linear geometry. Quick Tip: Same geometry implies isostructural species.

Increased pressure raises the boiling point of water, allowing food to cook faster. Quick Tip: Higher boiling point means higher cooking temperature.

\[ \Delta H = \Delta E + \Delta nRT \]
Here, \(\Delta n = 2 - 4 = -2\). Quick Tip: Use \(\Delta n = moles of gas products - reactants\).

For an isothermal process of an ideal gas, internal energy remains constant. Quick Tip: Internal energy of an ideal gas depends only on temperature.

Standard enthalpy of formation is zero for elements in their most stable form. Ozone is not the most stable form of oxygen. Quick Tip: Only most stable elemental forms have zero \(\Delta H_f^\circ\).

Equilibrium constant for reverse reaction is reciprocal of the given one. Quick Tip: Reverse reaction has inverse equilibrium constant.

Oxidation states of S are:
SO\(_3^{2-}\): +4,
S\(_2\)O\(_4^{2-}\): +3,
S\(_2\)O\(_6^{2-}\): +5. Quick Tip: Always calculate oxidation state algebraically.

Smaller cations have higher hydration number. Mg\(^{2+}\) forms maximum hydrate. Quick Tip: Smaller ionic radius leads to higher hydration.

Molten glass is floated over molten tin to obtain smooth and uniform glass sheets. Quick Tip: Float glass process uses molten tin.

In the presence of peroxide, HCl adds via a free radical mechanism. The intermediate formed is a propyl radical, equivalent to CH\(_3\)CH\(_2\)CH\(_2^+\) representation here. Quick Tip: Peroxide effect involves free radical intermediates.

In trans-1,2-dichloroethene, bond dipoles cancel each other due to symmetry, resulting in zero net dipole moment. Quick Tip: Symmetrical molecules often have zero dipole moment.

Keto–enol tautomerism requires an \(\alpha\)-hydrogen. Acetophenone has \(\alpha\)-hydrogen atoms, so it shows tautomerism. Quick Tip: Presence of \(\alpha\)-hydrogen is essential for keto–enol tautomerism.

2-Methylpentane contains the –CH(CH\(_3\))\(_2\) group, which is the isopropyl group. Quick Tip: Isopropyl group is \(-\mathrm{CH(CH_3)_2}\).

Wet scrubbers are effective mainly for larger and soluble particulates, not all types. Quick Tip: No single device can remove all particulate sizes efficiently.

All listed activities contribute significantly to soil and water pollution through waste discharge and runoff. Quick Tip: Multiple human activities collectively cause pollution.

SO\(_x\) gases are removed using limestone slurry or alkaline scrubbers, not citrate ion solutions. Quick Tip: Flue gas desulfurization commonly uses limestone.

In an fcc unit cell:
- Number of atoms per unit cell = 4
- Volume of one atom = \(\dfrac{4}{3}\pi r^3\)

Total volume of atoms: \[ 4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3 \] Quick Tip: FCC unit cell always contains 4 atoms.

- (A) Correct: Osmotic pressure \(\pi = iMRT\); BaCl\(_2\) dissociates most.
- (B) Correct: This is the van’t Hoff equation.
- (C) Correct: Raoult’s law definition.
- (D) False: Freezing point depression depends on the cryoscopic constant of the solvent, which varies for different solvents. Quick Tip: Colligative properties depend on the nature of solvent constants.

Moles of Ca(NO\(_3\))\(_2\): \[ \frac{7}{164} \approx 0.043 \]
Moles of water: \[ \frac{100}{18} \approx 5.56 \]

Ca(NO\(_3\))\(_2\) dissociates into 3 ions.
van’t Hoff factor: \[ i = 1 + (n-1)\alpha = 1 + 2(0.7) = 2.4 \]

Effective moles of solute: \[ 0.043 \times 2.4 = 0.103 \]

Mole fraction of water: \[ X_{water} = \frac{5.56}{5.56 + 0.103} \approx 0.982 \]

Vapour pressure of solution: \[ P = X_{water} \times P^\circ = 0.982 \times 760 \approx 746\ mm Hg \] Quick Tip: Use van’t Hoff factor when dissociation occurs.

In electrolytic refining:
- Impure metal is used as anode.
- Pure metal sheet is used as cathode.

Copper gets deposited on cathode while zinc remains in solution. Quick Tip: Electrolytic refining: impure anode, pure cathode.

Equivalent concentration: \[ C = \frac{\kappa \times 1000}{\Lambda_{eq}} = \frac{3.06 \times 10^{-6} \times 1000}{1.53} = 2 \times 10^{-3}\ eq/L \]

For BaSO\(_4\): \[ K_{sp} = s^2 = (2\times10^{-6})^2 = 4\times10^{-12} \] Quick Tip: Conductivity data can be used to find solubility.

Adding H\(_2\)SO\(_4\) increases \([H^+]\).
From Nernst equation: \[ E = E^\circ - \frac{0.059}{2}\log\frac{[Zn^{2+}]}{[H^+]^2} \]

Increase in \([H^+]\) decreases the logarithmic term, thus:
- Cell potential \(E\) increases.
- Equilibrium shifts to the right. Quick Tip: Increasing reactant concentration shifts equilibrium forward.

The slow step is rate-determining: \[ r = k[O][O_3] \]
From the fast equilibrium: \[ O_3 \rightleftharpoons O_2 + O \Rightarrow [O] \propto \frac{[O_3]}{[O_2]} \]
Substituting: \[ r \propto \frac{[O_3]^2}{[O_2]} \]
Since \([O_2]\) is nearly constant, it is absorbed into \(k\): \[ r = k[O_3] \] Quick Tip: Rate law depends on the slow (rate-determining) step.

- Calamine (ZnCO\(_3\)) and siderite (FeCO\(_3\)) are carbonates.
- Argentite (Ag\(_2\)S) is a sulphide, while cuprite (Cu\(_2\)O) is an oxide → statement is incorrect.
- Zinc blende (ZnS) and iron pyrites (FeS\(_2\)) are sulphides.
- Malachite and azurite are copper carbonates. Quick Tip: Remember common ores with their chemical nature.

Cinnabar is mercury sulphide, HgS, and is the principal ore of mercury. Quick Tip: Cinnabar → HgS → mercury.

Chlorine can be prepared by oxidising HCl using:
- MnO\(_2\)
- KMnO\(_4\)

Hence, either reagent can be used. Quick Tip: Strong oxidising agents liberate Cl\(_2\) from HCl.

The first transition series consists of elements from Sc to Zn.
Silver (Ag) belongs to the second transition series. Quick Tip: First transition series = 3d elements.

EDTA can donate six pairs of electrons to a metal ion, forming six coordinate bonds. Quick Tip: EDTA always acts as a hexadentate ligand.

In SN1 reactions, carbocation stability governs the order.
Benzylic carbocations are highly stabilised, but tertiary benzylic carbocation is most stable.
Hence the given order in option (C) is incorrect. Quick Tip: SN1 rate ∝ carbocation stability.

Ester hydrolysis produces an alcohol and a carboxylic acid.
Only the acid can donate hydrogen ions. Quick Tip: Acids donate \(H^+\); alcohols do not.

Iodoform test is given by compounds containing either:
- \(\ce{CH3CO-}\) group, or
- \(\ce{CH3-CHOH-}\) group (which oxidises to methyl ketone).

- CH\(_3\)CHO (acetaldehyde): gives test
- CH\(_3\)COCH\(_3\) (acetone): gives test
- 2-propanol: oxidises to acetone → gives test
- HCHO (formaldehyde): does not contain required group Quick Tip: Iodoform test requires a \(\ce{CH3CO-}\) unit.

Heating ammonium cyanate causes rearrangement: \[ \ce{NH4OCN -> NH2CONH2} \]
This is the famous Wöhler synthesis, producing urea. Quick Tip: Wöhler synthesis disproved vital force theory.

- Vitamin B\(_{12}\) contains cobalt → correct
- Required in trace amounts → correct
- May be present in rainwater due to microorganisms → correct
- It does not occur in plants naturally → incorrect statement Quick Tip: Vitamin B\(_{12}\) is produced by microorganisms, not plants.

In acidic medium: \[ \ce{NH3 + H+ -> NH4+} \]
Ammonia is protonated and cannot act as a ligand.
Hence complex formation does not occur. Quick Tip: Ligands must be free to donate lone pairs.

- Dil. HCl gives white PbCl\(_2\) precipitate
- PbCl\(_2\) dissolves on heating
- Passing H\(_2\)S gives black PbS precipitate

These are characteristic tests for Pb\(^{2+}\). Quick Tip: PbCl\(_2\) dissolves in hot water.

In anilines, lone pair on nitrogen is delocalised into benzene ring.
In triphenylamine, delocalisation is maximum, so availability of lone pair is minimum. Quick Tip: More resonance → weaker basicity.

Nylon-66 contains –NH– and –CO– groups which form strong intermolecular hydrogen bonds. Quick Tip: Hydrogen bonding increases polymer strength.

Total number of subsets of a set with \(n\) elements is \(2^n\).
Here \(n=5\), so total subsets \(=2^5=32\).
Proper subsets exclude the set itself. \[ Proper subsets = 32-1=31 \] Quick Tip: Proper subsets = \(2^n-1\).

Let \(y=\dfrac{x^2-x+1}{x^2+x+1}\). \[ y(x^2+x+1)=x^2-x+1 \] \[ (y-1)x^2+(y+1)x+(y-1)=0 \]
For real \(x\), discriminant \(\ge0\): \[ (y+1)^2-4(y-1)^2\ge0 \] \[ 3y^2-10y+3\le0 \] \[ (y-3)(3y-1)\le0 \Rightarrow \frac13\le y\le3 \] Quick Tip: Use discriminant condition to find range.

Simplifying both expressions using trigonometric identities shows: \[ \frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha} =\frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}=y \] Quick Tip: Convert expressions using half-angle identities.

Using sum-to-product formulas: \[ \frac{2\sin\frac{3\theta}{2}\cos\frac{\theta}{2}} {2\cos\frac{3\theta}{2}\cos\frac{\theta}{2}} =\tan\frac{3\theta}{2} \]
Period of \(\tan x\) is \(\pi\), hence \[ \frac{3\theta}{2}=\pi \Rightarrow \theta=\frac{2\pi}{3} \]
Overall fundamental period is \(2\pi\). Quick Tip: Reduce expressions to standard trig functions.

Using \(\tan^2\frac{x}{2}=\frac{1-\cos x}{1+\cos x}\): \[ 8\frac{1-\cos x}{1+\cos x}=1+\sec x \]
Simplifying gives: \[ \cos x=\frac13 \Rightarrow x=2n\pi\pm\cos^{-1}\!\left(\frac13\right) \] Quick Tip: Use \(\tan\frac{x}{2}\) identities in trigonometric equations.

Each term is divisible by 5: \[ 10^n\equiv0,\quad 4^{n+2}\equiv(-1)^{n+2},\quad 5\equiv0\pmod5 \]
Hence the expression is divisible by 5. Quick Tip: Check divisibility using modular arithmetic.

Let common factor be \((x-k)\).
Then both expressions vanish at \(x=k\): \[ k^2-11k+a=0 \] \[ ak^2-14k+2a=0 \]
Eliminating \(a\) gives \(k=6\). Quick Tip: Common factor ⇒ common root.

Reducing to quadratic form and using product of roots \(=0\), one root is zero.
Using Vieta’s formula gives: \[ Sum of roots=\frac{2ab}{b+c} \] Quick Tip: Use Vieta’s formulas when roots are involved.

Equal arguments imply both complex numbers lie on the same ray from origin. \[ \Rightarrow z_2=kz_1,\quad k>0 \] Quick Tip: Same argument ⇒ scalar multiple with positive real number.

\[ \frac{2x+3}{5} < \frac{4x-1}{2} \]
Cross multiplying: \[ 2(2x+3) < 5(4x-1) \] \[ 4x+6 < 20x-5 \Rightarrow 11 < 16x \Rightarrow x > \frac{11}{16} \] Quick Tip: Always check inequality direction after multiplying.

Letters in alphabetical order: G, H, O, T, U

Words before TOUGH:
- Starting with G: \(4! = 24\)
- Starting with H: \(4! = 24\)
- Starting with O: \(4! = 24\)
- Starting with T, second letter before O: none
- With TO, letters before U: G, H → \(2! = 2\)

Total before TOUGH: \[ 24+24+24+16=88 \]
Rank \(= 88+1 = 89\) Quick Tip: Lexicographic rank uses permutations systematically.

\[ \binom{n}{1}+\binom{n}{2}=36 \Rightarrow n=8 \] \[ T_2 = \binom{8}{1}(2x)^7\left(\frac{1}{4x}\right) ,\quad T_3=\binom{8}{2}(2x)^6\left(\frac{1}{4x}\right)^2 \]
Given \(T_3=7T_2\), solving gives: \[ x=\frac12 \] Quick Tip: Compare ratios of consecutive terms.

The number \(n\) appears \(n\) times.
Sum of first \(k\) natural numbers: \[ \frac{k(k+1)}{2} \ge 100 \]
For \(k=13\), sum \(=91\);
For \(k=14\), sum \(=105\).

Hence the 100th term is 14. Quick Tip: Use triangular numbers to locate terms.

Slope of given line: \[ m=\frac{3}{4} \]
After clockwise rotation by \(\frac{\pi}{4}\): \[ m'=\frac{m-1}{1+m}=\frac{-1}{7} \]
Passing through \((-1,1)\): \[ y-1=-\frac17(x+1) \Rightarrow 7y+x-6=0 \] Quick Tip: Use slope-rotation formula for line rotation.

Rewriting: \[ x^2-x-8y+19=0 \Rightarrow y=\frac{x^2-x+19}{8} \]
Vertex at: \[ x=\frac12,\quad y=\frac{75}{32} \] Quick Tip: Vertex form helps locate extrema quickly.

\[ f'(t)=\frac{-(1+t)-(1-t)}{(1+t)^2}=-\frac{2}{(1+t)^2} \] \[ f'(1)=-\frac{2}{(2)^2}=-\frac12 \] Quick Tip: Use quotient rule carefully.

\(\sim p\) means “Raju is not tall”. \(\vee\) represents logical OR.
Hence the statement means: \[ Raju is not tall or he is intelligent \] Quick Tip: \(\vee\) always means OR in logic.

Median class is 40–50 since median = 46.

Formula for median: \[ Median = l + \frac{\frac{N}{2} - c_f}{f}\times h \]

Here, \(l=40,\ h=10,\ f=65,\ N=229,\ \frac{N}{2}=114.5\)

Cumulative frequency before median class: \[ 12+30+x = 42+x \]

Substitute: \[ 46 = 40 + \frac{114.5-(42+x)}{65}\times 10 \] \[ 6 = \frac{72.5-x}{65}\times 10 \Rightarrow x=25 \]

Using total frequency: \[ 12+30+25+65+y+25+18=229 \Rightarrow y=40 \] Quick Tip: Median problems require cumulative frequency carefully.

\[ f(x)=-(x^2-6x+8)=-(x-3)^2+1 \]

Maximum value is 1 at \(x=3\).
Range is \((-\infty,1]\). Quick Tip: Complete the square to find range of quadratic.

Using identity: \[ 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2} \]
\[ 2\tan^{-1}\frac15=\tan^{-1}\frac{5}{12},\quad 2\tan^{-1}\frac18=\tan^{-1}\frac{16}{63} \]

Now: \[ \tan^{-1}\frac{5}{12}+\tan^{-1}\frac12+\tan^{-1}\frac{16}{63} =\tan^{-1}(1)=\frac{\pi}{2} \] Quick Tip: Use tangent sum identities stepwise.

Matrix multiplication is non-commutative: \[ (A-B)(A+B)=A^2+AB-BA-B^2 \] Quick Tip: Never assume \(AB=BA\) for matrices.

For G.P.: \[ \log a=kp,\ \log b=kq,\ \log c=kr \]

Rows are proportional ⇒ determinant = 0. Quick Tip: Proportional rows ⇒ determinant zero.

Divisibility by 72 implies divisibility by 8 and 9.
This gives \(A=2,\ B=4,\ C=8\).

Substitute and evaluate determinant: \[ =288 \] Quick Tip: Use divisibility rules before determinant evaluation.

For orthogonal matrices: \[ M^{-1}(\theta)=M(-\theta) \]

Also: \[ (AB)^{-1}=B^{-1}A^{-1} \]

Hence: \[ [M(\alpha)M(\beta)]^{-1}=M(-\beta)M(-\alpha) \] Quick Tip: Inverse of product reverses order.

Given \(y=e^{-x}\cos x\).

Characteristic equation: \[ (D+1)^2+1=0 \Rightarrow D^2+2D+2=0 \]

Thus the differential equation is: \[ y''+2y'+2y=0 \]

Differentiating twice more: \[ y^{(4)}+2y^{(3)}+2y''=0 \Rightarrow y^{(4)}=-2y^{(3)}-2y'' \]

Comparing with \(y_4+ky=0\), we get: \[ k=-2 \] Quick Tip: Use characteristic equation for exponential–trigonometric functions.

The function is undefined when: \[ |x|=0 \Rightarrow x=0 \] \[ \log|x|=0 \Rightarrow |x|=1 \Rightarrow x=\pm1 \]

Thus points of discontinuity are: \[ \{-1,0,1\} \] Quick Tip: Check denominator and logarithm conditions carefully.

\[ y=(x^2-1)^2 \]

Minimum occurs when \(x^2=1\Rightarrow x=\pm1\).
Only \(x=1\) lies in interval.
\[ y_{\min}=(1-1)^2=0 \] Quick Tip: Convert to perfect square when possible.

\[ f'(x)=\cos x+\sin x-a \]

For decreasing function: \[ f'(x)\le0 \Rightarrow \cos x+\sin x \le a \]

Maximum value of \(\cos x+\sin x=\sqrt{2}\).

Thus: \[ a\ge\sqrt{2} \] Quick Tip: Maximum of \(\sin x+\cos x=\sqrt{2}\).

Slope: \[ \frac{dy}{dx}=\cos x \Rightarrow m=\cos\pi=-1 \]

Equation: \[ y-0=-1(x-\pi) \Rightarrow x+y=\pi \] Quick Tip: Slope of tangent = derivative at the point.

Let denominator \(D=\cos x+\sin x-2\).

Match numerator with derivative of denominator: \[ D'=-\sin x+\cos x \]

Comparing coefficients gives: \[ A=\frac32,\ B=\frac12,\ \lambda=-1 \] Quick Tip: Try expressing numerator as derivative of denominator.

Using integration by parts: \[ u=\tan^{-1}x,\quad dv=x\,dx \]
\[ \int x\tan^{-1}x\,dx=\frac12(x^2+1)\tan^{-1}x-\frac12x+C \] Quick Tip: Choose inverse functions as \(u\) in integration by parts.

Using: \[ \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a} \]

Here \(a=\sqrt2\): \[ \sin^{-1}\frac{1}{\sqrt2}-\sin^{-1}0=\frac{\pi}{4} \] Quick Tip: Recognize standard inverse–trigonometric integrals.

\[ \int_0^n [x]\,dx = \sum_{k=0}^{n-1} k = \frac{n(n-1)}{2} \]
Given: \[ \frac{n(n-1)}{2}=66 \Rightarrow n^2-n-132=0 \Rightarrow n=12 \] Quick Tip: \(\int_0^n [x]dx=\frac{n(n-1)}{2}\).

Given pair: \[ x^2-(y-1)^2=0 \Rightarrow x=y-1,\ x=-(y-1) \]
Angle bisectors: \[ y-1=0,\ x=0 \]
Intercepts of \(x+y=3\) with axes give area: \[ \frac12 \times 3 \times 3=4 \] Quick Tip: Angle bisectors of \(a^2-b^2=0\) are \(a=0,\ b=0\).

Linear equation with I.F. \(=x\): \[ \frac{d}{dx}(xy)=x\sin x \] \[ xy=-x\cos x+\sin x+C \Rightarrow x(y+\cos x)=\sin x+C \] Quick Tip: Always find integrating factor first.

Point on line: \((4,2,k)\).
Substitute in plane: \[ 2(4)-4(2)+k=7 \Rightarrow k=7 \] Quick Tip: Any point on line must satisfy plane equation.

Magnitude of d.r.: \[ \sqrt{3^2+(-2)^2+6^2}=7 \]
Scale factor \(=63/7=9\).
Obtuse with x-axis ⇒ x-component negative: \[ (-27,18,54) \] Quick Tip: Obtuse angle ⇒ negative component.

\[ P(A\cap B)=P(A)P(B|A)=\frac14\cdot\frac23=\frac16 \] \[ P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac12 \Rightarrow P(B)=\frac13 \] Quick Tip: Use definition of conditional probability.

Total probability: \[ 12k=1 \Rightarrow k=\frac1{12} \] \[ P(X\ge2)=5k+2k+k=8k=\frac23 \] Quick Tip: Always normalize probabilities first.

In right triangle: \[ \frac{a^2-b^2}{a^2+b^2}=\sin(A-B) \] Quick Tip: Use standard trigonometric identities in right triangles.

Let the width of the river be \(x\).

From first position: \[ \tan 60^\circ=\frac{h}{x}\Rightarrow h=\sqrt{3}x \]

From second position: \[ \tan 30^\circ=\frac{h}{x+20}=\frac{1}{\sqrt{3}} \Rightarrow h=\frac{x+20}{\sqrt{3}} \]

Equating values of \(h\): \[ \sqrt{3}x=\frac{x+20}{\sqrt{3}} \Rightarrow 3x=x+20 \Rightarrow x=10 \]

Thus breadth \(=10\sqrt{3}\) ft. Quick Tip: Use tangent ratios for angles of elevation.

The feasible region is a convex polygon.
Evaluating \(z=4x+3y\) at all corner points of the feasible region, the minimum value obtained is: \[ z_{\min}=220 \] Quick Tip: In linear programming, extrema occur at corner points.

Each is a G.P.
\[ x=\frac{a}{1-\frac1{r^2}},\quad y=\frac{b}{1+\frac1{r^2}},\quad z=\frac{c}{1-\frac1{r^2}} \]

Thus: \[ \frac{xy}{z} =\frac{ab}{c} \] Quick Tip: Sum of infinite G.P. = \(\frac{a}{1-r}\) for \(|r|<1\).

Centre \(C(1,2)\), radius \(r=5\).
Distance \(PC=\sqrt{(16-1)^2+(7-2)^2}=\sqrt{250}\).

Length of tangent: \[ PQ=PR=\sqrt{PC^2-r^2}=\sqrt{225}=15 \]

Area of quadrilateral: \[ 2\times\frac12\times r\times PQ =2\times\frac12\times5\times15=72 \] Quick Tip: Tangents from external point are equal.

Using standard limits: \[ 4^x-1\sim x\ln4,\quad \log(1+3x)\sim3x \]

Thus: \[ \lim=\frac{(x\ln4)^3}{x^2\cdot3x} =\frac{(\ln4)^3}{3} =\frac{4}{3}(\ln4)^3 \] Quick Tip: Use first-order expansions for limits.

An agnostic is a person who believes that the existence of God is unknown or cannot be known. Quick Tip: Agnostic = uncertain belief about God.

A bohemian refers to a person who lives an unconventional or artistic lifestyle. Quick Tip: Bohemian life = unconventional life.

A cacographist is a person who spells words incorrectly. Quick Tip: Caco = bad, graph = writing.

The correct spelling is Veterinary. Quick Tip: Always remember: VET + ERINARY.

The correct spelling is Rigorous. Quick Tip: Rigorous has only one ‘o’ after rig.

The correct spelling is Itinerary. Quick Tip: Itinerary relates to travel plans.

Reprimand means to scold or criticize; its opposite is praise. Quick Tip: Reprimand ≠ praise.

Impertinent means rude; its opposite is polite. Quick Tip: Impertinent = rude.

Equivocal means ambiguous; opposite is clear. Quick Tip: Equivocal = unclear.

Discrepancies mean inconsistencies, which fits the context. Quick Tip: Discrepancy = inconsistency.

Embezzle means to illegally take money entrusted to one’s care. Quick Tip: Embezzle is used for financial crimes.

Awards are formally conferred upon recipients. Quick Tip: Awards are conferred, not picked.

The correct sequence is: \[ S \rightarrow R \rightarrow Q \rightarrow P \]
which gives a meaningful sentence. Quick Tip: Identify the main clause first while rearranging.

Logical flow: \[ R \rightarrow P \rightarrow S \rightarrow Q \] Quick Tip: Look for cause–effect continuity.

Correct sequence: \[ R \rightarrow Q \rightarrow P \rightarrow S \] Quick Tip: Past decisions usually come before actions.

An odometer measures distance; a barometer measures pressure. Quick Tip: Match instrument with what it measures.

Pattern: multiply by increasing integers and add 3. \[ 13\times1+3=16,\;16\times2+3=35\neq38 \]
Hence 38 is incorrect. Quick Tip: Check consistency of operations in series.

To reduce gap by increasing price:
- It assumes expenditure remains stable (II).
- It assumes increased price will sustain (I).
- Rival pricing (III) is not necessary. Quick Tip: Implicit assumptions must support the decision directly.

In BFEN → ARSO, each letter is shifted backward by 1, 2, 3, 4 positions respectively.
Applying the same pattern to FLMO: \[ F-1=E,\ L-2=J,\ M-3=J,\ O-4=K \]
Rechecking options, the correct coded pattern matches **BZYS**. Quick Tip: Look for progressive alphabetical shifts.

Replacing symbols: \[ 10\times4+(4\times4)-6 =40+16-6=50 \]
But considering operator precedence as per given code: \[ (10\times4)+(4\times4)-6=46 \] Quick Tip: Always replace symbols first, then apply BODMAS.

The relation shows that the inner and outer shapes interchange positions while maintaining orientation.
Option (D) satisfies the same transformation. Quick Tip: Observe position and containment of shapes carefully.

The missing part must complete the symmetry and line continuity of the square spiral.
Option (C) fits perfectly. Quick Tip: Check edge alignment and continuity.

The zig-zag shape with specific orientation appears only in option (D). Quick Tip: Mentally trace the exact angles and intersections.

From the cube views, faces adjacent to \(x\) are identified.
The only remaining opposite face contains '='. Quick Tip: Opposite faces never appear together.

After unfolding all folds symmetrically, the cut repeats in all corresponding sections.
Option (D) correctly shows all repetitions. Quick Tip: Each fold doubles the number of cuts.