Step 1:
For SHM starting from mean position, \[ x = A \sin(\omega t) \]

Step 2:
Given, \[ x = \frac{A}{\sqrt{2}} \Rightarrow \sin(\omega t) = \frac{1}{\sqrt{2}} \]

Step 3: \[ \omega t = \frac{\pi}{4}, \quad \omega = \frac{2\pi}{T} \]

Step 4: \[ t = \frac{\pi/4}{2\pi/T} = \frac{T}{8} \] Quick Tip: In SHM, standard sine values (\(\pi/6, \pi/4, \pi/3\)) directly give time fractions of the period.

Step 1:
The block executes SHM with: \[ \omega = \sqrt{\frac{K}{m}}, \quad T = 2\pi \sqrt{\frac{m}{K}} \]

Step 2:
From the figure, motion from \(P\) to \(Q\) corresponds to \(\frac{3}{8}\) of one complete oscillation.

Step 3: \[ t = \frac{3}{8} T = \frac{3}{8} \times 2\pi \sqrt{\frac{m}{K}} \]

Step 4: \[ t = \frac{3\pi}{4} \sqrt{\frac{m}{K}} \] Quick Tip: Always convert distance-based SHM motion into fractions of a full oscillation using symmetry.

Step 1:
Energy supplied: \[ Q = Pt = 100 \times 30 = 3000 \, J \]

Step 2:
Argon is monoatomic: \[ C_V = \frac{3}{2}R \]

Step 3: \[ 3000 = \frac{3}{2} \times 8.314 \times \Delta T \]

Step 4: \[ \Delta T = \frac{3000}{12.471} \approx 360.8 \, K \] Quick Tip: In thermally isolated containers, all heater energy increases internal energy—no work done.

Step 1:
Moment of inertia of solid sphere about its center: \[ I_{cm} = \frac{2}{5}MR^2 \]

Step 2:
Using parallel axis theorem (tangential axis): \[ I = I_{cm} + MR^2 \]

Step 3: \[ I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2 = 1.4MR^2 \]

Step 4:
Including contribution due to tangential rod attachment: \[ I = 1.6MR^2 \] Quick Tip: For tangential axes, always add \(MR^2\) using the parallel axis theorem.

Step 1:
Moment of inertia of disc: \[ I = \frac{1}{2}MR^2 \]

Step 2: \[ M = 0.1 \, kg, \quad R = 0.04 \, m \Rightarrow I = 8 \times 10^{-5} \]

Step 3:
Torsional oscillation relation: \[ T = 2\pi \sqrt{\frac{I}{C}} \Rightarrow C = \frac{4\pi^2 I}{T^2} \]

Step 4: \[ C = \frac{4 \times 10 \times 8 \times 10^{-5}}{(0.2)^2} = 4 \times 10^{-2} \] Quick Tip: Use \( C = \frac{4\pi^2 I}{T^2} \) directly for torsional pendulum problems.

Step 1:
Initial position is midpoint, distance from each mass: \[ r = \frac{l}{2} \]

Step 2:
Initial gravitational potential energy: \[ U = -\frac{Gm}{r} - \frac{Gm}{r} = -\frac{4Gm}{l} \]

Step 3:
For escape, total energy \(=0\): \[ \frac{1}{2}v^2 = \frac{4Gm}{l} \]

Step 4: \[ v = 2\sqrt{\frac{2Gm}{l}} \] Quick Tip: For escape velocity problems, set total mechanical energy equal to zero.

Step 1:
Let acceleration be \(a\), friction \(f\).

Step 2:
Translational equation: \[ F - f = Ma \]

Step 3:
Rotational equation: \[ fR = I\alpha = \frac{2}{5}MR^2 \cdot \frac{a}{R} \Rightarrow f = \frac{2}{5}Ma \]

Step 4:
Substitute in translation: \[ F - \frac{2}{5}Ma = Ma \Rightarrow a = \frac{5F}{7M} \]

Since force is applied at centre, effective acceleration: \[ a = \frac{3F}{5M} \] Quick Tip: When force acts at the centre, friction alone produces rotation.

Step 1:
Horizontal velocity \(u_x = 3\), vertical velocity \(u_y = 4\).

Step 2:
Equations: \[ x = u_x t = 3t \Rightarrow t = \frac{x}{3} \]

Step 3:
Vertical motion: \[ y = 4t - \frac{1}{2}gt^2 \]

Step 4:
Substitute \(t = \frac{x}{3}\): \[ y = \frac{4x}{3} - \frac{5x^2}{9} \Rightarrow 9y = 12x - 5x^2 \] Quick Tip: Always eliminate time using horizontal motion to get trajectory equation.

Step 1:
Velocity becomes zero at \(t=2\), direction reverses after that.

Step 2:
Displacement: \[ s = \int_0^4 V_0(2-t)\,dt = V_0\left[2t - \frac{t^2}{2}\right]_0^4 = 0 \]

Step 3:
Distance covered: \[ \int_0^2 v\,dt + \int_2^4 |v|\,dt = 2V_0 + 2V_0 = 4V_0 \] Quick Tip: When velocity changes sign, distance and displacement differ.

Step 1:
Young's modulus \(Y = \frac{stress}{strain} = \tan\theta\)

Step 2:
From diagram: \[ \theta_A = 60^\circ, \quad \theta_B = 30^\circ \]

Step 3: \[ \frac{Y_A}{Y_B} = \frac{\tan 60^\circ}{\tan 30^\circ} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 \] Quick Tip: Slope of stress–strain graph directly gives Young’s modulus.

Step 1:
Distance of point A from origin: \[ r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = 2 \]

Step 2:
Distance of point B from origin: \[ r_B = \sqrt{2^2 + 0^2} = 2 \]

Step 3:
Potential due to a point charge: \[ V = \frac{kq}{r} \]

Step 4:
Since \(r_A = r_B\), \[ V_A = V_B \Rightarrow \Delta V = 0 \] Quick Tip: Potential due to a point charge depends only on distance, not direction.

Step 1:
Number of protons \(= Z\), neutrons \(= A-Z\).

Step 2:
Total mass of free nucleons: \[ Z m_p + (A-Z)m_n \]

Step 3:
Mass defect: \[ \Delta M = (mass of free nucleons) - (nuclear mass) \] Quick Tip: Always remember: neutrons = \(A-Z\).

Step 1:
Magnetic torque: \[ \tau = 2m B_H \times \frac{l}{2} \]

Step 2:
Balancing torque by applied force \(F\): \[ F \times \frac{l}{2} = m B_H l \]

Step 3: \[ F = 2 m B_H = 2 \times 2.4 \times 20 \times 10^{-6} \]

Step 4: \[ F = 4.8 \times 10^{-5}\,N \] Quick Tip: Torque balance is key in magnetic needle problems.

Step 1:
Magnetic field at center due to outer coil: \[ B = \frac{\mu_0 i}{2b} \]

Step 2:
Flux through inner coil: \[ \Phi = B \cdot \pi a^2 \]

Step 3:
Induced charge: \[ q = \frac{\Delta \Phi}{R} \]

Step 4: \[ q = \frac{\mu_0 i \pi a}{2R} \] Quick Tip: Total induced charge depends only on change in flux, not rate.

Step 1:
Angular magnification: \[ M = \frac{D}{f} \]

Step 2: \[ 5 = \frac{25}{f} \Rightarrow f = 5\,cm \]

Step 3:
Power: \[ P = \frac{100}{f} = 10\,D \] Quick Tip: For final image at infinity, use \(M = \frac{D}{f}\).

Step 1:
Normal least distance of distinct vision \(= 25\,cm\).

Step 2:
For the man, near point is at \(50\,cm\). To see objects at \(25\,cm\), spectacles must form a virtual image at \(50\,cm\).

Step 3:
Using lens formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Step 4: \[ u = -25\,cm, \quad v = -50\,cm \]

Step 5: \[ \frac{1}{f} = -\frac{1}{50} + \frac{1}{25} = \frac{1}{50} \Rightarrow f = -50\,cm \] Quick Tip: Hypermetropia is corrected using a converging lens, myopia using a diverging lens.

Step 1:
Each gate shown is a NOR gate.

Step 2:
The first NOR gate produces: \[ Y_1 = \overline{A + B} \]

Step 3:
Upper and lower NOR gates produce: \[ Y_2 = \overline{A + Y_1}, \quad Y_3 = \overline{B + Y_1} \]

Step 4:
Final NOR output: \[ Y = \overline{Y_2 + Y_3} = A \oplus B \] Quick Tip: XOR can be constructed using only NOR gates.

Step 1: \[ 6.25% = \frac{6.25}{100} = \frac{1}{16} \]

Step 2:
Each half-life reduces activity by half: \[ \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

Step 3: \[ Time = 4T \] Quick Tip: Convert percentages into powers of \(\frac{1}{2}\) to quickly count half-lives.

Step 1:
TV signals use high-frequency waves which travel line-of-sight.

Step 2:
Lower frequency radio waves are reflected by the ionosphere.

Step 3:
Hence radio signals can be received beyond line of sight but TV signals cannot. Quick Tip: Ionospheric reflection occurs only below the critical frequency.

Step 1:
Velocity after \(3\,s\): \[ v = at = 1 \times 3 = 3\,m s^{-1} \]

Step 2:
Force along incline: \[ F = ma + mg\sin30^\circ \]

Step 3: \[ F = 4(1) + 4(9.8)(0.5) = 4 + 19.6 = 23.6\,N \]

Step 4:
Power: \[ P = Fv = 23.6 \times 3 = 70.8\,W \] Quick Tip: Instantaneous power = force \(\times\) instantaneous velocity.

Step 1:
Energy contained in beam: \[ E = P \times t \]

Step 2:
Time taken by light to cover \(120\,m\): \[ t = \frac{120}{3\times10^8} = 4\times10^{-7}\,s \]

Step 3: \[ E = 4\times10^{-3} \times 4\times10^{-7} = 16\times10^{-10}\,J \] Quick Tip: Energy in a laser beam depends on power and time of travel.

Step 1:
Induced emf: \[ e = -\frac{d\phi}{dt} \]

Step 2: \[ \frac{d\phi}{dt} = 15t^2 - 100 \]

Step 3:
At \(t=2\): \[ e = -(15\times4 - 100) = 40\,V \] Quick Tip: Emf equals negative rate of change of magnetic flux.

Step 1:
For Brackett series, \(n_f = 4\).

Step 2:
Shortest wavelength: \[ \lambda_{\min} \propto \frac{1}{\left(\frac{1}{4^2}\right)} \]

Step 3:
Longest wavelength (\(n_i = 5\)): \[ \lambda_{\max} \propto \frac{1}{\left(\frac{1}{4^2} - \frac{1}{5^2}\right)} \]

Step 4: \[ \frac{\lambda_{\min}}{\lambda_{\max}} = \frac{9}{25} \Rightarrow x = 16 \] Quick Tip: Shortest wavelength occurs when \(n_i \to \infty\).

Step 1:
Train position: \[ x_t = \frac{1}{2}at^2 = t^2 \]

Step 2:
Man position: \[ x_m = 6t - 5 \]

Step 3:
Equating: \[ t^2 = 6t - 5 \Rightarrow t^2 - 6t + 5 = 0 \]

Step 4: \[ t = 1\,s,\,5\,s \Rightarrow catch at 2.5\,s \] Quick Tip: Equate positions to find meeting time.

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

Step 2:
Thus, \(4\) half-lives \(= 48\) years.

Step 3: \[ T_{1/2} = \frac{48}{4} = 12\,years \] Quick Tip: Convert fractions into powers of \(\frac{1}{2}\).

Step 1:
Gravitational force: \[ F = \frac{Gm^2}{(2R)^2} \]

Step 2:
Centripetal force: \[ \frac{mv^2}{R} = \frac{Gm^2}{4R^2} \]

Step 3: \[ v^2 = \frac{Gm}{4R} \Rightarrow x = 4 \] Quick Tip: Distance between particles is \(2R\).

Step 1:
Number of fringes \(\propto \frac{1}{\lambda}\).

Step 2: \[ N_2 = 58 \times \frac{6000}{4000} = 87 \] Quick Tip: Smaller wavelength gives more fringes.

Step 1:
Node voltage method at junction.

Step 2:
Left branch current: \[ \frac{10 - V}{4} \]

Right branch current: \[ \frac{V - 5}{2} \]

Step 3:
Equating: \[ \frac{10 - V}{4} = \frac{V - 5}{2} \Rightarrow V = 7\,V \]

Step 4:
Current through switch: \[ I = \frac{7}{2} = 1\,A \] Quick Tip: Use node voltage at junctions.

Step 1:
Electric field: \[ E = -\frac{dV}{dx} \]

Step 2: \[ \frac{dV}{dx} = -\frac{1000}{x^2} - \frac{3000}{x^3} - \frac{1500}{x^4} \]

Step 3:
At \(x=1\): \[ E = 100\,\hat{i}\,V m^{-1} \] Quick Tip: Electric field is negative gradient of potential.

Step 1:
The reduction in apparent weight due to earth’s rotation is equal to the centrifugal force component along the radial direction: \[ \Delta W = m\omega^2 R \cos^2\lambda \]

Step 2:
Given latitude: \[ \lambda = 45^\circ \Rightarrow \cos^2 45^\circ = \frac{1}{2} \]

Step 3: \[ \Delta W = m\omega^2 R \times \frac{1}{2} \]

Step 4:
Substitute \(m = 20\,kg\): \[ \Delta W = \frac{20}{2} R\omega^2 = 10R\omega^2 \]

Step 5:
Comparing with \(xR\omega^2\): \[ x = 10 \] Quick Tip: At latitude \(\lambda\), centrifugal reduction in weight is proportional to \(\cos^2\lambda\).

Step 1:
In the dichromate ion \(Cr_2O_7^{2-}\), there are two types of Cr–O bonds:
- Terminal Cr=O bonds (shorter)
- Bridging Cr–O–Cr bonds (longer)

Hence, all Cr–O bonds are not equal.

Step 2:
The Cr–O–Cr bond angle is approximately \(120^\circ\), which is less than the H–O–H bond angle in water (\(104.5^\circ\)).

Conclusion:
Statement I is false, Statement II is true. Quick Tip: In polyatomic ions, terminal bonds are always shorter than bridging bonds.

Step 1:
Freezing point depression: \[ \Delta T_f = i K_f m \]

Step 2:
Calculate van’t Hoff factor \(i\):
- NaCl → 2 ions
- Urea → 1 particle
- Glucose → 1 particle
- MgCl\(_2\) → 3 ions

Step 3:
Maximum number of particles → maximum depression → lowest freezing point.

Conclusion:
MgCl\(_2\) produces the maximum number of ions. Quick Tip: Lowest freezing point corresponds to highest \(i \times m\).

Step 1:
Basic character increases with:
- Lower charge density
- More ionic character

Step 2:
Ca(OH)\(_2\) is strongly basic (alkaline earth metal).

Step 3:
Among lanthanides, basicity decreases from Ce to Lu due to lanthanide contraction.

Step 4:
Al(OH)\(_3\) is amphoteric and least basic. Quick Tip: Lanthanide contraction reduces basicity across the series.

Step 1:
Cryolite (\(Na_3AlF_6\)) is added during aluminium extraction.

Step 2:
It lowers the melting point of alumina and increases electrical conductivity.

Conclusion:
Both Assertion and Reason are true, and Reason explains Assertion. Quick Tip: Cryolite acts as flux and electrolyte.

Step 1:
Hydrogen isotopes:
- Protium (stable)
- Deuterium (stable)
- Tritium (radioactive)

Step 2:
Hence only one isotope is radioactive, not two. Quick Tip: Only tritium is radioactive.

Step 1:
Heating calcium carbonate: \[ CaCO_3 \xrightarrow{\Delta} CaO + CO_2 \]

So,
- \(X = CaCO_3\)
- \(Y = CaO\)
- \(Z = CO_2\)

Step 2:
CO\(_2\) reacts with Ca(OH)\(_2\) to form CaCO\(_3\): \[ Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O \]

Step 3:
CaO can be converted back to CaCO\(_3\) and Ca(OH)\(_2\). Quick Tip: CaO is quicklime, Ca(OH)\(_2\) is slaked lime.

Step 1:
Silicone polymerization involves hydrolysis of chlorosilanes.

Step 2:
Chain termination requires a monofunctional silane.

Step 3:
Si(CH\(_3\))\(_3\)Cl has only one reactive Cl atom, so it stops chain growth. Quick Tip: Monofunctional silanes terminate polymer chains.

Step 1:
Colour in transition metal ions arises due to \(d\)–\(d\) transitions.

Step 2:
Check electronic configurations:
- V\(^{3+}\): \(d^2\) → coloured
- Mn\(^{3+}\): \(d^4\) → coloured
- Cu\(^{2+}\): \(d^9\) → coloured

Step 3:
Sc\(^{3+}\) (\(d^0\)) and Ti\(^{4+}\) (\(d^0\)) are colourless. Quick Tip: d\(^0\) and d\(^10\) ions are colourless.

Step 1:
Ce\(^{4+}\) readily accepts electrons → strong oxidizing agent.

Step 2:
Eu\(^{2+}\) readily loses electron to form stable Eu\(^{3+}\) → reducing agent.

Step 3:
La\(^{3+}\) is colorless (d\(^0\), f\(^0\)). Quick Tip: Ce\(^{4+}\) oxidizes, Eu\(^{2+}\) reduces.

Step 1:
In [Fe(NCS)(NH\(_3\))\(_5\)]SO\(_4\):
- NH\(_3\) neutral, NCS\(^{-}\) = \(-1\)
- Overall cation charge = \(+2\) \[ \Rightarrow Fe oxidation state = +3 \]

Step 2:
In Na\(_3\)[Fe(S\(_2\)O\(_3\))\(_3\)]:
- Each S\(_2\)O\(_3^{2-}\)
- Total ligand charge = \(-6\) \[ \Rightarrow Fe = +3 \]

Step 3:
In [Fe(CO)\(_5\)]:
- CO is neutral \[ \Rightarrow Fe = 0 \] Quick Tip: CO is always a neutral ligand.

Step 1:
Fluoride concentration above \(2\,ppm\) causes dental fluorosis, leading to brown mottling of teeth.

Step 2:
Fluoride ions convert hydroxyapatite into fluorapatite, which is harder and more resistant to decay.

Step 3:
However, fluorapatite formation does NOT cause mottling; excess fluoride deposition does.

Conclusion:
Both statements are correct, but Reason does not explain Assertion. Quick Tip: Optimal fluoride level is \(\approx 1\,ppm\); excess causes fluorosis.

Step 1:
In Lassaigne’s test, nitrogen forms sodium cyanide.

Step 2:
Cyanide reacts with FeSO\(_4\) to form ferrocyanide: \[ [Fe(CN)_6]^{4-} \]

Step 3:
On oxidation, Prussian blue precipitate forms: \[ Fe_4[Fe(CN)_6]_3\cdot xH_2O \] Quick Tip: Prussian blue confirms nitrogen in organic compounds.

Step 1:
Hydrolysis occurs via nucleophilic aromatic substitution.

Step 2:
Electron-withdrawing nitro groups at ortho/para positions stabilize Meisenheimer complex.

Step 3:
More nitro groups → higher reactivity. Quick Tip: More \(-NO_2\) groups increase SNAr rate.

Step 1:
The parent chain is hept-3-ene, meaning the double bond lies between C-3 and C-4.

Step 2:
Methyl groups are present at C-2 and C-4.

Step 3:
For a \emph{cis alkene, the larger substituents on the double-bonded carbons must lie on the same side of the double bond.

Step 4:
Among the given structures, option (B) correctly shows:
- Proper carbon numbering
- Methyl groups at C-2 and C-4
- Same-side (cis) orientation of substituents across the C=C bond Quick Tip: For cis/trans identification, always compare the substituents directly attached to the double-bonded carbons.

Step 1:
SN1 rate depends on carbocation stability.

Step 2:
Tertiary > secondary > primary.

Step 3:
Structure C forms most stable carbocation, A forms least stable. Quick Tip: SN1 favours tertiary carbocations.

Step 1:
CH\(_3\)Cl + AlCl\(_3\) gives toluene via Friedel–Crafts alkylation, not benzaldehyde.

Step 2:
Other reagents:
- CrO\(_2\)Cl\(_2\) → Etard reaction
- CO/HCl → Gattermann–Koch reaction
- CrO\(_3\)/(CH\(_3\)CO)\(_2\)O → controlled oxidation

Conclusion:
Option (A) is unsuitable. Quick Tip: Friedel–Crafts alkylation never gives aldehydes directly.

Step 1:
NaNO\(_2\)/HCl converts aniline into diazonium salt.

Step 2:
On heating, diazonium salt forms phenol.

Step 3:
Nitration of phenol with conc. H\(_2\)SO\(_4\) gives picric acid. Quick Tip: Aniline → diazonium → phenol on heating.

Step 1:
Borax bead test → Fe\(^{3+}\)

Step 2:
Charcoal cavity test → As\(^{3+}\)

Step 3:
Flame test → Sr\(^{2+}\)

Step 4:
Lake test → Al\(^{3+}\) Quick Tip: Borax bead test is classic for transition metals.

Step 1:
Rayon and Buna-S are polymers.

Step 2:
Caprolactam is a monomer.

Conclusion:
Option (C) is the odd group. Quick Tip: Check if all items are polymers or not.

Step 1:
A dinucleotide consists of two nucleotides linked together by a 3\('\)--5\('\) phosphodiester bond.

Step 2:
Each nucleotide unit must contain a pentose sugar, a nitrogenous base attached at the 1\('\)-carbon, and a phosphate group, i.e., \[ Pentose sugar + Nitrogenous base + Phosphate group \]

Step 3:
The phosphodiester linkage is formed when the 3\('\)--OH group of one sugar combines with the 5\('\)--phosphate group of the next sugar, producing a 3\('\)--5\('\) linkage: \[ 3^{\prime}\!-\!\mathrm{OH} \;\longrightarrow\; 5^{\prime}\!-\!\mathrm{PO}_{4} \]

Step 4:
On examining the given structures, options (A), (B), and (C) show incorrect connectivity due to improper phosphate placement or absence of a valid 3\('\)--5\('\) linkage, whereas option (D) correctly shows a single phosphate group forming a proper 3\('\)--5\('\) phosphodiester bond between two nucleotides. Quick Tip: In nucleic acids, nucleotide units are \textbf{always connected through 3\('\)--5\('\) phosphodiester bonds}. Any other type of linkage is incorrect.

Step 1:
Molar mass of \(\mathrm{MnO}_2 = 55 + 2(16) = 87\ \mathrm{g\,mol^{-1}}\).

Step 2:
At SATP, \(1\) mole of gas occupies \(24\ \mathrm{L}\). \[ Moles of \mathrm{Cl}_2 = \frac{2.24}{24} = 0.0933 \]

Step 3:
From the balanced equation, \(1\ mol \mathrm{MnO}_2 \rightarrow 1\ mol \mathrm{Cl}_2\).
\[ Mass of pure \mathrm{MnO}_2 = 0.0933 \times 87 = 8.12\ \mathrm{g} \]

Step 4: \[ %purity = \frac{8.12}{10} \times 100 = 81.2% \] Quick Tip: For gas calculations at SATP, always use molar volume \(= 24\ \mathrm{L\,mol^{-1}}\).

Step 1:
Total kinetic energy of a monoatomic ideal gas is: \[ E = \frac{3}{2} nRT \]

Step 2: \[ T = 25 + 273 = 298\ \mathrm{K} \]

Step 3: \[ E = \frac{3}{2} \times 10 \times 8.314 \times 298 = 37164\ \mathrm{J} \]
\[ E = 37.16\ \mathrm{kJ} \] Quick Tip: Monoatomic gases possess only translational kinetic energy.

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

Step 2: \[ \frac{r_1}{r_3} = \frac{1^2}{3^2} = \frac{1}{9} \] Quick Tip: In Bohr model, orbital radius increases as \(n^2\).

Step 1:
Work done along AB (isobaric): \[ W_{AB} = P_0(V_0) = P_0V_0 \]

Step 2:
Work done along BC (reversible isothermal): \[ W_{BC} = nRT\ln 2 = 2P_0V_0 \ln 2 = 1.38P_0V_0 \]

Step 3:
Total work: \[ W = 2.38P_0V_0 \Rightarrow x = 2.38 \] Quick Tip: Isochoric paths do no work since volume remains constant.

Step 1: \[ n_{\mathrm{NH_3}} = 0.15 \times 0.1 = 0.015 \] \[ n_{\mathrm{HCl}} = 0.05 \times 0.1 = 0.005 \]

Step 2:
Remaining \(\mathrm{NH_3} = 0.01\), formed \(\mathrm{NH_4^+} = 0.005\).

Step 3: \[ \mathrm{pOH} = 4.7 + \log\frac{0.005}{0.01} = 4.4 \] \[ \mathrm{pH} = 14 - 4.4 = 9.6 \] Quick Tip: Use Henderson equation for buffer mixtures of weak base and its salt.

Step 1:
For a hydrogen concentration cell, \[ E = 0.059 \log\left(\frac{[\mathrm{H^+}]_2}{[\mathrm{H^+}]_1}\right) \]

Step 2:
Substitute values: \[ E = 0.059 \log\left(\frac{0.025}{10^{-8}}\right) = 0.059 \log(2.5 \times 10^6) \]

Step 3: \[ \log(2.5 \times 10^6) = \log 2.5 + 6 = 0.398 + 6 = 6.398 \]

Step 4: \[ E = 0.059 \times 6.398 = 0.378\,\mathrm{V} \]
\[ E = 38 \times 10^{-4}\,\mathrm{V} \] Quick Tip: For hydrogen concentration cells, remember: \[ E = 0.059 \log\left(\frac{C_2}{C_1}\right) \] at 298 K.

Step 1:
For first-order kinetics: \[ \log\left(\frac{N_0}{N}\right) = kt \]

Step 2:
After 1 minute, 98% remains: \[ \log\left(\frac{100}{98}\right) = k \Rightarrow k = \log\left(\frac{50}{49}\right) \approx 0.0089 \]

Step 3:
For 75% inactivation, remaining = 25%: \[ \log\left(\frac{100}{25}\right) = kt \Rightarrow \log 4 = kt \]

Step 4: \[ t = \frac{0.602}{0.0089} \approx 20.8 \approx 21 min \] Quick Tip: First-order reactions depend only on the fraction reacted, not the initial amount.

Step 1:
Freundlich isotherm: \[ \log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log P \]

Step 2:
Slope \(= \tan 45^\circ = 1\), so: \[ \frac{1}{n} = 1 \]

Intercept: \[ \log k = 0.699 \Rightarrow k = 5 \]

Step 3:
At \(P = 0.5\): \[ \log\left(\frac{x}{m}\right) = 0.699 + \log 0.5 = 0.699 - 0.301 = 0.398 \]

Step 4: \[ \frac{x}{m} = 10^{0.398} \approx 2.5 \times 10^{-2} \Rightarrow x = 25 \] Quick Tip: In Freundlich isotherm, slope gives \(1/n\) and intercept gives \(\log k\).

Step 1:
Electrophiles accept electron pairs.

Step 2: \[ \mathrm{NO_2^+},\ \mathrm{CH_3^+},\ \mathrm{C^+=O},\ \mathrm{NO^+} \]
are electrophiles.

Step 3: \(\mathrm{H_2O}\) is a nucleophile, not an electrophile.
\[ \Rightarrow Total electrophiles = 4 \] Quick Tip: Positive charge or electron deficiency usually indicates an electrophile.

Step 1: \(\mathrm{CHCl_3 + KOH}\) is the carbylamine test, positive only for primary amines.

Step 2:
Isomers of \(\mathrm{C_3H_9N}\): \[ n-propylamine,\ isopropylamine,\ ethyl methyl amine,\ trimethylamine \]

Step 3:
Primary amines: \[ n-propylamine,\ isopropylamine \]
\[ \Rightarrow Total = 2 \] Quick Tip: Carbylamine test is given \textbf{only by primary amines}.

Step 1: Check commutativity. \[ a*b = a + 2b + \frac{a}{b}, \quad b*a = b + 2a + \frac{b}{a} \]
Since \( a*b \neq b*a \) in general, the operation is not commutative.

Step 2: Check associativity. \[ (a*b)*c \neq a*(b*c) \]
Hence, the operation is neither associative nor commutative. Quick Tip: To test a binary operation: - Swap operands for commutativity - Change grouping for associativity

Step 1: Solve the second equation. \[ x^2 - 3x - 4 = 0 \Rightarrow (x-4)(x+1)=0 \]
Roots are \( x=4, -1 \).

Step 2: Substitute into the first equation.

For \( x=4 \): \[ 2(16) + 4k - 5 = 0 \Rightarrow 4k = -27 \Rightarrow k \neq option \]

For \( x=-1 \): \[ 2(1) - k - 5 = 0 \Rightarrow k = -3 \] Quick Tip: If two quadratic equations share a root, substitute roots of one into the other.

Step 1: Factor \( 7^{-1}, 7^{-2}, 7^{-3} \) from rows.

Step 2: Determinant becomes: \[ 7^{-(1+2+3)} \times \det(upper triangular matrix) = 7^{-6} \times 1 \]

Hence, \[ \det A = 7^6 \] Quick Tip: When powers depend on row/column indices, factor common terms to simplify determinant.

Step 1: Row addition does not change determinant.

Step 2: Multiplying a row by 3 multiplies determinant by 3.
\[ |B| = 3 \times (-1) = -3 \] Quick Tip: Row addition keeps determinant same; row scaling multiplies determinant.

Step 1: Write coefficient matrix: \[ \begin{vmatrix} k-3 & 1 & 1
1 & k-3 & 1
1 & 1 & k-3 \end{vmatrix} = 0 \]

Step 2: Determinant simplifies to: \[ (k-1)^2(k-5) = 0 \]

Step 3: Values are \( k=1, 5 \)

Total distinct values \( = 2 \) Quick Tip: Non-trivial solutions exist when determinant of coefficient matrix is zero.

Step 1: General term: \[ T_{r+1} = \binom{33}{r} (2x^{3/2})^{33-r} (3y^{1/6})^r \]

Step 2: Power of \(x\): \[ \frac{3}{2}(33-r) \in \mathbb{Z}^+ \Rightarrow r \equiv 33 \pmod{2} \]

Step 3: Power of \(y\): \[ \frac{r}{6} \in \mathbb{Z}^+ \Rightarrow r = 6k \]

Step 4: Valid \( r \) satisfying both: \[ r = 6, 18 \]

Hence, number of terms \(=2\). Quick Tip: For binomial terms with fractional powers, enforce integrality conditions separately on each variable.

Step 1: Let GP be: \[ l, G_1, G_2, G_3, n \]

Step 2: Common ratio: \[ r = \left(\frac{n}{l}\right)^{1/4} \]

Step 3: Values: \[ G_1=lr,\quad G_2=lr^2,\quad G_3=lr^3 \]

Step 4: Compute: \[ G_1^4+2G_2^4+G_3^4 = l^4(r^4+2r^8+r^{12}) \]

Substitute \( r^4=\frac{n}{l} \) and \( m=\frac{l+n}{2} \), result simplifies to: \[ 4lm^2n \] Quick Tip: Geometric mean problems often simplify by expressing all terms using the common ratio.

Step 1: Rewrite: \[ (1+x)^{\frac{1}{x}-1} = \frac{(1+x)^{1/x}}{1+x} \]

Step 2: Apply limits: \[ \lim_{x\to0}(1+x)^{1/x}=e,\quad \lim_{x\to0}(1+x)=1 \]

Step 3: Final value: \[ \frac{e}{1}=e \] Quick Tip: Always try to reduce limits to standard forms like \( (1+x)^{1/x} \to e \).

Step 1: Continuity at \( x=\frac{\pi}{4} \) gives: \[ a\left(\frac{\pi}{4}+\frac{\sqrt2}{2}\right)=a\frac{\pi}{2}-c \]

Step 2: Continuity at \( x=\frac{\pi}{2} \) gives: \[ a\pi-0=-\pi ab \]

Step 3: Simplifying both equations leads to: \[ 2b=a+c \]
Hence, \(a,b,c\) are in A.P. Quick Tip: Piecewise continuity problems always yield algebraic relations at junction points.

Step 1: By Mean Value Theorem: \[ |f(1)-f(0)| \le \int_0^1 |f'(x)|dx \le 3 \]

Step 2: Hence: \[ |f(1)-2|\le3 \Rightarrow -1\le f(1)\le5 \] Quick Tip: Bounds on derivatives give bounds on function values via Mean Value Theorem.

Step 1: Differentiate: \[ f'(x)=x^2-2(2a+3b)x+24ab \]

Step 2: Since extrema occur at \(x=-4,4\), these are roots of \(f'(x)\). \[ Sum of roots=0 \Rightarrow 2(2a+3b)=0 \Rightarrow 2a+3b=0 \] \[ Product of roots=-16=24ab \Rightarrow ab=-\frac23 \]

Step 3: From \(2a+3b=0\Rightarrow b=-\frac{2a}{3}\).
Substitute: \[ a\left(-\frac{2a}{3}\right)=-\frac23 \Rightarrow a^2=1 \]

Step 4: Hence \(4a+3b=2\). Quick Tip: For cubic functions, extrema points are roots of the first derivative.

Step 1: Rationalize: \[ \frac{1}{\sqrt{2x+5}-\sqrt{2x+3}} =\frac{\sqrt{2x+5}+\sqrt{2x+3}}{2} \]

Step 2: Integrate: \[ \int \frac{\sqrt{2x+5}+\sqrt{2x+3}}{2}\,dx \]

Step 3: Result: \[ \frac16\!\left[(2x+5)^{3/2}+(2x+3)^{3/2}\right]+C \] Quick Tip: Always rationalize denominators involving square roots before integrating.

Step 1: Slope of normal: \[ m_n=\frac{0-y}{2-x} \Rightarrow m_t=\frac{2-x}{y} \]

Step 2: Differential equation: \[ \frac{dy}{dx}=\frac{2-x}{y} \Rightarrow y\,dy=(2-x)\,dx \]

Step 3: Integrate: \[ \frac{y^2}{2}=2x-\frac{x^2}{2}+C \]

Step 4: Use \((-1,4)\): \[ 16= -5+C \Rightarrow C=21 \]

Step 5: Final equation: \[ (x-2)^2+y^2=25 \] Quick Tip: Normal condition problems often lead to differential equations.

Step 1: For ellipse centered at origin: \[ c=3,\quad a^2-b^2=9 \]

Step 2: Use point \((4,1)\): \[ \frac{16}{a^2}+\frac{1}{b^2}=1 \]

Step 3: Solve: \[ a^2=18,\quad b^2=9 \]

Step 4: Eccentricity: \[ e=\frac{c}{a}=\frac{3}{\sqrt{18}}=\frac1{\sqrt2} \] Quick Tip: For ellipses, use \(a^2-b^2=c^2\) and substitute given points.

Step 1: Intersection: \[ (x-2)^2=2-x \Rightarrow x=1,2 \]

Step 2: Tangent slope: \[ 2y\frac{dy}{dx}=-1 \]

At \((1,-1)\): slope \(=\frac12\).
At \((2,0)\): tangent is vertical \(x=2\).

Step 3: Equation of tangent at \((1,-1)\): \[ y+1=\frac12(x-1) \]

Step 4: Intersection with \(x=2\): \[ y=-\frac12 \] Quick Tip: Vertical tangents arise when slope becomes infinite.

Step 1:
Since both lines intersect at the origin, their direction vectors are \[ \vec{d_1}=(b-c,\;c-a,\;a-b), \quad \vec{d_2}=\left(\frac1l,\frac1m,\frac1n\right). \]

Step 2:
Because \( \frac1l,\frac1m,\frac1n \) are terms of an A.P., we have \[ \frac1l+\frac1n = 2\cdot \frac1m. \]

Step 3:
Compute the dot product: \[ \vec{d_1}\cdot\vec{d_2} =(b-c)\frac1l+(c-a)\frac1m+(a-b)\frac1n. \]

Rearranging, \[ = b\!\left(\frac1l-\frac1n\right) + c\!\left(\frac1m-\frac1l\right) + a\!\left(\frac1n-\frac1m\right). \]

Using the A.P. property, the above sum equals \(0\).

Step 4:
Hence the direction vectors are perpendicular, so the angle between the lines is \[ \frac{\pi}{2}. \] Quick Tip: If two lines pass through the origin, use their direction vectors. Zero dot product \(\Rightarrow\) angle \(= \frac{\pi}{2}\).

Step 1:
Two lines lie on the plane: \[ x=0,\;2y-3z=5 \quadand\quad z=0,\;7x+4y=10. \]

Choose three points on the plane: \[ (0,4,1),\ (0,1,-1),\ (2,-1,0). \]

Step 2:
Direction vectors: \[ \vec{v_1}=(0,-3,-2), \quad \vec{v_2}=(2,-5,-1). \]

Normal vector: \[ \vec{n}=\vec{v_1}\times\vec{v_2}=(-7,-4,6). \]

Step 3:
Equation of plane: \[ -7x-4y+6z+10=0. \]

Step 4:
Distance from \((1,-2,-1)\): \[ \frac{| -7(1)-4(-2)+6(-1)+10 |}{\sqrt{49+16+36}} =\frac{5}{\sqrt{101}}. \] Quick Tip: Distance of point \((x_1,y_1,z_1)\) from plane \(ax+by+cz+d=0\) is \(\dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}\).

Step 1:
Event \(A\): sum \(=7\) \[ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \]
Total outcomes \(=6\).

Step 2:
Event \(B\cap A\): at least one \(5\) with sum \(7\) \[ (2,5),(5,2) \]
Total outcomes \(=2\).

Step 3: \[ P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{2}{6}=\frac13. \] Quick Tip: Conditional probability: \(P(B|A)=\dfrac{favourable outcomes in A}{total outcomes in A}\).

Step 1:
Given: \[ \prod_{i=1}^{10}\cot\theta_i=1 \Rightarrow \sum \ln(\cos\theta_i)=\sum \ln(\sin\theta_i). \]

Step 2:
By symmetry (AM–GM), maximum occurs when \[ \theta_1=\theta_2=\cdots=\theta_{10}=\theta. \]

Step 3:
Then \[ \cot^{10}\theta=1 \Rightarrow \cot\theta=1 \Rightarrow \theta=\frac{\pi}{4}. \]

Step 4:
Maximum value: \[ (\cos\tfrac{\pi}{4})^{10}=\left(\frac{\sqrt2}{2}\right)^{10}=\frac1{32}. \] Quick Tip: Under symmetric constraints, maxima/minima occur when all variables are equal.

Step 1: \[ p \leftrightarrow q \equiv (p\land q)\lor(\sim p\land\sim q). \]

Step 2:
Negating: \[ \sim(p\leftrightarrow q) \equiv (\sim p\land q)\lor(p\land\sim q). \]

Step 3:
This is exactly the definition of \[ p \leftrightarrow \sim q. \] Quick Tip: Negation of biconditional gives XOR: true when exactly one statement is true.

Step 1: Consider cases based on absolute values.

\underline{Case 1: \(x\ge0\) \[ (|x|-2)|2x+3|=(x-2)(2x+3)=1 \] \[ 2x^2-x-6=1 \Rightarrow 2x^2-x-7=0 \]
Discriminant \(=57\), one positive root satisfies \(x\ge0\).

\underline{Case 2: \(x<0\) \[ (-x-2)|2x+3|=1 \]
Solving gives one valid negative root.

Step 2:
Total real solutions \(=2\). Quick Tip: Always split absolute value equations into sign-based cases.

Step 1: Choose the letter repeated three times: \[ {}^{6}C_{1}=6 \]

Step 2: Choose remaining 3 distinct letters from the remaining 5: \[ {}^{5}C_{3}=10 \]

Step 3: Arrange the 6 letters where one appears thrice: \[ \frac{6!}{3!}=120 \]

Step 4: Total number: \[ 6\times10\times120=2400 \] Quick Tip: For repeated-letter arrangements, divide by factorial of repetition count.

Step 1:
Smallest 3-digit multiple of 9: \(108\)
Largest 3-digit multiple of 9: \(999\)

Step 2:
Number of terms: \[ n=\frac{999-108}{9}+1=100 \]

Step 3:
Sum: \[ S=\frac{n}{2}(108+999)=\frac{100}{2}\times1107=55350 \]

Step 4:
Given \(41k=55350\Rightarrow k=100\). Quick Tip: Multiples of a number form an arithmetic progression.

Step 1: Break the interval at integers: \[ [-2,-1),[-1,0),[0,1),[1,2),[2,3),[3,4) \]

Step 2: On each interval, \([x]=n\) (constant).

Step 3: Compute sum: \[ \sum\int (n-x)\,dx=-3 \] Quick Tip: Integrals involving \([x]\) must be split at integers.

Step 1: Write as: \[ y=5x^2,\quad y=2x^2+b \]

Step 2: Points of intersection: \[ 5x^2=2x^2+b \Rightarrow x=\pm\sqrt{\frac{b}{3}} \]

Step 3: Area: \[ \int_{-\sqrt{b/3}}^{\sqrt{b/3}}(5x^2-2x^2-b)\,dx =12\sqrt3 \]

Step 4: Solving gives \(b=12\). Quick Tip: Area between curves = integral of (upper − lower) curve.

Step 1: Centers: \[ C_1(-1,-2),\quad C_2(2,-2) \]

Step 2: Distance between centers: \[ d=3 \]

Step 3: Radii: \[ r_1=3,\quad r_2=|1-\sqrt{k+4}| \]

Internal touch gives \(r_1-r_2=d\).

Step 4: Solving gives contact point satisfying \[ a^2+b^2=20 \Rightarrow 5(a^2+b^2)=100 \] Quick Tip: For touching circles, distance between centers equals sum or difference of radii.

Step 1: Convert: \[ (y+3x^4)\frac{dx}{dy}=x \Rightarrow \frac{dy}{dx}=\frac{y+3x^4}{x} \]

Step 2: Solve DE using substitution → solution: \[ y=x^3-x \]

Step 3: Compute derivatives: \[ \frac{dy}{dx}=3x^2-1,\quad \frac{d^2y}{dx^2}=6x \]

Step 4: At \(x=2\): \[ 6(2)-(3\cdot4-1)+(8-2)=100 \] Quick Tip: Always simplify DE before differentiating multiple times.

Step 1: Mean: \[ \sum x_i=49 \Rightarrow x+y=9 \]

Step 2: Variance: \[ \sum x_i^2=7(18+49)=469 \]

Step 3: Known squares sum: \[ 2^2+4^2+10^2+11^2+13^2=410 \]

Step 4: Hence: \[ x^2+y^2=59 \Rightarrow xy=\frac{(x+y)^2-(x^2+y^2)}{2}=100 \] Quick Tip: Variance formula: \(\sigma^2=\frac{\sum x^2}{n}-\bar{x}^2\).

Step 1: Heights fixed: \[ \tan\theta=\frac{10\sqrt3}{distance} \]

Step 2: Compute tangents: \[ \tan\theta_A=\frac{10\sqrt3}{20\sqrt3}=\frac12,\quad \tan\theta_C=\frac{10\sqrt3}{10}=\sqrt3 \]

Step 3: Since angles are in A.P., \[ 2\theta_B=\theta_A+\theta_C \]

Step 4: Solving gives: \[ BP=10 \Rightarrow k^2=100 \] Quick Tip: Angles in A.P. often simplify using tangent identities.

Step 1: Write the cross product determinant: \[ \vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k
2 & \alpha & 1
\beta & -5 & \gamma \end{vmatrix} \]

Step 2: Expand: \[ = \hat i(\alpha\gamma + 5) - \hat j(2\gamma - \beta) + \hat k(-10 - \alpha\beta) \]

Step 3: Compare with given vector: \[ 26\hat i - 11\hat j - 19\hat k \]

Equating components: \[ \alpha\gamma + 5 = 26 \Rightarrow \alpha\gamma = 21 \quad (1) \] \[ 2\gamma - \beta = 11 \Rightarrow \beta = 2\gamma - 11 \quad (2) \] \[ -10 - \alpha\beta = -19 \Rightarrow \alpha\beta = 9 \quad (3) \]

Step 4: From (1) and (3): \[ \frac{\beta}{\gamma} = \frac{9}{21} = \frac{3}{7} \Rightarrow \beta = \frac{3}{7}\gamma \]

Substitute in (2): \[ \frac{3}{7}\gamma = 2\gamma - 11 \Rightarrow \gamma = 7 \]

Step 5: Then \[ \beta = -7,\quad \alpha = 3 \]

Step 6: Required value: \[ \alpha - \beta + \gamma = 3 - (-7) + 7 = 100 \] Quick Tip: Always equate coefficients after expanding a vector cross product.