BITSAT 2016 Question Paper with Answer Key PDF

Step 1: Let the angular velocity of earth be \( \omega \).
The apparent weight at the equator is: \[ W' = mg - m\omega^2 R \]

Step 2: Given that the weight becomes \( \frac{3}{5} \) of the original weight: \[ mg - m\omega^2 R = \frac{3}{5}mg \]

Step 3: Simplifying: \[ m\omega^2 R = mg - \frac{3}{5}mg = \frac{2}{5}mg \] \[ \omega^2 = \frac{2g}{5R} \]

Step 4: Substituting values \( g = 9.8\,m/s^2 \), \( R = 6.4 \times 10^6\,m \): \[ \omega^2 = \frac{2 \times 9.8}{5 \times 6.4 \times 10^6} = 6.125 \times 10^{-7} \] \[ \omega = \sqrt{6.125 \times 10^{-7}} \approx 7.8 \times 10^{-4}\,rad/s \] Quick Tip: At the equator, apparent weight decreases due to centrifugal force. Always use: \( W = mg - m\omega^2 R \) for rotating frames.

Step 1: Component of weight along incline for A: \[ mg\sin45^\circ = \frac{mg}{\sqrt{2}}, \quad f_A = \mu_A mg\cos45^\circ = \frac{2}{3}\cdot\frac{mg}{\sqrt{2}} \]

Step 2: Net force on A along plane: \[ \frac{mg}{\sqrt{2}} - \frac{2mg}{3\sqrt{2}} = \frac{mg}{3\sqrt{2}} \]

Step 3: For block B: \[ 2mg\sin45^\circ - \mu_B (2mg\cos45^\circ) = \frac{2mg}{\sqrt{2}} - \frac{2mg}{3\sqrt{2}} = \frac{4mg}{3\sqrt{2}} \]

Step 4: The forces balance through the string, hence no net acceleration occurs. Quick Tip: In pulley–incline systems, always compare net driving forces on both sides before assuming motion.

Step 1: Electric field on the axis of a charged disc: \[ E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) \]

Step 2: At \(x = R\): \[ E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{1}{\sqrt{2}}\right) \]

Step 3: Fractional reduction: \[ \frac{1}{\sqrt{2}} \approx 0.707 \Rightarrow reduction = 29.3% \] Quick Tip: For discs, remember: centre field is maximum and decreases gradually along the axis.

Step 1: RMS velocity depends only on temperature: \[ v_{rms} \propto \sqrt{T} \]

Step 2: Convert temperatures to Kelvin: \[ T_1 = 300\,K, \quad T_2 = 400\,K \]

Step 3: \[ v_2 = 200 \sqrt{\frac{400}{300}} = 200\sqrt{\frac{4}{3}} = 100\sqrt{2} \] Quick Tip: RMS speed is independent of pressure; it depends only on absolute temperature.

Step 1: Equivalent resistance in series with inductor: \[ R = R_1 + R_2 = 4\Omega \]

Step 2: Time constant: \[ \tau = \frac{L}{R} = \frac{0.4}{4} = 0.1\,s \]

Step 3: Voltage across inductor: \[ V_L = V_0 e^{-t/\tau} = 6e^{-5t}\,V \] Quick Tip: In RL circuits, voltage across the inductor decays exponentially after switching.

Step 1: Extension of a wire: \[ \Delta x = \frac{FL}{AY} \]

Step 2: Same volume implies: \[ A_1L_1 = A_2L_2 \Rightarrow L_2 = \frac{L_1}{3} \]

Step 3: For same extension: \[ \frac{F_1L_1}{A_1Y} = \frac{F_2L_2}{A_2Y} \Rightarrow F_2 = 9F_1 \] Quick Tip: For same material, extension depends on \( \frac{FL}{A} \).

Step 1: Heat flow rate: \[ \frac{Q}{t} \propto \frac{kA}{d} \]

Step 2: Time for melting: \[ t \propto \frac{d}{kA} \]

Step 3: Using given ratios and times: \[ \frac{k_L}{k_S} = \frac{t_S}{t_L} \cdot \frac{d_L}{d_S} \cdot \frac{A_L}{A_S} = \frac{16}{25} \cdot \frac{1}{4} \cdot 4 = \frac{5}{4} \] Quick Tip: Melting time increases with thickness and decreases with thermal conductivity.

Step 1: Lens maker formula (assuming glass): \[ \frac{1}{f} = \frac{1}{20} + \frac{1}{20} = \frac{1}{10} \Rightarrow f = 10\,cm \]

Step 2: Lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow v = 15\,cm \]

Step 3: Magnification: \[ m = \frac{v}{u} = \frac{15}{30} = \frac{1}{2} \Rightarrow h_i = 1\,cm \] Quick Tip: For convex lenses, objects beyond focus form real, inverted images.

Step 1: In steady state, capacitors act as open circuits.

Step 2: Voltage divides equally across equal resistances.

Step 3: Total voltage \(=100\,V\): \[ V_{AB} = V_{BC} = 50\,V \] Quick Tip: In steady state DC, capacitors block current completely.

Step 1: Let initial atoms be \(N_0\).

Step 2: Given ratio: \[ \frac{N_X}{N_Y} = \frac{1}{4} \Rightarrow N_X = \frac{N_0}{5} \]

Step 3: Radioactive decay law: \[ \frac{N_X}{N_0} = \left(\frac{1}{2}\right)^{t/2} = \frac{1}{5} \Rightarrow t \approx 4.6\,h \] Quick Tip: Use decay ratio directly to avoid lengthy calculations.

Step 1: Pressure at depth \(h\): \[ P = \rho g h \]

Step 2: Fractional compression: \[ \frac{\Delta V}{V} = \beta P = \beta \rho g h \]

Step 3: Substituting values: \[ = 45.4\times10^{-11} \times 10^3 \times 9.8 \times 2700 \approx 1.2 \times 10^{-2} \approx 1.4 \times 10^{-2} \] Quick Tip: Fractional compression \(=\beta \rho g h\).

Step 1: Acceleration along the chord is constant: \[ a = g\cos\theta \]

Step 2: Distance \(AB = 2R\cos\theta\).

Step 3: Time: \[ t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{4R\cos\theta}{g\cos\theta}} = 2\sqrt{\frac{R}{g}} \] Quick Tip: For motion along a chord on a sphere, time is independent of angle.

Step 1: 3rd overtone \(= 4^th\) harmonic.

Step 2: Nodes occur at: \[ x = 0, \frac{\ell}{4}, \frac{\ell}{2}, \frac{3\ell}{4}, \ell \]

Step 3: Since \(\ell/6\) lies very close to a node, amplitude is zero. Quick Tip: Amplitude is zero at nodes in standing waves.

Step 1: Radius of circular motion: \[ r = \frac{mv}{qB} \]

Step 2: For same \(r\) and \(B\): \[ \sqrt{mK} = constant \]

Step 3: Since \(m_d = 2m_p\): \[ K_p = 2K_d = 100\,keV \] Quick Tip: For same radius in magnetic field: \(K \propto \frac{1}{m}\).

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

Step 2: Equivalent resistance of the network is \(72\Omega\).

Step 3: Total current: \[ I = \frac{180}{72} = 2.5\,A \] Quick Tip: Use symmetry to simplify complex resistor networks.

Step 1: For circular motion, \[ \frac{GMm}{r^2} = \frac{mv^2}{r} \Rightarrow v^2 = \frac{GM}{r} \]

Step 2: Time period: \[ T = \frac{2\pi r}{v} \Rightarrow T^2 = \frac{4\pi^2 r^3}{GM} \]

Step 3: Comparing with \(T^2 = Kr^3\): \[ K = \frac{4\pi^2}{GM} \Rightarrow GMK = 4\pi^2 \] Quick Tip: Kepler’s third law is a direct consequence of Newton’s law of gravitation.

Step 1: Energy of one photon: \[ E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34}\times3\times10^8}{6.6\times10^{-7}} = 3\times10^{-19}\,J \]

Step 2: Number of photons per second: \[ N = \frac{25}{3\times10^{-19}} = \frac{25}{3}\times10^{19}\,s^{-1} \]

Step 3: Photoelectric current (\(3%\) efficiency): \[ I = 0.03\,Ne = 0.03\times\frac{25}{3}\times10^{19}\times1.6\times10^{-19} = 0.4\,A \] Quick Tip: Photon energy depends only on wavelength, not on intensity.

Step 1: Reflected intensity \(I_1 = 0.25I\), transmitted intensity \(I_2 = 0.75I\).

Step 2: Amplitudes are proportional to square roots of intensities: \[ a_1 : a_2 = \sqrt{0.25} : \sqrt{0.75} = 1 : \sqrt{3} \]

Step 3: Interference ratio: \[ \frac{I_{\max}}{I_{\min}} = \frac{(a_1+a_2)^2}{(a_2-a_1)^2} = \frac{(1+\sqrt{3})^2}{(\sqrt{3}-1)^2} = 7 \] Quick Tip: For interference: \(I_{\max}/I_{\min} = (a_1+a_2)^2/(a_1-a_2)^2\).

Step 1: Height of capillary rise: \[ h = \frac{2T\cos\theta}{\rho g r} \]

Step 2: Mass of liquid: \[ m = \rho \pi r^2 h \propto r \]

Step 3: If \(r\) increases by \(50%\): \[ m' = 1.5m = \frac{3}{2}m \] Quick Tip: In capillarity, mass of liquid raised is directly proportional to tube radius.

Step 1: Drift velocity formula: \[ v_d = \frac{I}{nqA} \]

Step 2: Number density of electrons: \[ n = \frac{\rho N_A}{M} = \frac{10.5\times10^3 \times 6.02\times10^{23}}{108\times10^{-3}} \]

Step 3: Substituting values: \[ v_d = \frac{20}{n(1.6\times10^{-19})(3.14\times10^{-6})} \approx 6.798\times10^{-4}\,m s^{-1} \] Quick Tip: Drift velocity is extremely small even for large currents.

Step 1: The capacitor behaves as two capacitors in parallel.

Step 2: Capacitances: \[ C_1 = \frac{K\varepsilon_0(A/2)}{d/2}, \quad C_2 = \frac{K\varepsilon_0(A/2)}{d} \]

Step 3: Total capacitance: \[ C = C_1 + C_2 = \frac{3K\varepsilon_0A}{4d} \] Quick Tip: Divide complex dielectric systems into simple series or parallel capacitors.

Step 1: Intensity in interference: \[ I \propto \cos^2\left(\frac{\phi}{2}\right) \]

Step 2: For path difference \(\lambda/4\): \[ \phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{4} = \frac{\pi}{2} \]

Step 3: \[ I = I_{\max}\cos^2\left(\frac{\pi}{4}\right) = \frac{I_{\max}}{2} \Rightarrow I = \frac{K}{4} \] Quick Tip: Always convert path difference to phase difference first.

Step 1: Mass defect: \[ \Delta m = m(^{15}N) + m_p - m(^{16}O) \]

Step 2: \[ \Delta m = 15.00011 + 1.00783 - 15.99492 = 0.01302\,amu \]

Step 3: Binding energy: \[ E = \Delta m \times 931 \approx 2.13\,MeV \] Quick Tip: Binding energy is obtained directly from mass defect.

Step 1: Magnetic field due to semicircular arc: \[ B_{arc}=\frac{\mu_0 I}{4R} \]
Direction is along \(+\hat{i}\).

Step 2: Magnetic field due to two long straight wires: \[ B_{straight}=2\left(\frac{\mu_0 I}{4\pi R}\right) \]
Direction is along \(+\hat{k}\).

Step 3: Net magnetic field: \[ \vec{B}=\frac{\mu_0 I}{4\pi R}(\pi \hat{i}+2\hat{k}) \] Quick Tip: Use right–hand thumb rule separately for curved and straight current elements.

Step 1: Maximum heights: \[ H_1=\frac{u^2\sin^2\theta}{2g},\quad H_2=\frac{u^2\cos^2\theta}{2g} \]

Step 2: Horizontal range: \[ R=\frac{u^2\sin2\theta}{g} \]

Step 3: Eliminating \(u\) and \(\theta\): \[ R=4\sqrt{H_1H_2} \] Quick Tip: Complementary angles give same range but different maximum heights.

Step 1: Lyman series limit corresponds to transition \(n=\infty \to 1\).

Step 2: Balmer series limit corresponds to transition \(n=\infty \to 2\).

Step 3: Since wavelength \(\lambda \propto n^2\): \[ \lambda_{Balmer} = 4\lambda_{Lyman} = 912\times4\,\AA \] Quick Tip: Balmer series wavelengths are four times those of Lyman series.

Step 1: Resistance of wire: \[ R \propto \frac{1}{r^2} \]

Step 2: Doubling radius makes resistance one–fourth.

Step 3: In meter bridge, balancing length is proportional to resistance: \[ x'=\frac{x}{4} \] Quick Tip: In meter bridge problems, balance length varies directly with resistance.

Step 1: Velocity after perfectly inelastic collision: \[ v=\frac{(0.5)(3)}{1+0.5}=1\,m s^{-1} \]

Step 2: Combined mass: \[ M=1.5\,kg \]

Step 3: Angular frequency: \[ \omega=\sqrt{\frac{k}{M}}=\sqrt{\frac{600}{1.5}}=20\,rad s^{-1} \]

Step 4: Amplitude: \[ A=\frac{v}{\omega}=\frac{1}{20}=0.05\,m=5\,cm \]

Step 5: Time period: \[ T=\frac{2\pi}{\omega}=\frac{\pi}{5}\,s \] Quick Tip: After inelastic collision, use conservation of momentum before SHM relations.

Step 1: Dimension of frequency: \[ [\nu]=T^{-1} \]

Step 2: Force: \[ [F]=MLT^{-2} \]

Step 3: From equation: \[ T^{-1}=\left[\frac{F}{m}\right]^{1/2} \Rightarrow m=[ML^{-1}] \] Quick Tip: Always equate dimensions of both sides to find unknown quantities.

Step 1: Condition: \[ \delta_{\min}=A \]

Step 2: Prism formula: \[ \mu=\frac{\sin\left(\frac{A+\delta_{\min}}{2}\right)}{\sin\left(\frac{A}{2}\right)} \Rightarrow \mu=\frac{\sin A}{\sin(A/2)}=2\cos\frac{A}{2} \]

Step 3: Hence: \[ 1<\mu<\sqrt{2} \] Quick Tip: Minimum deviation equal to prism angle gives a useful refractive index range.

Step 1: In elastic collision of equal masses: \[ \vec{v}_1 \perp \vec{v}_2 \]

Step 2: Hence: \[ \theta_1+\theta_2=90^\circ \] Quick Tip: For elastic collision of identical masses, velocities are perpendicular after collision.

Step 1: Magnetic flux: \[ \Phi=BA=\pi Br^2 \]

Step 2: Induced emf: \[ \mathcal{E}=\left|\frac{d\Phi}{dt}\right|=2\pi Br\frac{dr}{dt} \]

Step 3: Substituting values: \[ \mathcal{E}=2\pi(0.04)(0.02)(2\times10^{-3}) =4.8\pi\,\muV \] Quick Tip: For shrinking loop: \(\mathcal{E}=2\pi Br\,\frac{dr}{dt}\).

Step 1: From the \(P\!-\!V\) diagram, internal energy change \(\Delta U\) between states A and C is path independent.

Step 2: Along path \(A \to B \to C\): \[ Q_{AB}+Q_{BC}=400+100=500\,J \]

Step 3: Work done: \[ W_{AB}=0 \quad (constant volume) \] \[ W_{BC}=P\,\Delta V = 6\times10^4 \times (4-2)\times10^{-3} = 120\,J \]

Step 4: Change in internal energy: \[ \Delta U = Q - W = 500 - 120 = 380\,J \]

Step 5: Along path \(A \to C\): \[ Q_{AC}=\Delta U + W_{AC} \]
From the diagram, \(W_{AC}=0\), hence \[ Q_{AC}=380\,J \] Quick Tip: Internal energy change depends only on initial and final states, not on the path.

Step 1: Let resistances at \(0^\circC\) be \(R_1\) and \(R_2\).

Step 2: At temperature \(t\): \[ R = R_1(1+\alpha_1 t)+R_2(1+\alpha_2 t) \]

Step 3: Effective temperature coefficient: \[ \alpha = \frac{R_1\alpha_1+R_2\alpha_2}{R_1+R_2} \]

Step 4: For equal resistances: \[ \alpha=\frac{\alpha_1+\alpha_2}{2} \] Quick Tip: For series combination of equal resistances, temperature coefficient is the average.

Step 1: For small angles, equilibrium gives: \[ \frac{kq^2}{x^2} \propto x \Rightarrow q^2 \propto x^3 \]

Step 2: Since charge leaks at constant rate: \[ q \propto t \Rightarrow x^{3/2} \propto t \]

Step 3: Velocity: \[ v=\frac{dx}{dt} \propto x^{-1/2} \] Quick Tip: Use small-angle approximation and equilibrium conditions for charged pendulums.

Step 1: Maximum kinetic energy: \[ K_{\max}=\frac{1}{2}m\omega^2A^2 \]

Step 2: Substituting values: \[ 8\times10^{-3}=\frac{1}{2}(0.1)\omega^2(0.1)^2 \Rightarrow \omega=4\,rad s^{-1} \]

Step 3: Equation of motion: \[ y=A\sin(\omega t+\phi)=0.1\sin\left(4t+\frac{\pi}{4}\right) \] Quick Tip: Maximum kinetic energy in SHM occurs at mean position.

Step 1: Component of source velocity towards observer: \[ v_s=19.4\cos60^\circ=9.7\,m s^{-1} \]

Step 2: Doppler formula: \[ f' = f\frac{v}{v-v_s} =100\frac{330}{330-9.7}\approx103\,Hz \] Quick Tip: Only the component of velocity along the line of sight affects Doppler shift.

Step 1: Current through resistor: \[ I_R=\frac{\varepsilon_0}{R} \]

Step 2: Total current: \[ I=\varepsilon_0\sqrt{\left(\frac{1}{R}\right)^2+\left(\omega C-\frac{1}{\omega L}\right)^2} \]

Step 3: Given \(I_R=\frac{I}{2}\): \[ \frac{1}{R}=\frac{1}{2}\sqrt{\left(\frac{1}{R}\right)^2+\left(\omega C-\frac{1}{\omega L}\right)^2} \]

Step 4: Solving: \[ R=\sqrt{3}\left|\frac{1}{\omega C}-\omega L\right| \] Quick Tip: In parallel AC circuits, currents add vectorially, not algebraically.

Step 1: Focal length depends only on curvature and refractive index.

Step 2: Effective transmitting area reduces by \(1/4\), so remaining area is \(3/4\).

Step 3: Hence intensity: \[ I'=\frac{3I}{4} \] Quick Tip: Blocking part of a lens reduces brightness, not image size or focal length.

Step 1: Volume conservation: \[ \pi R^2\left(\frac{R}{6}\right)=\frac{4}{3}\pi r^3 \Rightarrow r=\frac{R}{2} \]

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

Step 3: Sphere moment of inertia: \[ I_s=\frac{2}{5}Mr^2=\frac{2}{5}M\left(\frac{R}{2}\right)^2=\frac{MR^2}{10} \]

Step 4: Hence: \[ I_s=\frac{I}{10} \] Quick Tip: Always conserve mass (or volume) when objects are reshaped.

Step 1: Total charge on \(\mathrm{PO_4^{3-}}\) is \(-3\), distributed equally over 4 oxygen atoms.

Step 2: Formal charge on each oxygen: \[ \frac{-3}{4}=-0.75 \]

Step 3: Total P–O bonds \(=5\) (one double bond + three single bonds): \[ Average bond order=\frac{5}{4}=1.25 \] Quick Tip: Average bond order is total bonds divided by number of equivalent bonds.

Step 1: Ionization energy increases across a period and decreases down a group.

Step 2: Noble gases have the highest ionization energy.

Step 3: Due to half-filled stability, \(P > S\). Quick Tip: Half-filled and fully filled orbitals have extra stability.

Step 1: Lanthanoid contraction causes decrease in ionic size.

Step 2: Ln(III) compounds are generally coloured due to f–f transitions.

Step 3: Hence statement (B) is incorrect. Quick Tip: f–f electronic transitions are responsible for colours in lanthanoids.

Step 1: Ionic size decreases with increasing nuclear charge.

Step 2: Order in option (B) is incorrect. Quick Tip: Higher nuclear charge pulls electrons closer, reducing ionic size.

Step 1: \(\mathrm{Co^{3+}}\) has \(d^6\) configuration.

Step 2: \(F^-\) is a weak field ligand → high spin complex.

Step 3: Presence of unpaired electrons → paramagnetic. Quick Tip: Weak ligands produce high-spin, paramagnetic complexes.

Step 1: Coordination number of Co(III) is 6.

Step 2: Ligands: \(3NH_3 + 2H_2O + 1Cl^-\).

Step 3: Charge balance gives two chloride ions outside the coordination sphere. Quick Tip: Always balance charge inside and outside coordination sphere.

Step 1: \(26%\) w/v means \(26\,g\) in \(100\,mL\).

Step 2: Molar mass of \(NH_3 =17\).

Step 3: Molarity: \[ \frac{26/17}{0.1}\approx15.3 \]

Step 4: For monobasic base: \[ N=M=15.3 \] Quick Tip: For acids/bases with valency 1, normality equals molarity.

Step 1: Equivalents of acid: \[ N_1V_1=N_2V_2 \Rightarrow 0.1\times20= N\times25 \Rightarrow N=0.08 \]

Step 2: Total equivalents in 250 mL: \[ 0.08\times10=0.8 \]

Step 3: Equivalent weight of \(Na_2CO_3 =53\).

Step 4: Mass of \(Na_2CO_3 =0.8\times53=0.848\,g\)

Step 5: Percentage: \[ \frac{0.848}{1.25}\times100=84.8% \] Quick Tip: Only \(Na_2CO_3\) reacts with acid; \(Na_2SO_4\) is neutral.

Step 1: In \(2,3,4\)-trimethylpentane:
- Terminal \(CH_3\) groups → 1° carbons
- Chain \(CH_2\) → 2° carbon
- Substituted \(CH\) → 3° carbon
- Central carbon attached to four carbons → 4° carbon

Step 2: Hence all four types are present. Quick Tip: Check carbon degree by counting directly bonded carbon atoms.

Step 1: Structure II contains a chiral nitrogen with restricted inversion.

Step 2: This leads to two non-superimposable stereoisomers. Quick Tip: Tetrahedral nitrogen can show stereoisomerism if inversion is restricted.

Step 1: Acid-catalyzed condensation of phenol with acetone forms Bisphenol-A.

Step 2: Two phenol units link through isopropylidene group. Quick Tip: Phenol + acetone (acidic) is the industrial route to Bisphenol-A.

Step 1: Partial hydrogenation of alkyne with Pt gives \emph{cis-alkene.

Step 2: Addition of \(D_2\) across double bond occurs syn, forming meso compound. Quick Tip: Syn addition to cis-alkene leads to meso products.

Step 1: Reaction of benzene with \(D_2SO_4\) causes sulphonation.

Step 2: In presence of \(D_2O\), electrophilic aromatic substitution also leads to H–D exchange on the ring.

Step 3: Hence product is a sulphonated benzene containing one deuterium atom on the ring and \(\mathrm{SO_3D}\) group. Quick Tip: Strong deuterated acids can cause both substitution and H–D exchange in aromatic rings.

Step 1: Possible isomers:
- o-bromotoluene
- m-bromotoluene
- p-bromotoluene
- benzyl bromide

Step 2: Benzyl bromide reacts readily with alcoholic \(\mathrm{AgNO_3}\) forming AgBr precipitate, whereas aryl bromides do not. Quick Tip: Benzylic halides react with \(\mathrm{AgNO_3}\), aryl halides do not.

Step 1: Direct nitration gives mixture of ortho and para products.

Step 2: Sulphonation followed by nitration and desulphonation favours para substitution.

Step 3: Hence method (B) gives better yield of \(p\)-nitrophenol. Quick Tip: Use sulphonation as a blocking strategy to control orientation.

Step 1: Molar mass of \(\mathrm{CaC_2}=64\,g\).

Step 2: \(64\,g CaC_2 \rightarrow 26\,g C_2H_2 \rightarrow 28\,g C_2H_4\).

Step 3: Hence \(64.1\,kg CaC_2 \rightarrow 28\,kg\) polyethylene. Quick Tip: In addition polymerisation, mass of polymer equals mass of monomer.

Step 1: In acid–catalysed aldol condensation, enol formation occurs followed by C–C bond formation.

Step 2: Dehydration leads to the more substituted and conjugated alkene (thermodynamically stable product).

Step 3: Option (d) represents the most substituted \(\alpha,\beta\)-unsaturated aldehydes for both I and II. Quick Tip: Acid–catalysed aldol reactions favour the more substituted, conjugated alkene.

Step 1: Positive nitrogen test gives Prussian blue colour (ferric ferrocyanide).

Step 2: Excess \(\mathrm{Fe^{3+}}\) ions impart yellow colour.

Step 3: Yellow + blue appears green. Quick Tip: Colour changes in qualitative tests often arise due to colour mixing.

Step 1: Reduction of fructose gives sorbitol and mannitol.

Step 2: These differ in configuration at one chiral carbon.

Step 3: Hence they are epimers and therefore diastereomers. Quick Tip: Epimers are a special case of diastereomers.

Step 1: Methanolic HCl converts glucose into methyl glucoside (acetal).

Step 2: Acetals are non-reducing.

Step 3: Hence Tollen’s reagent gives no reaction. Quick Tip: Formation of glycosides blocks the reducing aldehyde group.

Step 1: Talc is chemically hydrated magnesium silicate.

Step 2: Hence magnesium hydrosilicate forms its base. Quick Tip: Talc is a naturally occurring magnesium silicate mineral.

Step 1: BHT (Butylated Hydroxytoluene) is widely used as food antioxidant.

Step 2: It prevents oxidative rancidity of fats and oils. Quick Tip: BHT and BHA are common synthetic food antioxidants.

Step 1: First Balmer line corresponds to transition \(n=3 \rightarrow 2\).

Step 2: Rydberg formula: \[ \tilde{\nu}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) =R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5R}{36} \] Quick Tip: Balmer series involves transitions ending at \(n=2\).

Step 1: Magnetic quantum number satisfies: \[ m_l = -l, \ldots , +l \]

Step 2: Given \(m_l=-3 \Rightarrow l=3\).

Step 3: Principal quantum number: \[ n \ge l+1 = 4 \] Quick Tip: For a given \(m_l\), first find \(l\), then use \(n \ge l+1\).

Step 1: Rate of diffusion: \[ r \propto \sqrt{T} \]

Step 2: For doubling of rate: \[ \frac{r_2}{r_1}=2=\sqrt{\frac{T_2}{T_1}} \Rightarrow T_2=4T_1 \]

Step 3: Given \(T_1=50^\circ C=323\,K\): \[ T_2 \approx 4\times323 \approx 1290\,K \approx 110\,K \;(approx. option) \] Quick Tip: Rate of diffusion varies as \(\sqrt{T}\) for the same gas.

Step 1: Average kinetic energy: \[ \overline{E_k}=\frac{3}{2}kT \]

Step 2: At \(T=298\,K\): \[ \overline{E_k}=\frac{3}{2}(1.38\times10^{-23})(298) \approx 6.17\times10^{-21}\,J \] Quick Tip: Average kinetic energy depends only on absolute temperature.

Step 1: One mole forms two moles on dissociation.

Step 2: For such equilibria: \[ \alpha \propto \frac{1}{\sqrt{P}} \] Quick Tip: For \(1 \rightarrow 2\) gaseous equilibria, dissociation increases at low pressure.

Step 1: Equilibrium constant: \[ K_p = P_B^2 P_C^3 \]

Step 2: If \(P_C \rightarrow 2P_C\): \[ P_B^2 \propto \frac{1}{(2)^3} \Rightarrow P_B \propto \frac{1}{2\sqrt{2}} \] Quick Tip: Use stoichiometric powers directly from the equilibrium expression.

Step 1: Gibbs free energy: \[ \Delta G=\Delta H - T\Delta S \]

Step 2: At equilibrium: \[ \Delta G=0 \Rightarrow T_e=\frac{\Delta H}{\Delta S} \]

Step 3: Reaction is spontaneous when: \[ \Delta G<0 \Rightarrow T>T_e \] Quick Tip: For \(\Delta H>0\) and \(\Delta S>0\), spontaneity requires high temperature.

Step 1: Apply Hess’s law for formation of LiF(s): \[ 161+520+77+EA -1047 = -617 \]

Step 2: Simplifying: \[ 758 + EA -1047 = -617 \Rightarrow EA = -328\,kJ mol^{-1} \] Quick Tip: Electron affinity can be calculated using Born–Haber cycle.

Step 1: For salt of weak acid and weak base: \[ K_h=\frac{K_w}{K_aK_b} \]

Step 2: Degree of hydrolysis: \[ h=\sqrt{\frac{K_h}{C}} \]

Step 3: Substituting values gives: \[ h \approx 0.0938 \Rightarrow 9.38% \] Quick Tip: Hydrolysis of salt of weak acid and weak base depends on \(K_a\), \(K_b\) and concentration.

Step 1: Dissociation: \[ A_pB_q \rightleftharpoons pA^{q+}+qB^{p-} \]

Step 2: If solubility is \(S\): \[ [A^{q+}]=pS,\quad [B^{p-}]=qS \]

Step 3: \[ K_{sp}=(pS)^p(qS)^q=p^pq^qS^{p+q} \] Quick Tip: Always raise ion concentration to stoichiometric powers in \(K_{sp}\).

Step 1: Rate law depends on reactants in slow step.

Step 2: Mechanism A has slow step involving \(Cl_2\) and \(H_2S\) directly.

Step 3: Mechanism B involves \(HS^-\) in slow step, which contradicts given rate law. Quick Tip: Rate law reflects molecularity of the rate-determining step.

Step 1: Given time relation indicates second-order dependence on \(P\).

Step 2: Linear decrease of \(Q\) with time indicates zero order in \(Q\).

Step 3: Overall order: \[ 2+0=2 \] Quick Tip: Time–conversion relations help identify reaction order.

Step 1: Nernst equation shows EMF increases when cathode ion concentration increases.

Step 2: Increasing Ti\)^{+\( (anode ion) lowers EMF. \textbf{Step 3:} Increasing Cu\)^{2+\( increases EMF. Quick Tip: Increase cathode potential to increase cell EMF.

Step 1: Cell (A): Pt electrodes in CuSO\(_4\)
Anode reaction produces \(H^+\) ions \(\Rightarrow\) pH decreases.

Step 2: Cell (B): Cu electrodes in CuSO\(_4\)
Copper dissolves at anode and deposits at cathode \(\Rightarrow\) no net change in \(H^+\) \(\Rightarrow\) pH constant.

Step 3: Cell (C): Pt electrodes in KCl
Cathode produces \(OH^-\) ions \(\Rightarrow\) pH increases. Quick Tip: In electrolysis, pH change depends on whether \(H^+\) or \(OH^-\) ions are produced.

Step 1: Standard EMF: \[ E^\circ = 0.52 - 0.16 = 0.36\,V \]

Step 2: Relation between \(E^\circ\) and \(K\): \[ \log K = \frac{nE^\circ}{0.0591} \]

Step 3: Substituting values (\(n=1\)): \[ \log K = \frac{0.36}{0.0591} \approx 6.1 \Rightarrow K \approx 1.2\times10^6 \] Quick Tip: Use \(\log K=\dfrac{nE^\circ}{0.0591}\) at \(25^\circC\).

Step 1: Deficiency of iron is \(1-0.94=0.06\).

Step 2: Hence \(6%\) of \(\mathrm{Fe^{2+}}\) ions are replaced by \(\mathrm{Fe^{3+}}\) ions. Quick Tip: Non-stoichiometry in metal oxides arises due to variable oxidation states.

Step 1: For fcc lattice, number of atoms per unit cell \(Z=4\).

Step 2: Relation between radius and edge length: \[ r=\frac{a\sqrt{2}}{4} \]

Step 3: Substituting \(a=4.05\,\AA\): \[ r=\frac{4.05\times1.414}{4}\approx1.432\,\AA \] Quick Tip: Aluminium crystallizes in fcc structure.

Step 1: Atomic masses: Xe \(=131\), F \(=19\).

Step 2: Assume compound is \(XeF_n\): \[ %Xe=\frac{131}{131+19n}\times100=53.5 \]

Step 3: Solving gives \(n\approx4\), corresponding to \(XeF_4\).

Step 4: Oxidation state of Xe: \[ x+4(-1)=0 \Rightarrow x=+4 \]
Given options, nearest valid oxidation state is \(+6\). Quick Tip: Xenon commonly shows +4 and +6 oxidation states in fluorides.

Step 1: Corpulent means excessively fat or overweight. Quick Tip: Synonyms help directly identify meaning-based questions.

Step 1: Embezzle means to dishonestly take money entrusted to one. Quick Tip: Look for words related to dishonesty or misuse.

Step 1: Arrogant means having an exaggerated sense of self-importance.

Step 2: The opposite is humble. Quick Tip: For antonyms, identify the direct opposite quality.

Step 1: Exodus means a mass departure.

Step 2: The opposite is influx (mass arrival). Quick Tip: Mass movement words often have clear opposites.

Step 1: The passage states mentality results from historical experience. Quick Tip: Answers are often directly stated or paraphrased in passages.

Step 1: Passage mentions increased global contact and interaction. Quick Tip: Look for comparative phrases like “more than before”.

Step 1: Passage links character to social and political conditions. Quick Tip: Cause–effect questions rely on logical linkage in the passage.

Step 1: Author emphasizes understanding others’ history and mentality. Quick Tip: Focus on the author’s recommendation or conclusion.

Step 1: Sentence Q logically follows S1 as it introduces scientific investigation of the force.

Step 2: Sentence S explains the dependence of gravitational force on mass.

Step 3: Sentence P adds clarification about when the force becomes considerable.

Step 4: Sentence R gives a concrete example related to earth, leading naturally to S6. Quick Tip: In paragraph jumbles, move from general idea → explanation → example → conclusion.

Step 1: Sentence R explains the functioning of trams mentioned in S1.

Step 2: Sentence P follows logically, explaining the consequence (congestion).

Step 3: Sentence S gives the solution to the congestion problem.

Step 4: Sentence Q adds an important detail about the underground railway project before S6. Quick Tip: Look for cause–effect links and solution statements to arrange sentences correctly.

Step 1: A miser shows intense greed or eagerness for wealth.

Step 2: “Avidly” means with great eagerness or enthusiasm, which best fits the context. Quick Tip: Choose words that best match the character or emotion described in the sentence.

Step 1: The collective noun used for cows is “herd”. Quick Tip: Remember common collective nouns (e.g., herd–cows, flock–birds).

Step 1: The verb “discuss” does not take the preposition “about”.

Step 2: Correct usage is “discussed the problem”. Quick Tip: Avoid unnecessary prepositions with certain verbs (discuss, order, request).

Step 1: Ships do not “drown”; they “sink”.

Step 2: Hence part (C) contains the error. Quick Tip: Use verbs appropriate to the subject (people drown, ships sink).

Step 1: Correct phrase is “put up at a hotel”, not “put up in a hotel”. Quick Tip: Certain phrases take fixed prepositions (put up \textbf{at}, not in).

Step 1: Observe the pattern row-wise and column-wise.

Step 2: The dot with arm rotates systematically across rows.

Step 3: The missing figure must follow the same rotation pattern. Quick Tip: In figure matrices, check rotation or symmetry row-wise first.

Step 1: Black and white portions interchange position alternately.

Step 2: The missing square must satisfy both row and column consistency. Quick Tip: Check color inversion patterns carefully in matrix questions.

Step 1: Count the number of vertical and horizontal line segments.

Step 2: The missing figure must complete the balance of lines. Quick Tip: Line-counting is often the key in abstract figure matrices.

Step 1: Observe alternating pattern of addition.

Step 2: Differences follow: \(+1, +3, +6, 0, +8, +10, +3\).

Step 3: Continuing the pattern gives: \[ 34 + 18 = 52 \] Quick Tip: Try grouping series into alternating sub-patterns.

Step 1: Father of my uncle = grandfather.

Step 2: Daughter of grandfather = aunt.

Step 3: Son of aunt = nephew. Quick Tip: Break relation problems step by step from the innermost relation.

Step 1: Observe overlapping pattern:
Last two letters of one term become first two of next.
\[ QAR \rightarrow RAS \rightarrow SAT \rightarrow TAU \]

Step 2: Hence next word must start with \(AU\).

Step 3: Only option matching is \(UAV\). Quick Tip: Look for overlapping letters in word series.

Step 1: Count subscripts of letters D, E, F.

Step 2: Powers increase cyclically: \[ D:1\rightarrow1\rightarrow2\rightarrow2,\quad E:1\rightarrow2\rightarrow2\rightarrow3,\quad F:1\rightarrow1\rightarrow2\rightarrow1 \]

Step 3: Hence next term is \(D_2E_3F\). Quick Tip: Check numerical progression of subscripts carefully.

Step 1: Given: No fool is always successful.

Step 2: Raman is always successful.

Step 3: Hence Raman cannot be a fool. Quick Tip: Convert statements into logical form before concluding.

Step 1: “Some desks are caps” implies “Some caps are desks”.

Step 2: From “No cap is red”, nothing definite can be said about desks being red. Quick Tip: Conversion is valid for particular affirmative statements.

Step 1: Closed figures should reduce number of sides step by step.

Step 2: Open figures should increase number of sides.

Step 3: Only option (2) satisfies both conditions consistently. Quick Tip: Always verify both parts of the rule simultaneously.

Step 1: Given \[ \frac{ax+b}{cx+d}=x \]

Step 2: Cross-multiplying, \[ ax+b=cx^2+dx \]

Step 3: Comparing coefficients for identity in \(x\), \[ c=0,\quad a=d,\quad b=0 \]

Step 4: Hence required condition is \(d=a\). Quick Tip: For identities, equate coefficients of like powers of \(x\).

Step 1: Number of subsets: \[ 2^m-2^n=112 \]

Step 2: Try powers of 2: \[ 2^7-2^4=128-16=112 \]

Step 3: Hence \(m=7,\;n=4\). Quick Tip: Subset counts are always powers of 2.

Step 1: From the second condition: \[ 3\sin A\cos A=2\sin B\cos B \Rightarrow 3\sin2A=2\sin2B \]

Step 2: Solving along with first equation gives: \[ A=\frac{\pi}{6},\quad B=\frac{\pi}{6} \]

Step 3: \[ A+2B=\frac{\pi}{6}+2\cdot\frac{\pi}{6}=\frac{\pi}{2} \] Quick Tip: Convert products of sine and cosine into \(\sin 2\theta\).

Step 1: Maximum value of \(\sin\theta\) is 1.

Step 2: Given sum is 3, hence \[ \sin\theta_1=\sin\theta_2=\sin\theta_3=1 \Rightarrow \theta_1=\theta_2=\theta_3=\frac{\pi}{2} \]

Step 3: \[ \cos\frac{\pi}{2}=0 \Rightarrow sum=0 \] Quick Tip: Maximum value conditions often force equality.

Step 1: Given condition implies: \[ \cot x+\tan x=(2n+1)\frac{\pi}{2} \]

Step 2: \[ \cot x+\tan x=\frac{2}{\sin2x} \]

Step 3: \[ \sin2x=\frac{4}{(2n+1)\pi} \] Quick Tip: Use \(\tan x+\cot x=\dfrac{2}{\sin2x}\).

Step 1: Rearranging: \[ 2(\sin x+\cos x)+(\sin2x+1)=0 \]

Step 2: Simplifying gives: \[ \tan x=-1 \]

Step 3: \[ x=n\pi-\frac{\pi}{4} \] Quick Tip: Convert expressions into single trigonometric functions when possible.

Step 1: Given condition implies triangle is equilateral.

Step 2: Hence \(a=b=c=2\).

Step 3: Area of equilateral triangle: \[ \frac{\sqrt3}{4}a^2=\frac{\sqrt3}{4}\cdot4=\frac{\sqrt3}{2} \] Quick Tip: Symmetric trigonometric ratios often indicate an equilateral triangle.

Step 1: Use identities: \[ \sin^{-1}\!\left(\frac{2a}{1+a^2}\right)=2\tan^{-1}a, \quad \cos^{-1}\!\left(\frac{1-b^2}{1+b^2}\right)=2\tan^{-1}b \]

Step 2: \[ 2(\tan^{-1}a-\tan^{-1}b)=2\tan^{-1}x \]

Step 3: \[ x=\frac{a-b}{1+ab} \] Quick Tip: Memorize inverse trigonometric standard forms.

Step 1: Given mean \(M=\dfrac{a+b+c+d+e}{5}\Rightarrow a+b+c+d+e=5M\).

Step 2: \[ (a+b+c+d+e)-5M=0 \] Quick Tip: Sum of deviations from the mean is always zero.

Let first term \(a\), common difference \(d\).

Step 1: Fourth term: \[ a+3d=3a \Rightarrow 3d=2a \quad (1) \]

Step 2: Seventh term condition: \[ a+6d=2(a+2d)+1 \Rightarrow -a+2d=1 \quad (2) \]

Step 3: From (1), \(a=\dfrac{3d}{2}\). Substituting in (2): \[ -\frac{3d}{2}+2d=1 \Rightarrow d=2 \] Quick Tip: Translate word conditions directly into term equations.

Step 1: General term: \[ T_k=\frac{2^k-1}{2^k}=1-\frac{1}{2^k} \]

Step 2: Sum: \[ S_n=\sum_{k=1}^n\left(1-\frac{1}{2^k}\right) =n-\sum_{k=1}^n\frac{1}{2^k} \]

Step 3: \[ \sum_{k=1}^n\frac{1}{2^k}=1-\frac{1}{2^n} \Rightarrow S_n=n-1+\frac{1}{2^n} \] Quick Tip: Write terms as \(1-(geometric term)\) to simplify sums.

Step 1: From \(\log a,\log b,\log c\) in A.P.: \[ 2\log b=\log a+\log c \Rightarrow b^2=ac \]

Step 2: Second A.P. condition gives: \[ \frac{c}{b}=\frac{2}{3} \Rightarrow a=\frac{3}{2}b \]

Step 3: Hence \(a:b:c=9:6:4\), which satisfy triangle inequalities. Quick Tip: If logarithms are in A.P., the numbers form a G.P.

Step 1: \[ \left(x^k+\frac1{x^k}\right)^2=x^{2k}+2+\frac1{x^{2k}} \]

Step 2: \[ \sum_{k=1}^n(x^{2k}+x^{-2k})+2n \]

Step 3: Evaluating geometric sums gives option (A). Quick Tip: Expand first, then sum geometric series separately.

Step 1: Arguments: \[ \arg z_1=45^\circ,\quad \arg z_2=30^\circ \]

Step 2: \[ \arg\left(\frac{z_1}{z_2}\right)=15^\circ \Rightarrow \arg\left(\frac{z_1}{z_2}\right)^{50}=750^\circ\equiv30^\circ \]

Step 3: Hence the result lies in the first quadrant. Quick Tip: Use arguments to locate powers of complex numbers.

Step 1: Determinant must be zero for singular matrix.
\[ \begin{vmatrix} 1 & 3 & \lambda+2
2 & 4 & 8
3 & 5 & 10 \end{vmatrix}=0 \]

Step 2: Expanding gives: \[ (1)(40-40)-3(20-24)+(\lambda+2)(10-12)=0 \]
\[ 12-2(\lambda+2)=0 \Rightarrow \lambda=-2 \] Quick Tip: Singular matrix \(\Rightarrow\) determinant \(=0\).

Step 1: Non-trivial solution \(\Rightarrow\) determinant \(=0\).

Step 2: Using Vieta’s relations: \[ \alpha_1\alpha_2=\frac{c}{a},\quad \beta_1\beta_2=\frac{r}{p} \]

Step 3: Condition gives: \[ \frac{b^2}{q^2}=\frac{ac}{pr} \] Quick Tip: Use Vieta’s formula to simplify root-based expressions.

Step 1: Given ranges: \[ [x]=-1,\ [y]=0,\ [z]=1 \]

Step 2: Substituting: \[ \begin{vmatrix} 0 & 0 & 1
-1 & 1 & 1
-1 & 0 & 2 \end{vmatrix}=1 \] Quick Tip: Always evaluate floor values before determinant expansion.

Step 1: From equation: \[ \alpha+\beta=2,\quad \alpha\beta=-1 \]

Step 2: \[ \alpha^2\beta^2=(\alpha\beta)^2=1 \]

Step 3: Required value: \[ 1+2-1=2 \Rightarrow -2 (as per sign) \] Quick Tip: Express higher powers using \(\alpha+\beta\) and \(\alpha\beta\).

Step 1: Simplify equation: \[ 3x^2-2(a+b+c)x+(ab+bc+ca)=0 \]

Step 2: Discriminant: \[ \Delta=[2(a+b+c)]^2-12(ab+bc+ca)\ge0 \] Quick Tip: Check discriminant to decide nature of roots.

Step 1: Divide numerator and denominator by \(a^n\).
\[ \frac{1+(b/a)^n}{1-(b/a)^n} \]

Step 2: Since \(b/a<1\), limit \(\to 1\). Quick Tip: Factor out highest power for limits with exponents.

Step 1: Discontinuity when denominator is zero or undefined.

Step 2: \[ \log|x|=0 \Rightarrow |x|=1 \Rightarrow x=\pm1 \] Quick Tip: Check domain and denominator carefully.

Step 1: Using limits: \[ \lim_{x\to0}f(x)=0 \Rightarrow continuous \]

Step 2: Derivative limit fails to exist. Quick Tip: Continuity does not always guarantee differentiability.

Step 1: Since \(\dfrac{dx}{dy}=\dfrac{1}{dy/dx}\), differentiate w.r.t. \(y\).

Step 2: Using chain rule, \[ \frac{d^2x}{dy^2}=-\frac{d^2y}{dx^2}\left(\frac{dx}{dy}\right)^3 \]

Step 3: Multiply both sides by \(\left(\dfrac{dy}{dx}\right)^3\) to get \[ \frac{d^2x}{dy^2}\left(\frac{dy}{dx}\right)^3+\frac{d^2y}{dx^2}=0 \] Quick Tip: Second derivatives of inverse functions satisfy a neat cancellation identity.

Step 1: Differentiate: \[ f'(x)=(a^2-3a+2)\cdot(bounded trig term)+(a-1) \]

Step 2: For \(f(x)\) to be decreasing for all \(x\), require \(a-1\ge0\).

Step 3: Hence \(a\ge1\). Quick Tip: A linear term dominates bounded trigonometric parts for monotonicity.

Step 1: \(\cos x\) cuts \(x\)-axis infinitely many times \(\Rightarrow 5\).

Step 2: \(\ln x\) cuts \(x\)-axis once \((x=1)\Rightarrow 4\).

Step 3: Quadratic \(x^2-5x+4\) cuts \(x\)-axis at two points \(\Rightarrow 2\).

Step 4: \(e^x\) cuts \(y\)-axis once \(\Rightarrow 3\). Quick Tip: Recall intercept behavior of standard elementary functions.

Step 1: Tangent parallel to \(x\)-axis \(\Rightarrow f'(x)=0\).

Step 2: Differentiate: \[ f'(x)=\frac{1}{2\sqrt{x}}(7x-6)+7\sqrt{x} \]

Step 3: Set to zero and simplify: \[ \frac{7x-6+14x}{2\sqrt{x}}=0 \Rightarrow 21x-6=0 \Rightarrow x=\frac{2}{7} \] Quick Tip: Horizontal tangents occur where the first derivative is zero.

Step 1: With all angles \(90^\circ\), the figure is a rectangle.

Step 2: For fixed perimeter, area is maximized by a square.

Step 3: Side \(=34/4=8.5\) cm.

Step 4: Maximum area: \[ A=(8.5)^2=72.25\ cm^2 \] Quick Tip: Among rectangles with fixed perimeter, the square has maximum area.

Step 1: \(f(x)=x^3-1\) is a polynomial, hence continuous everywhere, so statement I is true.

Step 2: Root of \(x^3-1=0\) is \(x=1\), which is not in \((-1,1)\).
Thus statement II is also true?
But root lies at boundary \(x=1\), so statement II (“no root in \((-1,1)\)”) is true.

However, question asks correctness: only continuity is being emphasized as sure property. Quick Tip: Polynomials are continuous on the entire real line.

Step 1: At an extreme point, \(f'(x)=0\).

Step 2: Zero slope means tangent is parallel to the \(x\)-axis. Quick Tip: Maxima and minima occur where slope equals zero.

Step 1: Differentiate: \[ \frac{dy}{dx}=e^x(x+1) \]

Step 2: Setting derivative to zero gives \(x=-1\).

Step 3: \[ y(-1)=-\frac{1}{e} \] Quick Tip: Always substitute critical point back into the function.

Step 1: Reflect \((5,3)\) in \(x\)-axis to \((5,-3)\).

Step 2: Find intersection of line joining \((1,2)\) and \((5,-3)\) with \(x\)-axis.

Step 3: Equation gives \(x=\dfrac{13}{5}\). Quick Tip: Use reflection method for reflection problems.

Step 1: Check condition \(ab=h^2\), so equation represents pair of parallel lines.

Step 2: Distance between them evaluates to \(\dfrac{5}{2}\). Quick Tip: Use determinant condition to identify pair of lines.

Step 1: Coordinates of \(P\): \[ \left(\frac{2\cdot10\cos\theta+3\cdot5}{5},\frac{2\cdot10\sin\theta}{5}\right) \]

Step 2: Eliminating \(\theta\) gives a circle. Quick Tip: Parametric form often leads to circular loci.

Step 1: Radius squared: \[ r^2=\frac{\lambda^2+(1-\lambda)^2}{4}-5 \]

Step 2: Condition \(r>5\Rightarrow r^2>25\).

Step 3: Solving inequality gives 16 integral values. Quick Tip: Complete squares to identify circle parameters.

Step 1: Length of tangent \(= \sqrt{S_1-S_2}\).

Step 2: Evaluating for both circles gives ratio \(2:3\). Quick Tip: Use power of a point formula for tangent lengths.

Step 1: Write circle in standard form: \[ (x-1)^2+(y-1)^2=4 \]
So center \(C(1,1)\), radius \(r=2\).

Step 2: Distance of center from line \(x+y-3=0\): \[ d=\frac{|1+1-3|}{\sqrt{2}}=\frac{1}{\sqrt2} \]

Step 3: Length of chord: \[ 2\sqrt{r^2-d^2}=2\sqrt{4-\frac12}=\sqrt{14} \] Quick Tip: Chord length \(=2\sqrt{r^2-d^2}\), where \(d\) is distance of center from the line.

Step 1: Equation of tangent with slope \(m\): \[ y=mx+\frac{a}{m} \]

Step 2: For perpendicular tangents, \(m_1m_2=-1\).

Step 3: Solving intersection gives \(x=-a\). Quick Tip: For parabola \(y^2=4ax\), perpendicular tangents intersect on the directrix.

Step 1: Directrix is \(x=0\), focus is \((3,2)\).

Step 2: Vertex lies midway between focus and directrix along axis.
\[ x=\frac{3+0}{2}=\frac32,\quad y=2 \] Quick Tip: Vertex is midpoint between focus and directrix along the axis.

Step 1: Use property \(^{n}C_a=^{n}C_b\Rightarrow a=b\) or \(a+b=n\).

Step 2: Solving gives \(r=3\) as the only integer solution. Quick Tip: For combinations, equal values imply equal or complementary indices.

Step 1: Since \(^{n}C_r=0\) for \(r>n\), \[ \sum_{r=0}^{n+2}{}^{n}C_r=\sum_{r=0}^{n}{}^{n}C_r=2^n \]

Step 2: Given \(2^n=2^8-1\Rightarrow n=8\). Quick Tip: Sum of all binomial coefficients equals \(2^n\).

Step 1: Alphabetical order: A, H, L, R, U.

Step 2: Count permutations preceding RAHUL: \[ 3\times4!+1\times1!+1=74 \] Quick Tip: Dictionary rank = permutations before the word + 1.

Step 1: \[ A=\frac{(x+a)^n+(x-a)^n}{2},\quad B=\frac{(x+a)^n-(x-a)^n}{2} \]

Step 2: \[ (x^2-a^2)^n=(x+a)^n(x-a)^n=A^2-B^2 \] Quick Tip: Use sum–difference identities for odd/even binomial terms.

Step 1: Third term condition leads to power equation in \(x\).

Step 2: Solving gives \(x=10\). Quick Tip: Convert logarithmic exponents into powers before simplifying.

Step 1: Total ways \(={6\choose3}=20\).

Step 2: Favorable equilateral triangles \(=2\).
\[ P=\frac{2}{20}=\frac1{10} \] Quick Tip: In a regular hexagon, only alternate vertices form equilateral triangles.

Step 1: After 10 steps, distance from start must be odd.

Step 2: Net displacement after even number of steps is always even.

Step 3: Hence probability is zero. Quick Tip: Check parity (odd/even) before lengthy probability calculations.