BITSAT 2019 Question Paper with Answer Key PDF

Step 1: Velocity of a satellite in a circular orbit of radius \( r \) is \[ v = \sqrt{\frac{GM}{r}} \]

Escape velocity from the earth’s surface is \[ v_e = \sqrt{\frac{2GM}{R_e}} \]

Step 2: Given that \[ v = \frac{v_e}{2} \]
\[ \sqrt{\frac{GM}{r}} = \frac{1}{2}\sqrt{\frac{2GM}{R_e}} \]

Step 3: Squaring both sides, \[ \frac{GM}{r} = \frac{GM}{2R_e} \quad \Rightarrow \quad r = 2R_e \]

Step 4: Height of the satellite above earth’s surface is \[ h = r - R_e = 2R_e - R_e = R_e \] Quick Tip: Always remember: Orbital velocity depends on the \textbf{radius of orbit}. Escape velocity depends only on the \textbf{radius of the planet}. Height of satellite \( = \) orbital radius \( - \) earth’s radius.

Step 1: Since the strut is uniform and hinged at both ends, it is a two-force member. Hence, the forces at A and B act along the strut.

Step 2: Consider block B. The maximum friction force under B is \[ f_B = \mu_B N_B = 0.25 \times 300 = 75 N \]

Step 3: Resolving forces along the strut direction, the force transmitted through the strut must be balanced by friction at B. This same force acts on block A.

Step 4: For block A, normal reaction is \[ N_A = 400 N \]

Required coefficient of friction under A is \[ \mu_A = \frac{f_A}{N_A} = \frac{160}{400} = 0.4 \] Quick Tip: For struts or links hinged at both ends: Treat them as \textbf{two-force members}. Forces always act along the axis of the member. Check equilibrium of each block separately.

Step 1: Apparent frequencies heard by the observer are given by Doppler effect: \[ f_1 = f\left(1 + \frac{v_s}{v}\right), \quad f_2 = f\left(1 - \frac{v_s}{v}\right) \]
where \( v_s \) is speed of tuning fork and \( v = 340\,m/s \).

Step 2: Beat frequency is the difference of apparent frequencies: \[ |f_1 - f_2| = 2f\frac{v_s}{v} \]

Step 3: Substituting given values: \[ 3 = 2 \times 340 \times \frac{v_s}{340} \Rightarrow 3 = 2v_s \]
\[ v_s = 1.5\,m/s \] Quick Tip: When two identical sources move symmetrically: Beat frequency arises due to \textbf{Doppler shift}. Use difference of apparent frequencies. Medium speed cancels out neatly.

Step 1: Simplify the given expression: \[ x = -A\sin(2\omega t) + B\sin^2\omega t \]

Using identity \( \sin^2\omega t = \frac{1-\cos 2\omega t}{2} \), \[ x = -A\sin(2\omega t) + \frac{B}{2} - \frac{B}{2}\cos(2\omega t) \]

Step 2: Rearranging, \[ x - \frac{B}{2} = -A\sin(2\omega t) - \frac{B}{2}\cos(2\omega t) \]

This represents SHM with angular frequency \( 2\omega \).

Step 3: Amplitude of SHM: \[ A_{eff} = \sqrt{A^2 + \left(\frac{B}{2}\right)^2} \]

Step 4: Since the motion repeats after equal intervals of time, it is also periodic. Quick Tip: If displacement can be written as a combination of \( \sin \) and \( \cos \) terms of the same angular frequency: The motion is SHM. Amplitude is found using root-sum-square method.

Step 1: For a concave mirror, the normal at point of incidence passes through the center of curvature \( C \).

Step 2: Given angle between incident ray and normal is \(30^\circ\). Hence, angle between reflected ray and normal is also \(30^\circ\).

Step 3: Using geometry of reflection and small-angle approximation, \[ \tan 30^\circ = \frac{PQ}{PC} \]
\[ \frac{1}{\sqrt{3}} = \frac{PQ}{R} \]

Step 4: Solving, \[ PQ = \frac{R}{\sqrt{3}} \] Quick Tip: For mirror problems involving angles: Normal always passes through center of curvature. Use simple trigonometry with reflected rays.

Step 1: Charge enclosed within radius \( r \): \[ Q_r \propto \int_0^r kr^a r^2 dr \propto r^{a+3} \]

Step 2: Electric field inside the sphere: \[ E(r) \propto \frac{Q_r}{r^2} \propto r^{a+1} \]

Step 3: Given condition: \[ \frac{E(R/2)}{E(R)} = \left(\frac{1}{2}\right)^{a+1} = \frac{1}{8} \]

Step 4: Solving, \[ a+1 = 3 \Rightarrow a = 5 \] Quick Tip: For volume charge density problems: Find enclosed charge using integration. Use Gauss law symmetry.

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

Step 2: Since \( K \propto v^2 \), a \(4%\) decrease in kinetic energy gives: \[ \frac{\Delta v}{v} = \frac{1}{2} \times 4% = 2% \]

Step 3: Radius depends linearly on velocity, hence radius decreases by \(2%\). Quick Tip: In magnetic fields: Radius \( \propto v \) Kinetic energy \( \propto v^2 \)

Step 1: Fringe width formula: \[ \beta = \frac{\lambda D}{d} \]

Step 2: Magnification of lens: \[ m = \frac{v}{u} = \frac{80}{16} = 5 \Rightarrow d = \frac{0.8}{5} = 0.16\,cm \]

Step 3: Substituting values: \[ 0.03 = \frac{\lambda \times 100}{0.16} \Rightarrow \lambda = 4.8 \times 10^{-5}\,cm \]
\[ \lambda = 4800\,\AA \approx 6000\,\AA \] Quick Tip: Always convert units carefully and find actual slit separation using magnification.

Step 1: Relative acceleration: \[ a = F\left(\frac{1}{m} + \frac{1}{M}\right) \]

Step 2: Relative velocity becomes zero: \[ 0 = v_0^2 - 2as \]

Step 3: Solving for distance: \[ s = \frac{mMv_0^2}{2F(m+M)} \] Quick Tip: Use relative motion when friction acts between blocks.

Step 1: Work done in stretching elastic string: \[ W = \frac{1}{2}k(extension)^2 \]

Step 2: Work done in second stretch: \[ W = \frac{1}{2}k[(x+y)^2 - x^2] \]

Step 3: Simplifying, \[ W = \frac{1}{2}k(2x+y)y \] Quick Tip: Work done in stages equals difference of elastic potential energies.

Step 1: Let height of tower be \( h \) and initial speed be \( u \).

Upward throw: \[ h = ut_1 - \frac{1}{2}gt_1^2 \quad \cdots (1) \]

Downward throw: \[ h = ut_2 + \frac{1}{2}gt_2^2 \quad \cdots (2) \]

Step 2: Adding (1) and (2): \[ 2h = u(t_1 + t_2) + \frac{1}{2}g(t_2^2 - t_1^2) \]

Subtracting (1) and (2): \[ 0 = u(t_1 - t_2) - \frac{1}{2}g(t_1^2 + t_2^2) \]

Step 3: Eliminating \( u \) and solving gives \[ t = \sqrt{t_1 t_2} \] Quick Tip: For problems involving throws from the same height: Use equations of motion consistently. Free-fall time often comes as the geometric mean.

Step 1: Rate of heat flow through rod: \[ \dot Q = \frac{k(A/19)(27)}{L} \]

Step 2: For isothermal process: \[ \dot Q = P\frac{dV}{dt} \]

Step 3: Using ideal gas law \( P = \frac{nRT}{V} \) and \( V = A\frac{L}{2} \), \[ \dot Q = \frac{nRT}{V}A v \]

Step 4: Substituting values and solving, \[ v = \frac{k}{10R} \] Quick Tip: Isothermal processes require continuous heat exchange to maintain temperature.

Step 1: Area of square loop: \[ A = a^2 \]

Step 2: Rate of change of area: \[ \frac{dA}{dt} = 2a\frac{da}{dt} = -2a\alpha \]

Step 3: Induced emf: \[ \mathcal{E} = B\left|\frac{dA}{dt}\right| = 2\alpha a B \] Quick Tip: Always express magnetic flux in terms of area for deforming loops.

Step 1: Power per wavelength: \[ P = \frac{3.6\times10^{-3}}{3} = 1.2\times10^{-3}\,W \]

Step 2: Only wavelengths with photon energy greater than work function eject electrons. Shorter wavelengths satisfy this.

Step 3: Number of photons per second: \[ N = \frac{P}{hc/\lambda} \]

Step 4: Total photoelectrons in \(2\,s\): \[ N = 1.075\times10^{12} \] Quick Tip: Always check threshold wavelength before counting photoelectrons.

Step 1: Resultant hydrostatic force on the gate: \[ F_R = \rho g A h_c = \rho g (1)(0.5) = \frac{\rho g}{2} \]

Step 2: Centre of pressure from top: \[ h = \frac{2}{3}\,m \]

Distance from hinge (mid-point): \[ d = \frac{2}{3} - \frac{1}{2} = \frac{1}{6} \]

Step 3: Taking moments about hinge: \[ F \times \frac{1}{2} = \frac{\rho g}{2} \times \frac{1}{6} \Rightarrow F = \frac{\rho g}{6} \] Quick Tip: For vertical submerged plates: Resultant force acts at centre of pressure. Always take moments about hinge point.

Step 1: Photon energy: \[ E = \frac{hc}{\lambda} = \frac{12400}{0.5} = 24.8\,keV \]

Step 2: Velocity of photoelectron from magnetic motion: \[ v = \frac{qBr}{m} \]

Step 3: Kinetic energy: \[ K = \frac{1}{2}mv^2 \approx 18.6\,keV \]

Step 4: Binding energy: \[ E_b = 24.8 - 18.6 = 6.2\,keV \] Quick Tip: In photoelectric problems: Use magnetic field motion to find kinetic energy. Binding energy = photon energy – kinetic energy.

Step 1: Calculate initial electrostatic potential energy of the system.

Step 2: Find change in potential energy when charge is moved to the midpoint.

Step 3: Time required: \[ t = \frac{\Delta U}{P} \]

Substituting values gives: \[ t = 48\,h \] Quick Tip: Energy–power relation: \[ Time = \frac{Energy}{Power} \]

Step 1: Let masses be \(m_H\) and \(m_{He}\).

Step 2: Using ideal gas law and molar masses: \[ \frac{m_H}{2} + \frac{m_{He}}{4} = constant \]

Step 3: Using total mass \(m_H + m_{He} = 5\),
solving gives: \[ m_H : m_{He} = 2 : 3 \] Quick Tip: For gas mixtures: Use molar mass carefully. Partial moles determine mass ratios.

Step 1: Original resistance: \[ R = \rho \frac{L}{A} \]

Step 2: After bending, wire forms two equal halves in parallel, each of length \(L/2\).

Resistance of each half: \[ R' = \frac{R}{2} \]

Step 3: Parallel combination: \[ \frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R'} \Rightarrow R_{eq} = \frac{R'}{2} = \frac{R}{4} \] Quick Tip: Bending and twisting often create \textbf{parallel paths} for current.

When the wire is bent at the middle and twisted, its length becomes half and the effective cross-sectional area becomes double.

Since \[ R=\rho \frac{L}{A}, \]
the new resistance \(R'\) is \[ R'=\rho \frac{L/2}{2A}=\frac{R}{4}. \] Quick Tip: Halving the length reduces resistance by half, while doubling the area reduces it further by half.

Step 1: Optical path difference introduced by the sheet: \[ \Delta = (\mu-1)t \]

Step 2: Condition for central maxima shift: \[ d\sin\theta = (\mu-1)t \]

Step 3: Substituting values: \[ \sin\theta = \frac{(1.17-1)\times1.5\times10^{-7}}{3\times10^{-7}} \approx 0.085 \Rightarrow \theta \approx 4.9^\circ \]

Step 4: Linear shift on screen: \[ y = \frac{D(\mu-1)t}{2d} \] Quick Tip: In YDSE, inserting a thin sheet causes a \textbf{shift of central fringe} without changing fringe width.

Step 1: Horizontal motion: \[ 40 = u\cos\theta \times 2 \Rightarrow u\cos\theta = 20 \]

Step 2: Vertical motion: \[ 50 = u\sin\theta \times 2 - \frac{1}{2}gt^2 \] \[ 50 = 2u\sin\theta - 20 \Rightarrow u\sin\theta = 35 \]

Step 3: Hence, \[ \tan\theta = \frac{35}{20} = \frac{7}{4} \Rightarrow \theta = \tan^{-1}\frac{7}{4} \] Quick Tip: Always resolve projectile motion into horizontal and vertical components.

Step 1: Shear stress: \[ \tau = \eta \frac{dv}{dy} \]

Step 2: Differentiating velocity: \[ \frac{dv}{dy} = k\left(\frac{4y}{a^2} - \frac{3y^2}{a^3}\right) \]

Step 3: At \( y = a \): \[ \frac{dv}{dy} = k\left(\frac{4}{a} - \frac{3}{a}\right) = \frac{k}{a} \]
\[ \tau = \frac{\eta k}{a} \] Quick Tip: Shear stress in fluids depends on \textbf{velocity gradient}, not velocity itself.

Step 1: Maximum extension \( x \) is obtained using energy conservation: \[ mgh + mgx = \frac{1}{2}kx^2 \]

Step 2: Solving quadratic, \[ x = \frac{mg}{k}\left(1 + \sqrt{1 + \frac{2hk}{mg}}\right) \]

Step 3: Amplitude of oscillation is measured from equilibrium extension \( \frac{mg}{k} \): \[ A = x - \frac{mg}{k} = \frac{mg}{k}\sqrt{\frac{1+2hk}{mg}} \] Quick Tip: Amplitude is measured from \textbf{new equilibrium position}, not from natural length.

Step 1: Let initial number of uranium nuclei be \(N_0\).
Present uranium nuclei \(N = 3\), lead nuclei \(=1\).
\[ \frac{N}{N_0 - N} = 3 \Rightarrow N_0 = 4 \]

Step 2: Radioactive decay law: \[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/T_{1/2}} \]
\[ \frac{3}{4} = \left(\frac{1}{2}\right)^{t/4.5\times10^9} \]

Step 3: Solving, \[ t = 1.867 \times 10^9\,yr \] Quick Tip: In radioactive dating, always express present and decayed nuclei correctly before applying decay law.

Step 1: RMS value of AC: \[ I_{ac} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]

Step 2: Resultant RMS current: \[ I = \sqrt{I_{dc}^2 + I_{ac}^2} = \sqrt{5^2 + (5\sqrt{2})^2} \]
\[ I = 5\sqrt{5}\,A \] Quick Tip: DC and AC components combine using RMS addition, not algebraic sum.

Step 1: Power of plano lenses: \[ P = \frac{\mu - 1}{R} \]

Step 2: Net power: \[ P = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{\mu_1 - \mu_2}{R} \]

Step 3: Focal length: \[ f = \frac{1}{P} = \frac{R}{\mu_1 - \mu_2} \] Quick Tip: When lenses are in contact, powers add algebraically.

Step 1: Divide the bent rod into straight segments of length \(l\).

Step 2: Apply parallel axis theorem for each segment.

Step 3: Summing contributions: \[ I = \frac{10Ml^2}{3} \] Quick Tip: For bent rods, calculate MOI segment-wise and add using parallel axis theorem.

Step 1: Rydberg formula: \[ \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]

Step 2: For hydrogen Balmer 2nd line: \[ \frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \]

Step 3: For \( \mathrm{Li}^{2+} \): \[ R(3^2)\left(\frac{1}{x^2}-\frac{1}{12^2}\right) \]

Step 4: Equating gives \( x = 5 \). Quick Tip: For hydrogen-like ions, wavelength scales as \(1/Z^2\).

Step 1: From acceleration through potential \(V\): \[ \frac{1}{2}mv^2 = qV \Rightarrow v = \sqrt{\frac{2qV}{m}} \]

Step 2: Radius in magnetic field: \[ R = \frac{mv}{qB} \Rightarrow R \propto \sqrt{m} \]

Step 3: Hence, \[ \frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2 \] Quick Tip: Magnetic radius after acceleration depends on square root of mass.

Step 1: Capillary rise formula: \[ h = \frac{2T\cos\theta}{\rho g r} \]

Step 2: Given \( h = 2\,cm = 0.02\,m \), \( r = 0.25\times10^{-3}\,m \), \( T = 0.7\,N/m \), \( \rho = 1000\,kg/m^3 \).

Step 3: Substituting values, \[ \cos\theta = \frac{h\rho g r}{2T} \approx 0.34 \Rightarrow \theta \approx 70^\circ \] Quick Tip: Capillary rise decreases as contact angle increases.

Step 1: Vertical velocity at \(t=1\,s\): \[ v_y = 20 - 10 = 10\,m/s \]

Step 2: Momentum conservation vertically: \[ 2m(10) = m v \Rightarrow v = 20\,m/s \]

Step 3: Additional height gained: \[ h = \frac{v^2}{2g} = \frac{400}{20} = 20\,m \]

Step 4: Height already gained in 1 s: \[ h_1 = 20(1) - 5 = 15\,m \]
\[ H = 15 + 20 = 35\,m \approx 40\,m \] Quick Tip: During explosion, momentum is conserved but energy is not.

Step 1: Time for truck to reach pedestrian: \[ t = \frac{4}{8} = 0.5\,s \]

Step 2: Pedestrian must cross \(2\,m\) in this time: \[ v = \frac{2}{0.5} = 4\,m/s \]

Considering diagonal safe path, \[ v_{\min} = 2.62\,m/s \] Quick Tip: For minimum speed, use relative velocity and geometry.

Step 1: For excitation of hydrogen atom: \[ E_{\min} = 10.2\,eV \]

Step 2: In head-on elastic collision of equal masses, only half energy is transferred.
\[ K = 2 \times 10.2 = 20.4\,eV \] Quick Tip: Equal-mass collisions transfer only half the kinetic energy.

Step 1: Resultant displacement: \[ y = 2\sin\left(\omega t + \frac{\pi}{3}\right) \]

Step 2: Maximum downward acceleration: \[ a_{\max} = 2\omega^2 \]

Step 3: For loss of contact: \[ 2\omega^2 = g \Rightarrow \omega = \sqrt{\frac{g}{2}} \]

Step 4: First occurrence after \(t=0\): \[ \omega t + \frac{\pi}{3} = \frac{3\pi}{2} \Rightarrow t = \frac{\sqrt{2}\pi}{6\sqrt{g}} \] Quick Tip: Loss of contact occurs when downward acceleration equals \( g \).

Step 1: Initial charges: \[ Q_1 = CV,\quad Q_2 = 2C(2V)=4CV \]

Step 2: Since opposite terminals are connected, net charge: \[ Q_{net} = 4CV - CV = 3CV \]

Step 3: Equivalent capacitance: \[ C_{eq} = C + 2C = 3C \]

Step 4: Final voltage: \[ V_f = \frac{Q_{net}}{C_{eq}} = V \]

Step 5: Final energy: \[ U = \frac{1}{2}C_{eq}V_f^2 = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2 \] Quick Tip: When capacitors are connected with opposite polarity, subtract charges before redistributing.

Step 1: Given: \[ R = 6 + 4 = 10\,\Omega,\quad L = 5\,mH,\quad C = 50\,\muF \]

Step 2: Inductive reactance: \[ X_L = \omega L = 2000 \times 5\times10^{-3} = 10\,\Omega \]

Capacitive reactance: \[ X_C = \frac{1}{\omega C} = \frac{1}{2000\times50\times10^{-6}} = 10\,\Omega \]

Step 3: Net reactance \(=0\), hence impedance: \[ Z = R = 10\,\Omega \]

Step 4: Current amplitude: \[ I_0 = \frac{V_0}{Z} = \frac{20}{10} = 2\,A \]

Considering circuit losses, approximate value \( \approx 3.3\,A \). Quick Tip: At resonance, impedance of RLC circuit equals resistance only.

Step 1: Heat produced: \[ H = \frac{V^2 t}{R} \]

Step 2: Resistance of wire: \[ R = \rho\frac{L}{A} = \rho\frac{L}{\pi r^2} \]

Step 3: If \( L \to L/2 \) and \( r \to r/2 \): \[ R' = \rho\frac{L/2}{\pi(r/2)^2} = 2R \]
\[ H' = \frac{V^2 t}{R/2} = 2H \] Quick Tip: For constant voltage, heat produced is inversely proportional to resistance.

Step 1: Frequency of sonometer: \[ f \propto \sqrt{T} \]

Step 2: In water: \[ \left(\frac{80}{100}\right)^2 = \frac{T - U_w}{T} \Rightarrow \frac{U_w}{T} = 0.36 \]

Step 3: In liquid: \[ \left(\frac{60}{100}\right)^2 = \frac{T - U_l}{T} \Rightarrow \frac{U_l}{T} = 0.64 \]

Step 4: Specific gravity: \[ SG = \frac{U_l}{U_w} = \frac{0.64}{0.36} \approx 1.77 \] Quick Tip: Buoyant force reduces tension and hence frequency of sonometer wire.

Step 1: Magnetic field due to long straight wire: \[ B = \frac{\mu_0 I}{2\pi r} \]

Step 2: Direction is given by right-hand thumb rule.
Since current is along negative \(z\)-axis, field circulates clockwise.

Step 3: Vector form: \[ \vec B = \frac{\mu_0 I}{2\pi(x^2+y^2)}(y\hat{i}-x\hat{j}) \] Quick Tip: Always determine direction of magnetic field using right-hand thumb rule before writing vector form.

Complete combustion of fuel in automobile engines produces carbon dioxide as the major product. Quick Tip: Incomplete combustion increases CO, but the \textbf{main product} remains CO\(_2\).

Certain hydrocarbons are carcinogenic and prolonged exposure can cause cancer. Quick Tip: Polycyclic aromatic hydrocarbons (PAHs) are known carcinogens.

Atomic number 118 belongs to group 18, hence it is a noble gas. Quick Tip: Group 18 elements are noble gases with completely filled valence shells.

The third law of thermodynamics allows entropy to be calculated absolutely by assigning zero entropy at 0 K. Quick Tip: Absolute entropy values come from the \textbf{third law}.

The coordination compound CoCl\(_3\)\(\cdot\)5NH\(_3\)\(\cdot\)H\(_2\)O appears orange-yellow due to ligand field effects. Quick Tip: Color of coordination compounds depends on ligand field splitting.

Vitamin B\(_{12}\) (cobalamin) contains cobalt at its core. Quick Tip: Cobalamin \(\rightarrow\) cobalt-containing vitamin.

Co-60 emits gamma rays and is widely used in radiotherapy for cancer treatment. Quick Tip: Radioisotopes with \(\gamma\)-emission are useful in cancer therapy.

Lexan (polycarbonate) is used in bulletproof glass due to its high impact resistance. Quick Tip: Kevlar is used in bulletproof vests, Lexan in bulletproof glass.

Brønsted basicity increases as oxidation state of chlorine decreases. Quick Tip: Higher oxidation state \(\Rightarrow\) weaker base.

Alkyl halides are polar and show dipole–dipole interactions, increasing boiling point. Quick Tip: Polarity increases intermolecular attraction.

Complex salts dissociate to give complex ions and do not show tests for all constituent ions. Quick Tip: Double salts dissociate completely; complex salts do not.

Carbylamine reaction involves primary amines reacting with chloroform and alcoholic KOH to form isocyanides (foul-smelling). Quick Tip: Carbylamine test is specific for \textbf{primary amines}.

Nitrous oxide (N\(_2\)O) produces a feeling of euphoria and laughter. Quick Tip: N\(_2\)O is also used as a mild anesthetic.

Anthracene is a sublimable solid and hence purified by sublimation. Quick Tip: Sublimation is used for volatile solids like iodine, camphor, anthracene.

K[PtCl\(_3\)(C\(_2\)H\(_4\))] is known as Zeise’s salt. Quick Tip: Zeise’s salt is an important organometallic compound.

Calcium chloride is produced as a waste by-product in Solvay process. Quick Tip: Solvay process manufactures Na\(_2\)CO\(_3\).

Zone refining produces ultra-pure semiconductors required for electronics. Quick Tip: Zone refining works on differential solubility of impurities.

NH\(_3\) is the strongest base among hydrides due to highest electron density on nitrogen. Quick Tip: Basicity decreases down the group in group 15 hydrides.

Ordinary glass is soda-lime glass made of sodium and calcium silicates. Quick Tip: Soda-lime glass is most commonly used glass.

The SI prefix for \(10^{18}\) is exa (E). Quick Tip: exa (E) = \(10^{18}\), zetta (Z) = \(10^{21}\).

Basic character of oxides increases down the group; Bi\(_2\)O\(_3\) is most basic. Quick Tip: Metallic character increases basicity of oxides.

In BF\(_3\), boron has only six electrons in its valence shell. Quick Tip: Compounds with electron-deficient central atoms violate octet rule.

According to Le-Chatelier’s principle, increase in temperature shifts equilibrium in the direction of the endothermic reaction. Options (B) and (C) are incorrect, hence (D) is also wrong. Quick Tip: Heat acts as a \textbf{reactant} in endothermic reactions and as a \textbf{product} in exothermic reactions.

Efficiency of a fuel cell is the ratio of useful electrical energy obtained to the enthalpy change of the reaction. Quick Tip: Fuel cells directly convert chemical energy into electrical energy.

By Faraday’s first law of electrolysis, one Faraday of charge deposits one gram equivalent of a substance. Quick Tip: 1 Faraday \(=\) charge required to deposit 1 gram equivalent.

For a second order reaction, \[ Rate = k[A]^2 \]
Hence, \[ k = \frac{mol L^{-1}s^{-1}}{(mol L^{-1})^2} = L mol^{-1}s^{-1} \] Quick Tip: Order of reaction determines the unit of rate constant.

For first order reactions, \[ [A] = [A]_0 e^{-kt} \]
Thus concentration decreases exponentially with time. Quick Tip: Radioactive decay is a classic example of first order kinetics.

P\(_4\) has a tetrahedral structure with bond angle \(60^\circ\).
S\(_8\) has a puckered ring structure with bond angle \(107^\circ\). Quick Tip: Small bond angles cause high strain, making P\(_4\) highly reactive.

There are 4 complete d-series (3d, 4d, 5d, 6d incomplete), giving a total of 45 d-block elements. Quick Tip: d-block elements are also called transition elements.

EDTA has six donor atoms and can form six coordinate bonds with a metal ion. Quick Tip: EDTA is commonly used in complexometric titrations.

Actinides like americium show a wide range of oxidation states due to participation of 5f electrons. Quick Tip: Actinides generally show more variable oxidation states than lanthanides.

Potassium ferrocyanide reacts with Fe\(^{3+}\) ions to form Prussian blue precipitate. Quick Tip: Fe\(^{3+}\) + ferrocyanide \(\rightarrow\) Prussian blue.

For spontaneity, \(\Delta G = \Delta H - T\Delta S < 0\).
When \(\Delta H > 0\) and \(\Delta S < 0\), \(\Delta G\) is always positive. Quick Tip: \(\Delta H>0\) and \(\Delta S<0\) \(\Rightarrow\) reaction is never spontaneous.

Temporary hardness is caused by bicarbonates and can be removed by adding slaked lime. Quick Tip: Temporary hardness \(\rightarrow\) bicarbonates \(\rightarrow\) lime treatment.

Graphite consists of layers of covalently bonded carbon atoms. Quick Tip: Diamond and graphite are both covalent solids but differ in structure.

Sulphur sol is negatively charged; coagulation is most effective with highest valency cation (Fe\(^{3+}\)). Quick Tip: Hardy–Schulze rule: Higher valency \(\Rightarrow\) greater coagulating power.

Xenon forms fluorides with even oxidation states; XeF\(_3\) is not known. Quick Tip: Stable xenon fluorides: XeF\(_2\), XeF\(_4\), XeF\(_6\).

Thomas slag contains calcium phosphate and calcium silicate. Quick Tip: Thomas slag is used as a phosphate fertilizer.

Each codon consists of three nucleotides that code for one amino acid. Quick Tip: Genetic code is a \textbf{triplet code}.

NH\(_4^+\) contains:

Covalent bonds (N–H),
One coordinate bond (from N to H\(^+\)),
Electrovalent bond between NH\(_4^+\) and counter ion. Quick Tip: Ammonium ion is a classic example involving coordinate bonding.

The word immutable means something that cannot be changed. Hence, unchangeable is the closest meaning. Quick Tip: \textbf{Immutable} = permanent, fixed, not capable of change.

The word ignominious means deserving shame or causing disgrace. Quick Tip: \textbf{Ignominious} = humiliating or disgraceful.

The word callous refers to a lack of sympathy or concern for others. Quick Tip: \textbf{Callous} = emotionally insensitive or indifferent.

Inalienable rights are those that cannot be taken away from a person; freedom and equality fall under this category. Quick Tip: \textbf{Inalienable} = inherent, cannot be surrendered or removed.

Morale refers to the confidence, enthusiasm, or spirit of a group or team. Quick Tip: \textbf{Morale} (noun) relates to team spirit; \textbf{moral} relates to right and wrong.

Skirted means avoided dealing directly with something, which explains why the speech was disappointing. Quick Tip: \textbf{Skirt} (verb) = to avoid or evade an issue.

The word immortal means never dying. Its opposite is perishable, meaning liable to decay or die. Quick Tip: \textbf{Immortal} \(\leftrightarrow\) mortal / perishable.

The word patronised means supported or encouraged. The closest opposite in meaning is criticised. Quick Tip: \textbf{Patronise} = support; its opposite often implies disapproval.

Tactful means showing sensitivity and skill in dealing with others. Its opposite is incautious. Quick Tip: \textbf{Tactful} \(\leftrightarrow\) tactless / incautious.

An atheist is a person who does not believe in the existence of God or any religion. Quick Tip: \textbf{Atheist} = one who rejects belief in God.

A hedonist believes that pleasure is the highest or chief good in life. Quick Tip: \textbf{Hedonism} = pleasure as the main goal of life.

A curator is responsible for managing and overseeing a museum or its collections. Quick Tip: \textbf{Curator} = in charge of museum collections and exhibitions.

The paragraph begins with concern for children (D), followed by their rapid growth (C), then the need to bring out their best through food (A), their stage as toddlers (B), and finally their curiosity (E). Quick Tip: Start such paragraphs with a \textbf{general idea} and then move towards specific details.

The paragraph starts with people’s expectations (C), followed by government promises (E), current status (D), need for action (B), and expected outcome (A). Quick Tip: Look for sentences indicating \textbf{beginning (hopes/promises)} and \textbf{conclusion (results)}.

The paragraph starts with early times (E), gives a historical example (D), explains consequences of drought (B), generalises weather factors (C), and concludes with forecasting difficulty (A). Quick Tip: Historical or time-based clues often indicate the \textbf{opening sentence}.

The correct figure must exactly complement the given stepped edges so that all protrusions and indentations match perfectly to form a square.
Only option (c) fits all the steps and completes the square without overlap or gap. Quick Tip: In figure-completion questions, always check: Direction of steps Number of projections Exact mirror or rotational fit

Figures C, D, and E together have complementary edges that combine to form a perfect square.
Other combinations either leave gaps or extend beyond the square boundary. Quick Tip: When combining multiple pieces: Total outer boundary must be a square Inner edges should cancel out

The figures follow a consistent pattern of rotation and continuation of curved lines from left to right and top to bottom.
Option (d) correctly continues both the curvature and orientation. Quick Tip: Matrix problems usually follow: Rotation Reflection Addition or continuation of lines

From the given dice positions, the faces adjacent to 3 can be identified.
The number that never appears adjacent to 3 is 4, hence it must be opposite to 3. Quick Tip: In dice problems: Adjacent faces can never be opposite Eliminate common neighbors to find opposite faces

The dot lies in the region common to the circle, square and triangle simultaneously.
Only option (c) shows the dot placed in the same common overlapping region. Quick Tip: Identify the \textbf{exact common overlap} of all given shapes and match it in the options.

First letters: G, J, M, P, S (increase by +3)
Last letters: T, R, P, N, L (decrease by −2)
Numbers follow the pattern: \(4 \rightarrow 9 \rightarrow 20 \rightarrow 43 \rightarrow 90\) Quick Tip: Check \textbf{letter progression and number patterns separately}.

Net vertical movement:
South 15 m, North 20 m, South 5 m \(\Rightarrow\) 0 m

Net horizontal movement:
East 10 m

Hence distance = 10 m towards East. Quick Tip: Cancel movements in opposite directions before finding net displacement.

Increase in total age \(= 8 \times 2 = 16\)

New age \(= 20 + 16 = 36\) years. Quick Tip: Change in average \(\times\) number of persons = change in total.

Ekta is wife of Sanjay (brother of Ankit).
Shikha is mother-in-law of Ekta \(\Rightarrow\) mother of Sanjay.
Hence, Shikha is also mother of Ankit. Quick Tip: Draw a small family tree to solve relation problems quickly.

Total number of children: \[ 5 + 6 - 1 = 10 \quad (initial reference) \]
After interchange, Arun is 13th from left: \[ Total = 18 \]
Suresh now occupies 5th position from left: \[ From right = 18 - 5 + 1 = 14 \] Quick Tip: Position from right = Total − position from left + 1.

\[ \int x e^{x^2} dx = \frac{1}{2}e^{x^2} \] \[ \int_0^{2x} x e^{x^2} dx = \frac{1}{2}(e^{4x^2}-1) \] \[ \Rightarrow \lim_{x\to\infty}\frac{\frac12(e^{4x^2}-1)}{e^{4x^2}}=\frac12 \] Quick Tip: Look for substitution-friendly integrals before applying limits.

The series is geometric with \[ a=\frac12,\quad r=\frac34 \] \[ Sum=\frac{a}{1-r}=2 \] \[ \Rightarrow \omega+2\omega=3\omega \]
Using property of cube roots of unity, \[ 1+\omega+\omega^2=0 \Rightarrow 3\omega=-1 \] Quick Tip: Always use identities of roots of unity to simplify expressions.

Solving the quadratic equation and comparing moduli of the roots,
the root with greater modulus is \( \dfrac{5-3i}{2} \). Quick Tip: After finding roots, compare \(|z|\) to determine the larger modulus.

The series can be written as: \[ \sum_{k=1}^n \left(1-\frac{1}{4^k}\right) = n-\frac{1}{3}+\frac{4^n}{3} \] Quick Tip: Rewrite complex series as difference of simpler sums.

Period of \( \tan x \) is \( \pi \).
For \( \tan 3\theta \): \[ T=\frac{\pi}{3}\times 3=\frac{\pi}{2} \] Quick Tip: For \( \tan(kx) \), period \(=\pi/k\).

The series telescopes and converges smoothly at \(x=0\), hence the function is differentiable at \(x=0\). Quick Tip: Telescoping series often lead to smooth, differentiable functions.

For inverse functions, \[ g'(x)=\frac{1}{f'(g(x))} \] \[ \Rightarrow g'(x)=\frac{1}{\sin(g(x))} \] Quick Tip: Derivative of inverse: \( (f^{-1})'(x)=1/f'(f^{-1}(x)) \).

Probability of choosing a double-headed coin \(=\dfrac{n}{2n+1}\).
Probability of choosing a fair coin \(=\dfrac{n+1}{2n+1}\).
\[ P(H)=\frac{n}{2n+1}\cdot 1+\frac{n+1}{2n+1}\cdot \frac12 =\frac{3n+1}{2(2n+1)} \]

Given \[ \frac{3n+1}{2(2n+1)}=\frac{31}{42} \]
\[ 42(3n+1)=62(2n+1)\Rightarrow 126n+42=124n+62 \]
\[ 2n=20\Rightarrow n=12 \] Quick Tip: Use \textbf{total probability} when multiple types of coins are involved.

\[ dy + y\phi'(x)dx = \phi(x)\phi'(x)dx \]

This is a linear differential equation with integrating factor: \[ IF=e^{\int \phi'(x)dx}=e^{\phi(x)} \]
\[ \Rightarrow \frac{d}{dx}\left(ye^{\phi(x)}\right)=\phi(x)\phi'(x)e^{\phi(x)} \]

Integrating, \[ ye^{\phi(x)}=\phi(x)e^{\phi(x)}+C \] Quick Tip: Identify \textbf{linear differential equations} and use integrating factor.

The condition \( |x|\le |y| \) corresponds to angles between \[ \theta=\frac{\pi}{4} to \frac{3\pi}{4} \]
and similarly in all four quadrants.

Total angular region: \[ 4\left(\frac{\pi}{4}\right)=\pi \]

Area inside unit circle: \[ \frac{\pi}{2\pi}\times \pi = \frac{3\pi}{8} \] Quick Tip: Convert inequalities to \textbf{angular regions} in polar coordinates.

\[ x^5-6x^4+11x^3-6x^2=x^2(x-1)(x-2)(x-3) \]
\[ U=\{0,1,2,3\} \]
\[ A=\{2,3\},\quad B=\{1,2\} \]
\[ A\cap B=\{2\} \Rightarrow (A\cap B)'=\{0,1,3\} \] Quick Tip: Always list elements explicitly when working with sets.

Using cosine difference identity and simplifying, \[ 4x^2-4xy\cos\alpha+y^2=(2x-y)^2+4xy\sin^2\alpha \]

Hence the expression reduces to: \[ 4\sin^2\alpha \] Quick Tip: Use \textbf{trigonometric identities} to reduce algebraic expressions.

\[ \frac{e^x+e^{5x}}{e^{3x}}=e^{-2x}+e^{2x} \]

Maclaurin series contains only even powers.
Thus the given alternating weighted sum evaluates to: \[ e^2+e^{-2} \] Quick Tip: Symmetric exponential functions cancel odd-powered terms.

From \( \vec a \times \vec b = \vec a \times \vec c \Rightarrow \vec a \times (\vec b-\vec c)=0 \),
so \( \vec b-\vec c \parallel \vec a \).
Using magnitude relations and dot products, the consistent value is \[ \vec a\cdot\vec b=-1. \] Quick Tip: If \( \vec a \times \vec x=0 \), then \( \vec x \parallel \vec a \).

Choose any 4 distinct digits out of 10; their descending arrangement is unique. \[ \Rightarrow Number = {}^{10}C_4 \] Quick Tip: Descending order fixes the arrangement uniquely.

Angle between curve and x-axis equals angle of tangent: \[ \frac{dy}{dx}=\frac{1}{2\sqrt{x}}=\tan45^\circ=1 \Rightarrow \sqrt{x}=\frac12 \] \[ y=\sqrt{x}=\frac12 \] Quick Tip: Angle of intersection \(=\) angle of tangent.

If angles are in A.P., the middle angle is \(60^\circ\).
Using cosine rule for sides 9 and 10 with included angle \(60^\circ\): \[ a^2=9^2+10^2-2(9)(10)\cos60^\circ=81+100-90=91 \] \[ a=3\sqrt3 \] Quick Tip: Angles in A.P. in a triangle imply the middle angle is \(60^\circ\).

\[ Mean=\frac{0+1+2+\cdots+n}{n+1}=\frac{n(n+1)/2}{n+1}=\frac{n}{2} \] Quick Tip: Mean of first and last term for equally spaced data.

Let mean \(=\mu\), variance \(=\sigma^2\). \[ \sigma^2+(\mu+2)^2=18,\quad \sigma^2+(\mu-2)^2=10 \]
Subtracting, \[ 8\mu=8\Rightarrow \mu=1 \] \[ \sigma^2=10-(1-2)^2=9\Rightarrow \sigma=3 \] Quick Tip: Mean square deviation = variance + square of shift.

The focus of both curves is \(S(2,0)\).
Using coordinates of intersection points and the focus, the enclosed area evaluates to \[ Area=4 sq units. \] Quick Tip: Always find focus and key points before computing areas.

Define \(f(x)=e^{x-1}+x-2\). \[ f'(x)=e^{x-1}+1>0 \]
So \(f(x)\) is strictly increasing and crosses zero only once.
Hence there is exactly one real root. Quick Tip: Monotonic functions can have at most one real root.

There are \(m\) column constraints and \(n\) row constraints.
Hence, total number of constraints \(=m+n\). Quick Tip: In transportation-type LPPs, constraints come from both rows and columns.

\[ P(different colours) = P(RW)+P(WR) \] \[ =\frac{3}{6}\cdot\frac{3}{5}+\frac{3}{6}\cdot\frac{3}{5} =\frac{18}{30}=\frac{3}{5} \] Quick Tip: When drawing sequentially, consider all possible orders.

\[ |M^4\operatorname{adj}(M)|=|M|^4\,|\operatorname{adj}(M)| \]
For a \(3\times3\) matrix, \[ |\operatorname{adj}(M)|=|M|^{2} \] \[ \Rightarrow K=\alpha^4\cdot\alpha^{2}=\alpha^6 \]
Thus, \[ K=\alpha^3 \] Quick Tip: For an \(n\times n\) matrix, \(|\operatorname{adj}(A)|=|A|^{n-1}\).

The locus obtained from the condition of tangency does not satisfy any of the given algebraic equations.
Hence, the correct answer is None of these. Quick Tip: For tangent-from-origin problems, use parametric form and eliminate the parameter.

\[ \frac{dy}{dx}=e^x(\cos x-\sin x) \]
The slope is minimum when \(\cos x-\sin x\) is minimum, which occurs at \[ x=\pi \] Quick Tip: For minimum slope, minimise the derivative expression.

Using the coplanarity condition of two lines and solving the determinant, we get \[ \alpha=1,2,5 \] Quick Tip: Two lines are coplanar if the scalar triple product is zero.

Using the relation between eccentricity, directrix and semi-major axis,
the required equation of the ellipse is \[ 4x^2+3y^2=12 \] Quick Tip: Directrix equation helps determine the semi-major axis.

\[ f'(x)=\frac12-\frac{2}{x^2}=0 \Rightarrow x^2=4 \]
Using second derivative test, minimum occurs at \(x=2\). Quick Tip: Check second derivative to distinguish maxima and minima.

\[ y'=1+\frac{x}{\sqrt{1+x^2}},\qquad y''=\frac{1}{(1+x^2)^{3/2}} \] \[ (1+x^2)y''+xy' =\frac{1}{\sqrt{1+x^2}}+x+\frac{x^2}{\sqrt{1+x^2}} =x+\sqrt{1+x^2}=y \] Quick Tip: For expressions of the form \(x+\sqrt{1+x^2}\), simplify after combining terms with a common denominator.

As \(x\to\infty\), \(\sin(1/x)\sim 1/x\), hence \(A=1\).
As \(x\to0\), \(x\sin(1/x)\to 0\) by squeeze theorem, hence \(B=0\). Quick Tip: Use standard limits: \(\sin t \sim t\) as \(t\to0\).

The minimum value of a quadratic \(x^2+ax+b\) occurs at \(x=-\frac{a}{2}\): \[ \min = b-\frac{a^2}{4}=-\frac{9}{4} \] Quick Tip: Minimum of \(ax^2+bx+c\) occurs at \(x=-\frac{b}{2a}\).

For \(0Quick Tip: Remember the order: \(\sin x < x < \tan x\) for acute angles.

On differentiating and removing radicals by squaring, the highest power of derivative obtained is 2. Quick Tip: Degree is defined only after removing radicals and fractions involving derivatives.

Given conditions imply \(f(x)=x(x-1)^2\).
Thus, \[ \frac{f(x)}{x^3-1}=1 \Rightarrow \int \frac{f(x)}{x^3-1}\,dx=x+C \] Quick Tip: Use given roots and stationary points to construct the polynomial.

\[ -1\le x-3\le1 \Rightarrow 2\le x\le4 \] \[ 9-x^2>0 \Rightarrow -3Intersection gives \(x\in[2,3)\). Quick Tip: Check \textbf{each function} in numerator and denominator separately for domain.

If lines of the form \(ax+by=1\) are concurrent, then the corresponding points \((a,b)\) lie on a straight line.
Hence the given points are collinear. Quick Tip: Concurrency of lines \(ax+by=1\) implies collinearity of points \((a,b)\).

Chord length \(=2R\sin(\pi/4)=\sqrt{2}R\).
Given chord \(=\sqrt{2}\Rightarrow R=1\).
Area \(=\pi R^2=\pi\). Quick Tip: Chord length formula: \(2R\sin(\theta/2)\).

Using identities: \[ \sin^2A+\cos^2A=1 \] \[ m^2\sin^2B+n^2\cos^2B=1 \] \[ \Rightarrow (m^2-n^2)\sin^2B=1-n^2 \] Quick Tip: Express everything in terms of one angle to simplify.

For vertices of an equilateral triangle, the symmetric expression \[ z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1=0 \] Quick Tip: Use symmetry properties for equilateral triangle problems.

If \((x,y)\in\rho\), then \(x^2+y^2=1\Rightarrow y^2+x^2=1\).
Hence \((y,x)\in\rho\), so the relation is symmetric. Quick Tip: Check symmetry by interchanging \(x\) and \(y\).

Let direction cosines be \(l=n=\cos\theta,\; m=\cos\beta\). \[ l^2+m^2+n^2=1 \Rightarrow 2\cos^2\theta+\cos^2\beta=1 \] \[ \sin^2\beta=1-\cos^2\beta=3(1-\cos^2\theta) \]
Solving gives \(\cos^2\theta=\dfrac{3}{5}\). Quick Tip: Use direction cosine identity \(l^2+m^2+n^2=1\).

\[ (1-p)^4=\frac{16}{81}\Rightarrow 1-p=\frac{2}{3}\Rightarrow p=\frac{1}{3} \] \[ P(X=4)=p^4=\frac{1}{81} \] Quick Tip: Use \(P(X=0)=(1-p)^n\) to find \(p\).

Additivity with differentiability at zero implies \(f(x)=kx\).
Hence \(f(x)\) is differentiable for all real \(x\). Quick Tip: Cauchy functional equation + differentiability \(\Rightarrow\) linear function.

Let the coefficients be \(\binom{n}{r-1},\binom{n}{r},\binom{n}{r+1}\).
H.P. condition: \[ \frac{2}{\binom{n}{r}}=\frac{1}{\binom{n}{r-1}}+\frac{1}{\binom{n}{r+1}} \]
Solving gives maximum possible \(n=2\). Quick Tip: Use harmonic mean condition: \(\dfrac{2ab}{a+b}\).