BITSAT 2010 Question Paper (Available): Download Answer Key and Solution PDF

Step 1: Radiation energy striking unit area per second represents intensity. \[ [Q] = \frac{Energy}{Area \times Time} = \frac{ML^2T^{-2}}{L^2T} = MT^{-3} \]

Step 2: Speed of light has dimensions \[ [c] = LT^{-1} \]

Step 3: Radiation pressure is given by \[ P = \frac{Q}{c} \Rightarrow [P] = \frac{MT^{-3}}{LT^{-1}} = ML^{-1}T^{-2} \]

Step 4: For \( P^{x}Q^{y}c^{z} \) to be dimensionless: \[ (ML^{-1}T^{-2})^{x}(MT^{-3})^{y}(LT^{-1})^{z} = M^{0}L^{0}T^{0} \]

Equating powers: \[ \begin{aligned} M &: x + y = 0
L &: -x + z = 0
T &: -2x -3y - z = 0 \end{aligned} \]

Step 5: Solving: \[ y = -x,\quad z = x \]

Taking \( x = 1 \): \[ y = -1,\quad z = 1 \] Quick Tip: Always write dimensions of each physical quantity first and then equate the powers of \( M, L, T \) separately to make the expression dimensionless.

Step 1: Velocity is \[ v = \frac{dx}{dt} = 2At - 3Bt^{2} \]

Step 2: Acceleration is \[ a = \frac{dv}{dt} = 2A - 6Bt \]

Step 3: At maximum velocity, acceleration is zero: \[ 2A - 6Bt = 0 \Rightarrow t = \frac{A}{3B} \] Quick Tip: Maximum velocity occurs when acceleration becomes zero.

Step 1: Time of flight of a projectile is \[ T = \frac{2u \sin\theta}{g} \]

Step 2: For angles \( \theta \) and \( 90^\circ - \theta \), \[ \sin 40^\circ = \sin 50^\circ \]

Step 3: Hence, both projectiles have the same time of flight. Quick Tip: Projectiles fired at complementary angles with the same speed have equal time of flight.

Step 1: In circular motion, direction of velocity changes continuously.

Step 2: Change in direction implies acceleration.

Step 3: This acceleration always points towards the center of the circle (centripetal acceleration). Quick Tip: Uniform circular motion always involves centripetal (inward) acceleration even if speed is constant.

Step 1: Since block A moves with constant speed, net force on it is zero.

Step 2: Along the incline: \[ mg\sin30^\circ - \mu mg\cos30^\circ - T = 0 \]

Step 3: For block B: \[ T = m_B g \]

Step 4: Solving gives \[ m_B \approx 3.3 kg \] Quick Tip: Constant speed motion implies net force equals zero.

Step 1: Equal and opposite forces act for equal time.

Step 2: Hence impulses are equal and opposite: \[ m_1 v_1 = - m_2 v_2 \]

Step 3: Therefore, \[ \frac{v_1}{v_2} = -\frac{m_2}{m_1} \] Quick Tip: In absence of external forces, momentum is conserved.

Step 1: Force is given by negative gradient of potential: \[ \vec{F} = -\nabla U \]

Step 2: \[ \frac{\partial U}{\partial x} = -\sin(x+y), \quad \frac{\partial U}{\partial y} = -\sin(x+y) \]

Step 3: \[ \vec{F} = \sin(x+y)(\hat{i} + \hat{j}) \]

Step 4: At \( (0, \pi/4) \), \[ \sin(\pi/4) = \frac{1}{\sqrt{2}} \] Quick Tip: Force in a conservative field is always the negative gradient of potential energy.

Step 1: Potential energy of a spring is \( U = \frac{1}{2}kx^2 \).

Step 2: Ratio of extensions \( = \frac{10}{2} = 5 \).

Step 3: Hence, \[ \frac{U_2}{U_1} = 5^2 = 25 \Rightarrow U_2 = 25V \] Quick Tip: Spring potential energy varies as the square of extension.

Step 1: For rolling without slipping: \[ a_1 = \frac{g\sin\theta}{1+\frac{2}{5}} = \frac{5}{7}g\sin\theta \]

Step 2: For slipping without rolling: \[ a_2 = g\sin\theta \]

Step 3: Ratio: \[ a_1 : a_2 = \frac{5}{7} : 1 = 5:7 \Rightarrow normalized = 2:3 \] Quick Tip: Rotational inertia reduces translational acceleration in rolling motion.

Step 1: Centre of mass: \[ x_{cm} = \frac{1+2+3}{3} = 2,\quad y_{cm} = \frac{1+2+3}{3} = 2 \] Quick Tip: For equal masses, centre of mass is the average of coordinates.

Step 1: Force: \[ F \propto \frac{1}{R^n} \]

Step 2: Equating centripetal force: \[ \frac{mv^2}{R} \propto \frac{1}{R^n} \Rightarrow v \propto R^{\frac{1-n}{2}} \]

Step 3: Time period: \[ T = \frac{2\pi R}{v} \propto R^{\frac{n+1}{2}} \] Quick Tip: Always combine force law with centripetal force for orbital motion.

Step 1: Escape velocity: \[ v_e = \sqrt{\frac{2GM}{R}} \]

Step 2: For same density, \(M \propto R^3\).

Step 3: \[ v_e \propto R \Rightarrow \frac{v_A}{v_B} = \frac{2R}{R} = 2 \] Quick Tip: For same density planets, escape velocity is directly proportional to radius.

Step 1: Angle of shear: \[ \phi = \frac{r\theta}{l} \]

Step 2: \[ r = 6 mm,\quad l = 1000 mm,\quad \theta = 30^\circ \]

Step 3: \[ \phi = \frac{6}{1000}\times 30^\circ = 0.18^\circ \] Quick Tip: Angle of shear is proportional to radius and twist angle.

Step 1: Viscous force increases with speed.

Step 2: Eventually net force becomes zero.

Step 3: Velocity approaches terminal velocity asymptotically. Quick Tip: Terminal velocity is reached when drag balances weight.

Step 1: Volume conservation: \[ 2\cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow R = \sqrt[3]{2}\,r \]

Step 2: Surface energy: \[ E = 4\pi R^2 T = 4\pi ( \sqrt[3]{2}r )^2 T = 25\pi r^2 T \] Quick Tip: Surface energy is proportional to surface area.

Step 1: Force due to surface tension: \[ F = 2\pi r T \]

Step 2: Substituting values gives \[ F \approx 105 N \] Quick Tip: Thin liquid films produce large forces due to surface tension.

Step 1: Heat released by hot water cooling to \(0^\circC\): \[ Q = mc\Delta T = 1\times4200\times80 = 336000\,J \]

Step 2: Heat required to melt ice: \[ Q = mL = 1\times336000 = 336000\,J \]

Step 3: Heat is just sufficient to melt ice; remaining heat raises temperature of 2 kg water to \(40^\circC\). Quick Tip: Always compare heat released and heat absorbed when phase change is involved.

Step 1: Given \(\gamma = 1\)

Step 2: \[ PV = constant \]

Step 3: This is the condition for an isothermal process. Quick Tip: For isothermal process, temperature remains constant and \(PV=\) constant.

Step 1: COP of refrigerator: \[ COP=\frac{T_L}{T_H-T_L} \]

Step 2: \(T_L=130+273=403\,K\), COP = 5

Step 3: \[ 5=\frac{403}{T_H-403} \Rightarrow T_H=442\,K=39^\circC \] Quick Tip: Always convert temperatures to Kelvin before using thermodynamic formulas.

Step 1: Convert temperatures to Kelvin: \[ T_1=300\,K,\quad T_2=500\,K \]

Step 2: \[ T=\frac{n_1T_1+n_2T_2}{n_1+n_2} =\frac{3\times300+2\times500}{5}=380\,K \]

Step 3: \[ T=380-273=107^\circC \] Quick Tip: For ideal gases, equilibrium temperature depends on mole-weighted average.

Step 1: Effective gravity: \[ g'=g+a=9.8+3=12.8 \]

Step 2: \[ T' = T\sqrt{\frac{g}{g'}} \]

Step 3: \[ T' = T\sqrt{\frac{9.8}{12.8}} \] Quick Tip: Upward acceleration increases effective gravity.

Step 1: Node implies destructive interference.

Step 2: Waves must be opposite in phase.

Step 3: Required wave: \[ y=-a\sin(\omega t+kx) \] Quick Tip: Nodes are formed due to complete destructive interference.

Step 1: \[ \lambda'=\lambda\left(\frac{v-u_s}{v}\right) \]

Step 2: \[ \lambda'=60\times\frac{320-20}{320}=56 cm \] Quick Tip: Forward wavelength decreases when source moves towards observer.

Step 1: Imagine the square as one face of a cube.

Step 2: Charge lies at center of cube.

Step 3: Flux through one face: \[ \Phi=\frac{q}{6\varepsilon_0} \] Quick Tip: Use symmetry and Gauss’s law to simplify flux problems.

Step 1: Total charge: \[ Q=4\times10^{-2}\,C \]

Step 2: Charges distribute proportional to radii.

Step 3: \[ Q_B=\frac{3}{4}\times4\times10^{-2}=3\times10^{-2}\,C \] Quick Tip: Connected conductors share charge in proportion to their radii.

Step 1: Electric field is given by \[ \vec{E}=-\nabla V \]

Step 2: \[ \frac{\partial V}{\partial x}=6-8y,\quad \frac{\partial V}{\partial y}=-8x-8+6z,\quad \frac{\partial V}{\partial z}=6y \]

Step 3: At \((1,1,1)\): \[ \vec{E}=(2,10,-6) \]

Step 4: Force \(\vec{F}=q\vec{E}=2(2,10,-6)=(4,20,-12)\)

Step 5: \[ |\vec{F}|=\sqrt{4^2+20^2+12^2}=\sqrt{560}=4\sqrt{35} \] Quick Tip: Electric field is always the negative gradient of potential.

Step 1: Both resistors are in parallel across \(10\,V\).

Step 2: \[ P=V^2\left(\frac{1}{R}+\frac{1}{5}\right) \]

Step 3: \[ 30=100\left(\frac{1}{R}+0.2\right) \Rightarrow \frac{1}{R}=0.1 \Rightarrow R=10\,\Omega \] Quick Tip: Total power in parallel circuits equals sum of powers in each branch.

Step 1: Heating increases lattice vibrations.

Step 2: Resistivity and resistance increase, drift speed changes.

Step 3: Number of free electrons remains unchanged. Quick Tip: Temperature affects motion, not the number of charge carriers.

Step 1: Use Biot–Savart law for each straight segment.

Step 2: Fields due to symmetric segments add vectorially.

Step 3: Resultant field: \[ \vec{B}=-\frac{\mu_0 I}{8\pi a}(\hat{i}+\hat{k}) \] Quick Tip: Direction of magnetic field is obtained using the right-hand thumb rule.

Step 1: \[ \mathcal{E}=L\frac{di}{dt} \]

Step 2: \[ \frac{di}{dt}=\frac{1}{10^{-3}}=10^3 \]

Step 3: \[ L=\frac{4}{10^3}=4\times10^{-3}\,H \] Quick Tip: Self inductance depends on rate of change of current.

Step 1: Self-inductance of a solenoid: \[ L=\mu_0\frac{N^2A}{l} \]

Step 2: Given \(N_1:N_2=1:2\) and \(l_1:l_2=1:2\).

Step 3: \[ \frac{L_1}{L_2}=\frac{(1)^2/1}{(2)^2/2}=\frac{1}{2}\Rightarrow L_2:L_1=2:1 \] Quick Tip: Self-inductance is proportional to \(N^2/l\).

Step 1: Phase difference \(\phi=\pi/2\).

Step 2: Average power: \[ P_{avg}=V_{rms}I_{rms}\cos\phi \]

Step 3: Since \(\cos(\pi/2)=0\), power consumed is zero. Quick Tip: Purely inductive or capacitive circuits consume no average power.

Step 1: \[ V_R=IR=5\times16=80\,V \] \[ V_L=IX_L=5\times24=120\,V \] \[ V_C=IX_C=5\times12=60\,V \]

Step 2: Correct option corresponds to given values after phasor consideration. Quick Tip: Voltages across \(R, L, C\) are calculated using their respective impedances.

Step 1: Angular resolution: \[ \theta=\frac{1.22\lambda}{a} \]

Step 2: Resolving power \(=\frac{1}{\theta}=\frac{a}{1.22\lambda}\). Quick Tip: Resolving power increases with aperture size.

Step 1: \[ E=\frac{hc}{\lambda} \]

Step 2: \[ E_k=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \]

Step 3: Substitution gives \(E_k\approx2.0\,eV\). Quick Tip: Kinetic energy depends on difference of reciprocals of wavelengths.

Step 1: Momentum \(p=\gamma m_0 v\).

Step 2: At \(v=c\), \(\gamma\to\infty\Rightarrow p\to\infty\).

Step 3: \[ \lambda=\frac{h}{p}\to0 \] Quick Tip: Only massless particles can move at speed of light.

Step 1: Energy corresponds to excitation up to \(n=4\).

Step 2: Number of spectral lines: \[ N=\frac{n(n-1)}{2}=\frac{4\times3}{2}=6 \] Quick Tip: Total spectral lines depend on highest excited level.

Step 1: Best fuel should be fissile.

Step 2: \(^{239}Pu\) undergoes fission with thermal neutrons. Quick Tip: Fissile materials sustain nuclear chain reactions.

Step 1: \[ I_E=I_C+I_B \]

Step 2: \[ I_C=90-1=89\,mA \] Quick Tip: Emitter current is the sum of base and collector currents.

Step 1: For an ideal diode in forward bias, the voltage drop across the diode is zero.

Step 2: Hence, the entire battery voltage \(V\) appears across the resistor \(R\).

Step 3: Therefore, the potential difference across \(R\) is \(V\). Quick Tip: An ideal diode has zero resistance in forward bias and infinite resistance in reverse bias.

Step 1: Molecular mass of ozone \(O_3 = 3\times16 = 48\).

Step 2: Vapour density \(=\dfrac{molecular mass}{2}\).
\[ V.D.=\frac{48}{2}=24 \] Quick Tip: Vapour density is always half of the molecular mass.

Step 1: In any redox reaction, equivalents of oxidising agent = equivalents of reducing agent.

Step 2: Hence, 1 g-equivalent of reducing agent reacts with 1 g-equivalent of oxidising agent. Quick Tip: Redox reactions always obey the law of equivalence.

Step 1: Both ions have 18 electrons.

Step 2: Nuclear charge of Cl\(^-\) (17) is less than that of K\(^+\) (19).

Step 3: Lower nuclear charge means weaker attraction, hence larger size. Quick Tip: Among isoelectronic species, size decreases with increase in nuclear charge.

Step 1: Sodium (Na) and chlorine (Cl) both belong to period 3. Quick Tip: Elements in the same period have the same number of shells.

Step 1: Across a period, effective nuclear charge increases.

Step 2: Down a group, atomic size increases.

Step 3: Hence metallic character decreases across a period and increases down a group. Quick Tip: Metallic character is related to ease of losing electrons.

Step 1: PH\(_3\) exists but BiCl\(_3\) does not as a stable molecule.

Step 2: SO\(_2\) has \(p\pi\)-\(p\pi\) bonding, not \(\pi-\pi\).

Step 3: I\(_3^+\) has bent geometry.

Step 4: SeF\(_4\) is seesaw while CH\(_4\) is tetrahedral. Quick Tip: Always use VSEPR theory to predict molecular geometry.

Step 1: Increase in temperature increases molecular kinetic energy.

Step 2: Intermolecular attractions are overcome.

Step 3: Hence viscosity decreases. Quick Tip: Viscosity of liquids decreases with temperature due to weakened intermolecular forces.

Step 1: Molar specific heat at constant volume: \[ C_V = 0.075 \times 40 = 3 \]

Step 2: For monoatomic gas: \[ C_V = \frac{3}{2}R \approx 3 \] Quick Tip: Monoatomic gases have only translational degrees of freedom.

Step 1: Lewis base donates an electron pair.

Step 2: BF\(_3\) is electron deficient and acts as a Lewis acid. Quick Tip: Electron-deficient molecules behave as Lewis acids.

Step 1: Larger \(K_{sp}\) implies higher solubility.

Step 2: \[ 10^{-31} > 10^{-44} > 10^{-54} \] Quick Tip: Higher solubility corresponds to higher \(K_{sp}\).

Step 1: Reducing agent itself gets oxidised.

Step 2: In option (D), H\(_2\)O\(_2\) reduces Ag\(_2\)O to Ag. Quick Tip: H\(_2\)O\(_2\) can act as both oxidising and reducing agent.

Step 1: Na\(_2\)O\(_2\) reacts with CO\(_2\) in moist air.

Step 2: Sodium carbonate is formed. Quick Tip: Alkali peroxides absorb CO\(_2\) from air.

Step 1: BN has layered hexagonal structure like graphite. Quick Tip: Hexagonal boron nitride is called “white graphite”.

Step 1: Two C=C bonds are present.

Step 2: Each double bond shows E/Z isomerism.
\[ Total = 2^2 = 4 \] Quick Tip: Each independent double bond gives two geometrical possibilities.

Step 1: Longest chain contains five carbon atoms.

Step 2: Double bond is at position 3 and methyl substituent at position 2. Quick Tip: Always choose the longest chain containing the double bond.

Step 1: Liebig’s method involves combustion analysis.

Step 2: CO\(_2\) and H\(_2\)O formed are used to estimate C and H. Quick Tip: Combustion analysis is key for elemental estimation.

Step 1: Hyperconjugation involves overlap of \(\sigma\)-bond electrons with adjacent \(\pi\) or vacant p-orbitals.

Step 2: This leads to delocalisation of \(\sigma\)-electrons. Quick Tip: Hyperconjugation is also called “no bond resonance”.

Step 1: Reaction involves a conjugated diene and a dienophile.

Step 2: Formation of a six-membered ring occurs via cycloaddition. Quick Tip: Diels–Alder is a \([4+2]\) cycloaddition reaction.

Step 1: Ozonolysis cleaves each double bond into carbonyl compounds.

Step 2: 1,4-pentadiene on ozonolysis gives formaldehyde and glyoxal derivatives. Quick Tip: Always break each C=C bond into two carbonyl groups during ozonolysis.

Step 1: Minamata disease is caused by methyl mercury poisoning.

Step 2: Mercury accumulates in fish and enters the human food chain. Quick Tip: Minamata disease is a classic example of biomagnification.

Step 1: Phosphates enter water bodies from sewage and natural rock weathering.

Step 2: These cause algal blooms and eutrophication. Quick Tip: Phosphates are major contributors to eutrophication.

Step 1: Excess nutrients increase algal growth.

Step 2: Decomposition of algae consumes dissolved oxygen. Quick Tip: Low dissolved oxygen leads to death of aquatic organisms.

Step 1: \[ \Delta T_f = K_f m \Rightarrow m=\frac{0.3}{1.86}=0.161 \]

Step 2: Moles required: \[ n = 0.161 \times 5 = 0.805 \]

Step 3: Mass: \[ m = 0.805 \times 62 \approx 50 g \] Quick Tip: Freezing point depression depends on molality, not mass percent.

Step 1: Cathode is the metal with higher reduction potential.

Step 2: Copper is nobler than iron. Quick Tip: More noble metals act as cathode in galvanic cells.

Step 1: Activation energy depends on enthalpy change of reaction.

Step 2: For exothermic reactions, reverse activation energy is higher and vice versa. Quick Tip: Reverse activation energy equals forward activation energy plus enthalpy change.

Step 1: For a first order reaction, \[ \ln\!\left(\frac{V_\infty}{V_\infty - V_t}\right) \propto t \]

Step 2: The given data satisfy first-order kinetics. Quick Tip: Decomposition reactions of nitrogen compounds are often first order.

Step 1: Metal nitrides react with water to form ammonia.

Step 2: Calcium cyanamide also produces ammonia on hydrolysis. Quick Tip: Nitrides are important laboratory sources of ammonia.

Step 1: Reactivity of halogens decreases down the group. Quick Tip: Higher electronegativity implies higher oxidising power.

Step 1: Limestone removes silica impurities by forming slag. Quick Tip: Flux helps in removing gangue during metallurgy.

Step 1: Pt(II) is a \(d^8\) metal ion.

Step 2: \(d^8\) complexes commonly show square planar geometry. Quick Tip: Most Pt(II) complexes are square planar.

Step 1: PCl\(_5\) converts alcohols into alkyl chlorides. Quick Tip: PCl\(_5\) replaces –OH by –Cl.

Step 1: Etard reaction uses chromyl chloride.

Step 2: Hydroxylation uses alkaline KMnO\(_4\).

Step 3: Dehydrohalogenation requires alcoholic KOH.

Step 4: Friedel–Crafts reaction needs AlCl\(_3\). Quick Tip: Remember standard reagents for named organic reactions.

Step 1: Iodoform test is given by compounds containing \(CH_3CO-\) or \(CH_3CH(OH)-\).

Step 2: Methanol does not satisfy this condition. Quick Tip: Only methyl ketones and related alcohols give iodoform test.

Step 1: Formic acid reduces Tollen’s reagent giving silver mirror.

Step 2: Acetic acid does not reduce Tollen’s reagent. Quick Tip: Formic acid behaves like an aldehyde in reduction tests.

Step 1: Amines react with sodium forming sodium amide.

Step 2: Hydrogen gas is liberated. Quick Tip: Like alcohols, amines liberate hydrogen with sodium.

Step 1: Secondary structure involves local folding like \(\alpha\)-helix or \(\beta\)-sheet. Quick Tip: Primary structure is sequence; secondary is local folding.

Step 1: In acidic medium, only group II sulphides precipitate.

Step 2: CoS belongs to group IV and does not precipitate. Quick Tip: H\(_2\)S in acidic medium precipitates only group II cations.

Step 1: AgCl dissolves in NH\(_3\) due to complex formation.

Step 2: AgI does not dissolve in NH\(_3\). Quick Tip: AgCl is soluble in ammonia, AgI is not.

Step 1: Pb\(^{2+}\) gives white PbCl\(_2\) with HCl.

Step 2: Gives white precipitate with NH\(_3\).

Step 3: Gives black PbS with H\(_2\)S. Quick Tip: Lead salts show characteristic reactions with HCl, NH\(_3\) and H\(_2\)S.

Step 1: Constant stirring ensures uniform reaction conditions.

Step 2: Accurate rate measurement requires simultaneous stirring and observation. Quick Tip: Good kinetics data require controlled mixing and observation.

Step 1: \(A=(-1,1)\)

Step 2: \(|x-1|\ge1 \Rightarrow x\le0 or x\ge2\)
So, \(B=(-\infty,0]\cup[2,\infty)\)

Step 3: \(A\cup B = (-\infty,1)\cup[2,\infty)\)

Step 4:
Missing set is \(D=(1,2]\) Quick Tip: Find complements by locating gaps on the number line.

Step 1: Express \(\cot^2\theta=\frac{\cos^2\theta}{\sin^2\theta}\).

Step 2: Solve the resulting quadratic in \(\cos\theta\).

Step 3: Corresponding \(\sin\theta\) values are \(\frac{3}{5}\) and \(1\). Quick Tip: Convert trigonometric equations to a single function.

Step 1: Use identity: \[ \tan A+\tan B+\tan A\tan B \tan(A+B)=\tan(A+B) \]

Step 2:
Here \(A=20^\circ, B=40^\circ\) and \(\tan60^\circ=\sqrt{3}\). Quick Tip: Look for standard angle identities.

Step 1: Discriminant: \[ D=(2\sqrt{2})^2-4=8-4=4>0 \]

Step 2: Roots are real and unequal. Quick Tip: Nature of roots depends on discriminant.

Step 1: Multiply numerator and denominator by conjugate.

Step 2: Result is a complex number of unit modulus.
\[ A^2+B^2=1 \] Quick Tip: \(|z|=1\Rightarrow A^2+B^2=1\).

Step 1: Number of intersection points: \[ \binom{n}{4}=70 \Rightarrow n=8 \]

Step 2: Diagonals: \[ \frac{n(n-3)}{2}=\frac{8\times5}{2}=20 \] Quick Tip: Each intersection is formed by choosing 4 vertices.

Step 1: Choose 2 different vowels for middle: \[ {}^5P_2=20 \]

Step 2: Choose consonants for first and last places: \[ 17\times17 \]

Step 3: \[ 17^2\times20=5780 \] Quick Tip: Middle positions fixed → arrange vowels first.

Step 1: General term: \[ T_{r+1} = {^{15}C_r}(x^4)^{15-r}\left(-\frac{1}{x^3}\right)^r \]

Step 2: Power of \(x\): \[ x^{60-7r} \]

Step 3: \[ 60-7r=32 \Rightarrow r=4 \]

Step 4: Coefficient: \[ {^{15}C_4}(-1)^4 = -15C_2 \] Quick Tip: Always equate power of \(x\) with required power.

Step 1: Let common difference be \(d\).

Step 2: \[ A_7 = 1+7d,\quad A_{m-1}=1+(m-1)d \]

Step 3: \[ \frac{1+7d}{1+(m-1)d}=\frac{5}{9} \]

Step 4: Solving gives \(m=20\). Quick Tip: Use general term of A.P. for mean problems.

Step 1: Use reflection formula: \[ (x',y')=\left(x-\frac{2a(ax+by+c)}{a^2+b^2},y-\frac{2b(ax+by+c)}{a^2+b^2}\right) \]

Step 2: Substituting values gives \((-1,-14)\). Quick Tip: Reflection formula is faster than geometry.

Step 1: Find midpoints of opposite sides.

Step 2: Diagonals pass through midpoints. Quick Tip: Diagonals of parallelogram bisect each other.

Step 1: Compare with standard form \(y^2=4ax\).

Step 2: Directrix is \(x=-a\Rightarrow k=6\). Quick Tip: Always reduce to standard parabola form.

Step 1: Tangency condition for ellipse. Quick Tip: Tangency gives discriminant zero condition.

Step 1: Center of circle is \((4,0)\).

Step 2: Required chord is perpendicular to radius.

Step 3: Equation is \(y=3\). Quick Tip: Chord bisector passes through midpoint perpendicular to radius.

Step 1: Rationalize numerator.

Step 2: Limit evaluates to 2. Quick Tip: Use rationalization for square root limits.

Step 1: Wrong total: \[ 25\times78.4=1960 \]

Step 2: Correct total: \[ 1960+(96-69)=1987 \]

Step 3: Correct mean: \[ \frac{1987}{25}=79.48 \] Quick Tip: Always correct total before recalculating mean.

Step 1: Pearson’s coefficient: \[ Sk=\frac{Mean-Mode}{SD} \]

Step 2: Substitution gives \(\frac{2}{3}\). Quick Tip: Positive skewness means tail on right side.

Step 1: Total outcomes \(=36\).

Step 2: Count cases where black die \(>\) twice white die.

Step 3: Favorable outcomes \(=6\).
\[ P=\frac{6}{36}=\frac{1}{6} \] Quick Tip: Always list outcomes systematically in dice problems.

Step 1: Total functions \(=2^4=16\).

Step 2: Non-onto functions: all map to 1 or all to 2 \(\Rightarrow 2\).

Step 3: \[ Onto = 16-2=14 \] Quick Tip: Onto = total functions − non-onto functions.

Step 1: \[ f(f(x))=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}} \]

Step 2: \[ =\frac{x}{\sqrt{1+2x^2}} \] Quick Tip: Be careful while simplifying composite functions.

Step 1: Put \(x=\cos\theta\).

Step 2: Expression simplifies to \(\theta + (\frac{\pi}{3}-\theta)\). Quick Tip: Substitution \(x=\cos\theta\) simplifies inverse trigonometric sums.

Step 1: \[ D=A+B \]

Step 2: Adding matrices gives option (B). Quick Tip: If \(A+B-D=0\Rightarrow D=A+B\).

Step 1: Expand determinant using row/column operations.

Step 2: Determinant evaluates to 39. Quick Tip: Use row operations to simplify determinants.

Step 1: For differentiability, function must be continuous.

Step 2: \[ a+1=a(1)^2+1 \Rightarrow a=1 \]

Step 3: Left and right derivatives are equal for \(a=1\). Quick Tip: Check continuity before differentiability.

Step 1: Area of circle \(A=\pi r^2\)

Step 2: Differentiate w.r.t. time: \[ \frac{dA}{dt}=2\pi r\frac{dr}{dt} \]

Step 3: Given \(\frac{dA}{dt}=3c\frac{dr}{dt}\)
\[ 2\pi r=3c \Rightarrow c=\frac{2\pi r}{3} \]

Step 4: For \(r=6\), \[ c=2\pi \] Quick Tip: Always relate rates using differentiation.

Step 1: \[ f'(x)=\sec^2x-4 \]

Step 2: \[ \sec^2x<4 \Rightarrow \cos^2x>\frac14 \Rightarrow |x|<\frac{\pi}{3} \] Quick Tip: Strictly decreasing \(\Rightarrow f'(x)<0\).

Step 1: Differentiate implicitly: \[ 4x-6y\frac{dy}{dx}=0 \]

Step 2: \[ \frac{dy}{dx}=\frac{2x}{3y} \]

Step 3: At \((3,2)\), \[ \frac{dy}{dx}=-1 \] Quick Tip: Use implicit differentiation for conics.

Step 1: Use product-to-sum identities.

Step 2: Integrate term by term. Quick Tip: Reduce products of trigonometric functions before integration.

Step 1: Use recurrence relation: \[ I_m = (-1)^m m! \]

Step 2: Substituting gives value \(e\). Quick Tip: Definite integrals with \(\ln x\) often follow recurrence relations.

Step 1: Function is odd and symmetric.

Step 2: \[ Area=2\int_0^1 x^2 dx=\frac{2}{3} \] Quick Tip: Use symmetry to simplify area calculations.

Step 1: Integrating factor \(=e^{2x}\)

Step 2: \[ y=\frac{1-e^{-2x}}{2} \] Quick Tip: Always apply initial conditions after solving.

Step 1: Rearrange equation: \[ \frac{dy}{dx}-\frac{y}{2x}=\frac{3}{2x} \]

Step 2: Solution is linear in \(x\) and \(y\). Quick Tip: Linear differential equations represent straight-line families.

Step 1: Identity: \[ (\vec a \times \vec b)^2 + (\vec a \cdot \vec b)^2 = a^2 b^2 \]

Step 2: \[ a^2(2)^2 = 676 \Rightarrow a^2 = 169 \]

Step 3: \[ |\vec a| = 13 \] Quick Tip: Use vector identity to avoid lengthy calculations.

Step 1: Required direction is \(\vec a \times \vec b\).

Step 2: \[ \vec a \times \vec b = 2(\hat i+\hat j) \]

Step 3: Unit vector: \[ \frac{\hat i+\hat j}{\sqrt2} \] Quick Tip: Perpendicular direction is given by cross product.

Step 1: \[ \vec a+\vec b+\vec c=(\hat i+\hat j-\hat k) \]

Step 2: Magnitude \(=\sqrt3\)

Step 3: Unit vector: \[ \dfrac{1}{\sqrt3}(\hat i+\hat j-\hat k) \] Quick Tip: Add vectors component-wise before normalising.

Step 1: Position vector of centroid remains same for medial triangle.

Step 2: \[ \frac{A+B+C}{3}=\frac23(\hat i+\hat j+\hat k) \] Quick Tip: Centroid of medial triangle equals centroid of original triangle.

Step 1: Use distance formula: \[ d=\frac{|\vec{AP}\times\vec d|}{|\vec d|} \]

Step 2: Substitution gives \(d=7\). Quick Tip: Point-to-line distance uses cross product.

Step 1: Plane containing line ⇒ normal vector ⟂ direction vector.

Step 2: \[ a\ell+b m+c n=0 \] Quick Tip: Direction ratios of line are perpendicular to plane’s normal.

Step 1: Mean \(\lambda=2\).

Step 2: \[ P(X>1.5)=P(X\ge2)=1-[P(0)+P(1)] \]

Step 3: \[ P(0)=e^{-2},\quad P(1)=2e^{-2} \]

Step 4: \[ P(X\ge2)=1-3e^{-2} \] Quick Tip: For Poisson distribution, probabilities are defined only at integers.

Step 1: \[ P(B)=P(A\cup B)-P(A)+P(A\cap B)=\frac23-\frac13+\frac16=\frac12 \]

Step 2: \[ P(A)P(B)=\frac13\times\frac12=\frac16=P(A\cap B) \] Quick Tip: If \(P(A\cap B)=P(A)P(B)\), events are independent.

Step 1: \[ \tan\theta=\frac{height}{shadow}=\frac{6}{2\sqrt3}=\sqrt3 \]

Step 2: \[ \theta=60^\circ \] Quick Tip: Use \(\tan\theta=\frac{opposite}{adjacent}\) in height–shadow problems.

Step 1: Profit per quintal:
Rice = ₹40, Wheat = ₹25

Step 2:
Objective function: \[ Z=40x+25y \] Quick Tip: Objective function always represents profit or cost to be optimised.

Step 1: By AM–GM inequality.

Step 2: Minimum occurs at \(x=y=z\).

Step 3: Substituting gives \(8\sqrt2\). Quick Tip: Symmetric expressions attain extrema when variables are equal.

Step 1: For continuity: \[ \lim_{x\to0}f(x)=12 \]

Step 2: Using standard limits: \[ \frac{(e^x-1)^2}{x^2}\to1,\; \frac{\sin(x/a)}{x}\to\frac1a,\; \frac{\log(1+x/4)}{x}\to\frac14 \]

Step 3: \[ \frac{1}{(1/a)(1/4)}=12 \Rightarrow a=2 \] Quick Tip: Use standard limits to evaluate continuity at zero.

Step 1: Check left and right derivatives at \(x=0\).

Step 2: Only option (D) has equal derivatives. Quick Tip: Differentiability requires LHD = RHD.

Step 1: Borrowing ideas or words and presenting them as one’s own is called plagiarism. Quick Tip: Plagiarism means intellectual theft.

Step 1: Archives are places where official or historical records are stored. Quick Tip: Archives store documents of permanent value.

Step 1: To give back something lost means to restore. Quick Tip: Restore means bring back to original condition.

Step 1: Paramour means a secret or illicit lover. Quick Tip: Paramour is associated with romantic relationship.

Step 1: Refectory means a dining hall, especially in institutions. Quick Tip: Refectory is commonly used in monasteries or colleges.

Step 1: Assent means expression of approval or agreement. Quick Tip: Assent implies consent.

Step 1: Showing one’s teeth means behaving aggressively or threateningly. Quick Tip: Idioms rarely have literal meanings.

Step 1: Pouring oil on troubled water means calming a tense situation. Quick Tip: This idiom is opposite of provoking conflict.

Step 1: Sentence (E) introduces the GST council’s proposal.

Step 2: Sentence (A) logically explains the purpose of this proposal. Quick Tip: Look for sentences explaining the objective right after the introduction.

Step 1: Sentence (F) works as the concluding sentence.

Step 2: Sentence (D) naturally precedes the conclusion as it mentions official support. Quick Tip: The sentence before last often provides endorsement or justification.

Step 1: Sentence (C) provides background context about India’s taxation system.

Step 2: It helps complete the logical flow of the paragraph. Quick Tip: Background or context-setting sentences often complete the idea.

Step 1: After the proposal and its objective, sentence (B) explains its economic impact. Quick Tip: Impact-related sentences usually follow explanation of policy.

Step 1: Sentence (E) clearly introduces the GST council’s proposal. Quick Tip: Opening sentences usually introduce the main idea or proposal.

Step 1: The phrase “novel of real ambition” fits naturally and conveys scope and seriousness. Quick Tip: Check collocation: ambition commonly pairs with creative works.

Step 1: “Associated with” is grammatically and contextually correct. Quick Tip: Use verb–preposition combinations carefully.

Step 1: Each letter in the first group is shifted forward to obtain the second group.

Step 2: Applying the same forward shift pattern to EFLK gives STXY. Quick Tip: In letter analogy questions, check consistent forward or backward shifts.

Step 1: The relation is \textit{person : birthplace.

Step 2: Jawaharlal Nehru was born in Allahabad. Quick Tip: Identify the common relationship before choosing the option.

Step 1: The statement mentions heavy burden on limited national resources.

Step 2: It concludes that education should be restricted to a few.

Step 3: Hence both assumptions are implied. Quick Tip: Assumptions are ideas taken for granted in the statement.

Step 1: Each letter of the word is coded using a fixed positional shift.

Step 2: Applying the same letter-wise coding rule to STRAY gives EFVWT. Quick Tip: Coding questions often follow a fixed positional or alternating pattern.

Step 1: Observe the pattern: \[ 2\times1+1=3,\; 3\times2+1=7,\; 7\times3+1=22,\; 22\times7+1=155 \]

Step 2: \[ 155\times22+1=3411 \] Quick Tip: Look for multiplication with the previous term followed by addition.

Step 1: Black and White are subsets of Colour.

Step 2: They are distinct but both lie completely within Colour. Quick Tip: When two categories are subsets of a larger one but not of each other, use two separate circles inside one big circle.

Step 1: Observe the exact orientation and relative position of all line segments in figure (X).

Step 2: Only option (B) contains the same arrangement as a subfigure. Quick Tip: Do not rotate or flip the given figure unless explicitly allowed.

Step 1: Identify the fold line and note symmetry.

Step 2: Reflect the cut portions across the fold.

Step 3: Option (A) matches the correct symmetrical expansion. Quick Tip: Always mirror the cuts across each fold line.

Step 1: Observe the progression of intersecting lines.

Step 2: The missing figure must continue the same directional and density pattern. Quick Tip: Look for continuation of both orientation and number of lines.

Step 1: Count and match all given pieces without overlap or distortion.

Step 2: Only option (C) uses all pieces exactly once. Quick Tip: All pieces must be used fully—no gaps and no overlaps.