Exhaustive Physics, Chemistry, and Mathematics Study Notes

Mechanics, Kinematics, and Measurement

  • Slopes of Motion Graphs: The slope of a time-speed graph is defined as the acceleration of the body (a=dvdta = \frac{dv}{dt}).
  • Relative Velocity: For two bodies A and B moving in the same direction with velocities VA=20i^m/s\mathbf{V_A} = 20\hat{i}\,m/s and VB=15i^m/s\mathbf{V_B} = 15\hat{i}\,m/s, the relative velocity of A with respect to B (VAB\mathbf{V_{AB}}) is calculated as VAVB=5i^m/s\mathbf{V_A} - \mathbf{V_B} = 5\hat{i}\,m/s.
  • Kinematic Equations: For a bicycle starting from rest (u=0u = 0) and accelerating at 1.5m/s21.5\,m/s^2 for 44 seconds, the total distance covered is found using s=ut+12at2s = ut + \frac{1}{2}at^2. Calculation: s=0+12(1.5)(4)2=12ms = 0 + \frac{1}{2}(1.5)(4)^2 = 12\,m.
  • Work Done by Gravity: A man weighing 70kg70\,kg carries a 30kg30\,kg box to the top of a building of height 20m20\,m. The total mass moved is 100kg100\,kg. The work done is W=mgh=100kg×9.8m/s2×20m=19600JW = mgh = 100\,kg \times 9.8\,m/s^2 \times 20\,m = 19600\,J.
  • Dimensional Analysis: To convert a velocity of 72kmh172\,kmh^{-1} into ms1ms^{-1}: 72×518=20ms172 \times \frac{5}{18} = 20\,ms^{-1}.
  • Unit Analysis: In the equation S=a+bt+ct2S = a + bt + ct^2, where SS is in metres and tt is in seconds, the unit of the constant cc must be ms2ms^{-2} to satisfy dimensional homogeneity.
  • Errors in Measurement:     * Maximum Error in Density: Density ρ=ML3\rho = \frac{M}{L^3}. The maximum relative error is Δρρ=ΔMM+3ΔLL\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 3\frac{\Delta L}{L}. Given mass error 1.5%1.5\% and length error 1%1\%, the maximum error is 1.5+3(1)=4.5%1.5 + 3(1) = 4.5\%.     * Percentage Error in Length: A rod measured as 5.6cm5.6\,cm with a ruler having a smallest division (least count) of 0.1cm0.1\,cm has a percentage error of 0.15.6×1001.79%\frac{0.1}{5.6} \times 100 \approx 1.79\%.
  • Screw Gauge Calculations: Diameter is given by Main scale reading+(Circular scale reading×Least Count)\text{Main scale reading} + (\text{Circular scale reading} \times \text{Least Count}). If 1mm=1001\,mm = 100 divisions, Least Count=0.01mm\text{Least Count} = 0.01\,mm. For a reading of 0mm0\,mm and 5252 divisions: 0+52(0.01)=0.52mm=0.052cm0 + 52(0.01) = 0.52\,mm = 0.052\,cm.
  • Circular Motion: For a body moving in a circular path with constant speed, the work done by the net force (centripetal force) is zero because the force is always perpendicular to the displacement.
  • Projectile and Path Equations: A particle with coordinates x=asin(ωt)x = a\sin(\omega t) and y=acos(ωt)y = a\cos(\omega t) follows a circular path because x2+y2=a2x^2 + y^2 = a^2.
  • Frame of Reference: A coin dropped in an elevator reaches the floor in time t1t_1 at rest and t2t_2 moving uniformly. In both cases, the acceleration relative to the elevator is gg, so t1=t2t_1 = t_2.
  • Relative Motion Crossing: A train 150m150\,m long moving North at 10m/s10\,m/s and a parrot flying South at 5m/s5\,m/s. Relative speed is 10+5=15m/s10 + 5 = 15\,m/s. Time to cross = 15015=10s\frac{150}{15} = 10\,s.
  • Contact Forces: Frictional force is a contact force, unlike gravitational, magnetic, or electrostatic forces.

Thermodynamics and Kinetic Theory

  • Specific Heat Calculations: To raise the temperature of 22 moles of ideal gas at constant pressure (ΔT=5C\Delta T = 5^\circ C) requires 70cal70\,cal. Since Qp=nCpΔTQ_p = nC_p\Delta T, 70=2(Cp)(5)Cp=7calmol1K170 = 2(C_p)(5) \Rightarrow C_p = 7\,cal\,mol^{-1}K^{-1}. Using Mayer's Relation CpCv=RC_p - C_v = R (where R2calmol1K1R \approx 2\,cal\,mol^{-1}K^{-1}), Cv=5calmol1K1C_v = 5\,cal\,mol^{-1}K^{-1}. Heat needed at constant volume (Qv=nCvΔTQ_v = nC_v\Delta T) is 2(5)(5)=50cal2(5)(5) = 50\,cal.
  • First Law of Thermodynamics: Along path iafiaf, Q=50calQ = 50\,cal, W=20calΔU=30calW = 20\,cal \Rightarrow \Delta U = 30\,cal. Along path ibfibf, if Q=36calQ = 36\,cal, then W=QΔU=3630=6calW = Q - \Delta U = 36 - 30 = 6\,cal.
  • Kinetic Theory of Gases: The root mean square (rms) speed of a gas molecule of mass mm at temperature TT is given by vrms=3kTmv_{rms} = \sqrt{\frac{3kT}{m}}.
  • Sound Propagation: When sound waves travel in a gaseous medium, the process is considered adiabatic because the compressions and rarefactions happen too rapidly for heat exchange.
  • Adiabatic Relations: For a monoatomic gas (where γ=5/3\gamma = 5/3), the adiabatic relation is PV5/3=constantPV^{5/3} = \text{constant}.

Optics and Waves

  • Lens Maker's Formula: For a double convex lens with radii R1=15cmR_1 = 15\,cm and R2=30cmR_2 = -30\,cm and n=1.5n = 1.5: 1f=(n1)(1R11R2)=(1.51)(115130)=0.5(330)=120\frac{1}{f} = (n-1)(\frac{1}{R_1} - \frac{1}{R_2}) = (1.5 - 1)(\frac{1}{15} - \frac{1}{-30}) = 0.5(\frac{3}{30}) = \frac{1}{20}. Thus, f=20cmf = 20\,cm.
  • Young’s Double Slit Experiment (YDSE): The number of fringes nn and wavelength λ\lambda in a set segment are related by n1λ1=n2λ2n_1\lambda_1 = n_2\lambda_2. Given n1=12n_1 = 12, λ1=600nm\lambda_1 = 600\,nm, λ2=400nm\lambda_2 = 400\,nm: 12(600)=n2(400)n2=1812(600) = n_2(400) \Rightarrow n_2 = 18.
  • Interference Intensity: If the ratio of maximum to minimum intensity is ImaxImin=16\frac{I_{max}}{I_{min}} = 16, then the ratio of amplitudes a1+a2a1a2=16=4\frac{a_1+a_2}{a_1-a_2} = \sqrt{16} = 4. Solving gives a1a2=53\frac{a_1}{a_2} = \frac{5}{3}. The intensity ratio is I1I2=(a1a2)2=25:9\frac{I_1}{I_2} = (\frac{a_1}{a_2})^2 = 25:9.
  • Vibrations and Sound:     * Tuning Fork: A vibrating tuning fork on a table creates forced vibrations in the table board.     * Phase Difference: For a wave with T=0.05sT = 0.05\,s and v=300m/sv = 300\,m/s, λ=vT=15m\lambda = vT = 15\,m. The path difference Δx=1510=5m\Delta x = 15 - 10 = 5\,m. Phase difference Δϕ=2πλΔx=2π15(5)=2π3\Delta \phi = \frac{2\pi}{\lambda}\Delta x = \frac{2\pi}{15}(5) = \frac{2\pi}{3}.
  • Wave Equations: For Y=0.5sin[π(400tx)]Y = 0.5\sin[\pi(400t - x)], the velocity is v=ωk=400ππ=400m/sv = \frac{\omega}{k} = \frac{400\pi}{\pi} = 400\,m/s.
  • Resolving Power: The resolving power of an optical microscope is inversely proportional to wavelength. Ratio RP1:RP2=λ2:λ1=6000:4000=3:2RP_1 : RP_2 = \lambda_2 : \lambda_1 = 6000:4000 = 3:2.

Electricity and Magnetism

  • Electrostatic Work and Potential:     * Work to carry 6μC6\,\mu C charge through a 9V9\,V battery is W=qV=6×106×9=54×106JW = qV = 6 \times 10^{-6} \times 9 = 54 \times 10^{-6}\,J.     * Potential from field E=25i^+30j^N/C\mathbf{E} = 25\hat{i} + 30\hat{j}\,N/C: V=Edr=(25x+30y)V = -\int \mathbf{E} \cdot d\mathbf{r} = -(25x + 30y). At (2,2)(2,2), V=(25(2)+30(2))=110VV = -(25(2) + 30(2)) = -110\,V.
  • Circuits and Resistors:     * Potentiometer: E1E2=l1l2\frac{E_1}{E_2} = \frac{l_1}{l_2}. 1.5E2=2754E2=3V\frac{1.5}{E_2} = \frac{27}{54} \Rightarrow E_2 = 3\,V.     * Color Code: For a resistor with code Yellow (4), Violet (7), Brown (10110^1), Gold (5%5\%), the value is 470Ω±5%470\,\Omega \pm 5\%.     * Specific Resistance: Increases with an increase in temperature for conductors.     * Parallel resistors: Thermal energy ratio for RR and 3R3R in parallel is P=V2RP = \frac{V^2}{R}. Ratio P1:P2=V2/RV2/3R=3:1P_1:P_2 = \frac{V^2/R}{V^2/3R} = 3:1.     * Capacitor Heat: Energy stored U=12CV2=12(4×106)(400)2=0.32JU = \frac{1}{2}CV^2 = \frac{1}{2}(4 \times 10^{-6})(400)^2 = 0.32\,J. This energy is released as heat in the resistor.
  • Magnetism:     * Magnetic Force: F=q(v×B)\mathbf{F} = q(\mathbf{v} \times \mathbf{B}). If a particle moves along a magnetic field line, v\mathbf{v} is parallel to B\mathbf{B}, so the force is zero.     * Stored Energy: For a coil in a steady state DC circuit, the magnetic energy is U=12LI2U = \frac{1}{2}LI^2. In Question 34, calculation leads to 25J25\,J.     * Magnetic Field at Center: B0=μ0nI2RB_0 = \frac{\mu_0 n I}{2R}. If rewound to 3 turns (n=3n=3, R=R/3R' = R/3), the new field B=μ0(3)I2(R/3)=9B0B = \frac{\mu_0(3)I}{2(R/3)} = 9B_0.
  • Galvanometer: Torque τ=NIAB=Cθ\tau = NIAB = C\theta. For N=175N=175, A=104m2A=10^{-4}\,m^2, C=106N-m/radC=10^{-6}\,N\text{-}m/rad, I=103AI=10^{-3}\,A, and θ=π180\theta = \frac{\pi}{180}, the magnetic field BB is approximately 103T10^{-3}\,T.

Modern Physics and Nuclear Chemistry

  • Duality: Kinetic energy ratio for electron and proton with same de Broglie wavelength: K=h22mλ2KeKp=mpme1836K = \frac{h^2}{2m\lambda^2} \Rightarrow \frac{K_e}{K_p} = \frac{m_p}{m_e} \approx 1836. Ratio is 1:18361: 1836 (A).
  • Bohr Model:     * Energy in nn-th orbit En=13.6Z2n2eVE_n = -13.6 \frac{Z^2}{n^2}\,eV. For helium (Z=2Z=2), EHe=4EnE_{He} = 4 E_n.     * Spectral lines for transition to n=4n=4: (42)=6\binom{4}{2} = 6.     * Wave number for n=4n=2n=4 \rightarrow n=2: νˉ=R(122142)=R(14116)=3R16\bar{\nu} = R(\frac{1}{2^2} - \frac{1}{4^2}) = R(\frac{1}{4} - \frac{1}{16}) = \frac{3R}{16}.
  • Nuclear Decay: Process A(180,72)αA1(176,70)βA2(176,71)αA3(172,69)γA4(172,69)A(180, 72) \xrightarrow{\alpha} A_1(176, 70) \xrightarrow{\beta} A_2(176, 71) \xrightarrow{\alpha} A_3(172, 69) \xrightarrow{\gamma} A_4(172, 69).
  • Uncertainty Principle: ΔEΔth4π\Delta E \Delta t \geq \frac{h}{4\pi}. High certainty in measurement of energy implies high uncertainty in time.
  • X-Rays: Moseley's Law 1λ(Z1)2\frac{1}{\lambda} \propto (Z-1)^2. For Molybdenum (Z=42Z=42) and Zinc (Z=30Z=30), the wavelength of Zinc is determined as 1.3872…1.3872\,\text{…}.

General Chemistry

  • Thermochemistry:     * Enthalpy of formation for CH4CH_4 calculated via Hess's Law using provided combustion data: 75kJ/mol-75\,kJ/mol.     * Heat evolved for toluene (46g46\,g) combustion: Toluene molar mass is 92g/mol92\,g/mol. Since 46g46\,g is 0.5mol0.5\,mol, heat evolved = 0.5×3910.3=1955.15kJ0.5 \times 3910.3 = 1955.15\,kJ. Converting to kCal: 1955.154.1=477kCal\frac{1955.15}{4.1} = 477\,kCal.
  • Chemical Equilibrium:     * If reaction proceeds as 2aA+2bB2cC+2dD2aA + 2bB \rightleftharpoons 2cC + 2dD, the new equilibrium constant is K2K^2.     * Le Chatelier's: In endothermic water splitting (2H2O2H2+O22H_2O \rightleftharpoons 2H_2 + O_2), increasing temperature at constant pressure shifts equilibrium forward, decreasing water.
  • Solutions:     * Mole fraction of glucose (10moles10\,moles) = 0.500.50. 1010+nwater=0.5nwater=10moles\frac{10}{10 + n_{\text{water}}} = 0.5 \Rightarrow n_{\text{water}} = 10\,moles. Mass of water = 10×18=180g10 \times 18 = 180\,g (or 180mL180\,mL).     * Henry's Law: p=KHX760=4.27×105XX1.78×103p = K_H X \Rightarrow 760 = 4.27 \times 10^5 X \Rightarrow X \approx 1.78 \times 10^{-3}.
  • Electrochemistry:     * Molar conductivity of NH4OHNH_4OH at infinite dilution is calculated as Λ0(NH4Cl)+Λ0(NaOH)Λ0(NaCl)=149.7+248.1126.5=271.3Ω1cm2mol1\Lambda^0(NH_4Cl) + \Lambda^0(NaOH) - \Lambda^0(NaCl) = 149.7 + 248.1 - 126.5 = 271.3\,\Omega^{-1}cm^2mol^{-1}.
  • Kinetics:     * For first-order reaction t75%=2×t50%t_{75\%} = 2 \times t_{50\%}. Given t75%=60mint_{75\%} = 60\,min, then t50%=30mint_{50\%} = 30\,min.     * Radioactive disintegration is always a first-order process.

Organic Chemistry

  • Reactions:     * Hoffmann Elimination: 2-bromopentane with sodium tert-butoxide favors the less substituted alkene (1-pentene).     * Iodoform Test: Acetophenone reacts with I2/NaOHI_2/NaOH to give yellow precipitate of iodoform (CHI3CHI_3) and sodium benzoate (which yields benzoic acid PhCO2HPhCO_2H upon acidification).     * Tollens Negative: Ketones (like cyclohexanone) and certain alcohols test negative.     * Hydrolysis: Sucrose yields glucose and fructose.
  • Functional Groups and Polymers:     * Schiff Base: Formed by the reaction of primary amines with carbonyl compounds (imine).     * Polyurethane: Monomer is diisocyanate.     * Amino Acids: Cysteine contains a sulfur atom.     * Formaldehyde: Cyclic trimer is 1,3,5-trioxane.

Mathematics

  • Functions and Calculus:     * Function f(x)=x2f(x) = x^2 on [0,)[0, \infty) is one-one but not onto if range is not specified. Range of f(x)=x21+x2f(x) = \frac{x^2}{1+x^2} is [0,1)[0, 1).     * f(x)=xxf(x) = \frac{x}{|x|} is not continuous at x=0x=0.     * Integral of sec4/3xcsc2/3xdx=3tan1/3x+c\sec^{4/3}x \csc^{2/3}x \,dx = 3\tan^{1/3}x + c.     * Limit: limn[1n+1n+1++12n]=ln(2)\lim_{n\rightarrow \infty} [\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}] = \ln(2).
  • Probability and Stats:     * Probability of getting a 4 given an even number on a die: P(4even)=13P(4|\text{even}) = \frac{1}{3}.     * Mean of defective bolts (n=400,p=0.1n=400, p=0.1) is np=40np = 40. Standard deviation npq=400(0.1)(0.9)=6\sqrt{npq} = \sqrt{400(0.1)(0.9)} = 6.
  • Algebra and Matrices:     * Permutations of 'EULER': Total letters =5= 5, duplicate 'E's =2= 2. Ways =5!2!=60= \frac{5!}{2!} = 60.     * Zeros in 100!100!: Calculated by 100/5+100/25=20+4=24\lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20 + 4 = 24.     * The sum of roots α5+β5\alpha^5 + \beta^5: For x2x+2=0x^2 - x + 2 = 0, use recurrence or power sums to find value 6464.
  • Geometry:     * Distance between parallel lines 2x+y+4=02x + y + 4 = 0 and 2x+y+8=02x + y + 8 = 0: d=8422+12=45d = \frac{|8-4|}{\sqrt{2^2+1^2}} = \frac{4}{\sqrt{5}}.     * Shortest distance between yx=1y-x=1 and y=x2y=x^2 occurs where slope of tangent to $y=x^2isis1..2x=1 \Rightarrow x=0.5.Distanceis. Distance is\frac{3\sqrt{2}}{8}$$.

Answer Key Reference

  • Question Summary: Total of 225 questions across basic and advanced Physics, Chemistry, and Mathematics.
  • Key Trends: 1-75 (Physics focus), 76-135 (Chemistry focus), 136-225 (Mathematics focus).