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The questions your board exam loves to ask
800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.
PhysicsClass 112 markseasy
Mechanical Properties of Solids
Name the three moduli of elasticity and state what each measures.
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Young's modulus (Y) measures resistance to change in length (longitudinal stress/strain). Bulk modulus (B) measures resistance to change in volume (volume stress/strain) under uniform pressure. Shear (rigidity) modulus (G) measures resistance to change in shape (shearing stress/strain).
Marking-scheme points
- ✓Young's Y: change in length
- ✓Bulk B: change in volume
- ✓Shear/rigidity G: change in shape
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Thermal Properties of Matter
State Stefan's law and Wien's displacement law of black-body radiation.
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Stefan's (Stefan-Boltzmann) law: the total energy radiated per unit area per unit time by a black body is proportional to the fourth power of its absolute temperature, E = sigma T^4. Wien's displacement law: the wavelength at which the emission is maximum is inversely proportional to the absolute temperature, lambda_max * T = constant (= 2.9 x 10^-3 m K).
E = sigma T^4 ; lambda_max T = b
Marking-scheme points
- ✓Stefan: E = sigma T^4
- ✓Wien: lambda_max T = constant
- ✓Hotter body -> shorter peak wavelength
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Kinetic Theory
Define mean free path of a gas molecule. State two factors it depends on.
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The mean free path is the average distance a gas molecule travels between two successive collisions. It increases when the number density of molecules is low and when the molecular diameter is small; that is, it is inversely proportional to the number density and to the square of the molecular diameter.
lambda = 1 / (sqrt(2) pi d^2 n)
Marking-scheme points
- ✓Average distance between collisions
- ✓Inversely proportional to number density
- ✓Inversely proportional to (diameter)^2
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System of Particles and Rotational Motion
Define the centre of mass of a system of particles. Write its position for a two-particle system.
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The centre of mass is the point at which the entire mass of the system can be considered to be concentrated, and where an applied external force produces the same acceleration as on the whole system. For two particles of masses m1 and m2 at positions x1 and x2, x_cm = (m1 x1 + m2 x2) / (m1 + m2).
x_cm = (m1 x1 + m2 x2) / (m1 + m2)
Marking-scheme points
- ✓Point where total mass seems concentrated
- ✓Moves as if all mass and external force act there
- ✓x_cm = (m1 x1 + m2 x2)/(m1 + m2)
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Gravitation
How does the acceleration due to gravity vary with depth below the Earth's surface? What is its value at the centre?
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With depth d, g decreases as g' = g (1 - d/R), assuming uniform density, because only the mass within the smaller inner sphere attracts the body. At the centre (d = R) the value becomes zero, since the mass is symmetrically distributed all around.
g' = g (1 - d/R)
Marking-scheme points
- ✓g' = g (1 - d/R)
- ✓g decreases with depth
- ✓g = 0 at the centre
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Motion in a Plane
What is meant by resolution of a vector? Write the rectangular components of a vector A making angle theta with the x-axis.
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Resolution of a vector is the process of splitting it into two or more components, usually along mutually perpendicular directions. For a vector A at angle theta to the x-axis, the rectangular components are Ax = A cos(theta) along the x-axis and Ay = A sin(theta) along the y-axis, with A = sqrt(Ax^2 + Ay^2).
Ax = A cos(theta) ; Ay = A sin(theta)
Marking-scheme points
- ✓Splitting a vector into components
- ✓Ax = A cos(theta), Ay = A sin(theta)
- ✓A = sqrt(Ax^2 + Ay^2)
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Thermodynamics
What is a quasi-static process? Why is it important?
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A quasi-static process is one carried out infinitely slowly so that the system stays in thermal and mechanical equilibrium with its surroundings at every stage. It is important because only for such (reversible) processes can the state variables (P, V, T) be defined throughout, and the work done can be represented as an area on a P-V diagram.
Marking-scheme points
- ✓Infinitely slow, equilibrium at every step
- ✓System properties well-defined throughout
- ✓Basis of reversible processes / P-V work
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Gravitation
State Newton's universal law of gravitation and write its mathematical form.
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Every particle of matter attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, directed along the line joining them: F = G m1 m2 / r^2, where G is the universal gravitational constant (6.67 x 10^-11 N m^2 / kg^2).
F = G m1 m2 / r^2
Marking-scheme points
- ✓F proportional to m1 m2
- ✓F inversely proportional to r^2
- ✓F = G m1 m2 / r^2
Still unsure? Ask the AI tutor →ChemistryClass 112 markseasy
Some Basic Concepts of Chemistry
State the law of conservation of mass and the law of definite proportions.
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Law of conservation of mass: matter can neither be created nor destroyed in a chemical reaction; the total mass of reactants equals the total mass of products. Law of definite proportions: a given chemical compound always contains the same elements combined in the same fixed proportion by mass, irrespective of its source or method of preparation.
Marking-scheme points
- ✓Conservation of mass: mass of reactants = mass of products
- ✓Definite proportions: fixed ratio of elements by mass
- ✓Example: water always 1:8 H:O by mass
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Some Basic Concepts of Chemistry
Calculate the number of moles present in 11 g of carbon dioxide (CO2).
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Molar mass of CO2 = 12 + 2(16) = 44 g/mol. Number of moles = given mass / molar mass = 11 / 44 = 0.25 mol.
n = m / M
Marking-scheme points
- ✓Molar mass of CO2 = 44 g/mol
- ✓moles = mass / molar mass
- ✓n = 11/44 = 0.25 mol
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Some Basic Concepts of Chemistry
Define one mole and state the value of Avogadro's number.
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One mole is the amount of a substance that contains as many elementary entities (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12. This number of entities is Avogadro's number, N_A = 6.022 x 10^23 mol^-1.
1 mol = 6.022 x 10^23 particles
Marking-scheme points
- ✓Mole = amount containing N_A entities
- ✓Reference: atoms in 12 g of C-12
- ✓N_A = 6.022 x 10^23 per mole
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Some Basic Concepts of Chemistry
Define molarity and molality. State their units.
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Molarity (M) is the number of moles of solute dissolved per litre of solution; unit mol/L (or M). Molality (m) is the number of moles of solute dissolved per kilogram of solvent; unit mol/kg (or m). Molality is independent of temperature since it uses mass, whereas molarity changes with temperature because volume changes.
M = n_solute / V(L); m = n_solute / mass_solvent(kg)
Marking-scheme points
- ✓Molarity = moles of solute / litre of solution (mol/L)
- ✓Molality = moles of solute / kg of solvent (mol/kg)
- ✓Molality is temperature independent
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Some Basic Concepts of Chemistry
Define mole fraction and mass percent of a component in a solution.
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Mole fraction of a component = number of moles of that component / total number of moles of all components in the solution; it is dimensionless and the sum of mole fractions equals 1. Mass percent of a component = (mass of the component / total mass of solution) x 100.
x_A = n_A / n_total; mass% = (mass component / total mass) x 100
Marking-scheme points
- ✓Mole fraction x_A = n_A / (n_A + n_B); sum = 1
- ✓Dimensionless quantity
- ✓Mass percent = (mass of component / total mass) x 100
Still unsure? Ask the AI tutor →ChemistryClass 112 marksmedium
Structure of Atom
Calculate the energy of a photon of light of wavelength 4000 Angstrom. (h = 6.626 x 10^-34 J s, c = 3 x 10^8 m/s)
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Wavelength = 4000 Angstrom = 4000 x 10^-10 m = 4 x 10^-7 m. Energy E = hc/lambda = (6.626 x 10^-34 x 3 x 10^8) / (4 x 10^-7) = 1.988 x 10^-25 / 4 x 10^-7 = 4.97 x 10^-19 J.
E = hc / lambda
Marking-scheme points
- ✓Convert 4000 Angstrom = 4 x 10^-7 m
- ✓E = hc/lambda
- ✓E = 4.97 x 10^-19 J
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Structure of Atom
State Heisenberg's uncertainty principle and give its mathematical form.
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It is impossible to determine simultaneously and with absolute accuracy both the position and the momentum (or velocity) of a microscopic particle such as an electron. The product of the uncertainties in position (delta x) and momentum (delta p) is at least of the order of h/4pi.
delta x . delta p >= h / 4pi
Marking-scheme points
- ✓Cannot measure position and momentum exactly at once
- ✓Applies to microscopic particles like electrons
- ✓delta x . delta p >= h/4pi
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Structure of Atom
Calculate the wave number of the spectral line when an electron in a hydrogen atom jumps from n = 3 to n = 2. (Rydberg constant R = 1.097 x 10^7 m^-1)
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Wave number (nu bar) = R (1/n1^2 - 1/n2^2) = 1.097 x 10^7 (1/2^2 - 1/3^2) = 1.097 x 10^7 (1/4 - 1/9) = 1.097 x 10^7 x (5/36) = 1.523 x 10^6 m^-1. This is the H-alpha line of the Balmer series.
nu bar = R (1/n1^2 - 1/n2^2)
Marking-scheme points
- ✓nu bar = R(1/n1^2 - 1/n2^2), n1=2, n2=3
- ✓1/4 - 1/9 = 5/36
- ✓nu bar = 1.523 x 10^6 m^-1 (Balmer series)
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Structure of Atom
State Pauli's exclusion principle and Hund's rule of maximum multiplicity.
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Pauli's exclusion principle: no two electrons in an atom can have the same set of all four quantum numbers; an orbital can hold at most two electrons with opposite spins. Hund's rule of maximum multiplicity: electron pairing in orbitals of the same subshell (degenerate orbitals) does not occur until each orbital is singly occupied, and all singly filled orbitals have parallel spin.
Marking-scheme points
- ✓Pauli: no two electrons share all four quantum numbers
- ✓Max 2 electrons per orbital, opposite spins
- ✓Hund: singly fill degenerate orbitals first, parallel spins
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Structure of Atom
State the Aufbau principle and the (n + l) rule for filling of orbitals.
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Aufbau principle: in the ground state of an atom, electrons are filled into orbitals in order of increasing energy, i.e. the lowest energy orbital is filled first. (n + l) rule: the orbital with the lower (n + l) value has lower energy and is filled first; if two orbitals have the same (n + l) value, the one with the lower n is filled first (e.g. 4s (n+l=4) is filled before 3d (n+l=5)).
energy order by increasing (n + l)
Marking-scheme points
- ✓Fill lowest energy orbitals first
- ✓Lower (n + l) = lower energy = filled first
- ✓Equal (n+l): lower n filled first (4s before 3d)
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Classification of Elements and Periodicity
State the modern periodic law. How does it differ from Mendeleev's periodic law?
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Modern periodic law: the physical and chemical properties of elements are a periodic function of their atomic numbers. Mendeleev's law was based on atomic mass, whereas the modern law is based on atomic number (number of protons), which removed anomalies such as the position of argon and potassium.
Marking-scheme points
- ✓Properties are periodic function of atomic number
- ✓Mendeleev: based on atomic mass
- ✓Atomic number basis removes mass-order anomalies
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Classification of Elements and Periodicity
Why is the first ionization enthalpy of nitrogen greater than that of oxygen?
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Nitrogen has the configuration 1s2 2s2 2p3 with a half-filled 2p subshell, which is extra stable due to symmetry and exchange energy, so removing an electron requires more energy. Oxygen (1s2 2s2 2p4) has one paired electron in 2p; removing it relieves electron-electron repulsion and gives a stable half-filled configuration, so less energy is needed. Hence IE1 of nitrogen > oxygen.
Marking-scheme points
- ✓N has stable half-filled 2p3 configuration
- ✓O 2p4 has electron-pair repulsion, easier to remove
- ✓So IE1(N) > IE1(O)
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