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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 122 marksmedium
Nuclei
State Einstein's mass-energy relation. What is the energy equivalent of 1 atomic mass unit (u)?
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Einstein's mass-energy relation is E = m c^2, which states that mass and energy are interconvertible, where c is the speed of light. Using this relation, the energy equivalent of 1 atomic mass unit (1 u = 1.66 x 10^-27 kg) is about 931 MeV (mega electron volt). This relation explains the large amount of energy released in nuclear reactions such as fission and fusion.
E = m c^2; 1 u = 931 MeV
Marking-scheme points
- ✓E = m c^2 (mass and energy are interconvertible)
- ✓1 u is equivalent to about 931 MeV
- ✓Explains energy released in nuclear reactions
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Nuclei
Define mass defect and binding energy of a nucleus.
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The mass defect is the difference between the sum of the masses of the individual protons and neutrons (nucleons) and the actual mass of the nucleus; the actual nuclear mass is always less than the sum. This missing mass (delta m) is converted into energy that binds the nucleons together. The binding energy is the energy equivalent of the mass defect, BE = (delta m) c^2; it is the energy required to break the nucleus into its constituent nucleons.
BE = (delta m) c^2
Marking-scheme points
- ✓Mass defect = (sum of nucleon masses) - (actual nuclear mass)
- ✓This mass is converted into binding energy
- ✓Binding energy = (delta m) c^2
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Nuclei
Name the three types of radioactive radiations and state their nature.
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The three types of radioactive radiations are: (1) alpha rays, which are helium nuclei (2 protons + 2 neutrons), positively charged and with low penetrating power; (2) beta rays, which are fast-moving electrons, negatively charged and with greater penetrating power than alpha rays; and (3) gamma rays, which are high-energy electromagnetic waves (photons), electrically neutral and with very high penetrating power. In a magnetic field, alpha and beta rays are deflected in opposite directions while gamma rays are undeflected.
Marking-scheme points
- ✓Alpha: helium nuclei, positive, low penetration
- ✓Beta: fast electrons, negative, moderate penetration
- ✓Gamma: high-energy EM waves, neutral, high penetration
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Nuclei
What is nuclear fission? Give one example.
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Nuclear fission is the process in which a heavy nucleus (such as uranium-235) splits into two lighter nuclei of comparable masses, with the release of a few neutrons and a large amount of energy. For example, when a uranium-235 nucleus captures a slow neutron, it splits into barium and krypton nuclei plus three neutrons and energy. The released neutrons can cause further fissions, leading to a chain reaction, which is used in nuclear reactors and atom bombs.
Marking-scheme points
- ✓Heavy nucleus splits into two lighter nuclei with energy release
- ✓Example: U-235 + neutron -> lighter nuclei + neutrons + energy
- ✓Released neutrons can cause a chain reaction
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Nuclei
What is nuclear fusion? Why does it require very high temperature?
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Nuclear fusion is the process in which two light nuclei (such as isotopes of hydrogen) combine to form a heavier nucleus, with the release of an enormous amount of energy. It is the source of energy of the sun and stars, where hydrogen nuclei fuse to form helium. It requires very high temperature (millions of degrees) because the positively charged nuclei must overcome their strong electrostatic repulsion to come close enough to fuse.
Marking-scheme points
- ✓Two light nuclei combine into a heavier nucleus with energy release
- ✓Source of energy of the sun and stars (hydrogen to helium)
- ✓Needs very high temperature to overcome electrostatic repulsion
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Semiconductor Electronics
Distinguish between intrinsic and extrinsic semiconductors.
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An intrinsic semiconductor is a pure semiconductor (such as pure silicon or germanium) with no added impurity; its conductivity is low and is due to the equal number of electrons and holes generated thermally. An extrinsic semiconductor is one to which a small amount of a suitable impurity has been added (doping); this greatly increases its conductivity. Extrinsic semiconductors are of two types, n-type and p-type.
Marking-scheme points
- ✓Intrinsic: pure semiconductor, low conductivity, equal electrons and holes
- ✓Extrinsic: doped with impurity, higher conductivity
- ✓Extrinsic types: n-type and p-type
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Semiconductor Electronics
How are n-type and p-type semiconductors formed? Name the majority charge carriers in each.
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An n-type semiconductor is formed by doping a pure semiconductor (silicon) with a pentavalent impurity (such as phosphorus or arsenic), which donates free electrons; the majority carriers are electrons and the minority carriers are holes. A p-type semiconductor is formed by doping with a trivalent impurity (such as boron or aluminium), which creates holes; the majority carriers are holes and the minority carriers are electrons. Both types are electrically neutral overall.
Marking-scheme points
- ✓n-type: pentavalent doping (phosphorus); majority carriers = electrons
- ✓p-type: trivalent doping (boron); majority carriers = holes
- ✓Both are electrically neutral overall
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Semiconductor Electronics
What is a depletion region and potential barrier in a p-n junction?
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When a p-n junction is formed, electrons from the n-side diffuse into the p-side and holes from the p-side diffuse into the n-side, and they recombine near the junction. This leaves a region near the junction that has no free charge carriers but has immobile charged ions; this region is called the depletion region (or depletion layer). The immobile ions set up an internal electric field that opposes further diffusion; the potential difference developed across the depletion region is called the potential barrier.
Marking-scheme points
- ✓Depletion region: layer near the junction with no free carriers
- ✓Formed by diffusion and recombination of electrons and holes
- ✓Potential barrier: potential difference across the depletion region
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Semiconductor Electronics
Distinguish between forward biasing and reverse biasing of a p-n junction diode.
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In forward biasing, the p-side is connected to the positive terminal and the n-side to the negative terminal of the battery; this reduces the width of the depletion region and the potential barrier, so a large current flows and the diode conducts. In reverse biasing, the p-side is connected to the negative terminal and the n-side to the positive terminal; this increases the width of the depletion region and the potential barrier, so only a very small (negligible) current flows and the diode does not conduct.
Marking-scheme points
- ✓Forward bias: p to +, n to -; barrier reduced, diode conducts
- ✓Reverse bias: p to -, n to +; barrier increased, negligible current
- ✓Diode acts as a one-way valve for current
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Semiconductor Electronics
What is a Zener diode? State its main use.
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A Zener diode is a special heavily doped p-n junction diode designed to operate in the reverse breakdown region without being damaged. In this region, the voltage across it remains almost constant (equal to its Zener voltage) even when the current through it changes over a wide range. Because of this property, its main use is as a voltage regulator, that is, to provide a constant output voltage to a load in spite of changes in the input voltage or load current.
Marking-scheme points
- ✓Heavily doped diode that works in reverse breakdown safely
- ✓Voltage across it stays constant (Zener voltage)
- ✓Main use: voltage regulator
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The Solid State
Distinguish between crystalline and amorphous solids with one example each.
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Crystalline solids have a regular, long-range ordered arrangement of particles, sharp melting points, definite geometrical shapes and are anisotropic (properties differ with direction); e.g. sodium chloride and diamond. Amorphous solids have only a short-range order (irregular arrangement), no sharp melting point (they soften over a range), no definite shape and are isotropic (same properties in all directions); e.g. glass and rubber.
Marking-scheme points
- ✓Crystalline: long-range order, sharp melting point, anisotropic (NaCl)
- ✓Amorphous: short-range order, no sharp melting point, isotropic (glass)
- ✓Amorphous solids soften over a range of temperature
The concept
Crystalline solids have a long-range ordered arrangement of particles; amorphous solids are disordered.
Why it matters
It decides material properties - why diamond has sharp edges and a fixed melting point but glass does not.
💡 Memory trick
Crystalline = neat rows (sharp melting point); Amorphous = messy (softens over a range). Salt vs glass.
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The Solid State
What is a unit cell? Distinguish between a primitive and a centred unit cell.
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A unit cell is the smallest repeating three-dimensional portion of a crystal lattice which, when repeated in different directions, generates the entire crystal. In a primitive (simple) unit cell, the constituent particles are present only at the corners of the unit cell. In a centred unit cell, particles are present at positions other than the corners as well, such as body-centred (one at the centre of the body), face-centred (one at the centre of each face) or end-centred.
Marking-scheme points
- ✓Unit cell = smallest repeating unit of a crystal lattice
- ✓Primitive: particles only at the corners
- ✓Centred: extra particles (body-, face- or end-centred)
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The Solid State
Distinguish between Schottky and Frenkel defects.
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Schottky defect is a vacancy defect in which an equal number of cations and anions are missing from their lattice sites, so the density of the solid decreases; it occurs in ionic solids with high coordination number and similar cation and anion sizes (e.g. NaCl, KCl). Frenkel defect is a dislocation defect in which an ion (usually the smaller cation) leaves its lattice site and occupies an interstitial site, so the density remains unchanged; it occurs in solids with a large difference in the sizes of the ions (e.g. AgCl, ZnS).
Marking-scheme points
- ✓Schottky: equal cations and anions missing; density decreases (NaCl)
- ✓Frenkel: smaller ion shifts to an interstitial site; density unchanged (AgCl)
- ✓Both are point defects in ionic solids
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The Solid State
State the packing efficiency and coordination number of simple cubic, bcc and fcc structures.
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Packing efficiency is the percentage of the total space occupied by the particles. Simple cubic: packing efficiency 52.4 percent, coordination number 6. Body-centred cubic (bcc): packing efficiency 68 percent, coordination number 8. Face-centred cubic (fcc, also ccp/hcp): packing efficiency 74 percent (the highest), coordination number 12.
Marking-scheme points
- ✓Simple cubic: 52.4 percent, coordination number 6
- ✓bcc: 68 percent, coordination number 8
- ✓fcc/ccp/hcp: 74 percent (highest), coordination number 12
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Solutions
Define molarity, molality and mole fraction.
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Molarity (M) is the number of moles of solute per litre of solution (mol/L). Molality (m) is the number of moles of solute per kilogram of solvent (mol/kg); it is independent of temperature. Mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all components in the solution; it is dimensionless and the sum of the mole fractions equals 1.
M = n/V(L); m = n/mass of solvent(kg)
Marking-scheme points
- ✓Molarity = moles of solute/litre of solution (mol/L)
- ✓Molality = moles of solute/kg of solvent (mol/kg), temperature independent
- ✓Mole fraction = moles of component/total moles (dimensionless)
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Solutions
Calculate the molality of a solution containing 18 g of glucose (molar mass 180 g/mol) dissolved in 500 g of water.
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Moles of glucose = mass/molar mass = 18/180 = 0.1 mol. Mass of solvent (water) = 500 g = 0.5 kg. Molality = moles of solute/mass of solvent in kg = 0.1/0.5 = 0.2 mol/kg (0.2 m).
molality = moles of solute/mass of solvent (kg)
Marking-scheme points
- ✓Moles of glucose = 18/180 = 0.1 mol
- ✓Mass of water = 0.5 kg
- ✓Molality = 0.1/0.5 = 0.2 m
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Solutions
State Henry's law. Give one application.
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Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the liquid. Mathematically, p = KH x (mole fraction of gas), where KH is Henry's law constant. Applications: it explains why soft drinks (aerated under high pressure) fizz when opened, and why deep-sea divers can suffer from bends (nitrogen dissolving in blood under high pressure).
p = KH x (mole fraction)
Marking-scheme points
- ✓Solubility of a gas is proportional to its partial pressure
- ✓p = KH x (mole fraction of gas)
- ✓Application: aerated drinks, deep-sea diving (bends)
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Solutions
State Raoult's law for a solution of two volatile liquids.
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Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution. For components A and B, pA = pA(pure) x xA and pB = pB(pure) x xB, and the total vapour pressure = pA + pB. For a solution of a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute.
pA = pA(pure) x xA
Marking-scheme points
- ✓Partial pressure of each component is proportional to its mole fraction
- ✓pA = pA(pure) x xA
- ✓Total pressure = pA + pB
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Solutions
What is the Van't Hoff factor? What does its value indicate for association and dissociation?
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The Van't Hoff factor (i) is the ratio of the observed (experimental) value of a colligative property to the calculated (theoretical) value assuming no association or dissociation; equivalently, i = (actual number of particles after association/dissociation)/(number of particles before). For a solute that dissociates (like NaCl), i is greater than 1; for a solute that associates (like acetic acid in benzene), i is less than 1; and for a solute that neither associates nor dissociates, i = 1.
i = observed value/calculated value
Marking-scheme points
- ✓i = observed colligative property/calculated value
- ✓Dissociation: i greater than 1 (e.g. NaCl)
- ✓Association: i less than 1 (e.g. acetic acid in benzene)
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Solutions
Distinguish between ideal and non-ideal solutions.
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An ideal solution is one that obeys Raoult's law over the entire range of concentration; the enthalpy of mixing and the volume change on mixing are zero (e.g. benzene and toluene). A non-ideal solution does not obey Raoult's law; it shows either positive deviation (when A-B interactions are weaker than A-A and B-B, giving higher vapour pressure, e.g. ethanol and water) or negative deviation (when A-B interactions are stronger, giving lower vapour pressure, e.g. chloroform and acetone).
Marking-scheme points
- ✓Ideal: obeys Raoult's law; delta H(mix) = 0 and delta V(mix) = 0 (benzene-toluene)
- ✓Non-ideal positive deviation: higher vapour pressure (ethanol-water)
- ✓Non-ideal negative deviation: lower vapour pressure (chloroform-acetone)
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