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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 marksmedium
Gravitation
Write the expression for the gravitational potential energy of a mass m at a distance r from the centre of the Earth (mass M), and explain why it is negative.
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U = - G M m / r. It is negative because the gravitational force is attractive: zero potential energy is taken at infinity, and as the mass is brought closer, work is done by gravity, lowering the energy below zero. The negative sign shows the mass is in a bound state.
U = - G M m / r
Marking-scheme points
- ✓U = - G M m / r
- ✓Reference: U = 0 at infinity
- ✓Negative => attractive, bound system
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Units and Measurements
State any two uses and two limitations of dimensional analysis.
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Uses: (i) to check the dimensional consistency of an equation; (ii) to convert a quantity from one system of units to another; (iii) to derive the relation between physical quantities. Limitations: (i) it cannot find dimensionless constants (like 1/2 or 2 pi); (ii) it cannot be used if a quantity depends on more than three others, or on trigonometric/exponential/logarithmic functions.
Marking-scheme points
- ✓Uses: check equations, convert units, derive relations
- ✓Limits: cannot give numerical constants
- ✓Fails for trig/exp/log functions
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Laws of Motion
Why are roads banked at curves? Write the expression for the ideal speed on a frictionless banked road of angle theta and radius r.
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Roads are banked at curves so that the horizontal component of the normal reaction provides the centripetal force needed to turn, reducing reliance on friction and the risk of skidding. On a frictionless banked road, the ideal (safe) speed is v = sqrt(r g tan(theta)).
v = sqrt(r g tan(theta))
Marking-scheme points
- ✓Banking supplies centripetal force via normal reaction
- ✓Reduces dependence on friction
- ✓v = sqrt(r g tan theta)
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Work, Energy and Power
Define the coefficient of restitution. What are its values for a perfectly elastic and a perfectly inelastic collision?
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The coefficient of restitution (e) is the ratio of the relative velocity of separation after collision to the relative velocity of approach before collision: e = (velocity of separation) / (velocity of approach). For a perfectly elastic collision e = 1; for a perfectly inelastic collision e = 0.
e = (v2 - v1) / (u1 - u2)
Marking-scheme points
- ✓e = separation velocity / approach velocity
- ✓Perfectly elastic: e = 1
- ✓Perfectly inelastic: e = 0
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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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Mechanical Properties of Fluids
Write the expression for the terminal velocity of a small sphere falling through a viscous fluid and name the quantities.
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v_t = (2 r^2 (rho - sigma) g) / (9 eta), where r is the radius of the sphere, rho its density, sigma the density of the fluid, g the acceleration due to gravity and eta the coefficient of viscosity. It is obtained by balancing the weight against the buoyant force and the viscous (Stokes) drag.
v_t = 2 r^2 (rho - sigma) g / (9 eta)
Marking-scheme points
- ✓v_t = 2 r^2 (rho - sigma) g / (9 eta)
- ✓From weight = buoyancy + viscous drag
- ✓v_t proportional to r^2
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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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Thermodynamics
What is a refrigerator in thermodynamic terms? Write the expression for its coefficient of performance.
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A refrigerator is a heat engine working in reverse: it uses external work W to extract heat Q_c from a cold body and reject a larger heat Q_h to the hot surroundings, so Q_h = Q_c + W. Its coefficient of performance is beta = Q_c / W = Q_c / (Q_h - Q_c). A good refrigerator has a high beta.
beta = Q_c / (Q_h - Q_c)
Marking-scheme points
- ✓Reverse heat engine, uses work W
- ✓Q_h = Q_c + W
- ✓beta = Q_c / W = Q_c/(Q_h - Q_c)
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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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Oscillations
For a particle in SHM given by x = A sin(omega t), write the expressions for its velocity and acceleration, and state where each is maximum.
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Velocity v = dx/dt = A omega cos(omega t), with magnitude v = omega sqrt(A^2 - x^2); it is maximum (A omega) at the mean position (x = 0) and zero at the extremes. Acceleration a = dv/dt = - A omega^2 sin(omega t) = - omega^2 x; its magnitude is maximum (A omega^2) at the extreme positions (x = +/- A) and zero at the mean position.
v = omega sqrt(A^2 - x^2) ; a = - omega^2 x
Marking-scheme points
- ✓v = A omega cos(omega t), max at mean position
- ✓a = - omega^2 x, max at extremes
- ✓|v| = omega sqrt(A^2 - x^2)
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Waves
Write Newton's formula for the speed of sound in a gas and state Laplace's correction to it.
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Newton assumed sound propagation in a gas is isothermal, giving v = sqrt(P / rho), where P is pressure and rho density; this gave a value about 15% lower than the measured value. Laplace corrected this by treating the process as adiabatic (compressions and rarefactions are too rapid for heat exchange), giving v = sqrt(gamma P / rho), where gamma is the ratio of specific heats; this matches experiment.
v = sqrt(gamma P / rho)
Marking-scheme points
- ✓Newton (isothermal): v = sqrt(P/rho) - too low
- ✓Laplace (adiabatic): v = sqrt(gamma P/rho)
- ✓Correction factor sqrt(gamma)
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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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Laws of Motion
A person of mass 50 kg stands in a lift. Find the apparent weight when the lift accelerates upward at 2 m/s^2. Take g = 10 m/s^2.
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When the lift accelerates upward, the apparent weight (normal reaction) is N = m (g + a) = 50 (10 + 2) = 50 * 12 = 600 N. (At rest it would be m g = 500 N, so the person feels heavier.)
N = m (g + a)
Marking-scheme points
- ✓Upward acceleration: N = m(g + a)
- ✓= 50 * 12
- ✓N = 600 N (feels heavier)
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Work, Energy and Power
Write the condition (minimum speed) for a body to just complete a vertical circle of radius r at the highest point, and the corresponding minimum speed at the lowest point.
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At the highest point, gravity alone must provide the centripetal force, so m g = m v_top^2 / r, giving the minimum speed at the top v_top = sqrt(g r). Using energy conservation between the lowest and highest points, the minimum speed at the bottom is v_bottom = sqrt(5 g r).
v_top = sqrt(g r) ; v_bottom = sqrt(5 g r)
Marking-scheme points
- ✓At top: mg = m v^2/r => v_top = sqrt(g r)
- ✓Energy conservation over height 2r
- ✓v_bottom = sqrt(5 g r)
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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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Waves
Compare the harmonics produced in an open organ pipe and a closed organ pipe.
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In an open pipe (open at both ends) both even and odd harmonics are present; the fundamental is f1 = v/(2L) and the harmonics are f1, 2 f1, 3 f1, ... In a closed pipe (closed at one end) only odd harmonics are present; the fundamental is f1 = v/(4L) and the harmonics are f1, 3 f1, 5 f1, ... So an open pipe of the same length gives a higher fundamental and a richer set of harmonics.
Open: fn = n v/2L ; Closed: fn = (2n-1) v/4L
Marking-scheme points
- ✓Open pipe: all harmonics, f1 = v/2L
- ✓Closed pipe: only odd harmonics, f1 = v/4L
- ✓Closed-pipe fundamental is half that of the open pipe
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Motion in a Straight Line
A body starts from rest and moves with a uniform acceleration of 2 m/s^2. Find the distance travelled by it during the 5th second.
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Distance in the nth second: s_n = u + (a/2)(2n - 1). With u = 0, a = 2, n = 5: s_5 = 0 + (2/2)(2*5 - 1) = 1 * 9 = 9 m.
s_n = u + (a/2)(2n - 1)
Marking-scheme points
- ✓s_n = u + (a/2)(2n - 1)
- ✓u = 0, a = 2, n = 5
- ✓s_5 = 9 m
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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
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