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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.

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PhysicsClass 122 marksmedium

Current Electricity

State Kirchhoff's two laws for electrical circuits.

Reveal model answer + marking points

Kirchhoff's junction (current) law states that the algebraic sum of currents meeting at a junction is zero, that is, the total current entering a junction equals the total current leaving it; it is based on conservation of charge. Kirchhoff's loop (voltage) law states that the algebraic sum of the changes in potential around any closed loop of a circuit is zero; it is based on conservation of energy.

sum(I) at junction = 0; sum(V) around loop = 0

Marking-scheme points

  • Junction law: sum of currents at a junction = 0 (charge conservation)
  • Loop law: sum of potential changes around a loop = 0 (energy conservation)
  • Used to analyse complex circuits
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PhysicsClass 123 marksmedium

Current Electricity

Draw (in words) and derive the balance condition of a Wheatstone bridge.

Reveal model answer + marking points

A Wheatstone bridge has four resistances P, Q, R and S arranged in a quadrilateral, with a galvanometer between the junctions of P-Q and R-S, and a cell across the other diagonal. When the bridge is balanced, no current flows through the galvanometer, so the junction connected by the galvanometer are at the same potential. Applying Kirchhoff's laws, the potential drops give P/Q = R/S. This is the balance condition; if three resistances are known, the fourth can be found.

P/Q = R/S

Marking-scheme points

  • Four resistors P, Q, R, S with galvanometer across one diagonal
  • At balance, no current through galvanometer
  • Balance condition: P/Q = R/S
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PhysicsClass 123 marksmedium

Current Electricity

A cell of emf 2 V and internal resistance 0.5 ohm is connected to an external resistance of 4.5 ohm. Find the current in the circuit and the terminal potential difference.

Reveal model answer + marking points

Given emf E = 2 V, internal resistance r = 0.5 ohm, external resistance R = 4.5 ohm. Current I = E/(R + r) = 2/(4.5 + 0.5) = 2/5 = 0.4 A. Terminal potential difference V = I R = 0.4 x 4.5 = 1.8 V (equivalently V = E - I r = 2 - 0.4 x 0.5 = 1.8 V).

I = E/(R + r); V = E - I r

Marking-scheme points

  • I = E/(R + r) = 2/5 = 0.4 A
  • Terminal V = I R = 0.4 x 4.5 = 1.8 V
  • Check: V = E - I r = 1.8 V
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PhysicsClass 122 marksmedium

Current Electricity

Write the effective emf and internal resistance when n identical cells are connected in series.

Reveal model answer + marking points

When n identical cells, each of emf E and internal resistance r, are connected in series (all in the same direction), the effective emf is n E and the total internal resistance is n r. The current through an external resistance R is I = n E/(R + n r). Series grouping is advantageous when the external resistance is much larger than the internal resistance.

I = nE/(R + nr)

Marking-scheme points

  • Series: effective emf = nE, internal resistance = nr
  • Current I = nE/(R + nr)
  • Useful when external R is much greater than internal r
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PhysicsClass 122 marksmedium

Current Electricity

How does the resistance of a metallic conductor vary with temperature? Write the relevant relation.

Reveal model answer + marking points

The resistance of a metallic conductor increases with a rise in temperature, because increased thermal motion of the atoms causes more frequent collisions of electrons, increasing the resistance. The relation is R(t) = R0 (1 + alpha (delta T)), where R0 is the resistance at the reference temperature, alpha is the temperature coefficient of resistance and delta T is the rise in temperature.

R = R0 (1 + alpha delta T)

Marking-scheme points

  • Resistance of a metal increases with temperature
  • More atomic vibration -> more electron collisions
  • R = R0 (1 + alpha delta T)
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PhysicsClass 122 markseasy

Moving Charges and Magnetism

Write the expression for the magnetic Lorentz force on a charge moving in a magnetic field. When is it maximum and when zero?

Reveal model answer + marking points

A charge q moving with velocity v in a magnetic field B experiences a magnetic force F = q v B sin theta, where theta is the angle between v and B; in vector form F = q(v x B). The force is maximum (F = qvB) when the charge moves perpendicular to the field (theta = 90 deg), and it is zero when the charge moves parallel or antiparallel to the field (theta = 0 or 180 deg). The force is always perpendicular to the velocity, so it does no work.

F = q v B sin theta

Marking-scheme points

  • F = q v B sin theta = q(v x B)
  • Maximum (qvB) when v perpendicular to B
  • Zero when v parallel to B; force does no work
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PhysicsClass 123 marksmedium

Moving Charges and Magnetism

A straight conductor of length 0.5 m carrying a current of 4 A is placed perpendicular to a magnetic field of 0.2 T. Calculate the force on the conductor.

Reveal model answer + marking points

The force on a current-carrying conductor is F = B I L sin theta. Here B = 0.2 T, I = 4 A, L = 0.5 m and theta = 90 deg (so sin theta = 1). F = 0.2 x 4 x 0.5 x 1 = 0.4 N. The direction of the force is given by Fleming's left-hand rule.

F = B I L sin theta

Marking-scheme points

  • F = B I L sin theta
  • theta = 90 deg so sin theta = 1
  • F = 0.2 x 4 x 0.5 = 0.4 N
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PhysicsClass 122 marksmedium

Moving Charges and Magnetism

State the Biot-Savart law for the magnetic field due to a current element.

Reveal model answer + marking points

The Biot-Savart law states that the magnetic field dB due to a small current element I dl at a point P at distance r is directly proportional to the current I, the length dl and the sine of the angle theta between the element and the line joining it to P, and inversely proportional to the square of the distance r. Mathematically, dB = (mu0/4 pi) (I dl sin theta)/r^2, and its direction is perpendicular to the plane containing dl and r.

dB = (mu0/4 pi)(I dl sin theta)/r^2

Marking-scheme points

  • dB proportional to I dl sin theta and to 1/r^2
  • dB = (mu0/4 pi)(I dl sin theta)/r^2
  • Direction perpendicular to plane of dl and r
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PhysicsClass 123 marksmedium

Moving Charges and Magnetism

Derive the expression for the magnetic field at the centre of a circular current-carrying loop.

Reveal model answer + marking points

Consider a circular loop of radius R carrying current I. By the Biot-Savart law, each element I dl is perpendicular to the line joining it to the centre (theta = 90 deg), so dB = (mu0/4 pi)(I dl)/R^2. All elements produce fields in the same direction (perpendicular to the plane of the loop) at the centre, so they simply add. Integrating dl around the loop gives the circumference 2 pi R: B = (mu0/4 pi)(I/R^2)(2 pi R) = mu0 I/(2R). For N turns, B = mu0 N I/(2R).

B = mu0 I/(2R)

Marking-scheme points

  • Each element: dB = (mu0/4 pi)(I dl)/R^2 (theta = 90 deg)
  • All contributions add; integral of dl = 2 pi R
  • B = mu0 I/(2R); for N turns, mu0 N I/(2R)
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PhysicsClass 123 marksmedium

Moving Charges and Magnetism

State Ampere's circuital law and write the expression for the magnetic field inside a long solenoid.

Reveal model answer + marking points

Ampere's circuital law states that the line integral of the magnetic field B around any closed loop is equal to mu0 times the total current enclosed by the loop: the integral of B.dl = mu0 I(enclosed). Applying it to a long solenoid with n turns per unit length carrying current I, the magnetic field inside is uniform and given by B = mu0 n I, directed along the axis; the field outside an ideal long solenoid is nearly zero.

B = mu0 n I

Marking-scheme points

  • Integral of B.dl around a closed loop = mu0 I(enclosed)
  • Solenoid field is uniform inside: B = mu0 n I
  • n = number of turns per unit length; field outside is nearly zero
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PhysicsClass 123 marksmedium

Moving Charges and Magnetism

A charged particle moves in a circular path in a uniform magnetic field. Derive the expression for the radius of its path and its time period.

Reveal model answer + marking points

When a charged particle of charge q and mass m enters a magnetic field B perpendicularly with speed v, the magnetic force qvB provides the centripetal force m v^2/r. Equating: q v B = m v^2/r, so the radius r = m v/(q B). The time period T = 2 pi r/v = 2 pi m/(q B), which is independent of the speed and radius. This principle is used in the cyclotron.

r = mv/(qB); T = 2 pi m/(qB)

Marking-scheme points

  • Magnetic force provides centripetal force: qvB = m v^2/r
  • Radius r = m v/(q B)
  • Time period T = 2 pi m/(q B), independent of speed
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PhysicsClass 122 marksmedium

Moving Charges and Magnetism

Write the expression for the force per unit length between two long parallel current-carrying wires and use it to define the ampere.

Reveal model answer + marking points

Two long parallel wires separated by distance d carrying currents I1 and I2 experience a force per unit length F/L = (mu0 I1 I2)/(2 pi d); the force is attractive if the currents are in the same direction and repulsive if opposite. The ampere is defined as that steady current which, when maintained in two infinitely long parallel wires of negligible cross-section placed 1 metre apart in vacuum, produces a force of 2 x 10^-7 newton per metre of length between them.

F/L = mu0 I1 I2/(2 pi d)

Marking-scheme points

  • F/L = mu0 I1 I2/(2 pi d)
  • Same direction currents attract, opposite repel
  • 1 ampere gives 2 x 10^-7 N/m between wires 1 m apart
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PhysicsClass 123 marksmedium

Moving Charges and Magnetism

Explain the principle of a moving coil galvanometer. How is it converted into an ammeter and a voltmeter?

Reveal model answer + marking points

A moving coil galvanometer works on the principle that a current-carrying coil placed in a magnetic field experiences a torque. The deflection of the coil is directly proportional to the current passing through it (I is proportional to the deflection). To convert a galvanometer into an ammeter, a small resistance (shunt) is connected in parallel with it, so most of the current bypasses the galvanometer. To convert it into a voltmeter, a high resistance is connected in series with it, so that only a small current flows and it can measure potential difference.

Marking-scheme points

  • Principle: current-carrying coil in a field experiences a torque; deflection proportional to current
  • Ammeter: connect a low resistance (shunt) in parallel
  • Voltmeter: connect a high resistance in series
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PhysicsClass 122 markseasy

Magnetism and Matter

Write the expression for the magnetic dipole moment of a current-carrying loop. State its SI unit.

Reveal model answer + marking points

A current loop behaves as a magnetic dipole. The magnetic dipole moment of a coil of N turns each of area A carrying current I is m = N I A, and it is directed perpendicular to the plane of the loop (given by the right-hand rule). Its SI unit is ampere metre squared (A m^2). The torque on it in a field B is m x B.

m = N I A

Marking-scheme points

  • Magnetic moment m = N I A
  • Directed perpendicular to the plane of the loop
  • SI unit: ampere metre squared (A m^2)
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PhysicsClass 122 marksmedium

Magnetism and Matter

Distinguish between diamagnetic, paramagnetic and ferromagnetic substances with one example each.

Reveal model answer + marking points

Diamagnetic substances are weakly repelled by a magnetic field and move from stronger to weaker regions; they have a small negative susceptibility (e.g. bismuth, copper). Paramagnetic substances are weakly attracted by a magnetic field and have a small positive susceptibility (e.g. aluminium, sodium). Ferromagnetic substances are strongly attracted and can be permanently magnetised; they have a large positive susceptibility (e.g. iron, cobalt, nickel).

Marking-scheme points

  • Diamagnetic: weakly repelled, small negative susceptibility (bismuth)
  • Paramagnetic: weakly attracted, small positive susceptibility (aluminium)
  • Ferromagnetic: strongly attracted, large positive susceptibility (iron)
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PhysicsClass 122 markseasy

Magnetism and Matter

State any four properties of magnetic field lines.

Reveal model answer + marking points

(1) Magnetic field lines are continuous closed curves that pass from the south to the north pole inside the magnet and from the north to the south pole outside it. (2) The tangent drawn at any point on a field line gives the direction of the magnetic field at that point. (3) Two field lines never intersect each other (as the field can have only one direction at a point). (4) The lines are crowded where the field is strong and spread apart where it is weak.

Marking-scheme points

  • Continuous closed loops (S to N inside, N to S outside)
  • Tangent gives field direction; no two lines intersect
  • Crowded where field is strong
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PhysicsClass 123 marksmedium

Magnetism and Matter

Name and define the three elements of the earth's magnetic field.

Reveal model answer + marking points

The three elements of the earth's magnetism are: (1) Magnetic declination - the angle between the geographic meridian and the magnetic meridian at a place. (2) Magnetic dip or inclination - the angle made by the earth's total magnetic field with the horizontal at a place. (3) Horizontal component of the earth's field (BH) - the component of the earth's total magnetic field in the horizontal direction, given by BH = B cos(dip). These three completely specify the earth's field at a place.

BH = B cos(dip)

Marking-scheme points

  • Declination: angle between geographic and magnetic meridians
  • Dip/inclination: angle of total field with the horizontal
  • Horizontal component BH = B cos(dip)
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PhysicsClass 122 marksmedium

Magnetism and Matter

Define magnetic susceptibility and relative permeability. Write the relation between them.

Reveal model answer + marking points

Magnetic susceptibility (chi) is the ratio of the intensity of magnetisation (M) produced in a material to the magnetising field (H): chi = M/H; it measures how easily a material can be magnetised. Relative permeability (mu_r) is the ratio of the permeability of the material to that of free space. The relation between them is mu_r = 1 + chi.

mu_r = 1 + chi

Marking-scheme points

  • Susceptibility chi = M/H (ease of magnetisation)
  • Relative permeability mu_r = mu/mu0
  • Relation: mu_r = 1 + chi
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PhysicsClass 122 markseasy

Electromagnetic Induction

State Faraday's laws of electromagnetic induction.

Reveal model answer + marking points

Faraday's first law states that whenever the magnetic flux linked with a closed circuit changes, an emf is induced in the circuit, and it lasts as long as the flux is changing. Faraday's second law states that the magnitude of the induced emf is equal to the rate of change of magnetic flux linked with the circuit: e = -N (d(flux)/dt), where N is the number of turns. The negative sign is due to Lenz's law.

e = -N d(flux)/dt

Marking-scheme points

  • Changing magnetic flux induces an emf
  • Induced emf = rate of change of flux: e = -N d(flux)/dt
  • Negative sign from Lenz's law
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PhysicsClass 122 marksmedium

Electromagnetic Induction

State Lenz's law. Which conservation principle does it represent?

Reveal model answer + marking points

Lenz's law states that the direction of the induced current (or emf) is always such that it opposes the change in magnetic flux that produces it. For example, if a magnet is pushed towards a coil, the induced current opposes its approach. Lenz's law is a consequence of the law of conservation of energy, because work has to be done against the opposing force, and this work appears as electrical energy.

Marking-scheme points

  • Induced current opposes the change in flux causing it
  • Gives the direction of the induced current
  • Consequence of conservation of energy
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