Class 12 Physics — Important Board Questions with Answers

Everything the Class 12 Physics (Plus Two) board paper tends to ask, in one place — 100 most-asked questions across 14 chapters, each with a model answer and the exact marking-scheme points examiners reward. Revise chapter by chapter, and walk in sure of yourself.

100 questions+2 · Plus Two14 chaptersModel answersCBSE · ISC · State boards

Electric Charges and Fields8 questions

2 markseasyCoulomb's law

State Coulomb's law of electrostatics and write its mathematical form.

Reveal model answer + marking points

Coulomb's law states that the force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, acting along the line joining them. Mathematically, F = k q1 q2 / r^2, where k = 1/(4 pi epsilon0) = 9 x 10^9 N m^2 C^-2 in free space.

F = k q1 q2 / r^2

Marking-scheme points

  • F is proportional to product of charges and to 1/r^2
  • F = k q1 q2 / r^2 along the line joining them
  • k = 1/(4 pi epsilon0) = 9 x 10^9 N m^2 C^-2
3 marksmediumCoulomb's law calculation

Two point charges of 2 microcoulomb and 3 microcoulomb are placed 30 cm apart in air. Calculate the electrostatic force between them.

Reveal model answer + marking points

Given q1 = 2 x 10^-6 C, q2 = 3 x 10^-6 C, r = 30 cm = 0.3 m. Using F = k q1 q2 / r^2 = (9 x 10^9)(2 x 10^-6)(3 x 10^-6)/(0.3)^2 = (9 x 10^9 x 6 x 10^-12)/0.09 = 0.054/0.09 = 0.6 N. The force is repulsive since both charges are positive.

F = k q1 q2 / r^2

Marking-scheme points

  • Convert r = 30 cm = 0.3 m
  • F = (9e9 x 2e-6 x 3e-6)/(0.3)^2
  • F = 0.6 N (repulsive)
2 markseasyElectric field intensity

Define electric field intensity at a point. State its SI unit.

Reveal model answer + marking points

The electric field intensity at a point is the force experienced by a unit positive test charge placed at that point. It is a vector quantity given by E = F/q0, where q0 is the small positive test charge. Its SI unit is newton per coulomb (N/C) or equivalently volt per metre (V/m).

E = F/q0

Marking-scheme points

  • E = force per unit positive test charge = F/q0
  • Vector quantity, directed along the force on a positive charge
  • SI unit: N/C or V/m
3 marksmediumField due to a point charge

Calculate the electric field intensity at a point 30 cm from a point charge of 5 microcoulomb in air.

Reveal model answer + marking points

Given Q = 5 x 10^-6 C, r = 0.3 m. The field due to a point charge is E = kQ/r^2 = (9 x 10^9)(5 x 10^-6)/(0.3)^2 = (45000)/0.09 = 5 x 10^5 N/C. The field is directed radially outward since the charge is positive.

E = kQ/r^2

Marking-scheme points

  • E = kQ/r^2
  • = (9e9 x 5e-6)/(0.3)^2 = 45000/0.09
  • E = 5 x 10^5 N/C, directed radially outward
3 marksmediumElectric dipole

Define electric dipole moment. Derive the expression for the electric field at a point on the axial line of a short dipole.

Reveal model answer + marking points

An electric dipole is a pair of equal and opposite charges (+q and -q) separated by a small distance 2a. The dipole moment p = q x 2a, directed from the negative to the positive charge. On the axial line at distance r from the centre, the field due to +q is k q/(r-a)^2 (away) and due to -q is k q/(r+a)^2 (towards). Net E = kq[1/(r-a)^2 - 1/(r+a)^2] = kq[(4ar)/((r^2-a^2)^2)]. For a short dipole (r much greater than a), E = k(2p)/r^3, directed along the dipole moment.

E(axial) = 2kp/r^3

Marking-scheme points

  • Dipole moment p = q x 2a (from -q to +q)
  • Axial field = kq[1/(r-a)^2 - 1/(r+a)^2]
  • For short dipole: E(axial) = 2kp/r^3
2 marksmediumTorque on a dipole

Derive the expression for the torque acting on an electric dipole placed in a uniform electric field.

Reveal model answer + marking points

When a dipole of moment p is placed in a uniform electric field E at an angle theta, the two charges experience equal and opposite forces qE, forming a couple. The magnitude of the torque = force x perpendicular distance = qE x (2a sin theta) = (q x 2a) E sin theta = pE sin theta. In vector form, torque = p x E. The torque tends to align the dipole with the field.

torque = pE sin theta = p x E

Marking-scheme points

  • Equal and opposite forces qE form a couple
  • Torque = pE sin theta (magnitude)
  • Vector form: torque = p x E; aligns dipole with field
3 marksmediumGauss's law

State Gauss's law in electrostatics and write its mathematical form.

Reveal model answer + marking points

Gauss's law states that the total electric flux through any closed surface is equal to 1/epsilon0 times the total charge enclosed by that surface. Mathematically, the surface integral of E over the closed surface = q(enclosed)/epsilon0. The closed surface over which the flux is calculated is called a Gaussian surface. The law is useful for finding the electric field of symmetric charge distributions.

flux = q(enclosed)/epsilon0

Marking-scheme points

  • Total electric flux through a closed surface = q(enclosed)/epsilon0
  • Integral of E.dA over closed surface = q/epsilon0
  • Used for symmetric charge distributions (Gaussian surface)
3 markshardField of a line charge

Using Gauss's law, derive the expression for the electric field due to an infinitely long straight uniformly charged wire.

Reveal model answer + marking points

Consider an infinitely long straight wire with linear charge density lambda. Choose a coaxial cylindrical Gaussian surface of radius r and length L. By symmetry, the field E is radial and uniform over the curved surface. The flux through the curved surface = E x (2 pi r L); the flat ends contribute no flux. Charge enclosed = lambda L. By Gauss's law, E x 2 pi r L = lambda L / epsilon0, so E = lambda/(2 pi epsilon0 r). The field is directed radially and varies as 1/r.

E = lambda/(2 pi epsilon0 r)

Marking-scheme points

  • Use a coaxial cylindrical Gaussian surface
  • Flux = E x 2 pi r L; charge enclosed = lambda L
  • E = lambda/(2 pi epsilon0 r), directed radially

Electrostatic Potential and Capacitance8 questions

2 markseasyElectric potential

Define electric potential at a point. State its SI unit and write the expression for the potential due to a point charge.

Reveal model answer + marking points

The electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field. It is a scalar quantity and its SI unit is the volt (V), where 1 volt = 1 joule per coulomb. The potential due to a point charge Q at distance r is V = kQ/r = Q/(4 pi epsilon0 r).

V = kQ/r

Marking-scheme points

  • Work done per unit positive charge from infinity to the point
  • Scalar quantity; SI unit volt (1 V = 1 J/C)
  • V = kQ/r for a point charge
3 marksmediumPotential due to a point charge

Calculate the electric potential at a point 9 cm away from a point charge of 4 microcoulomb in air.

Reveal model answer + marking points

Given Q = 4 x 10^-6 C, r = 9 cm = 0.09 m. The potential due to a point charge is V = kQ/r = (9 x 10^9)(4 x 10^-6)/0.09 = (36000)/0.09 = 4 x 10^5 V. Thus the potential at the point is 4 x 10^5 volt.

V = kQ/r

Marking-scheme points

  • V = kQ/r (note r, not r^2, for potential)
  • = (9e9 x 4e-6)/0.09 = 36000/0.09
  • V = 4 x 10^5 V
2 marksmediumRelation between E and V

Write the relation between electric field and electric potential. What does the negative sign indicate?

Reveal model answer + marking points

The electric field is the negative gradient of the electric potential: E = -dV/dr. This means the field points in the direction in which the potential decreases most rapidly. The negative sign indicates that the electric field is directed from a region of higher potential to a region of lower potential, that is, potential decreases along the direction of the field.

E = -dV/dr

Marking-scheme points

  • E = -dV/dr (field = negative potential gradient)
  • Field points towards decreasing potential
  • Negative sign: E directed from high to low potential
2 markseasyCapacitance

Define capacitance of a conductor. State its SI unit.

Reveal model answer + marking points

The capacitance of a conductor is the ratio of the charge given to it to the resulting rise in its potential: C = Q/V. It is a measure of the ability of the conductor to store charge. Its SI unit is the farad (F), where 1 farad = 1 coulomb per volt. Capacitance depends on the size and shape of the conductor and the surrounding medium.

C = Q/V

Marking-scheme points

  • C = Q/V (charge stored per unit potential)
  • SI unit: farad (1 F = 1 C/V)
  • Depends on size, shape and surrounding medium
3 marksmediumParallel plate capacitor

Derive the expression for the capacitance of a parallel plate capacitor with air between the plates.

Reveal model answer + marking points

Let a parallel plate capacitor have plate area A, separation d, and charge Q with surface charge density sigma = Q/A. The uniform field between the plates is E = sigma/epsilon0 = Q/(A epsilon0). The potential difference between the plates is V = E d = Q d/(A epsilon0). Hence capacitance C = Q/V = Q/(Q d/(A epsilon0)) = epsilon0 A/d. So the capacitance increases with plate area and decreases with plate separation.

C = epsilon0 A/d

Marking-scheme points

  • Field between plates E = sigma/epsilon0 = Q/(A epsilon0)
  • V = E d = Q d/(A epsilon0)
  • C = Q/V = epsilon0 A/d
3 marksmediumCombination of capacitors

Three capacitors of 2 microfarad, 3 microfarad and 6 microfarad are connected first in series and then in parallel. Find the equivalent capacitance in each case.

Reveal model answer + marking points

In series: 1/Cs = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so Cs = 1 microfarad. In parallel: Cp = 2 + 3 + 6 = 11 microfarad. Thus the series combination gives 1 microfarad and the parallel combination gives 11 microfarad.

series: 1/Cs = sum(1/Ci); parallel: Cp = sum(Ci)

Marking-scheme points

  • Series: 1/Cs = 1/2 + 1/3 + 1/6 = 1 -> Cs = 1 microfarad
  • Parallel: Cp = 2 + 3 + 6 = 11 microfarad
  • Series capacitance is the smallest, parallel the largest
2 marksmediumEnergy stored in a capacitor

Write the expression for the energy stored in a charged capacitor in three equivalent forms.

Reveal model answer + marking points

The energy stored in a charged capacitor is the work done in charging it. It can be written in three equivalent forms: U = (1/2) C V^2 = (1/2) Q V = Q^2/(2C), where C is the capacitance, Q the charge and V the potential difference. This energy is stored in the electric field between the plates.

U = (1/2) C V^2 = (1/2) QV = Q^2/(2C)

Marking-scheme points

  • U = (1/2) C V^2
  • U = (1/2) Q V = Q^2/(2C)
  • Energy stored in the electric field between plates
2 marksmediumEffect of dielectric

How does the introduction of a dielectric slab between the plates of a capacitor affect its capacitance? Explain.

Reveal model answer + marking points

When a dielectric of dielectric constant K is fully inserted between the plates, the capacitance increases K times: C = K C0, where C0 is the capacitance with air. This is because the dielectric gets polarised and sets up an internal field opposite to the applied field, reducing the net field and hence the potential difference for the same charge; since C = Q/V, a smaller V means a larger C.

C = K C0

Marking-scheme points

  • Dielectric increases capacitance: C = K C0
  • Dielectric polarises and reduces the net field
  • Lower V for same Q -> higher C

Current Electricity9 questions

2 markseasyOhm's law

State Ohm's law. Define resistance and give its SI unit.

Reveal model answer + marking points

Ohm's law states that, at constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends, so V = IR, where R is a constant called the resistance. Resistance is the opposition offered by a conductor to the flow of current and is defined as R = V/I. Its SI unit is the ohm.

V = IR

Marking-scheme points

  • At constant temperature, V is proportional to I (V = IR)
  • Resistance R = V/I = opposition to current
  • SI unit: ohm
2 marksmediumResistivity

Define resistivity of a material. How does the resistance of a wire depend on its length and area of cross-section?

Reveal model answer + marking points

Resistivity (specific resistance) is the resistance of a conductor of unit length and unit area of cross-section; it depends on the material and temperature but not on its dimensions. The resistance of a wire is R = rho L/A, so it is directly proportional to its length L and inversely proportional to its area of cross-section A. The SI unit of resistivity is the ohm metre.

R = rho L/A

Marking-scheme points

  • Resistivity = resistance of unit length and unit area
  • R = rho L/A
  • R is proportional to L and inversely proportional to A
3 marksmediumDrift velocity

What is drift velocity? Derive the relation between current and drift velocity.

Reveal model answer + marking points

Drift velocity is the small average velocity with which free electrons move in a conductor under an applied electric field, opposite to the field. Consider a conductor of area A with n free electrons per unit volume moving with drift velocity vd. In time t, electrons in a length vd t cross a section, so charge crossing = (n)(A vd t)(e). Current I = charge/time = n A vd e t / t = n A e vd. Hence I = n A e vd.

I = n A e vd

Marking-scheme points

  • Drift velocity = average velocity of electrons under the field
  • Charge crossing in time t = n A vd t e
  • I = n A e vd
3 marksmediumCombination of resistances

Three resistors of 2 ohm, 3 ohm and 6 ohm are connected in parallel. Find their equivalent resistance. What would it be in series?

Reveal model answer + marking points

In parallel: 1/Rp = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so Rp = 1 ohm. In series: Rs = 2 + 3 + 6 = 11 ohm. Thus the parallel combination gives 1 ohm (less than the smallest resistance) and the series combination gives 11 ohm.

parallel: 1/Rp = sum(1/Ri); series: Rs = sum(Ri)

Marking-scheme points

  • Parallel: 1/Rp = 1/2 + 1/3 + 1/6 = 1 -> Rp = 1 ohm
  • Series: Rs = 2 + 3 + 6 = 11 ohm
  • Parallel resistance is less than the smallest resistor
2 marksmediumKirchhoff's laws

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
3 marksmediumWheatstone bridge

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
3 marksmediumEMF and internal resistance

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
2 marksmediumCombination of cells

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
2 marksmediumTemperature dependence of resistance

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)

Moving Charges and Magnetism8 questions

2 markseasyLorentz force

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
3 marksmediumForce on a conductor

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
2 marksmediumBiot-Savart law

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
3 marksmediumField at centre of a loop

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)
3 marksmediumAmpere's law and solenoid

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
3 marksmediumCharged particle in a magnetic field

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
2 marksmediumForce between parallel wires

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
3 marksmediumMoving coil galvanometer

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

Magnetism and Matter5 questions

2 markseasyMagnetic dipole moment

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)
2 marksmediumMagnetic materials

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)
2 markseasyMagnetic field lines

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
3 marksmediumElements of earth's magnetism

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)
2 marksmediumSusceptibility and permeability

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

Electromagnetic Induction7 questions

2 markseasyFaraday's laws

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
2 marksmediumLenz's law

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
3 marksmediumMotional emf

Derive the expression for the motional emf induced in a conducting rod moving in a uniform magnetic field.

Reveal model answer + marking points

Consider a conducting rod of length l moving with velocity v perpendicular to a uniform magnetic field B. In time dt the rod sweeps an area dA = l (v dt), so the change in flux is d(flux) = B dA = B l v dt. By Faraday's law, the induced emf e = d(flux)/dt = B l v. Alternatively, the free electrons in the rod experience a force qvB that pushes them to one end, setting up an emf e = B v l across the rod.

e = B l v

Marking-scheme points

  • Area swept in dt = l v dt, so d(flux) = B l v dt
  • e = d(flux)/dt = B l v
  • Also from force on electrons qvB
2 marksmediumSelf-inductance

Define self-inductance of a coil. State its SI unit.

Reveal model answer + marking points

Self-inductance is the property of a coil by virtue of which it opposes any change in the current flowing through it, by inducing an opposing emf (back emf). It is defined as the flux linkage per unit current (N flux = L I) or from the induced emf e = -L (dI/dt), where L is the self-inductance. Its SI unit is the henry (H).

e = -L dI/dt

Marking-scheme points

  • Coil opposes change in its own current (back emf)
  • N flux = L I, or e = -L dI/dt
  • SI unit: henry (H)
2 marksmediumMutual inductance

Define mutual inductance between two coils. On what factors does it depend?

Reveal model answer + marking points

Mutual inductance is the property by which a change of current in one coil (primary) induces an emf in a neighbouring coil (secondary) due to the change in flux linkage. It is defined by e2 = -M (dI1/dt), where M is the mutual inductance, whose SI unit is the henry. It depends on the number of turns of the coils, their geometry (area and length), the distance and orientation between them, and the permeability of the core material.

e2 = -M dI1/dt

Marking-scheme points

  • Change of current in one coil induces emf in another
  • e2 = -M dI1/dt (SI unit henry)
  • Depends on turns, geometry, separation and core material
3 marksmediumInduced emf calculation

A coil of 500 turns has the magnetic flux through it changing from 0.01 Wb to 0.05 Wb in 0.1 s. Calculate the induced emf.

Reveal model answer + marking points

Given N = 500, initial flux = 0.01 Wb, final flux = 0.05 Wb, so change in flux = 0.05 - 0.01 = 0.04 Wb, and time = 0.1 s. The induced emf (magnitude) = N (change in flux)/time = 500 x 0.04/0.1 = 500 x 0.4 = 200 V.

e = N (change in flux)/time

Marking-scheme points

  • Change in flux = 0.05 - 0.01 = 0.04 Wb
  • e = N (change in flux)/time = 500 x 0.04/0.1
  • e = 200 V
2 marksmediumEddy currents

What are eddy currents? State two applications.

Reveal model answer + marking points

Eddy currents are the circulating induced currents produced in the body of a conductor when the magnetic flux linked with it changes. They flow in closed loops within the conductor and generally cause heating and energy loss. Applications: they are used in induction furnaces (to melt metals by the heat produced), in electromagnetic braking of trains, in electric (induction) meters, and in induction cooktops. Laminating the cores of transformers reduces energy loss due to eddy currents.

Marking-scheme points

  • Circulating induced currents in a conductor due to changing flux
  • Cause heating and energy loss
  • Applications: induction furnace, electromagnetic braking, induction cooktop

Alternating Current7 questions

2 markseasyRMS value

Define the root mean square (RMS) value of alternating current. Write its relation with the peak value.

Reveal model answer + marking points

The RMS (root mean square) value of an alternating current is that value of steady direct current which produces the same heating effect in a given resistance in the same time as the alternating current does. For a sinusoidal current of peak value I0, the RMS value is Irms = I0/sqrt(2) = 0.707 I0. Similarly Vrms = V0/sqrt(2). AC meters read RMS values.

Irms = I0/sqrt(2)

Marking-scheme points

  • RMS = equivalent DC giving the same heating effect
  • Irms = I0/sqrt(2) = 0.707 I0
  • Vrms = V0/sqrt(2); AC meters read RMS
2 marksmediumReactance

Define inductive reactance and capacitive reactance. Write their expressions.

Reveal model answer + marking points

Inductive reactance is the opposition offered by an inductor to the flow of alternating current, given by XL = omega L = 2 pi f L; it increases with frequency. Capacitive reactance is the opposition offered by a capacitor to alternating current, given by XC = 1/(omega C) = 1/(2 pi f C); it decreases with frequency. Both are measured in ohm. For direct current (f = 0), XL = 0 and XC is infinite.

XL = omega L; XC = 1/(omega C)

Marking-scheme points

  • Inductive reactance XL = omega L = 2 pi f L (increases with f)
  • Capacitive reactance XC = 1/(omega C) (decreases with f)
  • Both measured in ohm
3 marksmediumImpedance of LCR circuit

Write the expression for the impedance of a series LCR circuit and the phase angle between voltage and current.

Reveal model answer + marking points

In a series LCR circuit connected to an AC source, the total opposition to current is called impedance Z, given by Z = sqrt(R^2 + (XL - XC)^2), where XL is the inductive reactance and XC is the capacitive reactance. The phase angle phi between the applied voltage and the current is given by tan(phi) = (XL - XC)/R. The current is I = V/Z.

Z = sqrt(R^2 + (XL - XC)^2)

Marking-scheme points

  • Z = sqrt(R^2 + (XL - XC)^2)
  • Phase angle: tan(phi) = (XL - XC)/R
  • Current I = V/Z
3 marksmediumResonance in LCR circuit

What is resonance in a series LCR circuit? Derive the expression for the resonant frequency.

Reveal model answer + marking points

Resonance in a series LCR circuit occurs when the inductive reactance equals the capacitive reactance (XL = XC). At resonance the impedance is minimum (Z = R), the current is maximum, and the circuit behaves as purely resistive. The condition XL = XC gives omega L = 1/(omega C), so omega^2 = 1/(LC), and the resonant frequency f = 1/(2 pi sqrt(LC)). Such a circuit is used for tuning radios and televisions.

f = 1/(2 pi sqrt(LC))

Marking-scheme points

  • Resonance when XL = XC (impedance minimum = R, current maximum)
  • omega L = 1/(omega C) -> omega = 1/sqrt(LC)
  • Resonant frequency f = 1/(2 pi sqrt(LC))
2 marksmediumPower factor

Define power factor of an AC circuit. What is wattless current?

Reveal model answer + marking points

The average power in an AC circuit is P = Vrms Irms cos(phi), where cos(phi) is called the power factor and phi is the phase difference between voltage and current. Thus the power factor is the ratio of true power to apparent power (cos phi = R/Z). Wattless current is the component of the AC current (Irms sin phi) that is 90 degrees out of phase with the voltage; it consumes no average power, so it is called the idle or wattless current.

P = Vrms Irms cos(phi)

Marking-scheme points

  • Power P = Vrms Irms cos(phi); cos(phi) = power factor = R/Z
  • Ratio of true power to apparent power
  • Wattless current (Irms sin phi) consumes no average power
3 marksmediumTransformer

State the principle of a transformer and write the transformer equation. Distinguish step-up and step-down transformers.

Reveal model answer + marking points

A transformer works on the principle of mutual induction: an alternating current in the primary coil produces a changing flux that induces an emf in the secondary coil. For an ideal transformer, Vs/Vp = Ns/Np = Ip/Is, where V is voltage, N is number of turns and I is current. A step-up transformer has more turns in the secondary (Ns > Np) and increases the voltage; a step-down transformer has fewer turns in the secondary (Ns < Np) and decreases the voltage.

Vs/Vp = Ns/Np = Ip/Is

Marking-scheme points

  • Works on mutual induction (needs AC)
  • Vs/Vp = Ns/Np = Ip/Is (ideal transformer)
  • Step-up: Ns > Np raises voltage; step-down: Ns < Np lowers it
3 marksmediumTransformer calculation

A step-down transformer converts 2200 V to 220 V. If the primary has 5000 turns, find the number of turns in the secondary.

Reveal model answer + marking points

For an ideal transformer, Ns/Np = Vs/Vp. Given Vp = 2200 V, Vs = 220 V, Np = 5000. So Ns = Np x (Vs/Vp) = 5000 x (220/2200) = 5000 x 0.1 = 500 turns. Since the secondary has fewer turns than the primary, it is a step-down transformer, as expected.

Ns/Np = Vs/Vp

Marking-scheme points

  • Ns/Np = Vs/Vp
  • Ns = 5000 x (220/2200) = 5000 x 0.1
  • Ns = 500 turns

Electromagnetic Waves4 questions

2 marksmediumDisplacement current

What is displacement current? How did it complete Ampere's law?

Reveal model answer + marking points

Displacement current is the current that arises due to a changing electric field (or changing electric flux) between the plates of a capacitor, even though no charge actually flows across the gap. It is given by Id = epsilon0 (d(electric flux)/dt). Maxwell introduced it to make Ampere's law consistent while charging a capacitor: the total current (conduction current plus displacement current) is continuous, so the modified Ampere-Maxwell law is the integral of B.dl = mu0 (I + Id).

Id = epsilon0 d(electric flux)/dt

Marking-scheme points

  • Current due to a changing electric field/flux (no charge flows)
  • Id = epsilon0 d(electric flux)/dt
  • Makes conduction + displacement current continuous (Ampere-Maxwell law)
2 markseasyProperties of EM waves

State any four properties of electromagnetic waves.

Reveal model answer + marking points

(1) Electromagnetic waves are transverse in nature, with the electric field E and magnetic field B oscillating perpendicular to each other and to the direction of propagation. (2) They do not require a material medium and can travel through vacuum. (3) They travel through vacuum with the speed of light, c = 3 x 10^8 m/s. (4) They carry energy and momentum, and the ratio of the amplitudes of E and B equals c (E0/B0 = c).

c = E0/B0

Marking-scheme points

  • Transverse; E and B perpendicular to each other and to propagation
  • Do not need a medium; travel through vacuum
  • Speed c = 3 x 10^8 m/s; E0/B0 = c
3 marksmediumElectromagnetic spectrum

Name the parts of the electromagnetic spectrum in order of increasing frequency and give one use of each.

Reveal model answer + marking points

In order of increasing frequency (decreasing wavelength): (1) Radio waves - used in radio and television communication. (2) Microwaves - used in radar and microwave ovens. (3) Infrared - used in remote controls and thermal imaging. (4) Visible light - used for vision and photography. (5) Ultraviolet - used to sterilise water and in detecting forgery. (6) X-rays - used in medical imaging and detecting fractures. (7) Gamma rays - used in cancer treatment (radiotherapy).

Marking-scheme points

  • Order of increasing frequency: radio, microwave, infrared, visible, UV, X-ray, gamma
  • Radio: communication; microwave: radar/oven; infrared: remote control
  • X-ray: medical imaging; gamma ray: cancer treatment
2 marksmediumSpeed of EM waves

Write the expression for the speed of electromagnetic waves in vacuum in terms of mu0 and epsilon0.

Reveal model answer + marking points

The speed of electromagnetic waves in vacuum is given by c = 1/sqrt(mu0 epsilon0), where mu0 is the permeability and epsilon0 the permittivity of free space. Substituting mu0 = 4 pi x 10^-7 and epsilon0 = 8.85 x 10^-12 gives c = 3 x 10^8 m/s, which equals the measured speed of light, showing that light is an electromagnetic wave. In a medium the speed is v = 1/sqrt(mu epsilon).

c = 1/sqrt(mu0 epsilon0)

Marking-scheme points

  • c = 1/sqrt(mu0 epsilon0)
  • Gives 3 x 10^8 m/s (speed of light)
  • Shows light is an electromagnetic wave

Ray Optics and Optical Instruments10 questions

2 markseasyMirror formula

Write the mirror formula and the expression for linear magnification produced by a spherical mirror.

Reveal model answer + marking points

The mirror formula relates the object distance u, image distance v and focal length f of a spherical mirror: 1/v + 1/u = 1/f. The linear magnification is m = -v/u = height of image/height of object. The focal length f = R/2, where R is the radius of curvature. The New Cartesian sign convention is used for the distances.

1/v + 1/u = 1/f

Marking-scheme points

  • Mirror formula: 1/v + 1/u = 1/f
  • Magnification m = -v/u = h(image)/h(object)
  • f = R/2; use sign convention
3 marksmediumConcave mirror calculation

An object is placed 20 cm in front of a concave mirror of focal length 15 cm. Find the position and nature of the image.

Reveal model answer + marking points

Using the sign convention, u = -20 cm and f = -15 cm. From the mirror formula 1/v = 1/f - 1/u = 1/(-15) - 1/(-20) = -1/15 + 1/20 = (-4 + 3)/60 = -1/60, so v = -60 cm. The magnification m = -v/u = -(-60)/(-20) = -3. The image is real (v negative), inverted (m negative) and magnified three times, formed 60 cm in front of the mirror.

1/v + 1/u = 1/f

Marking-scheme points

  • u = -20 cm, f = -15 cm; 1/v = 1/f - 1/u
  • v = -60 cm; m = -v/u = -3
  • Image is real, inverted and magnified 3 times
2 markseasyLaws of refraction

State the laws of refraction of light (Snell's law).

Reveal model answer + marking points

The laws of refraction are: (1) The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane. (2) For a given pair of media and a given colour of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, called the refractive index: sin i/sin r = n (Snell's law). Refraction occurs because light travels at different speeds in different media.

sin i/sin r = n

Marking-scheme points

  • Incident ray, refracted ray and normal lie in one plane
  • Snell's law: sin i/sin r = n (constant)
  • Caused by change of speed of light between media
3 marksmediumTotal internal reflection

What is total internal reflection? State the conditions required for it and one application.

Reveal model answer + marking points

Total internal reflection is the phenomenon in which a ray of light travelling from a denser to a rarer medium is completely reflected back into the denser medium when the angle of incidence exceeds a certain angle called the critical angle. Conditions: (1) light must travel from a denser to a rarer medium; (2) the angle of incidence must be greater than the critical angle C, where sin C = 1/n. Applications: optical fibres, sparkling of diamonds, and mirages.

sin C = 1/n

Marking-scheme points

  • Complete reflection back into the denser medium
  • Conditions: denser to rarer medium and angle of incidence > critical angle
  • sin C = 1/n; application: optical fibres
3 marksmediumCritical angle calculation

The refractive index of glass is 1.5. Calculate the critical angle for the glass-air interface.

Reveal model answer + marking points

The critical angle C is given by sin C = 1/n, where n is the refractive index of the denser medium (glass). So sin C = 1/1.5 = 0.667, giving C = sin inverse (0.667) = 41.8 degrees (approximately 42 degrees). For angles of incidence greater than this, total internal reflection occurs.

sin C = 1/n

Marking-scheme points

  • sin C = 1/n = 1/1.5 = 0.667
  • C = sin inverse (0.667)
  • Critical angle = about 41.8 degrees
3 marksmediumLens maker's formula

Write the lens maker's formula and explain the meaning of each term.

Reveal model answer + marking points

The lens maker's formula is 1/f = (n - 1)(1/R1 - 1/R2), where f is the focal length of the lens, n is the refractive index of the lens material with respect to the surrounding medium, and R1 and R2 are the radii of curvature of the two surfaces of the lens (taken with proper sign convention). It shows that the focal length depends on the material and the shape (curvature) of the lens.

1/f = (n - 1)(1/R1 - 1/R2)

Marking-scheme points

  • 1/f = (n - 1)(1/R1 - 1/R2)
  • n = refractive index of lens material relative to surroundings
  • R1, R2 = radii of curvature of the two surfaces (with sign convention)
3 marksmediumConvex lens calculation

An object is placed 30 cm from a convex lens of focal length 20 cm. Find the position and nature of the image.

Reveal model answer + marking points

Using the sign convention, u = -30 cm and f = +20 cm. From the lens formula 1/v - 1/u = 1/f, we get 1/v = 1/f + 1/u = 1/20 + 1/(-30) = 1/20 - 1/30 = (3 - 2)/60 = 1/60, so v = +60 cm. The magnification m = v/u = 60/(-30) = -2. The image is real (v positive for a lens), inverted (m negative) and magnified twice, formed 60 cm on the other side of the lens.

1/v - 1/u = 1/f

Marking-scheme points

  • u = -30 cm, f = +20 cm; 1/v = 1/f + 1/u
  • v = +60 cm; m = v/u = -2
  • Image is real, inverted and magnified 2 times
2 markseasyPower of a lens

Define the power of a lens. State its SI unit and the formula for the power of two thin lenses in contact.

Reveal model answer + marking points

The power of a lens is a measure of its ability to converge or diverge light and is defined as the reciprocal of its focal length in metres: P = 1/f (f in metres). Its SI unit is the dioptre (D). A converging (convex) lens has positive power and a diverging (concave) lens has negative power. For two thin lenses in contact, the total power is the sum: P = P1 + P2.

P = 1/f; P = P1 + P2

Marking-scheme points

  • P = 1/f (f in metres); SI unit dioptre (D)
  • Convex lens: positive power; concave lens: negative power
  • Lenses in contact: P = P1 + P2
3 marksmediumPrism formula

Write the prism formula relating refractive index to the angle of the prism and the angle of minimum deviation.

Reveal model answer + marking points

When a ray passes through a prism, it is deviated; the deviation is least at the angle of minimum deviation Dm, when the ray passes symmetrically through the prism. The refractive index of the prism material is given by n = sin((A + Dm)/2)/sin(A/2), where A is the refracting angle of the prism and Dm is the angle of minimum deviation. This formula is used to determine the refractive index of a transparent material.

n = sin((A + Dm)/2)/sin(A/2)

Marking-scheme points

  • Minimum deviation Dm occurs for symmetric passage
  • n = sin((A + Dm)/2)/sin(A/2)
  • A = angle of prism; used to find refractive index
3 marksmediumAstronomical telescope

Write the expression for the magnifying power of an astronomical telescope in normal adjustment and state the required focal lengths.

Reveal model answer + marking points

An astronomical telescope has an objective lens of large focal length fo and an eyepiece of small focal length fe. In normal adjustment (final image at infinity), the magnifying power is M = fo/fe, and the length of the telescope is fo + fe. For high magnification, the objective should have a large focal length (and large aperture) and the eyepiece a small focal length. The final image is inverted.

M = fo/fe

Marking-scheme points

  • Magnifying power (normal adjustment): M = fo/fe
  • Length of telescope = fo + fe
  • Objective: large fo; eyepiece: small fe; image inverted

Wave Optics8 questions

2 marksmediumHuygens' principle

State Huygens' principle of secondary wavelets.

Reveal model answer + marking points

Huygens' principle states that: (1) every point on a given wavefront acts as a source of new disturbance called secondary wavelets, which spread out in all directions with the speed of the wave; and (2) the new wavefront at a later instant is the forward envelope (tangential surface) of all these secondary wavelets. This principle is used to explain the laws of reflection and refraction and the propagation of light as a wave.

Marking-scheme points

  • Every point on a wavefront is a source of secondary wavelets
  • Wavelets travel with the speed of the wave
  • New wavefront = forward envelope of the secondary wavelets
3 marksmediumYoung's double slit experiment

Write the expression for the fringe width in Young's double slit experiment and explain the terms.

Reveal model answer + marking points

In Young's double slit experiment, two coherent sources produce alternate bright and dark fringes on a screen. The fringe width (the distance between two consecutive bright or dark fringes) is beta = lambda D/d, where lambda is the wavelength of light used, D is the distance between the slits and the screen, and d is the separation between the two slits. All fringes are of equal width, and the width increases with wavelength and D but decreases as d increases.

beta = lambda D/d

Marking-scheme points

  • Fringe width beta = lambda D/d
  • lambda = wavelength, D = slit-to-screen distance, d = slit separation
  • Fringes are equally spaced; beta increases with lambda and D
3 marksmediumFringe width calculation

In Young's double slit experiment, light of wavelength 600 nm is used with a slit separation of 1 mm and a screen 1 m away. Calculate the fringe width.

Reveal model answer + marking points

Given lambda = 600 nm = 600 x 10^-9 m, d = 1 mm = 1 x 10^-3 m, D = 1 m. Fringe width beta = lambda D/d = (600 x 10^-9 x 1)/(1 x 10^-3) = 600 x 10^-6 = 6 x 10^-4 m = 0.6 mm.

beta = lambda D/d

Marking-scheme points

  • Convert units: lambda = 6e-7 m, d = 1e-3 m
  • beta = lambda D/d = (6e-7 x 1)/1e-3
  • beta = 6 x 10^-4 m = 0.6 mm
2 marksmediumInterference conditions

State the conditions for constructive and destructive interference in terms of path difference.

Reveal model answer + marking points

For constructive interference (bright fringe), the path difference between the two interfering waves must be an integral multiple of the wavelength: path difference = n lambda, where n = 0, 1, 2, ... For destructive interference (dark fringe), the path difference must be an odd multiple of half the wavelength: path difference = (2n - 1) lambda/2. Equivalently, the phase difference is 2n pi for constructive and (2n - 1) pi for destructive interference.

constructive: n lambda; destructive: (2n-1) lambda/2

Marking-scheme points

  • Constructive: path difference = n lambda
  • Destructive: path difference = (2n - 1) lambda/2
  • Phase difference 2n pi (bright) or (2n-1) pi (dark)
2 marksmediumCoherent sources

What are coherent sources? State the conditions for obtaining sustained interference of light.

Reveal model answer + marking points

Coherent sources are two sources of light that emit waves of the same frequency (or wavelength) and have a constant phase difference between them. Conditions for sustained (steady) interference: (1) the two sources must be coherent; (2) they must have the same frequency and nearly equal amplitudes; and (3) they must be narrow and close together, and the light should preferably be monochromatic. In practice, coherent sources are obtained from a single source (for example, using two slits).

Marking-scheme points

  • Coherent sources: same frequency and constant phase difference
  • Need equal frequency and nearly equal amplitude
  • Obtained from a single source (e.g. two slits)
3 marksmediumSingle slit diffraction

What is diffraction of light? Write the condition for minima in single slit diffraction.

Reveal model answer + marking points

Diffraction is the bending of light around the edges of an obstacle or aperture and its spreading into the geometrical shadow region. In diffraction at a single slit of width a, a central bright maximum is flanked by alternate dark and bright fringes. The condition for the dark fringes (minima) is a sin theta = n lambda, where n = 1, 2, 3, ... and theta is the angle of diffraction. The central maximum is the brightest and widest.

a sin theta = n lambda

Marking-scheme points

  • Diffraction: bending/spreading of light around obstacles/apertures
  • Single slit minima: a sin theta = n lambda (n = 1, 2, 3...)
  • Central maximum is the brightest and widest
2 marksmediumInterference versus diffraction

State two differences between interference and diffraction of light.

Reveal model answer + marking points

(1) Interference is due to the superposition of waves from two (or more) different coherent sources, whereas diffraction is due to the superposition of secondary wavelets coming from different parts of the same wavefront. (2) In interference all bright fringes are of equal intensity and equal width, whereas in diffraction the central maximum is the brightest and the intensity of the secondary maxima decreases rapidly on either side.

Marking-scheme points

  • Interference: two coherent sources; diffraction: parts of the same wavefront
  • Interference fringes: equal width and intensity
  • Diffraction: central maximum brightest, others decrease
2 marksmediumPolarisation and Brewster's law

What is polarisation of light? State Brewster's law.

Reveal model answer + marking points

Polarisation is the phenomenon of restricting the vibrations of the electric field of a light wave to a single plane perpendicular to the direction of propagation; it shows that light is a transverse wave. Brewster's law states that when unpolarised light is incident on a transparent surface at a particular angle called the polarising angle (theta_p), the reflected light is completely plane-polarised, and the refractive index of the medium is n = tan(theta_p).

n = tan(theta_p)

Marking-scheme points

  • Polarisation restricts vibrations to one plane (light is transverse)
  • At the polarising angle, reflected light is fully plane-polarised
  • Brewster's law: n = tan(theta_p)

Dual Nature of Radiation and Matter6 questions

2 markseasyPhotoelectric effect

What is the photoelectric effect?

Reveal model answer + marking points

The photoelectric effect is the phenomenon of emission of electrons (called photoelectrons) from the surface of a metal when light of suitable frequency (usually ultraviolet or visible for some metals) falls on it. The emitted electrons carry kinetic energy. The effect occurs only when the frequency of the incident light is greater than a certain minimum value called the threshold frequency, and it provided evidence for the particle (photon) nature of light.

Marking-scheme points

  • Emission of electrons from a metal when light falls on it
  • Occurs only above the threshold frequency
  • Evidence for the particle (photon) nature of light
3 marksmediumEinstein's photoelectric equation

Write Einstein's photoelectric equation and explain each term.

Reveal model answer + marking points

Einstein's photoelectric equation is h f = W0 + KE(max), where h f is the energy of the incident photon (h is Planck's constant and f the frequency of light), W0 is the work function (the minimum energy needed to eject an electron from the metal surface), and KE(max) is the maximum kinetic energy of the emitted photoelectron. It expresses conservation of energy: the photon's energy is used partly to free the electron and the rest appears as its kinetic energy. Thus KE(max) = h f - W0.

h f = W0 + KE(max)

Marking-scheme points

  • h f = W0 + KE(max)
  • h f = photon energy; W0 = work function
  • KE(max) = h f - W0 (energy conservation)
2 marksmediumWork function and threshold frequency

Define work function and threshold frequency.

Reveal model answer + marking points

The work function (W0) of a metal is the minimum energy required to just remove an electron from the surface of the metal without giving it any kinetic energy. The threshold frequency (f0) is the minimum frequency of the incident light below which no photoelectric emission takes place, however intense the light may be. They are related by W0 = h f0, where h is Planck's constant.

W0 = h f0

Marking-scheme points

  • Work function W0 = minimum energy to free an electron
  • Threshold frequency f0 = minimum frequency for emission
  • Relation: W0 = h f0
3 marksmediumPhotoelectric calculation

Light of wavelength 400 nm is incident on a metal of work function 2 eV. Calculate the maximum kinetic energy of the emitted photoelectrons. (Use hc = 1240 eV nm)

Reveal model answer + marking points

The energy of the incident photon E = hc/lambda = 1240/400 = 3.1 eV. Using Einstein's equation, the maximum kinetic energy KE(max) = E - W0 = 3.1 - 2.0 = 1.1 eV. Since the photon energy (3.1 eV) is greater than the work function (2 eV), emission occurs and the photoelectrons have a maximum kinetic energy of 1.1 eV.

KE(max) = hc/lambda - W0

Marking-scheme points

  • Photon energy E = hc/lambda = 1240/400 = 3.1 eV
  • KE(max) = E - W0 = 3.1 - 2.0
  • KE(max) = 1.1 eV
2 marksmediumLaws of photoelectric effect

State any two laws of the photoelectric effect.

Reveal model answer + marking points

(1) For a given metal, photoelectric emission occurs only if the frequency of the incident light is greater than a certain minimum value (threshold frequency), whatever the intensity. (2) The maximum kinetic energy of the emitted photoelectrons depends on the frequency of the incident light and the nature of the metal, but is independent of the intensity of the light. (3) The number of photoelectrons emitted per second (photoelectric current) is directly proportional to the intensity of the incident light. (4) The emission is instantaneous, with no measurable time lag.

Marking-scheme points

  • Emission only above the threshold frequency
  • Max KE depends on frequency, not on intensity
  • Number of photoelectrons is proportional to intensity; emission is instantaneous
2 marksmediumde Broglie wavelength

What is the de Broglie hypothesis? Write the expression for the de Broglie wavelength.

Reveal model answer + marking points

The de Broglie hypothesis states that every moving particle has a wave associated with it, called a matter wave. The de Broglie wavelength is lambda = h/p = h/(m v), where h is Planck's constant, p is the momentum, m is the mass and v the velocity of the particle. In terms of kinetic energy, lambda = h/sqrt(2 m KE). This shows the dual (wave-particle) nature of matter; the wavelength is significant only for very small particles like electrons.

lambda = h/(m v)

Marking-scheme points

  • Every moving particle has an associated matter wave
  • lambda = h/p = h/(m v)
  • In terms of KE: lambda = h/sqrt(2 m KE)

Atoms5 questions

2 marksmediumRutherford's experiment

State the main conclusions of Rutherford's alpha-particle scattering experiment.

Reveal model answer + marking points

From the scattering of alpha particles by a thin gold foil, Rutherford concluded that: (1) most of the atom is empty space, since most alpha particles passed straight through; (2) the entire positive charge and almost all the mass of the atom are concentrated in a very small central region called the nucleus, since a few alpha particles were deflected through large angles; and (3) the electrons revolve around the nucleus, and the size of the nucleus is very small compared with the size of the atom.

Marking-scheme points

  • Most of the atom is empty space
  • Positive charge and mass concentrated in a tiny nucleus
  • Electrons revolve around the nucleus
3 marksmediumBohr's postulates

State the postulates of Bohr's model of the hydrogen atom.

Reveal model answer + marking points

Bohr's postulates are: (1) The electron revolves around the nucleus only in certain fixed circular orbits called stationary states, in which it does not radiate energy. (2) Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of h/(2 pi), that is, m v r = n h/(2 pi) (quantisation of angular momentum). (3) Energy is emitted or absorbed only when the electron jumps from one orbit to another, the energy of the emitted or absorbed photon being h f = E2 - E1.

m v r = n h/(2 pi)

Marking-scheme points

  • Electrons revolve in fixed stationary orbits without radiating
  • Angular momentum quantised: m v r = n h/(2 pi)
  • Energy change on jump: h f = E2 - E1
2 marksmediumEnergy levels of hydrogen

The energy of an electron in the ground state of hydrogen is -13.6 eV. Calculate the energy of the electron in the second orbit (n = 2).

Reveal model answer + marking points

The energy of the electron in the nth orbit of hydrogen is En = -13.6/n^2 eV. For n = 2, E2 = -13.6/2^2 = -13.6/4 = -3.4 eV. The negative sign shows that the electron is bound to the nucleus, and the energy increases (becomes less negative) as n increases.

En = -13.6/n^2 eV

Marking-scheme points

  • En = -13.6/n^2 eV
  • E2 = -13.6/4
  • E2 = -3.4 eV
3 marksmediumHydrogen spectrum

Calculate the energy of the photon emitted when an electron in a hydrogen atom jumps from n = 3 to n = 2. (Use En = -13.6/n^2 eV)

Reveal model answer + marking points

Energy of the electron in n = 3: E3 = -13.6/9 = -1.51 eV. Energy in n = 2: E2 = -13.6/4 = -3.4 eV. The energy of the emitted photon = E3 - E2 = -1.51 - (-3.4) = 1.89 eV. This corresponds to the H-alpha line of the Balmer series (visible red light).

E(photon) = E(higher) - E(lower)

Marking-scheme points

  • E3 = -13.6/9 = -1.51 eV; E2 = -13.6/4 = -3.4 eV
  • Photon energy = E3 - E2 = 1.89 eV
  • This is the H-alpha (Balmer series) line
2 marksmediumLimitations of Bohr's model

State two limitations of Bohr's model of the atom.

Reveal model answer + marking points

(1) Bohr's model applies successfully only to hydrogen and hydrogen-like single-electron atoms; it fails to explain the spectra of atoms having more than one electron. (2) It could not explain the fine structure of spectral lines or the relative intensities of the lines, and it does not account for the splitting of spectral lines in electric and magnetic fields (the Stark and Zeeman effects). Also, it arbitrarily assumes quantisation without explaining it (later explained by de Broglie).

Marking-scheme points

  • Works only for hydrogen/single-electron atoms
  • Cannot explain fine structure or relative intensities of lines
  • Cannot explain Zeeman/Stark effects; quantisation assumed arbitrarily

Nuclei7 questions

2 marksmediumMass-energy relation

State Einstein's mass-energy relation. What is the energy equivalent of 1 atomic mass unit (u)?

Reveal model answer + marking points

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
2 marksmediumMass defect and binding energy

Define mass defect and binding energy of a nucleus.

Reveal model answer + marking points

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
3 marksmediumBinding energy calculation

The mass defect of a helium nucleus is 0.0304 u. Calculate its binding energy. (1 u = 931 MeV)

Reveal model answer + marking points

The binding energy is the energy equivalent of the mass defect. BE = (mass defect in u) x 931 MeV = 0.0304 x 931 = 28.3 MeV. Thus the binding energy of the helium nucleus is about 28.3 MeV, and the binding energy per nucleon = 28.3/4 = 7.1 MeV.

BE (MeV) = (delta m in u) x 931

Marking-scheme points

  • BE = (mass defect) x 931 MeV
  • = 0.0304 x 931
  • BE = 28.3 MeV (about 7.1 MeV per nucleon)
2 marksmediumRadioactivity

Name the three types of radioactive radiations and state their nature.

Reveal model answer + marking points

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
3 marksmediumRadioactive decay law

State the radioactive decay law. What fraction of a radioactive sample remains after 3 half-lives?

Reveal model answer + marking points

The radioactive decay law states that the rate of decay of a radioactive sample is directly proportional to the number of undecayed nuclei present at that instant: N = N0 e^(-lambda t), where lambda is the decay constant. The half-life T is related to it by T = 0.693/lambda. After each half-life, half of the sample remains, so after 3 half-lives the fraction remaining = (1/2)^3 = 1/8 of the original sample.

N = N0 (1/2)^(t/T)

Marking-scheme points

  • Decay law: N = N0 e^(-lambda t); half-life T = 0.693/lambda
  • Fraction after n half-lives = (1/2)^n
  • After 3 half-lives: (1/2)^3 = 1/8 remains
2 marksmediumNuclear fission

What is nuclear fission? Give one example.

Reveal model answer + marking points

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
2 marksmediumNuclear fusion

What is nuclear fusion? Why does it require very high temperature?

Reveal model answer + marking points

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

Semiconductor Electronics8 questions

2 markseasyIntrinsic and extrinsic semiconductors

Distinguish between intrinsic and extrinsic semiconductors.

Reveal model answer + marking points

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
2 marksmediumn-type and p-type semiconductors

How are n-type and p-type semiconductors formed? Name the majority charge carriers in each.

Reveal model answer + marking points

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
2 marksmediump-n junction

What is a depletion region and potential barrier in a p-n junction?

Reveal model answer + marking points

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
2 marksmediumBiasing of a diode

Distinguish between forward biasing and reverse biasing of a p-n junction diode.

Reveal model answer + marking points

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
3 marksmediumHalf-wave rectifier

Explain how a p-n junction diode works as a half-wave rectifier.

Reveal model answer + marking points

A rectifier converts alternating current (AC) into direct current (DC). In a half-wave rectifier, a single diode is connected in series with the AC source and a load resistor. During the positive half-cycle of the AC input, the diode is forward biased and conducts, so current flows through the load. During the negative half-cycle, the diode is reverse biased and does not conduct, so no current flows. As a result, output is obtained only during one half of each cycle, giving a pulsating DC. Its efficiency is low because half the input is wasted.

Marking-scheme points

  • Rectifier converts AC into DC; uses one diode
  • Positive half-cycle: diode forward biased -> conducts
  • Negative half-cycle: diode reverse biased -> no output (pulsating DC)
3 marksmediumFull-wave rectifier

Explain the working of a full-wave rectifier using two diodes.

Reveal model answer + marking points

A full-wave rectifier converts both halves of the AC input into DC. It uses two diodes with a centre-tapped transformer and a load resistor. During the positive half-cycle, one diode is forward biased and conducts while the other is reverse biased; during the negative half-cycle, the second diode conducts while the first does not. In both half-cycles, the current through the load flows in the same direction, so output is obtained during the whole cycle. This gives a smoother, more efficient pulsating DC than a half-wave rectifier.

Marking-scheme points

  • Uses two diodes and a centre-tapped transformer
  • Each diode conducts during one half-cycle
  • Current through load is in the same direction for both halves (full-wave DC)
2 marksmediumZener diode

What is a Zener diode? State its main use.

Reveal model answer + marking points

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
3 marksmediumLogic gates

Write the truth tables of the OR, AND and NOT logic gates for inputs A and B.

Reveal model answer + marking points

A logic gate is a digital circuit that gives an output based on its inputs following a logical rule (using binary 0 and 1). OR gate (output Y = A + B): Y = 1 if any input is 1; for inputs (0,0),(0,1),(1,0),(1,1) the outputs are 0,1,1,1. AND gate (Y = A.B): Y = 1 only if both inputs are 1; outputs are 0,0,0,1. NOT gate (Y = not A): it has a single input and inverts it, so input 0 gives output 1 and input 1 gives output 0.

OR: Y = A + B; AND: Y = A.B; NOT: Y = not A

Marking-scheme points

  • OR (Y = A + B): output 1 if any input is 1 -> 0,1,1,1
  • AND (Y = A.B): output 1 only if both inputs 1 -> 0,0,0,1
  • NOT: single input, inverts it (0 -> 1, 1 -> 0)

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