Board Boosters

The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questionsModel answersMarking-scheme pointsEvery chapterCBSE · ISC · State boards

93 questions · clear filters

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
Still unsure? Ask the AI tutor →
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
Still unsure? Ask the AI tutor →
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)
Still unsure? Ask the AI tutor →
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
Still unsure? Ask the AI tutor →
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
Still unsure? Ask the AI tutor →
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
Still unsure? Ask the AI tutor →
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)
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Electromagnetic Induction

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Electromagnetic Induction

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Alternating Current

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Alternating Current

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))
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Alternating Current

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Alternating Current

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Electromagnetic Waves

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Ray Optics and Optical Instruments

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Ray Optics and Optical Instruments

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Ray Optics and Optical Instruments

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Ray Optics and Optical Instruments

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)
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Ray Optics and Optical Instruments

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
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Ray Optics and Optical Instruments

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
Still unsure? Ask the AI tutor →
← PrevPage 4 of 5Next →

You are more ready than you feel.

One question at a time is how every topper started. Bookmark this, revise a few each day, and watch the fear shrink. And if a friend is stressing about boards — send this their way. You both win.