Form 6 Physics Practice Paper: Electromagnetism, Optics And Modern Physics (12 Questions, 100 Marks, 75 Minutes)

Form 6 Physics Practice Paper: Electromagnetism, Optics and Modern Physics (12 questions, 100 marks, 75 minutes)

2026-11-05

Form 6 Physics – 12 questions – 100 marks – 75 minutes

Form 6 Physics practice paper: 12 questions, 100 marks, 75 minutes on electromagnetism, optics and modern physics, with full worked answers.

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Paper details　Form 6 Physics practice paper with 12 original questions on electromagnetism, optics and modern physics, plus full worked answers.

Section A: Electrostatics and capacitance (Q1-Q4)

Write the formula, substitute in SI units and state the direction or unit; give three significant figures.

#	Question	Answer and marking notes
1	Two point charges of +4.0 uC and +6.0 uC are 0.30 m apart. Calculate the magnitude of the electrostatic force between them. (k = 8.99 x 10^9 N m2 C-2)	Answer: 2.40 N. F = kq1q2 / r2 = (8.99 x 10^9)(4.0 x 10^-6)(6.0 x 10^-6) / 0.302 = 2.40 N. Both charges are positive, so the force is repulsive. Marking: formula 2, substitution 2, answer with unit 2, direction 2.
2	A point charge of +5.0 nC is placed in air. Calculate the electric field strength at a distance of 0.20 m from it.	Answer: 1.1 x 10^3 N/C (1124 N/C), directed radially outwards. E = kQ / r2 = (8.99 x 10^9)(5.0 x 10^-9) / 0.202 = 1.1 x 10^3 N/C. Marking: formula 3, substitution 3, direction 2.
3	The plates of a parallel-plate capacitor have area 0.020 m2 and are separated by 2.0 mm. Calculate its capacitance. (e0 = 8.85 x 10^-12 F/m)	Answer: 88.5 pF. C = e0A / d = (8.85 x 10^-12)(0.020) / 2.0 x 10^-3 = 8.85 x 10^-11 F. Marking: formula 3, unit conversion 2, answer 3; the separation must be converted from mm to m.
4	A 100 uF capacitor is charged to 12 V. Calculate the energy it stores.	Answer: 7.2 mJ. U = 0.5CV2 = 0.5 x 100 x 10^-6 x 144 = 7.2 x 10^-3 J. Marking: formula 3, capacitance conversion 2, answer with unit 3.

Section B: Magnetic fields, induction and AC (Q5-Q8)

Name the rule used (Fleming left-hand rule, Faraday law or Lenz law) and state the current direction.

#	Question	Answer and marking notes
5	A 0.50 m wire carries a current of 3.0 A and is placed perpendicular to a uniform magnetic field of 0.50 T. Calculate the magnetic force on the wire.	Answer: 0.75 N. F = BILsin(theta) = 0.50 x 3.0 x 0.50 x sin 90 = 0.75 N, perpendicular to both the wire and the field by Fleming left-hand rule. Marking: formula 3, substitution 2, answer 2, direction 1.
6	An electron enters a uniform magnetic field of 0.010 T at 2.0 x 10^6 m/s perpendicular to the field. Calculate the radius of its circular path. (electron mass 9.11 x 10^-31 kg, charge 1.60 x 10^-19 C)	Answer: 1.1 mm. The magnetic force provides the centripetal force: evB = mv2 / r, so r = mv / (eB) = (9.11 x 10^-31)(2.0 x 10^6) / (1.60 x 10^-19 x 0.010) = 1.1 x 10^-3 m. Marking: centripetal equation 3, rearrangement 2, substitution and answer 3.
7	A 0.25 m conducting rod cuts a 0.40 T magnetic field at 2.0 m/s. The total resistance of the circuit is 0.50 ohm. Calculate the induced emf and the induced current.	Answer: 0.20 V and 0.40 A. emf = BLv = 0.40 x 0.25 x 2.0 = 0.20 V; I = emf / R = 0.20 / 0.50 = 0.40 A. Marking: induced emf formula 3, answer 2, Ohm law 2, direction by Lenz law 1.
8	An ideal transformer has 1,000 turns in its primary connected to 220 V and 50 turns in its secondary. The secondary current is 4.0 A. Calculate the secondary voltage and the primary current.	Answer: secondary 11 V, primary 0.20 A. Vs = Vp x Ns / Np = 220 x 50 / 1000 = 11 V; for an ideal transformer the power is unchanged, so Ip = Is x Ns / Np = 4.0 x 50 / 1000 = 0.20 A. Marking: turns ratio 3, power conservation 3, answers 2.

Section C: Optics and modern physics (Q9-Q12)

Quote the formula, convert units (nm to m, eV) and give angles to one decimal place.

#	Question	Answer and marking notes
9	Light travels from air into water (refractive index 1.33) at an angle of incidence of 40 degrees. Calculate the angle of refraction.	Answer: 28.9 degrees. n1sin(theta1) = n2sin(theta2), so sin(theta2) = sin 40 / 1.33 = 0.483 and theta2 = 28.9 degrees; the ray bends towards the normal. Marking: Snell law 3, substitution 2, inverse sine 2, direction note 1.
10	Light travels from glass (refractive index 1.50) towards air. Calculate the critical angle for total internal reflection.	Answer: 41.8 degrees. sin C = n2 / n1 = 1.00 / 1.50 = 0.667, so C = 41.8 degrees; beyond this angle no light is refracted out. Marking: formula 4, calculation 2, condition stated 2.
11	Light of wavelength 400 nm falls on a metal surface with a work function of 2.10 eV. Calculate the maximum kinetic energy of the emitted photoelectrons. (hc = 1,240 eV nm)	Answer: 1.00 eV. Photon energy E = hc / wavelength = 1,240 / 400 = 3.10 eV; maximum kinetic energy = E – W = 3.10 – 2.10 = 1.00 eV. Marking: photon energy 3, subtracting work function 3, answer with unit 2.
12	In a double-slit experiment the slit separation is 0.20 mm, the screen distance is 1.5 m and the wavelength is 600 nm. Calculate the fringe spacing.	Answer: 4.5 mm. Fringe spacing = wavelength x D / d = (600 x 10^-9)(1.5) / (0.20 x 10^-3) = 4.5 x 10^-3 m. Marking: formula 3, unit conversion 3, answer 2.

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