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ZIMSEC A Level · 6032/2 · N2025

Physics Paper 2 November 2025

Questions
33
Total marks
60
Syllabus code
6032/2

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Questions
33
Pass mark
20
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[2 marks]couples; resultant of coplanar forces; gravitational potential; geostationary orbit
Two forces acting on an object form a couple. Which statement gives the conditions the two forces must satisfy?
  1. AThey are equal in magnitude and opposite in direction, but share the same line of action.
  2. BThey are equal in magnitude and opposite in direction, and act along parallel lines a distance apart.
  3. CThey are unequal in magnitude and opposite in direction, and act along parallel lines a distance apart.
  4. DThey are equal in magnitude and act in the same direction, along parallel lines a distance apart.

Question 102

[2 marks]couples; resultant of coplanar forces; gravitational potential; geostationary orbit
Two forces of 12 N and 10 N act at the same point on an object, with an angle of 60 degrees between their lines of action. Calculate the magnitude of their resultant force.

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Question 103

[1 marks]couples; resultant of coplanar forces; gravitational potential; geostationary orbit
Two forces of 12 N and 10 N act at the same point on an object, with an angle of 60 degrees between their lines of action. At what angle to the 12 N force does their resultant act?
  1. A35.8 degrees to the 12 N force
  2. B24.6 degrees to the 12 N force
  3. C27.0 degrees to the 12 N force
  4. D30.0 degrees to the 12 N force

Question 104

[1 marks]couples; resultant of coplanar forces; gravitational potential; geostationary orbit
Gravitational potential at a point in a gravitational field is defined as
  1. Athe work done per unit mass in bringing a small test mass from infinity to that point.
  2. Bthe energy stored per unit volume of the field in the region around that point.
  3. Cthe force per unit mass acting on a small test mass placed at that point.
  4. Dthe gravitational force with which the earth attracts a mass of one kilogram there.

Question 105

[1 marks]couples; resultant of coplanar forces; gravitational potential; geostationary orbit
State the SI unit of gravitational potential.

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Question 106

[3 marks]couples; resultant of coplanar forces; gravitational potential; geostationary orbit
A geostationary satellite orbits the earth, of radius R = 6 400 km, with period T. Its orbital radius r satisfies r3/T2=gR2/(4π2)r^3/T^2 = gR^2/(4\pi^2), where g = 9.81 m s^-2. Taking the geostationary period as 24 hours, calculate the height of the satellite above the earth's surface.

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Question 201

[1 marks]damping and resonance; diffraction grating
In the study of oscillations, the term damping refers to
  1. Athe transfer of energy from a driver to an oscillator that is free to vibrate at any frequency.
  2. Bthe change in the natural frequency of a system when extra mass is attached to it.
  3. Cthe loss of energy from an oscillating system to resistive forces, so its amplitude falls.
  4. Dthe rise in amplitude of an oscillating system when it is driven at its natural frequency.

Question 202

[2 marks]damping and resonance; diffraction grating
Suspension bridges are more prone to collapse in high winds than beam and pier bridges. The main reason is that
  1. Aa suspension bridge carries a far greater load than a beam bridge does, so a strong wind adds more stress to it.
  2. Ba suspension bridge is built from steel cable rather than concrete, and steel is weakened faster by wind.
  3. Ca suspension bridge has a natural frequency far above any frequency the wind can supply, so wind energy builds in it.
  4. Da suspension bridge is flexible and lightly damped with a low natural frequency, so wind drives it into resonance.

Question 203

[1 marks]damping and resonance; diffraction grating
A diffraction grating has 500 lines per millimetre. Calculate the spacing d between adjacent lines of the grating.

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Question 204

[2 marks]damping and resonance; diffraction grating
Monochromatic light of wavelength 600 nm falls normally on a diffraction grating with 500 lines per millimetre. Determine the highest order number that can be observed.

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Question 205

[2 marks]damping and resonance; diffraction grating
Monochromatic light of wavelength 600 nm falls normally on a diffraction grating with 500 lines per millimetre. Calculate the angle of diffraction of the third order maximum.

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Question 301

[1 marks]electric field strength; velocity selector; r.m.s. voltage; power transmission
Electric field strength at a point in an electric field is defined as
  1. Athe force per unit positive charge acting on a small test charge placed at that point.
  2. Bthe work done per unit positive charge in bringing a test charge from infinity to that point.
  3. Cthe charge stored per unit potential difference between two chosen points in the field.
  4. Dthe potential difference per unit charge measured between two chosen points in the field.

Question 302

[1 marks]electric field strength; velocity selector; r.m.s. voltage; power transmission
A beam of electrons travels horizontally from left to right through the gap between two parallel plates, the upper plate positive and the lower plate negative. A magnetic field perpendicular to the electric field keeps the beam undeflected. In which direction does the magnetic field point?
  1. AVertically downwards, running from the positive plate towards the negative plate.
  2. BVertically upwards, running from the negative plate towards the positive plate.
  3. CInto the page, perpendicular to both the beam and the electric field.
  4. DOut of the page, perpendicular to both the beam and the electric field.

Question 303

[2 marks]electric field strength; velocity selector; r.m.s. voltage; power transmission
A beam of electrons passes undeflected through crossed fields: an electric field of 500 kV m^-1 and a magnetic field of 50 mT perpendicular to it. Calculate the speed of the electrons.

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Question 304

[1 marks]electric field strength; velocity selector; r.m.s. voltage; power transmission
The root mean square value of an alternating voltage is
  1. Athe steady direct voltage that would dissipate the same average power in a given resistor.
  2. Bthe peak value of the alternating voltage divided by two over one complete cycle.
  3. Cthe average of all the instantaneous values of the alternating voltage over one cycle.
  4. Dthe largest instantaneous value the alternating voltage reaches during one complete cycle.

Question 305

[2 marks]electric field strength; velocity selector; r.m.s. voltage; power transmission
An alternating supply is rated 230 V, 50 Hz, where 230 V is the root mean square voltage. Calculate the peak voltage.

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Question 306

[1 marks]electric field strength; velocity selector; r.m.s. voltage; power transmission
An alternating supply has a frequency of 50 Hz. Calculate its angular frequency.

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Question 307

[1 marks]electric field strength; velocity selector; r.m.s. voltage; power transmission
One advantage of transmitting electrical energy as alternating current rather than direct current is that
  1. Aits voltage can be stepped up or down efficiently by a transformer at each end of the line.
  2. Bit dissipates less power in the cables than a direct current at exactly the same voltage.
  3. Cit can be carried by a thinner cable than a direct current at exactly the same voltage.
  4. Dit travels along a transmission line faster than a direct current of the same size does.

Question 308

[1 marks]electric field strength; velocity selector; r.m.s. voltage; power transmission
One advantage of transmitting electrical energy at high voltage is that
  1. Aa higher voltage pushes the current along the cables at a greater speed, so the energy arrives with less delay.
  2. Ba higher voltage lowers the resistance of the transmission cables themselves, so far less energy is wasted as heat.
  3. Ca higher voltage means a smaller current for the same power, so the heating loss in the line is smaller.
  4. Da higher voltage means a larger current for the same power, so more of the energy reaches the consumer.

Question 401

[3 marks]logic gates and truth tables; digital transmission; resistor networks and internal resistance
In a logic circuit, inputs A and B feed a NAND gate whose output is C. A second NAND gate takes A and C as its inputs and gives output D. A third NAND gate takes C and B as its inputs and gives output E. Finally D and E feed an AND gate whose output is F. For which input combinations is the output F equal to 1?
  1. AWhen A = 0, B = 0 and when A = 1, B = 1, that is whenever the two inputs are the same.
  2. BWhen A = 0, B = 1 and when A = 1, B = 0, that is whenever the two inputs are different.
  3. CWhen A = 1, B = 1 only, that is whenever both of the two inputs are at logic 1.
  4. DWhen A = 0, B = 0 only, that is whenever both of the two inputs are at logic 0.

Question 402

[1 marks]logic gates and truth tables; digital transmission; resistor networks and internal resistance
In a logic circuit, inputs A and B feed a NAND gate whose output is C; a second NAND gate takes A and C and gives D; a third NAND gate takes C and B and gives E; D and E feed an AND gate whose output is F. Working through the four input rows gives F = 1 only when A and B are the same. Name the single logic gate that this whole circuit is equivalent to.

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Question 403

[2 marks]logic gates and truth tables; digital transmission; resistor networks and internal resistance
An advantage of transmitting data digitally rather than in analogue form is that
  1. Aan analogue signal cannot be carried along an optic fibre at all, while a digital signal can.
  2. Ba digital signal is attenuated far less by a copper cable than an analogue signal of the same power.
  3. Ca digital signal needs no repeater amplifiers however long the transmission route becomes.
  4. Da digital signal can be regenerated exactly at a repeater, so the noise picked up is removed.

Question 404

[2 marks]logic gates and truth tables; digital transmission; resistor networks and internal resistance
A 3 ohm resistor and a 6 ohm resistor are connected in parallel, and that pair is in series with an 8 ohm resistor across a 4 V battery of negligible internal resistance. Calculate the charge that passes through the battery in 5 minutes.

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Question 405

[2 marks]logic gates and truth tables; digital transmission; resistor networks and internal resistance
A 3 ohm resistor and a 6 ohm resistor in parallel are in series with an 8 ohm resistor across a battery of e.m.f. 4 V and internal resistance r. In 5 minutes the charge that actually flows is measured as 105 C. Calculate r.

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Question 501

[2 marks]hysteresis in rubber; ideal gas law; internal energy
A stress-strain graph for a sample of rubber shows a loading curve and a lower unloading curve that together enclose a shaded loop. The area of that shaded loop represents
  1. Athe work done per unit volume in stretching the rubber from its natural length to its full extension.
  2. Bthe difference in the Young modulus of the rubber between the loading and the unloading stage.
  3. Cthe elastic potential energy per unit volume that the rubber gives back when the load is removed.
  4. Dthe energy per unit volume dissipated as heat in the rubber during each loading and unloading cycle.

Question 502

[2 marks]hysteresis in rubber; ideal gas law; internal energy
A conveyor belt is made of rubber whose loading and unloading stress-strain curves enclose a loop. Why is a smaller enclosed loop better for the belt?
  1. AA smaller loop means the belt has a lower Young modulus, so a less powerful drive motor is needed.
  2. BA smaller loop means less energy is lost as heat each cycle, so the belt runs cooler and lasts longer.
  3. CA smaller loop means the rubber is working closer to its breaking stress, so it can carry a heavier load.
  4. DA smaller loop means the belt stretches further under load, so it grips the drive rollers more firmly.

Question 503

[1 marks]hysteresis in rubber; ideal gas law; internal energy
A fixed mass of an ideal gas at 400 K has a volume of 1.00 x 10^-4 m^3. Its temperature is raised to T2 and its volume increases to 1.25 x 10^-4 m^3, the pressure staying at 4 000 Pa throughout. Calculate T2.

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Question 504

[3 marks]hysteresis in rubber; ideal gas law; internal energy
A fixed mass of an ideal monatomic gas is heated at a constant pressure of 4 000 Pa. Its volume increases from 1.00 x 10^-4 m^3 at 400 K to 1.25 x 10^-4 m^3 at 500 K. Taking the internal energy as U = (3/2)nRT with R = 8.31 J K^-1 mol^-1, calculate the change in internal energy.

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Question 601

[2 marks]Millikan's oil drop experiment; photoelectric effect; attenuation and optical fibres
In a Millikan oil drop experiment the charges measured on twelve different drops, in non S.I. units, were 14.22, 56.88, 28.44, 42.66, 85.32, 71.10, 33.18, 14.22, 56.88, 80.58, 42.66 and 99.54. Every reading must be a whole number multiple of the elementary charge. Determine the magnitude of the electron charge in the same units.

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Question 602

[2 marks]Millikan's oil drop experiment; photoelectric effect; attenuation and optical fibres
Which observation from the photoelectric effect shows that electromagnetic radiation behaves as particles rather than as a continuous wave?
  1. AThe photoelectric current increases steadily as the illuminated area of the metal is made larger.
  2. BBelow a threshold frequency no electrons are emitted at all, however intense the light is made.
  3. CThe number of electrons emitted each second rises as the intensity of the incident light is raised.
  4. DThe electrons are emitted from the surface of the metal rather than from deep inside the metal.

Question 603

[1 marks]Millikan's oil drop experiment; photoelectric effect; attenuation and optical fibres
Attenuation of a signal travelling along a coaxial cable is caused mainly by
  1. Atotal internal reflection of the signal at the boundary around the inner conductor.
  2. Bscattering of the signal by impurities in the glass from which the core has been drawn.
  3. Cthe signal spreading out into an ever wider beam as it travels further along the cable.
  4. Dthe resistance of the copper conductors, which turns part of the signal energy into heat.

Question 604

[1 marks]Millikan's oil drop experiment; photoelectric effect; attenuation and optical fibres
Attenuation of a light signal travelling along an optic fibre is caused mainly by
  1. Areflection of the signal back towards the transmitter each time the fibre is made to bend.
  2. Babsorption by impurities in the glass together with scattering by irregularities within it.
  3. Cthe cladding having a higher refractive index than the core, which lets the light leak away.
  4. Dthe resistance of the glass core, which turns part of the signal into heat as a current flows.

Question 605

[3 marks]Millikan's oil drop experiment; photoelectric effect; attenuation and optical fibres
A cable 70 km long has an attenuation per unit length of 6.2 dB km^-1. Twelve repeater amplifiers, each of gain 36 dB, are connected along the cable. A signal of input power 200 mW is transmitted along it. Calculate the power at the receiver end.

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