Salam Academy
Practise

Question bank

Original exam-style questions, mapped to your syllabus. Answers are checked for mathematical equivalence and every attempt updates your mastery.

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2730 questions · page 75 of 137 · choose a topic above to narrow them down

SA-9702-EC1-0032 marksL2 StandardCalculatorDrift velocity
A copper wire of cross-sectional area 1.0×10−61.0\times10^{-6} m² carries a current of 2.02.0 A. Copper has 8.5×10288.5\times10^{28} free electrons per m³. Calculate the drift velocity, in mm s⁻¹.

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Salam Academy Original Practice · Proposed mark scheme
SA-9702-EC1-0051 markL2 StandardCalculatorCharge & current
Which charge is not possible on an oil drop?
Choose one option

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SA-9702-EC2-0042 marksL2 StandardCalculatorElectrical power
A heater is rated 10001000 W on a 230230 V supply. Calculate its resistance.

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SA-9702-EC3-0022 marksL2 StandardCalculatorResistivity
A wire 0.500.50 m long with cross-sectional area 1.0×10−71.0\times10^{-7} m² has resistance 4.04.0 Ω. Calculate its resistivity, in units of 10−710^{-7} Ω m.

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SA-9702-EC3-0032 marksL2 StandardCalculatorResistivity
Copper has resistivity 1.7×10−81.7\times10^{-8} Ω m. Calculate the length of copper wire of cross-sectional area 1.0×10−61.0\times10^{-6} m² that has a resistance of 1010 Ω.

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SA-9702-EF1-0032 marksL2 StandardCalculatorUniform fields
Calculate the acceleration of an electron in a field of 2.5×1042.5\times10^4 V m⁻¹, in units of 101510^{15} m s⁻². (me=9.11×10−31m_e=9.11\times10^{-31} kg)

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SA-9702-EF2-0012 marksL2 StandardCalculatorCoulomb's law
Calculate the electric force between a proton and an electron 5.3×10−115.3\times10^{-11} m apart, in units of 10−810^{-8} N. (14πε0=8.99×109\tfrac1{4\pi\varepsilon_0}=8.99\times10^9 N m² C⁻²)

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SA-9702-EF2-0052 marksL2 StandardCalculatorPoint charge field
Charges of +4.0+4.0 μC and +1.0+1.0 μC are 6.06.0 cm apart. Calculate the distance from the 4.04.0 μC charge to the neutral point between them, in cm.

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SA-9702-EF3-0022 marksL2 StandardCalculatorElectric potential
An α-particle (charge +2e+2e) has kinetic energy 8.0×10−138.0\times10^{-13} J and travels head-on towards a gold nucleus (charge +79e+79e). Calculate the distance of closest approach, in units of 10−1410^{-14} m. (14πε0=8.99×109\tfrac1{4\pi\varepsilon_0}=8.99\times10^9 N m² C⁻²; e=1.60×10−19e=1.60\times10^{-19} C)

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SA-9702-EF3-0051 markL2 StandardCalculatorElectric potential
A +3.0+3.0 μC charge is moved from a point at 5050 V to a point at 250250 V. Calculate the work done on it, in mJ.

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SA-9702-FD1-0032 marksL2 StandardCalculatorMoments & couples
A uniform beam 2.02.0 m long and of weight 100100 N is hinged at one end. It is held horizontal by a vertical force at the other end. Calculate this force.

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SA-9702-FD1-0042 marksL2 StandardCalculatorMoments & couples
A 4040 N force acts on a bar 0.500.50 m from the pivot, at 30∘30^\circ to the bar. Calculate its moment about the pivot.

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SA-9702-FD2-0052 marksL2 StandardCalculatorUpthrust
The object in the previous question has a mass of 5.05.0 kg. Calculate its apparent weight when fully submerged.

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SA-9702-GF1-0012 marksL2 StandardCalculatorLaw of gravitation
Earth (mass 6.0×10246.0\times10^{24} kg) and the Moon (mass 7.3×10227.3\times10^{22} kg) are 3.8×1083.8\times10^8 m apart. Calculate the gravitational force between them, in units of 102010^{20} N. (G=6.67×10−11G=6.67\times10^{-11} N m² kg⁻²)

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SA-9702-GF1-0022 marksL2 StandardCalculatorField strength
Calculate the gravitational field strength at a height of 400400 km above Earth's surface. (Earth: mass 6.0×10246.0\times10^{24} kg, radius 6.4×1066.4\times10^6 m; G=6.67×10−11G=6.67\times10^{-11} N m² kg⁻²)

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SA-9702-GF2-0012 marksL2 StandardCalculatorGravitational potential
Calculate the gravitational potential at Earth's surface, in units of 10710^7 J kg⁻¹. (Earth: mass 6.0×10246.0\times10^{24} kg, radius 6.4×1066.4\times10^6 m; G=6.67×10−11G=6.67\times10^{-11} N m² kg⁻²)

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SA-9702-GF2-0022 marksL2 StandardCalculatorOrbits
Calculate the speed of a satellite in a circular orbit of radius 7.0×1067.0\times10^6 m around Earth. (GM=4.0×1014GM=4.0\times10^{14} N m² kg⁻¹)

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SA-9702-GF2-0041 markL2 StandardCalculatorGravitational potential
Gravitational potential is negative because
Choose one option

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SA-9702-IG1-0012 marksL2 StandardCalculatorIdeal gas equation
A room of volume 5050 m³ contains air at 1.0×1051.0\times10^5 Pa and 2020 °C. Calculate the number of moles of air. (R=8.31R=8.31 J mol⁻¹ K⁻¹)

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SA-9702-IG1-0052 marksL2 StandardCalculatorThe mole
Calculate the number of moles in 8888 g of carbon dioxide (molar mass 4444 g mol⁻¹), and hence the number of molecules, in units of 102410^{24}.

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Salam Academy Original Practice · Proposed mark scheme