Michel · ICS Mushrif · Test on 30 October 2026 · 41 questions
Which of the following is an SI base unit?
Answer: C
Of these, only the kilogram is one of the seven SI base units (m, kg, s, A, K, mol, cd). The newton and joule are derived; the gram is not the base unit of mass.
Write the following in scientific notation with SI units: (a) 4.7 km (b) 350 µs (c) 0.026 MA
Answer: (a) 4.7 × 10³ m (b) 3.5 × 10⁻⁴ s (c) 2.6 × 10⁴ A
k = 10³, µ = 10⁻⁶, M = 10⁶. 350 × 10⁻⁶ = 3.5 × 10⁻⁴; 0.026 × 10⁶ = 2.6 × 10⁴.
A wavelength is 650 nm. In metres this is
Answer: B
n = 10⁻⁹, so 650 × 10⁻⁹ m = 6.5 × 10⁻⁷ m.
A car travels at 72 km h⁻¹. Convert this to m s⁻¹.
Answer: 20 m s⁻¹
72 km h⁻¹ = 72 × 1000 m ÷ 3600 s = 20 m s⁻¹ (shortcut: ÷ 3.6).
A sheet of glass has an area of 2.5 cm². Express this in m².
Answer: 2.5 × 10⁻⁴ m²
1 cm = 10⁻² m, so 1 cm² = (10⁻²)² = 10⁻⁴ m². 2.5 cm² = 2.5 × 10⁻⁴ m². (Square the prefix too!)
The density of iron is 7.9 g cm⁻³. Convert this to kg m⁻³.
Answer: 7.9 × 10³ kg m⁻³ (7900 kg m⁻³)
7.9 g = 7.9 × 10⁻³ kg; 1 cm³ = 10⁻⁶ m³. 7.9 × 10⁻³ ÷ 10⁻⁶ = 7.9 × 10³ kg m⁻³.
Which of these is the unit of acceleration written in SI base units in the correct format?
Answer: D
a = Δv/Δt → (m s⁻¹)/s = m s⁻². Units are written with a space and negative indices.
Estimate the order of magnitude of the ratio (height of the Burj Khalifa ≈ 830 m) ÷ (thickness of a sheet of paper ≈ 0.1 mm).
Answer: 10⁷
830 ÷ 1 × 10⁻⁴ = 8.3 × 10⁶ ≈ 10⁷ (8.3 rounds up to the next power of ten since log₁₀8.3 > 0.5).
A 1.5 t car moves at 90 km h⁻¹. Using KE = ½mv², find its kinetic energy in J, then express it with a suitable prefix.
Answer: 4.7 × 10⁵ J = 0.47 MJ (≈ 470 kJ)
m = 1.5 t = 1500 kg; v = 90 ÷ 3.6 = 25 m s⁻¹. KE = ½ × 1500 × 25² = 468 750 J ≈ 4.7 × 10⁵ J = 470 kJ.
A student repeats a reading many times and gets values that are very close together but far from the true value. The readings are
Answer: A
Close together = precise (small random error). Far from true value = inaccurate (systematic error).
Which error can be reduced by taking repeated readings and averaging?
Answer: B
Repeating and averaging reduces random error. Systematic errors (zero error, parallax done the same way every time) are not reduced by repeating.
State the reading uncertainty of: (a) a measuring cylinder with 2 mL divisions (b) a digital balance reading 52.37 g.
Answer: (a) ±1 mL (b) ±0.01 g
Analogue: half the smallest division → 2/2 = 1 mL. Digital: the smallest division the display can show → 0.01 g.
How many significant figures are in: (a) 0.00450 m (b) 3.020 × 10⁴ s (c) 600.0 g?
Answer: (a) 3 (b) 4 (c) 4
Leading zeros never count; trailing zeros after a decimal point do count. 4,5,0 → 3; 3,0,2,0 → 4; 6,0,0,0 → 4.
A metre rule with 1 mm divisions is used to measure a rod. The left end is at 3.0 cm and the right end at 72.4 cm. Write the length with its absolute uncertainty.
Answer: 69.4 ± 0.1 cm
Each end has reading uncertainty ±0.05 cm (half of 1 mm). Length uses two readings → uncertainties add: ±0.1 cm. L = 72.4 − 3.0 = 69.4 cm.
A pendulum period is timed five times with a stopwatch reading to 0.01 s: 3.42 s, 3.58 s, 3.47 s, 3.51 s, 3.62 s. Find the mean with its uncertainty (half-range plus reading error) and the percentage uncertainty.
Answer: 3.52 ± 0.11 s, ≈ 3.1 %
Mean = 17.60/5 = 3.520 s. Half range = (3.62 − 3.42)/2 = 0.10 s; + 0.01 reading error = 0.11 s. % = 0.11/3.52 × 100 = 3.1 %.
Two masses are (5.6 ± 0.3) kg and (2.1 ± 0.2) kg. Find (a) their total and (b) their difference, each with absolute uncertainty.
Answer: (a) 7.7 ± 0.5 kg (b) 3.5 ± 0.5 kg
Addition AND subtraction: add absolute uncertainties, 0.3 + 0.2 = 0.5 kg. 5.6 + 2.1 = 7.7; 5.6 − 2.1 = 3.5.
An experiment gives g = 9.62 m s⁻². The accepted value is 9.81 m s⁻². Find the percentage difference. Does this tell you about accuracy or precision?
Answer: 1.9 % (below the true value) — it measures accuracy
% difference = |9.62 − 9.81| / 9.81 × 100 = 1.9 %. Percentage difference compares with the true value, so it is about accuracy.
A cart travels d = (24.0 ± 0.5) m in t = (1.5 ± 0.1) s. Find the average speed v = d/t with percentage and absolute uncertainty.
Answer: 16 ± 1 m s⁻¹ (≈ 8.8 %)
v = 24.0/1.5 = 16.0 m s⁻¹. % = 0.5/24.0 + 0.1/1.5 = 2.1 % + 6.7 % = 8.8 %. Δv = 0.0875 × 16 = 1.4 → 1 m s⁻¹ (1 s.f.). Answer 16 ± 1 m s⁻¹.
The radius of a disc is (2.50 ± 0.05) cm. Find its area A = πr² with absolute uncertainty.
Answer: 19.6 ± 0.8 cm²
A = π × 2.50² = 19.63 cm². % in r = 0.05/2.50 = 2 %; r² → ×2 = 4 % (π has no uncertainty). ΔA = 0.04 × 19.63 = 0.79 ≈ 0.8 cm². Answer 19.6 ± 0.8 cm².
A quantity is calculated from Z = x³/y. x has 2 % uncertainty and y has 3 % uncertainty. The percentage uncertainty in Z is
Answer: D
Powers multiply the % uncertainty: 3 × 2 % = 6 %. Division adds: 6 % + 3 % = 9 %.
On a v–t graph with error bars, the steepest acceptable line passes through (0.0 s, 0.6 m s⁻¹) and (4.0 s, 9.0 m s⁻¹). The least steep acceptable line passes through (0.0 s, 1.4 m s⁻¹) and (4.0 s, 8.2 m s⁻¹). Find the acceleration with its uncertainty.
Answer: 1.9 ± 0.2 m s⁻²
Max gradient = (9.0 − 0.6)/4.0 = 2.1 m s⁻². Min gradient = (8.2 − 1.4)/4.0 = 1.7 m s⁻². Best ≈ average = 1.9 m s⁻²; uncertainty = half the difference = (2.1 − 1.7)/2 = 0.2 m s⁻². Gradient of v–t = acceleration.
Which list contains only vector quantities?
Answer: B
Displacement, velocity, acceleration and force all have direction. Speed, distance, time, mass and energy are scalars.
A student walks 30 m east then 30 m west. State (a) the distance travelled and (b) the displacement.
Answer: (a) 60 m (b) 0 m
Distance is a scalar: 30 + 30 = 60 m. Displacement is a vector: +30 + (−30) = 0 m (back at the start).
A boat moves at 6.0 m s⁻¹ east relative to the water, in a current flowing at 8.0 m s⁻¹ north. Find the magnitude and direction of the resultant velocity.
Answer: 10 m s⁻¹ at 53° north of east
Perpendicular → Pythagoras: √(6.0² + 8.0²) = 10 m s⁻¹. Angle: tan θ = 8.0/6.0 → θ = 53° north of east.
A rope pulls a box with a force of 50 N at 30° above the horizontal. Resolve the force into horizontal and vertical components.
Answer: Horizontal 43 N, vertical 25 N
Fx = 50 cos 30° = 43.3 N ≈ 43 N; Fy = 50 sin 30° = 25 N.
Two forces of 5 N and 12 N act on an object. Which value could NOT be the magnitude of their resultant?
Answer: D
The resultant lies between |12 − 5| = 7 N (opposite) and 12 + 5 = 17 N (same direction). 18 N is impossible.
A ball is kicked at 20 m s⁻¹ at 40° above the horizontal. Find its horizontal and vertical velocity components.
Answer: vx ≈ 15 m s⁻¹, vy ≈ 13 m s⁻¹
vx = 20 cos 40° = 15.3 m s⁻¹; vy = 20 sin 40° = 12.9 m s⁻¹.
A drone flies 40 m due east, then 30 m in a direction 60° north of east. Find the magnitude and direction of its total displacement by resolving into components.
Answer: ≈ 61 m at 25° north of east
East: 40 + 30 cos 60° = 40 + 15 = 55 m. North: 30 sin 60° = 26.0 m. R = √(55² + 26.0²) = 60.8 m ≈ 61 m. θ = tan⁻¹(26.0/55) = 25° north of east.
A puck moving at 4.0 m s⁻¹ east is struck and moves off at 3.0 m s⁻¹ north. Find the magnitude of the change in velocity Δv = v − v₀.
Answer: 5.0 m s⁻¹ (directed north-west, 37° north of west)
Δv = v + (−v₀): add 3.0 north and 4.0 west. |Δv| = √(3.0² + 4.0²) = 5.0 m s⁻¹; tan θ = 3/4 → 37° north of west.
On a position–time (x–t) graph, a horizontal straight line means the object is
Answer: A
Gradient of x–t = velocity. Horizontal → gradient 0 → stationary.
The area under a velocity–time graph represents
Answer: C
Area under v–t = displacement. (Gradient of v–t = acceleration; area under a–t = change in velocity.)
A scooter starts from rest and accelerates uniformly at 3.2 m s⁻² for 7.5 s. Find its final velocity.
Answer: 24 m s⁻¹
v = v₀ + at = 0 + 3.2 × 7.5 = 24 m s⁻¹.
A train slows from 30 m s⁻¹ to 12 m s⁻¹ in 4.5 s. Find its acceleration.
Answer: −4.0 m s⁻²
a = (v − v₀)/t = (12 − 30)/4.5 = −4.0 m s⁻² (negative: deceleration).
A cyclist moving at 8.0 m s⁻¹ accelerates at 1.5 m s⁻² for 6.0 s. How far does she travel in this time?
Answer: 75 m
x − x₀ = v₀t + ½at² = 8.0(6.0) + ½(1.5)(6.0²) = 48 + 27 = 75 m.
A car travelling at 25 m s⁻¹ brakes with a constant acceleration of −6.25 m s⁻². Find the stopping distance.
Answer: 50 m
No time given → use v² = v₀² + 2a(x − x₀): 0 = 25² + 2(−6.25)Δx → Δx = 625/12.5 = 50 m.
A stone is dropped from a cliff 20 m high (ignore air resistance, g = 9.8 m s⁻²). Find (a) the time to reach the ground (b) its speed on impact.
Answer: (a) 2.0 s (b) 20 m s⁻¹ (19.8)
(a) 20 = ½(9.8)t² → t = √(40/9.8) = 2.02 s. (b) v = gt = 9.8 × 2.02 = 19.8 m s⁻¹ (or v² = 2gh).
A sprinter starts from rest and accelerates uniformly, covering 100 m in 10 s. Find (a) the acceleration (b) the final velocity (c) the average velocity.
Answer: (a) 2.0 m s⁻² (b) 20 m s⁻¹ (c) 10 m s⁻¹
(a) Δx = ½at² → 100 = ½a(10²) → a = 2.0 m s⁻². (b) v = at = 20 m s⁻¹. (c) v̄ = Δx/t = 100/10 = 10 m s⁻¹ (= (0 + 20)/2 ✓).
A v–t graph for a lift: v rises uniformly from 0 to 12 m s⁻¹ in the first 4.0 s, stays at 12 m s⁻¹ until t = 10 s, then falls uniformly to 0 at t = 13 s. Find (a) the acceleration in each stage (b) the total displacement (c) the average velocity.
Answer: (a) 3.0, 0, −4.0 m s⁻² (b) 114 m (c) 8.8 m s⁻¹
(a) 12/4 = 3.0 m s⁻²; 0; (0 − 12)/3 = −4.0 m s⁻². (b) Area = ½(4)(12) + 6(12) + ½(3)(12) = 24 + 72 + 18 = 114 m. (c) 114/13 = 8.8 m s⁻¹.
A ball is thrown vertically upwards at 14.7 m s⁻¹ from ground level. Find (a) the time to reach the top (b) the maximum height (c) the total time before it lands.
Answer: (a) 1.5 s (b) 11 m (c) 3.0 s
Up = +, a = −9.8 m s⁻². (a) 0 = 14.7 − 9.8t → t = 1.5 s. (b) 0 = 14.7² − 2(9.8)h → h = 11.0 m ≈ 11 m. (c) Symmetric flight: 2 × 1.5 = 3.0 s.
A truck passes a parked police car at a constant 20 m s⁻¹. At that moment the police car starts from rest with acceleration 2.5 m s⁻². Find (a) the time for the police car to catch the truck (b) the distance covered (c) the police car's speed then.
Answer: (a) 16 s (b) 320 m (c) 40 m s⁻¹
Same position: 20t = ½(2.5)t² → t = 40/2.5 = 16 s. Distance = 20 × 16 = 320 m. v = 2.5 × 16 = 40 m s⁻¹.
A stone is thrown straight down from a bridge at 12 m s⁻¹ and hits the water 2.5 s later. Find (a) the height of the bridge (b) the stone's speed on hitting the water.
Answer: (a) 61 m (b) 36.5 m s⁻¹
Down = +. (a) Δx = 12(2.5) + ½(9.8)(2.5²) = 30 + 30.6 = 60.6 m ≈ 61 m. (b) v = 12 + 9.8(2.5) = 36.5 m s⁻¹.