Chapter 1 — Units, measurement, motion and force
What this chapter is for
A police-exam physics question often hides a simple relationship under familiar words. A vehicle travels a distance; an object changes speed; a force acts in one direction while friction acts in another. Before using a formula, decide what is being measured and in which unit. This chapter develops that habit through units, graphs, worked calculations and force diagrams. Its examples are synthetic practice situations, not measurements of actual traffic or policing.
A measurement has a number and a unit
“Twenty” alone is incomplete as a length. Twenty metres and twenty kilometres describe very different journeys. In the International System of Units, length uses the metre (m), mass the kilogram (kg) and time the second (s). Current uses the ampere (A), thermodynamic temperature the kelvin (K), amount of substance the mole (mol), and luminous intensity the candela (cd). Speed is a derived quantity measured in metres per second (m/s); force is measured in newtons (N), with 1 N = 1 kg·m/s². The unit itself can catch an incorrect operation: distance divided by time has the unit m/s, while distance multiplied by time has m·s and cannot be a speed.
Convert before calculating. One kilometre is 1,000 metres and one hour is 3,600 seconds. Therefore 1 km/h = 1,000/3,600 m/s = 5/18 m/s. A speed of 72 km/h is 72×5/18=20 m/s. To reverse the conversion, multiply m/s by 18/5. Write the unit at the end of every line; this is a practical way to detect a factor-of-3.6 error. If a question reports 60 km in 1.5 hours, average speed is 60/1.5=40 km/h. Converting both distance and time first gives about 11.1 m/s, the same speed expressed in another unit.
Scientific notation is useful when values are very large or small. For example, 0.00025 m is 2.5×10⁻⁴ m. The exponent records place value; it does not create a new unit. A prefix changes the scale: 1 millimetre is 10⁻³ metre and 1 centimetre is 10⁻² metre. Do not confuse milli (one thousandth) with micro (one millionth). In routine MCQs, the safest method is to write the conversion equality and then substitute numbers instead of recalling an isolated multiplier.
Check a calculation by its dimensions
Dimensional checking is a fast way to reject a formula without remembering its name. Speed has dimensions length divided by time, written L/T. Acceleration has L/T². Force is mass times acceleration, so its dimensions are ML/T². The expression v²/r has dimensions (L/T)²/L=L/T² and can therefore represent an acceleration; v/r has dimensions 1/T and cannot. Matching dimensions is necessary, but it is not proof that a formula is physically correct: v²/r and 2v²/r have the same dimensions, although only one may fit a specified motion model. Use dimensions first to eliminate impossible options, then use the stated physical relationship to choose among the survivors.
Suppose an item asks for the distance travelled from rest under a constant acceleration of 4 m/s² for 3 s. The expression at has units m/s, so it gives a speed change. The expression ½at² has units m and can give a distance: ½×4×3²=18 m. The squared time is essential. If the options include 12 m/s, 18 m, 36 N and 9 s, a unit check already isolates the plausible kind of answer before the arithmetic begins. A numerical answer without the requested unit is incomplete in written work, even when a multiple-choice option supplies it.
Measurements also have sensible precision. A stopwatch reading to the nearest tenth of a second does not justify a final speed quoted to eight decimal places. For exam arithmetic, calculate with enough digits to avoid premature rounding and round the final result to match the precision of the options. If two options are close, retain an extra digit until the comparison is complete. Precision should not be confused with accuracy: repeated measurements can agree closely with one another yet all be displaced by a badly calibrated instrument. This distinction matters in experiment-based questions, even when no formal uncertainty calculation is required.
Distance, displacement, speed and velocity
Distance measures the length of the path travelled. Displacement joins the starting point to the ending point with a direction. If a person walks 3 km east and then 3 km west, distance is 6 km but displacement is zero. If the trip takes 2 hours, average speed is 6/2=3 km/h, while average velocity is zero displacement divided by 2 hours, or zero. This does not mean the person stood still; it means the final position matches the initial position. An exam distractor often uses distance where displacement is needed.
Speed is a scalar: a magnitude is enough. Velocity is a vector: magnitude and direction matter. A change in direction is a change in velocity even if speed stays constant. A vehicle moving around a circular track at constant speed continuously changes direction and therefore accelerates. This distinction explains why “constant speed” does not always mean “zero acceleration.” In a straight-line question where direction never changes, the numerical values of speed and velocity can coincide, but their meanings remain different.
Average speed for several legs equals total distance divided by total time. Do not take the simple mean of leg speeds unless the time spent at each speed is equal. A traveller covers 60 km at 30 km/h and another 60 km at 60 km/h. The first leg takes 2 hours, the second 1 hour, so average speed is 120/3=40 km/h, not 45 km/h. If the legs instead lasted one hour each, the speeds could be averaged directly because the times are equal. Ask which quantity is equal before choosing a shortcut.
Route problems without a hidden shortcut
For two perpendicular legs, displacement can be found from a right triangle. A patrol vehicle moves 3 km east and then 4 km north. Total path length is 7 km; the straight-line displacement has magnitude √(3²+4²)=5 km and points northeast. If the journey takes half an hour, average speed is 7/0.5=14 km/h. Average-velocity magnitude is 5/0.5=10 km/h, with a northeast direction. The values 7 and 5 answer different questions. If the route instead had a 3 km east leg followed by 4 km west, the displacement would be 1 km west, not 7 km and not 5 km. Draw the final position before invoking a square root.
The familiar equal-distance average-speed shortcut can be derived rather than memorised. Let each of two legs have distance d, travelled at speeds v₁ and v₂. The total time is d/v₁+d/v₂, so average speed is 2d/(d/v₁+d/v₂)=2v₁v₂/(v₁+v₂). At 30 and 60 km/h this is 2×30×60/90=40 km/h, agreeing with the direct calculation above. This formula is valid for two equal distances. If each speed is maintained for an equal time instead, average speed is the arithmetic mean, 45 km/h in that example. When an option uses the wrong average, the stem usually contains the clue “equal distances” or “equal times.”
Relative speed questions also depend on direction. In a straight line, two vehicles approaching one another at 40 and 50 km/h close the gap at 90 km/h. If they travel in the same direction, the faster gains on the slower at 10 km/h. A 180 km separation closes in 2 hours in the first case. A 20 km lead would be erased in 2 hours in the same-direction case. These are different initial separations, deliberately chosen to demonstrate the rate. Do not add speeds simply because two vehicles appear in a question; decide whether the distance between them is shrinking by the sum or the difference.
Acceleration and motion graphs
Acceleration is the rate at which velocity changes: a=(v−u)/t for a constant change over time t. If a cart's straight-line velocity rises from 5 m/s to 17 m/s in 4 seconds, a=12/4=3 m/s². The squared time unit is not optional: acceleration is a change in m/s per second. A negative answer for an object moving in the positive direction can represent slowing down, but a sign alone does not tell you whether a vehicle is moving backwards. Define the positive direction before interpreting signs.
A distance–time graph shows how accumulated path length changes with time. Its slope represents speed. A flat section means the distance did not increase during that interval, so the object was stationary. A steeper upward section means greater speed. On a displacement–time graph, a downward slope can mean movement in the negative direction; distance itself cannot decrease merely because the traveller turns around. First check which vertical quantity the graph displays.
A velocity–time graph gives acceleration through its slope. For constant acceleration, the area under the graph gives displacement over the interval, with signed area if velocity crosses zero. Suppose velocity rises uniformly from 0 to 12 m/s over 4 seconds. The graph forms a triangle with base 4 s and height 12 m/s, so displacement is ½×4×12=24 m. The slope is 12/4=3 m/s². The same graph yields two different answers depending on whether the question asks for area or slope. The accompanying original diagram labels both quantities.
Read a motion graph in intervals
Many graph items are easier when split into rectangles and triangles. Imagine a velocity–time graph in which velocity rises uniformly from 0 to 10 m/s during the first 5 s, stays at 10 m/s for the next 3 s, and then falls uniformly to zero over 2 s. Displacement is the sum of three areas: ½×5×10=25 m, 3×10=30 m, and ½×2×10=10 m. The total is 65 m over 10 s, giving an average speed of 6.5 m/s because velocity never becomes negative. Acceleration is 2 m/s² during the first part, zero during the middle part, and −5 m/s² during the final part. A single average acceleration over all 10 s would be (0−0)/10=0, even though the object accelerated and decelerated within the journey. This is why a whole-trip average cannot describe every segment.
If a velocity graph crosses below the time axis, the signed area below it counts as negative displacement. For example, 20 m of positive area followed by 8 m of negative area gives net displacement +12 m but total distance 28 m. A speed–time graph has no negative speed region: its area gives distance directly. The vertical-axis label, not the shape alone, decides whether to add absolute areas or signed areas. A decreasing positive velocity is still forward motion with negative acceleration; it is not backward motion until velocity itself becomes negative under the chosen direction convention.
Uniform-acceleration equations are useful when a graph is absent. In one dimension, v=u+at, s=ut+½at² and v²=u²+2as. State the sign convention first. A vehicle at +20 m/s brakes uniformly to rest over 50 m. With u=+20, v=0 and s=+50, the third relation gives 0=400+100a, so a=−4 m/s². The stopping time follows from 0=20−4t: t=5 s. The negative acceleration means it points against the original motion. Had the problem asked for the magnitude of deceleration, the answer would be 4 m/s², a positive magnitude. A negative sign is information, not an arithmetic mistake.
Newton's laws as a reasoning framework
Newton's first law says that a body continues at rest or in uniform straight-line motion unless a net external force changes that state. It is a statement about net force. Several forces may act while their vector sum is zero. A book resting on a table has weight downward and a supporting normal force upward; in a simple stationary model these balance. Friction can oppose relative motion or a tendency to slip. It is not automatically equal to the object's weight.
The second law connects net force, mass and acceleration: F=ma for constant mass in an inertial frame. If a 2 kg object has a net force of 6 N in one direction, its acceleration is 3 m/s² in that direction. If two 6 N forces act in opposite directions, their net force is zero and this calculation must not be made with just one of them. Draw arrows and combine forces before substituting into F=ma. Mass is resistance to acceleration and is measured in kilograms; weight is a gravitational force measured in newtons. Near Earth's surface, weight is approximately mg, where g is about 9.8 m/s². A 2 kg mass therefore weighs about 19.6 N in that approximation.
Newton's third law says that interacting bodies exert equal and opposite forces on one another. These two forces act on different bodies, so they do not cancel as forces on the same object. A foot pushes the ground backward and the ground pushes the foot forward. Confusing the third-law pair with two balancing forces on one body is a common exam mistake. Identify the body receiving each force before describing a pair.
Draw the body before listing forces
A free-body diagram represents the forces on one chosen body. For a block resting on a horizontal surface, draw its weight vertically downward and the normal support vertically upward. If the block is stationary and no other vertical force acts, their magnitudes match. If a hand pushes the block horizontally while static friction prevents movement, the friction force opposes that push and can match it up to its limiting value. It is incorrect to say that friction must always equal μN: the familiar μN expression is a limiting or kinetic-friction model under specified conditions, not the actual static friction in every situation. In a school-level question that explicitly gives a friction magnitude, use the given value and direction.
For a vehicle moving right while braking, the horizontal road force on the tyres can point left, producing a leftward net force and a reduction in rightward velocity. Motion direction and force direction need not match. By contrast, if net force is zero, velocity is constant in an inertial frame; it need not be zero. A bus moving straight at steady speed illustrates zero acceleration, not necessarily an absence of individual forces. Engine driving force and resistive forces can balance. These examples make the first and second laws consistent rather than contradictory.
The normal force and weight on a stationary book are a balanced pair on the book. They are not a Newton third-law pair. The third-law partner to the table's upward force on the book is the book's downward force on the table. The third-law partner to Earth's downward gravitational force on the book is the book's upward gravitational force on Earth. Those paired forces act on different objects. In an MCQ, write “force on A by B” beside each arrow; if both arrows are on A, they are not a third-law pair.
Force calculations should follow a three-step routine: choose the body, draw and label all relevant force directions, and resolve the net force along each requested axis. A 5 kg crate receives a 25 N horizontal pull right and a 10 N resistive force left. Horizontal net force is 15 N right, so horizontal acceleration is 15/5=3 m/s² right. Its weight is about 49 N downward, but that weight is balanced by a vertical support force in the simple level-floor case. Adding 49 N to the 15 N horizontal net force as ordinary numbers would mix perpendicular directions. A question asking only horizontal acceleration needs the horizontal net force.
Worked mixed example
A trolley of mass 4 kg is pulled right by 18 N while friction acts left with 6 N. Net horizontal force is 18−6=12 N right. Acceleration is 12/4=3 m/s² right. If it starts from rest and acceleration remains constant for 5 seconds, final velocity is u+at=0+3×5=15 m/s. The distance in that interval is ut+½at²=0+½×3×25=37.5 m. Each answer has its own unit: 12 N, 3 m/s², 15 m/s and 37.5 m. If a distractor offers 18/4=4.5 m/s², it has ignored friction. If it offers 3×25=75 m, it has omitted the one-half in the displacement formula for motion from rest.
The formula above depends on constant acceleration. If the pulling force or friction changes with time, that simple substitution no longer gives the exact distance. Exam questions usually say “uniform acceleration” or provide a graph when they expect a direct calculation. Read that condition rather than applying a remembered formula to every moving object.
Recall and error check
Before opening the practice set, try these prompts without options. What is the SI unit of acceleration? Why can a moving object have zero average velocity over a full trip? What does the slope of a velocity–time graph represent? What does its area represent? Which forces should be combined before using F=ma? On which bodies do the two forces in a third-law pair act? If you can answer those questions and preserve units in the worked example, the chapter's factual MCQs become easier to reason through rather than merely recognise.
Figure 1. Original chapter diagram.
Practice 36 questions
Choose one answer for each item. The answer key and explanations follow this chapter set.
1. The relation F = ma expresses which law of motion?
A. Newton's first law B. Newton's third law C. Law of universal gravitation D. Newton's second law
2. What is the SI unit of acceleration?
A. Metre per second squared (m/s²) B. Metre per second (m/s) C. Kilogram metre per second (kg m/s) D. Newton (N)
3. What is the SI base unit for the amount of substance?
A. Candela B. Mole C. Kelvin D. Ampere
4. In the usual school-level comparison under the same load and comparable contact conditions, arrange these resistances in decreasing order: A. rolling resistance; B. maximum (limiting) static friction; C. kinetic sliding friction.
A. B, C, A B. C, B, A C. A, C, B D. B, A, C
5. In which work did Isaac Newton first publish his three laws of motion?
A. Philosophiæ Naturalis Principia Mathematica B. Opticks C. Dialogue Concerning the Two Chief World Systems D. Astronomia Nova
6. Giving vehicles, ships and aircraft a streamlined shape mainly reduces:
A. Static friction B. Fluid friction (drag) C. Rolling friction D. Gravitational pull
7. For an artificial satellite orbiting the Earth, the necessary centripetal force is provided by:
A. Centrifugal force B. Air friction C. Earth's gravitational pull D. Earth's magnetic field
8. A body moves along a circular path with constant speed. Which statement is correct?
A. Both its speed and velocity are constant B. Its velocity is constant but speed changes C. Its acceleration is zero D. Its velocity changes, so it is accelerated
9. The launching of a rocket is based on which law or principle?
A. Newton's third law of motion B. Newton's first law of motion C. Bernoulli's principle D. Pascal's law
10. Match List I (law/concept) with List II (everyday example) and select the correct code.
List I:
A. Newton's first law
B. Newton's second law
C. Newton's third law
D. Centripetal force
List II:
1. Recoil of a gun on firing
2. Passenger falls forward when a bus brakes
3. Same push accelerates an empty trolley more than a loaded one
4. Stone whirled on a string
A. A-3, B-2, C-1, D-4 B. A-2, B-1, C-3, D-4 C. A-1, B-3, C-2, D-4 D. A-2, B-3, C-1, D-4
11. A 20 g bullet leaves a 4 kg gun at 300 m/s. What is the magnitude of the gun's recoil velocity?
A. 0.15 m/s B. 15 m/s C. 6 m/s D. 1.5 m/s
12. A car covers the first half of a journey at 40 km/h and the second half (the same distance) at 60 km/h. What is its average speed for the whole journey?
A. 50 km/h B. 52 km/h C. 45 km/h D. 48 km/h
13. A car moving at 72 km/h is brought uniformly to rest in 5 seconds. How far does it travel while stopping?
A. 100 m B. 50 m C. 25 m D. 360 m
14. A runner goes from one end of the diameter of a semicircular track of radius 7 m to the other end along the curved path. Taking π = 22/7, what are the distance covered and the magnitude of displacement?
A. 22 m and 14 m B. 14 m and 22 m C. 44 m and 14 m D. 22 m and 22 m
15. A ball of mass 500 g moves with a velocity of 20 m/s. What is its momentum?
A. 10,000 kg m/s B. 100 kg m/s C. 10 kg m/s D. 1 kg m/s
16. What force is needed to give a car of mass 1500 kg an acceleration of 2 m/s²?
A. 750 N B. 1502 N C. 300 N D. 3000 N
17. For the same change in a cricket ball's momentum, why does a fielder pull his hands backward while catching it?
A. Increases the time of impact and so reduces the average force on the hands B. Reduces the momentum of the ball to zero instantly C. Increases the force on the hands D. Increases the mass of the ball
18. Which is the SI unit of luminous intensity?
A. Lumen B. Lux C. Candela D. Watt
19. When a moving bus stops suddenly, standing passengers lurch forward. This is due to:
A. Inertia of rest B. Newton's third law C. Inertia of motion D. Conservation of energy
20. When the branch of a mango tree is shaken vigorously, ripe mangoes fall off. This is an example of:
A. Inertia of motion B. Newton's third law C. Centripetal force D. Inertia of rest
21. One newton of force is equal to how many dyne?
A. 10³ dyne B. 10⁷ dyne C. 10⁵ dyne D. 10² dyne
22. The inertia of a body depends on its:
A. Velocity B. Mass C. Shape D. Colour
23. Which statement about the action and reaction forces of Newton's third law is correct?
A. They act on the same body and cancel each other B. They are equal and opposite and act on two different bodies, so they do not cancel C. Reaction is always smaller than action D. Reaction begins only after the action ends
24. A stone tied to a string is whirled in a horizontal circle.
Statement I: If the string snaps, the stone's initial horizontal direction of motion is tangent to the circle at that instant.
Statement II: If the string snaps, the stone's initial horizontal direction of motion is radially outward from the centre.
Which of the statements is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
25. Consider the following statements under a usual comparison for the same contacting surfaces.
Statement I: The maximum (limiting) static friction is generally greater than kinetic sliding friction.
Statement II: Action and reaction forces act on the same body.
Which of the above statement(s) is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
26. Consider the following statements:
Statement I: The magnitude of displacement can be greater than the distance travelled.
Statement II: Speed is a vector quantity.
Which of the above statement(s) is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
27. Which of the following does NOT help to reduce friction between moving parts or with the surrounding fluid?
A. Applying oil or grease between machine parts B. Using ball bearings in place of sliding contact C. Giving vehicles and aircraft a streamlined shape D. Making the sliding surfaces rougher
28. Which of the following is a dimensionless quantity?
A. Density B. Specific gravity C. Pressure D. Momentum
29. Which of the following is NOT a vector quantity?
A. Velocity B. Speed C. Momentum D. Acceleration
30. Arrange the following SI prefixes in increasing order of the multiplying factor they represent:
A. micro
B. pico
C. milli
D. nano
A. D, B, A, C B. B, D, C, A C. C, A, D, B D. B, D, A, C
31. A micrometer screw gauge works on which principle?
A. Principle of moments B. Archimedes' principle C. Pascal's law D. Principle of the screw — each rotation advances a fixed small distance
32. Which instrument is mainly used to measure the curvature of a spherical surface such as a lens?
A. Screw gauge B. Vernier caliper C. Spherometer D. Odometer
33. What is the dimensional formula of work (or energy)?
A. [M L² T⁻²] B. [M L T⁻²] C. [M L² T⁻³] D. [M L⁻¹ T⁻²]
34. Match List I (physical quantity) with List II (SI unit) and select the correct code.
List I:
A. Pressure
B. Frequency
C. Electric charge
D. Electric resistance
List II:
1. coulomb
2. ohm
3. pascal
4. hertz
A. A-4, B-3, C-1, D-2 B. A-3, B-4, C-2, D-1 C. A-3, B-4, C-1, D-2 D. A-1, B-4, C-3, D-2
35. A vernier caliper has 1 main-scale division = 1 mm and 10 vernier divisions span 9 main-scale divisions. What is its least count?
A. 0.001 cm B. 0.01 cm C. 0.1 cm D. 0.0001 cm
36. A 100 W bulb glows for exactly one hour. How much energy does it consume?
A. 3,600 J B. 1 × 10² J C. 3.6 × 10⁵ J D. 6,000 J
Answers and explanations
1. D. F = ma is Newton's second law: for a given mass a larger force gives a larger acceleration, and for a given force a larger mass gives a smaller acceleration. The first law describes inertia and the third describes equal and opposite action and reaction.
2. A. Acceleration is the rate of change of velocity, so its unit is (m/s) ÷ s = metre per second squared (m/s²). Metre per second is the unit of speed and velocity, kg m/s of momentum and the newton of force.
3. B. The mole (symbol mol) is the SI base unit of the amount of substance. Candela measures luminous intensity, kelvin measures temperature and ampere measures electric current; with metre, kilogram and second these make up the seven SI base units.
4. A. The limiting static friction is the largest resistance just before sliding begins. Kinetic sliding friction is usually lower under the stated comparison, and rolling resistance is usually lower still. The order is B, C, A. Ordinary static friction can take many values below its limit, so the word 'maximum' matters.
5. A. Newton's laws of motion and his law of gravitation appeared in the Principia Mathematica (1687). Opticks (1704) was Newton's book on light, the Dialogue was Galileo's, and Astronomia Nova was Kepler's.
6. B. A streamlined shape lets air or water flow smoothly around the body, reducing fluid friction or drag. Rolling friction is reduced by wheels and bearings, and streamlining has no effect on gravity.
7. C. Earth's gravity pulls the satellite toward Earth's centre and acts as the centripetal force that keeps it on its curved orbit. Centrifugal force is only an apparent effect, and there is almost no air at orbital heights.
8. D. In uniform circular motion the speed stays the same but the direction changes continuously, so velocity (a vector) changes. A changing velocity means acceleration, directed toward the centre — centripetal acceleration.
9. A. A rocket expels hot gases downward at great speed (action) and the gases push the rocket upward with an equal and opposite force (reaction) — Newton's third law. Bernoulli's principle explains lift on wings, Pascal's law hydraulic machines.
10. D. First law (inertia) — passenger lurching forward; second law (F = ma) — lighter trolley accelerates more for the same push; third law — gun recoil; centripetal force — tension in the string keeps a whirled stone on its circle. Code: A-2, B-3, C-1, D-4.
11. D. Momentum is conserved and starts at zero, so the gun's backward momentum equals the bullet's: 0.02 kg × 300 m/s = 6 kg m/s. Recoil velocity = 6 ÷ 4 = 1.5 m/s. The gun moves slowly because its mass is far greater than the bullet's.
12. D. Let each half be d. Time = d/40 + d/60 = 5d/120 = d/24. Total distance is 2d, so average speed = 2d ÷ (d/24) = 48 km/h. The simple mean, 50 km/h, is wrong because more time is spent at the lower speed.
13. B. 72 km/h = 72 × 5/18 = 20 m/s. For uniform retardation the average speed is (20 + 0)/2 = 10 m/s, so distance = 10 × 5 = 50 m (the retardation is 20/5 = 4 m/s²). Using 20 m/s for the whole time would wrongly give 100 m.
14. A. Distance is the path length: half the circumference = πr = (22/7) × 7 = 22 m. Displacement is the straight-line shift between the two ends of the diameter = 2r = 14 m. Distance is never less than displacement.
15. C. Momentum p = mv. Convert mass to kilograms: 500 g = 0.5 kg. p = 0.5 × 20 = 10 kg m/s. Forgetting to convert grams gives the wrong value 10,000.
16. D. By Newton's second law, F = ma = 1500 kg × 2 m/s² = 3000 N. Dividing (1500 ÷ 2 = 750) or adding the numbers (1502) are common errors.
17. A. The ball's momentum must drop to zero either way; stretching that change over a longer time lowers the average force (force = change in momentum ÷ time). Crumple zones and padded mats use the same idea.
18. C. Luminous intensity is one of the seven SI base quantities and its unit is the candela (cd). Lumen is the unit of luminous flux, lux of illuminance, and watt of power.
19. C. The passenger's body was moving with the bus and tends to keep moving (inertia of motion, Newton's first law) even when the bus's floor stops. Inertia of rest explains falling backward when a stationary bus starts suddenly.
20. D. The branch moves suddenly but the mango, due to its inertia of rest, tends to stay where it was, so the stem snaps and the fruit falls. Beating a carpet to remove dust works the same way (Newton's first law).
21. C. 1 N = 1 kg·m/s² = (1000 g) × (100 cm)/s² = 10⁵ g·cm/s² = 10⁵ dyne. The dyne is the CGS unit of force. Note 10⁷ is the joule-to-erg factor, a common trap.
22. B. Inertia is the tendency of a body to resist change in its state of rest or motion, and mass is its measure: a loaded truck has more inertia than an empty auto-rickshaw. Velocity, shape and colour do not determine inertia.
23. B. Action and reaction are equal in magnitude, opposite in direction, simultaneous, and act on different bodies. Because they act on different bodies they never cancel; that is why a push on a wall is felt back by your own body.
24. A. At the instant the string snaps, the stone retains its tangential horizontal velocity. The radial tension is gone, so Statement I is correct and Statement II is false. Gravity can bend the later three-dimensional path; the statement concerns only the initial horizontal direction.
25. A. Statement I compares the limiting value of static friction with kinetic friction and is generally correct under the stated comparison. Ordinary static friction can be smaller than its maximum. Statement II is false: Newton's third-law pair acts on two different bodies.
26. D. Both are wrong. Displacement (shortest straight-line shift) is always less than or equal to distance (total path length). Speed has only magnitude, so it is a scalar; velocity, which has direction, is the vector.
27. D. Lubricants (oil, grease) separate the surfaces, ball bearings replace sliding friction by the much smaller rolling friction, and a streamlined shape cuts fluid friction (drag). Making surfaces rougher increases friction, so it is the odd one out.
28. B. Specific gravity (relative density) is the ratio of two densities, so M, L and T cancel and it is dimensionless. Density [M L⁻³], pressure [M L⁻¹ T⁻²] and momentum [M L T⁻¹] all have dimensions.
29. B. Speed has only magnitude, so it is a scalar. Velocity, momentum (mass × velocity) and acceleration all have both magnitude and direction and are vectors.
30. D. pico = 10⁻¹² < nano = 10⁻⁹ < micro = 10⁻⁶ < milli = 10⁻³. Each step up is a thousand times larger, so the increasing order is pico, nano, micro, milli, i.e. B, D, A, C.
31. D. A screw gauge converts rotation into tiny linear movement: one full turn advances the spindle by the pitch, so it reads to about 0.01 mm. Moments apply to levers and balances, Archimedes to buoyancy, Pascal to fluid pressure.
32. C. A spherometer measures the radius of curvature of spherical surfaces like lenses; it works on the screw principle. A screw gauge measures thin-wire diameter, a vernier caliper length/depth, and an odometer distance travelled.
33. A. Work = force × displacement = [M L T⁻²] × [L] = [M L² T⁻²]. [M L T⁻²] is force, [M L² T⁻³] is power (work/time), and [M L⁻¹ T⁻²] is pressure (force/area).
34. C. Pressure is measured in pascal (N/m²), frequency in hertz (1/s), electric charge in coulomb (A·s) and electric resistance in ohm (V/A). So the code is A-3, B-4, C-1, D-2.
35. B. One vernier division is 9/10 mm = 0.9 mm. Least count = 1 main-scale division − 1 vernier division = 1 − 0.9 = 0.1 mm = 0.01 cm. Different caliper designs can have different least counts, so the scale geometry is specified.
36. C. Energy = power × time. A 100 W bulb uses 100 J every second, so in one hour (3,600 s) it uses 100 × 3,600 = 3,60,000 J = 3.6 × 10⁵ J, which is 0.1 kWh. Watt is a rate; joule is the total amount.