Physics Class 9 Chapter 1: Physical Quantities and Measurement

Free chapter

Class 9
Physics
Unit 1
FBISE SSC-I

Practice this chapter Take a test Physics page

This is the free study page for Physics Class 9, Chapter 1: Physical Quantities and Measurement. It follows the Federal Board (FBISE) SSC-I syllabus. This chapter has 13 learning outcomes, and Parhayi.com has 317 MCQs, 107 short questions and 17 long questions for it. Read the topics and the key questions with their answers below, then tap Practice this chapter to try the questions yourself.

Physics explains how the world works, from motion and forces to heat, light and magnets. Learn each idea in simple steps, check the units, and practise numericals every day.

What you will learn

Differentiate Between physical and non-physical quantities
Explain with examples that physics is based on physical…
Differentiate between base and derived physical…
Apply the seven units of System International (SI)…
Analyze and express numerical data using scientific…
Analyze and express numerical data, using prefixes…
Differentiate between scalar and vector quantities.…
Justify that distance, speed, time, mass, energy, and…
Justify that displacement, force; weight, velocity…
Determine, by calculation or graphically, the resultant…
Make reasonable estimates of physical quantities…
Critique and analyze experiments for sources of error…
Differentiate between precision and accuracy

Key questions and answers

These are the short questions of this chapter with their answers (107 questions).

Q1. Explain, with an example, how a systematic error differs from a random error in an experiment. (3 marks)

Systematic error: an error that occurs consistently in one direction (every reading too high or too low), e.g. a zero error of a balance that always reads 0.2 g extra.

Random error: an unpredictable error that changes from reading to reading, e.g. the reaction time of a person starting and stopping a stopwatch.

A systematic error can be corrected by calibration; a random error is reduced by taking many readings and finding their mean.

Q2. Differentiate between scalar and vector quantities. Give two examples of each. (3 marks)

Scalar: a quantity that has magnitude (with unit) only, e.g. mass, time.

Vector: a quantity that has both magnitude and direction, e.g. force, velocity.

Scalars are added by ordinary algebra; vectors by the head-to-tail rule.
A vector is represented by an arrow; a scalar is just a number with a unit.

Q3. Enlist the steps to represent a vector in a coordinate system. (3 marks)

Steps to represent a vector in a coordinate system:

Choose and draw a coordinate system (x-axis and y-axis, or N, E, S, W).
Select a suitable scale (e.g. 1 cm = 10 N).
Draw a line in the given direction whose length equals the magnitude according to the scale.
Put an arrow head on the line to show the direction of the vector.

Q4. What are the base quantities? Enlist any two base quantities with SI units along with their symbols. (3 marks)

Base quantities are the simplest physical quantities, chosen so that all other quantities can be derived from them.

Base quantity	SI unit	Symbol
Length	metre	m
Mass	kilogram	kg

Q5. Determine the magnitude and direction of a force F from its perpendicular components. Draw diagram. (3 marks)

Let the perpendicular components be F_x (along the x-axis) and F_y (along the y-axis).

Magnitude: F = √(F_x² + F_y²)

Direction: tan θ = F_y / F_x, so θ = tan⁻¹(F_y / F_x) from the x-axis.

Q6. Justify that displacement is a vector quantity. Also draw diagram. (3 marks)

A vector needs magnitude and direction. Displacement is the shortest (straight-line) distance from the starting point to the end point, and its direction is from start to end.

Example: a man walks 30 km from A towards the east and reaches B. His displacement is 30 km east. If he said only 30 km, we would not know where he is, so the direction is also necessary. Hence displacement is a vector.

Q7. Compare scalar and vector quantities. (Any three points) (3 marks)

Scalar	Vector
Has magnitude only	Has magnitude and direction
Added by ordinary algebra	Added by head-to-tail (vector) rule
Examples: mass, time, energy	Examples: force, velocity, displacement
Not drawn as an arrow	Drawn as an arrow

Q8. Show that prefix micro is thousand times smaller than prefix milli. (3 marks)

micro (µ) = 10⁻⁶ and milli (m) = 10⁻³.

1 mm = 10⁻³ m and 1 µm = 10⁻⁶ m

1 mm ÷ 1 µm = 10⁻³ ÷ 10⁻⁶ = 10³ = 1000

So 1 µm = ¹/₁₀₀₀ mm: a micro unit is one thousand times smaller than a milli unit.

Q9. Differentiate scalar and vector quantities. Give one example each. (3 marks)

Scalar: has magnitude only, e.g. mass = 5 kg.

Vector: has magnitude and direction, e.g. force = 10 N towards the north.

Q10. Enlist any three SI base units along with their symbols. (3 marks)

Base quantity	SI base unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s

Q11. Vernier calipers or Screw gauge, which instrument gives more precise measurement? Give scientific reason for your choice. (3 marks)

The screw gauge gives the more precise measurement.

Its least count is 0.01 mm, which is smaller than the least count of a vernier caliper (0.1 mm).

An instrument with a smaller least count can show a smaller change in length, so repeated readings with it agree more closely. Hence the screw gauge is more precise.

Q12. What is meant by limitation of measuring instruments? Discuss the limitation of Vernier calipers. (3 marks)

Limitation of an instrument: every measuring instrument can measure only up to a certain smallest value called its least count; it cannot give readings finer than that, so every reading has some uncertainty.

Limitations of vernier calipers:

Its least count is 0.1 mm (for a 1 mm main-scale division and 10 vernier divisions), so lengths smaller than this cannot be measured.
If it has a zero error, every reading is wrong by the same amount unless corrected.
Readings are affected by parallax if the scale is read from an angle.

Q13. Estimate your age in minutes and seconds. (3 marks)

Take an age of 15 years as an example.

1 year ≈ 365 days = 365 × 24 × 60 min = 525 600 min

Age in minutes = 15 × 525 600 ≈ 7.9 × 10⁶ min

Age in seconds = 7.9 × 10⁶ × 60 ≈ 4.7 × 10⁸ s

Q14. What base quantities are involved in these derived physical quantities: force, pressure, power and charge? (3 marks)

Derived quantity	SI unit in base units	Base quantities involved
Force	kg m s⁻²	mass, length, time
Pressure	kg m⁻¹ s⁻²	mass, length, time
Power	kg m² s⁻³	mass, length, time
Charge	A s	electric current, time

Q15. Justify that displacement is a vector quantity while energy is a scalar quantity. (3 marks)

Displacement is the shortest distance from the starting point to the end point. To describe it we must say how far and in which direction, e.g. 5 km towards the east. It has magnitude and direction, so it is a vector.

Energy is completely described by a number and a unit, e.g. 100 J. Saying 'towards the east' has no meaning for energy. It has magnitude only, so it is a scalar.

Q16. How can random errors be reduced? (3 marks)

Random errors vary irregularly, so they are reduced by taking many readings of the same measurement and finding their mean (average).

Example: to find the period of a pendulum, measure the time for 10 oscillations several times, take the mean and divide by 10. The reaction-time error of about 0.2 s is then shared among 10 oscillations (0.02 s each).

Q17. Differentiate between precision and accuracy. (3 marks)

Precision	Accuracy
Closeness of repeated measurements to each other	Closeness of a measured value to the true (accepted) value
Shows consistency / repeatability	Shows absence of systematic error
A thermometer that always reads 2 °C high is precise	The same thermometer is not accurate

Q18. Define physical quantities and non-physical quantities. Give two examples of each. (3 marks)

Physical quantities are quantities that can be measured, e.g. length and mass.

Non-physical quantities cannot be measured, e.g. honesty and taste.

Q19. Why is 'honesty' not a physical quantity, while 'time' is a physical quantity? (3 marks)

A physical quantity can be measured with an instrument and expressed in a unit. Time can be measured with a clock and expressed in seconds.

Honesty cannot be measured with any instrument and has no unit, so it is a non-physical quantity.

Q20. A friend says: 'The tea is at 70 °C and it tastes too sweet.' Which part of the statement is a physical quantity? Give a reason. (3 marks)

The temperature (70 °C) is a physical quantity because it can be measured with a thermometer and has a unit.

'Too sweet' is a feeling (taste) which cannot be measured, so it is not a physical quantity.

Q21. What is meant by measurement? Name its two parts. (3 marks)

Measurement is the comparison of an unknown physical quantity with a standard (unit) to see how large or small it is compared with that standard.

Every measurement has two parts: the numerical magnitude (a number) and the unit, e.g. 1.72 m.

Q22. Why is a measurement without a unit meaningless? Explain with an example. (3 marks)

The number tells only how many times the standard is taken; the unit tells which standard is used.

Example: 'the length of a field is 120'. It could mean 120 m or 120 ft, which are very different lengths. When we write 120 m, the measurement is complete.

Q23. Explain with an example how physics is based on physical quantities. (3 marks)

The laws and principles of physics are written using measurable quantities. Example: the speed of a car is found from two measured quantities, distance and time (speed = distance ÷ time). If a car covers 120 km in 2 hours its speed is 60 km/h.

Without measuring physical quantities we could not state, test or use physical laws.

Q24. Differentiate between base quantities and derived quantities. Give two examples of each. (3 marks)

Base quantities	Derived quantities
Simplest quantities from which all others are obtained	Obtained by multiplying or dividing base quantities
Examples: length, mass, time	Examples: area, speed, force
Units: m, kg, s	Units: m², m s⁻¹, N

Q25. Name the base quantities that are involved in the derived quantities speed, density and force. (3 marks)

Derived quantity	Formula	Base quantities
Speed	length ÷ time	length, time
Density	mass ÷ volume (length³)	mass, length
Force	mass × acceleration	mass, length, time

Q26. Why is the newton called a derived unit? Express it in base units. (3 marks)

The newton is the unit of force. Force = mass × acceleration, so its unit is built from the base units of mass (kg), length (m) and time (s).

1 N = 1 kg × 1 m s⁻² = 1 kg m s⁻². It is given a special name 'newton' for convenience.

Q27. The unit of pressure is the pascal (Pa). Starting from pressure = force ÷ area, express the pascal in base units. (3 marks)

1 Pa = 1 N ÷ 1 m²

= (1 kg m s⁻²) ÷ m²

= 1 kg m⁻¹ s⁻²

Q28. Write the seven SI base quantities with their SI units and symbols. (3 marks)

Base quantity	SI unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	kelvin	K
Amount of substance	mole	mol
Light (luminous) intensity	candela	cd

Q29. What is meant by the SI system? Why is it used all over the world? (3 marks)

SI is the International System of Units (French: Système international d'unités), a complete set of units for all physical quantities, built on seven base units.

It is used worldwide because each unit is defined carefully, so measurements are accurate and reproducible, and scientists in all countries understand each other.

Q30. A student writes the temperature as 300 °K and the mass as 5 Kg. Correct the symbols and explain. (3 marks)

The SI base unit of temperature is the kelvin, whose symbol is K without the degree sign, so 300 K.

The symbol of kilogram is kg (small k, small g), so 5 kg.

Symbols of units must be written exactly as defined in SI.

Q31. Write the SI base unit and its symbol for (a) the temperature of a room, (b) the amount of salt in a solution, (c) the brightness of a lamp. (3 marks)

(a) temperature: kelvin (K)

(b) amount of substance: mole (mol)

Q32. What is scientific notation? Why is it useful? (3 marks)

Scientific (standard) notation writes a number as mantissa × 10^exponent, where the mantissa is between 1 and 10 and the exponent is a whole number.

It is useful because very large or very small numbers (e.g. 5.98 × 10²⁴ kg or 1.7 × 10⁻¹⁵ m) are written briefly and the chance of a mistake in counting zeros is small.

Q33. The width of the observable universe is 8.8 × 10²⁶ m. Identify the mantissa and the exponent and explain what the exponent shows. (3 marks)

Mantissa = 8.8 (it lies between 1 and 10); exponent = 26.

The exponent 26 shows that the decimal point must be moved 26 places to the right to write the number in full, i.e. 88 followed by 25 zeros.

Q34. Is 25 × 10⁴ m a correct scientific notation? If not, write it correctly. (3 marks)

No. The mantissa 25 is not between 1 and 10.

25 × 10⁴ = 2.5 × 10 × 10⁴ = 2.5 × 10⁵ m.

Q35. What are prefixes? Why are they used in measurement? (3 marks)

A prefix is a name (with a symbol) given to a power of ten, e.g. kilo (k) = 10³, milli (m) = 10⁻³.

Prefixes are used so that very large or very small quantities can be written conveniently, e.g. 5000 m = 5 km and 0.002 s = 2 ms.

Q36. Write the symbols and the powers of ten of the prefixes kilo, mega, giga, milli, micro and nano. (3 marks)

Prefix	Symbol	Power of ten
kilo	k	10³
mega	M	10⁶
giga	G	10⁹
milli	m	10⁻³
micro	µ	10⁻⁶
nano	n	10⁻⁹

Q37. Show that a nanometre is one thousand times smaller than a micrometre. (3 marks)

1 nm = 10⁻⁹ m and 1 µm = 10⁻⁶ m.

1 µm ÷ 1 nm = 10⁻⁶ ÷ 10⁻⁹ = 10³ = 1000.

So 1 nm = ¹/₁₀₀₀ µm.

Q38. Convert 1 gram into kilograms, milligrams and micrograms. (3 marks)

Step	Reason
Mass = 1 g	Write the given data
kilo = 10³, so 1 g = 10⁻³ kg	1 kg = 10³ g, so 1 g is one thousandth of a kg
milli = 10⁻³, so 1 g = 10³ mg	1 mg is one thousandth of a gram, so 1 g has 1000 mg
micro = 10⁻⁶, so 1 g = 10⁶ µg	1 µg is one millionth of a gram, so 1 g has 1 000 000 µg
1 g = 10⁻³ kg = 10³ mg = 10⁶ µg

Q39. Which unit with a suitable prefix would you use for (a) the thickness of a hair, (b) the distance between Lahore and Multan, (c) the mass of a mosquito? Give a reason. (3 marks)

(a) micrometre (µm) – a hair is only a fraction of a millimetre thick.

(b) kilometre (km) – a long distance is best given in a large unit.

Q40. How is a vector represented graphically? Explain the use of a scale with an example. (3 marks)

A vector is represented by an arrow. The length of the arrow (drawn to a chosen scale) shows the magnitude and the arrow head shows the direction.

Example: a force of 40 N towards the east, scale 1 cm = 10 N: the arrow is 40 ÷ 10 = 4 cm long and points east.

Q41. A force of 40 N acts towards the east. How will you draw it on a coordinate system if the scale is 1 cm = 10 N? (3 marks)

Length of arrow = 40 N ÷ (10 N per cm) = 4 cm.

Draw the x-axis towards the east, draw a 4 cm line from the origin along the +x direction and put an arrow head at its end. Label it 40 N.

Q42. Why cannot vectors be added by ordinary algebra? Illustrate with a 3 N force towards the east and a 4 N force towards the north. (3 marks)

Vectors have direction as well as magnitude, so their directions must be considered. Adding the numbers gives 3 + 4 = 7 N, which is wrong.

The forces are perpendicular, so the resultant is R = √(3² + 4²) = √25 = 5 N, which is less than 7 N.

Q43. State how scalars and vectors are added. Give one example of each. (3 marks)

Scalars are added by ordinary algebra: 5 kg + 3 kg = 8 kg.

Vectors are added by the head-to-tail rule (vector algebra): the tail of the second vector is placed at the head of the first and the resultant is drawn from the tail of the first to the head of the last. Example: 3 N east + 4 N north = 5 N.

Q44. Justify that mass is a scalar quantity. (3 marks)

A mass such as 5 kg is completely described by its magnitude and unit. It has no direction; it makes no sense to say '5 kg towards the north'.

Two masses are added by ordinary addition (5 kg + 3 kg = 8 kg). Hence mass is a scalar.

Q45. Name any six scalar quantities. Why do they not need a direction? (3 marks)

Distance, speed, time, mass, energy and temperature.

Each of them is completely described by a number with a unit (e.g. 30 s, 5 kg, 35 °C), so no direction is required.

Q46. A man walks 400 m towards the east and then 300 m towards the west. Find his total distance and his displacement. Which of the two is a scalar? (3 marks)

Step	Reason
Eastward walk = 400 m, westward walk = 300 m	Write the given data; the second walk is opposite to the first
Distance = d₁ + d₂	Distance is a scalar, so lengths of both paths are simply added
Distance = 400 m + 300 m = 700 m	Substitute and add
Displacement = 400 m − 300 m	Displacement is a vector, so the opposite direction is subtracted
Displacement = 100 m towards the east	Only the straight-line gap between start and end counts
Scalar quantity = distance	Distance has magnitude only; displacement needs direction
Distance = 700 m (scalar); displacement = 100 m towards the east

Q47. Justify that force is a vector quantity. (3 marks)

A force of 20 N pushing a box towards the east moves it east, while the same 20 N pushing it towards the north moves it north. The effect depends on the direction as well as the magnitude, so we must state both.

Hence force is a vector quantity.

Q48. Why is weight a vector while mass is a scalar? (3 marks)

Weight is the force of gravity on a body (W = mg). It acts towards the centre of the Earth, so it has a direction as well as magnitude: a vector.

Mass is the quantity of matter in a body and has magnitude only: a scalar.

Q49. List the vector quantities that you have studied in this chapter. (3 marks)

Displacement, force, weight, velocity, acceleration, momentum, electric field strength and gravitational field strength.

Q50. State the head-to-tail rule of vector addition. How is the resultant of three vectors found graphically? (3 marks)

Head-to-tail rule: draw the vectors to a common scale and place them one after another so that the tail of each vector touches the head of the previous one. The resultant is drawn from the tail of the first vector to the head of the last vector.

For three vectors A, B and C: place the tail of B at the head of A, the tail of C at the head of B, then join the tail of A to the head of C.

Q51. Write the formulae for the magnitude and direction of the resultant of two perpendicular vectors A and B. (3 marks)

Magnitude: R = √(A² + B²)

Direction: tan θ = B / A, where θ is the angle of R with vector A.

Q52. What are the components of a vector? Write the formulae for the components of a force F making angle θ with the x-axis. (3 marks)

The components of a vector are two perpendicular vectors (along the x- and y-axes) which together have the same effect as the original vector.

F_x = F cos θ (horizontal component)

F_y = F sin θ (vertical component)

Q53. Does the order in which two vectors are added matter? Explain with a diagram. (3 marks)

No. A + B gives the same resultant as B + A.

If A is placed first and B at its head, or B first and A at its head, the two paths form a parallelogram and both end at the same point, so the resultant is the same.

Q54. Estimate the number of times your heart beats in one day if the pulse rate is 70 beats per minute. (3 marks)

Step	Reason
Pulse rate = 70 beats per minute	Write the given data
Time = 1 day = 24 × 60 min	Convert one day into minutes
Time = 1440 min	Multiply to get the number of minutes
Total beats = rate × time	Beats in a day = beats per minute × minutes in a day
Total beats = 70 × 1440 = 100 800	Substitute and multiply
Total beats ≈ 1.0 × 10⁵	Estimate: round and write in scientific notation
Heart beats about 1.0 × 10⁵ times in one day

Q55. Why do we make estimates of physical quantities? Give an example. (3 marks)

An estimate gives a quick, approximate value of the right size (order of magnitude). It helps us plan experiments and check whether a calculated answer is reasonable.

Example: an adult's mass is about 65 kg; if a calculation gives 6500 kg we know that a power of ten has been wrongly used.

Q56. Define error. Name the two main types of errors in measurement. (3 marks)

Error is the uncertainty that arises during a measurement; all measurements are only approximate.

The two main types are systematic errors and random errors.

Q57. Name three sources of systematic errors. (3 marks)

Instrumental errors: imperfect design or calibration of the instrument, or zero error.
Imperfect experimental technique or changing external conditions such as temperature, humidity or wind.
Personal errors: observer's bias, improper set-up of apparatus or carelessness.

Q58. The first centimetre of a metre rule is broken, so a student's readings of lengths are all 1 cm too small. Name the type of error and explain how the readings can be corrected. (3 marks)

This is a systematic error (instrumental error) because every reading is wrong by the same amount in the same direction.

It is corrected by adding 1 cm to every reading, or by using the rule from the 1 cm mark (or by using a better instrument). Taking the mean of many readings does not remove it.

Q59. A balance reads +0.3 g when nothing is placed on it. It then reads 25.8 g for a metal piece. Find the true mass and name the error. (3 marks)

Step	Reason
Zero error = +0.3 g, reading = 25.8 g	Write the given data; the balance shows 0.3 g with nothing on it
Zero error is a systematic error	It is the same in every reading and always in one direction
True mass = reading − zero error	A positive zero error makes every reading too large, so subtract it
True mass = 25.8 g − 0.3 g = 25.5 g	Substitute and subtract
True mass = 25.5 g (zero error, a systematic error)

Q60. Explain with the help of a dart target the meaning of 'precise but not accurate'. (3 marks)

If all the darts are tightly grouped but far from the bullseye, the throws are consistent (precise) but not close to the target value (not accurate).

Example: a balance that always shows 0.5 kg too much gives close readings (precise) but not the true mass (not accurate).

Q61. A thermometer always reads 2 °C higher than the true temperature. Is it precise? Is it accurate? Explain. (3 marks)

Its readings are consistent with each other, so the thermometer is precise.

The readings are always 2 °C away from the true value, so it is not accurate. The consistent 2 °C difference is a systematic error.

Q62. Which is better: a result that is accurate but not precise, or one that is both accurate and precise? Explain with an example. (3 marks)

The best result is both accurate and precise: the readings agree with each other and are close to the true value.

Example: true mass 50.0 g; readings 50.1, 49.9, 50.0 g are both accurate and precise, while readings 46, 54, 50 g are accurate on average but not precise.

Q63. Show 2000 nano-grams in micro grams. (3 marks)

Step	Reason
Mass = 2000 ng	Write the given data
1 ng = 10⁻⁹ g	Prefix nano
1 µg = 10⁻⁶ g	Prefix micro
1 ng = 10⁻⁹10⁻⁶ µg = 10⁻³ µg	Divide to compare the two prefixes
2000 ng = 2000 × 10⁻³ µg	Replace each ng
2000 ng = 2 µg	Multiply
2000 ng = 2 µg

Q64. Show 4000 nano grams in milli grams. (3 marks)

Step	Reason
Mass = 4000 ng	Write the given data
4000 ng = 4000 × 10⁻⁹ g	Prefix nano is 10⁻⁹; change to grams first
4000 ng = 4 × 10⁻⁶ g	Write in scientific notation
1 mg = 10⁻³ g	Prefix milli
4 × 10⁻⁶ g = 4 × 10⁻⁶10⁻³ mg	Divide by the value of 1 mg
4 × 10⁻⁶ g = 4 × 10⁻³ mg = 0.004 mg	Subtract the powers: −6 − (−3) = −3
4000 ng = 0.004 mg

Q65. Express the following quantities in prefix micro: (a) 0.00002 g (b) 0.0000000081 m (c) 0.000004 s (3 marks)

Step	Reason
(a) 0.00002 g = 2 × 10⁻⁵ g	Write the first quantity in scientific notation
2 × 10⁻⁵ g = 20 × 10⁻⁶ g	Make the power −6, the power of micro
20 × 10⁻⁶ g = 20 µg	Replace 10⁻⁶ with µ
(b) 0.0000000081 m = 8.1 × 10⁻⁹ m	Second quantity in scientific notation
8.1 × 10⁻⁹ m = 0.0081 × 10⁻⁶ m	Make the power −6 by moving the decimal point
0.0081 × 10⁻⁶ m = 0.0081 µm	Replace 10⁻⁶ with µ
(c) 0.000004 s = 4 × 10⁻⁶ s = 4 µs	The power is already −6
(a) 20 µg (b) 0.0081 µm (c) 4 µs

Q66. Your hair grow at the rate of 1 mm per day. Find their growth rate in nm s⁻¹ and µm s⁻¹. (3 marks)

Step	Reason
Growth rate = 1 mm per day	Write the given data
1 mm = 10⁻³ m = 10⁶ nm	Convert millimetre into nanometre
1 mm = 10⁻³ m = 10³ µm	Convert millimetre into micrometre
1 day = 24 × 60 × 60 s = 86 400 s	Convert day into seconds
Rate = 10⁶ nm86 400 s = 11.57 nm s⁻¹	Divide length in nm by time in s
Rate = 10³ µm86 400 s = 1.157 × 10⁻² µm s⁻¹	Same division, now with length in µm
Growth rate = 11.57 nm s⁻¹ = 1.157 × 10⁻² µm s⁻¹

Q67. How many kilometers are there in 25 micrometers? (3 marks)

Step	Reason
Length = 25 µm	Write the given data
1 µm = 10⁻⁶ m	Prefix micro means 10⁻⁶
25 µm = 25 × 10⁻⁶ m = 2.5 × 10⁻⁵ m	First change to metre
1 km = 10³ m, so 1 m = 10⁻³ km	Prefix kilo means 10³; we need km next
2.5 × 10⁻⁵ m = 2.5 × 10⁻⁵ × 10⁻³ km	Replace each metre with 10⁻³ km
= 2.5 × 10⁻⁸ km	Add the powers: −5 + (−3) = −8
25 µm = 2.5 × 10⁻⁸ km

Q68. Express the following using prefixes: (a) 29.3 × 10⁻¹⁰ m (b) 0.00034 × 10⁻³ s (3 marks)

Step	Reason
(a) 29.3 × 10⁻¹⁰ m = 2.93 × 10⁻⁹ m	Make the number 2.93 by shifting the power
2.93 × 10⁻⁹ m = 2.93 nm	10⁻⁹ is the prefix nano
(b) 0.00034 × 10⁻³ s = 3.4 × 10⁻⁴ × 10⁻³ s	Write 0.00034 in scientific notation
= 3.4 × 10⁻⁷ s = 0.34 × 10⁻⁶ s	Add the powers, then make the power −6
= 0.34 µs	10⁻⁶ is the prefix micro
(a) 2.93 nm (b) 0.34 µs (= 340 ns)

Q69. Write the following in form of prefixes: (a) 0.00004 g (b) 0.0000000021 m (3 marks)

Step	Reason
(a) 0.00004 g = 4 × 10⁻⁵ g	Write the first quantity in scientific notation
4 × 10⁻⁵ g = 40 × 10⁻⁶ g	Make the power −6, which is the power of micro
40 × 10⁻⁶ g = 40 µg	Replace 10⁻⁶ with the prefix µ
(b) 0.0000000021 m = 2.1 × 10⁻⁹ m	Now the second quantity in scientific notation
2.1 × 10⁻⁹ m = 2.1 nm	10⁻⁹ is already the prefix nano, so replace it with n
(a) 40 µg (b) 2.1 nm

Q70. Express 300 micrometers in 'millimeters' and 'kilometers'. (3 marks)

Step	Reason
Length = 300 µm	Write the given data
300 µm = 300 × 10⁻⁶ m = 3 × 10⁻⁴ m	Change to metres first
1 m = 10³ mm	Prefix milli
3 × 10⁻⁴ m = 3 × 10⁻⁴ × 10³ mm = 0.3 mm	Multiply by 10³ for millimetres
1 m = 10⁻³ km	Prefix kilo
3 × 10⁻⁴ m = 3 × 10⁻⁴ × 10⁻³ km = 3 × 10⁻⁷ km	Multiply by 10⁻³ for kilometres
300 µm = 0.3 mm = 3 × 10⁻⁷ km

Q71. A force is acting on a body making an angle of 30° with x-axis. The vertical component of force is 25 N. Find the force. (3 marks)

Step	Reason
θ = 30°, F_y = 25 N, F = ?	Write the given data
F_y = F sin θ	The vertical component uses sine of the angle with the x-axis
F = F_ysin θ	Rearrange to find F
F = 25 Nsin 30°	Substitute the values
F = 25 N0.5	Put sin 30° = 0.5
F = 50 N	Divide
F = 50 N

Q72. Express the following quantities using suitable prefixes: (a) 4.0 × 10⁻⁵ m (b) 4.132 × 10¹⁶ m (c) 572 × 10⁻⁸ s (3 marks)

Step	Reason
(a) 4.0 × 10⁻⁵ m = 40 × 10⁻⁶ m	Make the power −6
40 × 10⁻⁶ m = 40 µm	Replace 10⁻⁶ with µ
(b) 4.132 × 10¹⁶ m = 41.32 × 10¹⁵ m	Make the power 15, the power of peta
41.32 × 10¹⁵ m = 41.32 Pm	Replace 10¹⁵ with P
(c) 572 × 10⁻⁸ s = 5.72 × 10⁻⁶ s	Move the decimal point to make the power −6
5.72 × 10⁻⁶ s = 5.72 µs	Replace 10⁻⁶ with µ
(a) 40 µm (b) 41.32 Pm (c) 5.72 µs

Q73. The hair of a horse grow at the rate of 0.5 mm per day. Find their growth rate in nm s⁻¹. (3 marks)

Step	Reason
Growth rate = 0.5 mm per day	Write the given data
0.5 mm = 0.5 × 10⁶ nm = 5 × 10⁵ nm	1 mm = 10⁶ nm, so change the length into nm
1 day = 24 × 60 × 60 s = 86 400 s	Change the time into seconds
Rate = 5 × 10⁵ nm86 400 s	Length divided by time gives the rate
Rate = 5.79 nm s⁻¹	Divide
Growth rate = 5.79 nm s⁻¹

Q74. A force of 200 N makes an angle of 60° with x-axis. Resolve this force into rectangular components. (3 marks)

Step	Reason
F = 200 N, θ = 60°	Write the given data
F_x = F cos θ	The horizontal component uses cosine
F_x = 200 × cos 60° = 200 × 0.5	Substitute; cos 60° = 0.5
F_x = 100 N	Multiply
F_y = F sin θ	The vertical component uses sine
F_y = 200 × sin 60° = 200 × 0.866	Substitute; sin 60° = 0.866
F_y = 173.2 N	Multiply
F_x = 100 N and F_y = 173.2 N

Q75. Write the following numbers in scientific notation: (a) 1234 m (b) 0.000023 s (c) 469.3 × 10⁵ m (d) 0.00985 × 10⁷ s (3 marks)

Step	Reason
(a) 1234 m = 1.234 × 10³ m	Keep one digit before the decimal point
(b) 0.000023 s = 2.3 × 10⁻⁵ s	Move the decimal point right 5 places
(c) 469.3 × 10⁵ m = 4.693 × 10² × 10⁵ m	Write 469.3 in scientific notation first
= 4.693 × 10⁷ m	Add the powers: 2 + 5 = 7
(d) 0.00985 × 10⁷ s = 9.85 × 10⁻³ × 10⁷ s	Write 0.00985 in scientific notation first
= 9.85 × 10⁴ s	Add the powers: −3 + 7 = 4
(a) 1.234 × 10³ m (b) 2.3 × 10⁻⁵ s (c) 4.693 × 10⁷ m (d) 9.85 × 10⁴ s

Q76. Express the following measurements using prefixes: (a) 27.5 × 10⁻¹⁰ m (b) 0.00023 × 10⁻² s (3 marks)

Step	Reason
(a) 27.5 × 10⁻¹⁰ m = 2.75 × 10⁻⁹ m	Make the number 2.75 by shifting the power
2.75 × 10⁻⁹ m = 2.75 nm	10⁻⁹ is the prefix nano
(b) 0.00023 × 10⁻² s = 2.3 × 10⁻⁴ × 10⁻² s	Write 0.00023 in scientific notation
= 2.3 × 10⁻⁶ s	Add the powers: −4 − 2 = −6
= 2.3 µs	10⁻⁶ is the prefix micro
(a) 2.75 nm (b) 2.3 µs

Q77. A boy is 15 years 2 months and 10 days old. Convert his age into (a) seconds (b) milliseconds (c) mega seconds. (Take 1 year = 365 days and 1 month = 30 days.) (3 marks)

Step	Reason
Age = 15 years 2 months 10 days; 1 year = 365 days, 1 month = 30 days	Write the given data
Days = 15 × 365 + 2 × 30 + 10	Change the whole age into days
Days = 5475 + 60 + 10 = 5545 days	Add
1 day = 24 × 3600 s = 86 400 s	Days to seconds
(a) Seconds = 5545 × 86 400 = 4.791 × 10⁸ s	Multiply
(b) Milliseconds = 4.791 × 10⁸ × 10³ ms	1 s = 10³ ms
= 4.791 × 10¹¹ ms	Add the powers: 8 + 3 = 11
(c) Mega seconds = 4.791 × 10⁸10⁶ Ms	1 Ms = 10⁶ s
= 479.1 Ms	Subtract the powers: 8 − 6 = 2
(a) 4.791 × 10⁸ s (b) 4.791 × 10¹¹ ms (c) 479.1 Ms

Q78. Write in scientific notation: (a) 7 600 000 m (b) 0.000 034 s (c) 125.6 × 10⁵ g (3 marks)

Step	Reason
(a) 7 600 000 m = 7.6 × 10⁶ m	Move the decimal point left 6 places
(b) 0.000 034 s = 3.4 × 10⁻⁵ s	Move the decimal point right 5 places
(c) 125.6 × 10⁵ g = 1.256 × 10² × 10⁵ g	Write 125.6 in scientific notation first
= 1.256 × 10⁷ g	Add the powers: 2 + 5 = 7
(a) 7.6 × 10⁶ m (b) 3.4 × 10⁻⁵ s (c) 1.256 × 10⁷ g

Q79. Express using suitable prefixes: (a) 3.5 × 10⁻⁶ m (b) 2.7 × 10⁹ m (c) 6.0 × 10⁻³ g (3 marks)

Step	Reason
(a) 3.5 × 10⁻⁶ m	10⁻⁶ is the prefix micro
3.5 × 10⁻⁶ m = 3.5 µm	Replace 10⁻⁶ with µ
(b) 2.7 × 10⁹ m = 2.7 Gm	10⁹ is the prefix giga
(c) 6.0 × 10⁻³ g = 6.0 mg	10⁻³ is the prefix milli
(a) 3.5 µm (b) 2.7 Gm (c) 6.0 mg

Q80. A cyclist travels at 15 m s⁻¹. Express his speed in km h⁻¹, and express a car speed of 72 km h⁻¹ in m s⁻¹. (3 marks)

Step	Reason
Cyclist speed = 15 m s⁻¹	Write the first given data
1 m = 11000 km and 1 s = 13600 h	Write both units in the new units
15 m s⁻¹ = 15 × 1 km1000 × 36001 h	Replace m and s
15 m s⁻¹ = 15 × 3.6 = 54 km h⁻¹	Simplify: the factor is 3.6
Car speed = 72 km h⁻¹	Now the second given data
72 km h⁻¹ = 72 × 1000 m3600 s	Replace km and h
72 km h⁻¹ = 723.6 = 20 m s⁻¹	Divide by 3.6 (reverse of the first step)
15 m s⁻¹ = 54 km h⁻¹; 72 km h⁻¹ = 20 m s⁻¹

Q81. A gold bar has a mass of 12.5 kg. Express its mass in grams, in milligrams and in scientific notation in milligrams. (3 marks)

Step	Reason
Mass = 12.5 kg	Write the given data
1 kg = 10³ g	Prefix kilo means 10³
12.5 kg = 12.5 × 10³ g = 12 500 g	Change to grams
1 g = 10³ mg	Prefix milli means 10⁻³, so 1 g has 10³ mg
12 500 g = 12 500 × 10³ mg = 12 500 000 mg	Change to milligrams
12 500 000 mg = 1.25 × 10⁷ mg	Write in scientific notation
12 500 g; 12 500 000 mg; 1.25 × 10⁷ mg

Q82. A file of size 2 GB is downloaded at a rate of 5 MB per second. How long will the download take? (1 GB = 1000 MB) (3 marks)

Step	Reason
File size = 2 GB, rate = 5 MB per second	Write the given data
1 GB = 1000 MB	Given in the question
Size = 2 × 1000 MB = 2000 MB	Write the size in MB so both have the same unit
Time = sizerate	Time needed = amount to download divided by the rate
Time = 2000 MB5 MB s⁻¹	Substitute the values
Time = 400 s	Divide (this is 6 min 40 s)
Time = 400 s

Q83. The width of a DNA strand is 2 nm. Express it in metres, millimetres and micrometres. (3 marks)

Step	Reason
Width = 2 nm	Write the given data
1 nm = 10⁻⁹ m	Prefix nano means 10⁻⁹
2 nm = 2 × 10⁻⁹ m	In metres
1 m = 10³ mm	Prefix milli is 10⁻³
2 × 10⁻⁹ m = 2 × 10⁻⁹ × 10³ mm = 2 × 10⁻⁶ mm	In millimetres, multiply by 10³
1 m = 10⁶ µm	Prefix micro is 10⁻⁶
2 × 10⁻⁹ m = 2 × 10⁻⁹ × 10⁶ µm = 2 × 10⁻³ µm	In micrometres, multiply by 10⁶
2 × 10⁻⁹ m = 2 × 10⁻⁶ mm = 2 × 10⁻³ µm

Q84. The radius of the Earth is 6.38 × 10⁶ m. Express it in kilometres and in megametres. (3 marks)

Step	Reason
Radius r = 6.38 × 10⁶ m	Write the given data
1 km = 10³ m	Prefix kilo
r = 6.38 × 10⁶10³ km = 6380 km	Divide by 10³ to get km
1 Mm = 10⁶ m	Prefix mega
r = 6.38 × 10⁶10⁶ Mm = 6.38 Mm	Divide by 10⁶ to get Mm
r = 6380 km = 6.38 Mm

Q85. Light travels at 3 × 10⁸ m s⁻¹. Find the time it takes to travel 1 m. Write the answer in scientific notation and with a suitable prefix. (3 marks)

Step	Reason
Speed = 3 × 10⁸ m s⁻¹, distance = 1 m	Write the given data
Time = distancespeed	Time taken = distance divided by speed
Time = 1 m3 × 10⁸ m s⁻¹	Substitute the values
Time = 0.333 × 10⁻⁸ s	Divide
Time = 3.33 × 10⁻⁹ s	Scientific notation: one digit before the decimal
Time = 3.33 ns	10⁻⁹ is the prefix nano
Time = 3.33 × 10⁻⁹ s = 3.33 ns

Q86. One sheet of paper is 0.08 mm thick. Find the height of a pile of 250 sheets in millimetres, in centimetres and in metres. (3 marks)

Step	Reason
Thickness of one sheet = 0.08 mm, sheets = 250	Write the given data
Height = 250 × 0.08 mm	Pile height = number of sheets × thickness of one sheet
Height = 20 mm	In millimetres
Height = 2010 cm = 2 cm	1 cm = 10 mm, so divide by 10
Height = 201000 m = 0.02 m = 2 × 10⁻² m	1 m = 1000 mm, so divide by 1000
Height = 20 mm = 2 cm = 0.02 m

Q87. A hard disk has a capacity of 500 GB (1 GB = 10⁹ bytes). Express the capacity in bytes in scientific notation. (3 marks)

Step	Reason
Capacity = 500 GB, 1 GB = 10⁹ bytes	Write the given data
Capacity = 500 × 10⁹ bytes	Replace GB with 10⁹ bytes
Capacity = 5 × 10² × 10⁹ bytes	Write 500 in scientific notation
Capacity = 5 × 10¹¹ bytes	Add the powers: 2 + 9 = 11
Capacity = 5 × 10¹¹ bytes

Q88. The distance from the Earth to the Moon is 3.84 × 10⁸ m. Express it in kilometres (scientific notation) and in gigametres. (3 marks)

Step	Reason
Distance = 3.84 × 10⁸ m	Write the given data
1 km = 10³ m	Prefix kilo
Distance = 3.84 × 10⁸10³ km = 3.84 × 10⁵ km	Divide by 10³ to get km
1 Gm = 10⁹ m	Prefix giga
Distance = 3.84 × 10⁸10⁹ Gm = 0.384 Gm	Divide by 10⁹ to get Gm
3.84 × 10⁵ km = 0.384 Gm

Q89. The mass of the Sun is 1.99 × 10³⁰ kg and the mass of the Earth is 5.98 × 10²⁴ kg. How many times is the Sun heavier than the Earth? (3 marks)

Step	Reason
m_Sun = 1.99 × 10³⁰ kg, m_Earth = 5.98 × 10²⁴ kg	Write the given data
Ratio = m_Sunm_Earth	'How many times' means divide the two masses
Ratio = 1.99 × 10³⁰5.98 × 10²⁴	Substitute the values
Ratio = 1.995.98 × 10^{30 − 24} = 0.333 × 10⁶	Divide the numbers and subtract the powers
Ratio = 3.33 × 10⁵	Scientific notation
The Sun is 3.33 × 10⁵ times heavier than the Earth

Q90. A block of aluminium has a mass of 2.7 kg and a volume of 1.0 × 10⁻³ m³. Find its density and write the unit in base units. (3 marks)

Step	Reason
m = 2.7 kg, V = 1.0 × 10⁻³ m³	Write the given data
ρ = mV	Density is mass per unit volume
ρ = 2.7 kg1.0 × 10⁻³ m³	Substitute the values
ρ = 2700 kg m⁻³	Divide
Unit of density = kgm³ = kg m⁻³	It is made from the base units kg and m
Density = 2700 kg m⁻³

Q91. A force of 500 N acts on an area of 0.25 m². Find the pressure and express its unit in base units. (3 marks)

Step	Reason
F = 500 N, A = 0.25 m²	Write the given data
P = FA	Pressure is force per unit area
P = 500 N0.25 m²	Substitute the values
P = 2000 N m⁻² = 2000 Pa	Divide
1 N = 1 kg m s⁻²	Express newton in base units
1 Pa = 1 kg m s⁻² m⁻² = 1 kg m⁻¹ s⁻²	Put it in N m⁻² to get the base unit of pressure
Pressure = 2000 Pa = 2000 kg m⁻¹ s⁻²

Q92. A current of 2 A flows through a wire for 30 s. Find the charge that passes (charge = current × time) and give its unit in base units. (3 marks)

Step	Reason
I = 2 A, t = 30 s	Write the given data
Q = I × t	Charge = current × time (given in the question)
Q = 2 A × 30 s	Substitute the values
Q = 60 A s	Multiply; both A and s are base units
1 A s = 1 C	The coulomb is the name of ampere second
Charge = 60 A s = 60 C

Q93. Starting from work = force × distance, show that the unit of energy, the joule, is kg m² s⁻² in base units. (3 marks)

Work = force × distance

1 J = 1 N × 1 m = (1 kg m s⁻²) × 1 m

= 1 kg m² s⁻²

Q94. Two perpendicular forces of 12 N and 16 N act at a point. Find the magnitude of the resultant force and its angle with the 12 N force. (3 marks)

Step	Reason
A = 12 N, B = 16 N, at right angles	Write the given data
R = √(A² + B²)	For perpendicular vectors the resultant comes from Pythagoras theorem
R = √(12² + 16²)	Substitute the values
R = √(144 + 256) = √(400)	Square and add
R = 20 N	Take the square root
tan θ = BA = 1612 = 1.333	Now find the direction from the 12 N force
θ = tan⁻¹(1.333) = 53.1°	Use the inverse tangent
R = 20 N at 53.1° with the 12 N force

Q95. A man walks 600 m towards the east and then 800 m towards the north. Find (a) the total distance covered, (b) the magnitude of his displacement and (c) the direction of the displacement. (3 marks)

Step	Reason
East = 600 m, north = 800 m, at right angles	Write the given data
(a) Distance = 600 m + 800 m	Distance is a scalar, so add the path lengths
Distance = 1400 m	Add
(b) d = √(600² + 800²)	Displacement is the straight line start to end (Pythagoras theorem)
d = √(360 000 + 640 000) = √(1 000 000)	Square and add
d = 1000 m	Take the square root
(c) tan θ = 800600 = 1.333	Direction comes from the ratio of north to east
θ = tan⁻¹(1.333) = 53.1° north of east	Use the inverse tangent
(a) 1400 m (b) 1000 m (c) 53.1° north of east

Q96. A boat can move at 4 m s⁻¹ straight across a river. The river current flows at 3 m s⁻¹ along the bank. Find the resultant velocity of the boat and the angle it makes with the river bank. (3 marks)

Step	Reason
Boat velocity = 4 m s⁻¹ across, current = 3 m s⁻¹ along the bank	Write the given data; the two are at right angles
R = √(v₁² + v₂²)	Resultant of perpendicular vectors (Pythagoras theorem)
R = √(4² + 3²) = √(16 + 9)	Substitute and square
R = √(25) = 5 m s⁻¹	Add and take the square root
tan θ = 43 = 1.333	Angle with the river bank: across component over bank component
θ = tan⁻¹(1.333) = 53.1°	Use the inverse tangent
R = 5 m s⁻¹ at 53.1° with the river bank

Q97. A force of 80 N acts at an angle of 37° above the horizontal. Find its horizontal and vertical components. (cos 37° = 0.80, sin 37° = 0.60) (3 marks)

Step	Reason
F = 80 N, θ = 37°, cos 37° = 0.80, sin 37° = 0.60	Write the given data
F_x = F cos θ	Horizontal component uses cosine
F_x = 80 × 0.80	Substitute the values
F_x = 64 N	Multiply
F_y = F sin θ	Vertical component uses sine
F_y = 80 × 0.60 = 48 N	Substitute and multiply
F_x = 64 N and F_y = 48 N

Q98. An aeroplane flies at 240 km/h towards the north. A wind blows at 70 km/h towards the east. Find the resultant velocity of the aeroplane (magnitude and direction). (3 marks)

Step	Reason
Plane = 240 km/h north, wind = 70 km/h east	Write the given data; the two are at right angles
R = √(v₁² + v₂²)	Resultant of perpendicular vectors
R = √(240² + 70²) = √(57 600 + 4900)	Substitute and square
R = √(62 500) = 250 km/h	Add and take the square root
tan θ = 70240 = 0.2917	Direction from the north: east component over north component
θ = tan⁻¹(0.2917) = 16.3°	Use the inverse tangent
R = 250 km/h at 16.3° east of north

Q99. The string of a kite pulls with a force of 500 N at 60° above the horizontal. Find the horizontal and vertical components of the pull. (3 marks)

Step	Reason
F = 500 N, θ = 60°	Write the given data
F_x = F cos θ = 500 × cos 60°	Horizontal component uses cosine
F_x = 500 × 0.5 = 250 N	cos 60° = 0.5
F_y = F sin θ = 500 × sin 60°	Vertical component uses sine
F_y = 500 × 0.866 = 433 N	sin 60° = 0.866
F_x = 250 N and F_y = 433 N

Q100. The perpendicular components of a force are F_x = 5.0 N and F_y = 12 N. Find the force and its direction with the x-axis. (3 marks)

Step	Reason
F_x = 5.0 N, F_y = 12 N	Write the given data
F = √(F_x² + F_y²)	Components are perpendicular, so use Pythagoras theorem
F = √(5² + 12²) = √(25 + 144)	Substitute and square
F = √(169) = 13 N	Take the square root
tan θ = F_yF_x = 125 = 2.4	Direction with the x-axis
θ = tan⁻¹(2.4) = 67.4°	Use the inverse tangent
F = 13 N at 67.4° with the x-axis

Q101. The vertical component of a 100 N force is 50 N. Find the angle which the force makes with the horizontal, and its horizontal component. (3 marks)

Step	Reason
F = 100 N, F_y = 50 N	Write the given data
F_y = F sin θ	Vertical component uses sine
sin θ = F_yF = 50100 = 0.5	Rearrange to find the angle
θ = sin⁻¹(0.5) = 30°	Use the inverse sine
F_x = F cos θ	Now the horizontal component uses cosine
F_x = 100 × cos 30° = 100 × 0.866 = 86.6 N	Substitute and multiply
θ = 30° with the horizontal, F_x = 86.6 N

Q102. The time for 10 oscillations of a pendulum is 18.4 s. The reaction-time error in measuring this time is about 0.2 s. Find (a) the period of the pendulum and (b) the error in the period. (3 marks)

Step	Reason
Time for 10 oscillations = 18.4 s, error in this time = 0.2 s	Write the given data
(a) T = total timenumber of oscillations	Period is the time of one oscillation
T = 18.4 s10 = 1.84 s	Substitute and divide
(b) Error in T = 0.2 s10	The error is shared by the 10 oscillations, so divide it as well
Error in T = 0.02 s	Divide
T = 1.84 ± 0.02 s	Write the period with its error
(a) T = 1.84 s (b) error = 0.02 s

Q103. A student measures the length of a pencil five times: 5.12 cm, 5.10 cm, 5.14 cm, 5.12 cm and 5.13 cm. Find the mean length. Why is the mean a better value than a single reading? (3 marks)

Step	Reason
Readings: 5.12, 5.10, 5.14, 5.12, 5.13 cm	Write the given data
Sum = 5.12 + 5.10 + 5.14 + 5.12 + 5.13 = 25.61 cm	Add all the readings
Mean = sum of readingsnumber of readings	Mean value is the sum divided by the number of readings
Mean = 25.61 cm5 = 5.122 cm	Substitute and divide
Mean ≈ 5.12 cm	Round to the precision of the readings
The mean is better because high and low readings cancel	Random errors are reduced by taking the mean
Mean length = 5.12 cm

Q104. A thermometer reads 1.5 °C in melting ice (true 0 °C). It reads 101.0 °C in a certain hot water. Find the true temperature of the water and name the type of error. (3 marks)

Step	Reason
Reading in melting ice = 1.5 °C, true value = 0 °C, reading in hot water = 101.0 °C	Write the given data
Zero error = 1.5 − 0 = +1.5 °C	The thermometer reads 1.5 °C too high, always
This is a systematic error	It has the same size and direction in every reading
True temperature = reading − zero error	Subtract the error from the reading
True temperature = 101.0 − 1.5 = 99.5 °C	Substitute and subtract
True temperature = 99.5 °C (systematic error)

Q105. The true value of a length is 20.0 cm. Student X records 19.9, 20.1 and 20.0 cm, while student Y records 21.5, 21.6 and 21.4 cm. Find the mean of each student and state which student is accurate and which is precise. (3 marks)

Step	Reason
True value = 20.0 cm; X: 19.9, 20.1, 20.0 cm; Y: 21.5, 21.6, 21.4 cm	Write the given data
Mean of X = 19.9 + 20.1 + 20.03 = 60.03 = 20.0 cm	Sum of readings divided by their number
Mean of Y = 21.5 + 21.6 + 21.43 = 64.53 = 21.5 cm	Same for student Y
X: mean equals 20.0 cm and the readings are close together	Close to the true value means accurate; close to each other means precise
Y: readings close together but 1.5 cm above the true value	Y is precise but not accurate
Mean of X = 20.0 cm (accurate and precise); mean of Y = 21.5 cm (precise only)

Q106. A student is 14 years old. Estimate his age in seconds. (1 year = 365 days) (3 marks)

Step	Reason
Age = 14 years, 1 year = 365 days	Write the given data
1 year = 365 × 24 × 3600 s	Change one year into seconds
1 year = 3.15 × 10⁷ s	Multiply
Age = 14 × 3.15 × 10⁷ s	Multiply by the number of years
Age ≈ 4.4 × 10⁸ s	Estimate: round to two figures
Age ≈ 4.4 × 10⁸ s

Q107. Differentiate between base quantities and derived quantities with one example of each. (3 marks)

Base quantities: the fundamental quantities chosen as the basis of measurement; they cannot be expressed in terms of other quantities. Example: length (SI unit metre).

Derived quantities: quantities obtained by multiplying or dividing base quantities. Example: speed = length/time (SI unit m/s).

There are seven base quantities in SI, while the number of derived quantities is large.

Long questions with answers

Q1. What is meant by Base Physical Quantities and Derived Physical Quantities? Give three examples of each with symbol and their SI units. (5 marks)

Base physical quantities are the simplest quantities, chosen so that all other physical quantities can be derived from them. Derived physical quantities are obtained by multiplying or dividing base quantities.

Examples of base quantities:

Base quantity	Symbol of quantity	SI unit	Symbol of unit
Length	l	metre	m
Mass	m	kilogram	kg
Time	t	second	s

Examples of derived quantities:

Derived quantity	Symbol of quantity	SI unit	Symbol of unit
Area	A	square metre	m²
Speed	v	metre per second	m s⁻¹
Force	F	newton	N

Q2. What is scientific notation or standard form? Give at least five examples. (6 marks)

Scientific notation is a way of writing very large or very small numbers as mantissa × 10^exponent, where the mantissa is a number between 1 and 10 and the exponent is a whole number (positive or negative).

Examples:

Quantity	Ordinary number	Scientific notation
Mass of the Earth	5 980 000 000 000 000 000 000 000 kg	5.98 × 10²⁴ kg
Width of observable universe	88 followed by 25 zeros m	8.8 × 10²⁶ m
Diameter of hydrogen nucleus	0.0000000000000017 m	1.7 × 10⁻¹⁵ m
Seconds in a day	86 400 s	8.64 × 10⁴ s
Speed of light	300 000 000 m/s	3 × 10⁸ m/s

Rule: moving the decimal point to the left gives a positive exponent; to the right gives a negative exponent.

Q3. What is the International System of Units (SI)? Write the seven base quantities with their SI units and symbols. Express the units of force, pressure and energy in base units. (6 marks)

SI (International System of Units) is a complete system of units in which seven quantities are chosen as base quantities. The units are carefully defined so that measurements are accurate and reproducible all over the world.

Base quantity	SI unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	kelvin	K
Amount of substance	mole	mol
Light (luminous) intensity	candela	cd

Derived units in base units: force: N = kg m s⁻²; pressure: Pa = kg m⁻¹ s⁻²; energy: J = kg m² s⁻².

Q4. Define the resultant of vectors. Explain the head-to-tail rule and derive the magnitude and direction of the resultant of two perpendicular vectors A and B. Find the resultant of two perpendicular forces of 6 N and 8 N. (6 marks)

Resultant: a single vector that has the same effect as two or more vectors acting together.

Head-to-tail rule: draw the vectors to scale, place the tail of B at the head of A; the resultant R is drawn from the tail of A to the head of B.

Magnitude: in the right-angled triangle R² = A² + B², so R = √(A² + B²).

Direction: tan θ = B / A, where θ is the angle between R and A.

Example: R = √(6² + 8²) = √100 = 10 N; tan θ = 8/6 = 1.333, θ = 53.1° with the 6 N force.

Practice MCQs

Choose the answer in your mind first. The answer is shown below each question.

1. Which one of the following units is NOT a derived unit?

pascal
newton
kilogram
watt

Answer: c

Kilogram is an SI base unit; pascal, newton and watt are derived units.

2. Which of the following lists contains ONLY SI base units?

metre, newton, second
metre, kilogram, second
kilogram, litre, second
centimetre, gram, second

Answer: b

Newton and litre are not base units; centimetre and gram are not SI base units.

3. A scalar quantity has:

direction only
both magnitude and direction
neither magnitude nor direction
magnitude only

Answer: d

Scalars are described completely by a number and a unit.

4. A force makes an angle of 30° with the x-axis and its vertical component is 25 N. The force is:

50 N
12.5 N
25 N
43.3 N

Answer: a

F = Fy ÷ sin 30° = 25 ÷ 0.5 = 50 N.

5. Which row gives the SI base units of electric current, amount of substance and light intensity in this order?

V, mol, cd
A, g, cd
A, mol, lm
A, mol, cd

Answer: d

ampere (A), mole (mol), candela (cd).

Chapter 1 is free

You are reading a free chapter. The other chapters of this subject are Premium. Rs 3000 now, Rs 4000 from 15 Dec.

Get Premium How to get Premium Log in

All chaptersPhysics Next chapterKinematics