NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
Answers to Chapter 8, Measurement of Time and Motion (Curiosity, pages 105–120): Activities 8.1 to 8.4, the in-text questions, Think Like a Scientist!, Let Us Enhance Our Learning (Questions 1–11) and the Exploratory Projects, including the Olympic race-speed task. Each card first gives what to write; open Explanation for the reasons, extra examples and tips.
NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
| Chapter | Download | |
|---|---|---|
| Solutions PDF (English medium) | Measurement of Time and Motion | View PDF |
| Solutions PDF (Hindi medium) | समय एवं गति का मापन | View PDF |
NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion – Questions and Answers
All 36 questions of the chapter (textbook pages 105 – 120) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
In-text Question — Page 1068.1 Measurement of Time
By watching events in nature that repeat themselves after definite intervals of time — the rising and setting of the Sun, the phases of the Moon and the changing seasons. First, people made calendars (one day = one cycle of sunrise and sunset). To measure smaller intervals within a day they made sundials, water clocks, hourglasses and candle clocks:
| Device | What repeats or flows in it | How time was read |
|---|---|---|
| Sundial (Fig. 8.1) | The Sun’s daily journey across the sky | From the changing position of the shadow of an object cast by sunlight |
| Water clock, out-flow type (Fig. 8.2a) | Water flowing out of a marked vessel | From the falling water level against the markings for time |
| Water clock, floating-bowl type (Fig. 8.2b) | Water entering a bowl through a fine hole at its bottom | The bowl filled up in a fixed time and sank; it was lifted and floated again |
| Hourglass (Fig. 8.3) | Sand flowing from one bulb to another | From how much sand had run down |
| Candle clock (Fig. 8.4) | A candle burning down steadily | From the markings on the candle that the flame had reached |
Activity 8.1: Let us construct — Page 1078.1 Measurement of Time
Yes. Steps to make a simple water clock:
- Take a used transparent plastic bottle (½ litre or larger) with its cap and cut it into two, roughly in the middle [Fig. 8.5a].
- Make a small hole in the cap with a drawing pin [Fig. 8.5b].
- Place the upper part, with the cap on, upside down over the lower half [Fig. 8.5c].
- Fill the upper part with water coloured with a few drops of ink [Fig. 8.5d].
- As the water drips down, mark its level after every minute (use a watch) till all the water has dripped down.
Pour the water from the lower part back into the top part and watch the level of water dripping into the lower part. Every time the level touches a mark you had made, one more minute has passed.
In-text Question — Page 1098.1.1 A simple pendulum
Yes — it is the same thing. An eraser hung from a thread and set swinging is a simple pendulum; only the name and the bob are different.
| Grade 6 activity | Simple pendulum of this chapter |
|---|---|
| An eraser tied to a thread | A small metallic ball — the bob — tied to a thread |
| Held in the hand or hung from a hook | Hung from a rigid support |
| Moved to one side and released — it swings to and fro | Moved slightly to one side and released — it starts oscillatory motion |
| The swinging repeats again and again | The motion is periodic: it repeats its path after a fixed interval of time |
Activity 8.2: Let us experiment — Pages 109 & 1108.1.1 A simple pendulum
Yes. The bob swings from one side to the other and back again, again and again, along the same path — that is oscillatory motion, and it is periodic because the path repeats after a fixed interval of time.
| Table 8.1 Time period of a simple pendulum (Length of the string = 100 cm) | ||
|---|---|---|
| S.No. | Time taken for 10 oscillations (seconds) | Time period (seconds) |
| 1. | ||
| 2. | ||
| 3. | ||
Yes, the time period is almost the same every time. Sample readings for Table 8.1 (time period = time for 10 oscillations ÷ 10):
| Table 8.1 Time period of a simple pendulum (Length of the string = 100 cm) | ||
|---|---|---|
| S.No. | Time taken for 10 oscillations (seconds) | Time period (seconds) |
| 1. | 20.1 | 20.1 ÷ 10 = 2.01 |
| 2. | 19.9 | 19.9 ÷ 10 = 1.99 |
| 3. | 20.2 | 20.2 ÷ 10 = 2.02 |
Conclusion: the time period of a simple pendulum of a given length is constant at a place.
Think Like a Scientist! — Page 1108.1.1 A simple pendulum
What to investigate: does the time period depend on the length of the pendulum, the mass of the bob, or how far the bob is pulled aside? Does a longer pendulum swing more slowly? Does a heavier bob swing faster?
Would all pendulums have the same time period? No. Pendulums of different lengths have different time periods; pendulums of the same length have the same time period at a place.
How to test: change one thing at a time — the length (same bob), then the mass of the bob (same length) — and measure the time period each time.
Yes, the time period changes. The longer the pendulum, the greater its time period — a long pendulum swings more slowly than a short one. Sample readings with the same bob:
| Length of pendulum | Time for 10 oscillations | Time period | What we see |
|---|---|---|---|
| 25 cm | 10.0 s | 1.0 s | Short pendulum — swings quickly |
| 50 cm | 14.2 s | 1.42 s | Longer — slower |
| 100 cm | 20.0 s | 2.0 s | Longest — slowest |
No, there is no change. Sample readings with the length fixed at 100 cm:
| Bob used (length fixed at 100 cm) | Time for 10 oscillations | Time period |
|---|---|---|
| Small iron ball (light) | 20.0 s | 2.0 s |
| Heavy metal ball | 20.1 s | 2.01 s |
| Stone of about the same size | 19.9 s | 1.99 s |
The time period of a simple pendulum depends on its length but not on the mass of the bob.
Activity 8.3: Let us identify — Page 1128.1.2 SI unit of time
One second (1 s). The second hand moves from one small division of the dial to the next in 1 s (60 divisions in 60 s), and there are no marks in between. So one second is the smallest interval of time that we can measure using this clock.
In-text Questions — Pages 112 & 1138.2 Slow or Fast · 8.3 Speed
By comparing their speeds — the distance each runner covers in unit time (say, one second). Speed = distance ÷ time. The runner who covers more distance in unit time is faster, whatever the length of the race.
We mean a comparison of the distances moved in a given interval of time. Something is moving fast if it covers more distance in the same time than another object; it is slow if it covers less.
The one who is ahead of the others at that instant is running faster. All the runners started together from the same starting line, so they have all been running for the same time. The one who is ahead has therefore covered more distance in the same time.
It can be calculated if we know the total distance covered by the object and the time taken to cover it.
Since speed is distance ÷ time, its unit is a unit of length divided by a unit of time. The SI unit of speed is metre/second, written m/s.
Speed can also be expressed in kilometre/hour (km/h), when the distance is in kilometre and the time in hour.
Activity 8.4: Let us calculate — Pages 114 & 1158.3 Speed
Name of the railway station nearest to your place of stay ____________________.
| Table 8.2: Finding the speed of trains | |||||
|---|---|---|---|---|---|
| S.No. | Name of the train | Name of the next station | Distance till the next station (km) | Time taken till the next station (h) | Speed of the train between these two stations (km/h) |
| 1. | |||||
| 2. | |||||
| 3. | |||||
| 4. | |||||
| 5. | |||||
Use the railway timetable: note the distance to the next stopping station and the difference between the departure time from your station and the arrival time at the next one, then divide.
Sample answer (nearest railway station: Bhopal Junction; illustrative values — fill in the values from your own timetable in the same way):
| Table 8.2: Finding the speed of trains | |||||
|---|---|---|---|---|---|
| S.No. | Name of the train | Name of the next station | Distance till the next station (km) | Time taken till the next station (h) | Speed of the train between these two stations (km/h) |
| 1. | Superfast train A | Jhansi | 291 | 3.00 | 291 ÷ 3.00 = 97 km/h |
| 2. | Superfast train B | Jhansi | 291 | 3.50 | 291 ÷ 3.50 = 83.1 km/h |
| 3. | Mail/Express train | Bina | 139 | 2.25 | 139 ÷ 2.25 = 61.8 km/h |
| 4. | Intercity express | Vidisha | 54 | 0.75 | 54 ÷ 0.75 = 72 km/h |
| 5. | Passenger train | Vidisha | 54 | 1.50 | 54 ÷ 1.50 = 36 km/h |
The train which has covered the maximum distance in unit time is the fastest train — that is, the one with the highest speed. In the sample table above, that is Superfast train A at 97 km/h.
In-text Questions — Page 1168.4 Uniform and Non-uniform Linear Motion
The runners who kept the same speed were in uniform linear motion — they covered equal distances in equal intervals of time. The runners who sped up or slowed down were in non-uniform linear motion — they covered unequal distances in equal intervals of time. All of them ran along a straight road, so all the motions were linear.
Yes. In Grade 6 we learnt that when an object moves along a straight line, its motion is called linear motion — for example, a train moving on a straight track.
In-text Question — Page 1178.4 Uniform and Non-uniform Linear Motion
| Table 8.3: Distances travelled by two trains in equal time intervals of 10 minutes | ||||
|---|---|---|---|---|
| Time (AM) | Train X | Train Y | ||
| Position (km) | Distance (km) | Position (km) | Distance (km) | |
| 10:00 | 0 | 0 | 0 | 0 |
| 10:10 | 20 | 20 | 20 | 20 |
| 10:20 | 40 | 20 | 35 | 15 |
| 10:30 | 60 | 20 | 50 | 15 |
| 10:40 | 80 | 20 | 75 | 25 |
| 10:50 | 100 | 20 | 95 | 20 |
| 11:00 | 120 | 20 | 120 | 25 |
Train X. Train X covers equal distances in equal intervals of time, so it is in uniform linear motion, while Train Y is in non-uniform linear motion.
Let Us Enhance Our Learning — Pages 118 & 119Chapter exercises
= 150 m ÷ 10 s
= 15 m/s
Change m/s into km/h (1 m/s = 3.6 km/h):
15 m/s = 15 × 3.6 = 54 km/h
Second runner: speed = 400 m ÷ 45 s = 8.89 m/s (8 8/9 m/s) ≈ 8.9 m/s
The second runner has the greater speed — the one who took less time for the same distance.
The speed is in m/s, so first put the distance in metre:
Time = 3,60,000 m ÷ 25 m/s = 14,400 s
14,400 s ÷ 60 = 240 min = 240 ÷ 60 = 4 h
(i) km/h
(ii) m/s
(iii) What distance will it travel in 4 h if it maintains the same speed throughout the journey?
(i) In km/h
(ii) In m/s
Speed = 1,80,000 m ÷ 10,800 s = 16.67 m/s (that is 50/3 m/s ≈ 16.7 m/s)
Check: 60 ÷ 3.6 = 16.67 m/s ✔
(iii) Distance in 4 h at the same speed
= 60 km/h × 4 h = 240 km
Bring both speeds to the same unit before comparing.
Horse: 18 m/s = 18 × 3.6 = 64.8 km/h
The train is faster than the horse, by 20 − 18 = 2 m/s (that is 7.2 km/h).
| Point of difference | Car on an empty straight highway | Car in city traffic |
|---|---|---|
| Type of motion | Uniform linear motion | Non-uniform linear motion |
| Speed | Stays constant, say 60 km/h, because nothing forces the driver to change it | Changes constantly — slows at red lights and crossings, speeds up when the road clears |
| Distance in equal intervals of time | Equal: 1 km in every minute | Unequal: 1 km in one minute, then only 300 m in the next |
| Speedometer reading | Needle stays near one mark | Needle keeps moving up and down, and falls to zero at signals |
Both cars are moving along a straight road, so both are in linear motion. The difference lies only in whether the speed stays constant.
| Time (s) | 0 | 10 | 20 | 30 | ___ | 50 | ___ | 70 |
|---|---|---|---|---|---|---|---|---|
| Distance (m) | 0 | 8 | ___ | 24 | 32 | 40 | ___ | 56 |
Time for 32 m = 32 ÷ 0.8 = 40 s
Seventh column (between 50 s and 70 s) = 60 s, and distance = 0.8 × 60 = 48 m
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|---|---|
| Distance (m) | 0 | 8 | 16 | 24 | 32 | 40 | 48 | 56 |
No, the motion is not uniform. In equal intervals of time (one hour each) the car covers unequal distances — 60 km, 70 km and 50 km. In uniform motion, equal distances are covered in equal intervals of time.
Average speed:
Total time taken = 1 + 1 + 1 = 3 h
Average speed = 180 km ÷ 3 h = 60 km/h
In SI units: 60 ÷ 3.6 = 16.67 m/s
Non-uniform motion is far more common in daily life. Uniform linear motion is an idealisation; we seldom find objects moving with a constant speed over long distances or for long intervals of time.
- A school bus speeds up on open roads, slows down at speed breakers and stops at bus stops.
- Walking to school — we walk slowly, then faster when late, and stop at crossings.
- A ball rolled along the ground moves fast at first, then slows down and stops.
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Distance (m) | 0 | 6 | 10 | 16 | 21 | 29 | 35 | 42 | 45 | 55 | 60 |
Work out the distance covered in each 10-second interval by subtracting successive distances.
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Distance (m) | 0 | 6 | 10 | 16 | 21 | 29 | 35 | 42 | 45 | 55 | 60 |
| Covered in that interval (m) | — | 6 | 4 | 6 | 5 | 8 | 6 | 7 | 3 | 10 | 5 |
The intervals of time are all equal (10 s each), but the distances covered in them — 6, 4, 6, 5, 8, 6, 7, 3, 10, 5 m — are not equal.
So the speed of the object is non-uniform.
= 60 m ÷ 100 s
= 0.6 m/s
In km/h: 0.6 × 3.6 = 2.16 km/h
Time for the first 500 m = 500 m ÷ 10 m/s = 50 s
Time for the next 500 m = 500 m ÷ 5 m/s = 100 s
Remaining distance = 2000 m − 1000 m = 1000 m
Remaining time = 200 s − (50 s + 100 s) = 50 s
Required speed = 1000 m ÷ 50 s = 20 m/s
Exploratory Projects — Pages 119 & 120Interdisciplinary projects
Sample answer. Float a light bowl with a fine hole at the centre of its bottom on water in a bucket, start a watch, and note when the bowl fills and sinks. Repeat with holes and bowls of different sizes. Sample readings:
| Trial | Bowl | Hole made with | Sinking time | What to do next |
|---|---|---|---|---|
| 1 | Small katori | Thin nail (large hole) | 2 min | Sinks far too fast — use a smaller hole |
| 2 | Small katori | Drawing pin | 9 min | Still fast — use a bigger bowl |
| 3 | Large bowl | Drawing pin | 21 min | Very close — try a slightly finer hole |
| 4 | Large bowl | Sewing needle | ≈ 24 min | Target reached — this is one ghati |
Result: a bigger bowl takes longer to sink; a bigger hole makes it sink sooner. A large bowl with a very fine hole sinks in about 24 minutes — one ghati.
Part 1 — Pulse rate: place two fingers on the inside of a friend’s wrist, below the thumb, count the beats for 30 s and multiply by 2. Repeat three times and take the average. Record each friend’s pulse rate at rest and after exercise:
| Name | Beats in 30 s (at rest) | Pulse rate (per min) | Pulse rate after exercise |
|---|---|---|---|
| Sample: Meera | 39 | 78 | 116 |
| Sample: Arun | 36 | 72 | 124 |
Part 2 — Pulse as a clock: at 72 beats per minute, one beat ≈ 60 ÷ 72 = 0.83 s. So counting pulse beats can measure a short time, such as a friend’s 100 m race.
| Possible reason for the difference | How to control it |
|---|---|
| Reaction time — our hand starts and stops the watch a little late | Time 20 or 30 oscillations instead of 10 and divide; start the watch at the count ‘zero’ when the bob passes the mean position, where it moves fastest and the instant is easiest to judge |
| Miscounting the oscillations | Count only the passes through the mean position in one direction; have a second student count aloud |
| Push while releasing the bob, or too wide a swing | Hold the bob, let the thread become taut, and release without any push; keep the swing small |
| Slack or stretching thread, or a slipping knot — the length keeps changing | Use a thin, strong, non-elastic thread; clamp it firmly; measure the length up to the centre of the bob each time |
| A shaky support, or the bob swinging in a circle instead of a plane | Fix the support to a heavy table or a rigid stand; release the bob in one plane only |
| Draughts of air from a fan or window | Switch off the fan and close the window while taking readings |
| Least count of the watch (a wall clock reads only to 1 s) | Use a stopwatch or a mobile phone stopwatch that reads to 0.01 s |
Repeat the activity with these controls: the readings come much closer to one another.
Yes — a swing is a large pendulum. The chains are the thread, the seat with the child on it is the bob, and the frame overhead is the rigid support.
Sample readings:
| Swing | Child on the swing | Time for 10 oscillations | Time period |
|---|---|---|---|
| Long swing (about 2.5 m) | Light child | 31.6 s | 3.16 s |
| Long swing (about 2.5 m) | Heavier child | 31.7 s | 3.17 s |
| Long swing (about 2.5 m) | Two children together | 31.5 s | 3.15 s |
| Medium swing (about 1.6 m) | Light child | 25.4 s | 2.54 s |
| Short swing (about 1.0 m) | Light child | 20.0 s | 2.00 s |
Findings: children of different weights on the same swing give almost the same time period. A longer swing has a greater time period.
Holistic Lens — Page 120Interdisciplinary task
Speed = distance ÷ winning time. Sample, using the winning times of the Tokyo 2020 and Paris 2024 Games:
| Event | Games | Winning time (s) | Speed = distance ÷ time (m/s) | Speed (km/h) |
|---|---|---|---|---|
| 100 m, men | Tokyo 2020 | 9.80 | 100 ÷ 9.80 = 10.20 | 36.7 |
| 100 m, men | Paris 2024 | 9.79 | 100 ÷ 9.79 = 10.21 | 36.8 |
| 200 m, men | Tokyo 2020 | 19.62 | 200 ÷ 19.62 = 10.19 | 36.7 |
| 200 m, men | Paris 2024 | 19.46 | 200 ÷ 19.46 = 10.28 | 37.0 |
| 400 m, men | Tokyo 2020 | 43.85 | 400 ÷ 43.85 = 9.12 | 32.8 |
| 400 m, men | Paris 2024 | 43.40 | 400 ÷ 43.40 = 9.22 | 33.2 |
| 100 m, women | Tokyo 2020 | 10.61 | 100 ÷ 10.61 = 9.43 | 33.9 |
| 100 m, women | Paris 2024 | 10.72 | 100 ÷ 10.72 = 9.33 | 33.6 |
| 200 m, women | Tokyo 2020 | 21.53 | 200 ÷ 21.53 = 9.29 | 33.4 |
| 200 m, women | Paris 2024 | 21.83 | 200 ÷ 21.83 = 9.16 | 33.0 |
| 400 m, women | Tokyo 2020 | 48.36 | 400 ÷ 48.36 = 8.27 | 29.8 |
| 400 m, women | Paris 2024 | 48.17 | 400 ÷ 48.17 = 8.30 | 29.9 |
The speed was fastest in the men’s 200 m (Paris 2024, 10.28 m/s). The 100 m and 200 m speeds are nearly equal (men about 10.2 m/s, women about 9.2–9.4 m/s), and the 400 m is clearly the slowest (about 9.2 m/s for men and 8.3 m/s for women).
Chapter 8 of 12 – Science
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Also for This Chapter
NCERT Solutions Class 7 Science Chapter 8 Measurement of Time and Motion – Overview
| Class | Class 7 |
| Subject | Science |
| Chapter | 8 – Measurement of Time and Motion |
| Textbook pages | 105 – 120 |
| Questions solved | 36 |
| Session | 2027 |
| Format | Full answers on this page + PDF |
NCERT Solutions Class 7 Science Chapter 8 – FAQs
Where can I download NCERT Solutions Class 7 Science Chapter 8 Measurement of Time and Motion PDF?
Read every answer on this page, or download the free PDF from the table at the top.
Is this for the 2027 session?
Yes, it follows the latest NCERT Class 7 Science book for 2027.
How many questions are solved in NCERT Class 7 Science Chapter 8?
All 36 questions of the chapter (textbook pages 105 – 120) – the in-text questions, activities and the end-of-chapter exercises.
How should I use these solutions?
Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.