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NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion (PDF) – 2027

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.

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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

Q1How was time measured when there were no clocks and watches?
Answer

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:

DeviceWhat repeats or flows in itHow time was read
Sundial (Fig. 8.1)The Sun’s daily journey across the skyFrom 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 vesselFrom 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 bottomThe 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 anotherFrom how much sand had run down
Candle clock (Fig. 8.4)A candle burning down steadilyFrom the markings on the candle that the flame had reached

Activity 8.1: Let us construct — Page 1078.1 Measurement of Time

Activity
Q1Should we make a simple water clock?
Answer

Yes. Steps to make a simple water clock:

  1. 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].
  2. Make a small hole in the cap with a drawing pin [Fig. 8.5b].
  3. Place the upper part, with the cap on, upside down over the lower half [Fig. 8.5c].
  4. Fill the upper part with water coloured with a few drops of ink [Fig. 8.5d].
  5. As the water drips down, mark its level after every minute (use a watch) till all the water has dripped down.
Q2Your water clock is ready. Can you now guess how to use it?
Answer

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

Q1We did an activity in the chapter ‘Measurement of Length and Motion’ in the Grade 6 Science textbook Curiosity, where we observed the oscillatory motion of an eraser hung with a thread. Is the pendulum similar to that?
Answer

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 activitySimple pendulum of this chapter
An eraser tied to a threadA small metallic ball — the bob — tied to a thread
Held in the hand or hung from a hookHung from a rigid support
Moved to one side and released — it swings to and froMoved slightly to one side and released — it starts oscillatory motion
The swinging repeats again and againThe 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

Activity
Q1Gently hold the bob, move it slightly to one side and release it. Take care not to push the bob while releasing it and that the string is taut. Is your pendulum now oscillating?
Answer

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.

Q2Is the time period of your pendulum almost the same every time? What do you conclude from this observation?
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.
Answer

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)
1.20.120.1 ÷ 10 = 2.01
2.19.919.9 ÷ 10 = 1.99
3.20.220.2 ÷ 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

Think Like a Scientist!
Q1Suppose you were Galileo experimenting with pendulums, what all would you investigate? What questions could you ask? Would all pendulums have the same time period? How would you test this?
Answer

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.

Q2Repeat Activity 8.2 using the same bob but with pendulums of two or three different lengths. Does the time period change? If so, how does the length affect it?
Answer

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 pendulumTime for 10 oscillationsTime periodWhat we see
25 cm10.0 s1.0 sShort pendulum — swings quickly
50 cm14.2 s1.42 sLonger — slower
100 cm20.0 s2.0 sLongest — slowest
Q3If changing the length influences the time period, does the bob’s mass also matter? Test this by repeating Activity 8.2 with a fixed pendulum length but with bobs of different mass. Do you observe any change?
Answer

No, there is no change. Sample readings with the length fixed at 100 cm:

Bob used (length fixed at 100 cm)Time for 10 oscillationsTime period
Small iron ball (light)20.0 s2.0 s
Heavy metal ball20.1 s2.01 s
Stone of about the same size19.9 s1.99 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

Activity
Q1Look at the wall clock shown in Fig. 8.9 carefully. What is the smallest interval of time you can measure with it?
123456789101112
Fig. 8.9, page 112 — redrawn sketch of the wall clock.
Answer

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

Q1For races covering the same distance, we can tell who was faster by measuring time. But how can we tell that when comparing races for different distances?
Answer

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.

Q2What do we mean when we say something is moving fast or slow?
Answer

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.

Q3All the players begin from the starting line together but after sometime they are not running together (Fig. 8.10). How do you decide who is running faster amongst them?
Starting line
Fig. 8.10, page 113 — redrawn sketch: boys running a race on a straight track.
Answer

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.

Q4How can we determine the speed of an object?
Answer

It can be calculated if we know the total distance covered by the object and the time taken to cover it.

Speed = Total distance covered ÷ Total time taken
Q5What would be the unit of speed?
Answer

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

Activity
Q1Calculate the speed of the train between the two stations and record it in Table 8.2.

Name of the railway station nearest to your place of stay ____________________.

Table 8.2: Finding the speed of trains
S.No.Name of the trainName of the next stationDistance 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.
Answer

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.

Speed of the train = Distance till the next station ÷ Time taken till the next station

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 trainName of the next stationDistance 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 AJhansi2913.00291 ÷ 3.00 = 97 km/h
2.Superfast train BJhansi2913.50291 ÷ 3.50 = 83.1 km/h
3.Mail/Express trainBina1392.25139 ÷ 2.25 = 61.8 km/h
4.Intercity expressVidisha540.7554 ÷ 0.75 = 72 km/h
5.Passenger trainVidisha541.5054 ÷ 1.50 = 36 km/h
Q2Compare the speeds of the trains. Which is the fastest train?
Answer

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

Q1I once watched a part of marathon on a straight road stretch. I noticed that some people seemed to be running at the same speed during that distance while some people would speed up or slow down. How were their motion different?
Answer

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.

Q2Do you remember learning about linear motion in the chapter ‘Measurement of Length and Motion’ in the Grade 6 Science textbook Curiosity?
Answer

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

Q1Which of the two trains is in uniform linear motion between 10:00 AM and 11:00 AM?
Table 8.3: Distances travelled by two trains in equal time intervals of 10 minutes
Time (AM)Train XTrain Y
Position (km)Distance (km)Position (km)Distance (km)
10:000000
10:1020202020
10:2040203515
10:3060205015
10:4080207525
10:50100209520
11:001202012025
Answer

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

Let Us Enhance Our Learning
Q1Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.
Answer
Speed = Total distance covered ÷ Total time taken
= 150 m ÷ 10 s
= 15 m/s

Change m/s into km/h (1 m/s = 3.6 km/h):

1 m/s = 3.6 km/h
15 m/s = 15 × 3.6 = 54 km/h
Q2A runner completes 400 metres in 50 seconds. Another runner completes the same distance in 45 seconds. Who has a greater speed and by how much?
Answer
First runner: speed = 400 m ÷ 50 s = 8 m/s
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.

Difference = 8.89 m/s − 8 m/s = 0.89 m/s (exactly 8/9 m/s ≈ 0.9 m/s)
Q3A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?
Answer
Total time taken = Total distance covered ÷ Speed

The speed is in m/s, so first put the distance in metre:

360 km = 360 × 1000 m = 3,60,000 m
Time = 3,60,000 m ÷ 25 m/s = 14,400 s
14,400 s ÷ 60 = 240 min = 240 ÷ 60 = 4 h
Q4A train travels 180 km in 3 h. Find its speed in:
(i) km/h
(ii) m/s
(iii) What distance will it travel in 4 h if it maintains the same speed throughout the journey?
Answer

(i) In km/h

Speed = 180 km ÷ 3 h = 60 km/h

(ii) In m/s

180 km = 1,80,000 m  ·  3 h = 3 × 3600 s = 10,800 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

Total distance covered = Speed × Total time taken
= 60 km/h × 4 h = 240 km
Q5The fastest galloping horse can reach the speed of approximately 18 m/s. How does this compare to the speed of a train moving at 72 km/h?
Answer

Bring both speeds to the same unit before comparing.

Train: 72 km/h = 72 ÷ 3.6 = 20 m/s
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).

Q6Distinguish between uniform and non-uniform motion using the example of a car moving on a straight highway with no traffic and a car moving in city traffic.
Answer
Point of differenceCar on an empty straight highwayCar in city traffic
Type of motionUniform linear motionNon-uniform linear motion
SpeedStays constant, say 60 km/h, because nothing forces the driver to change itChanges constantly — slows at red lights and crossings, speeds up when the road clears
Distance in equal intervals of timeEqual: 1 km in every minuteUnequal: 1 km in one minute, then only 300 m in the next
Speedometer readingNeedle stays near one markNeedle 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.

Q7Data for an object covering distances in different intervals of time are given in the following table. If the object is in uniform motion, fill in the gaps in the table.
Time (s)0102030___50___70
Distance (m)08___243240___56
Answer
Speed = 8 m ÷ 10 s = 0.8 m/s (the same all through, because the motion is uniform)
Distance at 20 s = 0.8 × 20 = 16 m
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)010203040506070
Distance (m)08162432404856
Q8A car covers 60 km in the first hour, 70 km in the second hour, and 50 km in the third hour. Is the motion uniform? Justify your answer. Find the average speed of the car.
Answer

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 distance covered = 60 + 70 + 50 = 180 km
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
Q9Which type of motion is more common in daily life—uniform or non-uniform? Provide three examples from your experience to support your answer.
Answer

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.

  1. A school bus speeds up on open roads, slows down at speed breakers and stops at bus stops.
  2. Walking to school — we walk slowly, then faster when late, and stop at crossings.
  3. A ball rolled along the ground moves fast at first, then slows down and stops.
Q10Data for the motion of an object are given in the following table. State whether the speed of the object is uniform or non-uniform. Find the average speed.
Time (s)0102030405060708090100
Distance (m)06101621293542455560
Answer

Work out the distance covered in each 10-second interval by subtracting successive distances.

Time (s)0102030405060708090100
Distance (m)06101621293542455560
Covered in that interval (m)—64658673105

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.

Average speed = Total distance covered ÷ Total time taken
= 60 m ÷ 100 s
= 0.6 m/s
In km/h: 0.6 × 3.6 = 2.16 km/h
Q11A vehicle moves along a straight line and covers a distance of 2 km. In the first 500 m, it moves with a speed of 10 m/s and in the next 500 m, it moves with a speed of 5 m/s. With what speed should it move the remaining distance so that the journey is complete in 200 s? What is the average speed of the vehicle for the entire journey?
Answer
Total distance = 2 km = 2000 m
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
Average speed = Total distance ÷ Total time = 2000 m ÷ 200 s = 10 m/s

Exploratory Projects — Pages 119 & 120Interdisciplinary projects

Exploratory Projects
Q1Construct a floating bowl-type water clock. Experiment by using bowls of different sizes and making holes of different sizes in them so that the sinking time of the bowl can be close to 24 minutes.
Answer

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:

TrialBowlHole made withSinking timeWhat to do next
1Small katoriThin nail (large hole)2 minSinks far too fast — use a smaller hole
2Small katoriDrawing pin9 minStill fast — use a bigger bowl
3Large bowlDrawing pin21 minVery close — try a slightly finer hole
4Large bowlSewing needle≈ 24 minTarget 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.

Q2Design an activity for measuring the pulse rate (number of times the pulse of a person beats in 1 minute) of your friends. Think of an activity where you can use your pulse to measure time and develop a story over that idea.
Answer

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:

NameBeats in 30 s (at rest)Pulse rate (per min)Pulse rate after exercise
Sample: Meera3978116
Sample: Arun3672124

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.

Q3What might be the reasons for the slight differences in the time periods of a pendulum of a given length in different readings taken in Activity 8.2. Think of ways to control those and repeat the activity to check if the difference in readings is reduced.
Answer
Possible reason for the differenceHow to control it
Reaction time — our hand starts and stops the watch a little lateTime 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 oscillationsCount 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 swingHold 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 changingUse 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 planeFix the support to a heavy table or a rigid stand; release the bob in one plane only
Draughts of air from a fan or windowSwitch 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.

Q4Visit a playground with a few swings. Measure the time taken by a swing for 10 oscillations and calculate its time period. Repeat it a few times with children of different weights to find out if its time period is almost the same. Repeat this with swings of different lengths. Find out how the time period changes with increasing length of the swings. Is the swing also an example of a pendulum?
Answer

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:

SwingChild on the swingTime for 10 oscillationsTime period
Long swing (about 2.5 m)Light child31.6 s3.16 s
Long swing (about 2.5 m)Heavier child31.7 s3.17 s
Long swing (about 2.5 m)Two children together31.5 s3.15 s
Medium swing (about 1.6 m)Light child25.4 s2.54 s
Short swing (about 1.0 m)Light child20.0 s2.00 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

Holistic Lens
Q1Gather the timings of the winners of the races — 100 m, 200 m, and 400 m for men and women in the last two Olympic games. Calculate and compare their speeds. In which event is the speed the fastest?
Answer

Speed = distance ÷ winning time. Sample, using the winning times of the Tokyo 2020 and Paris 2024 Games:

EventGamesWinning time (s)Speed = distance ÷ time (m/s)Speed (km/h)
100 m, menTokyo 20209.80100 ÷ 9.80 = 10.2036.7
100 m, menParis 20249.79100 ÷ 9.79 = 10.2136.8
200 m, menTokyo 202019.62200 ÷ 19.62 = 10.1936.7
200 m, menParis 202419.46200 ÷ 19.46 = 10.2837.0
400 m, menTokyo 202043.85400 ÷ 43.85 = 9.1232.8
400 m, menParis 202443.40400 ÷ 43.40 = 9.2233.2
100 m, womenTokyo 202010.61100 ÷ 10.61 = 9.4333.9
100 m, womenParis 202410.72100 ÷ 10.72 = 9.3333.6
200 m, womenTokyo 202021.53200 ÷ 21.53 = 9.2933.4
200 m, womenParis 202421.83200 ÷ 21.83 = 9.1633.0
400 m, womenTokyo 202048.36400 ÷ 48.36 = 8.2729.8
400 m, womenParis 202448.17400 ÷ 48.17 = 8.3029.9

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

Also for This Chapter

NCERT Solutions Class 7 Science Chapter 8 Measurement of Time and Motion – Overview

ClassClass 7
SubjectScience
Chapter8 – Measurement of Time and Motion
Textbook pages105 – 120
Questions solved36
Session2027
FormatFull 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.

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