Table of Contents
NCERT Exemplar Class 7 Science Chapter –13 – Motion and Time(Solved with detailed explanation)
Question 1. Which of the following cannot be used for measurement of time?
(a) A leaking tap.
(b) Simple pendulum.
(c) Shadow of an object during the day.
(d) Blinking of eyes.
- Answer: (d) Blinking of eyes.
- Explanation: A leaking tap, a simple pendulum, and the shadow of an object during the day can all be used to measure time intervals, as they have periodic or predictable movements. The blinking of eyes, however, is irregular and varies greatly from person to person, thus it cannot be used as a reliable measure of time.
Question 2: Two clocks A and B are shown in Figure 13.1. Clock A has an hour and a minute hand whereas clock B has an hour hand, minute hand as well as a second hand. Which of the following statement is correct for these clocks?
(a) A time interval of 30 seconds can be measured by clock A.
(b) A time interval of 30 seconds cannot be measured by clock B.
(c) Time interval of 5 minutes can be measured by both A and B.
(d) Time interval of 4 minutes 10 seconds can be measured by clock A.
Answer: (c) Time interval of 5 minutes can be measured by both A and B.
Explanation:
Clock A is equipped with an hour and a minute hand, but lacks a second hand. As a result, it cannot accurately measure time intervals in seconds. However, it can measure time intervals in minutes, which makes it capable of measuring a 5-minute interval.
On the other hand, Clock B has an hour hand, a minute hand, and a second hand, providing the ability to measure time intervals as precisely as seconds. Therefore, it can certainly measure intervals in minutes, including a 5-minute interval.
Therefore, both clocks, A and B, are capable of measuring a time interval of 5 minutes, making option (c) the correct answer.
Also Check – Chapter 9 – Motion and Time – 4 Worksheets Solved and Unsolved
Question 3: Two students were asked to plot a distance-time graph for the motion described by Table A and Table B. The graph given in Figure 13.2 is true for
(a) both A and B.
(b) A only.
(c) B only.
(d) neither A nor B.
Answer: (a) both A and B.
Explanation:
The key to understanding this question lies in the characteristics of the distance-time graph. For motion that is constant, the distance-time graph should be a straight line. This straight-line graph indicates that the object is moving at a constant speed, covering equal distances in equal intervals of time.
Since both Table A and Table B describe motions with constant speeds, the distance-time graph for both would be represented by straight lines. Assuming that the graph shown in Figure 13.2 is a straight line, it would correctly represent the motion described by both Table A and Table B. Therefore, the correct answer is (a) both A and B, as they both show motion at a constant speed, which is represented by a straight line on a distance-time graph.
Question 4: A bus travels 54 km in 90 minutes. The speed of the bus is
(a) 0.6 m/s
(b) 10 m/s
(c) 5.4 m/s
(d) 3.6 m/s
Answer:
To find the speed of the bus, we use the formula: Speed = Distance / Time
First, convert the time from minutes to seconds, as the speed is often measured in metres per second (m/s).
90 minutes = 90 × 60 seconds = 5400 seconds
Now, convert the distance from kilometres to metres.
54 km = 54 × 1000 metres = 54000 metres
Calculate the speed using the formula.
Speed = 54000 metres / 5400 seconds
Speed = 10 metres/second
Therefore, the correct answer is (b) 10 m/s
Question 5: The odometer of a car reads 57321.0 km when the clock shows the time 08:30 AM. What is the distance moved by the car if at 08:50 AM the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time. Express the speed in km/h also.
Answer:
- Calculate the distance moved by the car:
Distance moved = Final odometer reading – Initial odometer reading
Distance moved = 57336.0 km – 57321.0 km
Distance moved = 15.0 km - Calculate the time in minutes:
Time taken = 08:50 AM – 08:30 AM
Time taken = 20 minutes - Calculate the speed in km/min:
Speed = Distance / Time
Speed = 15.0 km / 20 min
Speed = 0.75 km/min - Convert the speed to km/h:
Speed in km/h = Speed in km/min × 60 min/h
Speed in km/h = 0.75 km/min × 60 min/h
Speed in km/h = 45 km/h
Therefore, the car moved a distance of 15.0 km, with a speed of 0.75 km/min, which is equivalent to 45 km/h
Question 6: The time period of a simple pendulum is the time taken by it to travel from
(a) A to B and back to A.
(b) O to A, A to B and B to A.
(c) B to A, A to B and B to O.
(d) A to B.
Answer: (a) A to B and back to A.
Explanation:
The time period of a simple pendulum is defined as the duration it takes for the pendulum to complete one full oscillation. A full oscillation for a simple pendulum occurs when it swings from its initial position (let’s say A) to the opposite extreme (B) and then back to its starting position (A). This complete to-and-fro movement constitutes one oscillation. Therefore, the time period is the time taken by the pendulum to move from A to B and then back to A.
Question 7: Fig. 13.4 shows an oscillating pendulum. The time taken by the bob to move from A to C is t1 and from C to O is t2. The time period of this simple pendulum is
(a) (t1 + t2)
(b) 2 (t1 + t2)
(c) 3 (t1 + t2)
(d) 4 (t1 + t2)
Answer: (d) 4 (t1 + t2)
Explanation:
The time period of a simple pendulum is the total time taken for it to complete one full oscillation. In this context, one full oscillation includes the movement of the pendulum from one extreme (A), through the mean position (O), to the other extreme (C), and back to the initial extreme (A) via the mean position (O) again.
- The time taken from A to C is t1, and from C to O is t2.
- To complete one full oscillation, the pendulum must also move from O to A (another t2) and then from A to C (another t1).
Therefore, the total time for one full oscillation is t1 + t2 (from A to O) plus another t1 + t2 (from O back to A), which equals 2(t1 + t2). However, since the pendulum needs to return to its starting point (A), we need to consider the journey from A to C and back to A, which is double the time of a half oscillation, resulting in 4(t1 + t2).
Thus, the time period of the pendulum, i.e., the time taken to complete one full oscillation from A to B and back to A, is accurately represented by 4(t1 + t2)
Question 8: The correct symbol to represent the speed of an object is
(a) 5 m/s
(b) 5 mp
(c) 5 m/s-1
(d) 5 s/m
Answer: (a) 5 m/s
Explanation:
The correct symbol to represent the speed of an object is “m/s” which stands for metres per second. This unit is derived from the basic formula of speed, which is distance divided by time. Since distance is often measured in metres and time in seconds, the unit for speed becomes metres per second. Therefore, the correct answer is (a) 5 m/s
Question 9: Boojho walks to his school which is at a distance of 3 km from his home in 30 minutes. On reaching he finds that the school is closed and comes back by a bicycle with his friend and reaches home in 20 minutes. His average speed in km/h is
(a) 8.3
(b) 7.2
(c) 5
(d) 3.6
Answer: (b) 7.2
Explanation:
To calculate the average speed, we first find the total distance travelled and the total time taken.
- Total distance: Boojho travels 3 km to school and 3 km back home, making it a total of 3 km + 3 km = 6 km.
- Total time: He takes 30 minutes to walk to school and 20 minutes to cycle back, so the total time is 30 min + 20 min = 50 minutes.
Convert the total time into hours since speed is required in km/h.
50 minutes = 50/60 hours ≈ 0.83 hours
Now, calculate the average speed.
Average speed = Total distance / Total time
Average speed = 6 km / 0.83 hours ≈ 7.2 km/h
Therefore, the correct answer is (b) 7.2 km/h, which is Boojho’s average speed for the entire journey
Question 10: A simple pendulum oscillates between two points, A and B, as shown in Figure 13.5. Is the motion of the bob uniform or non-uniform?
Answer: The motion of the bob is non-uniform.
Explanation:
The motion of a simple pendulum is characterised as non-uniform because the speed of the bob changes throughout its oscillation. When a pendulum moves between two points (A and B in this case), its speed is not constant. It is fastest at the lowest point of its swing (the mean position) and gradually slows down as it reaches the extreme positions (A and B). At these extreme positions, the speed momentarily becomes zero before the pendulum swings back in the opposite direction. This variation in speed during the oscillation indicates that the motion is non-uniform.
Question 11: Paheli and Boojho have to cover different distances to reach their school but they take the same time to reach the school. What can you say about their speed?
Answer:
If Paheli and Boojho take the same time to reach school but cover different distances, it means they are travelling at different speeds. Speed is calculated as the distance travelled divided by the time taken. Since the time taken is the same for both but the distances are different, the one covering the longer distance in the same amount of time is travelling at a higher speed. Therefore, their speeds are not the same, and the one covering the greater distance has a higher speed.
Question 12: If Boojho covers a certain distance in one hour and Paheli covers the same distance in two hours, who travels at a higher speed?
Answer:
Speed is calculated as distance divided by time. Since both Boojho and Paheli cover the same distance but Boojho does it in less time, Boojho’s speed is higher. To illustrate:
- Boojho’s speed = Distance / 1 hour
- Paheli’s speed = Distance / 2 hours
Since the distance is the same for both, the speed is inversely proportional to the time taken. Therefore, Boojho, who takes less time to cover the same distance, travels at a higher speed.
Short Answer Questions
13. Complete the data of the table given below with the help of the distance-time graph given in Figure 13.6.
Soln:
Question 14: The average age of children of Class VII is 12 years and 3 months. Express this age in seconds.
Answer:
To convert the age of 12 years and 3 months into seconds, follow these steps:
- Convert years and months into seconds:
- There are 12 months in a year and 30 days in a month (approximately).
- 12 years = 12 × 365 days/year = 4380 days
- 3 months = 3 × 30 days/month = 90 days
- Total days = 4380 days + 90 days = 4470 days
- Convert days into seconds:
- 1 day = 24 hours
- 1 hour = 60 minutes
- 1 minute = 60 seconds
- Total seconds = 4470 days × 24 hours/day × 60 minutes/hour × 60 seconds/minute
Question 15: A spaceship travels 36000 km in one hour. Express its speed in km/s.
Answer:
To convert the speed from km/h to km/s, divide the speed in kilometres per hour by the number of seconds in an hour (3600 seconds).
Speed in km/s=Speed in km per h/ Number of seconds in an hour
Speed in km/s=36000 km per h/3600 seconds
Speed in km/s=10 km/s
Therefore, the speed of the spaceship is 10 kilometres per second.
16. Starting from A, Paheli moves along a rectangular path ABCD as shown in Figure 13.7. She takes 2 minutes to travel to each side. Plot a distance-time graph and explain whether the motion is uniform or non-uniform.
Soln:
Given that the distance traversed in each unit of time is not consistent throughout the journey, the motion is classified as non-uniform.
17. Plot a distance-time graph of the tip of the second hand of a clock by selecting 4 points on the x-axis and y-axis, respectively. The circumference of the circle traced by the second hand is 64 cm.
Soln:
Long Answer Questions
18. Given below as Figure 13.8 is the distance-time graph of the motion of an object.
(i) What will be the position of the object in the 20s?
After 20 seconds, the object will have moved 8 meters from its initial position.
(ii) What will be the distance travelled by the object in 12s?
Within 12 seconds, it covers a distance of 6 meters.
(iii) What is the average speed of the object?
The object’s average speedTotal distance
Total distance/Time taken=8m/20s=0.4m/s
19. Distance between Bholu’s and Golu’s houses is 9 km. Bholu has to attend Golu’s birthday party at 7 o’clock. He started from his home at 6 o’clock on his bicycle and covered a distance of 6 km in 40 minutes. At that point, he met Chintu, and he spoke to him for 5 minutes and reached Golu’s birthday party at 7 o’clock. With what speed did he cover the second part of the journey? Calculate his average speed for the entire journey.
Answer – The speed with which Bholu covered the second part of the journey
Soln:
20. Boojho goes to the football ground to play football. The distance-time graph of his journey from his home to the ground is given as Figure 13.9.
(a) What does the graph between points B and C indicate about the motion of Boojho?
Answer- (a) The section of the graph between points B and C shows that Boojho is stationary since the distance does not change over this time interval.
(b) Is the motion between 0 to 4 minutes uniform or nonuniform?
Answer-(b) The motion from 0 to 4 minutes is characterized as non-uniform, as indicated by the graph’s varying slope.
(c) What is his speed between 8 and 12 minutes of his journey?
Answer-(c) During the time frame from 8 to 12 minutes, Boojho’s speed is calculated to be 75/4 = 18.75 m/minute
Rapid Revision – Class 7 Science- Chapter 13 – Motion and Time
Chapter 13 – Motion and Time–Class 7 science- Question and Answer (Fill in the Blanks)
Non-Uniform and Uniform Motion – for Upper Primary School Students
Class 7 science -Chapter 13 – Motion and Time- Complete Notes
Chapter 13- Motion and Time–Class 7 science- Question and Answer (Solved MCQs)
Class 7 – Science- Chapter 13- Motion and Time- Question Answer (Long Question Answer)
Class 7 -Science- Chapter 13- Motion and Time Question Answer (Short Question Answer)
NCERT Solutions for Class 7 Science Chapter 13-Motion and Time
