Important Questions Class 7 Science Chapter 13

Important Questions Class 7 Science Chapter 13 – Motion and Time

Science is a basic foundational subject consisting of three branches: physics, biology, and chemistry. In today’s world, we are surrounded by a lot of Science-oriented developments. Hence, students need to strengthen the basic concepts of science and the application of those concepts in the practical world.

Chapter 13 of Science Class 7 deals with the concept of motion and time, an introductory chapter for the complex and advanced topics in the upper classes. The core topics covered in this chapter are:

• Concept of motion and its types
• Speed and its applications
• Measurement of time
• Measurement of speed
• Distance-time graphs for different kinds of motion

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As students move up to higher classes, Science becomes more challenging. It is critical for students to give importance to complex subjects and focus on understanding concepts and practising numerals. In the chapter motion and time, students must grasp the meaning of terms and their application in daily life for an easier retention of the concept. Students are advised to solve Important Questions Class 7 Science Chapter 13 to get easier access to all the important concepts at a single glance.

Important Questions Class 7 Science Chapter 13 – With Solutions

Practising Important Questions Class 7 Science Chapter 13 will aid students in ensuring the preparation of all the important topics for exams. While self accessing, if they realise the need to practise a topic or question, they can learn it while solving Important Questions Class 7 Science Chapter 13.

The following Chapter 13 Class 7 Science Important Questions are provided for students to help them better understand Chapter 13- motion and time.

Question 1: Classify the following motions as straight line, circular, or oscillatory:

(a) Hand motion while sprinting.

(b) The motion of a horse while pulling a cart along a straight road.

(c) Movement of a child on a merry-go-round.

(d) A child swinging on a see-saw.

(e) The motion of an electric bell’s hammer.

(f) The motion of a train on a straight bridge.

Answer 1:

 (a) Oscillatory motion- while moving, hands move back and forth across our bodies after a given interval of time. As a result, it oscillates.

(b) Straight line- as horse is moving cart on a straight road, hence motion is along a straight line

(c)Circular motion- Motion of a child in a merry-go-round is circular, as the merry-go-round also has a circular motion.

(d) Oscillatory motion- Motion of a child on a see-saw is oscillatory as the see-saw goes up and down while the child rides it.

(e) Oscillatory motion- the hammer hits the electric bell ,and vibrates to and fro rapidly, hence its oscillatory motion.

(f)Straight line- Motion of a train on a straight bridge is a straight line as the path is straight.

Question 2: It takes 32 seconds for a simple pendulum to complete 20 oscillations. What is the pendulum’s time period?

Answer 2: 

Number of oscillation completed= 20

Time taken to complete 20 oscillations= 32 sec

Time period of the pendulum :

T=total   time  taken number of oscillations

T= 3220

T=1.6 sec

Question 3: The distance-time graph for the motion of two vehicles, A and B, is shown in Figure. 

Which of them is moving the fastest?

Distance-time graph of two cars in motion.

Answer 3:  Vehicle A is moving faster than vehicle B.

As speed is directly proportional to distance covered and inversely proportional to time taken hence,

At a given time ‘t’ the distance covered by vehicle A is more than the distance covered by vehicle B. Hence, vehicle A is faster.

Question 4: The distance between two stations is 240 kilometres. 

This distance is covered in four hours by train. 

Calculate the train’s speed.

Answer 4: 

Distance between two stations= 240 km

Time taken to cover the distance= 4 hours

speed= distancetime=2404=60km/h

Question 5: When the clock reads 08:30 AM, the odometer of a car reads 57321.0 km. 

What is the distance travelled by the car if the odometer reads 57336.0 km at 08:50 AM? 

Calculate the car’s speed in kilometres per hour during this time. 

Express the speed in kilometres per hour as well.

Answer 5: 

The car’s odometer reads 57321.0 km at the start.

The car’s final odometer reading = 57336.0 km

Distance driven = Final reading of the car’s odometer + Initial reading of the car’s odometer

The car travelled 57336.0 57321.0 = 15 kilometres.

The specified car departs at 8:30 a.m.

and terminates at 8:50 a.m.

As a result, the time it takes the car to travel the distance is (8:50 8:30) min = 20 min.

The car travelled 15 kilometres.

Time taken by the car = 20 min

speed=distance  coveredtime taken=1520=0.75 km/min

60 min = 1 hour

20 min=16020= 13h

Time taken by car in hours = 1/3h

speed=distance  coveredtime taken=151/3=45 km/h

Question 6: The basic unit of speed is:

(a) km/min

(b) km/h

(c) m/min

(d) m/s

Answer 6: (d) m/s

speed=distance  coveredtime taken

Unit of distance is measured in metres (m).

Unit of time is measured in seconds (sec).

So, the basic unit of speed is m/s.
Question 7:A car travels at 40 km/h for 15 minutes, then at 60 km/h for the next 15 minutes. 

The car has travelled the following distance:

(a)15 km

(b) 25 km

(c) 100 km

(d) 10 km

Answer 7: (b) 25 km

Case I

The car’s maximum speed is 40 kilometres per hour.

Time spent = 15 minutes = 15/60 = 0.25 hours

speed=distance traveled/time spent

Distance travelled, d1 = Speed Time Required = 40 0.25 = 10 km

Case No. II

The car’s maximum speed is 60 kilometres per hour.

Time spent = 15 minutes = 15/60 = 0.25 hours

speed=distance traveled/time spent

Distance travelled, d2 = Speed Time Required = 60 0.25 = 15 km

d = d1 + d2 = 10 + 15 = 25 km total distance travelled by car

As a result, the total distance travelled by the car is 25 kilometres.

Question 8: Which of the following statements is incorrect?

(a) The fundamental unit of time is the second.

(b) Every object moves with a constant speed.

(c) Distances between cities are expressed in kilometres.

(d) A given pendulum time period is constant.

(e) The speed of a train is measured in metres per hour.

Answer 8: Incorrect statements are :

(b) Every object moves with a constant speed.

(d) The time period of a given pendulum is constant.

(e) The speed of a train is expressed in m/h.

Question 9: A bus travels 54 km in 90 minutes. The speed of the bus is

(a)5.4 m/s

(b) 10 m/s

(c) 3.6 m/s

(d) 0.6 m/s

Answer 9: (b) 10m/s

speed=distance  coveredtime taken

Distance = 54km= 54 X 1000 = 54000m

Time = 90 minutes = 90 X 60 = 5400sec

speed=distance  coveredtime taken=540005400= 10m/s

Question 10: Boojho walks to his school, which is 3 kilometres from his house, in 30 minutes. 

When he arrives, he discovers that the school is closed and returns home on his bicycle with a friend in 20 minutes. 

His average speed in kilometres per hour is: 

(a) 8

(b) 7.2

(c) 8.3

(d) 3.6

Answer 10: (b) 7.2

speed=distance  coveredtime taken= 65060=7.2km/h

Question 11: Given below as Figure 13.8 is the distance-time graph of the motion an object.

(a) What will be the position of the object in the 20s?

(b) What will be the distance travelled by the object in 12s?

(c) What is the average speed of the object?

Answer 11: 

(a) At 20sec the object will be 8 m away from the starting point.

(b) In 12 sec, the distance travelled by the object will be 6 metres.

(c) Average speed of the object is the total distance/total time taken = 8m/20 sec = 0.4m/s

Question 12:The distance between Bholu house and Golu’s house is 9 kilometres. 

At 7 p.m., Bholu must attend Golu’s birthday party. 

He set out on his bicycle at 6 a.m. and covered a distance of 6 km in 40 minutes. 

At that point, he met Chintu and spoke with him for 5 minutes before arriving at Golu’s birthday party at 7 p.m. 

How quickly did he complete the second leg of the journey? 

Calculate his average speed over the course of the journey.

Answer 12: 

The speed with which Bholu covered the second part of the journey = Distance  left  to  reach  Golu’s house Time left

Speed=9km – 6km1hour- 45 min=3km1/4h=12km/h

Average speed= Total distance travelledTotal time taken= 9km1h=9km/h

Question 13: Differentiate between distance and displacement.

Answer 13:  

Question 14: Define a) Non-uniform speed b) Instantaneous speed

Answer 14: 

a) Non-uniform speed: An object is said to be travelling with non-uniform speed if it covers equal distance in unequal intervals of time or unequal distance in equal interval of time.

b)Instantaneous speed: The speed of a moving object at any particular instant of time is called instantaneous speed.

Question 15: What do you mean by the oscillation of a pendulum?

Answer 15:Pendulum oscillation is defined as a full to and fro motion of the pendulum from its rest position.. The time taken by it to complete one oscillation is called the pendulum’s time period. It depends on the string of the pendulum.

Question 16: Speedometer records the reading in

a)m/h

b)km/s

c)m/s

d)km/h

Answer 16: d) km/h

Speedometer records the reading in km/hr.

Question 17: Minimum speed required for rockets to launch satellites into earth’s orbit is:

1. 11km/s
2. 8 km/s
3. 12 km/s
4. 10 km/s

Answer 17: b) 8 km/s

Minimum speed required for rockets to launch satellites into earth’s orbit to overcome the resisting force is 8 km/s.

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