Science Class 7 Chapter 13 -Motion and Time- NCERT Exercise Solutions (Question-Answer)

CBSE Science Class 7 Chapter 13 Motion and Time -NCERT Exercise Solutions (Question-Answer) NCERT Exercise Solution (Question-Answer) is given below. Total 13 Questions, all are solved here.

Q1. Classify the following as motion along a straight line, circular or oscillatory motion:

(i) Motion of your hands while running.

Answer: Oscillatory motion

(ii) Motion of a horse pulling a cart on a straight road.

Answer: Linear motion (motion along straight line)

(iii) Motion of a child in a merry-go-round.

Answer: Circular motion

(iv) Motion of a child on a see-saw.

Answer: Oscillatory motion

(v) Motion of the hammer of an electric bell.

Answer: Oscillatory motion

(vi) Motion of a train on a straight bridge.

Answer: Linear motion

Q2. Which of the following are not correct?

(i) The basic unit of time is second.

Answer: Correct

(ii) Every object moves with a constant speed.

Answer: Not correct

Correct: Object moves with constant or variable speed

(iii) Distances between two cities are measured in kilometres.

Answer: Correct

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

Answer: Not correct

Correct: The time period of the pendulum depends upon the length of the thread.

(v) The speed of a train is expresses in m/h.

Answer: Not correct

Correct: Speed of train is expressed in m/s or km/h

Q3. A simple pendulum takes 32s to complete 20 oscillations. What is the time period of the pendulum?

Answer: Given,

        Time taken by pendulum = 32 s

        No. of oscillation= 20

       Time period = total time taken/no. of oscillation

                              = 32/20

                             = 1.6 s

Q4. The distance between the two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.

Answer: Given,

Distance between stations = 240 km

Time taken by train = 4 hours

Speed of train = Distance /Time

                           = 240/4 km/h

                           =60 km/h

Q5. The odometer of a car reads 57321.0 km when the clock shows the time 08:30 AM. What is the distance moved by car, if 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: Given,

At 8:30 AM, odometer of car reads (initial reading) = 57321.0 km

At 8:50 AM, odometer of car reads (final reading) = 57336.0 km

Distance covered by car = Final reading – Initial reading

                                            = 57336.0 – 57321.0

                                            = 15 km

Time taken = 08:50 AM – 08:30 AM

          = 20 min

Speed of the car = Total Distance/Time taken

=15/20 km/min

                        =3/4 km/min

Now,

       60 min = 1 hr

20 min = ? hr

        ? hr = (1 x 20)/60 hr

                      = 1/3 hr

 Time taken by car = 1/3 hr

Speed of car = 15/(1/3)

                       =45 km/hr

Q6. Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the distance between her house and the school.

Answer: Time taken by Salma = 15 min

                                                     = 15 * 60 second

                                                     = 900 second

                 Speed of bicycle = 2 m/s

                Distance covered by her = speed * time

                                                             = 2*900

                                                            = 1800 m

= 1.8 km      (1 km = 1000 m)

Q7. Show the shape of the distance-time graph for the motion in the following cases:

(i) A car moving with a constant speed.

Answer:

(ii) A car parked on a side road.

Answer:

Q8. Which of the following relations is correct?

(i) speed= distance * time

(ii) speed= distance/time

(iii) speed= time/distance

(iv) speed= 1/ distance*time

Answer: (ii) speed= distance/time

Q9. The basic unit of speed is:

(i) km/min

(ii) m/min

(iii) km/h

(iv) m/s

Answer: (iv) m/s

Q10. A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:

(i) 100 km

(ii) 25 km

(iii) 15 km

(iv) 10 km

Answer: (ii) 25 km

Explanation:

For the first 15 min car moves with 40 km/h and the next 15 min car moves with 60 km/h

A/Q

 In case 1:

15 min = 15/60 hr

              =0.25 hr

Distance covered by car (d1) = speed * time

                                             = 40 * 0.25

                                             = 10 km

Similarly, in case 2:

Distance covered by car (d2) = 60 * 0.25

                                                    = 15 km

Hence, total distance covered by car = d1 + d2

                                                                   = 10 + 15

                                                                   = 25 km

Q11. Suppose the two photographs, shown in fig. 13.1 and fig. 13.2, had been taken at an interval of 10 seconds, if a distance of 100 meters is shown by 1 cm in these photographs, calculate the speed of the fastest car.

Answer: This answer may varry from people to people because here exact measuring is not done. Although, we will try to reach the correct answer.

After comparing these two figures, we can observe that the green car is moving fast. It moves about 1.2 cm as shown in the photograph above. If a distance of 100 metres is shown by 1 cm in these photographs, it means the green car moves 120 m.

So, the speed will be = Distance/time

= 120 m/10 s

=12 m/s

Q12. Fig. 13.15 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

Answer: Vehicle “A” is moving faster

Q13. Which of the following distance-time graph shows a truck moving with the speed which is not constant?

Answer: (iii)

Science Class 7 Chapter 12 Reproduction in Plants – NCERT Exercise Solutions (Question-Answer) ended here.👍👍👍

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