Multiple Choice Questions.......................
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.
Solution 1: (d) Blinking of eyes.
Blinking is not a periodic phenomenon of the eye. For a specified interval of time, the eyes do not blink. Blinking of the eyes should also not be used for time estimation.
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.
Solution 2: (c) Time interval of 5 minutes can be measured by both A and B.
Clock A does not have a hand for seconds. Seconds will thus not be determined by the clock A. By using both clocks, 5 minutes can be measured.
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.
Solution 3: (a) both A and B.
Since the speed is constant for A and B, as seen in the diagram, the graph for A and B is a straight line.
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
Solution 4: (b) 10 m/s
Speed = Distance / Time
Distance = 54km
= 54 x1000
= 54000m
Time = 90 minutes
= 90×60
= 5400s
Speed = 54000 m/5400 s
= 10 m/s
Question 5: If we denote speed by S, the distance by D and time by T, the relationship between these quantities is
(a) S = D × T
(b) T = S/D
(c) S = 1/SxD
(c) S = T/D
Solution 5: (c) S = 1/SxD
The correction equation is alternative c) since Speed = Distance/Time.
Question 6: Observe Figure 13.3.
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.
Solution 3: (a) both A and B.
Since the speed is constant for A and B, as seen in the diagram, the graph for A and B is a straight line.
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
Solution 4: (b) 10 m/s
Speed = Distance / Time
Distance = 54km
= 54 x1000
= 54000m
Time = 90 minutes
= 90×60
= 5400s
Speed = 54000 m/5400 s
= 10 m/s
Question 5: If we denote speed by S, the distance by D and time by T, the relationship between these quantities is
(a) S = D × T
(b) T = S/D
(c) S = 1/SxD
(c) S = T/D
Solution 5: (c) S = 1/SxD
The correction equation is alternative c) since Speed = Distance/Time.
Question 6: Observe Figure 13.3.
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.
Solution 6: (a) A to B and back to A.
The time period of a simple pendulum is the time taken by it to travel from A to B and back to A.
Question 7: Fig. 13.4 shows an oscillating pendulum
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 + t 2 )
(c) 3 (t1 + t 2 )
(d) 4 (t1 + t 2 )
Solution 7: (d) 4 (t1 + t 2)
The average time the bob takes to travel from A to 0 is (t1+ t2), which is 1/4th of the pendulum's maximum cycle time. A pendulum time period, i.e. time taken by the pendulum, would be 4 (t1 + t2) to complete one oscillation from A to B and back to A.
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
Solution 8: (a) 5 m/s
The speed unit is a meter/second, so the response 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 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
Solution 9: (b) 7.2
=7.2 km/h (⸪ 1h = 60 mins)
Very Short Answer Questions.........................
Question 10: A simple pendulum is oscillating between two points A and B as shown in Figure 13.5. Is the motion of the bob uniform or non-uniform?
Solution 10: The bob's motion is non-uniform since the bob's speed keeps shifting.
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?
Solution 11: Paheli and Boojho have to cover different distances to reach their school but they take the same time to reach the school because their speeds can vary from each other.
Question 12: If Boojho covers a certain distance in one hour and Paheli covers the same distance in two hours, who travels in a higher speed?
Solution 12: Boojho is going at a quicker pace and he crosses the gap early.
Short Answer Questions................
Question 13: Complete the data of the table given below with the help of the distance-time graph given in Figure 13.6.
Solution 13:
Question 14: The average age of children in Class VII is 12 years and 3 months. Express this age in seconds.
Solution 14: 12 years 3 months
A year has 365 days approximately. There would be 30 days approximately in a month.
= 12 × 365 + 3 × 30 = 4470 days
A day has 24 hours, and those 24 hours consist 60 minutes each. One minute is equals to 60 seconds.
= 4470 × 24 × 60 × 60 s = 386208000 seconds
Question 15: A spaceship travels 36,000 km in one hour. Express its speed in km/s.
Solution 15: 36000 km/h=36000 km/60∗60 s
Speed of spaceship is 10 km/s.
Question 16: Starting from A, Paheli moves along a rectangular path ABCD as shown in Figure 13.7. She takes 2 minutes to travel each side. Plot a distance-time graph and explain whether the motion is uniform or non-uniform.
Solution 16: Since the distance travelled per unit time is not the same for the whole distance covered, the motion is non-uniform.
Question 17: Plot a distance-time graph of the tip of the second hand of a clock by selecting 4 points on x-axis and y-axis respectively. The circumference of the circle traced by the second hand is 64 cm.
Solution 17:
Long Answer Questions.....................
Question 18: Given below Figure 13.8 is the distance-time graph of the motion an object.
(i) What will be the position of the object at the 20s?
(ii) What will be the distance travelled by the object in 12s?
(iii) What is the average speed of the object?
Solution 18:
(i) The object is going to be 8 m away from the starting point at 20 s.
(ii) The distance traveled by the object would be 6 m in 12 seconds.
(iii) Average speed of the object is Total distance/Time taken =8m/20s =0.4m/s.
Question 19: Distance between Bholu’s and Golu’s house 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. The speed with which Bholu covered the second part of the journey?
Solution 19:
Question 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 point B and C indicate about the motion of Boojho?
(b) Is the motion between 0 to 4 minutes uniform or non-uniform?
(c) What is his speed between 8 and 12 minutes of his journey?
Solution 20:
(a) Boojho's pace is zero, so he's going to relax.
(b) Movements between 0 and 4 minutes are non-uniform.
(c) Boojho’s speed would be
= 75 m (225 m – 150 m) /4 m (12 minutes – 8 minutes)
= 18.75 m/minute