Question 1

Question 1A car covers a distance of 160 km between two cities in 4 h. What is the average speed of the car?

Accepted Answer

Distance = 160 km;time = 4 hAverage Speed=Total distance travelledtotal time takenAverage Speed=1604=40km h−1\text{Average Speed} = \dfrac{\text{Total distance travelled}}{\text{total time taken}} \[1em] \text{Average Speed} = \dfrac{\text{160}}{\text{4}} = 40 \text{km h}^{-1}Average Speed=total time takenTotal distance travelled​Average Speed=4160​=40km h−1So, average speed of the car = 40 km h-1.

Question 2

Question 2A train travels a distance of 300 km with an average speed of 60 km h-1. How much time does it take to cover the distance?

Accepted Answer

Distance = 300 km;Average speed = 60 km h-1time = ?Average speed=Distance travelledtime60=300timetime=30060=5hr\text{Average speed} = \dfrac{\text{Distance travelled}}{\text{time}} \[1em] \text{60} = \dfrac{\text{300}}{\text{time}} \[1em] \text{time} = \dfrac{\text{300}}{\text{60}} = 5 \text{hr}Average speed=timeDistance travelled​60=time300​time=60300​=5hrSo, time taken = 5 hour.

Question 3

Question 3A boy travels with an average speed of 10 m s-1 for 20 min. How much distance does he travel?

Accepted Answer

Average speed = 10 m s-1;time = 20 min1 min = 60 sec20 min = 60 x 20 = 1200 secdistance = ?Average speed=distancetime10=distance1200distance=1200×10=12000m\text{Average speed} = \dfrac{\text{distance}}{\text{time}} \[1em] \text{10} = \dfrac{\text{distance}}{\text{1200}} \[1em] \text{distance} = 1200 \times 10 = 12000 \text{m}Average speed=timedistance​10=1200distance​distance=1200×10=12000mSo, distance covered = 12000 m or 12 km.

Question 4

Question 4A boy walks a distance of 30 m in 1 minute and another 30 m in 1.5 minute. Describe the type of motion of the boy and find his average speed in m s-1.

Accepted Answer

The motion of the boy is non-uniform.total distance travelled = 30 m + 30 m = 60 mtotal time taken = 1 min + 1.5 min = 2.5 min1 min = 60 sec2.5 min = 2.5 x 60 = 150 secAverage speed = ?Average Speed=Total distance travelledtotal time takenAverage Speed=60150=0.4m s−1\text{Average Speed} = \dfrac{\text{Total distance travelled}}{\text{total time taken}} \[1em] \text{Average Speed} = \dfrac{\text{60}}{\text{150}} = 0.4 \text{m s}^{-1}Average Speed=total time takenTotal distance travelled​Average Speed=15060​=0.4m s−1So, average speed of the boy = 0.4 m s-1.

Question 5

Question 5A cyclist travels a distance of 1 km in the first hour, 0.5 km in the second hour and 0.3 km in the third hour. Find the average speed of the cyclist in:(i) km h-1(ii) m s-1.

Accepted Answer

(i) total distance travelled = 1 km + 0.5 km + 0.3 km = 1.8 kmtotal time taken = 1 h + 1 h + 1 h = 3 hAverage Speed=Total distance travelledtotal time takenAverage Speed=1.83=0.6km h−1\text{Average Speed} = \dfrac{\text{Total distance travelled}}{\text{total time taken}} \[1em] \text{Average Speed} = \dfrac{\text{1.8}}{\text{3}} = 0.6 \text{km h}^{-1}Average Speed=total time takenTotal distance travelled​Average Speed=31.8​=0.6km h−1So, average speed of the cyclist = 0.6 km h-1.(ii) total distance travelled = 1 km + 0.5 km + 0.3 km = 1.8 km1 km = 1000 m1.8 km = 1.8 x 1000 = 1800 mtotal time taken = 1 h + 1 h + 1 h = 3 h1 h = 3600 sec3 h = 3 x 3600 = 10800 secAverage Speed=Total distance travelledtotal time takenAverage Speed=180010800=0.167m s−1\text{Average Speed} = \dfrac{\text{Total distance travelled}}{\text{total time taken}} \[1em] \text{Average Speed} = \dfrac{\text{1800}}{\text{10800}} = 0.167 \text{m s}^{-1}Average Speed=total time takenTotal distance travelled​Average Speed=108001800​=0.167m s−1So, average speed of the cyclist = 0.167 m s-1.

Question 6

Question 6A car travels with speed 30 km h-1 for 30 minutes and then with speed 40 km h-1 for 1 hour. Find:(a) the total distance travelled by the car.(b) the total time of travel, and(c) the average speed of the car.

Accepted Answer

(a) In first part of journeyspeed = 30 km h-1time = 30 min60 min = 1 h30 min = 12\dfrac{\text{1}}{\text{2}}21​ hSpeed=distancetime30=distance12distance=30×12=15km\text{Speed} = \dfrac{\text{distance}}{\text{time}} \[1em] \text{30} = \dfrac{\text{distance}}{\dfrac{1}{2}} \[1em] \text{distance} = 30 \times {\dfrac{1}{2}} = 15 \text{km}Speed=timedistance​30=21​distance​distance=30×21​=15kmIn second part of journeyspeed = 40 km h-1time = 1 hSpeed=distancetime40=distance1distance=40×1=40km\text{Speed} = \dfrac{\text{distance}}{\text{time}} \[1em] \text{40} = \dfrac{\text{distance}}{\text{1}} \[1em] \text{distance} = 40 \times 1 = 40 \text{km}Speed=timedistance​40=1distance​distance=40×1=40kmSo, total distance travelled = 15 km + 40 km = 55 km.(b) time taken in first part of journey = 30 min60 min = 1 h30 min = 12\dfrac{\text{1}}{\text{2}}21​ htime taken in second part of journey = 1 htotal time taken = 1 h + 12\dfrac{\text{1}}{\text{2}}21​ h = 1.5 hSo, total time taken = 1.5 h.(c) Average Speed=Total distance travelledtotal time takenAverage Speed=551.5=36.67km h−1\text{Average Speed} = \dfrac{\text{Total distance travelled}}{\text{total time taken}} \[1em] \text{Average Speed} = \dfrac{\text{55}}{\text{1.5}} = 36.67 \text{km h}^{-1}Average Speed=total time takenTotal distance travelled​Average Speed=1.555​=36.67km h−1So, average speed of the car = 36.67 km h-1.

Question 7

Question 7On earth the weight of a body of mass 1.0 kg is 10 N. What will be the weight of a boy of mass 37 kg in(a) kgf(b) N

Accepted Answer

(a) Mass of boy = 37 kgSo, Weight of boy in kgf = 37 kgf.(b) 1 kgf = 10 N37 kgf = 37 x 10 = 370 NSo, weight of boy in N = 370 N.

Question 8

Question 8The weight of a body of mass 6.0 kg on moon is 10 N. If a boy of mass 30 kg goes from earth to the moon surface, what will be his(a) mass,(b) weight?

Accepted Answer

(a) Mass of boy = 30 kg on moon surface as mass remains constant. Mass of a body does not change with place.(b) Weight of body of mass 6.0 kg on moon surface = 10 NWeight of body of mass 30 kg on moon surface = 106.0\dfrac{	ext{10}}{	ext{6.0}}6.010​ x 30 = 50 NSo, weight of boy of 30 kg on moon surface = 50 N.

Question 9

Question 4fSelect the correct alternative:A boy goes from his house to school by bus at a speed of 20 km h-1 and returns back through the same route at a speed of 30 km h-1. The average speed of his journey is:24 km h-125 km h-130 km h-120 km h-1

Accepted Answer

24 km h-1Reason — Suppose the distance between school and house is D km.In first part the boy goes to school by bus from home at speed (V1) = 20 km h-1.time taken (t1) = distancespeed\dfrac{\text{distance}}{\text{speed}}speeddistance​ =D20\dfrac{\text{D}}{\text{20}}20D​In second part the boy returns from school to home through same route at speed(V2) = 30 km h-1.time taken (t2) = distancespeed\dfrac{\text{distance}}{\text{speed}}speeddistance​ =D30\dfrac{\text{D}}{\text{30}}30D​Total distance = D + D = 2D kmTotal time taken = D20\dfrac{\text{D}}{\text{20}}20D​ + D30\dfrac{\text{D}}{\text{30}}30D​ = 5D60\dfrac{\text{5D}}{\text{60}}605D​Average speed = total distance travelledtotal time taken\dfrac{\text{total distance travelled}}{\text{total time taken}}total time takentotal distance travelled​Speed=distancetime=2D5D60=2×12=24km h−1\text{Speed} = \dfrac{\text{distance}}{\text{time}} \[1em] = \dfrac{2D}{\dfrac{5D}{60}} \[1em] = 2 \times 12 \[1em] = 24 \text{km h}^{-1}Speed=timedistance​=605D​2D​=2×12=24km h−1

Question 10

Question 10Explain oscillatory motion by giving one example.

Accepted Answer

The to and fro motion of a body from its rest position is called the oscillatory motion.Example — Motion of a pendulum of a wall clock. The bob of pendulum moves to side A from its rest position O, again comes back to position O and then moves to side B and then comes back to rest position O. This process is continuously repeated and this to and fro motion is called oscillatory motion.

Question 11

Question 11What is vibratory motion? Give one example.

Accepted Answer

The to and fro motion of a part of the body keeping its rest part fixed is called vibratory motion.Example — Vibration of the membrane of tabla.

Question 12

Question 12Differentiate between periodic and non-periodic motions by giving an example of each.

Accepted Answer

Periodic motionNon-periodic motionA motion which gets repeated after a regular interval of time is called a periodic motion.The motion which does not repeat itself after a regular interval of time is called non-periodic motion.Example — Pendulum of a wall clock repeats its motion after every 2 s.Example — A footballer running on a field.

Question 13

Question 13What is random motion? Give an example.

Accepted Answer

The motion of an object which takes path in no specific direction is called random motion.Example — Motion of molecules of a gas.

Question 14

Question 14Name the type/types of motion being performed by each of the following:(a) Vehicle on a straight road.(b) Blades of an electric fan in motion.(c) Pendulum of a wall clock.(d) Smoke particles from chimney.(e) Hands of a clock.(f) Earth around the Sun.(g) A spinning top.

Accepted Answer

(a) Rectilinear motion(b) Rotatory motion(c) Oscillatory motion and Periodic motion(d) Non-periodic motion(e) Circular motion(f) Periodic motion, circular motion(g) Rotatory motion

Question 15

Question 15Give two examples to illustrate that a body can have two or more types of motion simultaneously.

Accepted Answer

The two examples that show a body can have two or more types of motion (mixed motion) simultaneously are:The Earth rotates about its axis (rotatory motion) and at the same time, it revolves around the sun in a curved path (curvilinear or circular motion) in a fixed time interval (periodic motion).A ball rolling on the ground has rotatory and translatory motion as it moves on the ground.

Question 16

Question 16State the types of motion of the following:(a) The needle of a sewing machine(b) The wheel of a bicycle(c) The drill machine(d) The carpenter's saw

Accepted Answer

(a) Periodic motion(b) Translatory and rotatory motion(c) Translatory and rotatory motion(d) Translatory and oscillatory motion

Question 17

Question 17Distinguish between uniform and non-uniform motions, giving an example of each.

Accepted Answer

Uniform motionNon-uniform motionIf a moving body travels equal distances in equal intervals of time, its motion is said to be uniform.If a moving body travels unequal distances in equal intervals of time, its motion is said to be non-uniform.Example — A car covering same distance in same interval of time at constant speed in a straight line.Example — Motion of a bicycle in a busy road.

Question 18

Question 18How do you determine the average speed of a body in non-uniform motion?

Accepted Answer

The average speed is calculated by finding the ratio of the total distance travelled by the body to the total time taken in the journey.Average speed = total distance travelledtotal time taken\dfrac{	ext{total distance travelled}}{	ext{total time taken}}total time takentotal distance travelled​

Question 19

Question 19Define the term weight and state its S.I. unit.

Accepted Answer

The weight of the body on earth is the force with which earth attracts the body. The S.I. unit of weight is newton (N).

Question 20

Question 20How are the units of weight, kgf and newton related?

Accepted Answer

1 kgf = 9.8 (nearly 10) N

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