It is my home work page
And I do my homework regularly
Date : 05/04/2016
Chapter:1 Matter In Our Surrounding
Properties of particles
1. The particles of the matter very small. 2. Thee particles of the matter are continuously moving means they have kinetic energy .
3. The particles of the matter have space between them.Date:07/04/2016 Matter In Our Surrounding
Qns1. Why solids have fixed shape?
Ans1. Solids have fixed shape because in solids the atoms are bonded tightly and there is very very less space in between atoms. So they have a definite shape and size.
Date:12/04/2016Matter In Our Surrounding
Give Reasons
- Why the table is hard?
Answer:-The table is hard because in the table there are some particles and they are tightly packed .In this the inter molecular force of attraction is very high so they attract with each other. So this is the reason that's why the table is hard.
- Diver is able to cut the water?
Answer:-the particles of matter have space between them. This space is the maximum in the gases and the minimum in liquid. Thus, one cannot cut through a solid easily but a diver is able to cut easily through water in a swimming pool.
- How the smell of hot sizzling food is reached to our nose?
Answer:- when food is cooked, some of the substances in food release gases having the smell of food in them. The particles of these gases move very quickly and mix up with air by diffusion. When the air containing these gases reaches your nose, we get the smell of hot and sizzling food.
Homework- 18/April/2016
EXERCISE
Ans 1- (a) 298 K
(b) 646 K
Ans 2- (a) 20*c
(b) 197*c
(b) 197*c
Ans 3- (a) Naphthalene balls disappear with time as the temperature rises, it gets sublimated in gas without leaving any solid.
(b) The molecules of perfume gets mixed with air molecules which is the least densest matter , that is why we gets smell of perfume sitting several meters away.
(b) The molecules of perfume gets mixed with air molecules which is the least densest matter , that is why we gets smell of perfume sitting several meters away.
Ans 4- Oxygen -------> Water -------> Sugar
Ans 5- (a) Liquid State
(b) Solid State
(c) Gaseous State
(b) Solid State
(c) Gaseous State
Ans 6- (a) (i) At room temperature the force of attraction between the molecules of water is medium that is why it is in liquid form.
(ii) The room temperature (25*c) is higher than freezing point (0*c) which will make it solid and it is less than boiling point (100*c) which will make it gas.
(b) (i) At room temperature , the intermolecular force of solid is so high that its molecules gets packed together tightly.
(ii) It has fixed shape and fixed volume.
(ii) The room temperature (25*c) is higher than freezing point (0*c) which will make it solid and it is less than boiling point (100*c) which will make it gas.
(b) (i) At room temperature , the intermolecular force of solid is so high that its molecules gets packed together tightly.
(ii) It has fixed shape and fixed volume.
Ans 7- At 273 K or 0*c the ice will give more cooling than water because it can absorb more heat than water due to its latent heat of fusion.
Ans 8- Steam
Ans 9-
A- Melting
B- Evaporation
C- Condensation
D- Solidification
E- Sublimation
F- SublimationB- Evaporation
C- Condensation
D- Solidification
E- Sublimation
Tuesday, 26 April 2016.
CHAPTER 5.
FUNDAMENTAL UNIT OF LIFE
Q.What is cell ?
Ans.
- Cell is the structure and fundamental unit of life.
- Cell is discovered by Robert Hooke in 1665.
- Robert Hooke observed the thin slice of the cork tissue under the microscope and coined the term cell.
Q.Write down the definition of :
- Diffusion - It is the process of movement of particles from their higher concentration to their lower concentration. For eg. Exchange of gases by Stomata.
- Osmosis - It is the process of movement of molecules from their lower concentration to higher concentration in presence of semipermeable membrane. For eg. The movement of water in and out of the cell takes place by the process of osmosis.
- Cell membrane -
- Provides support and protection .
- Function as a cell.
- Think of our school doors and inter come system.
4. Cytoplasm -
- Jelly like material that fills the cell.
- Contain water and food for cell.
- Hold organelles in place.
- Think of our that fill up this entire school.
5. Nucleus -
- Nucleus contain DNA, which makes you who you are.
- Direct the activity of the cell. For eg. When it grows and divides.
- Think about the Principal's office as the nucleus and the student files in the office as DNA.
6. Endoplasmic Reticulum -
- The "highway" of the cell that moves material around the other part.
- Some part of the E.R contain ribosomes, which create proteins.
- Think of the school hallways where students and teachers.
- Packeges ,stores and secretes engergy of the cell.
- Golgi Complex plays a very important role in the formation of lysosomes and peroxisomes.
- It also involved in the formation of cell plate after the cell division in plants.
8. Centrosome -
- It is a clear area of cytoplasm near the nucleas.
- It help in cell division.
- It contains two genual like centriols.
9. Mitochondria -
- It is also known as the power house of the cell.
- The mitochondria is double membrane bound cell orgeneles.
- The respiration takes place in mitochondria and energy is released in the form APT.APT is the adenosine triphosphat.
10. Chloroplasts -
- The plastids which have chlorophyll in them are known as kitchen of the cell.
- They are presence only in plant cell.
- The photosynthesis takes place in chloroplast.
11. Lysosomes - Lysosomes are known as Suicidal Bags of the cell.These organelles have hydrolytic Enzymes and help in the digestion of the food.During the disturbance
in the metabolism these enzymes eat their own wall and destroy themselves.
Q. Why does the water come out from the cucumber when we add salt in it?
Ans, The water come out from the cucumber when we add salt because of hyper tonic and hypo tonic conditions. The outside environment is hyper tonic to the inside environment of the cucumber. Since there is less water outside or none and a high concentration of salt and there is a high concentration of water inside the cucumber and low salt or none at all. Through diffusion water will go from high to low concentration hence into to out. When water moves out its called osmosis.
DATE-07/04/2016
EXERCISE
Qns.Make a comparison and write down ways in which plant cells are different from animal's cell ?
Ans.
ANIMAL'S CELL. PLANT'S CELL
a) Animal's cell are generally small in. × a) Plant's cell are usually large in
size (5-10 mm). in size (10-100 mm).
b) Cell wall is absent. × b) Cell is present.
c) Plastics are absent. × c) Plastids are present.
d) Vacuoles are smaller in size. × d) Vacuoles are larger in size.
Qns.How is prokaryotic cell different from a eukaryotic cell.
Ans.
PROKARYOTIC CELL. EUKARYOTIC CELL
a) Most prokayotic cell are × a) Most eukaryotic cell are
unicellular. multicellular.
b) Size of the cell is generally × b) Size of the cell generally
small (0.5 - 5 mm). large (50 - 100 mm).
c) It contains a single chromoso- × c) It contains more than one
me. chromosomes.
d) Nucleas is absent. × d) Nucleas is present.
e) Membrane Bound Cell organe- × e) Membrane Bound Cell org-
lles is absent. nelles is present.
Qns.What would happen if the plasma membrane ruptures or breaks down ?
Ans.If the plasma membrane of the cell is ruptured,then the will die.The plasma membrane regulates the movement of substance in and out of the cell by diffusion or osmosis.Thus,if the plasma membrane is ruptured,then the cell might leak out its contents.
Qns.What would happen to the life of the cell if there was no Golgi Apparatus?
Ans.If the Golgi Apparatus is absent in the cell,then the process of storage,modification, and packaging of products will not be possible and the sinthesis of lysosomes or peroxisomes will not be possible in the cell.
Qns.Which organelle is known as the power house of the cell?Why?
Ans.Mitochondria is known as the power house of the cell.Mitochondria create energy for the cell known as Cellular Respiration.
Qns.Where do the Lipids and Proteins constituting the cell membrane he sinthesised?
Ans.Lipids and proteins constituting the cell membrane are synthesised in the endoplasmic reticulum.
Qns.How does an amoeba obtain its food?
Ans.Amoeba obtain its food through the process is endocytosis.
Qns.What is Osmosis?
Ans.The process of movement of molecules from their higher concentration to lower concentration in presence of Semipermeable Membrane.
Chapter- 8. Motion
1. Scalar Quantities:
Definition - A scalar is any quantity that only requires a magnitude or size to describe it completely. A scalar is any number that gives you the size or magnitude of a quantity, so a unit of measure must be attached to the number, like degrees or meters. Any random number is not a scalar.
Example - Temperature: The temperature at a given point is a single number. Velocity, on the other hand, is a vector quantity: Velocity in three- dimensional space is specified by three values; in a Cartesian coordinate system the values are the speeds relative to each coordinate axis. The associated fields describe the temperature and velocity in each point of some space. Considering the norms of the velocity vectors results in a scalar field of the speeds in each point of the space.
2.Displacement:
Definition - The shortest distance covered by an object from initial point to final point.
Distance and Displacement are same in magnitude.
Distance is the circumference of the semicircle but displacement is the diameter of the circle.
3. Velocity:
Velocity is denoted by ‘V’.
Velocity is a vector quantity. It is defined as displacement per unit time. v = s/t
SI unit is m/s.
4. Acceleration:
It is defined as the rate of change of velocity.
Acceleration is denoted by ‘s’.
a = v-u/t where ‘v’ is the final velocity and ‘u’ is the initial velocity.
SI unit of ‘a’ = m/s^-2
5. Retardation:
Retardation is the change in velocity per unit time when the initial velocity is more than the final velocity . (-a)
SI unit of retardation is m/s^2.
6. Average Speed:
It is the total distance covered by an object divided by total time taken.
7. Uniform Motion:
It is defined as the equal distance covered by an object in equal interval of time.
8. Non-Uniform:
When the object travels unequal distance in equal interval of time.
9.Equations of motion:
I. v = u+at
II. s = ut+1/2at^2
III. v^2-u^2 = 2as
Example:8.5
A train starting from roots attains a velocity of 72Km/h^-1 in 5 min.Assuming that the acceleration is uniform find 1)The acceleration and 2)The distance traveled by the train for attaining this velocity.
Solution:
We have given
u=0,v=72Km^h-1=20ms^-1 and t=5min=300sec.
1)a=v-u
t
2)2as=v^2-u^2=v^2-u s=v^2
2as
20ms^-1-0ms^-1 =[20ms^-1]^2
300sec 2x[1/15]ms^-2
1 ms^-2 3000m
15 3Km
The acceleration of train is 1 ms^-2and the distance travelled is 3 km
15
Example:8.6
A car accelerates uniformly from 18Km/h^-1 to 3Km/h^-1 in 5sec.Calculating 1)The accelaration and 2) The distance covered by the car in that time.
Solution:
u=18Km/h^-1=5ms^-1
v=36Km/h^-1=10ms^-1 and t=5s
1)a=v-u 2)s=ut+1at^2
t 2
10ms^-1 -5ms^-1 =5ms^-1x58+1x1ms^-2x(58)
5s 2
1ms^-2 25m+12.5m
=37.5m
The acceleration of the car is 1ms^-2 and the distance covered is 37.5m.
Exercice
2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer
The magnitude of average velocity of an object is equal to its average speed, only when an object is moving in a straight line.
3. What does the odometer of an automobile measure?
Answer
The odometer of an automobile measures the distance covered by an automobile.
4. What does the path of an object look like when it is in uniform motion?
Answer
An object having uniform motion has a straight line path.
5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s−1.
Answer
Speed= 3 × 108 m s−1
Time= 5 min = 5 x 60 = 300 secs.
Distance= Speed x Time
Distance= 3 × 108 m s−1 x 300 secs. = 9 x 1010 m
Page No: 103
1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer
(i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant.
2. A bus decreases its speed from 80 km h−1 to 60 km h−1 in 5 s. Find the acceleration of the bus.
Answer
3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1 in 10 minutes. Find its acceleration.
Answer
Page No: 107
1. What is the nature of the distance - 'time graphs for uniform and non-uniform motion of an object?
Answer
When the motion is uniform,the distance time graph is a straight line with a slope.
2. What can you say about the motion of an object whose distance - time graph is a straight line parallel to the time axis?
3. What can you say about the motion of an object if its speed - 'time graph is a straight line parallel to the time axis?
If speed time graph is a straight line parallel to the time axis, the object is moving uniformly.
4. What is the quantity which is measured by the area occupied below the velocity -time graph?
The area below velocity-time graph gives the distance covered by the object.
Page No: 109
1. A bus starting from rest moves with a uniform acceleration of 0.1 m s−2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer
Initial speed of the bus, u= 0
Acceleration, a = 0.1 m/s2
Time taken, t = 2 minutes = 120 s
(a) v= u + at
v= 0 + 0.1 × 120
v= 12 ms–1
(b) According to the third equation of motion:
v2 - u2= 2as
Where, s is the distance covered by the bus
(12)2 - (0)2= 2(0.1) s
s = 720 m
Speed acquired by the bus is 12 m/s.
Distance travelled by the bus is 720 m.
Page No: 110
2. A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.
Answer
Initial speed of the train, u= 90 km/h = 25 m/s
Final speed of the train, v = 0 (finally the train comes to rest)
Acceleration = - 0.5 m s-2
According to third equation of motion:
v2= u2+ 2 as
(0)2= (25)2+ 2 ( - 0.5) s
Where, s is the distance covered by the train
The train will cover a distance of 625 m before it comes to rest.
3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s−2. What will be its velocity 3 s after the start?
Answer
Initial Velocity of trolley, u= 0 cms-1
Acceleration, a= 2 cm s-2
Time, t= 3 s
We know that final velocity, v= u + at = 0 + 2 x 3 cms-1
Therefore, The velocity of train after 3 seconds = 6 cms-1
4. A racing car has a uniform acceleration of 4 m s - '2. What distance will it cover in 10 s after start?
Answer
Initial Velocity of the car, u=0 ms-1
Acceleration, a= 4 m s-2
Time, t= 10 s
We know Distance, s= ut + (1/2)at2
Therefore, Distance covered by car in 10 second= 0 × 10 + (1/2) × 4 × 102
= 0 + (1/2) × 4× 10 × 10 m
= (1/2)× 400 m
= 200 m
5. A stone is thrown in a vertically upward direction with a velocity of 5 m s−1. If the acceleration of the stone during its motion is 10 m s−2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer
Given Initial velocity of stone, u=5 m s-1
Downward of negative Acceleration, a= 10 m s-2
We know that 2 as= v2- u2
Page No: 112
Excercise
1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer
Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 x ( 22 / 7 ) x 100
Speed of the athlete (v) = Distance / Time
= (2 x 2200) / (7 x 40)
= 4400 / 7 × 40
Therefore, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 x 40) x (2 x 60 + 20)
= 4400 / ( 7 x 40) x 140
= 4400 x 140 /7 x 40
= 2200 m
Number of round in 40 s =1 round
Number of round in 140 s =140/40
=3 1/2
After taking start from position X,the athlete will be at postion Y after 3 1/2 rounds as shown in figure
= 200 m
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer
Total Distance covered from AB = 300 m
Total time taken = 2 x 60 + 30 s
=150 s
Therefore, Average Speed from AB = Total Distance / Total Time
=300 / 150 m s-1
=2 m s-1
Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s-1
=2 m s-1
Total Distance covered from AC =AB + BC
=300 + 200 m
Total time taken from A to C = Time taken for AB + Time taken for BC
= (2 x 60+30)+60 s
= 210 s
Therefore, Average Speed from AC = Total Distance /Total Time
= 400 /210 m s-1
= 1.904 m s-1
Displacement (S) from A to C = AB - BC
= 300-100 m
= 200 m
Therefore, Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m s-1
= 0.952 m s-1
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h−1. On his return trip along the same route, there is less traffic and the average speed is 40 km h−1. What is the average speed for Abdul’s trip?
Answer
The distance Abdul commutes while driving from Home to School = S
Let us assume time taken by Abdul to commutes this distance = t1
Distance Abdul commutes while driving from School to Home = S
Let us assume time taken by Abdul to commutes this distance = t2
Average speed from home to school v1av = 20 km h-1
Average speed from school to home v2av = 30 km h-1
Also we know Time taken form Home to School t1 =S / v1av
Similarly Time taken form School to Home t2 =S/v2av
Total distance from home to school and backward = 2 S
Total time taken from home to school and backward (T) = S/20+ S/30
Therefore, Average speed (Vav) for covering total distance (2S) = Total Distance/Total Time
= 2S / (S/20 +S/30)
= 2S / [(30S+20S)/600]
= 1200S / 50S
= 24 kmh-1
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s−2for 8.0 s. How far does the boat travel during this time?
Answer
Given Initial velocity of motorboat, u = 0
Acceleration of motorboat, a = 3.0 m s-2
Time under consideration, t = 8.0 s
We know that Distance, s = ut + (1/2)at2
Therefore, The distance travel by motorboat = 0 x 8 + (1/2)3.0 x 8 2
= (1/2) x 3 x 8 x 8 m
= 96 m
5. A driver of a car travelling at 52 km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Answer
As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh-1 and 3 kmh-1 respectively.
Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) x OR x OP
= (1/2) x 5 s x 52 kmh-1
= (1/2) x 5 x (52 x 1000) / 3600) m
= (1/2) x 5x (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) x OQ x OS
= (1/2) x 10 s x 3 kmh-1
= (1/2) x 10 x (3 x 1000) / 3600) m
= (1/2) x 10 x (5/6) m
= 5 x (5/6) m
= 25/6 m
= 4.16 m
6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
Answer
(a) Object B
(b) No
(c) 5.714 km
(d) 5.143 km
Therefore, Speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.
(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.
On the distance axis:
7 small boxes = 4 km
Therefore,1 small box = 4 / 7 Km
Initially, object C is 4 blocks away from the origin.
Therefore, Initial distance of object C from origin = 16 / 7 Km
Distance of object C from origin when B passes A = 8 km
Distance covered by C
Page No: 113
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s−2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer
Let us assume, the final velocity with which ball will strike the ground be 'v' and time it takes to strike the ground be 't'
Initial Velocity of ball, u =0
Distance or height of fall, s =20 m
Downward acceleration, a =10 m s-2
As we know, 2as =v2-u2
v2 = 2as+ u2
= 2 x 10 x 20 + 0
= 400
∴ Final velocity of ball, v = 20 ms-1
t = (v-u)/a
∴Time taken by the ball to strike = (20-0)/10
= 20/10
= 2 seconds
8. The speed-time graph for a car is shown is Fig. 8.12.
(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
Answer
(a)
The shaded area which is equal to 1 / 2 x 4 x 6 = 12 m represents the distance travelled by the car in the first 4 s.
(b)
The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.
9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer
(a) Possible
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.
(b) Possible
When a car is moving in a circular track, its acceleration is perpendicular to its direction.
10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer
Radius of the circular orbit, r= 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v= (2π r)/t
=[2× (22/7)×42250 × 1000] / (24 × 60 × 60)
=(2×22×42250×1000) / (7 ×24 × 60 × 60) m s-1
=3073.74 m s -1
Exercice
In Text Question
Page No: 100
1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer
Yes,an object can have zero displacement even when it has moved through a distance.This happens when final position of the object coincides with its initial position. For example,if a person moves around park and stands on place from where he started then here displacement will be zero.
2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer
Given, Side of the square field= 10m
Therefore, perimeter = 10 m x 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m
= 140 m / 40 m = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.
(b) Its magnitude is greater than the distance travelled by the object.
Answer
None of the statement is true for displacement First statement is false because displacement can be zero. Second statement is also false because displacement is less than or equal to the distance travelled by the object.
Page No: 102
1. Distinguish between speed and velocity.
Answer
Page No: 100
1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer
Yes,an object can have zero displacement even when it has moved through a distance.This happens when final position of the object coincides with its initial position. For example,if a person moves around park and stands on place from where he started then here displacement will be zero.
2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer
Given, Side of the square field= 10m
Therefore, perimeter = 10 m x 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m
Therefore, in 1s distance covered by farmer = 40 / 40 m = 1m
Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 mNow, number of rotation to cover 140 along the boundary= Total Distance / Perimeter= 140 m / 40 m = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.
Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from intial position.
3. Which of the following is true for displacement?
(a) It cannot be zero.(b) Its magnitude is greater than the distance travelled by the object.
Answer
None of the statement is true for displacement First statement is false because displacement can be zero. Second statement is also false because displacement is less than or equal to the distance travelled by the object.
Page No: 102
1. Distinguish between speed and velocity.
Answer
Speed
|
Velocity
|
| Speed is the distance travelled by an object in a given interval of time. | Velocity is the displacement of an object in a given interval of time. |
| Speed = distance / time | Velocity = displacement / time |
| Speed is scalar quantity i.e. it has only magnitude. | Velocity is vector quantity i.e. it has both magnitude as well as direction. |
2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer
The magnitude of average velocity of an object is equal to its average speed, only when an object is moving in a straight line.
3. What does the odometer of an automobile measure?
Answer
The odometer of an automobile measures the distance covered by an automobile.
4. What does the path of an object look like when it is in uniform motion?
Answer
An object having uniform motion has a straight line path.
5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s−1.
Answer
Speed= 3 × 108 m s−1
Time= 5 min = 5 x 60 = 300 secs.
Distance= 3 × 108 m s−1 x 300 secs. = 9 x 1010 m
Page No: 103
1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer
(i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant.
Answer
3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1 in 10 minutes. Find its acceleration.
Answer
Page No: 107
1. What is the nature of the distance - 'time graphs for uniform and non-uniform motion of an object?
Answer
When the motion is uniform,the distance time graph is a straight line with a slope.
When the motion is non uniform, the distance time graph is not a straight line.It can be any curve.
Answer
If distance time graph is a straight line parallel to the time axis, the body is at rest.
Answer
If speed time graph is a straight line parallel to the time axis, the object is moving uniformly.
4. What is the quantity which is measured by the area occupied below the velocity -time graph?
Answer
Page No: 109
1. A bus starting from rest moves with a uniform acceleration of 0.1 m s−2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer
Initial speed of the bus, u= 0
Time taken, t = 2 minutes = 120 s
(a) v= u + at
v= 0 + 0.1 × 120
v= 12 ms–1
(b) According to the third equation of motion:
v2 - u2= 2as
Where, s is the distance covered by the bus
(12)2 - (0)2= 2(0.1) s
s = 720 m
Speed acquired by the bus is 12 m/s.
Distance travelled by the bus is 720 m.
Page No: 110
2. A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.
Answer
Initial speed of the train, u= 90 km/h = 25 m/s
Final speed of the train, v = 0 (finally the train comes to rest)
Acceleration = - 0.5 m s-2
According to third equation of motion:
v2= u2+ 2 as
(0)2= (25)2+ 2 ( - 0.5) s
Where, s is the distance covered by the train
The train will cover a distance of 625 m before it comes to rest.
3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s−2. What will be its velocity 3 s after the start?
Answer
Initial Velocity of trolley, u= 0 cms-1
Acceleration, a= 2 cm s-2
Time, t= 3 s
We know that final velocity, v= u + at = 0 + 2 x 3 cms-1
Therefore, The velocity of train after 3 seconds = 6 cms-1
4. A racing car has a uniform acceleration of 4 m s - '2. What distance will it cover in 10 s after start?
Answer
Initial Velocity of the car, u=0 ms-1
Acceleration, a= 4 m s-2
Time, t= 10 s
We know Distance, s= ut + (1/2)at2
Therefore, Distance covered by car in 10 second= 0 × 10 + (1/2) × 4 × 102
= 0 + (1/2) × 4× 10 × 10 m
= (1/2)× 400 m
= 200 m
5. A stone is thrown in a vertically upward direction with a velocity of 5 m s−1. If the acceleration of the stone during its motion is 10 m s−2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer
Given Initial velocity of stone, u=5 m s-1
Downward of negative Acceleration, a= 10 m s-2
We know that 2 as= v2- u2
Page No: 112
Excercise
1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer
Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 x ( 22 / 7 ) x 100
Speed of the athlete (v) = Distance / Time
= (2 x 2200) / (7 x 40)
= 4400 / 7 × 40
Therefore, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 x 40) x (2 x 60 + 20)
= 4400 / ( 7 x 40) x 140
= 4400 x 140 /7 x 40
= 2200 m
Number of round in 40 s =1 round
Number of round in 140 s =140/40
=3 1/2
After taking start from position X,the athlete will be at postion Y after 3 1/2 rounds as shown in figure
Hence, Displacement of the athlete with respect to initial position at x= xy
= Diameter of circular track= 200 m
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer
Total Distance covered from AB = 300 m
Total time taken = 2 x 60 + 30 s
=150 s
=300 / 150 m s-1
=2 m s-1
Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s-1
=2 m s-1
Total Distance covered from AC =AB + BC
=300 + 200 m
Total time taken from A to C = Time taken for AB + Time taken for BC
= (2 x 60+30)+60 s
= 210 s
Therefore, Average Speed from AC = Total Distance /Total Time
= 400 /210 m s-1
= 1.904 m s-1
Displacement (S) from A to C = AB - BC
= 300-100 m
= 200 m
Time (t) taken for displacement from AC = 210 s
Therefore, Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m s-1
= 0.952 m s-1
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h−1. On his return trip along the same route, there is less traffic and the average speed is 40 km h−1. What is the average speed for Abdul’s trip?
Answer
The distance Abdul commutes while driving from Home to School = S
Let us assume time taken by Abdul to commutes this distance = t1
Distance Abdul commutes while driving from School to Home = S
Let us assume time taken by Abdul to commutes this distance = t2
Average speed from home to school v1av = 20 km h-1
Average speed from school to home v2av = 30 km h-1
Also we know Time taken form Home to School t1 =S / v1av
Similarly Time taken form School to Home t2 =S/v2av
Total distance from home to school and backward = 2 S
Total time taken from home to school and backward (T) = S/20+ S/30
Therefore, Average speed (Vav) for covering total distance (2S) = Total Distance/Total Time
= 2S / (S/20 +S/30)
= 2S / [(30S+20S)/600]
= 1200S / 50S
= 24 kmh-1
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s−2for 8.0 s. How far does the boat travel during this time?
Answer
Given Initial velocity of motorboat, u = 0
Acceleration of motorboat, a = 3.0 m s-2
Time under consideration, t = 8.0 s
We know that Distance, s = ut + (1/2)at2
Therefore, The distance travel by motorboat = 0 x 8 + (1/2)3.0 x 8 2
= (1/2) x 3 x 8 x 8 m
= 96 m
5. A driver of a car travelling at 52 km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Answer
As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh-1 and 3 kmh-1 respectively.
= (1/2) x OR x OP
= (1/2) x 5 s x 52 kmh-1
= (1/2) x 5 x (52 x 1000) / 3600) m
= (1/2) x 5x (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) x OQ x OS
= (1/2) x 10 s x 3 kmh-1
= (1/2) x 10 x (3 x 1000) / 3600) m
= (1/2) x 10 x (5/6) m
= 5 x (5/6) m
= 25/6 m
= 4.16 m
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d)How far has B travelled by the time it passes C?
(d)How far has B travelled by the time it passes C?
Answer
(a) Object B
(b) No
(c) 5.714 km
(d) 5.143 km
Therefore, Speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.
(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.
On the distance axis:
7 small boxes = 4 km
Therefore,1 small box = 4 / 7 Km
Initially, object C is 4 blocks away from the origin.
Therefore, Initial distance of object C from origin = 16 / 7 Km
Distance of object C from origin when B passes A = 8 km
Distance covered by C
Page No: 113
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s−2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer
Let us assume, the final velocity with which ball will strike the ground be 'v' and time it takes to strike the ground be 't'
Initial Velocity of ball, u =0
Distance or height of fall, s =20 m
Downward acceleration, a =10 m s-2
As we know, 2as =v2-u2
v2 = 2as+ u2
= 2 x 10 x 20 + 0
= 400
∴ Final velocity of ball, v = 20 ms-1
t = (v-u)/a
∴Time taken by the ball to strike = (20-0)/10
= 20/10
= 2 seconds
8. The speed-time graph for a car is shown is Fig. 8.12.
(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer
(a)
The shaded area which is equal to 1 / 2 x 4 x 6 = 12 m represents the distance travelled by the car in the first 4 s.
(b)
The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.
9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer
(a) Possible
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.
(b) Possible
When a car is moving in a circular track, its acceleration is perpendicular to its direction.
10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer
Radius of the circular orbit, r= 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v= (2π r)/t
=[2× (22/7)×42250 × 1000] / (24 × 60 × 60)
=(2×22×42250×1000) / (7 ×24 × 60 × 60) m s-1
=3073.74 m s -1
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