Class 9

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 8 Motion Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 8 Motion

PSEB 9th Class Science Guide Motion Textbook Questions and Answers

Question 1.
An athelete completes one round of circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 min 20 s?
Solution:

Diameter of circular track (d) = 200m
Radius of circular track (r) = \(\frac{d}{2}\) = \(\frac{200m}{2}\) = 100m
Length of circular track (circumference) = 2πr = 2 × \(\frac{22}{7}\) × 100
= \(\frac{4400}{7}\)m
Time taken to complete 1 round (t) = 40 s
Total time = 2 minutes 20 seconds
= (2 × 60 + 20) seconds
= (120 + 20) seconds
= 140 s.
Distance covered in 40 s = \(\frac{4400}{7}\) m = (Circumference of 1 complete circular track)
Distance covered in 1 s = \(\frac{4400}{7×40}\) m
Distance covered in 140 s = \(\frac{4400}{7×40}\) × 140 = 2200 m
An athelete starting from A and going in clockwise direction returns to point A in 3 rounds and reaches point B in 3.5 rounds.
∴ Displacement in 3.5 rounds = AB = shortest distance between initial and final position = 200 m from A to B.

Question 2.
Joseph jogs from end A to the other end B of a straight 300 m road in 2 minutes 50 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 (b) from A to C?
Solution:
(a) Length between end point A and end point B (AB) = 300 m
Time taken (t) = 2 min. 30 s
= (2 × 60 + 30) s
= (120 + 30) s
= 150 s.

Average speed = Average velocity
= \(\frac{Total distance between A and B(AB)}{Total time(t)}\)
= \(\frac{300m}{150s}\)
= 2ms^-1

(b) Length from end A to end B + Length on return from B to point C.
= AB + BC
= 300 m + 100 m
= 400 m
Total Time = 2 min 30 s + 1 min
= 3 min 30 s
= (3 × 60 + 30) s
= (180 + 30) s
= 210 s

Question 3.
Abdul while driving to school, computes the average speed for his trip to be 20 km h^-1. On his trip along the same route, there is less traffic and average speed is 40 km h^-1. What is the average speed for Abdul’s trip?
Solution:

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s^-2 for 8.0 s. How far does the boat travel during this time?
Solution:
Here, initial velocity of motorboat (u) = 0 [Starting from rest]
Acceleration (a) = 3.0 m s^-2
Time (t) = 8.0 s
Distance covered by the motorboat (S) =?
We know, S = ut + \(\frac{1}{2}\)at²
= 0 × 8 + \(\frac{1}{2}\) × 3 × (8)²
= 0 + \(\frac{1}{2}\) × 3 × 8 × 8
∴ S = 96 m.
In other words, the motorboat covers a distance (S) = 96 m.

Question 5.
A driver of a car travelling at 52 kmh^-1 applies the brakes and accelerates uniformly in opposite direction. The car stops in 5 s. Another driver going at 3 km h^-1 applies his brakes slowly and stops in 10 s. On the same graph paper plot the speed versus time graph for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
In the figure AB and CD represent velocity-time graphs of two cars which have their speeds 52 kmh^-1 and 30 kmh^-1 respectively.


In this way, after applying brakes the second car would cover more distance than the first car.

Question 6.
Fig shows the distance-time graphs 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?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
Solution:
(a) Velocity of A = Slope of PN
\(\frac{10-6}{1.1-0}\)
\(\frac{40}{11}\) = 3.63 kmh^-1


Because slope of object B is maximum of all therefore, it is moving fastest.
(b) Since all the three graphs do not intersect at any point therefore, all the three do not meet ever at the same point on the road.
(c) When the object B passes A at point E (at 1.4 hr) then at that time the object C will be at F i.e. 9.3 km away from the origin O.
(d) B passes C at G after covering 8 km.

Question 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 it will strike the ground?After what time will it strike the ground?
Solution:
u = 0 ms^-1
S = 20 m
a = 10 ms^-2
υ = ?
t = ?

Using υ² – u² = 2as
υ² – (0)² = 2 × 10 × 20
υ² = 4000
∴ υ = \(\sqrt{400}\)
= \(\sqrt{20 \times 20}\)
= 20 m s^-1
Now υ = u + at
20 = 0 + 10 × t
or t = \(\frac{20}{10}\)
∴ t = 2 s

Question 8.
Speed-time graph for a car is shown in the fig.

(a) Find how far the car travelled in first 4 s. Shade the area on the graph that represents the distance travelled by car during this period.
(b) Which part of the graph represents uniform motion of the car?
Solution:

(a) 5 small squares of x axis = 2s
3 small squares of y axis = 2 ms^-1
Area of 15 small squares = 2s × 2 ms^-1 = 4m
∴ Area of 1 small square = \(\frac{4}{15}\)
Area of velocity-time graph under 0 to 5s = 57 complete small squares + \(\frac{1}{2}\) × 6 small squares.
= (57 + 3) small squares
= 60 small squares.
Distance covered by car in 4 s = 60 × \(\frac{4}{15}\) m
= 16 m

(b) After 6 s the car has uniform motion.

Question 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) Yes, this situation is possible.
Example: When an object is projected upwards, its velocity at the maximum height is zero although acceleration on it is 9.8 ms^-2 i.e. equal to g.

(b) Yes, at the maximum height of projection the velocity is in the horizontal direction and its acceleration is perpendicular to the direction of motion as shown in figure.

Question 10.
An artificial satellite is moving in a circular path orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
Radius of circular path of artificial satellite (r) = 42,250 km
Angle formed (subtended) at the centre of earth (θ) = 2π radian
Time taken by the satellite to complete 1 revolution (t) = 24hrs
= 24 × 3600s
= 86400 s

Science Guide for Class 9 PSEB Motion InText Questions and Answers

Question 1.
An object has moved through a distance. Can it have zero displacements? If yes,support your answer with an example.
Answer:
Yes, a body can have zero displacement, if fhis body While moving occupies its final position coinciding with its initial position.
Example: Suppose a body starting its motion from initial position O covers some distance and reaches a position A. If this body while moving returns to its initial position O then in that situation its displacement will be zero.

But distance covered by the body = OA + AO
= 60 km + 60 km
= 120 km

Question 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?
Solution:
Total distance round the boundary of field once (i.e. circumference)
= AB + BC + CD + DA
= 10 m + 10 m + 10 m + 10 m = 40 m
Time taken to go round the field once = 40 s

Total time taken = 2minutes 20 seconds
= (2 × 60 + 20) seconds
= (120 + 20) seconds
= 140 seconds.
Time taken by fanner to complete 3 rounds of field = 3 × 40 s = 120 s
Time left after completing 3 rounds of field = (140 – 120)s = 20 s
∴ Distance covered by farmer in 40 s = 40 m
∴ Distance covered in 1 s = 1 m
Distance that would be covered in 20 s = 20 m
In other words farmer starting from point A and while going along the boundary of the field and after completing 3 rounds in 2 min 20 s would reach the point C.
∴ Displacement = AC
(the shortest distance between initial and final position)

Question 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.
(c) Its magnitude is less than or equal to distance travelled by the object.
Answer:
(c) Its magnitude is less than or equal to distance travelled by the object.

Question 4.
Distinguish between speed and velocity.
Answer:
Distinction between Speed and Velocity:

Speed	Velocity
1. It is defined as the rate of a change of a position of a body i.e. the distance covered by a body per unit time.	It is defined as the rate of change of displacement of a body. i.e. it is the speed in a particular direction.
2. It is a scalar quantity and can be completely represented by its magnitude only.	It is a vector quantity. To represent it completely it requires both magnitude and direction.
3. Speed of an object is always positive.	Velocity of an object can be both positive and negative.

Question 5.
Under what condition(s) is the magnitude of average velocity of an object is equal to its average speed?
Answer:
We know, Average speed = Total distance travelled / Total time taken
and Average velocity = Displacement /Total time
When a body travels in a straight line with variable motion in the same direction then total distance covered and displacement are equal in magnitude. In this case the average speed and average velocity are equal.

Question 6.
What does the odometer of an automobile measure?
Answer:
The odometer of an automobile measures the distance covered by it.

Question 7.
What does the path of an object look like when it is in uniform motion?
Answer:
When an object is in uniform motion, it moves along a straight line. But an object can also move with uniform motion along a circular path.

Question 8.
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 a speed of light that is 3 × 10s ms^-1.
Solution:
Time taken by the signal to reach the ground station from spaceship (t) = 5 min = 5 × 60 s = 300 s
Speed of Signal (υ) = Speed of light = 3 × 10⁸ ms^-1
Distance of the spaceship from earth (s) = ?
Distance of spaceship from ground (s) = speed of signal (υ) × Time (t)
= 3 × 10⁸ × 300
= 3 × 10⁸ × 3 × 10²
= 9 × 10⁸ × 10²
= 9 × 10¹⁰ m

Question 9.
When will you say a body is in:
1. uniform acceleration?
2. non-uniform acceleration?
Answer:
1. Uniform Acceleration. When a body travels in a straight line and its velocity changes by equal amounts in equal intervals of time then it is said to travel with uniform acceleration.
2. Non-Uniform Acceleration. When the velocity of a body changes by unequal amounts in equal intervals of time then the body is said to travel with non-uniform acceleration.

Question 10.
A bus decreases its speed from 80 km h^-1 to 60 km h^-1 in 5 s. Find the acceleration of the bus.
Solution:


Hence, the bus has negative acceleration (retardation).

Question 11.
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.
Solution:

Question 12.
What is the nature of the distance-time graphs (x – t) for uniform and non-uniform motion of an object?
Answer:
When a body covers equal distances in equal intervals of time, then it is said to travel with uniform motion. In this situation, the distance covered by the body is directly proportional to the time taken. Therefore, distance-time (x – t) graph for uniform motion is a straight line.

Distance – time (x – t) graph for non-uniform motion may be a curved graph of any shape because a body travels unequal distances in equal intervals of time.

Question 13.
What can you say about the motion of object whose distance – time graph is a straight line parallel to time axis?
Answer:

The object whose distance-time (x – t) graph is a straight line parallel to the time axis will be at rest with respect to the surroundings.

Question 14.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to time axis?
Answer:

The object whose speed – time (u – t) graph is a straight line parallel to time axis shows that it is in motion with uniform speed.

Question 15.
What is the quantity which is measured by the area occupied below velocity-time graph?
Answer:

The area occupied below velocity-time graph measures displacement of the body.

Question 16.
A bus starting from rest moves with a uniform acceleration of 0.1 ms^-2 for two minutes. Find (a) the speed acquired (b) the distance travelled.
Solution:
(a) Initial speed of the bus (u) = 0 (Starting from Rest)
Acceleration of the bus (a) = 0.1 m s^-2
Time taken (t) = 2 minutes
= 2 × 60 s
= 120 s
Final speed of the bus (υ) = ?
Distance travelled by the bus (S) =?
We know, υ = u + at
υ = 0 + 0.1 × 120
υ = 1 × 12
υ = 12 ms^-1

(b) Again, using S = ut + \(\frac{1}{2}\) at²
S = 0 × 120 + \(\frac{1}{2}\) × 0.1 × (120)²
= 0 + \(\frac{1}{2}\) × 0.1 × 120 × 120
= \(\frac{1}{2}\) × 1 × 12 × 120
= 720 m/s

Question 17.
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 ms^-2. Find how far the train will move before it is brought to rest?
Solution:
Initial speed of train (υ) = 90km h^-1
= 90 × \(\frac{5}{18}\) m s^-1
= 5 × 5 ms^-1
= 25 ms^-1
Uniform acceleration (a) = – 0.5m s^-2
Final speed of the train (υ) = 0
Distance moved by the train (S) =?
We know, υ² – u² = 2as
(0)² – (25)² = 2 × (-0.5) × S
– 25 × 25 = -1 × S
∴ S = 625 m

Question 18.
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?
Solution:
Here initial velocity of trolley (u) = 0 [∵ starting from rest]
Acceleration (a) = 2cm s^-2
Time (t) = 3 s
Final velocity of trolley (υ ) = ?
We know, υ = u + at
υ = 0 + 2 × 3
∴ Final velocity of trolley (υ) = 6 cm s^-1 Ans.

Question 19.
A racing car has uniform acceleration of 4 ms^-2. What distance will it cover in 10 s after start?
Solution:
Acceleration of racing car (a) = 4 ms^-2
Initial velocity of racing car (u) = 0
Time (t) = 10 s
Distance covered by the car (S) = ?
We know, S = ut + \(\frac{1}{2}\)at²
S = 0 × 10 + \(\frac{1}{2}\) × 4 × (10)²
S = 0 + 2 × 10 × 10
∴ Distance covered by racing car (S) = 200 m

Question 20.
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 downw ard direction. What will be the height attained by the stone and how much time will it take to reach there?
Solution:
Here, initial velocity (u) = 5 m s^-1
Acceleration (a) = – 10 ms^-2
[∵ it moves upward against the gravity]
Final velocity of stone (υ) = 0 [At the highest point it is brought to rest]
Height attained (S = h) = ?
Time taken (t) =?
We know,
υ = u + at
0 = 5 + (-10) × t
0 = 5 – 10 × t
10 × t = 5
or t = \(\frac{5}{10}\)
∴ Time taken (t) = 0.5 s
Again, using υ² – u² = 2as
(0)² – (5)² = 2 × -10 × h
– 5 × 5 = – 20 × h
or h = \(\frac{-25}{-20}\) = \(\frac{5}{4}\)
∴ Height attained (h) – 1.25 m

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 7 Diversity in Living Organisms Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms

PSEB 9th Class Science Guide Diversity in Living Organisms Textbook Questions and Answers

Question 1.
What are the advantages of classifying organisms?
Answer:
Advantages of classification

1. Classification makes the study of wide variety of organisms easy.
2. Classification of organisms is responsible for description of species.
3. It helps in understanding the interrelation among different groups of organisms.
4. Classification of living beings recognises the basic taxonomic units of species.
5. It helps in understanding the evolution of organisms.
6. Exact identification of insects helps in controlling epidemic diseases like malaria, filaria, dengue fever, kala azar etc.

Question 2.
How would you choose between two characteristics to be used for developing a hierarchy in classification?
Answer:
Characteristic is a particular form or function. Many inter-related characteristics are used in order to classify all living organisms.

• Nature of cell is basic characteristic of classification.
• Number of cells.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:

1. Nature of cell. Prokaryotic or eukaryotic.
2. Number of cells. Unicellular or multicellular.
3. Mode of nutrition. Absorptive, autotrophic or holozoic.

Question 4.
What are the major divisions in kingdom Plantae? What are the basis for these divisions?
Answer:
Major divisions of Kingdom Plantae.

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?
Answer:
Criteria for dividing plants: The eukaryotic multicellular organisms with cell wall and those which carry out photosynthesis are placed under kingdom plantae.

The first level of classification among plants depends on whether the plant body has well-differentiated, distinct components. The next level of classification is based on whether the differentiated plant body has special tissues for the transport of water and other substances within it. Further classification looks at tire ability to bear seeds and whether the seeds are enclosed within fruits.

Criteria for dividing animals into sub-groups: The eukaryotic multicellular heterotrophic and those which lack cell wall are placed in animal kingdom. They are further classified on the basis of extent and type of body design and differentiation presence.

Question 6.
Explain how animals in vertebrata are classified into further sub-groups.
Answer:
Classification of vertebrata

Science Guide for Class 9 PSEB Diversity in Living Organisms InText Questions and Answers

Question 1.
Why do we classify organisms?
Answer:
Classification helps in the study of broad group of organisms with a wide variety in a simple way.

Question 2.
Give three examples of the range of variations that you see in life forms around you.
Answer:
Different forms in which life occurs on earth

1. Size. Microscopic bacteria a few micrometres in size at one end of the size-scale and ~ 30 metre long blue whale and ~ 100 metre tall red wood trees of California.
2. Life Span. Pine trees live for thousands of years while insects such as mosquitoes die within few days.
3. Colour. Colourless or transparent worms to brightly coloured birds and flowers.

Question 3.
Which do you think is more basic characteristic for classifying organism and why;
(a) the place where they live?
(b) the kind of cells they are made of?
Answer:
The kind of cells is more basic characteristic for classification of organisms. The cells may be prokaryotic or eukaryotic. The presence or absence of nucleus, or membrane bound organelles would reflect on every aspect of cell design and capacity to make a multicellular body.

Question 4.
What is the primary characteristic on which the first division of organism is made?
Answer:
Nature of Cell. Prokaryotic cell or eukaryotic cell.

Question 5.
On what bases plants and animals put into different categories?
Answer:

1. Mode of nutrition. Plants prepare their own food by photosynthesis due to the presence of chlorophyll in chloroplasts and animals acquire ready made food.
2. Plants are fixed whereas animals are motile.
3. Plants show limited growth whereas animals stop growing after attaining a certain size.
4. Plant cells are surrounded by cell wall and animal cells lack cell wall.

Question 6.
Which organisms are called primitive and how are they different from the so called advanced organisms?
Answer:

• Primitive organisms: The organisms which have ancient body designs that have not changed very much. They are also called ‘lower’ organisms.
• Advanced organisms: The group of organisms that have acquired their particular body design relatively recently. They are also called ‘higher’ organisms.

Question 7.
Will advanced organisms be the same as complex organisms? Why?
Answer:
Yes, complexity in design will increase over evolutionary time, hence older (primitive) organisms are simple while younger organisms are more complex.

Question 8.
What is the criterion for classification of organisms as belonging to kingdom monera or protista?
Answer:
The criterion for classification of monera and protista Nature and number of cell. Prokaryotes belong to kingdom monera and single celled eukaryotes belong to kingdom protista.

Question 9.
In which kingdom will you place an organism which is single celled eukaryotic and photosynthetic?
Answer:
Protista.

Question 10.
In the hierarchy of classification, which grouping will have the smallest number of organisms with a maximum of characteristics in common and which will have the largest number of organisms?
Answer:

1. Species will have smallest number of organisms.
2. Kingdom will have largest number.

Question 11.
Which division among plants has simplest organisms?
Answer:
Algae (Thallophytes).

Question 12.
How are pteridophytes different from phanerogams?
Answer:

1. Phanerogams produce seeds while pteridophytes do not.
2. Reproductive organs are hidden in pteridophytes but well developed in phanerogams.
3. Pteridophytes have specialised tissue for the conduction of water whereas proper vascular tissues are present in phanerogams.

Question 13.
How do gymnosperms and angiosperms differ from each other?
Answer:
Differences between gymnosperms and angiosperms:

Gymnosperms	Angiosperms
1. Seeds are naked.	1. Seeds are enclosed in fruits.
2. Reproductive organs sporophylls form cones.	2. Reproductive organs are flowers.
3. Ovules not enclosed in ovary.	3. Ovules are enclosed in ovary.
4. Xylern lack vessels.	4. Xylem contain vessels.
5. Companion cells are absent in phloem.	5. Companion cells are present.

Question 14.
How do poriferans animals differ from coelenterate animals?
Answer:
Differences between poriferans and coelenterates:

Poriferans	Coelenterates
1. Numerous minute pores ostia for entry of water and single osculum for exit of water present.	1. The body bears a single pore.
2. Appendages absent.	2. Tentacles as appendages present.
3. Intercellular digestion.	3. Digestion is intracellular as well as intercellular.
4 Spicules or spongin fibres present.	4. Stinging cells called cnidoblasts present.

Question 15.
How do annelid animals differ from arthropods?
Answer:
Differences between annelids and arthropods

Annelids	Arthropods
1. Jointed appendages absent. 2. Body covered with cuticle. 3. True coelom as body cavity present. 4. Closed type of circulatory system present.	1. Body bears jointed appendages. 2. Body covered with chitinous cuticle forming exoskeleton. 3. Haemocoel present. 4. Open type of circulatory system present

Question 16.
What are the differences between amphibia and reptilia?
Answer:
Differences between amphibia and reptilia:

Amphibia	Reptilia
1. Body wall not covered with any kind of exoskeleton.	1. Scales or dermal plates present.
2. Body divided into head and trunk.	2. Body divided into head, neck, trunk and tail.
3. Fertilization is external.	3. Fertilization is internal.

Question 17.
What are the differences between animals belonging to the aves group and those in the mammalia group?
Answer:
Differences between aves and mammals:

Aves	Mammals
1. Body covered with feathers.	1. Body covered with hair.
2. Wings present.	2. Wings absent.
3. Pinna absent.	3. Pinna present.
4. Diaphragm absent.	4. Diaphragm present.

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 6 Tissues Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 6 Tissues

PSEB 9th Class Science Guide Tissues Textbook Questions and Answers

Question 1.
Define the term ’tissue’.
Answer:
Tissue: A group of cells that are similar in structure and/or work together a particular function is called tissue.

Question 2.
How many types of elements together make xylem tissue? Name them.
Answer:
Four types of elements make xylem tissue. They are:

1. Tracheids
2. Vessels
3. Xylem parenchyma
4. Xylem fibres

Question 3.
How are simple tissues different from complex tissues in plants?
Answer:
Differences between simple tissue and complex tissue in plants:

Simple tissue	Complex tissue
1. Similar types of cells which have common origin and function. 2. All cells are similar in origin and structure. 3. Parenchyma, Collenchyma and Sclerenchyma are the three types.	1. A group of more than one type of cells having common origin and working! together as a unit. 2. The cells have different origin and structure. 3. Xylem and phloem are the two main types.!

Complex tissue

Question 4.
Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell walls.
Answer:

Feature	Parenchyma	Collenchyma	Sclerenchyma
Cell wall	Thin walls	Thickened cell wall at comers	Thickened walls due to lignin.

Question 5.
What are functions of stomata?
Answer:
Functions of stomata:

• They are necessary for exchanging gases with the atmosphere.
• Transpiration also takes place through stomata.

Question 6.
Diagrammatically show the differences between three types of muscle fibres.
Answer:
Three Types of Muscle Fibres:

Question 7.
What is the specific function of cardiac muscles?
Answer:
Cardiac muscles undergo rhythmic contraction and relaxation. They are responsible for heart beat and thus plays a role in circulation (pumping) of blood in the body.

Question 8.
Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.
Answer:
Differences between Striated, Non-Striated and Cardiac Muscle Fibres:

Striated Muscle Fibres	Non-striated Muscle Fibres	Cardiac Muscle Fibres
Structure: 1. The fibres or cells are long and cylindrical.2. The fibres are unbran­ched.3. Sarcolemma is present.4. The cells are multinucleate.5. They bear striations or alternate light and dark bands.6. The ends are blunt.7.  They are capable of quick contraction. Location8. Occur in body wall, limbs, tongue, pharynx and beginning of oesophagus.	1. The fibres or cells are narrow and spindle­shaped. They are comparatively short. 2. The fibres are unbran­ched. 3. Sarcolemma is absent. 4. They are uninucleate. 5. Striations are absent. 6. The ends are tapering. 7. Contraction is slow. 8. Occur in walls of hollow visceral organs, iris of eye, and dermis of skin.	1. The cells are short but cylindrical. 2. They develop lateral outgrowths at places to form cross-connections. 3. Sarcolemma is present. 4. The cells are uninu­cleate. 5. Striations are present but slightly fainter than found in striated fibres. 6. The ends are blunt. 7. The fibres show rhythmic contractions. 8. Occur in the wall of heart.

Question 9.
Draw a labelled diagram of a neuron.
Answer:
Structure of neuron

Question 10.
Name the following:
(a) Tissue that forms inner lining of our mouth.
(b) Tissue that connects muscle to bone in humans.
(c) Tissue that transports food in plants.
(d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix.
(f) Tissue present in the brain.
Answer:
(a) Squamous epithelium
(c) Phloem
(e) Blood (Vascular tissue)

Question 11.
Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:

1. Skin
2. Bark of tree
3. Bone
4. Lining of kidney tubule
5. Vascular bundle

Question 12.
Name the regions in which parenchyma tissue is present.
Answer:
Pith and cortex of stem and root.

Question 13.
What is the role of epidermis in plants?
Answer:
Role of epidermis:

1. Protection of all parts of plant.
2. Secretion of waxy-water resistant layer on their outer surface.
3. Epidermis aids in protection against loss of water, mechanical injury and invasion by parasitic fungi.

Question 14.
How does cork act as a protective tissue?
Answer:
Cell of cork are dead and compactly arranged without intercellular spaces. They have a chemical called suberin in their walls that makes them impervious to gases and water. Thus cork acts as a protective tissue.

Question 15.
Complete the table:
Answer:

1. Parenchyma
2. Sclerenchyma
3. Phloem

Science Guide for Class 9 PSEB Tissues InText Questions and Answers

Question 1.
What is a tissue?
Answer:
Tissue: A group of cells that are similar in structure and/or work together a particular function is called tissue.

Question 2.
What is utility of tissues in multicellular organisms?
Answer:

• They provide protection and mechanical strength.
• Tissues provide highest possible efficiency of function.

Question 3.
Name types of simple tissues.
Answer:
Types of simple tissues:

1. Parenchyma
2. Collenchyma
3. Sclerenchyma.

Question 4.
Where is apical meristem found?
Answer:
Apical meristem is present at the growing tips of stems and roots of plants.

Question 5.
Which tissue makes up the husk of coconut?
Answer:
Sclerenchyma tissue.

Question 6.
What are constituents of phloem?
Answer:
Four types of elements constitute phloem:

1. Sieve tubes
2. Companion cells
3. Phloem fibers
4. Phloem parenchyma

Question 7.
Name the tissue responsible for the movement of our body.
Answer:
Muscular tissue.

Question 8.
What does a neuron look like?
Answer:
A neuron consists of cell body cyton with hair-like parts called dendrites and a long axon. Thus gives the appearance of a miniature tree.

Question 9.
Give three features of cardiac muscles.
Answer:

1. Heart muscles are cylindrical, branched, and uninucleate.
2. They are involuntary and undergo rhythmic contraction and relaxation.
3. Intercalated discs are present at the junction of two cells.

Question 10.
What are the functions of areolar tissue?
Answer:
Areolar tissue fills space inside the organ, supports internal organs, and helps in repairing the tissues.