CBSE Class 9

In this page, we are providing Motion Class 9 Extra Questions and Answers Science Chapter 8 pdf download. NCERT Extra Questions for Class 9 Science Chapter 8 Motion with Answers will help to score more marks in your CBSE Board Exams.

Class 9 Science Chapter 8 Extra Questions and Answers Motion

Extra Questions for Class 9 Science Chapter 8 Motion with Answers Solutions

Motion Class 9 Extra Questions Very Short Answer Type

Question 1.
Define the following terms:
(a) Distance
(b) Displacement
(c) Speed
(d) Velocity
(e) Acceleration
(f) Uniform motion
(g) Uniform circular motion
(h) Scalar quantity
(i) Vector quantity
Answer:
(a) Distance: The total path length traveled by a body in a given interval of time is called distance.
(b) Displacement: The shortest distance measured from the initial to the final position of an object is known as displacement.
(c) Speed: The speed of a body is defined as the distance traveled per unit time.
(d) Velocity: Velocity is defined as displacement per unit time.
(e) Acceleration: The rate of change of velocity with respect to time.
(f) Uniform motion: A body moving in a straight line has a uniform motion if it travels the equal distance in equal intervals of tune.
(g) Non-uniform motion: A body has a non-uniform motion if it travels unequal distances in equal intervals of time.
(h) Scalar quantity: A physical quantity which is described completely by its magnitude only, is called a scalar quantity.
(i) Vector quantity: A physical quantity that has magnitude as well as direction and obeys the vector addition is called a vector quantity.

Question 2.
Draw position-time graph of a body that started from a position other than origin and moving with uniform speed.
Answer:

Question 3.
Give one example where the displacement is zero but the distance traveled is not zero.
Answer:
When a body completes one rotation in a circular path its initial and final positions are the same and hence its displacement is zero.

Question 4.
The area under the velocity-time curve represents which physical quantity?
Answer:
Distance and displacement.

Question 5.
Under what conditions, distance, and magnitude of the displacement are equal?
Answer:
Distance and displacement are equal fan object move along a straight line in one direction.

Question 6.
What is the numerical ratio of average velocity to the average speed of an object when it is moving in a straight path?
Answer:
In this case, both are equal, so the ratio is 1.

Question 7.
What does the speedometer of an automobile measure?
Answer:
The speedometer of a vehicle measures its instantaneous speed.

Question 8.
Classify the following quantities on the basis of scalar quantities and vector quantities.
(a) Distance
(b) Displacement
(c) Speed
(d) Velocity
(e) Acceleration
Answer:
(a) Distance: Scalar quantity
(b) Displacement: vector quantity
(c) Speed: Scalar quantity
(d) Velocity: Vector quantity
(e) Acceleration: Vector quantity

Question 9.
An object starts with initial velocity y and attains a final velocity y. The velocity is changing at a uniform rate. What is the formula for calculating average speed in this situation?
Answer:

Question 10.
What will you say about the motion of an object if its distance-time graph is a straight line with having a constant angle with a time axis?
Answer:
The motion of the body is uniform motion (constant speed).

Question 11.
What will you say about the motion of a body if its velocity-time graph is a straight line parallel to the time axis?
Answer:
The velocity of the body is constant.

Question 12.
A physical quantity measured is -20 ms^-1 It is speed or velocity?
Answer:
It is velocity because velocity can be positive or negative while speed cannot be negative.

Question 13.
A motorcycle drives from A to B with a uniform speed of 30 kmh¹ and returns hack with a speed of 20 kmh¹. Find its average speed.
Answer:
Let distance AB = x km.
Time taken from going A to B; t₁ = \(\frac {x}{30}\) h
Time taken from goingBto A; t₂ = \(\frac {x}{20}\) h
Total time taken,t = t₁ + t₂ = \(\frac {x}{30}\) + \(\frac {x}{20}\) = \(\frac {x}{50}\) h
total distance = x + x = 2x km

∴ Average speed = 24 km/h

Question 14.
A man swims in a pool of width 300 m. He covers 600 m in 10 minutes by swimming from on end to the other, and back along the same path. Find the average speed and average velocity of the man for the entire journey.
Answer:
Total distance covered by the man in 10 minutes = OA + AB = 600 m
Displacement of man in 10 minutes = 0 m

Average speed = 1 m/s

Average velocity = 0 m/s

Question 15.
What is the acceleration of a body moving with uniform velocity along a straight line?
Answer:
Acceleration of the body is zero as the velocity is constant.

Question 16.
Give an example of a uniform circular motion.
Answer:
An athlete running on a circular track with constant speed.

Question 17.
Convert speed 54 km in meter per second.
Answer:
Speed = 54 km/h =54 x \(\frac{1000 m}{3600}\) = 15 m/s

Question 18.
What ¡s the nature of the distance-time graph for accelerated motion?
Answer:

Motion Class 9 Extra Questions Short Answer Type 1

Question 1.
Distinguish between speed and velocity.
Answer:
Speed:

• Speed is the ratio of distance and time.
• Speed is always positive
• Speed is a scalar quantity.

Velocity:

• Velocity is the ratio of displacement and time.
• Velocity may be negative or positive.
• Velocity is a vector quantity.

Question 2.
Distinguish between distance and displacement.
Answer:
Distance

• It is the actual length of the path covered by a moving body.
• It is always positive or zero.
• It is a scalar quantity.

Displacement:

• It is the shortest distance measured between the initial and final positions.
• It may be positive, negative, or zero.
• it is a vector quantity.

Question 3.
Write down the SI unit of the following quantities:
(a) Displacement
(b) Speed
(c) Velocity
(d) Acceleration
Answer:
(a) m
(b) m/s
(c) m/s
(d) m/s²

Question 4.
Distinguish between uniform motion and non-uniform motion.
Answer:
1. Uniform motion:
A body moving in a straight line has a uniform motion if it travels the equal distance in equal intervals of time

2. Non.uniform motion:
A body has a non-uniform motion if it travels the unequal distance in equal intervals of time

Question 5.
Distinguish speed at any instant and average speed.
Answer:
1. Instantaneous speed:
The speed at any particular instant is known as instantaneous speed.

2. Average speed:
Average speed is the ratio of total distance traveled by a body and time taken to travel that distance.

Question 6.
Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Answer:
velocity-time graph

Question 7.
What is uniform circular motion? How is uniform circular motion regarded as an acceleration motion? Explain.
Answer:
When an object is moving in a circular path with a constant speed, the motion of an object is said to be uniform circular motion. When a body has a uniform circular motion, its velocity changes due to the continuous change in the direction of its motion. Hence, the motion of the body is accelerated motion.

Motion Class 9 Extra Questions Numericals

Question 8.
A person travels a distance of 4.0 m towards the east, then turns left and travels 3.0 m towards the north.
1. What is the total distance traveled by the person? North
2. What is his displacement?
Answer:
1. Total distance = OA + AB
= 4m + 3m
Total distance = 7m

2. Total displacement = OB = \(\sqrt{(O A)^{2}+(A B)^{2}}\)
=\(\sqrt{(4)^{2}+(3)^{2}}\) = \(\sqrt{25}\) = 5
Displacement = 5 m

Question 9.
A person travels on a semi-circular track of radius 50 m during a morning walk. If he starts from one end of the track and reaches the other end, calculates the distance traveled and displacement of the person.
Answer:
Let the person start moving from A and reach B via O.
The distance travelled by the person
= Length of track = πr
= \(\frac{22}{7}\) x 50 m = 157.14m
Distance = 157.14 m
The displacement is equal to the diameter of the semi-circular track joining A to B via O.
= 2r = 2 x 50 m = 100m
∴ Displacement = 100 m

Question 10.
Dinesh takes 20 minutes to cover a distance of 5 km on a bicycle. Calculate his average speed in
1. m/s
2. km/h
Answer:
1. Average speed in metre/second
Distance travelled by Dmesh = 5 km = 5 x 1000 m = 5000 m
time taken (t) = 20 min = 20 x 60 seconds = 1200 seconds

2. Average speed in kilometre/hour
Average speed = \(\frac{5 \mathrm{km}}{\frac{20}{60} \mathrm{h}}\) = 15 km/h
∴  Average speed = 15 km/h

Question 11.
A train starting from a railway station and moving with uniform acceleration attains a speed of 20 m/s in 10 seconds. Find its acceleration.
Answer:
Given,
initial velocity, u = 0 m/s
Final velocity, υ = 20 m/s
time taken, t = 10 s
Acceleration, a = \(\frac{v-u}{t}=\frac{20-0}{10}\)
∴ a = 2 m/s²

Question 12.
A car moving with a speed of 72 km/h is brought to rest in 10 seconds by applying brakes. Find the magnitude of average retardation due to brakes and distance traveled by car after applying the brakes.
Answer:
Given, initial velocity = 72 km/h = 72 x \(\frac{5}{18}\) = 20 m/s
Final velocity = 0 m/s
Time taken = 10s.

1. Retardation
Acceleration, a = \(\frac{υ – u}{t}\)
a = \(\frac{0 – 20}{10}\) = -2m/s²
∴ Retardation = 2 m/s²

2. Distance travelled using equation, υ² – u² = 2as
(0)² – (20)² = 2 (-2)s
∴ s = 100 m

Question 13.
The velocity of a particle moving along a straight line in a certain time interval is shown below. What is the distance traveled during acceleration?

Answer:
Here particle accelerates for 2 seconds.
Distance traveled during acceleration = area under υ – t graph for 2 sec.
\(\frac{1}{2}\) x Base x Height = \(\frac{1}{2}\) x 2 x 20 = 20m

Question 14.
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement—
The time graph is shown in the figure. Plot a velocity-time graph for the same.
Answer:
Velocity-time graph

Question 15.
Obtain a relation for the distance traveled by an object moving with uniform acceleration in the time interval between the 4th and 5th seconds.
Answer:
Using equation, s = ut + \(\frac{1}{2}\) at²
Distance travelled in 5 s; s₁ = u x 5 + \(\frac{1}{2}\) a(5)²
s₁ = 5u + \(\frac{25a}{2}\)

Distance travelled in 4 s; s₂ = u x 4 + \(\frac{1}{2}\) a(4)²
s₂ = 4u x 8a
Distance travelled in the interval between 4^th and 5^th sec.
= s₁ – s₁ = (5u + \(\frac{25a}{2}\)) – (4u + 8a)
∴ Distance travelled = u + \(\frac{9a}{2}\)

Motion Class 9 Extra Questions Short Answer Type 2

Question 1.
Draw a position-time graph of for
(a) rest
(b) uniform motion
(c) Non-uniform motion
Answer:
(a) Position – time graph for rest

(b) Position – time graph for uniform motion

(c) Position – time graph for non-uniform motion

Question 2.
A draw velocity-time graph for
(a) Uniform motion
(b) Uniform acceleration
(c) Uniform retardation
Answer:
(a) Velocity – time graph for uniform motion (Const. speed)

(b) Velocity – time graph for uniform acceleration (Const. acceleration)

(c) Velocity – time graph for uniform retardation

Question 3.
Explain the difference between the two graphs.

Answer:
(a) In the first graph, the body is starting its motion from origin but in the second graph, the body is starting its motion from any other point.
(b) In the first graph, the body has zero initial velocity but in the second graph the body has non-zero initial velocity.

Question 4.
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of \(\frac{1}{2}\)
(Assume upward acceleration is – g and downward acceleration to be +g).
Answer:
The maximum height reached by the stone can be calculated by the formula
υ² – u² = 2as
Here, υ = 0 and a= – g
Applying this equation for the first ball

Motion Class 9 Extra Questions Numericals

Question 5.
Study the velocity-time graph of a traveling metro train as shown in the figure. What is the acceleration of the train
(a) during the first two seconds?
(b) between the second and tenth seconds?
(c) during the last two seconds?

Answer:
(a) Acceleration, a = \(\frac {v – u}{t}\)
Here, u = 0, u = 20m/s and t = 2s
a = \(\frac {20 – 0}{2}\) = 10 m/s²

(b) Here, u = 20m/s, υ = 20m/s and t = 8s
a = \(\frac {20 – 0}{2}\) = 0 m/s²

(c) Here, u = 20 m/s, u = 0 m/s and t = 2s
a = 0 – 20 = – 10 m/s²

Question 6.
An electron moving with a velocity of 5 x 10 ms¹ enters into a uniform electric field and acquires a uniform acceleration of 10 ms² in the direction of its initial motion.
1. Calculate the time in which the electron would acquire a velocity double its initial velocity.
2. How much distance would the electron cover in this time?
Answer:
Given, u =5 x 10 m/s²
a = 10⁴ m/s²

1. From υ = u + at
take υ = 2u
2u = u + at
u = at
5 x 10⁴ = 10⁴ x t
∴ t = 5s

2. From s = ut + \(\frac{1}{2}\) at²
s = 5 x 10⁴ x 5 + \(\frac{1}{2}\) x 10⁴ x (5)²
∴ s = 37.5 x 10⁴ m

Question 7.
The motion of a train is described by the velocity-time graph as shown in the figure given below.

Find distance travelled by car between
1. 0 to 20 s
2. 20 s to 50 s
Answer:
Distance travelled = Area of velocity – time curve
1. For (0 to 20 s)
Distance = Area of ΔOAC = \(\frac{1}{2}\) x OC x AC = \(\frac{1}{2}\) x 20 x 10 = 100 m

2. For (20 s to 50 s)
Distance = Area of rectangle ABDC + Area of ΔBED
Distance = AC x CD + \(\frac{1}{2}\) DE x DB = 10 x (40 – 20) + \(\frac{1}{2}\) x 10 x 5 = 225m

Motion Class 9 Extra Questions Long Answer Type

Question 1.
Derive an expression for three equations of motion for uniform accelerated motion graphically.
Answer:
Equation of motion by graphical method
Let us consider a body is moving with acceleration where u is initial velocity and u is final velocity, s is the displacement of object and t is a time interval.

1. υ = u + at
We know that slope of υ – t graph gives acceleration so slope
= a = \(\frac{v-u}{t-0}\)
a = \(\frac{v-u}{t}\)
∴ υ = u + at

2. s = ut + \(\frac{1}{2}\) at²
We know that area under u – t graph gives the displacement.
Area = s = area of triangle CDE + area of rectangle ABCE
s = ut + \(\frac{1}{2}\) x t x (υ – u) from (υ – u = at)
Putting the value of υ – u
s = ut + \(\frac{1}{2}\) at²

3. υ² – u² = 2as
We know that area under υ – t graph gives displacement
Area = s = area of trapezium ABDE
s = \(\frac{1}{2}\) x (υ+u) x t From I (t = \(\frac{υ – u}{a}\))
Putting the value of t.
υ² – u² = 2as

Motion Class 9 Extra Questions HOTS

Question 1.
The distance-time graph of two trains is given below. The trains start simultaneously in the
same direction.
1. How much ahead of A is B when the motion starts?
2. What is the speed of B ?
3. When and where will A catch B?
4. What is the difference between speeds of A and B?

5. Is the speed of both the trains uniform or non-uniform? Justify your answer.

Answer:

1. 100km
2. Speed of B, υ = \(\frac{x_{2}-x_{1}}{t}\) = \(\frac{150-100}{2-0}\) = 25 km/h
3. After 2 hours, A catches B at 2. i.e.,150 km from origin 0.
4. The slope of the distance-time graph of A is greater than that of B. Therefore, the speed of A is greater than
   speed B. Speed of A = Slope of OQ = \(\frac{150}{2}\) = 75 km/h
   Difference between speed of A and B =75 km/h – 25 km/h = 50 km/h
5. The speed of both trains is uniform.

Question 2.
A car starts from rest and moves along the x-axis with constant acceleration 5 ms2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?
Answer:
Given,
u = 0, a = 5ms^-2, t₁ = 8s and t₂ = 12s.
Total distance = distance travelled during acceleration + distance travelled during constant velocity
s = s₁ + s₂

For S₁
s₁ = u₁t₁ + \(\frac{1}{2}\) \(a t_{1}^{2}\)
s₁ = \(\frac{1}{2}\) x 5 x (8)²
s₁ = 160 m

For S₁
Applying υ = u + at
υ = 0 + (5 x 8)
u = 40 m/s
s₁ = ut₂
s₂ = 40 x 12
s₂ = 480 m
Total distance, s = s₁ + s₂
s = 160 + 480 = 640m

Question 3.
An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2s if both the objects drop with the same acceleration? How does the difference in heights vary with time?
Answer:
At t = 0, difference in heights of two stones A and B
= 150 – 100 = 50 m
In t = 2s, distance through which each stone falls is –
s = ut + \(\frac{1}{2}\) at² = 0 x \(\frac{1}{2}\) x 10 x (2)² = 20m
∴ Difference in heights of two stones A and B
= (150 – 20) – (100 – 20) = 50
The difference in heights will remain the same at all times during their fall.

Question 4.
An object starting from rest travels 20 m in the first 2s and 160 m in the next 4s. What will be the velocity after 7s from the start?
Answer:
Given, the object travels 20 m first 2 s and 160 m in the next 4s.
From s = ut + \(\frac{1}{2}\) at²
In first 2 s, s = 20 m
20 = 0 + \(\frac{1}{2}\) a(2)²
20 = 2 a
∴ a = 10 m/s²
Velocity at the end of 2s is υ = u + at
υ = 0 + 10 x 2 = 20 m/s

In next 4 s,
s = ut + \(\frac{1}{2}\) a’t²
160 = 20 x 4 + \(\frac{1}{2}\) a’(4)²
80 = 8 a’
a’ = 10 m/s²

Motion Class 9 Extra Questions Value Based (VBQs)

Question 1.
Sumeet and Jaikishan are close friends. Sumeet is a science graduate and Jaikishan is a commerce graduate. Sumeet finds that while driving on a clear highway, Jaikishan often exceeds the speed limit and argues that there is no harm in doing so when the road is clear. Sumeet does not agree with him and tells him that with an increase in speed, the stopping distance of the car would increase and he would not be able to manage if someone appears suddenly on the way.
Read the above passage and answer the following questions:

1. is Sumeet right in his statement?
2. What values are displayed by Sumeet in his statement?
3. How is stopping distance related to the speed of the vehicle?

Answer:

1. Yes, Sumeet is absolutely right.
2. Sumeet is an intelligent boy and he can apply his knowledge of science in daily life.
3. Stopping distance (velocity)²

Question 2.
Raju and Raman are close friends. One day they went for a picnic. Raju was driving the car. When Raman saw the speedometer, he found that the speed of the car was more than the allowed speed limit. Raman advised Raju to reduce speed to avoid the punishment of Overspeed.
Answer the following questions:

1. What is the function of the speedometer?
2. What values are displayed by Raman?

Answer:

1. Speedometer provides instantaneous speed. B
2. Raman is a careful and law-abiding person.

Question 3.
Three friends Ram, Mohan, and Roy have to reach point B from point A. The path taken by them is shown in the figure.
Answer the following questions:

1. Distinguish between distance and displacement.
2. If all of them travel at the same speed, who will reach first at point B and why?
   

Answer:
1. Distance:

1. It is the actual length of the path covered by a moving body.
2. It is always positive or zero.
3. It is a scalar quantity.

Displacement:

1. It is the shortest measured distance between the initial and final positions.
2. It may be positive, negative, or zero.
3. It is a vector quantity.

2.  Mohan will reach first as he is traveling the shortest path.

In this page, we are providing Diversity in Living Organisms Class 9 Extra Questions and Answers Science Chapter 7 pdf download. NCERT Extra Questions for Class 9 Science Chapter 7 Diversity in Living Organisms with Answers will help to score more marks in your CBSE Board Exams.

Class 9 Science Chapter 7 Extra Questions and Answers Diversity in Living Organisms

Extra Questions for Class 9 Science Chapter 7 Diversity in Living Organisms with Answers Solutions

Diversity in Living Organisms Class 9 Extra Questions Very Short Answer Type

Question 1.
Who proposed the two kingdom classification?
Answer:
Carolus Linnaeus.

Question 2.
What is biodiversity?
Answer:
The variety of life forms including plants, animals and microscopic organisms which inhabit this earth constitute biodiversity.

Question 3.
What is classification?
Answer:
The grouping of organisms on the basis of their similarities and differences is called classification.

Question 4.
What is the taxonomy?
Answer:
The branch of science which deals with the classification of organisms is called taxonomy.

Question 5.
Who proposed the five kingdom system of classification?
Answer:
R.H. Whittaker.

Question 6.
Name the fundamental unit of classification.
Or
Which is the lowermost category in the hierarchy of classification?
Answer:
Species.

Question 7.
Give one point of difference between gymnosperms and angiosperms.
Answer:
The seeds of gymnosperms are naked whereas the seeds of angiosperms are enclosed within a fruit.

Question 8.
Name the largest phylum of kingdom Animalia.
Answer:
Arthropoda.

Question 9.
Name the class to which the sea horse belongs.
Answer:
Sea horse belongs to class Pisces.

Question 10.
Which group of plants is referred to as vascular cryptogams?
Answer:
Pteridophytes.

Question 11.
Which phylum in animial kingdom consists of pseudocoelomate organisms?
Answer:
Nematoda

Question 12.
Which group of organisms are called as the ‘Amphibians of plant kingdom*?
Answer:
Bryophytes

Question 13.
Name the two classes of angiosperms with one example of each.
Answer:
The two classes are –

• Monocots: Wheat
• Dicots: Pea

Question 14.
Name the division of cryptogams to which algae belong.
Answer:
Thallophyta

Question 15.
Which is the highest unit of classification?
Answer:
Kingdom

Question 16.
What is the characteristic feature due to which echinoderms are named?
Answer:
Spiny skin (Echino- spiny; derm-skin)

Question 17.
Name the group which comprises of bacteria and blue green algae.
Answer:
Monera

Question 18.
Name an organism which is called saprophyte. Why is it called so?
Answer:
Yeast. It is called so as it feeds on dead and decaying matter to obtain its nutrition.

Question 19.
Identify the kingdom in which the organisms do not have a well-defined nucleus and are not able to show multicellular designs.
Answer:
Monera

Question 20.
Give reason, why blue green algae are classified along with bacteria and placed in the kingdom Monera.
Answer:
As the blue green algae are unicellular prokaryotes like bacteria.

Question 21.
Mention any one characteristic feature of saprophytes.
Answer:
The saprophytes feed on dead and decaying organic matter.

Question 22.
Write one point of difference between monocotyledonous and dicotyledonous plants.
Answer:
Monocots are plants which bear single cotyledon in their seeds. Dicots are the plants which bear two cotyledons in their seeds.

Question 23.
Poriferans have hole or pores all over the body that lead to a system that helps in circulating water to bring in food and oxygen. Name the system.
Answer:
The system is called as the canal system.

Question 24.
State the phylum to which liver fluke and Planaria belong.
Answer:
Phylum Platyhelminthes

Question 25.
Write the type of body cavity and symmetry possessed by nematodes.
Answer:
Body cavity is pseudocoelom and symmetry is bilateral symmetry.

Question 26.
Name one mammal that lays eggs.
Answer:
Platypus

Question 27.
Name the substance which Coelomic cavity of arthropods is filled with. What type of symmetry do they have?
Answer:
Coelomic cavity of arthropods is filled with blood. They show bilateral symmetry.

Question 28.
Name the phylum to which centipede and prawn belong.
Answer:
Phylum Arthropoda

Question 29.
Echinoderms are marine animals. What is their skeleton made up of?
Answer:
Their skeleton is made up of calcium carbonate.

Question 30.
Name one reptile with a four chambered heart.
Answer:
Crocodile

Question 31.
Shyam knew the correct scientific name of mango but did not follow the convention while writing it and wrote it as Mangifera Indica. Rewrite the scientific name as per the convention.
Answer:
Mangifera indica

Question 32.
Rewrite the scientific names correctly:
(a) Panthera tigris
(b) Periplaneta Americana
Answer:
(a) Panthera tigris
(b) Periplaneta americana

Diversity in Living Organisms Class 9 Extra Questions Short Answer Type 1

Question 1.
How can we say that classification of organisms is closely related to their evolution?
Answer:
We can say that classification of organisms is closely related to their evolution because the simple organisms have a primitive body design as they appeared earlier whereas the complex organisms have more advanced body designs as they are more recent. This shows that during the course of evolution more complex body designs were formed from simpler ones.

Question 2.
What is the difference between algae and fungi?
Answer:
Algae:

• Have chlorophyll.
• Autotrophic mode of nutrition.
• Cell wall made of cellulose.
• Food stored in the form of starch.
• Examples: Chlamydomonas, Spirogyra, Ulo-thrix

Fungi:

• Lack chlorophyll.
• Heterotrophic mode of nutrition.
• Cell wall made of chitin.
• Food stored in the form of glycogen.
• Examples: Rhizopus, Agaricus, Yeast

Question 3.
Pick the odd one out and justify your choice by giving reasons:
(a) Moss, Fern, Pinus, Spirogyra
(b) Sea cucumber, Octopus, Feather star, Star fish
Answer:
(a) The odd one out in this case is Pinus as it is a phanerogams having covered reproductive parts whereas the other three are cryptogams which bear hidden reproductive organs.

(b) The odd one out in this case is Octopus as it belongs to phylum Mollusca while others are the members of phylum Echinodermata.

Question 4.
Why were fungi and bacteria considered as plants even though they do not have chlorophyll?
Answer:
Fungi and bacteria were considered as plants as they have cell wall which is a characteristic feature of the plants. So, some earlier classification systems included them under plants.

Question 5.
Why do bryophytes and pteridophytes grow in moist and shady places?
Answer:
Bryophytes and pteridophytes grow in moist and shady places as they need water for their reproduction. Male gametes are carried towards the female gamete by water in order to bring about fertilisation in them.

Question 6.
Which divisions of the plant kingdom are called cryptogams? Why are they called so?
Answer:
Thallophyta, Bryophyla and Pteridophyta are considered as cryptogams. They are called so because they bear hidden and inconspicuous reproductive orgAnswer:

Question 7.
Characteristics of some organisms are given. Identify their group and give one example of each.
(а) Single celled, eukaryotic and photosynthetic
(b) The body is divided into segments, may be unisexual or hermaphrodite
Answer:
(a) Protista: Euglena
(b) Annelida: Earthworm

Question 8.
How do saprophytes get their food? Give one example of saprophyte.
Answer:
Saprophytes derive their nutrition from the dead and decaying materials. For example, Agaricus (Mushroom), Rhizopus (Bread mould), Yeast etc.

Question 9.
What are phanerogams? How are they classified?
Answer:
The phanerogams are the plants which produce seeds and have a well differentiated body with true roots, stem and leaves. They are advanced members of kingdom Plantae. It includes gymnosperms and angiosperms.

Question 10.
What are gymnosperms? Give two examples.
Answer:
The plants which bear naked seeds which are not enclosed in fruit are called gymnosperms. For example, Cycas, Pinus, etc.

Question 11.
Write two peculiar characters of sponges.
Answer:
The two peculiar features of sponges are:
(i) They have pores called ostia all over their body and a single large opening at the top, called osculum which helps to develop a canal system for water movement.

(ii) They have a skeleton made up of calcareous or siliceous spicules or spongin fibres which gives strength and support.

Question 12.
Classify the following in their respective phylum/class: Jellyfish, earthworm, cockroach, rat
Answer:

• Jellyfish: Phylum Coelenterata
• Earthworm: Phylum Annelida
• Cockroach: Phylum Arthropoda
• Rat: Phylum Vertebrata, Class: Mammalia

Question 13.
What are the two peculiar features of phylum Echinodermata?
Answer:
The phylum Echinodermata has organisms which have

• spiny skin
• water vascular system

Question 14.
What are the three germ layers present in the organisms? What are the two groups of organisms on the basis of germ layers?
Answer:
The three germ layers are ectoderm, mesoderm and endoderm. The groups are diploblastic and triploblastic.

Question 15.
How are vertebrates different from the other chordates?
Answer:
The notochord is present at any stage of their life cycle in the case of chordates. In the vertebrates the notochord gets replaced by the vertebral column.

Question 16.
How are pteridophytes are different form Phanerogams? Give one example for each group.
Answer:
Pteridophytes have hidden, inconspicuous reproductive organs example ferns.
Phanerogams have well differentiated reproductive organs which are not hidden, example rose, apple.

Question 17.
(a) Given below are few plant species. Identify the divisions to which they belong and write the major characteristic of each division.
(i) Spirogyra
(ii) Deodar
(iii) Moss
(b) What is the mode of nutrition for all of them?
Answer:
(a) (i) Thallophyta: Plant body is not well differentiated
(ii) Gymnosperms: have naked seeds
(iii) Bryophyta: have rhizoids for absorption of water, have stem-like and leaf-like structures.

(b) All of them are autotrophic organisms.

Question 18.
What do you understand by the term ‘naked embryo’? Name any two divisions in kingdom Plantae that have naked embryo. Give one example of each division.
Answer:
Naked embryo is the term which refers to an embryo which is not borne inside the seed. The pteridophytes and gymnosperms bear naked embryo. For example, Ferns and horsetail are pteridophytes. Pinus and Cycas are gymnosperms.

Question 19.
Write the difference between Gymnosperms and Angiosperms giving example of each type.
Or
What are gymnosperms? Give two characteristics.
Answer:
Gymnosperm:

• Bear naked seeds.
• Are woody, evergreen, perennials.
• Examples: Pinus and Cycas

Angiosperm:

• Bear seeds enclosed in fruit.
• Can be woody, non-woody annual, biennial or perennials.
• Examples: Mango, Neem

Question 20.
Thallophyta, bryophyta and pteridophyta are classified as cryptogams whereas gymnosperms and angiosperms are classified as phanerogams, why?
Answer:
Due to the presence of hidden and inconspicuous reproductive organs, thallophyta, bryophyta and pteridophyta are called as cryptogams. Gymnosperms and angiosperms are phanerogams as they have well developed and distinct reproductive organs, flowers, fruits and seeds.

Question 21.
State reasons for the following:
(а) Platyhelminthes are called so.
(b) Birds have hollow bones.
Answer:
(a) Platyhelminthes are called so because they have a dorsoventrally flattened body.
(b) Presence of hollow bones is an adaptation in birds which helps them to keep low body weight which is helpful in flight.

Question 22.
Identify the phylum of animals by the given characteristics and give an example of each.
(a) The coelomic cavity is blood-filled and the animals have jointed legs.
(b) The animals are called as flatworms and are either free living or parasitic.
Answer:
(a) Phylum arthropoda: eg., cockroach, butterfly, spider, etc.
(b) Phylum Platyhelminthes: Planaria, liver fluke, tapeworm, etc.

Question 23.
Write one point of difference between the following:
(а) Bilateral symmetry and radial symmetry
(b) Annelids and Nematodes
Answer:
(a) Bilateral symmetry – Body can be divided into two exact halves from one plane only.
Radial symmetry – Body can be divided into equal halves from any plane.

(b) Annelids – Have true coelom
Nematodes – Have pseudocoelom.

Question 24.
Give two examples of each:
(а) Egg laying mammals
(b) Organisms with open circulatory system.
Answer:
(а) Duck-billed platypus and Echidna are the egg laying mammals.
(b) Cockroach and Octopus have open circulatory system.

Question 25.
Select the odd one out with respect to classification. Also give reason for your choice: prawn, scorpion, octopus, butterfly.
Answer:
Octopus is the odd one out as it belongs to phylum Mollusca whereas others belong to phylum Arthropoda.

Question 26.
Given below are the two groups of organisms belonging to kingdom Animalia. Write the names of the phylum to which they belong.
(a) Octopus, Pila, Unio
(b) Centipede, prawn, scorpion
Answer:
(а) Octopus, Pila, Unio belong to Phylum Mollusca.
(b) Centipede, prawn, scorpion belong to phylum Arthropoda.

Question 27.
What is binomial nomenclature? Who introduced it?
Answer:
The system of scientific naming of the organism which consists of a generic name and a specific epithet is called as binomial nomenclature. It was introduced by Carolus Linnaeus.

Question 28.
Give two differences between bony fish and cartilaginous fish. Give one example of each.
Answer:
Bony Fish:

• Endoskeleton is made up of bones.
• Operculum covers the gill slits.
• Terminal mouth Example: Rohu, sea horse

Cartilaginous Fish:

• Endoskeleton is made up of cartilage.
• Operculum absent.
• Ventral mouth Example: Shark, saw fish

Question 29.
State any two characteristics of Mammalia. Name two egg laying mammals.
Answer:
Two characteristics of Mammalia are:

• Presence of mammary glands, four chambered heart and are warm-blooded
• Skin has hairs, sweat glands and oil glands Egg laying mammals are platypus and Echidna.

Question 30.
Write appropriate terms for the following.
(a) Animals which are able to maintain a certain body temperature over a wide range of temperature in the environment.
(b) Animals which have pseudocoelom.
Answer:
Kingdom Mammalia: Warm-blooded animals
Phylum Nematoda: Its members have pseudocoelom.

Diversity in Living Organisms Class 9 Extra Questions Short Answer Type 2

Question 1.
Explain the three basic features for grouping all organisms into five major kingdoms.
Answer:
The three basic features for grouping the organisms into five kingdoms are
(i) Cell structure: On the basis of this the two groups are prokaryotes and the eukaryotes which are distinguished on the basis of absence or presence of well defined nuclear membrane.

(ii) Thallus organisation: The organisms are grouped as unicellular or multicellular organisms on the basis of their being composed of a single cell or of many cells respectively.

(iii) Mode of nutrition: The organisms are grouped as autotrophs or heterotrophs on the basis of their ability to synthesise their own food or being dependent on other organisms for their food.

Question 2.
What are the steps in building a hierarchy of classification?
Answer:
The characteristics chosen for developing a hierarchy of classification should start with the characteristic which forms the broadest division and then the next characteristic chosen should be dependent on the previous characteristic and division, besides having its own new characteristic features.

This process should be continued for the next levels in order to build a hierarchy in classification. For example, the classification of organisms into two broad categories prokaryotes and eukaryotes forms the basis of further characteristics on which their classification is based.

Question 3.
Differentiate between Bryophyta and Pteridophyta. Give example of each group.
Or
Write four main features of pteridophyta and give two examples.
Answer:
Bryophyta:

• They are called the ‘amphibians of the plant kingdom’.
• They lack vascular tissues.
• Body is not well-differentiated into true root, stem or leaves.
• The dominant phase or the main plant body is gametophyte (haploid).
• Sporophyte depends upon gametophyte for its support and nutrition.
• Spores are formed in capsule of sporophyte.
• Examples: Liverworts, Mosses

Pteridophyta:

• They are the first land plants.
• They have vascular tissues xylem and phloem.
• Body is well-differentiated into true roots, stem and leaves.
• The dominant phase or the main plant body is sporophyte (diploid).
• Sporophyte and gametophyte are independent structures in them.
• Spores are produced inside the sporangia borne on leaves or cones.
• Examples: Ferns, Horsetail, Marsilea

Question 4.
Name three groups which are placed under Cryptogamae. State and explain two characteristics which are exhibited by each category of these plant bodies.
Answer:
The three groups placed under Cryptogamae are: Thallophyta, Bryophyta and Pteridophyta.

The two characteristic features of cryptogams are:

• They have inconspicuous and hidden reproductive organs.
• They produce naked embryos called spores.

Question 5.
Define the following: (a) Radial symmetry (b) Bilateral symmetry
Answer:
When any plane passing through the central axis can divide the organism into identical halves, it is called radial symmetry, example coelenterates and the adults of Echinoderms.

When only one plane can divide the body of the organism into identical right and left halves, the symmetry is called as bilateral symmetry, example annelids, arthropods and humans.

Question 6.
Give main features of phylum Chordata.
Answer:
The main features of phylum Chordata are presence of:

• Notochord
• Dorsal Central nervous system
• Pharynx perforated by gill slits
• Ventral heart
• A post-anal tail

Question 7.
(a) Write any two features that are present in all chordates.
(b) Write one difference between pseudocoelom and true coelom.
Answer:
(a) The features present in all chordates are:

• Dorsal central nervous system and dorsal nerve cord.
• Triploblastic, coelomate with organ system level of organisation.

(b) If the body cavity of an organism is not lined by mesoderm and the space is filled with vacuolated cells, then the body cavity called pseudocoelom.

Question 8.
Define classification. Give any two of its significance.
Answer:
Grouping of organisms on the basis of their similarities and differences is called classification. Classification is important as it helps to make the study of vast variety of organisms easier and also helps us to understand the inter-relationships which occur among the organisms.

Question 9.
Classify the following plants into different plant divisions. Spirogyra, Fern, Funaria, Pinus, Apple tree, Mustard plant
Answer:
Spirogyra: Thallophyta, Fern: Pteridophyta, Funaria: Bryophyta, Pinus: Gymnosperm, Apple tree: Angiosperm, Mustard plant: Angiosperm

Question 10.
To which group do the following organisms belong and give one reason for each.
(a) Cyanobacteria
(b) Euglena
(c) Ulothrix
Answer:
(a) Cyanobacteria: Kingdom Monera; does not have a well-defined nucleus and lacks membrane bound cell organelles, prokaryotes.

(b) Euglena: Kingdom Protista; are aquatic, unicellular eukaryotes with well-defined nucleus and membrane bound cell organelles.

(c) Ulothrix: Kingdom Plantae, Thallophyta; Have a thalloid body, photosynthetic and eukaryotic.

Question 11.
Why are bryophytes called as ‘Amphibians of plant kingdom9?
Answer:
Bryophytes are called as ‘amphibians of plant kingdom’ because:

• They live in moist, damp places in order to get water from soil either directly or with the help of their rhizoids.
• They require water for the transfer of their gametes and are dependent on water for sexual reproduction.

Question 12.
(а) What is coelom?
(b) Presence of ‘coelom’ in an animal is considered advantageous. Why?
Answer:
(a) Coelom is a body cavity lined by mesodermal cells and lies between the body wall and alimentary canal (gut) of the organism.

(b) The coelom is considered advantageous as it helps to accommodate the various organs of the body in a proper way and gives a greater flexibility to the body of the organism.

Question 13.
(а) Write one characteristic each of amphibia and aves.
(b) Write the name of the class to which following belong:
(i) Sea-horse
(ii) King cobra
Answer:
(a) Amphibia: Can live both on land and in water, slimy to touch, cold blooded with three chambered heart.
Aves: Have forelimbs modified into wings for flight. Body covered with feathers, are warm-blooded with four-chambered heart

(b) Sea-horse: Class Pisces;
King cobra: Class Reptilia

Question 14.
Give the three characteristic features of Class Mammalia.
Answer:
The characteristic features of Class Mammalia are:

• Presence of Mammary glands.
• The skin has hairs.
• Their body has sweat glands and oil glands.
• They have a four-chambered heart, are warm-blooded.

Question 15.
Distinguish between monocots and dicots.
Answer:
Monocots:

• Possess a single cotyledon.
• Have fibrous roots.
• Have parallel venation. Examples: Wheat, Maize

Dicots:

• Possess two cotyledons.
• Have tap root.
• Have reticulate venation. Examples: Pea, gram

Question 16.
(a) What is symbiotic relationship?
(b) Name a symbiotic life form that grows on the bark of tree as large coloured patches.
Answer:
(a) The mutually dependent relationship between two organisms where both are benefitted is called symbiotic relationship.

(b) Lichen is a symbiotic life form that grows on the bark of trees as large coloured patches. It is symbiotic association between fungi and algae.

Question 17.
State the bases of classifying plants and animals into different categories.
Answer:
The bases for classifying animals and plants into different categories are:

• Mode of nutrition – Plants are autotrophic whereas animals heterotrophic.
• Presence or absences of cell wall – Plant cells have cell wall whereas animal cells do not have a cell wall.

Question 18.
(a) Differentiate between Fungi and Plantae.
(b) Mention the basis of classification among plants to different levels.
Answer:
Fungi:

• They are non-chlorophyllous and heterotrophic.
• Their cell wall is made of chitin.
• They are mostly saprophytes or parasites.

Plantae:

• They have chlorophyll and are autotrophic.
• Their cell wall is made of cellulose.
• They are autotrophs.

(b) The basis of classification of plants to different levels is –

• Presence or absence of well-differentiated body tissues
• Presence or absence of a well developed transport system for water, minerals and organic substances.
• Seed bearing ability and presence of seeds enclosed in fruits or not.

Question 19.
Name the group which is called as Amphibian of the plant kingdom. Cite an example of this group, also mention one important feature of the same group.
Answer:
Bryophytes are called amphibians of the plant kingdom, example Marchantia, Funaria.

Characteristic feature: Plant body is thalloid, not well-differentiated and has root-like, stem-like and leaf-like structures.

Question 20.
Identify the following:
(a) Amphibians of the plant kingdom.
(b) Plants with hidden reproductive organs
(c) Mutually benefitted relationship between two organisms.
Answer:
(a) Bryophytes
(b) Phanerogams comprising of Algae, Bryophytes and Pteridophytes.
(c) Symbiotic relationship as shown by algae and fungi in Lichen.

Question 21.
Describe the following:
(a) Lichens
(b) Cryptogams
(c) Phanerogams
Answer:
(a) Lichen is a symbiotic relationship between an algae and fungi.
(b) Cryptogams have hidden or inconspicuous reproductive organs.
(c) Phanerogams have well differentiated reproductive tissues which result in formation of seed.

Question 22.
(а) Which group of plants is known as ‘flowering plants’?
(b) On the basis of seed how is a maize plant different from a pea plant?
Answer:
(a) Angiosperms are called the flowering plants.
(b) Maize is a monocot plant as it bears one cotyledon in its seed whereas pea is a dicot as it has two cotyledons in its seed.

Question 23.
Name the largest group of animals. Write the salient features of this group. Give two examples.
Answer:
Phylum Arthropoda is the largest group of animals. They have jointed legs, bilateral symmetry and a body cavity filled with fluid. For example, Cockroach, spider, butterfly, etc.

Question 24.
What is binomial nomenclature? Who gave it? Write its advantage.
Answer:
The system of scientific naming of organisms which consists of a generic name and a specific epithet is called binomial nomenclature. It was given by Carolus Linnaeus. Its advantage is that it helps in the systematic study and understanding of the living organisms.

Question 25.
List the conventions used for writing a scientific name. What is the importance of scientific name?
Answer:
Convention for writing the scientific names:

• The name of the genus begins with a capital letter.
• The name of the species begins with a small letter.
• When printed, the scientific name is given in italics.
• When written by hand, the genus name and the species name have to be underlined separately.

Common names cannot be used in the same way by the scientist world over and can often result in confusion. To avoid this, a system of scientific names has been proposed.

Question 26.
Write true (T) or false (F)
(а) Whittaker proposed five kingdom classification.
(b) Monera is divided into Archaebacteria and Eubacteria.
(c) Starting from Class, Species comes before the Genus.
(d) Anabaena belongs to the kingdom Monera.
(e) Blue-green algae belongs to the kingdom Protista.
(f) All prokaryotes are classified under Monera.
Answer:
(a) True
(b) True
(c) False
(d) True
(e) False
(f) True

Question 27.
Fill in the blanks:
(a) Fungi shows _______ mode of nutrition.
(b) Cell wall of fungi is made up of _______
(c) Association between blue-green algae and fungi is called as _______
(d) Chemical nature of chitin is _______
(e) _______ has smallest number of organisms with maximum number of similar characters.
(f) Plants without well differentiated stem, root and leaf are kept in _______
(g) _______ are called the amphibians of the plant kingdom.
Answer:
(a) saprophytic
(b) chitin
(c) lichens
(d) Carbohydrate
(e) Species
(f) Thallophyta
(g) Bryophytes

Question 28.
You are provided with the seeds of gram, wheat, rice, pumpkin, maize and pea. Classify them whether they are monocot or dicot.
Answer:
Gram – dicot
Wheat – monocot
Rice – monocot
Pumpkin – dicot
Maize – monocot
Pea -dicot

Question 29.
Match items of column (A) with items of column (B).

Answer:
(a) (ii)
(b) (i)
(c) (iv)
(d) (iii)
(e) (vi)
(f) (u)
(g) (vii)

Question 30.
Match items of column (A) with items of column (B).

Answer:
(a) (iii)
(b) (ii)
(c) (vi)
(d) (i)
(e) (v)
(f) (iv)

Question 31.
Classify the following organisms based on the absence/presence of true coelom (i.e., acoelomate, pseudocoelomate and coelomate) Spongilla, Sea anemone, Planaria, Liver fluke, Wuchereria, Ascaris, Nereis, Earthworm, Scorpion, Birds, Fishes, Horse.
Answer:

• Spongilla: Acoelomate
• Sea anemone: Acoelomate
• Planaria: Acoelomate
• Liver fluke: Acoelomate
• Wuchereria: Pseudocoelomate
• Ascaris: Pseudocoelomate
• Nereis: Coelomate
• Scorpion: Coelomate
• Earthworm: Coelomate
• Birds, Fishes and Horse: Coelomate

Question 32.
Endoskeleton of fishes are made up of cartilage and bone; classify the following fishes as cartilagenous or bony:
Torpedo, Sting ray, Dog fish, Rohu, Angler fish, Exocoetus.
Answer:

• Torpedo: cartilagenous
• Sting ray: cartilagenous
• Dog fish: cartilagenous
• Rohu: bony
• Angler fish: cartilagenous
• Exocoetus: bony

Question 33.
Classify the following based on number of chambers in their heart.
Rohu, Scoliodon, Frog, Salamander, Flying lizard, King Cobra, Crocodile, Ostrich, Pigeon, Bat, Whale
Answer:

• Rohu, Scoliodon: 2 chambered,
• Frog, Salamander, Flying lizard, King Cobra: 3 chambered,
• Crocodile, Ostrich, Pigeon, Bat, Whale: 4 chambered

Question 34.
Classify Rohu, Scolidon, Flying lizard, King Cobra, Frog, Salamander, Ostrich, Pigeon, Bat, Crocodile and Whale into the cold-blooded/warm-blooded animals.
Answer:

• Cold-blooded: Rohu, Scolidon, Flying lizard, King Cobra, Frog, Salamander Crocodile
• Warm-blooded: Ostrich, Pigeon, Bat, Whale

Question 35.
Name two egg laying mammals.
Answer:

• Platypus
• Echidna.

Question 36.
Fill in the blanks:
(а) Five kingdom classification of living organisms is given by _______
(b) Basic smallest unit of classification is _______
(c) Prokaryotes are grouped in Kingdom _______
(d) Paramecium is a Protista because it is an _______
(e) Fungi do not contain _______
(f) A fungus _______ can be seen without microscope.
(g) Common fungi used in preparing bread is _______
(h) Algae and fungi form symbiotic association called _______
Answer:
(a) Robert Whittaker
(b) species
(c) Monera
(d) eukaryotic unicellular organism
(e) chlorophyll
(f) mushrooms
(g) yeast
(h) lichens

Question 37.
Give True (T) and False (F).
(а) Gymnosperms differ from Angiosperms in having covered seed.
(b) Non flowering plants are called Cryptogamae.
(c) Bryophytes have conducting tissue.
(d) Funaria is a moss.
(e) Compound leaves are found in many ferns.
(f) Seeds contain embryo.
Answer:
(a) False
(b) True
(c) False
(d) True
(e) True
(f) True

Question 38.
Give examples for the following.
(a) Bilateral, dorsiventral symmetry is found in _______
(b) Worms causing the disease elephantiasis is _______
(c) Open circulatory system is found in _______ where coelomic cavity is filled with blood.
(d) _______ are known to have pseudocoelom.
Answer:
(a) Liver Fluke
(b) Filarial worms
(c) Arthropods
(d) Nematodes

Question 39.
Label a, b, c and d, given in the figure below. Give the function of (b).
Answer:

(a) Dorsal fin
(b) Caudal fin
(c) Pelvic fin
(d) Pectoral fin
Function of Caudal fins: Caudal fins help in streamlined movement in water.

Question 40.
Fill in the boxes given in figure with appropriate characteristics/plant groups.

(a) Thallophyta
(b) Without specialized vascular tissue
(c) Pteridophyta
(d) Phanerogams
(e) Bear naked seeds
(f) Angiosperm
(g) Have seeds Thallophyta with two cotyledons
(h) Monocots

Diversity in Living Organisms Class 9 Extra Questions Long Answer Type

Question 1.
Explain the various categories of taxonomical hierarchy.
Answer:
The categories in taxonomical hierarchy are

• Kingdom: The highest category of classification
• Phylum (for animals) / Division (for plants): Group of related classes
• Class: Group of related orders.
• Order: Group of related families.
• Family: Group of related genus.
• Genus: Group of related species.
• Species: Group of organisms which can interbreed among themselves to produce a fertile offspring.

Question 2.
(a) Draw labelled diagrams of three protozoa.
(b) Euglena is a dual organism. Why?
Answer:
(a)

(b) Euglena is a dual organism as:

• It does not have a cell wall and behaves as a heterotrophic organism in the absence of sunlight.
• It contains chloroplast and can perform photosynthesis and behaves as autotroph in the presence of sunlight.

Question 3.
Mention the class to which they belong and give one characteristic feature of each. Frog, fish, lizard, pigeon, bat.
Answer:
(i) Fish: Class Pisces; Aquatic, respiration with gills, moist scales present on the body, two-chambered heart, cold-blooded

(ii) Frog: Class Amphibia; Can live both on land and in water, cold-blooded, three-chambered heart, slimy to touch

(iii) Lizard: Class Reptilia; Cold-blooded, three-chambered heart, hard covering present on eggs to prevent from desiccation

(iv) Pigeon: Class Aves; Forelimbs modified into wings for flight and have feathers, have beak, warm-blooded, four-chambered heart

(v) Bat: Class Mammalia; has mammary glands, warm-blooded, four-chambered heart.

Question 4.
(i) Draw a neat labelled diagram of Hydra.
(ii) Label mesoglea and gastro-vascular cavity.
(iii) Name the group of animals it belongs to.
(iv) Name one species of this group that lives in colonies.
Answer:
(i) and (ii)-See figure

(iii) Phylum Coelenterata
(iv) Corals live in colonies.

Question 5.
Write one difference for each of the following pairs:
(i) Thallophyta and Bryophyta
(ii) Nematoda and Annelida
(iii) Amphibians and Reptiles
Answer:
(i)

• Thallophyta Bryophyta – Plant body is not well differentiated into root, stem and leaves. It is a thalloid example – Spirogyra.
• Bryophyta – Plant body slightly more differentiated than Thallophyta with root-like, stem-like and leaf-like structures, called as Amphibians of the plant kingdom.

(ii)

• Nematoda – Called as roundworms, bilaterally symmetrical having pseudocoelom, example Ascaris
• Annelida – Have a metamerically segmented body and a true coelom, example Earthworm

(iii) Amphibians

• Can live both in land as well as water.
• Lack scales on body and have slimy skin.
• Lay eggs in water.
• Eggs are devoid of tough covering.

Reptiles:

• Most of them are terrestrial animals.
• Have dry scales on body.
• Lay their eggs mostly on land.
• Eggs have a tough covering to protect from drying.

Question 6.
(a) In which two ways are amphibians different from fishes?
(b) Identify the phylum of organisms having the characteristics:
(i) Pore bearing animals and radial symmetry
(ii) Body spiny and radial symmetry
(c) Why do gymnosperms not require water for fertilisation?
Answer:
(a) Amphibians can live both on land as well as water, have three-chambered heart and have lungs for respiration in adult stage.
Fishes are aquatic, have two-chambered heart and have gills for respiration.

(b) (i) Phylum Porifera
(ii) Phylum Echinodermata

(c) Gymnosperms do not require water for fertilisation as they bear pollen grains which are carried away by agents like wind to cause pollination.

Question 7.
Give four features of phylum Coelenterata. Give two examples.
Answer:
The four distinguishing features of phylum Coelenterata are:

• They have a radially symmetrical body.
• The body is made of two layers of cells, so called diploblastic.
• Has tentacles which surround their mouth. Tentacles bear cnidoblasts, stinging cells.
• They have a characteristic cavity called as coelenteron.
• They are solitary like Hydra and can be colonial like Obelia. example Hydra, Obelia, Jelly fish.

Question 8.
List the convention that is followed while writing the scientific names. Give scientific name of Mango and Tiger.
Answer:
The conventions followed while writing the scientific names are:

• The name of the genus begins with a capital letter.
• The name of the species begins with a small letter.
• When printed, the scientific name is given in italics.
• When written by hand, the genus name and the species name have to be underlined separately.

Scientific name of Mango is Mangifera indica and of Tiger is Panthera tigris.

Question 9.
Enlist the features of organisms placed in Protista. Give two examples.
Answer:
The members of the kingdom Protista have the following features:

• All of them are unicellular and eukaryotic.
• Their mode of nutrition is either autotrophic or heterotrophic.
• They are usually aquatic but some of them are parasitic.
• Have cilia, flagella or pseudopodia for movement. example Amoeba, Plasmodium

Question 10.
Give the general characteristics of fungi. Give two examples.
Answer:
The general characteristics of fungi are:

• They are heterotrophic organisms which can occur as saprophytes, parasites or symbionts.
• Their cell wall is made up of chitin and lack chlorophyll.
• Food is stored in the form of glycogen in them instead of starch as stored in plants.
• They are eukaryotes as they have membrane bound nucleus.
• For example: Yeast, Rhizopus, Agaricus, Penicillium, etc.

Question 11.
What are the five kingdoms of Whittaker? Give the most important characteristic feature of each kingdom.
Answer:
The five kingdoms proposed by R. H. Whittaker comprises of – Monera, Protista, Fungi, Plantae and Animalia.

• Kingdom Monera: Unicellular, prokaryotes
• Kingdom Protista: Unicellular eukaryotes
• Kingdom Fungi: Non-photosynthetic, chlorophyll absent, heterotrophic nutrition
• Kingdom Plantae: Chlorophyll bearing, photosynthetic autotrophs
• Kingdom Animalia: Heterotrophic nutrition, cell wall absent

Question 12.
(a) To which group do algae belong? Write one characteristic of the division. Give two examples.
(b) Name the group:
(i) Which includes unicellular eukaryotic organisms
(ii) In which mode of nutrition is saprophytic.
(iii) In which seeds are not enclosed in fruits.
(c) Classify flowering plants on the basis of number of cotyledons present in the seed.
Answer:
(a) Algae are members of division Thallophyta. Characteristic – Aquatic, chlorophyll bearing autotrophs.

(b) (i) Protista
(ii) Fungi
(iii) Gymnosperms

(c) The flowering plants are classified as monocots and dicots on the basis of presence of one cotyledon or two cotyledon respectively.

Question 13.
List three groups of plants and tell which plants are referred to as vascular plants? Also mention out of these which group is further classified on the basis of number of cotyledon? State its two characteristics.
Answer:
The pteridophytes, gymnosperms and angiosperms are the vascular plants. Out of these three, angiosperms are classified on the basis of number of cotyledon as – monocots having one cotyledon and dicots having two cotyledons.

Characteristic features of angiosperms are:

• Their seeds are enclosed within fruit.
• Embryo bears the cotyledons also called seed leaves which provide nutrition to the young plant/seedling on the germination of the seed.

Question 14.
(a) On what basis does the embryo of cryptogam differ from that of phanerogams?
(b) Describe the feature that divides the angiosperms into two groups.
(c) State the two sub-groups of angiosperms.
Answer:
(a) Cryptogams bear naked embryos whereas the phanerogams have embryos which are enclosed in seed.
(b) The feature that divides the angiosperms into two groups is the number of cotyledons present in their embryo.
(c) The two sub-groups of angiosperms are monocots (bear single cotyledon) and dicots (have two cotyledons).

Question 15.
Draw a neat diagram of Spirogyra and label the following parts:
(a) Outermost layer of the cell
(b) Organelle that performs the function of photosynthesis
(c) Jelly-like substance in the cell where all organelles are suspended.
(d) Dark coloured and dot-like structure generally present in the centre of the cell.
Answer:

(а) Outermost layer of the cell – Cell wall
(b) Organelle that performs the function of photosynthesis – Chloroplast
(c) Jelly-like substance in the cell where all organelles are suspended – Cytoplasm
(d) Dark coloured and dot-like structure generally present in the centre of the cell – Nucleus

Question 16.
Thallophyta, bryophyta and pteridophyta are called as ‘Cryptogams’. Gymnosperms and Angiosperms are called as ‘phanerogams’. Discuss why. Draw one example of Gymnosperm.
Answer:
Thallophyta, bryophyta and pteridophyta are called as ‘Cryptogams’ because they have hidden or inconspicuous reproductive organs. Spores are formed in them instead of seeds. Gymnosperms and Angiosperms are called as ‘phanerogams’ as they have well differentiated reproductive tissue/organs. Seed harbours the embryo and provides it nourishment too.

Question 17.
Define the terms and give one example of each
(a) Bilateral symmetry
(b) Coelom
(c) Triploblastic
Answer:
(a) If the organism can be divided exactly into two halves from one median plane only, the symmetry is called bilateral symmetry, example liver fluke.

(b) The internal body cavity present between visceral organs and body wall in which well developed organs can be accommodated is called as coelom, example butterfly.

(c) The organisms who have three embryonic layers are called as triploblastic organisms example star fish.

Question 18.
You are given leech, Nereis, Scolopendra, prawn and scorpion; and all have segmented body organisation. Will you classify them in one group? If no, give the important characters based on which you will separate these organisms into different groups.
Answer:
No, all the organisms given in the question do not belong to one group.

Leech and Nereis belong to phylum Annelida because they have metamerically segmented body i.e., body is divided into many segments internally by septa. Body segments are lined up one after the other from head to tail.

The characteristic identifying feature of Scolopendra, prawn and scorpion are the jointed legs and open circulating system due to which they are placed in phylum Arthropoda.

Question 19.
Which organism is more complex and evolved among Bacteria, Mushroom and Mango tree? Give reasons.
Answer:
Mango tree is more complex and evolved because, it is eukaryotic, autotrophic, terrestrial sporophyte with covered seed. Bacteria is unicellular prokaryote and fungi are the heterotrophic, simple thallophyte with no tissue systems.

Question 20.
Differentiate between flying lizard and bird. Draw the diagram.
Answer:
Flying lizard belongs to the group reptiles and is characterised as cold-blooded, body covered with scales and have three-chambered heart, while birds belong to group aves and are characterised as warm¬blooded, having feather covered body, forelimbs modified as wings and having four-chambered heart.

Question 21.
List out some common features in cat, rat and bat.
Answer:
Bat, rat and cat belong to the class Mammalia and have following common features:
(а) All have notochord at some stage of their life cycle.
(b) All are warm-blooded:
(c) All have four-chambered heart.
(d) All have skin covered with hair and with sweat and oil glands.

Question 22.
Why do we keep both snake and turtle in the same class?
Answer:
Because both are –

• Cold-blooded
• Have scales
• Breathe through lungs
• Have three-chambered heart
• They lay eggs with tough covering.

Diversity in Living Organisms Class 9 Extra Questions HOTS

Question 1.
Which group among the Amphibia and Pisces is more advanced and why?
Answer:
The amphibians are more advanced than the Pisces as

• Amphibians can survive both on land as well as in water whereas the members of Pisces can survive only in water.
• Amphibians have a three-chambered heart compared to the two-chambered heart present in the Pisces.
• Amphibians have the ability to respire through lungs but fishes respire through gills.

Question 2.
The protozoans have been included in Protista and not in kingdom Animalia. Give reason.
Answer:
Protozoans are unicellular, eukaryotes so they have been kept in the Kingdom Protista comprising of only the unicellular, eukaryotic organisms. Kingdom Animalia consists of multicellular eukaryotes.

Question 3.
Three types of animal specimen were collected by Rajeev and labelled as A, B and C. The specimen A had slimy skin, respired through lungs, B had dry scales with eggs having tough covering and the specimen C had moist scales with terminal mouth. Identify the class to which the specimens belong.
Answer:
A: Amphibia
B: Reptilia
C: Pisces

Question 4.
While playing near a pond Anmol experienced a pain on his feet. He saw a black coloured organism with metamerically segmented body was clinging to his foot and trying to suck blood from his foot. On the basis of this identify the organism and the phylum to which it may belong to. Name two other members of the phylum to which this organism belongs.
Answer:

• Organism: Leech
• Phylum: Annelida
• Other members: Nereis, Earthworm

Question 5.
Shobha went for a school trip and was shown some organisms which were told by their teachers as the members of the second largest phylum of kingdom Animalia. The organisms had a shell on their body and moved around with their muscular foot. Name the phylum to which the mentioned organism will belong. Give one more characteristic feature and two examples of such organisms.
Answer:
Phylum : Mollusca
Feature : Reduced Coelomic cavity
Examples : Snails and mussels

Question 6.
Three types of plant specimen were observed by Renu and labelled as A, B and C. The specimen A had green colour with undifferentiated thalloid body, B had well differentiated body which formed spores and had a vascular system and the specimen C had cones which had seeds but no fruits were formed in them. Identify the class to which the specimens belong.
Answer:
A: Thallophyta
B: Pteridophyta
C: Gymnosperms

Diversity in Living Organisms Class 9 Extra Questions Value Based (VBQs)

Question 1.
Reena read an article in newspaper regarding the problems caused by the chemical fertilisers. She surfed through the internet and came to know that certain blue-green algae are a good source of fertilisers. She advised the farmers of the area to use them and helped them to stop using chemical fertilisers.
(a) Which group comprises of the blue-green algae?
(b) How do these blue-green algae increase soil fertility?
(c) What are the characteristic features of the group to which they belong?
(d) What values are shown by Reena by her work?
Answer:
(a) Kingdom Monera
(b) They fix atmospheric nitrogen and increase soil fertility
(c) They are unicellular prokaryotes which do not have a well defined nucleus.
(d) Values shown by Reena are concern for environment, helpfulness, eco-friendliness and scientific attitude.

Question 2.
During their trip to a hill station, Shiksha observed some thalloid plants having little differentiation of body which were growing on the moist and damp surface in the form of dense mats. She asked her teacher about them and their role. Her teacher told her which group of organism they were and said that these dense mats were helpful in preventing soil erosion and need water for fertilisation.
(a) What is the probable group of organisms which her teacher would have told her?
(b) What is the peculiar feature of this group of organisms?
(c) Give two examples of members of the group.
(d) What are the values shown by Shiksha?
Answer:
(a) Bryophytes
(b) Can live both on land and in water
(c) Marchantia, Funaria
(d) The values shown by Shiksha are keen observation and scientific attitude.

In this page, we are providing Tissues Class 9 Extra Questions and Answers Science Chapter 6 pdf download. NCERT Extra Questions for Class 9 Science Chapter 6 Tissues with Answers will help to score more marks in your CBSE Board Exams.

Class 9 Science Chapter 6 Extra Questions and Answers Tissues

Extra Questions for Class 9 Science Chapter 6 Tissues with Answers Solutions

Tissues Class 9 Extra Questions Very Short Answer Type

Question 1.
Name any two types of simple permanent plant tissues.
Answer:
The simple permanent tissues of plants are: Parenchyma, collenchyma and sclerenchyma.

Question 2.
What are blood platelets?
Answer:
Blood platelets are the cell fragments present in the plasma of blood which help in the clotting of blood.

Question 3.
Name the connective tissue that is found between skin and muscles.
Answer:
Areolar tissue.

Question 4.
Name the tissue present in the brain.
Answer:
Nervous tissue which comprises of its basic unit called neurons.

Question 5.
In which of the simple plant tissue, deposition of lignin is found?
Answer:
Sclerenchyma.

Question 6.
Name the basic packing tissue of plant.
Answer:
Parenchyma.

Question 7.
Name the tissue which is present in the veins of leaves.
Answer:
Sclerenchyma.

Question 8.
Why is cork impervious to gases and water?
Answer:
Due to presence of a chemical substance called suberin.

Question 9.
What is the function of phloem?
Answer:
Phloem helps in the transport of food from leaves to the various parts of the plant.

Question 10.
Which body cell provides resistance against infections?
Answer:
White blood cells (WBC) provide resistance against infections.

Question 11.
Which biochemicals compose the solid matrix of cartilage?
Answer:
Proteins and sugars make up the solid matrix of cartilage.

Question 12.
Name the connective tissue which helps in the repair of tissues.
Answer:
Areolar connective tissue helps in the repair of tissue.

Question 13.
Which connective tissue is specialised for fat storage and acts as heat insulator?
Answer:
Adipose tissue helps in storage of fats and acts as heat insulator.

Question 14.
Which muscle has spindle-shaped cells?
Answer:
Smooth muscle cells have spindle shaped cells.

Question 15.
Which meristem is present at growing tips of stems and roots?
Answer:
Apical meristem is present at the growing tips of the stem and roots.

Tissues Class 9 Extra Questions Short Answer Type 1

Question 1.
List any four salient features of meristematic tissue.
Answer:
The salient features of meristematic tissue are:

• This tissue consists of cells which continuously divide to produce new cells.
• The cells of this tissue lack vacuoles.
• The cells of this tissue have dense cytoplasm.
• The cells of this tissue have thin cellulosic cell walls and prominent nuclei.

Question 2.
Write the four elements of xylem.
Answer:
The four elements of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 3.
How is ligament different from tendons?
Or
Differentiate between tendon and ligament.
Answer:
Ligament is a connective tissue which joins bone to bone and is elastic in nature.

Tendons join bone to muscles and are less elastic as compared to the ligaments.

Question 4.
Write a short note on the different types of meristematic tissue with their location and functions in the plants.
Answer:
The meristematic tissues are classified as apical, lateral and intercalary meristematic tissue based on the region where they are present.

Apical meristem – It is present at the growing tips of stem and roots and results in increase in the length of the stem and the root. Lateral meristem (cambium): It is present on the lateral sides of stem and roots. It helps to increase the girth of the stem or root.

Intercalary meristem – It is present at the base of the leaves or internodes. It helps in the longitudinal growth of plants.

Question 5.
Show the diagrammatic representation of location of lateral meristem and intercalary meristem in plant body.
Answer:

Question 6.
Differentiate between chlorenchyma and aerenchyma.
Or
Write the difference between aerenchyma and chlorenchyma.
Answer:
Chlorenchyma:

• It is a type of parenchyma which contains chlorophyll.
• It helps to perform photosynthesis.
• It is present in green parts of plants like the leaves.

Aerenchyma:

• This type of parenchyma that has large air cavities in it.
• It helps to provide buoyancy to the plants.
• It is present in the aquatic plants, example in their floating leaves.

Question 7.
Write the functions of collenchyma in plants.
Answer:
Collenchyma allows easy bending in various parts of a plant (leaf, stem) without breaking. It also provides mechanical support to plants like in the leaf stalks below the epidermis.

Question 8.
What are the roles of epidermis in plants?
Answer:
The functions of epidermis are:

• Epidermis is usually made up of a single layer of cells and gives protection.
• The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.
• The epidermis of leaves have small pores called as stomata which help in gaseous exchange and transpiration.
• The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

Question 9.
Answer the following:
(i) How is the epidermis of the plants living in very dry habitats adapted?
(ii) Write functions of guard cells of stomata in the leaf.
Answer:
(i) The epidermis of plants living in dry habitats may be thicker or often secrete a waxy, water-resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.

(ii) The guard cells of stomata in the leaf help in gaseous exchange and transpiration.

Question 10.
What is the function of areolar tissues?
Answer:
The functions of areolar tissues are:

• It fills the space inside the organs
• It supports internal organs.
• It helps in repair of tissues.

Question 11.
Determine the location of the following tissues:

1. Unstriated muscle fibres
2. Cuboidal epithelium
3. Adipose tissue
4. Striated muscle fibres

Answer:

1. Unstriated muscle fibres: Present in iris of the eye, ureters, blood vessels, alimentary canal and bronchi of lungs.
2. Cuboidal epithelium: Present in lining of kidney tubules and ducts of salivary glands.
3. Adipose tissue: It is found below the skin and between internal orgAnswer:
4. Striated muscle fibres: It is present in muscles of our limbs

Tissues Class 9 Extra Questions Short Answer Type 2

Question 1.
Explain how the bark of a tree is formed. How does it act as a protective tissue?
Answer:
In the older stem, a strip of secondary meristem replaces the epidermis. The secondary meristem cuts off cells towards outside to form a several-layer thick tissue; This is called the cork or the bark of the tree.

Cells of cork or bark are dead, compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them impervious to gases and water. In this way it acts as a protective tissue.

Question 2.
Draw a diagrammatic labelled sketch of stem tip to show location of meristematic tissue. Mention the functions of different types of meristematic tissue.
Answer:

The types of meristematic tissue are:
(i) Apical meristem: It is present at the growing tips of stems and roots and results in increase in the length of the stem and the root.

(ii) Lateral meristem (cambium): It is present on the lateral sides of stems and roots. It helps to increase the girth of the stem or root.

(iii) Intercalary meristem: It is present at the base of the leaves or internodes. It helps in the longitudinal growth of plants.

Question 3.
What are the two main components of blood? Why is blood considered a type of connective tissue?
Answer:
Blood is a special connective tissue consisting of a fluid matrix, plasma, and formed elements. The formed elements are red blood cells (RBCs), white blood cells (WBCs) and blood platelets. Blood is considered as a type of connective tissue as they have the same origin as other types of connective tissue and helps to connect the different parts of the body to facilitate exchange of various components like nutrients and gases.

Question 4.
Give one function of each of the following.
(i) Stomata
(ii) Root nodules
(iii) Cardiac muscle fibres
Answer:
(i) Stomata: Help in exchange of gases in the plants.

(ii) Root nodules: In leguminous plants, the root nodules harbour nitrogen fixing bacteria which convert atmospheric nitrogen into nitrates.

(iii) Cardiac muscle fibres: They help in rhythmic contraction and relaxation of the heart.

Question 5.
Differentiate between bone and cartilage.
Or
Differentiate between bone and cartilage with respect to structure, function and location.
Answer:
Bone:

• Bones have a hard and non-pliable ground substance.
• Its matrix is rich in calcium salts and collagen fibres.
• It is the main tissue that provides structural frame to the body.
• The bone cells (Osteocytes) are present in the spaces called lacunae.
• Bones are present in the limbs and form main skeletal framework of the body.

Cartilage:

• Cartilage is pliable, flexible and resist compression.
• Its matrix is rich in protein called chondrin and sugars.
• It is present in bones of the vertebral column, limbs and hands in adults.
• Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix secreted by them.
• Cartilage is present in the tip of nose, outer ear joints, between adjacent bones of the vertebral column.

Question 6.
Explain the basic criteria for classification of permanent tissue in plants.
Answer:
The permanent tissues are classified on the basis of the following criteria:

1. Simple (made of one type of cell) or complex (made of more than one type of cells)
2. Cell wall: Thin or thick
3. Type of cell: living or dead
4. Type of function the tissue performs: epidermis is protective, parenchyma is packing or supportive tissue and sclerenchyma makes up conducting tissue.

Question 7.
Identify the given two slides A and B as a parenchyma or sclerenchyma. Sclerenchyma can be identified by which characteristic?

Answer:
Slide A is parenchyma and Slide B is sclerenchyma.
Sclerenchyma can be identified by the type of cells which are long and narrow as the walls are thickened due to presence of lignin.

Question 8.
(i) Identify the given figures.

(ii) Give any two major differences between the structures identified.
(iii) Describe the role performed by these two in the plant body.
Answer:
(i) Structure (A) is a tracheid and structure (B) is a vessel.

(ii) Tracheid:

• Tracheids are elongated or tube-like cells with thick and lignified walls and tapering ends.
• They are in the form of single cells.

Vessel:

• Vessel is a long cylindrical tube-like structure made up of many cells called vessel members.
• They are composed of a number of cells fused together.

(iii) Tracheids and vessels help in vertical transport of water and minerals in the plants. They also help to provide mechanical strength to the plants.

Question 9.
Draw a well labelled diagram of cardiac muscle found in the human body. Write two differences between striated and smooth muscles.
Answer:

Question 10.
Draw a labelled diagram of unstriated muscle tissue and mention its occurrence, features and functions.
Answer:

(i) The cells are long and spindle-shaped.
(ii) They do not have striations.
(iii) Involuntary in nature as they are not under control of our will.
(iv) The cells of smooth muscles are uninucleate.
(v) Smooth muscle fibres are present in iris of the eye, ureters, blood vessels, alimentary canal and bronchi of lungs.

Question 11.
Name the kinds of muscles found in your limbs and lungs. How do they differ from each other structurally and functionally?
Answer:
Striated muscle fibres are found in limbs whereas smooth muscle fibres are present in lungs. The differences in their structure are:
(i) Striated muscle fibres have alternate light and dark bands which are not present in the smooth muscle fibres.

(ii) Striated muscle fibres are cylindrical and multinucleate whereas the smooth muscle fibres are spindle-shaped and uninucleate.

(iii) Striated muscles are voluntary in nature (under control of our will) whereas the smooth muscle fibres are involuntary in nature (not under control of our will).

Question 12.
What are neurons? Where are they found in the body? What function do they perform in the body of an organism?
Answer:
The cells of nervous tissue are called nerve cells or neurons. Neurons are the structural and functional unit of the nervous system. They are found in the brain, spinal cord and nerves.

Their functions are:

• They are highly specialised for transmitting the stimulus from one place to another within the body on being stimulated.
• They help to coordinate the various functions of the body.

Question 13.
Animals of colder regions and fishes of cold water have thicker layer of subcutaneous fat. Describe why?
Answer:
The thick layer of subcutaneous fat acts as insulator and prevents the heat of the body to escape out. The layer of fat acts as a subcutaneous insulation of body for thermoregulation.

Question 14.
Match the column (A) with the column (B).

Answer:
(a) (v)
(b) (iv)
(c) (iii)
(d) (i)
(e) (ii)
(f) (vi)

Question 15.
Match the column (A) with the column (B).

Answer:
(a) (i)
(b) (c)
(d) (iii)
(e) (iv)

Question 16.
If a potted plant is covered with a glass jar, water vapours appear on the wall of glass jar. Explain why.
Answer:
The water is lost by the plant in the form of water vapour due to the process of transpiration. These water vapours appear on the wall of the glass jar.

Question 17.
Name the different components of xylem and draw a living component.
Answer:
Xylem consists of four elements which are:
(a) tracheids
(b) vessels
(c) xylem parenchyma
(d) xylem fibres
The only living component of xylem is xylem parenchyma whose basic structure is shown below:

Question 18.
Draw and identify different elements of phloem.
Answer:
Phloem has four elements called sieve tubes, companion cells, phloem fibres and the phloem parenchyma.

Question 19.
Write true (T) or false (F).
(a) Epithelial tissue is protective tissue in animal body.
(b) The lining of blood vessels, lung alveoli and kidney tubules are all made up of epithelial tissue.
(c) Epithelial cells have a lot of intercellular spaces.
(d) Epithelial layer is permeable layer.
(e) Epithelial layer does not allow regulation of materials between body and external environment.
Answer:
(a) True
(b) True
(c) False
(d) True
(e) False

Question 20.
Differentiate between voluntary and involuntary muscles. Give one example of each type.
Answer:
Voluntary muscles are present in our limbs as skeletal muscles and can be moved by our conscious will whenever we want. Involuntary muscles cannot function on their own. They cannot be controlled by our will or desire. The cardiac muscle and the smooth muscles are involuntary in nature.

Question 21.
Differentiate the following activities on the basis of voluntary (V) or involuntary (IV) muscles.
(a) Jumping of frog
(b) Pumping of the heart
(c) Writing with hand
(d) Movement of chocolate in your intestine
Answer:
(a) (V)
(b) (IV)
(c) (V)
(d) (IV)

Question 22.
Fill in the blanks.
(a) Lining of blood vessels is made up of _______
(b) Lining of small intestine is made up of _______
(c) Lining of kidney tubules is made up of _______
(d) Epithelial cells with cilia are found in _______ of our body.
Answer:
(a) Squamous epithelium
(b) Columnar epithelium
(c) Cuboidal epithelium
(d) Respiratory tract

Question 23.
Water hyacinth floats on water surface. Explain.
Answer:
The parenchyma present in the swollen petiole of water hyacinth is called aerenchyma which has large cavities to provide buoyancy and help them float on the water surface.

Question 24.
Which structure protects the plant body against the invasion of parasites?
Answer:
The epidermis of plants has thick cuticle and waxy substances to prevent the invasion of parasites.

Question 25.
Fill in the blanks.
(а) Cork cells possesses _______on their walls that makes it impervious to gases and water.
(b) _______ have tubular cells with perforated walls and are living in nature.
(c) Bone possesses a hard matrix composed of and _______ and _______
Answer:
(a) suberin
(b) sieve tubes
(c) calcium and phosphorus

Question 26.
Why is epidermis important for the plants?
Answer:
The outermost layer of cells covering an organism is called epidermis. It is usually made up of a single layer of cells and gives protection.

The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.

The stomata present on the epidermis of leaves helps in gaseous exchange and the loss of water vapour by transpiration.

The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

Question 27.
Fill in the blanks.
(a) _______ are forms of complex tissue.
(b) _______ have guard cells.
(c) cells of cork contain a chemical called _______
(d) Husk of coconut is made of _______ tissue.
(e) _______ gives flexibility in plants.
(f) _______ and _______ are both conducting tissues.
(g) Xylem transports and _______ and _______ from soil.
(h) Phloem transport from _______ and _______ to other parts of the plant.
Answer:
(a) Xylem and phloem
(b) Stomata
(c) suberin
(d) sclerenchyma
(e) Collenchyma
(g) water; minerals
(h) Food; leaf

Tissues Class 9 Extra Questions Long Answer Type

Question 1.
Differentiate between
(i) Xylem and phloem
(ii) Vessel and sieve tube
(iii) Tracheid and vessel
Answer:
(i) Xylem and phloem –
Xylem:

• Xylem consists of tracheids, vessels, xylem parenchyma and xylem fibres.
• All the cells of xylem except the xylem parenchyma are dead.
• Xylem helps to transport water and minerals.
• The transport is unidirectional through xylem.

Phloem:

• Phloem has four elements called sieve tubes, companion cells, phloem fibres and the phloem parenchyma.
• All cells of phloem are living except the phloem fibres.
• Phloem transports food from leaves to other parts of the plant.
• The transport is bidirectional through the phloem.

(ii) Vessel and sieve tube –
Vessel:

• They are tubular structures having a hollow lumen and composed of dead cells.
• Vessel helps to conduct water and minerals in plants.
• The walls of vessels are lignified.
• They also provide mechanical strength to the plants.
• Their end walls are completely dissolved.

Sieve Tube:

• They are tubular structures having vacuolated cytoplasm and composed of living cells.
• They help to transport food from leaves to other parts of the plant.
• Their walls are not lignified.
• They do not provide mechanical strength to the plants.
• Their end walls have perforations in form of sieve plate.

(iii) Tracheid and vessel _
Tracheid:

• Tracheids are elongated or tube-like cells with thick and lignified walls and tapering ends.
• They are in the form of single cells.
• The inner layers of the cell walls are more thickened.
• They have narrow lumen.
• They have pointed ends.

Vessel:

• Vessel is a long cylindrical tube-like structure made up of many cells called vessel members.
• They are composed of a number of cells fused together.
• Their walls are less thickened.
• They have wide lumen.
• They have blunt ends.

Question 2.
Differentiate between striated, unstriated and cardiac muscle fibres.
Answer:

Question 3.
(i) What is nervous tissue?
(ii) Draw a well labelled diagram of neuron. (Label any 4 parts)
Answer:
Nervous tissue is a tissue made of neurons. It is divided into two parts: the central nervous system (CNS) consisting of the brain and spinal cord; and the peripheral nervous system (PNS) which regulates and controls the various functions and activities of the body.

Question 4.
Write the differences between animal tissue and plant tissue.
Answer:
Plant Tissue:

• The tissue is well differentiated into meristematic tissue and permanent tissue.
• The tissue can grow throughout life due to activity of meristematic tissue.
• They are autotrophic in nature.
• The tissue has more amount of dead tissue which provides mechanical strength to the plants.
• The tissue organisation is comparatively simple.

Animal Tissue:

• The tissue is not much differentiated like the plant tissue.
• The tissue does not show growth throughout life.
• They are heterotrophic in nature.
• The tissue has more amount of living tissue than dead tissue.
• The tissue is complex as it is organised into organs and organ systems.

Question 5.
Write a note on the protective tissue in plants. (Give appropriate diagram also)
Answer:
The protective tissues in plants are epidermis and the cork.
(i) Epidermis: The outermost layer of cells covering an organism is called epidermis. It is usually made up of a single layer of cells and gives protection.

The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin to prevent water loss.

The epidermis of leaves have small pores called stomata which are enclosed by two kidney-shaped cells called guard cells. Stomata help in gaseous exchange and transpiration.

The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

(ii) Cork: A strip of secondary meristem replaces the epidermis of the older stem and cuts off cells towards compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them impervious to gases and water.

Question 6.
Explain the significance of the following:
(i) Hair-like structures on epidermal cells.
(ii) Epidermis has thick waxy coating of cutin in desert plants.
(iii) Small pores in epidermis of leaf.
(iv) Numerous layers of epidermis in cactus.
(v) Presence of a chemical suberin in cork cells.
Answer:
(i) To increase the total absorptive surface area for absorption of water.
(ii) To prevent water loss by transpiration and protection from pathogens.
(iii) To help in gaseous exchange and transpiration.
(iv) To prevent water loss by transpiration.
(v) To make tissue impervious to gases and water.

Question 7.
Differentiate between sclerenchyma and parenchyma tissues. Draw well labelled diagram.
Answer:
Parenchyma:

• Cells are thin walled and thickened with cellulose.
• It is made up of living cells.
• Cells are usually loosely pac ked with large intercellular spaces.
• Helps to store nutrients and water in stem and roots.
• It is called chlorenchyma if it contains chlorophyll and performs photos ynthesis. The parenchyma of
• aquatic plants have large cavities to provide buoyancy to the plants to help them float, it is then called aerenchyma.

Sclerenchyma:

• Cells are thick and thickened with lignin.
• This tissue is made up of dead cells.
• There are no intercellular spaces between the cells.
• Provides strength to the various parts of the plant.
• The cells are long and narrow, make the plant hard and stiff. This tissue provides strength to the plants and is present in stems, around vascular bundles, in the veins of leaves and in the hard covering of seeds and nuts.

Question 8.
Describe the structure and function of different types of epithelial tissues. Draw diagram of each type of epithelial tissue.
Answer:
Epithelial tissues are the covering or protective tissues and cover most organs and cavities in the animal body. These cells are tightly packed, form a continuous sheet and are almost without any intercellular spaces between them. E.g., skin, the lining of the mouth, the lining of blood vessels, lung alveoli and kidney tubules are all made of epithelial tissue.

All epithelium is usually separated from the underlying tissue by an extracellular fibrous basement membrane. The types of epithelium on the basis of their structure and functions are:

(a) Squamous epithelium: Consists of flattened cells. Present in oesophagus and lining of the mouth. Skin epithelial cells are arranged in many layers to prevent wear and tear and are called stratified squamous epithelium.

(b) Columnar epithelium: Has tall or ‘pillar-like’ cells. It forms the inner lining of the intestine.

(c) Cuboidal epithelium: Has cube-shaped cells. It forms the lining of kidney tubules and ducts of salivary glands, where it provides mechanical support.

(d) Ciliated epithelium: Have cilia on the outer surfaces of epithelial cells. The cilia can move and their movement pushes the mucus in the respiratory tract forward to clear it.

(e) Glandular epithelium: Has gland cells which secrete substances at the epithelial surface.

Question 9.
Give reasons for
(a) Meristematic cells have a prominent nucleus and dense cytoplasm but they lack vacuole.
(b) Intercellular spaces are absent in sclerenchymatous tissues.
(c) We get a crunchy and granular feeling, when we chew pear fruit.
(d) Branches of a tree move and bend freely in high wind velocity.
(e) It is difficult to pull out the husk of a coconut tree.
Answer:
(a) Because the meristematic cells are actively dividing cells and there is no need of storage.
(b) Because they have a thick deposition of lignin in them.
(c) Due to the presence of stone cells (sclerenchyma) in the pear fruit.
(d) Due to the presence of collenchyma which provides flexibility to the various parts of the plant.
(e) Due to the sclerenchyma present in the husk of the coconut.

Question 10.
List the characteristics of cork. How are they formed? Mention their role.
Answer:
The characteristics of cork are:

• Cells of cork are dead at maturity.
• These cells are compactly arranged.
• Cells do not possess intercellular spaces.
• Cells possess a chemical substance suberin in their walls.
• They are several layers thick.

A strip of secondary meristem replaces the epidermis of the older stem and cuts off the outside cells to form a several-layer thick cork or the bark of the tree. Cells of cork are dead, compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them protective in function and impervious to gases and water.

Question 11.
(a) Differentiate between meristematic and permanent tissues in plants.
(b) Define the process of differentiation.
(c) Name any two simple and two complex permanent tissues in plants.
Answer:
(a) Meristematic tissue:

• This tissue consists of cells which continuously divide to produce new cells.
• They are located at specific regions of the plant, i.e., apical, lateral and intercalary.
• The cells of this tissue are very active, lack vacuoles, have dense cytoplasm, thin cell walls and prominent nuclei.

Permanent tissue:

• Consists of cells which have taken up a specific role and lost the ability to divide.
• They are distributed throughout the plant body.
• They are vacuolated, vary in shape and size. Their cell wall may be thick.

(b) The process of taking up a permanent shape, size, and a function by the cells is called differentiation.

(c) Simple: Parenchyma/collenchyma/sclerenchyma Complex: Phloem/xylem

Tissues Class 9 Extra Questions HOTS

Question 1.
The walls of the sclerenchymatous cells are thickened and have narrow lumen. Which substance thickens it and what is its role?
Answer:
The walls of the sclerenchymatous cells are thickened due to presence of lignin. It helps in providing mechanical strength to the various parts of the plant.

Question 2.
Which type of muscle fibres will contract to remove your hands instantly when you touch a hot object?
Answer:
Striated muscle fibres will contract to remove our hands instantly when we touch a hot object.

Question 3.
Which tissue helps the leaves of lotus plant to float on water? Why?
Answer:
Aerenchyma helps the leaves of lotus plant to float on water. Aerenchyma has large cavities to provide buoyancy to the parts of aquatic plants.

Question 4.
A tissue present in plants helps in storing food and sideways conduction of water. Identify the type of tissue.
Answer:
The tissue is xylem parenchyma.

Question 5.
Which tissue enables the heart to pump blood to various parts of the body? Why?
Answer:
The cardiac muscles help the heart to pump blood to various parts of the body as they show rhythmic contraction and relaxation throughout their life.

Question 6.
What will be the consequence of
(i) removal of blood platelets from blood?
(ii) removal of cutin from the layer of epidermis?
Answer:
(i) Removal of blood platelets from blood will inhibit clotting of blood if an injury occurs and the person may bleed to death.

(ii) Removal of cutin would increase the amount of water loss from the leaves of the plants.

Question 7.
Some actions of our body are under our control but many of them are not under our control. Why is it so?
Answer:
The actions of our body are controlled by our muscles. The voluntary actions are under the control of our will and are caused by the activity of striated muscles, eg., movement of our limbs. The involuntary . actions are not under the control of our will and are performed by the smooth muscles, e.g., the activity of bronchi of lungs. Even the activity of cardiac muscles which helps in the rhythmic contraction and relaxation of heart are involuntary in nature.

Question 8.
Which kind of meristem can help grasses to regenerate parts removed by the grazing herbivores?
Answer:
Intercalary meristem can help grasses to regenerate parts removed by the grazing herbivores.

Question 9.
Name the tissue which replaces the epidermal tissue in older stem and is rich in suberin. What is the function of suberin?
Answer:
Cork is the tissue which replaces the epidermal tissue in older stem and is rich in suberin. Suberin present in the walls of cork cells makes them impervious to gases and water.

Question 10.
The process of transpiration does not occur properly when the leaves are covered by a layer of oily substance. Why? Which other functions will get affected due to this covering?
Answer:
The layer of oily substance will close the stomata present in leaves and this would decrease the rate of transpiration. The rate of exchange of gases decreases and consequently the rate of photosynthesis would also decrease.

Tissues Class 9 Extra Questions Value Based (VBQs)

Question 1.
Raman got injured while playing football. His injured leg started bleeding and his friends immediately rushed to take him to the doctor to give him first aid. The blood flowing from the wound stopped after some time and the doctor applied antiseptic on the wound.
(i) Why did the blood stop flowing after some time from the wound?
(ii) What kind of tissue is blood? Why?
(iii) What values are shown by Raman’s friends?
Answer:
(i) The blood stopped flowing after some time from the wound as the blood platelets present in blood helped in clotting of blood.

(ii) Blood is a connective tissue. Blood is considered as a type of connective tissue as they have the same origin as other types of
connective tissue and helps to connect the different parts of the body.

(iii) The values shown by Raman’s friends are presence of mind, helpful and a caring nature.

Question 2.
During a sports event, Shivani suffered a sprain due to which she was not able to run. Her teacher gave her support and told her that it was due to a ligament tear. She also called the doctor to give treatment to Shivani.
(i) What is a ligament? What kind of tissue is it?
(ii) Which type of fibrous tissue has great strength, limited flexibility and is similar to ligament?
(iii) What values are shown by Shivani’s teacher?
Answer:
(i) Ligament is the connective tissue which connects two bones. It is a kind of connective tissue.

(ii) Tendon is a fibrous tissue that has great strength, limited flexibility and is similar to ligament.

(iii) The values shown by her teacher are knowledge, scientific approach and a caring nature.

Question 3.
Rishi brought an aquatic plant which was floating on the surface to the science laboratory of water. He cut a section of the leaf of the plant and saw a tissue with lot of air cavities in it. He went to his teacher and discussed about the role of the air cavities in the leaves of the aquatic plant.
(i) Which type of tissue present in plants has air cavities?
(ii) What is the role of large air cavities in the leaves of such plants?
(iii) What values are shown by Rishi?
Answer:
(i) Aerenchyma is the tissue present in the plants and has large air cavities.

(ii) The large air cavities in the leaves of such plants help in providing buoyancy to the leaves to help them float on water.

(iii) Rishi shows a scientific attitude, inquisitive nature and empirical approach.

In this page, we are providing The Fundamental Unit of Life Class 9 Extra Questions and Answers Science Chapter 5 pdf download. NCERT Extra Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life with Answers will help to score more marks in your CBSE Board Exams.

Class 9 Science Chapter 5 Extra Questions and Answers The Fundamental Unit of Life

Extra Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life with Answers Solutions

The Fundamental Unit of Life Class 9 Extra Questions Very Short Answer Type

Question 1.
What process is involved in the movement of O₂ in and out of the cell?
Answer:
Diffusion.

Question 2.
Which process is involved in the movement of water from outside into the cell?
Answer:
Osmosis.

Question 3.
Mention the process by which CO₂ and water move in and out of the cell.
Answer:
Movement of CO₂ occurs by diffusion whereas that of water occurs by osmosis.

Question 4.
Who discovered cells in living organisms? Give an example of a unicellular organism.
Answer:
A.V. Leeuwenhoek observed the.cells in living organisms. Example of a unicellular organism is Amoeba, Paramoecium, etc.

Question 5.
Give two examples of prokaryotic organisms.
Answer:
All the bacteria and cyanobacteria are prokaryotic. Example: Nostoc, Oscillatoria, etc.

Question 6.
Name the plastid which stores starch, oils and protein granules.
Answer:
Leucoplast

Question 7.
List the constituents of plasma membrane.
Answer:
The plasma membrane is composed of lipids and proteins.

Question 8.
Where do lipids and proteins constituting the cell membrane get synthesised?
Answer:
The lipids are synthesised in smooth endoplasmic reticulum (SER) and the proteins are synthesised in the rough endoplasmic reticulum (RER).

Question 9.
Name two unicellular organisms.
Answer:
Amoeba and Paramoecium are unicellular organisms.

Question 10.
Name the process by which unicellular freshwater organisms and most plant cells tend to . gain water.
Answer:
Endosmosis.

The Fundamental Unit of Life Class 9 Extra Questions Short Answer Type 1

Question 1.
Why is plasma membrane called selectively permeable membrane?
Answer:
Plasma membrane is called selectively permeable because it prevents the entry of some substances and allows the entry and exit of only some substances.

Question 2.
Define osmosis.
Answer:
The spontaneous movement of water molecules from a region of its high concentration to the region of its low concentration through a selectively permeable membrane is called osmosis.

Question 3.
What is membrane biogenesis?
Answer:
Some proteins and lipids made by smooth endoplasmic reticulum (SER) help in building the cell membrane of the cell and this process of making of plasma membrane is known as membrane biogenesis.

Question 4.
Why are mitochondria are referred to as strange organelles?
Answer:
Mitochondria are referred to as strange organelles as they have their own DNA and ribosomes so and can make some of their own proteins.

Question 5.
What is the function of Golgi body?
Answer:
The functions of Golgi body are: storage, modification and packaging of products in vesicles and the formation of lysosomes.

Question 6.
(i) In what form do mitochondria release energy? Write its full form.
(ii) The inner membrane of mitochondria is deeply folded. What is the advantage of these folds?
Answer:
(i) The energy is released by mitochondria in form of ATP. Full form of ATP is adenosine triphosphate.

(ii) The folds present on the inner membrane of mitochondria help to provide a large surface area for the ATP generation reactions.

Question 7.
Give three differences between plasma membrane and the cell wall.
Answer:
Plasma Membrane:

• It is the outermost covering of the cell which separates the contents of the cell from its external environment.
• It is mainly composed of lipids and proteins.
• It is living, thin and elastic.

Cell Wall:

• It is a rigid outer covering which lies outside the plasma membrane.
• It is made of cellulose which provides structural strength to plants.
• It is non-living and rigid.

Question 8.
Differentiate between nucleus and nucleoid.
Answer:
Nucleus:

• It has a well defined nuclear membrane around it.
• It has chromatin network and contains more than one chromosome.

Nucleoid:

• It lacks a nuclear membrane.
• It has a single chromosome only.

Question 9.
Write two statements to show that lysosomes are called as suicidal bags of the cell.
Answer:
Lysosomes are called as the suicidal bags of the cell as:

1. They contain hydrolytic enzymes which can breakdown the organic material.
2. If the cell gets damaged during disturbance in cellular metabolism, the lysosomes may burst and its enzymes digest their own cell.

Question 10.
Describe the structure of mitochondria with special reference to its membrane covering.
Answer:
Mitochondrion is a double-membrane structure whose outer membrane is very porous while the inner membrane is deeply folded to form cristae. Cristae are folds which create a large surface area for ATP- generating chemical reactions.

The Fundamental Unit of Life Class 9 Extra Questions Short Answer Type 2

Question 1.
Describe an activity to demonstrate endosmosis and exosmosis. Draw a diagram also.
Answer:
Put dried raisins or apricots in plain water and leave them for some time. Then place them into a concentrated solution of sugar or salt. You will observe the following:

(a) Each of the raisins or apricots gains water and swells when placed in water.
Reason: The raisins or apricots swell up as water moves inside them from outside because the water concentration is less inside the cell as compared to the solution outside. Hence, water moves inside the cell by endosmosis.

(b) However, when placed in the concentrated solution they lose water, and consequently shrink.
Reason: The raisins or apricots shrink as water moves outside from them because the water concentration is more inside the cell as compared to the solution outside. Hence, water moves out of the cell by exosmosis.

Question 2.
Differentiate between diffusion and osmosis. Write any two examples where a living organism uses osmosis to absorb water.
Answer:
Diffusion – The movement of a substance from a region of its high concentration to the region of its low concentration is called diffusion.

Osmosis – The spontaneous movement of water molecules from a region of its high concentration to the region of its low concentration through a selectively permeable membrane is called osmosis.

Examples where a living organism uses osmosis to absorb water are:

• Roots of plants absorb water by osmosis.
• Unicellular organisms like Amoeba absorb water by osmosis.

Question 3.
Differentiate between a prokaryotic and eukaryotic cell.
Answer:
Prokaryotic cell:

• Organisms whose cells lack a well defined nuclear membrane.
• They lack membrane bound cell organelles.
• Size is generally small (1-10 pm).
• Have a single chromosome.

Eukaryotic cell:

• Organisms with cells having a well defined nuclear membrane.
• They have membrane bound cell organelles.
• Size is generally large (5-100 pm).
• Have more than one chromosome.

Question 4.
Which organisms are called as the (i) powerhouse of the cell (ii) suicide bags of the cell (iii) kitchen of the cell?
Answer:

• Mitochondria
• Lysosomes
• Chloroplasts

Question 5.
(a) What would happen to the life of the cell if there was no Golgi apparatus?
(b) Which cell organelle detoxifies poisons and drugs in the liver of vertebrates?
Answer:
(a) If there was no golgi apparatus, then the following processes carried out by it would get affected:

• The storage, modification and packaging of products in vesicles.
• The packaging and dispatch of the material synthesised near the ER to various targets inside and outside the cell.
• The formation of lysosomes.
• The formation of cell plate during cell division.

So, in the absence of the Golgi apparatus, most of the functions of the cell would get disrupted.

(b) Smooth endoplasmic reticulum helps to detoxify poisons and drugs in the liver of vertebrates.

Question 6.
What are the functions of the Golgi apparatus?
Answer:
The Golgi apparatus is an important cell organelle as it performs several functions in the cell.

1. Its functions include the storage, modification and packaging of products in vesicles.
2. Golgi apparatus packages and dispatches the material synthesised near the ER to various targets inside and outside the cell.
3. It is also involved in the formation of lysosomes.

Question 7.
What types of enzymes are present in the lysosomes? What is their function? Which cell organelle manufactures these enzymes?
Answer:
Lysosomes are membrane-bound sacs filled with hydrolytic and digestive enzymes. They help to keep the cell clean by digesting any foreign material as well as worn-out cell organelles. Rough endoplasmic reticulum (RER) manufactures these enzymes.

Question 8.
What are chromoplasts and leucoplasts? Give an example of chromoplast which has green pigment.
Answer:
The coloured plastids which have pigments of different colours are called chromoplasts. The colourless plastids which help to store starch, oils or protein granules are called leucoplasts. The green coloured plastid is called chloroplast and it contains a green coloured pigment called chlorophyll.

Question 9.
Write the functions of:
(i) Inner membrane of mitochondria
(ii) Nucleus of the cell
(iii) Ribosomes present in active cells.
Answer:
The functions are:
(i) Inner membrane of mitochondria are the site for the formation of ATP.

(ii) Nucleus of the cell are the control centre of the cell as it controls all the activities of the cell, plays a central role in cellular reproduction and contains DNA which helps to transfer the genetic information from parents to their offsprings.

(iii) Ribosomes present in active cells are the sites for protein synthesis.

Question 10.
What are the different types of endoplasmic reticulum? Write the functions of each.
Answer:
There are two types of endoplasmic reticulum – Rough endoplasmic reticulum and smooth endoplasmic reticulum. The functions of these are:

Rough endoplasmic reticulum – It looks rough as it has particles called ribosomes attached to its surface. Ribosomes are the site of protein synthesis.

Smooth endoplasmic reticulum:

• It helps in the manufacture of fat molecules, or lipids, important for cell function.
• Some proteins and lipids made by SER help in building the cell membrane and this process is known as membrane biogenesis.
• Helps in detoxifying many poisons and drugs in the liver cells of the group of vertebrates.

Question 11.
Draw a neat and labelled diagram of a typical prokaryotic cell.
Answer:

Question 12.
Why do plant cells possess large sized vacuole?
Answer:
Plant cells possess large sized vacuoles as the vacuoles help to store many important substances. They contain cell sap that gives turgidity and rigidity to the plant cell.

Question 13.
Describe the structural features of cell membrane and cell wall. Why is cell membrane called selectively permeable membrane?
Answer:
Cell wall:

• It is a rigid outer covering which lies outside the plasma membrane.
• It is made of cellulose which provides structural strength to plants.
• The shrinkage or contraction of the contents of the cell away from the cell wall when a living plant cell loses water through osmosis is known as plasmolysis.

Cell membrane:

• It is the outermost covering of the cell.
• It separates the contents of the cell from its external environment.
• It is mainly composed of lipids and proteins.

The cell membrane is called selectively permeable as it permits the entry and exit of only some materials in and out of the cell. It prevents the movement of the contents of the cell out of the cell.

Question 14.
Explain in detail what do you know about the structure of the nucleus.
Answer:
The nucleus was discovered by Robert Brown in 1831. The structure and the features of nucleus are:

• It is a dark coloured, spherical or oval, dot-like structure near the centre of each cell.
• It is the control centre of the cell as it controls all the activities of the cell.
• It has a double layered covering called nuclear membrane.
• The nuclear membrane has pores which allow the transfer of materials from inside the nucleus to the cytoplasm.
• The nucleus plays a central role in cellular reproduction (process by which a single cell divides and forms two new cells).
• Nucleus along with the environment directs the chemical activities of the cell to determine the way the cell will develop and the form it will exhibit at maturity.
• Nuclear region of the cell may be poorly defined due to the absence of a nuclear membrane in some organisms like bacteria. Such an undefined nuclear region containing only nucleic acids is called a nucleoid.

Question 15.
Differentiate between rough and smooth endoplasmic reticulum. How is endoplasmic reticulum important for membrane biogenesis?
Answer:
The differences between smooth endoplasmic reticulum and the rough endoplasmic reticulum are:

Smooth Endoplasmic Reticulum (SER):

• SER looks smooth as it does not have ribosomal particles on the surface.
• SER helps in the manufacture of lipids and fat molecules and also in detoxifying many poisons and drugs in the liver cells of the group of vertebrates.

Rough Endoplasmic Reticulum (RER):

• RER appears rough as it has particles of ribosomes on the surface.
• Ribosomes present on RER are the sites of protein synthesis.

Ribosomes are the sites of protein synthesis in all active cells. Endoplasmic reticulum helps in transporting these proteins to various places. Some proteins and lipids made by SER help in building the cell membrane, this process is known as membrane biogenesis.

Question 16.
Describe the phenomenon of membrane biogenesis. Give one function of ER.
Answer:
Some proteins and lipids made by SER help in building the cell membrane and this process is known as membrane biogenesis.

ER serves as channels for the transport of materials (especially proteins) between various regions of the cytoplasm or between the cytoplasm and the nucleus. It also functions as a cytoplasmic framework providing a surface for some of the biochemical activities of the cell.

Question 17.
(i) Why are lysosomes known as ‘scavengers of the cell’?
(ii) Lysosomes are self-destructive. (True/False). Give reason.
Answer:
(i) The lysosomes are waste disposal system of the cell as they remove the debris and help to keep the cell clean by digesting any foreign material as well as worn-out cell organelles. Thus, they are known as the ‘scavengers of the cell’.

(ii) It is true that the lysosomes are self-destructive. They contain digestive enzymes. If the cell gets damaged during disturbance in cellular metabolism, the lysosomes may burst and its enzymes digest their own cell.

Question 18.
Name the organelle of the cell, which has membrane bound sac filled with powerful digestive enzymes. Write any four common functions it performs inside the cell.
Answer:
Lysosome is a membrane bound sac filled with powerful digestive enzymes.

The functions of lysosomes are:

• They are waste disposal system of the cell as they help to keep the cell clean by digesting any foreign material as well as worn-out cell organelles.
• They have digestive enzymes which are capable of breaking down all organic materials.
• If any disturbance occurs in the cellular metabolism or the cell gets damaged, the lysosomes burst open and its digestive enzymes digest all the contents of the cell. So, they are also called as the ‘suicide bags’ of the cell.
• During starvation they digest the food contents of the cell and provide energy to the cell.

Question 19.
State reasons for the following:
(i) Mitochondria are known as the powerhouse of the cell.
(ii) Plastids are able to make their own protein.
(iii) Plant cell shrinks when kept in hypertonic solution.
Answer:
(i) Mitochondria are known as the powerhouse of the cell as they are the site of storage of ATP which helps to provide energy for all the activities of the cell.

(ii) Plastids are able to make their own protein as they have their own DNA and ribosomes.

(iii) Hypertonic solution has a lower concentration of water than the cell. Due to this, the water moves out of the plant cell by exosmosis and results in shrinkage of the plant cell.

Question 20.
List the specific functions of the following:
(i) Endoplasmic reticulum
(ii) Golgi apparatus
(iii) Lysosomes
(iv) Plastids
(v) Mitochondria
(vi) Vacuoles.
Answer:
The specific functions are:
(i) Endoplasmic reticulum: Site of protein synthesis.

(ii) Golgi apparatus: Storage, modification and packaging of products in vesicles.

(iii) Lysosomes: Waste disposal system of the cell as they help to keep the cell clean by digesting any foreign material as well as worn-out cell organelles.

(iv) Plastids: Green coloured plastids are the site of photosynthesis.

(v) Mitochondria: They are called the powerhouse of the cell as they store energy required for the activities of the cell in the form of ATP.

(vi) Vacuoles: Provide turgidity and rigidity to the cell in plant cells. They also help in digestion of food and expulsion of excess water and waste from the cell in unicellular organisms like Amoeba.

Question 21.
Which types of plastids stores starch, oil and proteins?
Answer:
The leucoplasts help to store starch, oil and proteins in the amyloplasts, elaioplasts and aleuroplasts respectively.

Question 22.
How many membranes are present in mitochondria? Give the characteristic features of these membranes. What is the advantage of such features?
Answer:
Mitochondrion is a double-membrane structure whose outer membrane is very porous while the inner membrane is deeply folded to form cristae. The porous membrane helps in getting oxygen and food. Cristae present on the inner membrane are the folds which create a large surface area for ATP- generating chemical reactions.

Question 23.
Why are lysosomes known as ‘suicide-bags’ of a cell?
Answer:
Lysosomes are also known as the ‘suicide bags’ of a cell because if the cell gets damaged during disturbance in cellular metabolism, the lysosomes may burst and its enzymes digest their own cell.

Question 24.
Do you agree that “A cell is a building unit of an organism”. If yes, explain why?
Answer:
Yes, a cell is the building unit or the fundamental unit of an organism as the cells get organised to form tissues which in turn get organised into organs and further into organ system which organise to form an organism. It can be represented as:
Cell → tissue → organ → organ system organism

Question 25.
Why does the skin of your finger shrink when you wash clothes for a long time?
Answer:
The skin of our finger shrinks when we wash clothes for a long time because soap solution is a hypertonic solution i.e., very concentrated solution, so water moves out of our finger cells by the process of osmosis (called as exosmosis).

Question 26.
Why is endocytosis found in animals only?
Answer:
The plant cells have a rigid structure due to the presence of cell wall around their cell membrane whereas in the animals the cell membrane does not have a cell wall. So, endocytosis can occur only in the animal cells.

Question 27
A person takes concentrated solution of salt, after sometime, he starts vomiting. What is the phenomenon responsible for such situation? Explain.
Answer:
A concentrated solution of salt is hypertonic solution and causes dehydration. Water moves out of the cells and exosmosis in the parts of stomach and intestine make the person feel uncomfortable. Excessive dehydration results in anti-peristalsis due to which person starts vomiting.

Question 28.
Name any cell organelle which is non membranous.
Answer:
Ribosomes are non membranous organelles.

Question 29.
We eat food composed of all the nutrients like carbohydrates, proteins, fats, vitamins, minerals and water. After digestion, these are absorbed in the form of glucose, amino acids, fatty acids, glycerol, etc. What mechanisms are involved in absorption of digested food and water?
Answer:
Absorption of the nutrients takes place by diffusion and that of water occurs by the process of osmosis.

Question 30.
If you are provided with some vegetables to cook. You generally add salt into the vegetables during cooking process. After adding salt, vegetables release water. What mechanism is responsible for this?
Answer:
Adding salt into vegetables creates a hypertonic medium around them which results in exosmosis. Exosmosis causes release of water form vegetables.

Question 31.
If cells of onion peel and RBC are separately kept in hypotonic solution, what among the following will take place? Explain the reason for your answer.
(a) Both the cells will swell.
(b) RBC will burst easily while cells of onion peel will resist the bursting to some extent.
(c) a and b both are correct.
(d) RBC and onion peel cells will behave similarly.
Answer:
(b) RBC will burst easily while cells of onion peel will resist the bursting to some extent. This happens because the cells of onion peel have cell wall which prevents them from bursting. The RBC, being animal cells are devoid of cell wall and burst easily in hypotonic solution.

Question 32.
Bacteria do not have chloroplast but some bacteria are photo-autotrophic in nature and perform photosynthesis. Which part of bacterial cell performs this?
Answer:
The photo-autotrophic bacteria have small vesicles associated with plasma membrane which help in the process of photosynthesis.

Question 33.
Match the following A and B

Answer:
(a) (iv)
(b) (v)
(c) (iii)
(d) (i)
(e) (ii)

Question 34.
Write the name of different plant parts in which chromoplast, chloroplast and leucoplast are present. Chromoplasts are present in the flower and fruit.
Answer:
Chloroplasts are present in the leaves of the plant. Leucoplasts are present in the root of the plant.

Question 35.
Name the organelles which show the analogy written as under
(a) Transporting channels of the cell _______
(b) Powerhouse of the cell _______
(c) Packaging and dispatching unit of the cell _______
(d) Digestive bag of the cell _______
(e) Storage sacs of the cell _______
(f) Kitchen of the cell _______
(g) Control room of the cell _______
Answer:
(a) Transporting channels of the cell _______ Endoplasmic reticulum
(b) Powerhouse of the cell _______ Mitochondria
(c) Packaging and dispatching unit of the cell _______ Golgi body
(d) Digestive bag of the cell _______ Lysosome
(e) Storage sacs of the cell _______ Vacuole
(f) Kitchen of the cell _______ Chloroplast
(g) Control room of the cell _______ Nucleus

Question 36.
How is a bacterial cell different from an onion peel cell?
Answer:
Bacterial cell is a prokaryote so, they are smaller in size, their nucleus does not have a well defined nuclear membrane, possess a single chromosome and the cell is devoid of any cell organelles. Onion peel cell is an eukaryotic cell which is comparatively bigger in size, nucleus has a well defined nuclear membrane, possess more than one chromosome and have well defined cell organelles.

Question 37.
How do substances like carbon dioxide (CO₂) and water (H₂O) move in and out of the cell?
Answer:
The movement of carbon dioxide occurs by the process of diffusion in and out of the cell whereas the movement of water in and out of the cell occurs by the process of osmosis.

Question 38.
How does Amoeba obtain its food?
Answer:
Amoeba takes in food with the help of finger like extensions called pseudopodia by the process called endocytosis. Pseudopodia help to engulf the food which gets enclosed in a food vacuole. The complex particles of the food get broken down into simpler substances inside the food vacuole and diffuse into the cytoplasm of Amoeba. The undigested food particles are removed from the cell by exocytosis.

Question 39.
Name the two organelles in a plant cell that contain their own genetic material and ribosomes.
Answer:
Mitochondria and plastids (e.g. chloroplasts) contain their own genetic material and ribosomes.

Question 40.
Why are lysosomes also known as “scavengers of the cells”?
Answer:
The lysosomes are waste disposal system of the cell as they remove the debris and help to keep the cell clean by digesting any foreign material as well as worn-out cell organelles.

Question 41.
Which cell organelle controls most of the activities of the cell?
Answer:
Nucleus is the control centre of the cell and controls most of the activities of the cell.

Question 42.
Which kind of plastid is more common in
(a) roots of the plant
(b) leaves of the plant
(c) flowers and fruits
Answer:
The plastid that is more common in
(a) roots of the plant is leucoplast.
(b) leaves of the plant is chloroplast.
(c) flowers and fruits is chromoplast.

Question 43.
Why do plant cells possess large sized vacuole?
Answer:
The plant cells possess large sized vacuoles as the vacuoles help to store many important substances. They contain cell sap that gives turgidity and rigidity to the plant cell.

Question 44.
How are chromatin, chromatid and chromosomes related to each other?
Answer:
The chromatin material forms chromatids, i.e., chromatids are made up of chromatin. The chromosomes are made up of thread like structures called as chromatids.

Question 45.
What are the consequences of the following conditions?
(a) A cell containing higher water concentration than the surrounding medium
(b) A cell having low water concentration than the surrounding medium.
(c) A cell having equal water concentration to its surrounding medium.
Answer:
The consequences of the following conditions are:
(a) A cell containing higher water concentration than the surrounding medium will undergo exosmosis.

(b) A cell having low water concentration than the surrounding medium will undergo endosmosis.

(c) A cell having equal water concentration to its surrounding medium will not be affected as there won’t be any net movement of water into or outside the cell.

The Fundamental Unit of Life Class 9 Extra Questions Long Answer Type

Question 1.
Distinguish between hypotonic solution, isotonic solution and hypertonic solution.
Answer:
The differences between the three kinds of solutions are:

Question 2.
Describe the roles played by lysosomes. Why are they termed as suicidal bags? How do they perform their function?
Answer:
Roles played by lysosomes are:
(i) They are waste disposal system of the cell as they help to keep the cell clean by digesting any foreign material as well as worn-out cell organelles.

(ii) Lysosomes have powerful digestive enzymes capable of breaking down all organic materials. Lysosomes are also known as the ‘suicide bags’ of a cell because if the cell gets damaged during disturbance in cellular metabolism, the lysosomes may burst and its enzymes digest their own cell. They perform this function as they are membrane-bound sacs filled with hydrolytic and digestive enzymes made by rough endoplasmic reticulum (RER).

Question 3.
(a) Name the organelle which provides turgidity and rigidity to the plant cell. Name any two substances which are present in it.
(b) How are they useful in unicellular organisms?
Answer:
(a) The organelle which provides turgidity and rigidity to the plant cell is the large central vacuole which is filled with cell sap. The substances present in it are amino acids, sugars, various organic acids and some proteins.

(b ) They are very useful for the unicellular organisms in the form of food vacuole and contractile vacuole which perform the following functions:

Food vacuole found in Amoeba contains the food items that the Amoeba has consumed.
Contractile vacuole found in some unicellular organisms help in expelling excess water and some wastes from the cell.

Question 4.
What will happen if:
(i) Ribosomes are removed from the cell?
(ii) Golgi apparatus is removed from the cell?
(iii) Plasma membrane ruptures?
Answer:
(i) If ribosomes are removed from the cell, protein synthesis will not take place, enzymes will not be formed and ultimately death of cell will occur.

(ii) If golgi apparatus is removed from the cell, storage, modification and the packaging of substances in vesicles will not occur and lysosomes will not be formed.

(iii) Plasma membrane is a selectively permeable membrane, so in its absence, the entry and exit of the substances will not be regulated, contents of the cell will leak out and the cell will die.

Question 5.
Draw a plant cell and label the parts which
(a) determines the function and development of the cell
(b) package materials coming, from the endoplasmic reticulum
(c) provides resistance to microbes to withstand hypotonic external media without bursting
(d) is site for many biochemical reactions necessary to sustain life
(e) is a fluid contained inside the nucleus
Answer:
(a) Nucleus
(b) Golgi apparatus
(c) Cell wall
(d) Cytoplasm
(e) Nucleoplasm

Question 6.
Illustrate only a plant cell as seen under electron microscope. How is it different from an animal cell?
Answer:

Plant Cell:

• Cell wall is present.
• Plastids are present.
• A large central vacuole is present.
• Centrioles are absent.

Animal Cell:

• Cell wall is absent.
• Plastids are absent.
• Vacuoles are either absent or very small.
• Centrioles are present.

Question 7.
Draw a neat labelled diagram of an animal cell.
Answer:

Question 8.
Draw a well labelled diagram of an eukaryotic nucleus. How is it different from a nucleoid?
Answer:

It is different from a nucleoid as it is a membrane bound organelle.

Question 9.
Differentiate between rough and smooth endoplasmic reticulum. How is endoplasmic reticulum important for membrane biogenesis?
Answer:
Smooth Endoplasmic Reticulum (SER):

• SER looks smooth as it does not have ribosomal particles on its surface.
• SER helps in the manufacture of lipids and fat molecules and also in detoxifying many poisons and drugs in the liver cells of the group of vertebrates.

Rough Endoplasmic Reticulum (RER):

• RER appears rough as it has particles of ribosome on its surface.
• Ribosomes present on RER are the sites of protein synthesis.

Ribosomes are the sites of protein synthesis in all active cells. Endoplasmic reticulum helps in transporting these proteins to various places. Some proteins and lipids made by SER help in building the cell membrane, this process is known as membrane biogenesis.

Question 10.
In brief state what happens when
(a) dry apricots are left for sometime in pure water and later transferred to sugar solution?
(b) a red blood cell is kept in concentrated saline solution?
(c) the plasma-membrane of a cell breaks down?
(d) Rhoeo leaves are boiled in water first and then a drop of sugar syrup is put on it?
(e) golgi apparatus is removed from the cell?
Answer:
(a) If dry apricots are left for sometime in pure water they swell up due to endosmosis and later when they are transferred to sugar solution exosmosis occurs and they shrink.

(b) If a red blood cell is kept in concentrated saline solution it will lose water and shrink.

(c) If the plasma-membrane of a cell breaks down, the cell will die.

(d) If Rhoeo leaves are boiled in water first and then a drop of sugar syrup is put on it, no plasmolysis occurs as the cells get killed on boiling.

(e) If the golgi apparatus is removed from the cell, all sorts of vesicle formation in the cell get stopped.

Question 11.
Draw a neat diagram of plant cell and label any three parts which differentiate it from animal cell.
Answer:
The plant cells have chloroplast, large central vacuole and the cell wall which are not present in animal cells.

The Fundamental Unit of Life Class 9 Extra Questions HOTS

Question 1.
Unicellular organisms like Amoeba and Paramoecium survive as a single cell. How?
Answer:
In Amoeba and Paramoecium all the activities like digestion, absorption, assimilation, etc., which are necessary for life are carried out by a single cell. This enables them to survive even as a single cell.

Question 2.
Explain how cell walls permit the cells of fungi to withstand very dilute external media without bursting.
Answer:
The fungal cells swell up in very dilute external media as they take in water by the process of endosmosis. The plasma membrane of the cell exerts pressure on the cell wall and the cell wall also exerts an equal pressure on the membrane which helps to prevent the cell from bursting.

Question 3.
Give the roles of the following:
(i) Lipids and proteins of plasma membrane
(ii) Cellulose in the cell wall
(iii) Flexible nature of the cell membrane
(iv) Hydrolytic enzymes of lysosomes
(v) Ribosomes present on the RER
Answer:
The roles are:
(i) Lipids and proteins of plasma membrane: Provide flexibility to cell membrane and make it selectively permeable in nature.

(ii) Cellulose in the cell wall: Provides structural strength to the plants and prevents bursting of the cell when the cell swells in a hypotonic solution.

(iii) Flexible nature of the cell membrane: It helps organisms like Amoeba to engulf food by endocytosis.

(iv) Hydrolytic enzymes of lysosomes: Helps in the process of cell digestion and act as ‘scavengers of cell’.

(v) Ribosomes present on the RER: Are the site of protein synthesis.

Question 4.
Which cell organelle will play a major role when a cell is damaged due to disturbance in cellular metabolism? How?
Answer:
Lysosomes will play a major role when a cell is damaged due to disturbance in cellular metabolism. The lysosomes burst open and its hydrolytic digestive enzymes digest their own cell contents. Thus, lysosomes are also called as the ‘suicidal bags’.

Question 5.
What will be the effect on the cell when:
(a) The cell is placed in a medium having lower water concentration.
(b) The cell is placed in a medium having higher water concentration.
(c) The cell is placed in a medium having water concentration which is equal to that inside the cell.
Answer:
The consequences of the following conditions are:
(a) The cell having low water concentration than the surrounding medium will undergo endosmosis.
(b) The cell containing higher water concentration than the surrounding medium will undergo exosmosis.
(c) The cell having equal water concentration to its surrounding medium will not be affected as there won’t be any net movement of water into or outside the cell.

The Fundamental Unit of Life Class 9 Extra Questions Value Based (VBQs)

Question 1.
Rajni was told by his teacher that cell is the structural and fundamental unit of life. She asked her teacher to give reasons to support her statement. She felt very happy when her teacher showed her the slide of onion peel with the help of a microscope. She was able to identify the various parts of the cells visible in the slide.
(i) Why is the cell called the structural and functional unit of life?
(ii) With the help of a suitable diagram show the cells observed in onion peel and label them.
(iii) What values are shown by Rajni?
Answer:
(i) The body of all living organisms is composed of one or more cells, so cell is called the structural unit of life. Also, the various life processes like digestion, respiration, excretion, etc., are performed by the cells, so the cell is called the functional unit of life.

(ii) The cells of onion peel are shown in the diagram below.

(iii) Values shown by Rajni are curiosity, scientific temper, knowledge and sincerity towards her work.

In this page, we are providing Structure of the Atom Class 9 Extra Questions and Answers Science Chapter 4 pdf download. NCERT Extra Questions for Class 9 Science Chapter 4 Structure of the Atom with Answers will help to score more marks in your CBSE Board Exams.

Class 9 Science Chapter 4 Extra Questions and Answers Structure of the Atom

Extra Questions for Class 9 Science Chapter 4 Structure of the Atom with Answers Solutions

Structure of the Atom Class 9 Extra Questions Very Short Answer Type

Question 1.
Is it possible for the atom of an element to have one electron, one proton and no neutron? If so, name the element. [NCERT Exemplar]
Answer:
Yes, it is true for hydrogen atom which is represented as \(_{ 1 }^{ 1 }{ H }\)

Question 2.
Will ³⁵Cl and ³⁵Cl have different valencies? Justify your answer. [NCERT Exemplar]
Answer:
No, ³⁵Cl and ³⁵Cl are isotopes of an element.

Question 3.
Why did Rutherford select a gold foil in his a-ray scattering experiment? [NCERT Exemplar]
Answer:
Rutherford selected a gold foil in his a-ray scattering experiment because gold has high malleability.

Question 4.
One electron is present in the outermost shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outermost shell? [NCERT Exemplar]
Answer:
+ 1.

Question 5.
Write any two observations which support the fact that atoms are divisible. [NCERT Exemplar]
Answer:
Discovery of electrons and protons.

Question 6.
Write down the electron distribution of chlorine atom. How many electrons are there in a L-shell? (Atomic number of chlorine is 17). [NCERT Exemplar]
Answer:
The electron distribution of chlorine atom is 2, 8, 7 and the L-shell has 8 electrons.

Question 7.
In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed? [NCERT Exemplar]
Answer:
– 2.

Question 8.
Helium atom has 2 electrons in its valence shell but its valency is not 2, Explain.
Answer:
Helium atom has 2 electrons in its outermost shell and its duplet is complete. Hence the valency is zero.

Question 9.
An element X has a mass number 4 and atomic number 2. Write the valency of this element.
Answer:
Valency is zero, as K-shell is completely filled.

Question 10.
Who identified the sub-atomic particle electron?
Answer:
J.J. Thomson.

Question 11.
Name the rays discovered by E. Goldstein.
Answer:
Canal rays.

Question 12.
Give two examples from everyday life where we use cathode ray tubes.
Answer:
Television picture tube and Fluorescent light tubes.

Question 13.
What are the actual values of charge and mass of the electron?
Answer:
Charge = 1.60 x 10^-19 C (one unit negative charge)
Mass = 9.11 x 10^-31 kg (1/1840th of the mass of hydrogen atom).

Question 14.
Compare the radius of the nucleus with that of the atom.
Answer:
Radius of the atom is of the order of 10^-10 m while that of the nucleus is of the order of 10^-15 m. Thus, nucleus is 10^-15/10^-10 i.e., 1/100,000^th of the size of the atom.

Question 15.
Why are Bohr’s orbits called stationary states?
Answer:
According to Bohr, the orbits in which the electrons revolve have fixed energies. Hence, they are called stationary states (stationary means fixed).

Question 16.
Who discovered neutron?
Answer:
Chadwick.

Question 17.
What are nucleons? What is their number called?
Answer:
Protons and neutrons present in the nucleus are collectively called nucleons. Their total number is called ‘mass number’.

Question 18.
What do you mean by ‘Atomic number’?
Answer:
Atomic number of an element is the number of protons present in the nucleus of the atom of that element.

Question 19.
Who discovered the nucleus of the atom?
Answer:
Rutherford.

Question 20.
What is the charge on alpha particle?
Answer:
+ 2.

Question 21.
The mass number of an element is 18. It contains 7 electrons. What is the number of protons and neutrons in it?
Answer:
Number of protons = 7 Number of neutrons = 11 ,

Question 22.
What is the maximum number of electrons that can be present in M-shell?
Answer:
M-shell means 3rd shell for which n = 3. Hence, maximum number of electrons that can be present in M-shell = 2n² = 2 x 3² = 18.

Question 23.
Identify the isotopes out of A, B, C and D? ³³A₁₇, ⁴⁰B₂₀, ³⁷C₁₇, ³⁸D₁₉.
Answer:
³³A₁₄ and ³⁷C₁₇? are isotopes.

Question 24.
Represent the three isotopes of hydrogen and give their names.
Answer:
\(_{ 1 }^{ 1 }{ H }\) (Protium), \(_{ 2 }^{ 1 }{ H }\) (Deuterium), \(_{ 3 }^{ 1 }{ H }\) (Tritium).

Question 25.
How many electrons are present in the species He²⁺ ion? Suggest another name for it.
Answer:
He²⁺ ion has no electrons. It is also called alpha (α) particle.

Question 26.
What is the reason for the identical chemical properties of all the isotopes of an element?
Answer:
Isotopes have identical electronic configuration.

Question 27.
Give one similarity and one difference between a pair of isotopes.
Answer:
They have same number of protons but different number of neutrons in their nuclei.

Question 28.
Give the number of protons and neutrons in an atom of uranium (U-235) used in a nuclear reactor.
Answer:
In nuclear reactor U-235 is used. Atomic number of Uranium is 92.
∴ Number of protons = Atomic number = 92
Number of neutrons = Mass number – Atomic number = 235 – 92 = 143.

Question 29.
Out of proton and neutron, which is heavier?
Answer:
Neutron is slightly heavier (1.675 x 10^-27 kg) than proton (1.67 x 10^-27 kg).

Question 30.
If K and L shells of an atom are completely filled what will be its name?
Answer:
The atom will belong to the noble gas element neon (Ne).

Question 31.
Do isobars have also identical chemical characteristic like isotopes?
Answer:
No, these are not identical because the isobars have different atomic number as well as different electronic configuration.

Question 32.
What is the number of electrons in the valence shell of chlorine (Z = 17)?
Answer:
The electronic distribution of the element is:
K(2), L(8), M(7). This means that the valence shell of chlorine has 7 electrons.

Question 33.
Which radioisotope is used for the treatment of cancer?
Answer:
Radioisotope Co-60 is used for the treatment of cancer.

Question 34.
Out of C-12 and C-14 isotopes of carbon, which is of radioactive nature?
Answer:
C-14 isotope is of radioactive nature.

Question 35.
The electron configuration of an element is: 2(K), 8(L), 5(M). Predict its valency.
Answer:
The valency of the element is 3. It is calculated as: 8 – 5 = 3.

Question 36.
Will \(_{ 12 }^{ 6 }{ C }\) and \(_{ 14 }^{ 6 }{ C }\) have different valencies?
Answer:
Both the species are the isotopes of the same element i.e., the carbon. Since their atomic numbers Eire same, their electronic configurations as well as valencies will also be the same.

Question 37.
Do the elements \(_{ 3 }^{ 1 }{ X }\) and \(_{ 3 }^{ 2 }{ Y }\) represent pair of isotopes?
Answer:
No, they do not because the two elements differ in their atomic numbers. Isotopes have the same atomic numbers.

Structure of the Atom Class 9 Extra Questions Short Answer Type 1

Question 1.
In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer. [NCERT Exemplar]
Answer:
No, the statement is incorrect. In an atom the number of protons and electrons is always equal.

Question 2.
Calculate the number of neutrons present in the nucleus of an element X which is represented as \(_{ 31 }^{ 15 }{ X }\). [NCERT Exemplar]
Answer:
Mass number = No. of protons + No. of neutrons = 31
∴ Number of neutrons = 31 – number of protons
= 31 – 15
= 16

Question 3.
The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements? [NCERT Exemplar]
Answer:
∴ They are isobars because isobars have different atomic numbers but same mass number.

Question 4.
Why do Helium, Neon and Argon have a zero valency?
Answer:
Helium has two electrons in its only energy shell, while Argon and Neon have 8 electrons in their valence shells. As these have maximum number of electrons in their valence shells, they do not have any tendency to combine with other elements. Hence, they have a valency equal to zero.

Question 5.
What are isobars? Give one example.
Answer:
Isobars are the atoms of different elements which have different atomic numbers but same mass number.
Example is \(_{ 40 }^{ 18 }{ Ar }\), \(_{ 40 }^{ 20 }{ Ca }\) .

Question 6.
What is the number of valence electron in:
(i) Sodium ion (Na⁺)
(ii) Oxide ion (O^2-)?
Answer:
(i) Sodium ion (Na⁺)
No. of electrons: (11 – 1); Electronic configuration = 2, 8.
∴ Na⁺ ion has 8 valence electrons.

(ii) Oxide ion (O^2-)
No. of electrons: (8 + 2) = 10; Electronic configuration = 2, 8.
∴ O^2- ion has 8 valence electrons.

Question 7.
The number of electrons in the outermost ‘L’ shell of an atom is 5.
(a) Write its electronic configuration.
(b) What is its valency and why?
Answer:
(a) 2, 5
(b) 3, because it needs three more electrons to complete its octet.

Question 8.
Write two difference between isobars and isotopes.
Answer:
Isobars

• Same mass number but different atomic numbers.
• Different number of protons.

Isotopes:

• Same atomic number but different mass numbers.
• Same number of protons.

Question 9.
Mention the postulates Neils Bohr put forth to overcome the objections raised against Rutherford’s atomic model.
Answer:
Bohr put forward the following postulates in order to overcome the objections raised against Rutherford s atomic method: .

Only certain special orbits known as discrete orbits of electrons are allowed inside the atom.
While revolving in discrete orbits the electrons do not radiate energy. These orbits or shells are called energy levels.

Question 10.
Given that natural sample of iron has isotopes \(_{ 54 }^{ 28 }{ Fe }\), \(_{ 58 }^{ 28 }{ Fe }\) and \(_{ 57 }^{ 28 }{ Fe }\) in the ratio of 5%, 90% and 5% respectively. What will be the average atomic mass of iron (Fe)?
Answer:
Average atomic mass
= \(\frac { 54 × 5 + 56 × 90 + 57 × 5 }{ 100 }\)
= \(\frac { 270+5040+285 }{ 100 }\)
= 55.95 u.

Question 11.
(a) What are canal rays? Who discovered them?
What is the charge and mass of canal ray?
(b) How are the canal rays different from electron in terms of charge and mass?
Answer:
(a) New radiations in a gas discharge tube which are positively charged. E. Goldstein discovered them. Charge on canal rays is positive and mass is one unit.

(b) Electrons are negatively charged and their mass is approximately 1/2000 of that of canal rays.

Question 12.
The average atomic mass of a sample of an element y is 35.5 u. What are the percentages of isotopes \(_{ 37 }^{ 17 }{ y }\) and \(_{ 35 }^{ 17 }{ y }\), in the sample?
Answer:
Let the % composition of \(_{ 37 }^{ 17 }{ y }\) be x.
Then the % composition of \(_{ 37 }^{ 17 }{ y }\) will be 100 – x
Now,

Question 13.
(i) Write the postulates of Bohr model of atom.
(ii) Draw a sketch of Bohr model of an atom with atomic number 15.
Answer:
(i) Only certain special orbits-discrete orbits of electrons are allowed.

(ii) While revolving in discrete orbits, the electrons do not radiate energy.

Question 14.
List the main differences between an atom and an ion.
Answer:
Atom:

• It is electrically natural.
• The valence shell of atom may or may not have 8 electrons.
• The atoms are less stable.
• It may or may not exist freely in solution.

Ion:

• It has positive or negative charge.
• The valence shell of an ion has 8 electrons.
• The ions are more stable.
• It can exist freely in solution.

Question 15.
An ion has one neutron more than the number of protons but has one electron less than the number of protons. Which two ions are possible which satisfy this condition? Explain.
Answer:
(i) Na⁺ ion, because it has atomic number = 11 and mass number = 23. Thus, no. of protons = 11, no. of neutrons = 23 – 11 = 12 and number of electrons = 11 – 1 = 10.

(ii) K⁺ ion, because it has atomic number = 19 and mass number = 39. Thus, no. of protons = 19, No. of neutrons = 39 – 19 = 20 and no. of electrons = 19 – 1 = 18.

Question 16.
Give the main points of Dalton’s atomic theory.
Answer:

• All matter is made up of a large number of extremely small indivisible particles called atoms.
• Atoms of the different elements are different.
• Atom is the smallest unit of matter which takes part in a chemical reaction.
• Atoms can neither be created nor be destroyed.

Question 17.
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign proper symbol of the species.
Answer:
Number of electrons present = 18
Number of neutrons present = 16
Atomic number of the element = 16
The element with atomic number 16 is sulphur (S). Since it has 18 electrons and not 16, it is therefore, an anion (S^2- ion).

Question 18.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. [NCERT Exemplar]
Answer:
We know
Mass number = No. of protons + No. of neutrons = 81
i.e., p + n = 81
Let number of protons = x

Structure of the Atom Class 9 Extra Questions Short Answer Type 2

Question 1.
Show diagrammatically the electron distributions in a sodium atom and a sodium ion and also given their atomic number. [NCERT Exemplar]
Answer:

Since the atomic number of sodium atom is 11, it has 11 electrons. A positively charged sodium ion (Na⁺) is formed by the removal of one electron from a sodium atom. So, a sodium ion has 11 – 1 = 10 electrons. Thus, electronic distribution of sodium ion will be 2, 8. The atomic number of an element is equal to the number of protons in its atom. Since, sodium atom and sodium ion contain the same number of protons, therefore, the atomic number of both is 11.

Question 2.
The ratio of the radii of hydrogen atom and its nucleus is ~ 10⁵. Assuming the atom and the nucleus to be spherical. [NCERT Exemplar]
(i) What will be the ratio of their sizes?
(ii) If atom is represented by planet Earth ‘R_e’ = 6.4 x 10⁵ m, estimate the size of the nucleus.
Answer:
(i) Volume of the sphere = \(\frac { 4 }{ 3 }\) πr³
Let R be the radius of the atom and r be that of the nucleus.

(ii) If the atom is represented by the planet earth (R_e = 6.4 x 10⁶ m) then the radius of the nucleus
would be r_n = \(\frac{\mathrm{R}_{e}}{10^{5}}\)
r_n = \(\frac{6.4 \times 10^{6} \mathrm{m}}{10^{5}}\)
= 6.4 x 10 m = 64 m.

Question 3.
The number of protons, neutrons and electrons in particles from A to E are given below:

(i) Which one is a cation?
(ii) Which one is an anion?
(iii) Which represent pair of isotopes?
Answer:
(i) B is a monovalent cation (B⁺)
(ii) E is a monovalent anion (E^–)
(iii) A and D represent pair of isotopes.

Question 4.
Explain Bohr Bury rules for distribution of electrons into different shells. Write the distribution of electrons in sodium atom (Z = 11).
Answer:
Bohr Bury Rules:

• The maximum no. of electrons present in a shell is given by the formula 2n² (where n is shell number).
• The maximum number of electrons that can be accommodated in the outermost orbit is 8.
• Electrons are not accommodated in a given shell, unless the inner shells are filled.
• Electronic configuration of Na = 2, 8, 1.

Question 5.
Atom A has a mass number 238 and atomic number 92 and atom B has mass number 235 and atomic number 92.
(i) How many protons, atoms A and B have?
(ii) How many neutrons, atoms A and B have?
(iii) Are atoms A and B isotopes of the same element? How?
Answer:
(i) Protons in atoms A and B = 92
(ii) Neutrons in atoms A and B = 146 and 143
(iii) Yes, atom A and B are isotopes of the element since they have the same atomic number or same electronic configuration.

Question 6.
State three features of the nuclear model of an atom put forward by Rutherford.
Answer:

1. There is a positively charged centre in an atom called the nucleus. Nearly, all the mass of an atom resides in the nucleus.
2. The electrons revolve around the nucleus in well-defined orbits.
3. The size of the nucleus is very small as compared to the size of the atom.

Question 7.
The nucleus of an atom is found to have a total mass of nearly 20.088 x 10^-27 kg and a total charge of 9.612 x 10^-19 coulombs. Calculate the atomic number and mass number of the atom. Name the element.
Answer:
Mass of a nucleon (proton or neutron)
= 1.674 x 10^-27 kg
Total mass of nucleus = 20.088 x 10^-27 kg
∴ No. of nucleons (protons + neutrons) present in the nucleus
= \(\frac{20.088 \times 10^{-27}}{1.674 \times 10^{-27}}\) = 12
∴ Mass number = 12u.
Charge on one proton = 1.602 x 10^-19 coulombs; Total charge on the nuclear = 9.612 x 10^-19
∴ No.of protons in the nucleus
= \(\frac{9.612 \times 10^{-19}}{1.602 \times 10^{-19}}\) = 6
Atomic number = 6. Thus, the element is carbon.

Question 8.
Give reasons why?
(а) Atom is electrically neutral.
(b) Atom as a whole is an empty space.
(c) Rutherford model of atom could not provide stability to the nucleus.
Answer:
(a) An atom is electrically neutral because the number of protons and number of electrons in it are equal.

(b) According to Rutherford’s experiment, the size of the nucleus is very small as compared to the size of an atom, therefore, atom as a whole is an empty space.

(c) According to Rutherford, the protons are present inside the nucleus and electrons revolve around the nucleus. According to the elecromagnetic theory, a charged particle moving in a circular path continuously loses energy in the form of electromagnetic radiations and finally falls into the nucleus.

Question 9.
Write the complete symbol for
(i) the nucleus with atomic number 56 and mass number 138.
(ii) the nucleus with atomic number 26 and mass number 55.
(iii) the nucleus with atomic number 4 and mass number 9.
Answer:
(i) The element with atomic number 56 is Ba. Its symbol is \(_{ 138 }^{ 56 }{ Ba }\)

(ii) The element with atomic number 26 is Fe. Its symbol is \(_{ 56 }^{ 26 }{ Fe }\)

(iii) The element with atomic number 4 is Be. Its symbol for \(_{ 9 }^{ 4 }{ Be }\)

Question 10.
How many protons, electrons and neutrons are there in the following nuclei?
(i) \(_{ 17 }^{ 8 }{ O }\)
(ii) \(_{ 25 }^{ 12 }{ Mg }\)
(iii) \(_{ 80 }^{ 35 }{ Br }\)
Answer:
(i) \(_{ 35 }^{ 79 }{ Br }\)
Atomic number, Z = 8
Mass number, A = 17
No. of protons = No. of electrons = Z = 8
No. of neutrons + No. of protons = A
No. of neutrons + 8 = 17
or No. of neutrons = 17 – 8 = 9

(ii) \(_{ 25 }^{ 12 }{ Mg }\)
Atomic number, Z = 12
Mass number, A = 25
No. of protons = No. of electrons = Z = 12
No. of neutrons = A – No. of protons = 25 – 12 = 13.

(iii) \(_{ 80 }^{ 35 }{ Br }\)
Atomic number, Z = 35
Mass number, A = 80
No. of protons = No. of electrons = Z = 35
No. of neutrons = A – No. of protons = 80 – 35 = 45.

Question 11.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol for the ion.
Answer:
Since the iron carries 3 units of positive charge, it will have 3 electrons less than the number of protons. Let number of electrons = x
No. of protons = x + 3
No. of neutrons = x + \(\frac { x × 30.4 }{ 100 }\)
= x + 0.304x
= 1.304x
Now, No. of protons + No. of neutrons = 56
x + 3 + 1.304 x = 56
2.304 x = 53
x = \(\frac { 53 }{ 2.304 }\) = 23
No. of electrons = 23
No. of protons = 23 + 3 = 26
Symbol = \(_{ 56 }^{ 26 }{ Fe }\)³⁺.

Question 12.
The nuclear radius is of the order of 10^-13 cm while atomic radius is of the order 10^-8 cm. Assuming the nucleus and the atom to be spherical, what fraction of the atomic volume is occupied by the nucleus?
Answer:
The volume of a sphere = 4 πR^3/3 where R is the radius of the sphere.
∴ Volume of the nucleus = 4 πr^3/3 = 4π (10^-13)^3/3 cm³
Similarly,
Volume of the atom = 4 πR^3/3 = 4π (10^-8)^3/3 cm³
∴ Fraction of the volume of atom occupied by the nucleus
= \(\frac{4 \pi\left(10^{-13}\right)^{3} / 3 \mathrm{cm}^{3}}{4 \pi\left(10^{-8}\right)^{3} / 3 \mathrm{cm}^{3}}\)
= 10^-15.

Question 13.
Calculate the number of electrons, protons and neutrons in the following species:
(i) Phosphorus atom
(ii) Phosphide ion (P^3-)
(iii) Magnesium ion (Mg²⁺)
Mass numbers: P = 31; Mg = 24
Atomic numbers: P = 15, Mg = 12.
Answer:
(i) Phosphorus atom
Number of electrons = Atomic number = 15
Number of protons = Atomic number = 15
Number of neutrons = Mass number – Atomic number
= 31 – 15 = 16.

(ii) Phosphide ion (P^3-)
Phosphide ion (P^3-)
= Phosphorus atom + 3 electrons
P^3- = P + 3e^–
Thus, phosphide ion has same number of protons and neutrons as phosphorus atom but has three electrons more.
Number of electrons = 15 + 3 = 18
Number of protons = 15
Number of neutrons = 31 – 15 = 16

(iii) Magnesium ion (Mg²⁺)
Mg²⁺ ion is formed by loss of two electrons by Mg atom. Therefore, it has two electrons less than number of electrons in Mg atom.
Mg²⁺ = Mg – 2e^–
Number of electrons = 12 – 2 = 10
Number of protons = 12
Number of neutrons = (24 – 12) = 12

Structure of the Atom Class 9 Extra Questions Long Answer Type

Question 1.
(a) Enlist the conclusions drawn by Rutherford from his a-ray scattering experiment.
(b) In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?
Answer:
(a) Rutherford concluded from the a-particle scattering experiment that:

(i) Most of the space inside the atom is empty because the a-particles passed through the gold foil without getting deflected.

(ii) Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.

(iii) A very small fraction of a-particles were deflected by 180°, indicating that all the positive charges and mass of the gold atom were concentrated in a very small volume within the atom.

From the data he also calculated that the radius of the nucleus is about 105 times less than the radius of the atom.

(b) Rutherford proposed a model in which electrons revolve around the nucleus in well-defined orbits. There is a positively charged center in an atom called the nucleus. He also proposed that the size of the nucleus is very small as compared to the size of the atom and nearly all the mass of an atom is centred in the nucleus.

Whereas, Thomson proposed the model of an atom to be similar to a Christmas pudding. The electrons are studded like currants in a positively charged sphere like Christmas pudding and the mass of the atom was supposed to be uniformly distributed.

Question 2.
(a) What were the drawbacks of Rutherford’s model of an atom?
(b) What are the postulates of Bohr’s model of an atom?
Answer:
(a) The orbital revolution of the electron is not excepted to be stable. Any particle in a circular orbit would undergo an acceleration and the charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know.

(b) The postulates put forth by Neils Bohr’s about the model of an atom:
(i) Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.

(ii) While revolving in discrete orbits the electrons do not radiate energy.
These orbits are called energy levels. Energy levels in Em atom are shown by circles.
These orbits are represented by the letters K, L, M, N, …………….. or the numbers, n = 1, 2, 3, 4, ………………

Question 3.
An atom of an element has three electrons in the third shell which is the outermost shell. Write
(a) the electronic configuration
(b) the atomic number
(c) number of protons
(d) valency
(e) the name of the element
(f) its nature whether metal or non-metal.
Answer:
The third shell is M-shell. If the atom of the element has three electrons in the third shell, this means that K and L shells are already filled.
(a) Electronic configuration: 2, 8, 3
(b) Atomic number = No. of electrons = 13
(c) Number of protons = No. of electrons = 13
(d) Valency of the element = 3
(e) The element with Z = 13 is aluminium (Al)
(f) It is a metal.

Question 4.
Answer the following in one line or two:
(a) What is the maximum number of electrons that can be accommodated in the outermost energy shell in an atom?
(b) On the basis of Thomson’s model of an atom, explain how an atom is neutral as a whole.
(c) How many neutrons are present in hydrogen atom?
(d) Do isobars belong to the same element?
(e) An element has five electrons in the M shell which is the outermost shell. Write its electronic configuration.
Answer:
(a) The outermost energy shell in an atom can have a maximum of eight electrons.

(b) According to Thomson’s model of an atom, all the protons in an atom are present in the positively charged sphere while negatively charged electrons are studded in this sphere. Since the electrons and protons are equal in number each carrying one unit charge, the atom as a whole is electrically neutral.

(c) Hydrogen atom has no neutron.

(d) No, isolars belong to different elements since they differ in their atomic numbers.

(e) The electronic configuration of the element is K(2), L(8), M(5).

Question 5.
Explain why did Rutherford select a gold foil in his alpha particle scattering experiments? What observations in a-scattering experiment led Rutherford to make the following observations:
(i) Most of the space in an atom is empty.
(ii) Nucleus is positively charged.
Mention any two drawbacks of Rutherford’s model.
Answer:
Rutherford selected a gold foil because he wanted a very thin layer as possible. This gold foil was about 1000 atoms thick.

(i) As most of the alpha particles passed through the foil undeviated, it means that they did not come across any obstruction in their part. Thus, most of the space in an atom was thought to be empty.

(ii) Very few particles deviated by small angles from their path which suggested that nucleus is positively charged.

The revolution of the electron in any circular orbit is not expected to be stable. Any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy. Thus, the revolving electrons would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and matter would not exist in the form that we know. We know that atoms are quite stable.

Question 6.
(a) Why are anode rays called canal rays?
(6) Mention two postulates of J.J. Thomson’s model.
(c) Compare the properties of protons and electrons.
Answer:
(a) The anode rays produced at the anode of the discharge tube are called canal rays because they pass through the holes of the cathode.

(b)

• Atom consists of positively charged sphere and electrons are embedded in it.
• The negative and positive charges are equal in magnitude.

(c) Protons:

• Positively charged.
• Mass of 1 proton is equal to mass of H atom.

Electrons:

• Negatively charged.
• Mass of electron is 1/1840 times that of a proton.

Question 7.
(a) What are isobars?

(b) Atomic number of an element Y is 17.
(i) Write its electronic configuration.
(ii) What is the number of valence electron in Y?
(iii) How many electrons are needed to complete the octet of Y?
(iv) Is it a metal or non-metal?

(c) The valency of Na is 1 and not 7. Give reason.
Answer:
(a) Atoms of different elements with different atomic numbers which have same mass number are called isobars.

(b) (i) 2,8, 7
(ii) 7
(iii) 1
(iv) Non-metal

(c) Electronic configuration of Na is 2, 8,1. It can lose one electron to attain the electronic configuration of neon. Therefore, the valency of Na is 1. The valency of Na may be 7 when it gains 7 electrons in its valence shell, but gain ring 7 electrons is difficult. Therefore, the valency of Na is not 7.

Question 8.
An atom of an element has 2 electrons in the M-shell i.e., third shell. What will be the atomic number of this element? Name this element. Find the valency of this element. Also, find the number of neutrons in the atom of this element.
Answer:
(i) Atomic number is 12

(ii) Element is magnesium.
(iii) Valency is 2 (⁺²)
(iv) No. of neutrons = Atomic mass – No. of protons = 24 – 12 = 12.

Structure of the Atom Class 9 Extra Questions HOTS

Question 1.
What information do you get from the figure given below about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form. [NCERT Exemplar]

Answer:

Question 2.
In the Gold foil experiment of Geiger and Marsdein, that paved the way for Rutherford’s model of an atom, ~ 1.00% of the a-particles were found to deflect at angles > 50°. If one mole of α – particles were bombarded on the gold foil, compute the number of α – particles that would deflect at angles less than 50°. [NCERT Exemplar]
Answer:
% of α – particles deflected more than 50° = 1% of α – particles
% of α – particles deflected less than 50° = 100 – 1 = 99%
Number of α – particles bombarded = 1 mole = 6.022 x 10²³ particles
Number of particles that deflected at an angle less than 50°
= \(\frac { 99 }{ 100 }\) x 6.022 x 10²³
= \(\frac { 596.178 }{ 100 }\) x 10²³
= 5.96 x 10²³.

Question 3.
An ion Y^3- contains 18 electrons and 16 neutrons. Calculate the atomic number and mass number of the element Y. Name the element Y.
Answer:
Number of electrons in Y^3- ion = 18
Since negative charge is formed by gain of electrons by the neutral atom and the number of electrons gained is equal to the number of units of negative charge on the ion.
∴ Number of electrons in the neutral atom = 18-3 =15.
Now, for a neutral atom, Atomic number = Number of protons = Number of electrons.
∴ Atomic number of element Y = 15
Mass number of the element = Number of protons + Number of neutrons = 15 + 16 = 31
The given element Y with atomic number 15 is phosphorus.

Question 4.
The following data represent the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D.

Answer the following questions:
(a) Give the electronic distribution of element B.
(b) The valency of element A.
(c) The atomic number of element B.
(d) The mass number of element D.
Answer:
(а) Electronic distribution of elements B : 2, 8, 6.
(b) Valency of A is 1(-1).
(c) Atomic number of element B is 16.
(d) Mass number of element D is 39.

Question 5.
Give a reason to explain why:
(a) isotopes of an element show identical chemical properties?
(b) the atomic masses of elements are in fractions?
(c) atoms combine with other atoms?
Answer:
(a) Since all the isotopes of an element have identical electronic configuration containing the same number of valence electrons. Therefore all the isotopes of an element show identical chemical properties.

(b) The fractional atomic masses of elements are due to their isotopes having different masses.

(c) The atoms combine with other atoms to achieve the electronic configuration of the nearest noble gas and thus, become more stable.

Question 6.
(a) Describe the main features of Bohr’s model of an atom. Draw a neat and labelled diagram of energy levels.
(b) Which of the following pairs are isotopes and which are isobars?
(i) \(_{ 58 }^{ 26 }{ A }\), \(_{ 58 }^{ 28 }{ B }\)
(ii) \(_{ 79 }^{ 35 }{ X }\), \(_{ 80 }^{ 35 }{ Y }\)
Give reasons for your choice.
(c) Elements A and B have atomic numbers 18 and 16 respectively. Which of these two would be more reactive and why?
Answer:
(a) Features of Bohr’s Models:
(i) Electrons revolve in certain permitted orbits which are associated with fixed amount of energy. So they are called energy levels (K, L, M, N) or sub-shells.
(ii) As long as electron revolves in the same energy level, they do not lose energy.
(iii) The energy of orbit closest to the nucleus is lowest and farthest away is highest.
(iv) When we supply energy to an electron, it goes to a higher energy level and when it comes back to a lower level it radiates energy.

(b) \(_{ 79 }^{ 35 }{ X }\), \(_{ 80 }^{ 35 }{ Y }\) are isotopes because their atomic no. is same but mass no. is different. \(_{ 58 }^{ 26 }{ A }\), \(_{ 58 }^{ 28 }{ B }\) are iso¬bars, as mass no. is same but atomic no. is different.

(c) B would be more reactive as its electronic configuration is 2, 8, 6 and it requires two more electrons to complete its octet.

Question 7.
Choose the noble gases from the elements shown in the table below:

Answer:

A, B and D have completely filled valence shells, i.e., either 2 or 8 electrons in their valence shells.
∴ A, B and D are noble gases.

Question 8.
The mass number and electronic configuration of five elements A, B, C, D and E are as follows:

(а) Name the elements which has 22 neutrons in nucleus.
(b) What is atomic number of C?
(c) Name the elements which will form most stable ionic bond.
(d) Give the formation of the compound between B and D.
(e) Name the elements which will not take part in chemical combination.
Answer:
(a) E
(b) 11
(c) B and C
(d) DB₄
(e) E

Question 9.
Write the electronic configurations of the following elements and write the number of valence electrons present in it.
(a) \(_{ 14 }^{ 7 }{ N }\)
(b) \(_{ 28 }^{ 14 }{ Si }\)
(c) \(_{ 40 }^{ 20 }{ Ca }\)
(d) \(_{ 40 }^{ 18 }{ Ar }\)
(e) \(_{ 9 }^{ 4 }{ Be }\)
Answer:

Question 10.
Complete the following table.

Answer:
For F atom
Atomic no. = 9 (Give)
No. of electrons = 9
No. of protons = 9
No. of neutrons = 10 (Given)
Mass no. = Z + n = 9 + 10 = 19

For Mg²⁺ ion
No. of protons = 12 (Given)
Atomic no. = 12
No. of electrons = 12 – 2 = 10
Mass no. = 24 (Give)
No. of neutrons = A – Z = 24 – 12 = 12

For S atom
No. of protons = 16 (Given)
Atomic no. = 16
No. of protons = 16
Mass no. = 32 (Given)
No. of neutrons = A – Z = 32 – 16 = 16

For P^3- ion
No. of electrons = 18 (Given)
No. of protons = 18 – 3 = 15
Atomic no. = 15
No. of neutrons = 16 (Given)
Mass no. = Z + n = 15 + 16 = 31

The complete table is as follows:

Note: The value of positive charge on cation shows that the number of electrons in it are less than the number of protons by the same value. Similarly, the value of negative charge on anion shows that the number of electrons in it are more than the number of protons by the same value.