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Page No 376:

Question 1:

Give an example of prokaryotic cell.

Answer:

Bacterial cells are prokaryotic cells.

Page No 376:

Question 2:

What is the direction of air in coastal areas during the night?

Answer:

Air in coastal areas flow from land to the sea during night time.

Page No 376:

Question 3:

What type of motion, uniform or non-uniform, is exhibited by a freely falling body? Give reason for your answer.

Answer:

The motion exhibited by a freely falling body is non-uniform.

When a body falls freely, it starts from rest. When it falls down, the velocity increases as it reaches close to the ground. So, the motion exhibited by a freely falling body is non-uniform in nature. Because, the velocity of the body is not constant throughout the motion, instead it changes continuously.

Page No 376:

Question 4:

(i) Distinguish between longitudinal and transverse waves.
(ii) Are sound waves longitudinal or transverse?

Answer:

(i) The waves in which particles of the medium vibrate parallel to the direction of propagation of the wave are known as longitudinal waves. Sound is an example of the longitudinal wave.

Whereas the waves in which particles of the medium vibrate perpendicular to the direction of propagation of the wave are known as the transverse waves. Light waves are an example of the transverse waves.

(ii) When a sound wave travels in a medium, the disturbance created in the particles of the medium is parallel to the direction of propagation of the wave. Hence, sound waves are longitudinal waves. 
 

Page No 376:

Question 5:

What will happen in a cell if its nucleus is removed? Give reasons in support of your answer.

Answer:

In case the nucleus of a cell is removed, the cell will eventually die. This occurs because the nucleus is responsible for all the metabolic processes of a cell and in its absence, a cell cannot perform any metabolic activity.

Page No 376:

Question 6:

(i) What does the diffusion of gases tell us about their particles?
(ii) Give one example of diffusion of gases in a liquid

Answer:

(i) Intermixing of two gases to form a homogeneous mixture without any chemical change is called diffusion of gases.

Diffusion of gases tells about their particles as:
The particles move rapidly in all directions as gases have more space and no fixed shape or volume and nor forces of attractions. With the increase in temperature, the particles gain kinetic energy and move faster.
The particles are widely spaced and scattered at random throughout the container. Hence, gases have very low density.
Because of the 'empty' space between the particles, gases are readily compressed.

(ii) Diffusion of oxygen in water is an example of diffusion of gases in liquid. This is important for life forms under water.
 

Page No 376:

Question 7:

Calculate the molecular masses of the following compounds:
(i) Hydrogen sulphide, H2S
(ii) Carbon disulphide, CS2
(Atomic masses: H = 1 u; S = 32 u; C = 12 u)

Answer:

(i) The molecular mass of hydrogen sulphide,
=2×H+S=2×1+32=34 u

(ii) The molecular mass of carbon disulphide,
=C+2×S=12+2×32=76 u

Page No 376:

Question 8:

An element has atomic number 13 and atomic mass of 27.
(i) How many electrons are there in each atom of the element?
(ii) How are these electrons distributed in the various energy levels?

Answer:

The atomic number of the element is 13

(i) The electrons present in this element are 13.

(ii) Electronic configuration of this element is 2, 8, 3.
 Distribution of electrons in energy levels is K, L, M. 

Page No 376:

Question 9:

When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres.
(i) What was the initial speed of the ball?
(ii) How much time is taken by the ball to reach the highest point? (g = 10 m s−2)

Answer:

(i) Let initial velocity of the ball = u

Final velocity of the ball = 0 m/s

The maximum height reached by the ball = 5 m

Acceleration of the ball = -g (Ball is thrown upward)

Using the third equation of motion, v2=u2-2ghmax

0=u2-2×10×5u2=100u =10 m/s

Hence, the initial velocity of the ball is 10 m/s.

(ii) To calculate the time taken by the ball to reach the maximum height, we would use the first equation of motion.

v=u-gt0=10-10tt =1 s

Hence, the time taken by the ball to reach the maximum height is 1 second. 

Page No 376:

Question 10:

(i) Define the term 'energy' of a body. What is the SI unit of energy?
(ii) What are the various forms of energy?
(iii) Two bodies having equal masses are moving with uniform speeds of v and 2v, respectively. Find the ratio of their kinetic energies.

Answer:

(i) The energy of a body is defined as the capacity of the body to do work. Its SI unit is Joule and it is represented by the capital letter 'J'.

(ii) Following are some of the forms of energy:

Thermal energy, chemical energy, electric energy, electromagnetic energy, nuclear energy, magnetic energy, mechanical energy, sound energy, etc.

(iii) The kinetic energy of a body of mass 'm', moving with a velocity 'v' is given by 12mv2.

As both the bodies have equal masses, let's say 'M', then the ratio of their kinetic energies = K.E.1K.E.2=12M(v)212M(2v)2=v24v2=14

Hence, the ratio of kinetic energies of both bodies is 1:4.



Page No 377:

Question 11:

(i) Name four ways in which sound can be produced.
​(ii) Calculate the speed of a sound wave whose frequency is 2 kHz and wavelength 65 cm.

Answer:

(i) Four different ways to produce the sound:

(1) Sound can be produced by hitting a hard surface. i.e. Metals, walls, etc.

(2) Sound can be produced by generating vibrations, like in a guitar.

(3) Plucking the rubberband by the hand can produce sound.

(4) Sound can be produced by scratching hard or rough surfaces.


(ii) According to relation, Speed of the sound wave, v=f λ (Where 'f' is the frequency of the wave and 'λ' is the wavelength)

v=2×103 Hz×65×10-2 m=1300 m/s

Hence, the speed of the given sound wave is 1300 m/s. 

Page No 377:

Question 12:

Define: (i) Hypertonic solution; (ii) Hypotonic solution, (iii) Isotonic solution.

Answer:

(i) Hypertonic solution: A solution having a higher concentration of solute as compared to another solution is called hypertonic solution . In a hypertonic solution, a cell progressively loses water through exosmosis.

(ii) Hypotonic solutions: A solution having a lesser concentration of solute as compared to another solution is called hypotonic solution . In a hypotonic solution, a cell progressively gains water through endosmosis. 

(iii) Isotonic solutions: A solution having a concentration equal to that of another solution is called isotonic solution. In an isotonic solution, a cell neither gains nor loses water. 

Page No 377:

Question 13:


Label A, B, C and D in the given figure. Give the function of B.

Answer:

A represents the dorsal fin of the fish.

B represents caudal fin of the fish.

C represents ventral fin of the fish.

D represents pelvic fin of the fish.

Caudal fins act as the major locomotory organs in the fishes. These help in the propulsion of fishes through the water. The propulsive force is generated by the tail of the fish.

Page No 377:

Question 14:

It was diagnosed that the body of a patient has lost its power of fighting any infection. Name the disease he is suffering from. What type of microbe is responsible for this disease and how does it spread from one person to the other.

Answer:

The patient in the given case is suffering from Acquired Immuno Deficiency Syndrome (AIDS). In AIDS, the immune system of an individual is severely affected because of the progressive destruction of immune cells. The microbe responsible for AIDS is a virus called Human Immunodeficiency Virus (HIV). 

This virus can be transmitted from one organism to another in the following ways:
1. Unsafe sexual contact
2. Transfusion of contaminated blood
3. Use of infected syringes and needles
4. Vertical transmission from mother to the developing fetus

Page No 377:

Question 15:

Describe the major components of air pollution.

Answer:

Air pollution refers to the introduction of particulate materials in the air, which have an adverse effect on the quality of air. The different components of air pollution can be categorised into following types:
1. Particulate matter: These substances are also referred to as suspended particulate matter (SPM) because of their ability to remain suspended in the air for long periods. Example includes aerosols.
2. Gaseous pollutants: These include gaseous substances that have the potential to deteriorate the quality of air. Examples of gaseous pollutants are hydrogen sulphide, carbon monoxide and sulphur dioxide.

Page No 377:

Question 16:

(i) What is Brownian motion? Draw a diagram to show the movement of a particle (like a pollen grain) during Brownian motion.
​(ii) In a beam of sunlight entering a room, we can sometimes see dust particles moving in a haphazard way in the air. Why do these dust particles move?

Answer:

(i) Suspended particles in a liquid or a gas move in a random zig-zag motion. This type of motion is known as Brownian motion. Below is the diagram to show the Brownian motion of a particle (like a pollen grain):



(ii) The dust particles we see in a beam of sunlight, move in a haphazard way in the air because they are continuously hit by the fast-moving particles of air. 

Page No 377:

Question 17:

How is water purified on a large scale at water works? Explain with the help of a labelled diagram. Name the substance which is added to kill germs in the drinking water supply?

Answer:

The water is purified at a large scale in water treatment plants:
Following three steps are involved for purification of water at large scale:
1. Sedimentation: In this process, all the large impurities are removed from the water.
2. Filtration: In this process, the water is filtered through different types of filters that trap the minute impurities present in the water.
3. Disinfection: In this process, the water is disinfected to remove the remaining microorganisms from the water. 

Water purification on a large scale can be done as shown below:



Chlorine gas is passed through the water to kill germs in the drinking water supply.

Page No 377:

Question 18:

(i) Write the three equations of uniformly accelerated motion. Give the meaning of each symbol which occurs in them.
(ii) A car acquires a velocity of 72 km per hour in 10 seconds starting from rest. Find
​      (a) the acceleration, (b) the average velocity, and (c) the distance travelled in this time.

Answer:

(i) First equation of motion: v= u+at (Where 'u' is the initial velocity, 'v' is the final velocity, 'a' is the acceleration and 't' is the time taken)

Second equation of motion: s=ut+12at2 (Where 's' is the displacement, 'u' is the initial velocity, 'a' is the acceleration and 't' is the time taken)

Third Equation of motion: v2=u2+2as (Where 'u' is the initial velocity, 'v' is the final velocity, 'a' is the acceleration and 's' is the displacement)

(ii) (a) Initial velocity of the body = 0 m/s

Final velocity of the body = 72 km/h = 72× 1000 m3600 s=20 m/s

Time taken = 10 s

Acceleration = Final velocity - Initial velocityTime taken=20 m/s - 010 s=2 m/s2

Hence, the acceleration produced is 2 m/s2

(b) Using the second equation of motion, s = ut+12at2

s=0+12(2)(10)2= 100 ms =100 m

Total displacement = 100 m

 Average velocity = Total displacementTotal time taken=100 m10 s=10 m/s

(c) Here, the body is traveling in a straight line motion. So, displacement is equal to the distance. 

Hence, distance travelled by the body is the same as the displacement calculated above. i.e. 100 m.

Page No 377:

Question 19:

(i) If a balloon filled with air and its mouth untied, is released with its mouth in the downward direction, it moves upwards. Why?
​(ii) An unloaded truck weighing 2000 kg has a maximum acceleration of 0.5 m/s2. What is the maximum acceleration when it is carrying a load of 2000 kg?

Answer:

(a) The air filled in the balloon ejects at very high speed, which exerts an upward thrust force on the balloon to conserve the momentum of the whole system. Hence, the balloon moves in an upward direction. 

(b) Mass of the unloaded truck = 2000 kg

Maximum acceleration = 0.5 m/s2

The maximum force developed by the truck = mass×acceleration =2000 kg × 0.5 m/s2=1000 N

Now, the mass of the loaded truck = 4000 kg

Maximum acceleration of the loaded truck = Maximum ForceTotal Mass=1000 N4000 kg=0.25 m/s2

Hence, the maximum acceleration of the loaded truck is 0.25 m/s2
 

Page No 377:

Question 20:

Describe with a diagram the fluid mosaic organisation of the plasma membrane.

Answer:

The diagrammatic representation for the fluid-mosaic model is as follows:


The fluid mosaic model of the plasma membrane was given by Singer and Nicolson in 1972. According to this model, the plasma membrane is composed of a bilayer of lipid. This lipid bilayer encompasses two types of proteins called intrinsic proteins and extrinsic proteins. Intrinsic proteins span the lipid bilayer whereas, extrinsic proteins are attached to the lipid bilayer either from the inside or outside of the cell. These proteins are considered as floating in the lipids. Moreover, these proteins perform numerous functions such as catalysis, transportation and endocytosis.

Page No 377:

Question 21:

Define muscular tissue. Classify and explain different types of muscles with the help of suitable diagrams.

Answer:

Muscle tissue is one of the four basic tissues found in animals. It is a type of contractile tissue composed of muscle cells. These cells are also described as muscle fibers because of their large size and elongated nature. Muscle fibers are categorized into the following three types on the basis of their structure, location, and function:

(i) Striated muscles:


Striated muscle fibers are also called voluntary or skeletal muscles. These muscles are elongated, unbranched and multinucleated. Moreover, these muscles are attached to the bones and assist in body movements. 

(ii) Smooth muscles:


Smooth muscles are also called involuntary or unstriated muscles. These muscles are present as a fusiform elongated sheet. A single nucleus is present in these muscles and myofibrils (contractile threads) are present longitudinally in these cells.

(iii) Cardiac muscles:

Cardiac muscles are extensively branched and these branches join to form a compact network. Light and dark bands along with a single nucleus can be observed in the cells of these muscles. These muscles are found in the heart.



Page No 378:

Question 22:

Bromine and air take about 15 minutes to diffuse completely but bromine diffuses into a vacuum very rapidly. Why is this so?

Answer:

When bromine diffuses in the air, the bromine particles bump into the already present air particles due to which diffusion is slow but in vaccum, there are no air particles in the way of bromine particles, therefore bromine particles can diffuse fast.

Page No 378:

Question 23:

A solid mixture contains four constituents P, Q, R and S. P consists of tiny grains and it is mixed with cement for plastering the walls. Q is a white solid which is recovered on a large scale from sea water by the process of evaporation. R is in the form of tiny particles of a material whose corrosion is called rusting. And S is a white solid which is used in making ordinary dry cells.
(i) What could P, Q, R and S be?
(ii) How would you separate a mixture containing P, Q, R and S?

Answer:

(i) P is mixed with cement for plastering the wall, so P must be sand.
Q is a white solid recovered from seawater by the process of evaporation, so Q must be sodium chloride.
R is a material whose corrosion is called rusting, so R must be iron fillings.
S is a white solid used for making ordinary dry cells, so S must be ammonium chloride.

(ii) Iron filings can be separated by a magnet. The remaining mixture can be heated and ammonium chloride will sublime that can be collected by an inverted funnel. The other two can be mixed in water and then filtered. Salt being soluble in water will dissolve in it that can be obtained by evaporation of water. Sand can be collected as a residue of the filtration process.

Page No 378:

Question 24:

A body is moving uniformly in a straight line with a velocity of 5 m/s. Find graphically the distance covered by it in 5 seconds.

Answer:



A graph has been shown above for the given motion.

For a uniform motion in a straight line, the area under the velocity-time graph gives the distance covered by the body. 

The area under the graph for the given data = 5 m/s × 5 s=25 m

Hence, the distance covered by the body in 5 seconds is 25 m.

Page No 378:

Question 25:

A moving bicycle comes to rest after some time if we stop pedaling it. But Newton's first law of motion says that a moving body should continue to move forever unless some external force acts on it. How do you explain the bicycle case?

Answer:

If we stop pedaling then the bicycle comes to rest because of the frictional forces of the ground. The friction force of the ground is an external force which opposes the motion of the bicycle. Hence, after some time the bicycle stops.

Page No 378:

Question 26:

If Z = 3, what would be the valency of the element? Also name the element.

Answer:

If Z = 3, that means the atomic number of the element is 3. 
Its electronic configuration will be 2, 1.
Hence, the valency of the element is 1.

The element with Z = 3 is lithium(Li).

Page No 378:

Question 27:

What are meristematic tissues? Explain with the help of suitable diagram. Give their classification on the basis of their position in the plant body.

Answer:

Meristematic tissues are formed by cells which show continuous division. These tissues are responsible for increasing the length as well as the birth of a plant. 


Figure. 1. Different types of meristematic tissues.

Meristematic tissues are categorized into three types on the basis of their position in the plant. These are as follows:

(i) Apical meristem: These meristematic tissues are present at the growing tips of roots and stems, which constitute the shoot apex. These tissues are also present in the apical regions of the leaves.

(ii) Intercalary meristems: These meristematic tissues are found at the bases of leaves or internodes. In many cases, these tissues can also be found below the nodes.

(iii) Lateral meristems: These tissues are present below the cork cambium and can also be found in vascular cambium as well as vascular bundles of dicot roots.
 



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