P S Verma V K Aggarwal Biology%e2%80%932019 Solutions for Class 9 Science Chapter 10 Model Test Paper 4 are provided here with simple step-by-step explanations. These solutions for Model Test Paper 4 are extremely popular among Class 9 students for Science Model Test Paper 4 Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the P S Verma V K Aggarwal Biology%e2%80%932019 Book of Class 9 Science Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s P S Verma V K Aggarwal Biology%e2%80%932019 Solutions. All P S Verma V K Aggarwal Biology%e2%80%932019 Solutions for class Class 9 Science are prepared by experts and are 100% accurate.

Page No 379:

Question 1:

Which is the longest cell in the human body?

Answer:

The longest cell present in the human body is neuron.

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Question 2:

Who coined the term biodiversity?

Answer:

The term biodiversity was coined by Walter G. Rosen.

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Question 3:

A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? (g = 9.8 m/s2)

Answer:

The initial velocity of the ball, u = 15 m/s

Acceleration = -g

Final velocity, v = 0 m/s

Let the maximum height attained by the ball before it begins to fall back = H

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

0=(15)2-2×9.8×H19.6H=(15)2H=(15)219.6=22519.6=11.48 mH = 11.48 m

Hence, the maximum height attained by the ball is 11.48 m.

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Question 4:

A sound signal of 128 vibrations per second has a wavelength of 2.7 m. Calculate the speed with which the wave travels.

Answer:

Frequency of the given signal = 128 Hz (because it does 128 vibrations per second) 

The wavelength of the given signal = 2.7 m

According to relation, (Wave Speed) v =f (Frequency) × λ (wavelength)

v=fλ=128 Hz× 2.7 m=345.6 m/sv=345.6 m/s

Hence, the speed of the wave is 345.6 m/s.

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Question 5:

Why are plant and animal tissues different?

Answer:

Plant and animal tissues are different because plants tend to grow throughout their life, whereas animals grow up to a certain period. Moreover, plants do not show locomotion but animals show locomotion, which necessitates the presence of different types of tissues.

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Question 6:

State two characteristics of matter demonstrated by:
(i) diffusion
(ii) Brownian motion

Answer:

(i) Diffusion: 
a) Particles of matter are always in motion.
b) Particles of matter are made up of very small size.

(ii) Brownian motion:
a) Particles of matter are always in constant random motion.
b) Particles of matter have space between them.

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Question 7:

What is the significance of the formula of a substance? The molecular formula of glucose is C6H12O6. Calculate its molecular mass. (Atomic masses: C = 12 u; H = 1 u; O = 16 u)

Answer:

The significance of the formula of the substance is that it tells us about the components of the substance and tells about their chemical relationship.

The molecular mass of glucose,
=6×C+12×H+6×O=6×12+12×1+6×16=72+12+96=180 u

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Question 8:

(i) What is the nucleus of an atom and what is the nature of charge on it?
(ii) Name the scientist who discovered the nucleus of atom.

Answer:

(i) The nucleus of the atom is the central region which contains positively charged protons and neutral neutrons. Basically, the nucleus of the atom is positively charged.

(ii) Ernest Rutherford by his gold foil experiments discovered the nucleus of the atom. 
 

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Question 9:

(i) Write the formula for acceleration. Give the meaning of each symbol which occurs in it.
(ii) A train starting from Railway Station attains a speed of 21 m/s in one minute. Find its acceleration.

Answer:

(i) Acceleration is defined as the rate of change of velocity.

If 'v' is the final velocity attained by the body, 'u' is the initial velocity of the body and 't' is the time taken by the body to change its velocity from 'u' to 'v'.

Then Acceleration, a = v-ut

(ii) The initial speed of the train = 0 m/s

Final speed of the train = 21 m/s

Time taken by the train = 1 minute = 60 seconds

Using the relation, Acceleration, a = v-ut

a=21 m/s - 060 s=0.35 m/sa=0.35 m/s

Hence, the acceleration of the train is 0.35 m/s.
 

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Question 10:

On what factor does the inertia of a body depend? Which has more inertia, a cricket ball or a rubber ball of the same size? When a tree is shaken, its fruits and leaves fall down. Why?

Answer:

Inertia is defined as the ability or tendency of a body to resist any change in its current state of motion or of rest. Due to inertia, a body continues to exist in the same state unless an external force is applied to it. So, basically, Inertia depends upon the mass of the body. More the mass of the body more is the resistance to change the current state.

Out of cricket ball and rubber ball of the same size, the cricket ball has the more mass, so it has more inertia than the rubber ball.

When a tree is shaken, its fruits and leaves fall down. This is due to Inertia. When a tree at rest (at first) is shaken, the branches of the tree gains motion but the fruits and leaves tend to be in the state of rest due to the inertia of rest. Hence, they fall down.

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Question 11:

The frequency of the sound emitted by the loudspeaker is 1020 Hz. Calculate the wavelength of the sound wave in air in cm where its velocity is 340 m/s.

Answer:

Frequency of the wave = 1020 Hz 

The velocity of the wave = 340 m/s

According to relation, (Wave Speed) v =f (Frequency) × λ (wavelength)

v=fλ=1020 Hz×λ=340 m/sλ=340 m/s1020 Hz=0.33 m =33 cmλ=33 cm

Hence, the wavelength of the wave is 33 cm. 



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Question 12:

Plasma membrane is permeable to water. How does a cell show endosmosis or exosmosis?

Answer:

The movement of water into and outside the cell depends on the concentration of solutes present in the water. In case water inside the cell has low solute concentration than the water present outside the cell, the water will move from the cell to the outside. This is called exosmosis. In case the solute concentration outside the cell is lesser than the solute concentration inside the cell, water will move into the cell. This is called endosmosis.

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Question 13:

Observe the following figure and answer.

(i) What is shown by this figure?
​(ii) What is depicted by labels A, B, C, D and E.

Answer:

(i) The given figure shows different modes through which pathogens can be transmitted from one organism to another.

(ii) Label A represents the direct transmission of pathogens through air (airborne pathogens).
Label B represents the transmission of pathogens through droplets suspended in the air.
Label C represents the the transmission of airborne pathogens through hands.
Label D represent pathogens that can survive drying.
Label E represent transmission of pathogens through dust particles.

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Question 14:

What is bio magnification? Name two heavy metals which, when magnified resulted in diseases in humans.

Answer:

Bio magnification is a process in which the concentration of a non-biodegredable substance such as insecticide progressively increases from one trophic level to another. 

Typical example of bio magnification includes bio magnification of element mercury (Hg), which can lead to minamata disease in the human population. Moreover, the progressive bio accumulation of element cadmium (Cd) has resulted in itai itai disease in Japanese population.

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Question 15:

Give two examples each of
(i) Narrow leaved Rabi season weed
​(ii) Broad leaved Kharif season weed

Answer:

Weeds can be defined as the unwanted plants that grow in the field along with the crops. Most of the weed infestation is observed during the kharif season. Weeds are classified into narrow-leaf weeds and broad-leaf weeds on the basis of their morphology.

(i) Narrow-leaf weeds are a type of monocot weeds that can infest crops during kharif as well as rabi season. Typical examples of narrow leaved rabi season weeds include Wild oat (Jangali jaii) and Phalaris (Mandoosi).

(ii) Broad-leaf weeds are dicot weeds. Examples of broad-leaved kharif season weeds include Xanthium strumarium (Gokhroo) and Amaranthus viridis (Chaulii).

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Question 16:

(i) State the various postulates of Dalton's atomic theory of matter.
(ii) Which postulates of Dalton's atomic theory can explain the law of conservation of mass?
​(iii) Which postulate of Dalton's atomic theory can explain the law of constant proportions?

Answer:

(a) Postulates of Dalton's atomic theory:
(i) All matter is made up of very tiny indivisible particles called atoms.
(ii) All the atoms of a given element are identical in mass and chemical properties whereas those of different elements have different masses and chemical properties.
(iii) Atoms of different elements combine in a fixed whole number ratio to form compounds.
(iv) Chemical reactions involve reorganization of atoms. Atoms can neither be created nor be destroyed.

(b) Atoms can neither be created nor be destroyed in a chemical reaction, it can be changed from one form to another. So, this explains the law of conservation of mass.


(c) Atoms of different elements combine in a fixed whole number ratio to form compounds. So, this explains the law of constant proportions.

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Question 17:

(i) Describe the Rutherford's model of an atom. State one drawback of Rutherford's model of the atom.
​(ii) The mass number of an element is 23 and it contains 11 electrons. What is the number of protons and neutrons in it? What is the atomic number of the element?

Answer:

(i) Rutherford bombarded gold foil with alpha particles.
Set up for Rutherford's experiment:
1. A thin gold foil, approximately 1000 atoms thick, was taken. Gold was chosen for its high density.
2. A detector screen with a small slit was placed around the foil.
3. A source of alpha particles was kept in front of the foil.
4. The foil was bombarded with fast moving alpha particles.
The set up for the Rutherfords gold foil experiment is shown in the figure.

Rutherford's expectation and observations

Rutherford expected that alpha particles would pass straight through the foil and only a small fraction of alpha particles would be deflected. This expectation was in compliance with Thomson's atomic model.
Rutherfords findings were contrary to his expectation. He observes that:
1. Most of the fast-moving alpha particles passed straight through the gold foil.
2. Some particles were deflected through the foil by small angles.
3. One out of every 12000 particles rebounded, i.e. they got deflected by an angle of 180°.

Rutherford then carefully studied his observation and made the following conclusions:
1. Most alpha particles pass through the gold foil without any deflection. This indicates that most of the space inside an atom is empty.
2. Very few particles suffered a deflection from their path. This means that positive charge occupies very little space inside an atom.
3. Only a few fractions of particles underwent a 180° deflection. This shows that the entire positive charge and mass of an atom are present within a very small volume inside the atom.

The drawback of Rutherford's model of the atom:
a) Model is applicable for one electron system.
b) According to Rutherford as the electron revolves around the nucleus it liberates energy regularly so when the electron loses all its energy it should fall in the nucleus which was impossible.


(ii)
The mass number of the element = 23
Number of electrons = 11
Now,
Number of electrons = Number of protons
So, the number of protons = 11
Number of neutrons = Mass number of element - Number of protons
Number of neutrons = 23 - 11
Number of neutrons = 12

As the number of electrons are 11, so the atomic number of the element will be 11.
 

Page No 380:

Question 18:

(i) What do you understand by the term 'acceleration due to gravity of earth'?
(ii) What is the usual value of the acceleration due to gravity of earth?
​(iii) State the SI unit of acceleration due to gravity.

Answer:

(a) Due to the gravitational force of attraction, earth pulls other objects towards itself. The acceleration produced in a body due to the gravitational attraction of the earth is known as the acceleration due to gravity.

(b) The calculated value of acceleration due to the gravity of the earth is 9.8 m/s2.

(c) SI unit of acceleration due to gravity is the same as of the normal acceleration i.e. m/s2.



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Question 19:

(i) What happens to the work done when the displacement of a body is at right angle to the direction of force acting on it? Explain your answer.
​(ii) A force of 50 N acts on a body and moves it a distance of 4 m on a horizontal surface. Calculate the work done if the direction of force is at an angle of 60° to the horizontal surface.

Answer:

(a) Work is done when a body is displaced in the direction of the force. If force is perpendicular to the direction of displacement, then it simply means that no component of force is taking part in moving the body in the direction of given displacement.

Hence, the work done by a force is zero, when it is at the right angle to the direction of displacement of the body. 

(b)
 

Here, = 50 N, Displacement, d = 4 m

The angle between the force and the displacement = 60°

Work is done when the body is displaced in the direction of the force. 

So, the component of force in the direction of displacement = F cos60° = F2

Work done, W=F2×d=50 N2×4 m=100 J

Hence, the work done by the given force is 100 J.

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Question 20:

Explain general characteristics of phylum Platyhelminthes and Nematoda. Give two examples of each.

Answer:

The basic characteristics of phyum Platyhelminthes are as follows:
1. The body is bilaterally symmetrical and dorsoventrally flattened.
2. These animals are triploblastic and lack a body cavity (acoelomates).
3. Respiratory system, circulatory system and skeleton is absent in these organisms.
4.These organisms are hermaphrodites, which means that both female and male reproductive organs are present in same individual.

Typical examples of organism belonging to phyum Platyhelminthes are Schistosoma (blood-fluke) and Taenia solium (pork tape-worm).

The characteristics of phylum Nematoda are as follows:
1. These organism are unsegmented, triploblastic, bilaterally symmetrical and pseudocoelomate animals.
2. Cilia is absent and body is covered by a tough covering called cuticle.
3. Separate sexes can be observed in this phylum.
4. Body is flattened, cylindrical and worm-like.

Examples of organism belonging to phylum Nematoda include Ancylostoma (hook-worm) and Enterobius (pinworm of humans).

Page No 381:

Question 21:

Define diarrhoea. Give an account of occurrence, symptoms, prevention and control of this ailment.

Answer:

Diarrhoea can be described as a group of infections, which primarily affect the intestines. These infections are characterised by persistent discharge of fluidic or semisolid faeces.

Diarrhoea is an infectious disease and is caused due to bacterial and viral infections. Some of the causative agents of diarrhoea are Campylobacter jejuni, Enterovirus and Escherichia coli. This disease can spread through contaminated water, food and hand shakes.

The major symptoms of diarrhoea are as follows:
1. Recurrent vomiting and loose motions, which leads to dehydration.
2. Sunken eyes, irritability and dryness on the inner regions of cheek are common among patients.

Diarrhoea can be prevented by following ways:
1. Contamination can be prevented by covering the eatables.
2. Personal hygiene should be maintained and  hands should be properly washed with soap and water before having any meal.

The condition of diarrhoea can be controlled by taking following measures:
1. Oral Rehydration Solution (ORS) should be given to the patient, so that the extent of dehydration can be reduced.
2. For the treatment process, anti-bacterial drugs and anti-viral agents can be used.
3. The patient should take complete rest to make full recovery from the illness.

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Question 22:

Many indigestion mixtures are suspensions. What do the instructions written on the bottle of an indigestion mixture tell us before taking the mixture, and why?

Answer:

A suspension is a heterogeneous system in which the size of the solute particle is large. As a result, the solute particles do not dissolve in the solvent and settle down at the bottom in the solvent. Therefore, the solute particles in a suspension can be easily seen with a naked eye.

That's, why to shake well before use, is prescribed on the bottles, so that solute mixes with solvent before indigestion of mixture.
 

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Question 23:

Magnesium and oxygen combine in the ratio of 3 : 2 by mass to form magnesium oxide. What mass of oxygen gas would be required to react completely with 24 g of magnesium?

Answer:

Magnesium and oxygen combine to form magnesium oxide in the ratio 3:2
It follows the law of constant proportion,
3 g of magnesium combines with 2 g of oxygen
So,
24 g of magnesium combines = 23×24 g of oxygen=16 g of oxygen

So, 24 g of magnesium requires 16 g of oxygen to react completely.

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Question 24:

A stone and the earth attract each other with an equal and opposite force. Why then we see only the stone falling towards the earth but not the earth rising towards the stone?

Answer:

Stone and earth attract each other with an equal and opposite gravitational force. But, the mass of the earth is very large as compared to the mass of the stone. The gravitational force acting on the earth produces a negligible acceleration in the earth due to its high mass. Whereas mass of the stone is much smaller as compared to the earth, therefore, the same force produces a considerable acceleration in the stone. That's why we are able to see a stone moving towards earth but not earth moving towards the stone. 

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Question 25:

A man X to the top of a building by a vertical spiral staircase. Another man Y of the same mass goes to the top of the same building by a slanting ladder. Which of the two does more work against gravity and why?

Answer:

Work done against gravity for taking a mass 'm' to a height 'h'mgh (Where 'g' is the acceleration due to gravity) .....(1)

In the given question, the mass of man X and man Y are the same and they also go to the same height of the building. So, it does not matter what path they take to reach the top of the building. According to equation (1), the total work done by both of them will always be the same.


 

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Question 26:

A weed is growing on the border of your playing ground. How will you recognise it to be a dicot or monocot?

Answer:

A weed can be characterised as a monocot or dicot on the basis of differences in leaf venation, type of root system and type of flowers. A monocot weed has parallel ventation, fibrous root system and trimerous flowers. Dicot weed, on the other hand, has reticulate venation, tap root system and pentamerous or tetramerous flowers.

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Question 27:

A doctor/nurse/health-worker is exposed to more sick people then others in the community. Find out how she/he avoids getting sick herself/himself.

Answer:

A doctor/nurse/health-worker avoids getting sick by taking following measures:
1. Keeping workplace sterilized and clean by using chemical disinfectants.
2. Ensuring personal hygiene by washing hands after every examination of patient.



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