Lakhmir Singh Manjit Kaur Physics 2019 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 Lakhmir Singh Manjit Kaur Physics 2019 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 Lakhmir Singh Manjit Kaur Physics 2019 Solutions. All Lakhmir Singh Manjit Kaur Physics 2019 Solutions for class Class 9 Science are prepared by experts and are 100% accurate.

Page No 286:

Question 1:

Name the target organ of Japaneses encephalitis and AIDS virus respectively.

Answer:

Target organ of Japaneses encephalitis - brain 
Target organ of AIDS virus - lymph nodes

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Question 2:

Why nucleus is called director of the cell?

Answer:

Nucleus is called director of the cell as it controls all the functions of the cells. The nucleus contains DNA, which helps in the transmission of information from one generation to the next.

Page No 286:

Question 3:

Five electric fans of 120 watts each are used for 4 hours. Calculate the electrical energy consumed in kilowatt-hours.

Answer:

Energy consumed, E=P×t

E =5×120 W×4 h=2400 WhE =2.4 kWh

Page No 286:

Question 4:

A 60 g bullet fired from a 5 kg gun leaves with a speed of 500 m/s. Find the speed (velocity) with which the gun recoils (jerks backwards).

Answer:

Applying conservation of momentum, we can write here,

Momentum before the bullet leaves the gun =  Momentum after the bullet leaves the gun

Mass of the gun = 5 kg

Mass of the bullet = 60 g

The initial velocity of the gun = Initial velocity of the bullet = 0

The final velocity of the bullet = 500 m/s

The initial momentum of the system = Final momentum of the system

0 + 0 = 60 × 10-3 kg × 500 m/s + 5kg × velocity of the gun

The velocity of the gun = -305=-6 m/s

Hence, the recoil velocity of the gun is 6 m/s. 

Page No 286:

Question 5:

Write down the causal organisms of the following diseases:
Tuberculosis, Kala-azar, Malaria, Measles, Athlete's foot, Cholera.

Answer:

Tuberculosis - Mycobacterium tuberculosis
Kala-azar -  Leishmania donovani
Malaria - Plasmodium spp.
Measles - Paramyxovirus
Athlete's foot - Trichophyton
Cholera - Vibrio cholerae

Page No 286:

Question 6:

A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate:
(a) the velocity with which the gun recoils.
​(b) the force exerted on gunman due to recoil of the gun

Answer:

(a) Applying conservation of momentum, we can write here,

Momentum before bullet leaves the gun =  Momentum after bullet leaves the gun

Mass of the gun = 3 kg, Mass of the bullet = 30 g

The initial velocity of the gun = Initial velocity of the bullet = 0

The final velocity of the bullet = 100 m/s

The initial momentum of the system = Final momentum of the system

0 + 0 = 30 × 10-3 kg × 100 m/s + 3kg × velocity of the gun

The velocity of the gun = -33=-1 m/s

Hence, the recoil velocity of the gun is 1 m/s. 

(b) Change in the momentum of the gun after the bullet leaves the barrel = 3 kg × 1 m/s = 3 kg m/s

Time take by the bullet to leave the barrel = 0.003 s

Force exerted by the recoil of the gun on the gunman = Rate of change of momentum of the gun = 3 kg m/s0.003 s=1000 N

Page No 286:

Question 7:

A liquid compound X of molecular mass 18 u can be obtained from a number of natural sources. All the animals and plants need liquid X for their survival. When an electric current is passed through 200 grams of pure liquid X under suitable conditions, then 178 grams of gas Y and 22 grams of gas Z are produced. Gas Y is produced at the positive electrode whereas gas Z is obtained at the negative electrode. Moreover, gas Y supports combustion whereas gas Z burns itself causing explosions.
(a) Name (i) liquid X (ii) gas Y, and (iii) gas Z.
(b) What is the ratio of the mass of element Z to the mass of element Y in the liquid X?
(c) Which law of chemical combination is illustrated by this example?
(d) Name two sources of liquid X.
​(e) State an important use of Y in our life.

Answer:

Liquid compound X will be water(H2O).
When electrolysis of water is done it splits into hydrogen ion and oxygen ion. The ratio of hydrogen and oxygen is 2:1.

At cathode: 2H++2e-H2At anode: 2OH-H2O+12O2+2e-

(a) Liquid X is Water.
Liquid Y is oxygen.
Liquid Z is hydrogen.

(b) Electrolysis of 1 mole of water produces 2 moles of hydrogen gas and 1 mole of the oxygen gas. Hence, the Ratio of mass of gases, Z and Y is 4:32 or 1:8.

(c) Law of constant proportions is illustrated by the given example.

(d) Liquid X(water) can be found in Ponds and wells.

(e) Y(Oxygen) is required for breathing.

Page No 286:

Question 8:

(a) What are valence electrons? Where are valence electrons situated in an atom?
​(b) What is the number of valence electrons in the atoms of an element having atomic number 13? Name the valence shell of this atom.

Answer:

(a) Valence electrons are electrons present in the outermost orbital of the atom, basically, only valence electrons participate in bond formation.
Valence electrons are present on outermost orbit.

(b) Electronic configuration of the element with atomic number 13 is 2, 8, 3. So, the number of valence electrons present in the given element is 3. The electronic configuration of the given element is as follows:

2      8      3K     L     M

So, the valence shell will be M.



Page No 287:

Question 9:

Describe some flight adaptations of the birds.

Answer:

Some flight adaptations of birds are -

  • Their forelimbs are modified into wings which help them to fly.
  • Their wings are covered with feathers of different types which have their role in flight.
  • Their feathers are highly modified scales which provide the bird with required warmth at high altitudes during flight.
  • They possess long and hollow bones which are connected by air passages (reduces the weight of the body).
  • They have very large breast bone with strong muscles which help them in flight.
  • Their lungs have big air sacs to fill in lot of air to supply the bird with more oxygen.

Page No 287:

Question 10:

What are the causes of 'water pollution'? Discuss how you can contribute in reducing water pollution.

Answer:

a) Water pollution occurs due to a number of reasons such as:

(i) Dumping of domestic, industrial, and agricultural wastes in water

(ii) Occurrence of oil spills in the oceans

(iii) Washing clothes, bathing of animals, etc. in or near water sources


(b) Water pollution can be controlled by:

(i) Restricting ourselves from dumping domestic wastes into water bodies

(ii) Educating people properly about the ill-effects of water pollution

(iii) Controlling the over usage of fertilisers and pesticides

Page No 287:

Question 11:

Describe the structure of a nerve cell.

Answer:

Nerve cells are also known as neurons.

A neuron consists of 3 parts -

(i) Cell body − It contains a nucleus and cytoplasm.

(ii) Axon − It is a long part arising from the cell body. It transmits impulses away from the cell body.

(iii) Dendrites − These are short, branched parts arising from the cell body. They receive the nerve impulses.

Page No 287:

Question 12:

How will you separate common salt, sand and iron fillings from their mixture?

Answer:

We can use a magnet to separate the iron filings from the mixture of common salt, sand and iron filings. Sand and salt can be separated by adding them to water and allowing the salt to dissolve and the sand to settle. When the water is decanted(filtered), sand is left behind. By boiling the decanted water and letting it evaporate, salt can be obtained too.
 

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Question 13:

Write about three main criteria which should be considered while selecting the crops for rotation?

Answer:

Three main criteria which should be considered while selecting the crops for rotation are -

Nutritional requirement - The plants should be selected in such a way that there nutritional requirements are different. This helps in avoiding competition between the two for nutrition.

Root pattern - The plants should be selected in a manner where one is deep rooted while the other is short rooted. This allows maximum utilisation of nutrients and avoiding competition.

Crop maturation - There should be a difference in the time of maturity of the two plants. One should mature earlier than the other.

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Question 14:

(a) State the principle of flotation.
(b) A floating boat displaces water weighing 6000 newtons.
      (i) What is the buoyant force on the boat?
     â€‹ (ii) What is the weight of the boat?

Answer:

(a) According to the principle of floatation, an object will float in a liquid if the weight of an object is equal to the weight of the liquid displaced by it.

(b) Weight of the displaced liquid = 6000 N

(i) As we know that, when a body is immersed in a liquid it feels an upward buoyant force which is equal to the weight of the liquid displaced by the body. So, the buoyant force on the boat = weight of the liquid displaced by the boat = 6000 N.

(ii) As the boat is floating, so according to the law of floatation the weight of the boat must be equal to the weight of the liquid displaced by it. Hence, the weight of the boat is also 6000 N. 
 

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Question 15:

The following data was obtained for a body of mass 1 kg dropped from a height of 5 metres:

Distance above ground Velocity
5 m 0 m/s
3.2 m 6 m/s
0 m 10 m/s
Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance). (g = 10 m/s2).

Answer:

Law of conservation of energy says if there are no external forces working on the object then at any point of time the total energy of the object remains conserved.

Mass of the body = 1 kg

Case I:

Velocity of the body at height 5 m = 0 m/s

The potential energy at the height of 5 meters = mgh = 1 kg× 10 ms-2 × 5 m = 50 J

Kinetic energy at a height of 5 meters = 0

The total energy of the body at a height of 5  meters = 50 J +0 = 50 J


Case II:

The velocity of the body at the height of 3.2 meters = 6 m/s

The potential energy at the height of 3.2 meters = mgh = 1 kg× 10 ms-2 × 3.2 m = 32 J

Kinetic energy at a height of 5 meters = 12mv2=12×1×62=18 J

The total energy of the body at a height of 3.2 meters = 32 J + 18 J = 50 J


Case III:

The velocity of the body at the bottom = 10 m/s

The potential energy at the bottom = mgh = 1 kg× 10 ms-2 × 0 m = 0

Kinetic energy at the bottom = 12mv2=12×1×(10)2=50 J

The total energy of the body at the bottom = 0 J + 50 J = 50 J


Hence, we can see through all the three cases that the total energy of the body was conserved throughout the motion. 

 

Page No 287:

Question 16:

(a) State and explain the law of conservation of energy with an example.
​(b) Explain how, the total energy a swinging pendulum at any instant of time remains conserved. Illustrate your answer with the help of a labelled diagram.

Answer:

(a) The law of conservation of energy states that energy can neither be created nor be destroyed, but it can be transformed from one form to another. Example: When a body is thrown upward with some velocity, the total energy throughout the motion remains conserved. As the body moves upward, the kinetic energy of the body gradually gets converted into potential energy of the body and when the body falls back to the ground the potential energy of the body gets converted back into its kinetic energy and the body return back with the same velocity it was thrown upward.

(b)


A swinging pendulum is a perfect example to show the conservation of energy. The total energy of the bob of the pendulum is the sum of the potential energy and kinetic energy. Initially, the bob of the pendulum is at the mean position (B). When the bob is drawn to one side (Extreme position A), and raised to a little height, it gains some potential energy.

This is the energy transferred by work done by the hand. At the extreme position, the bob has only P.E. When it's left at the extreme position, it moves down. Then the P.E. decreases and K.E. increases. At the lowest (mean) position, the bob only has the K.E.

But as soon as it moves further to mean position, its K.E. decreases, and P.E. increases. At the extreme positions A and C, all energy is in the form of potential energy and therefore it tends to move down. At all other intermediate positions, the energy of the pendulum is partly potential and partly kinetic. But, the total energy of the pendulum remains conserved.

Page No 287:

Question 17:

Write short notes on- (a) Tuberculosis; and (b) Polio.

Answer:

Tuberculosis 
Tuberculosis is a bacterial disease which is caused by the bacterium Mycobacterium tuberculosis. This bacterium mainly affects the lungs but it can attack the brain, intestine, eyes etc. 

Modes of transmission -
It spreads through air.

Some of the symptoms of tuberculosis include − cough, low grade fever, loss of weight, chest pain, difficulty in breathing, etc.

Precautionary measures that must be taken to save oneself from the disease are−

  • Taking the B.C.G. vaccine
  • Isolation of the patient so that the disease does not spread.
  • Proper aeration and sanitary conditions

Polio
Polio is a disease caused by poliomyelitis virus and generally affects children. It generally occurs during warmer summer months and can cause paralysis of limbs.

Modes of transmission 
Spreads as droplet infection and through faeces and nasal secretion.

Symptoms - Fever, headache, stiffness or paralysis of limbs

Prevention
  • Administering oral polio vaccine to children in the age group of 18−24 months.

Page No 287:

Question 18:

With the help of a labelled diagram show
(a) Nitrogen cycle in nature
​(b) Describe briefly any two processes involved in the cycling of N2 in the environment.

Answer:

(a) Nitrogen-cycle is a biogeochemical cycle. It describes the transformation of atmospheric nitrogen to simpler molecules in soil and water that get converted to more complex molecules and eventually are converted back to the simple nitrogen molecule in the atmosphere. The given figure shows the nitrogen cycle.



(b) Two processes involved in the cycling of N2 in the environment are -

  1. Nitrogen fixation: N NO3 − or NH4

It is the process wherein atmospheric nitrogen is converted into water-soluble nitrates. This step is performed by organisms like RhizobiumAzotobacter and blue-green algae.

  1. Nitrification: NH3  NO2 −  NO3 −

It is the process in which ammonia is first converted into nitrites and then into nitrates. This step is performed by nitrifying bacteria like Nitrosomonas and Nitrobacter.

Page No 287:

Question 19:

(a) Define the following terms: (i) Echolocation (ii) Echocardiography, and (iii) Ultrasonography.
(b) Name an animal which navigates and finds food by echolocation.
(c) Which of the two produces ultrasonic waves: porpoise or whale?

Answer:

(a) (i) Echolocation is a method used by few animals like bats, porpoises and dolphins to locate the objects by hearing the echoes of their ultrasonic squeaks.

(ii) When ultrasound waves are used to create the images of the heart of our body, the technique used is known as the echocardiography.

(iii) The technique of obtaining the pictures of internal organs of the body by using echoes of ultrasound pulses is called ultrasonography.

(b) A bat navigates and finds food by echolocation.

(c) A porpoise produces ultrasonic waves. 



Page No 288:

Question 20:

(a) What is Brownian motion? Draw a diagram to show the movement of a particle (like a pollen grain) during Brownian motion.
​(b) 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:

(a) If we see a colloidal solution through a strong ultramicroscope, then we observe that the colloidal particles are in a state of continuous zig-zag motion. This type of motion is known as Brownian motion named after British botanist, Robert Brown.




(b) When a beam of light enters a room, the dust particles can be seen moving in a haphazard manner because they are constantly hit by the fast-moving particles of air. Those tiny dust particles scatter the light ray, that's why we are able to their random motion in the air.

Page No 288:

Question 21:

Three different atoms of oxygen are represented as:
O816,O817 and â€‹O818
(a) What do the subscripts (lower figures) and superscripts (upper figures) represent?
(b) What factor is responsible for the change in the superscripts 16, 17 and 18, though the element is the same?
(c) What is the usual name for such atoms of an element?
(d) Give the nuclear composition of O818​.

Answer:

(a) Subscripts represent the atomic number while superscripts represent the atomic mass of the atom.

(b) The number of neutrons is different for all the three given atoms, so the value of atomic mass is different too. That's why superscripts are different though the element is the same.

(c) The atoms which have the same atomic number but a different mass number are known as isotopes. 

(d) Nuclear composition of O818:
Number of electrons = 8
Number of protons = 8
Number of neutrons = Atomic mass - Number of protons
Number of neutrons = 18 - 8
Number of neutrons = 10
 

Page No 288:

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 a vacuum, there are no air particles in the way of bromine particles, therefore bromine particles can diffuse fast.

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Question 23:

Government has recently taken series of steps to minimise cigarette smoking and tobacco chewing by people as these are injurious to health.
(a) List at least three steps taken by government to aware people about harmful effects of cigarette smoking and tobacco chewing.
(b) What is passive smoking?
​(c) List at least one chronic disease that occurs due to continuous use of tobacco.

Answer:

(a) Three steps to minimise the use of cigarette smoking and tobacco chewing are -

  • Restricting the age of buying these products to 18 years
  • Advertisement in television regarding the side effects of use of these products.
  • Printing warnings such as "Smoking is injurious to health" and certain images on packets of such products.
(b) The involuntary inhalation of smoke by one person from another person's cigarettes, cigars, or pipes is called as passive smoking. 
(c) Continuous use of tobacco can lead to cancer of mouth.

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Question 24:

An element X forms the following compounds with hydrogen, carbon and oxygen:
H2X, CX2, XO2, XO3
​State the three valencies of element X which are illustrated by these compounds.

Answer:

Valency of X in H2X = -2
Valency of X in CX2 = -2
Valency of X in XO2  = +4
Valency of X in XO3 = +6

Hence, element X is showing three different valencies which are -2, +4 and +6.

Page No 288:

Question 25:

What will happen if
(a) ligament gets over stretched?
(b) heparin is absent in blood?
​(c) striated muscles contract rapidly for longer duration?

Answer:

(a) Ligaments belong to the group of connective tissues which connect bones to other bones at joints. It is made up of strong band of fibers and overstretching of ligaments can cause small breaks in the fiber or twisting and tearing of them. This can lead to severe pain and difficulty in performing daily chores.

(b) Heparin is an anticoagulant which is found in blood. It prevents the blood from clotting. If it is absent in blood, then the blood would coagulate inside the blood vessels.

(c) Rapid contraction of striated muscles for longer duration will cause fatigue of these muscles due to production of lactic acid.

Page No 288:

Question 26:

A ball hits a wall horizontally at 6.0 m s−1. It rebounds horizontally at 4.4  m s−1​. The ball is in contact with the wall for 0.040 s. What is the acceleration of the ball?

Answer:

Acceleration is simply the rate of change of velocity. 

The initial velocity of the ball =  6.0 ms−1

The final velocity of the ball = 4.4  ms−1

Change in velocity = 4.4  ms−1 - 6.0 ms−1-1.6 ms−1 

Time taken = 0.040 s

Acceleration of the ball = -1.6 ms-10.040 s=-40 ms-2



Page No 289:

Question 27:

In an experiment, Anhad studies sound waves. He sets up a loudspeaker to produce sound as shown below:



Anhad adjusts the signal to the loudspeaker to give a sound of frequency 200 Hz.
(a) What happens to the air in-between Anhad and the loudspeaker?
(b) Explain how Anhad receives sound in both ears.

Answer:

(a) When Anhad adjusts the signal to the loudspeaker to give a sound of frequency 200 Hz then the air between Anhad and the loudspeaker starts vibrating with the frequency of 200 Hz.

(b) Both the ear of Anhad receive the sound. His right ear receives the sound waves coming directly from the loudspeaker whereas his left ear receives the reflected sound waves from the wall of the classroom.



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