Lakhmir Singh & Manjit Kaur 2020 Solutions for Class 9 Science Chapter 10 - Model Test Paper 4

Answer:

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

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.

Answer:

Energy consumed, $E=P\times t$

$$\begin{gathered} E =5\times 120 \mathrm{W}\times 4 \mathrm{h}=2400 \mathrm{Wh} \\ \Rightarrow \mathrm{E} =2.4 \mathrm{kWh} \end{gathered}$$

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 $\times$ 10^-3 kg $\times$ 500 m/s + 5kg $\times$ velocity of the gun

The velocity of the gun = $-\frac{30}{5}=-6 \mathrm{m}/\mathrm{s}$

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

Answer:

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

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 $\times$ 10^-3 kg $\times$ 100 m/s + 3kg $\times$ velocity of the gun

The velocity of the gun = $-\frac{3}{3}=-1 \mathrm{m}/\mathrm{s}$

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 $\times$ 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 = $\frac{3 \mathrm{kg} \mathrm{m}/\mathrm{s}}{0.003 \mathrm{s}}=1000 \mathrm{N}$

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.

$$\begin{gathered} \mathrm{At} \mathrm{cathode}: 2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2 \\ \mathrm{At} \mathrm{anode}: 2{\mathrm{OH}}^{-}\to {\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2+2{\mathrm{e}}^{-} \end{gathered}$$

(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.

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:

$$\begin{gathered} 2 8 3 \\ \mathrm{K} \mathrm{L} \mathrm{M} \end{gathered}$$

So, the valence shell will be M.

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.

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

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.

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.
 

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.

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. 
 

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$\times$ 10 ms^-2 $\times$ 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$\times$ 10 ms^-2 $\times$ 3.2 m = 32 J

Kinetic energy at a height of 5 meters = $\frac{1}{2}m{v}^2=\frac{1}{2}\times 1\times 6^2=18 \mathrm{J}$

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$\times$ 10 ms^-2 $\times$ 0 m = 0

Kinetic energy at the bottom = $\frac{1}{2}m{v}^2=\frac{1}{2}\times 1\times (10{)}^2=50 \mathrm{J}$

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. 

 

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.

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

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

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: N₂  NO₃ ⁻ or NH₄

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

2. 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.

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. 

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.

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 ${}_8{}^{18}\mathrm{O}$:
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
 

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.

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.

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.

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.

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 = $\frac{-1.6 {\mathrm{ms}}^{-1}}{0.040 \mathrm{s}}=-40 {\mathrm{ms}}^{-2}$

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.