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#### Question 1:

Give one example each of kharif and Rabi Crops.

#### Answer:

A typical example of the Kharif crop is groundnut, whereas barley is an example of Rabi crop.

#### Question 2:

Mention one method by which living organisms influence the formation of soil.

#### Answer:

Living organisms perform biological weathering, which helps in the formation of soil.

#### Question 3:

A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 105 meters, calculate his speed. (Given $\mathrm{\pi }$= 22/7)

#### Answer:

Speed of the cyclist =

Distance travelled in one round = $2\pi R$

Time taken to complete one round = 2 minutes = 120 seconds

#### Question 4:

If a ringing bicycle bell is held tightly by hand, it stops producing sound. Why?

#### Answer:

A ringing bell produces the sound by vibrating. When we hold the bell by our hands tightly, the bell stops vibrating. Hence, it stops producing sound.

#### Question 5:

How do animals of Porifera differ from animals of Cnidaria?

#### Answer:

Animals belonging to Porifera show cellular level of organisation and intracellular digestion. Animals classified under Cnidaria, on the other hand, show tissue level of organisation and the digestion can be intracellular as well as extracellular.

#### Question 6:

(i) Explain why, steam at 100$°$C is better for heating purposes than boiling water at 100$°$C.
(ii) Which produces more severe burns: boiling water or steam? Why?
(iii) State one condition necessary to liquefy gases (other than applying high pressure.)

#### Answer:

(i) Steam has more energy than boiling water as it possesses the additional latent heat of vaporization. Therefore, steam at 100 $°$C is better for heating purposes.

(ii) Steam has more heat conserved as the latent heat of vaporization in comparison to boiling water, therefore steam produces more severe burns than boiling water.

(iii) The necessary condition to liquify gas is by compressing the gas at a temperature less than its critical temperature.

#### Question 7:

Which of the following can be separated by using a separating funnel and which cannot be separated by using a separating funnel?
(i) water and kerosene mixture
(ii) water and acetone mixture
(Give reasons for your answer.

Which method is better for recovering sugar from sugar solution: evaporation or crystallisation? Give reason for your answer.

#### Answer:

(i) Water and kerosene can be separated by using separating funnel as there is a difference in their densities due to which they can be easily separated. There is a difference in their colour also, so they can be easily separated.

(ii) Water and acetone mix together and form homogenous solution, so it is not possible to separate them using separating funnel.

Crystallization is a solid-liquid separation technique in which the solute transformed from the liquid state to the solid state. The solid state is obtained in the pure crystallization form. As compared to evaporation, it is considered as a better method to obtain sugar from sugar solution because:
1. The sugar will be obtained in pure crystalline form. In case of evaporation, impurities would be present.
2. When we obtain sugar from water, it may be possible that along with water molecules, sugar molecules may also be lost as the sugar may decompose on heating.

#### Question 8:

(i) Name the element used as standard for atomic mass scale.
(ii) Which particular atom of the above element is used for this purpose?
(iii) what value has been given to the mass of this reference atom?

#### Answer:

(i) Carbon is used as a standard for the atomic mass scale.

(ii) IUPAC adopted the mass of carbon (C12) isotope as the standard unit to measure.

(iii)  IUPAC adopted one-twelfth of the mass of carbon (C12) isotope as the reference atom.

#### Question 9:

(i) Explain Newton's second law of motion.
(ii) A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes.

#### Answer:

(i) Newton's second law explains that when a force is applied to a body, it changes the momentum of the body i.e. It changes the state of motion or of rest of the body. The rate of change of momentum of the body is directly proportional to the force applied to the body. If we increase the value of the force, it will create more change in the momentum of the body.

The mathematical form of Newton's second law,

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

Distance travelled by truck in coming down the hill = 400 m

Time taken by the truck = 20 s

Let the acceleration of the truck = a

Using the second equation of motion, we can write, $s=ut+\frac{1}{2}a{t}^{2}$

Mass of the body = 7 metric tons = 1000 kg

Force acting on the body =

#### Question 10:

(i) Which is more fundamental, the mass of a body or its weight? Why?
(ii) How much is the weight of an object on the moon as compared to its weight on the earth? Give a reason for your answer.
(iii) Can a body have mass but no weight? Give reasons for your answer.

#### Answer:

(i) Fundamental quantities are those quantities which cannot be expressed in the form of other quantities i.e. they are independent on their own. Example: Mass, Length, Time, etc.

Since weight is the amount of downward force a body exerts, therefore it is measured in Newton(N)
The unit 'Newton' can be expressed in terms of mass, length and time, so weight cannot be considered as a fundamental quantity.

Hence, mass is more fundamental.

(ii) Weight of an object on the moon is one-sixth of its weight on the earth. Mass and the radius of the earth and the moon are different in such a way that the earth's gravitational attraction is 6 times higher than the moon's gravitational attraction.

(iii) Weight of a body is simply the gravitational pull by any planet. In space, the force of gravity becomes zero. Therefore, the weight of the body becomes zero too. As the mass is a fundamental quantity, so a body can have zero weight but not zero mass.

#### Question 11:

In loading a truck, a man lifts boxes of 100 N each through a height of 1.5 m.
(i) How much work does he do in lifting one box?
(ii) How much energy is transferred when one box is lifted?
(iii) If the man lifts 4 boxes per minute, at what power is he working? (g = 10 m s-2)

#### Answer:

(a) Here the work is done against the force of gravity. F = Weight of the box

Work done to lift the one box up to a height of 1.5 m =

(b) The work done by the man against the force of gravity is stored in the box in the form of its potential energy. Hence, the energy transferred to lift a box is equal to the work done.

Energy transferred in lifting one box = 150 J

(c) Work done by the man in lifting one box = 150 J

Work done by the man in lifting 4 boxes = 150 J$×$4 = 600 J

Time taken by the man to lift the 4 boxes = 1 minute = 60 seconds

Power =

#### Question 12:

(i) Identify A - and B- cells.
(ii) What will happen if B-cells are kept in hypotonic solution
(iii) What will happen if A Cells are kept in hypertonic solution?

#### Answer:

(i) A cells represent a group of turgid cells. B cells represents a group of plasmolysed cells.

(ii) Water has a tendency to move from its higher concentration to its lower concentration. In the given case, higher concentration of water will be present in the hypotonic solution and lower concentration of water will be present within plasmolysed B cells. This difference in concentration of water will result in migration of water into the B-cells and these cells will become deplasmolysed.

(iii) Hypertonic solutions have lower concentration of water as compared to the hypotonic solutions. The tendency of water to flow from its higher concentration to lower concentration will lead to the movement of water outside the  A cell. This will result in gradual loss of water from A cells due to exosmosis and these cells will become plasmolysed cells.

#### Question 13:

Differentiate between parenchyma, collenchyma and sclerenchyma on the basic of their cell wall.

#### Answer:

The differences between parenchyma, collenchyma and sclerenchyma on the basis of cell wall are as follows:

 Parenchyma Collenchyma Sclerenchyma 1. Primary cell wall is present. 1. Primary cell wall is present. 1. Secondary cell wall is present. 2.Thin cell wall, primarily composed of cellulose is present. 2. Cell wall has localised thickening at the corners because of the deposition of cellulose. 2. Extremely thick cell wall is present because of the presence of lignin.

#### Question 14:

Define the given terms alongwith an example (a) Bilateral symmetry, (b) Coelom, (c) Triploblastic.

#### Answer:

(a) Bilateral symmetry refers to the presence of symmetrical left and right regions of the body. This type of symmetry can be observed in Flatworms.

(b) Coelom refers to the body cavity, which generally develops from mesodermal layers. This particular cavity is found in all members of coelomates.

(c) Triploblastic is used to define the presence of three germinal layers (ectoderm, mesoderm and endoderm). All the members of the phylum Nematoda are triploblastic.

#### Question 15:

(i) Differentiate between acute and chronic diseases.
(ii) Give one example each of acute and chronic diseases.
​(iii) Mention any two causes of baby's disease.

#### Answer:

(i) The differences between acute and chronic diseases have been summarised as follows:

 Acute diseases Chronic diseases 1. The effects of these diseases last for a relatively shorter period of time. 1. The effects of these diseases persist for a comparatively longer period of time. 2. The symptoms associated with acute diseases generally last for a short time. 2. The symptoms of chronic diseases can persist for a very long time.

(ii) An example of acute disease includes cholera. Cancer, on the other hand, is a typical chronic disease.

(iii) Baby's disease can be caused because of the following reasons:
1. Babies can inherit defective genes from their parents, which can lead to genetic anomalies.
2. Certain causative agents such as viruses can lead to the development of diseases in babies.

#### Question 16:

How can you obtain pure water from a salt-water mixture (or salt-solution)? Draw a neat ​and labelled diagram of the apparatus you would use to obtain pure water from salt-water mixture (or salt-solution).

#### Answer:

Simple distillation will separate water from salt water solution. In this, the solution is heated and which causes the water to evaporate. Next, the steam is collected, cooled and collected in a separate flask/container. This technique works because the boiling point of water is much lower than the salt.

Diagram showing the apparatus to obtain pure water from salt-water mixture:

#### Question 17:

(i) Describe Bohr's model of the atom. How did's Bohr explain the stability of atom?
​(ii) An element has an atomic number of 11 and its mass number is 23. What is the arrangement of electrons in the shells? State nuclear composition of an atom of the element.

#### Answer:

(a) Bohr's model of the atom:
Niels Bohr proposed the following postulates regarding the model of the atom.
(i) Only certain orbits known as discrete orbits of electrons are allowed inside the atom.
(ii) While revolving in these discrete orbits, the electrons do not radiate energy.
These discrete orbits or shells are shown in the following diagram

Bohr explained the stability of atom by proposing that electrons revolve around the nucleus in the definite circular paths having fixed energy and while moving in the same orbit they do not lose or gain energy.

(b)
The mass number of the element = 23
The atomic number of the element = 11
Number of electrons = 11
Number of protons = 11
Number of neutrons = 23 - 11
Number of neutrons = 12
Electronic configuration of an element is 2     8     1
And shells used are                                   K    L    M

#### Question 18:

(i) Explain the meaning of the following equation of motion: v = u + at, where symbols have their usual meanings.
(ii) A body starting from rest travels with uniform acceleration. If it travels 100 m in 5 s, What is the value of acceleration?

#### Answer:

(i) The given equation, v = u + at is known as the first equation of motion. This equation means that if a body changes its velocity from 'u' to a certain value at a rate 'a' in a time interval 't', then the final velocity 'v' of the body can be obtained by the given equation, v = u + at. The final velocity can be calculated from the equation if the other three quantities are known.

(ii) Distance travelled by the body = 100 m

Time taken = 5 s

Initial velocity = 0 m/s

Using the second equation of motion,

#### Question 19:

(i) Explain by an example of what is meant by potential energy. Write down the expression for the gravitational potential energy of a body of mass 'm' placed at a height 'h' above the surface of the earth.
(ii) What is the difference between potential energy and kinetic energy?
​(iii) A ball of mass 0.5 kg slows down from a speed of 5 m/s to that of 3 m/s. Calculate the change in kinetic energy of the ball. State your answer giving proper units.

#### Answer:

(i) The energy possessed by a body by virtue of its position with respect to the earth is known as the potential energy of the body.

Example: A body placed at a height with respect to the ground have the potential energy.

The potential energy of a body of mass 'm' placed at a height 'h' above the surface of the earth = mgh (Where 'g' is the acceleration due to gravity)

(ii) The energy possessed by a body by virtue of its position with respect to the ground is known as the potential energy of the body. Whereas, the energy possessed by a body by virtue of its motion is known as the kinetic energy.

(iii) Mass of the ball = 0.5 kg

Initial speed of the ball, u = 5 m/s

Initial kinetic energy = $\frac{1}{2}m{u}^{2}=\frac{1}{2}×0.5×\left(5{\right)}^{2}=\frac{25}{4}\mathrm{J}$

Final velocity of the ball, v = 3 m/s

Final kinetic energy of the ball = $\frac{1}{2}m{v}^{2}=\frac{1}{2}×0.5×\left(3{\right)}^{2}=\frac{9}{4}\mathrm{J}$

Change in kinetic energy = Initial kinetic energy $-$ Final kinetic energy =

#### Question 20:

In brief, state what happen when
(i) Dry apricot are left for some time in pure water and later transferred to sugar solution.
(ii) A red blood cell is kept in concentrated salt solution.
(iii) The plasma membrane of a cell breaks down.
(iv) Rheo leaves are boiled in water first and then a drop of sugar syrup is put on it.
​(v) Golgi apparatus is removed from the cell.

#### Answer:

(i) When dry apricots are left in pure water, the water will enter into the apricots through endosmosis, which will result in swelling of the apricots. These apricots will gradually lose water through exosmosis and shrink when they are transferred to a sugar solution.

(ii) A red blood cell kept in concentrated salt solution will lose water due to exosmosis and will shrink in size.

(iii) The breakdown of the plasma membrane will result in the expulsion or scattering of the cytoplasmic components of the cell.

(iv) When rheo leaves are boiled in water, the leaf cells die and are unable to perform osmosis. Thus, the introduction of these boiled leaves into sugar syrup solution will not induce osmosis.

(v) Golgi apparatus is required for the packaging, storage and transportation of proteins. Thus, removal of Golgi bodies will affect the storage, transport and packaging of proteins.

#### Question 21:

(i) Razia saw excessive growth of blue-green algae in nearby pond of her village.
(ii) Fish, which was previously in plenty were nowhere to be seen.
(iii) The pond has started stinking. Water of the pond changes its colour and is found to be not suitable for cattle. Some of the cattle who visited the pond for drinking and bathing have fallen sick.
​(iv) What explanation will Razia give for these anomalies to the paniced villagers?

#### Answer:

(i) The excessive growth of blue-green algae occurs when the water of the pond becomes polluted. These organisms secrete certain toxic substances, which cause a wide range of diseases.

(ii) Due to excessive algal growth, the dissolved oxygen of the pond water depletes. This depletion of oxygen results in the death of fishes, which depend on dissolved oxygen for survival.

(iii) The stinking of the pond is a result of the anaerobic metabolism, which could have been performed by microbes in the limited supply of dissolved oxygen. Moreover, the toxins produced by algae can cause diseases such as bovine blue-green algae toxicosis in animals.

(iv) In the given case, Razia can explain to the villagers that excessive growth of algae was caused due to the deposition of chemical fertilizers in the pond. This deposition was caused due to runoff of rainwater, which dissolved the chemical fertilizers used in the farms. This lead to the abrupt growth of algae which released toxins and depleted dissolved oxygen of the pond. This resulted in the death of fishes of the pond and it also sickened the cattle. Thus, chemical fertilizers should be used judiciously to prevent excessive algal growth.

#### Question 22:

An inflated balloon full of air goes down slowly (becomes smaller and smaller slowly) even though the knot at the mouth of the balloon is airtight. And after a week all the air has escaped from the balloon. Explain how the air particles got out of the balloon.

#### Answer:

This is due to the reason that balloons are not completely airtight. The air particles slowly diffuse through walls of the balloon or knot. Due to the high pressure inside the balloon, the flux is operating outside. This is the reason it slowly reduces and after a week all air goes out of the balloon.

#### Question 23:

Bromine occurs in nature mainly in the form of two isotopes ${}_{35}{}^{79}\mathrm{Br}$ and ${}_{35}{}^{81}\mathrm{Br}$. If the abundance of  ${}_{35}{}^{79}\mathrm{Br}$ is isotope is 49.7% and that of ${}_{35}{}^{81}\mathrm{Br}$​ isotope is 50.3% calculate the average atomic mass of bromine.

#### Question 24:

A boy of mass 50 kg running at 5 m/s jumps on to a 20 kg trolley travelling in the same direction at 1.5 m/s. What is their common velocity?

#### Answer:

Here the momentum of the system will remain conserved as no external forces are involved.

Total momentum before the boy jumps on the trolley =

Let the common velocity of the boy and the trolley is 'vCommon'.

Total momentum after the boy jumps on the trolley =

Accordding to momentum conservation, Initial momentum = Final momentum

#### Question 25:

A device called oscillator is used to send waves along a stretched string. The string is 20 cm long, and four complete waves fit along its length when the oscillator vibrates 30 time per second. For the waves on the string.
(i) What is their wavelength?
(ii) What is their frequency?
​(iii) What is their Speed?

#### Answer:

(i) Length of the string, L = 20 cm

As 4 complete waves fit along the length. Thus, the wavelength of single wave =

(ii) The given oscillator vibrates 30 times per second. Hence, the frequency of the oscillator = 30 Hz

(iii) Now the velocity of the waves generated will be, v = f λ = 30 Hz × 0.05 m = 1.5 m/s or  = 150 cm/s

#### Question 26:

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

#### Answer:

(i) The government of India has taken the following measures to minimise cigarette smoking and tobacco-chewing:
1. Cigarette packs have been labelled with a statutory warning “ smoking is injurious to health.”
2. The Indian government has also created no smoking zones.

(ii) Passive smoking refers to the unintended inhalation of cigarette smoke by a non-smoker person.

(iii) Continuous use of tobacco can lead to certain types of cancers such as throat cancer and lung cancer.

#### Question 27:

What is the need of crossing the exotic breeds of cattle with local (Indian) breed, when exotic cattle have higher yield as compared to the hybrid breed of cattle?

#### Answer:

The local (Indian) breeds of cattle have a relatively shorter lactation period because of the poor genetic constitution. In order to improve the milk production of local breeds, these organisms are crossed with exotic breeds through selective breeding. The hybrid produced by crossing local (Indian) breeds of cattle with exotic breeds has higher milk productivity and has better ability to survive in the local environment, unlike exotic breeds. The Indian breeds are chosen in this process so that the characteristics such as disease resistance against local pests can be acquired by the hybrid.

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