CBSE Sample Solved QP for Class 9 Science Set 1

Time Allowed: 3 hours

 

SECTION – A

Question 1:
If on a round trip you travel 6 km and then arrive back home, calculate your displacement after completing the trip.
Answer:
Displacement is zero.

 

Question 2:
The atomic number of three elements A, B and C are 9,10 and 13 respectively. Which of them will form a cation ?
Answer:
The element with atomic number 13 will form a cation. It has the electronic configuration 2, 8, 3. It loses three electrons to attain stable octet and forms the cation.

 

Question 3:
(i) What is the similarity in the electronic structure of the following set of atoms :
(a) Lithium
(b)Sodium
(c) Potassium
(ii) Which of the above elements is most reactive and why ?
Answer:
(i) Electronic configuration of the elements are given as under:

 

Question 4:
What are Phanerogams ? Point out differences between its two groups. Ans. Phanerogams are flowering plants. Its two groups are Gymnosperms and Angiosperms:
Answer:

Gymnosperms	Angiosperms
(i) They produce naked seeds. (ii) Considered more primitive in evolutionaiy term.	(i)   They produce well protected seeds, inside a fruit. (ii)   Considered most highly evolved in evolutionary terms.

 

Question 5:
Why are antibiotics effective against bacteria but not against viruses ?
Answer:
Antibiotics act by preventing one of the bio-chemical events towards formation of a new pathogen cell. Thus, most antibiotics prevent cell wall formation, which happens in bacterial cell only, thereby preventing bacterial diseases.

 

Question 6:
(a) Write four phenomena which were successfully explained using universal law of gravitation.
(b) The gravitational force between two objects is 100 N. How should the distance between the objects be changed so that force between them becomes 50 N ?
Answer:
(a) The universal law of gravitation successfully explained several phenomena. Some such phenomena are:
(i) the force which binds us to the earth
(ii) motion of moon around the earth
(iii) motion of planets around the sun
(iv) occurrence of sea tides due to attraction of moon and of sun.
(b) Let distance between the two objects be changed from r to r’ so that gravitational force between them changes from F = 100 N to F’ = 50 N.
Then as per inverse square law, we have

 

Question 7:
Give reason for the following :
(a) The sound of a thunder is heard a little later than the flash of light is seen,
(b) The ceilings of the concert hall are curved.
(c) Some animals get disturbed before the earthquakes.
Answer:
(a) In lightning process, flash and thunder are produced simultaneously. Flash is seen almost immediately because speed of light is extraordinarily large (\(c = 3 × 10^8 m s^{-1}\)). But thunder is heard a few seconds later because speed of sound is much less (about \(346 m s^{-1}\) at 25 °C) and requires time to cover up the distance from the site of thunder in sky to us.
(b) The ceilings of concert halls and conference halls are generally curved as shown in figure. It is done to ensure that the sound after reflection from the ceilings reaches all corners of the hall.
(c) Before earthquakes generally infrasonic shock waves are produced. Some animals, who are sensitive to infrasonic waves, get disturbed due to them.

 

Question 8:
The velocity-time graph of an object is shown in the figure.
(a) State the kind of motion that object has from AtoB and from B to C.
(b) Identify the part of graph where the object has zero acceleration. Give reason for your answer.
(c) Identify the part of graph where the object has negative acceleration. Give reason for your answer.
Answer:
(a) Region AB of graph shows uniform motion but the region BC shows non-uniform motion.
(b) In region AB of the graph the object has zero acceleration because velocity in this
region is constant at 40 m/s.
(c) In region BC the object has negative acceleration because the velocity is falling from 40 m/s at point B (time = 25 s) to zero at point C (time = 45 s).

 

Question 9:
(a) Suggest two ways to decrease pressure on a surface.
(b) Density of an object is 1.8 \(g/cm^3\). Express it in \(kg/m^3\).
Answer:
(a) Two suggested ways so as to decrease pressure on a surface are : decrease the force acting on the surface, and increase the surface area of given surface.
Density of an object in CGS system is d – 1.8 \(g/cm^3\) 1 \(g/cm^3\) = 103 \(kg/m^3\), hence density in SI system will be d = 1.8 x 103 \(kg/m^3\)

 

Question 10:
(a) Define atomicity.
(b) State the atomicity of the following molecules :
(i) Oxygen
(ii) Phosphorous
(iii)Sulphur
(iv) Argon
Answer:
(a) The number of atoms in one molecule of the substance is called atomicity. Atomicity of the substances is given as under:

OR

State three points of difference between anion and cation.
Answer:

Anion	Cation.
It is negatively charged. Anion is bigger than the atom from which it is formed. It is obtained by adding electrons to the atom. Non-metals generally form the anions.	It is positively charged. It is smaller than the atom from which it is formed It is obtained by removing electrons from the atom. Metals generally form the cations

 

Question 11:
An element X forms the following compounds with hydrogen, carbon and phosphorous : \(P_2X_3\), \(P_2X_5, H_2X_2, H_2X, CX_2, CX\). Find the valencies of X and other element present in the compound.
Answer:

 

Question 12:
(a) How are CFCs harmful for the environment and living beings ?
(b) Mention any two human activities which would lead to an increase in the carbon dioxide content of air.
Answer:
(a) CFCs result in damaging the ozone layer. Depletion of ozone layer in upper atmosphere of earth results in more UV radiations reaching the earth’s atmosphere. This results in global warming.
More UV radiations indicate more high energy radiations that can mutate our DNA. This has resulted in increased number of skin cancer.
(b) CO2 content can increase due to :
(i) burning of fossil fuel.
(ii) burning of dead leaves and other plant material.

 

Question 13:
(i) Write the definition of health as given by WHO.
(ii) List two factors that affect health.
Answer:
(i) According to WHO, health is a state of complete physical, mental and social well being, and not merely an absence of disease or infirmity.
(ii) Factors that affect health are –
(a) Clean physical environment – as it would result in pathogen-free environment.
(b) Malnourishment leads to inability to fight diseases or results in deficiency diseases.

OR

Explain the basis of Principle of Treatment for any disease.
Answer:
There are two ways to treat an infectious disease. They are as follows :
(i) Reduce the effect of disease : The principle involved is that the effect of disease is lessened without killing the infectious agent which is done either by taking medicines and antibiotics as well as taking appropriate rest so that the body heals.
(ii) Kill the microorganisms of infectious agents : The infectious agents like bacteria, viruses, fungi, helminths and protozoan’s have some essential biochemical life processes which are peculiar to this group. These processes may be pathways for respiration or synthesis of new substances. Drugs are available which block these processes and kill the infectious agents. Antibiotics are chemicals produced by microbes which kill or prevent the growth of other microbe by blocking life processes without harming human cells.

 

Question 14:
Differentiate between xylem and phloem tissues.
Answer:

Xylem	Phloem
It conducts water and inorganic solutes in vascular plants. Conduction usually unidirectional i.e.,  from roots upwards. Conducting channels arc tracheids and vessels, Tracheids, vessels, fibres and parenchyma are its components of which only parenchyma is living. It also provides mechanical strength.	It conducts organic solutes in vascular plants. Conduction in all directions Le.t from leaves to other plant parts. Conducting channels are sieve tubes. Sieve tubes, companion cells, fibres and parenchyma are its components of which only fibres are dead, It does not provide any mechanical strength.

 

Question 15:
Ravi is a poor labourer. For last few days he used to feel pain in his lower abdomen as well as in his back. When he went to a doctor, the doctor prescribed some clinical tests as well as ultrasonography of lower abdomen. When Rita, who lived nearby came to know of this, she decided to give financial help to Ravi. After tests and ultrasonography it was found that there was a stone in the left kidney of Ravi. Rita asked her neighbours to contribute for the operation. All the neighbours contributed for the operation as well as subsequent medical expenses. Now Ravi is absolutely fit.
(i) What is ultrasound ? Explain its working principle.
(ii) What values are shown by Rita and her neighbours ?
Answer:
(i) Ultrasound is a diagnostic technique, used to get images of internal organs. It is based on the principle called piezoelectric (pressure electricity) effect. The transducer probe is the main part of the ultrasound machine which makes sound waves and receives reflected echoes from the patient’s tissues.
(ii) Values shown by Rita and her neighbours are :
(a) Comradaric
(b) Concern
(c) Generosity
(d) Compassion.

 

Question 16:
(a) How would you arrive at a mathematical formula to measure force using second law of motion ? Define the unit of force using this formula.
(b) A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
(a) Newton’s second law of motion states that the rate of change of momentum of an object is directly proportional to the external unbalanced force applied on it and takes place in the direction of applied force.
Let an object of mass m was initially moving with a velocity u. It means that its initial
momentum
\(p_1 = mu\).
If on applying a constant force F for time t the velocity of the object changes to v then final momentum of given object \(p_2 = mv\).
Change in momentum = \(P_2 – P_1\) = mv – mu = m(v – u)
Rate of change of momentum = \({p_2−p_1\over t}={m(v−u)\over t}=ma\),
where a =(v−u)t = acceleration of the given object
According to statement of second law of motion, the rate of change of momentum is directly proportional to the applied force. Hence, we have
F ∝ \(p_2−p_1\over t\) or F ∝ ma
F = kma, where k is a constant of proportionality.
Generally the units of force are so selected that if m = 1 and a = 1, then force F = 1. In such a case value of constant k becomes 1 and the relation becomes :
F = ma
The SI unit of force is a newton (1 N), which is the force which produces an acceleration of 1 \(m s^{-2}\) when applied on an object of mass 1 kg.

and the force exerted by the wooden block on the bullet F = ma = (0.01) × (- 5000) = – 50 N
The -ve sign of force shows that the force is a retarding force or it is opposing the motion.

 

Question 17:
(a) Define work. Give SI unit of work done. Write an expression for positive work done.
(b) Calculate the work done in pushing a cart through a distance of 50 m against the force of friction equal to 250 N. Also state the type of work done.
(c) What will be the work done if displacement of the object is perpendicular to the direction of force ?
(d) When an object moves on a circular path, what will be the work done ?
Answer:
(a) Work done on an object is defined as the magnitude of the force acting on the object multiplied by the distance moved by the object in the direction of the applied force.
∴ Work (W) = Constant force applied (F) × Displacement along the direction of force (s).
SI unit of work is called joule (J or N m). Work is said to be 1 joule if under the influence of a force of 1 N the object moves through a distance of 1 m along the direction of applied force.
(b) Force applied in forward direction so as to overcome force of friction = F = 250 N and displacement in the direction of force s = 50 m
∴ Work done W = Fs = 250 N × 50 m = 12500 J
The work done is positive.
(c) If displacement is perpendicular to the direction of force the work done is zero.
(d) Work done on an object is zero when it moves on a circular path because displacement is perpendicular to the direction of force.

OR

(a) Define potential energy of an object. Give an expression for gravitational potential energy.
(b) A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 2 minutes. What is the power supplied by the crane ? (g = 9.8 \(m s^{-2}\))
Answer:
(a) The potential energy of an object is the energy present in it by virtue of its position or configuration or change thereoff.
If an object of mass m is situated at a height h from a reference ground level, then its gravitational potential energy is
Ep = mgh
(b) Here mass of car m = 2000 kg, vertical distance covered by car s = h = 30 m and time t = 2
minutes = 2 x 60 s = 120 s
∴ Total work done by crane W = mgh
∴ Power supplied by the crane t = \({w\over t} = {mgh\over t} = {2000×9.8×30\over120}\) = 4900 W or 4.9 kW

 

Question 18:
Wet clothes dry up similarly when we spill water on the floor it dries up after sometime. In both the cdses change of state from liquid to vapour takes place without reaching the boiling point.
(i) What is this phenomenon called ?
(ii) Explain how the change occurs at temperatures lower than the boiling point.
(iii) Mention three factors which determine the rate at which the change of state from water to vapours occurs at room temperature.
Answer:
(i) This phenomenon is called evaporation.
(ii) Molecules of water which are moving at speed higher than the average speed leave the surface and go into the atmosphere. This is how evaporation takes place.
(iii) Factors that determine the rate of evaporation :
(a) Surface Area – Greater the surface area, greater will be the rate of evaporation.
(b) Temperature – Higher the temperature, greater will be the rate of evaporation.

 

Question 19:
Rahul’s mother mixed oil and water in kitchen by mistake. Rahul told her that he can separate the mixture. Name the technique used by Rahul and explain how he will do. Draw the diagram and write the principle of this technique.
Answer:
The technique used by Rahul to separate the mixture of oil and water was to use separating funnel as shown in figure.
Kerosene oil does not mix with water and forms a separate layer. Oil being lighter than water, forms the upper layer. Steps involved in the separation of kerosene oil from water are:
(i) Pour the mixture of kerosene oil and water in the separating funnel.
Let it stand undisturbed for sometime so that separate
layers of oil and water are formed.
(ii) Open the stopcock of the separating funnel and pour out the lower layer of water carefully.
(iii) Close the stopcock of the separating funnel as the oil reaches the stopcock.
(iv) Transfer oil from the neck of the separating funnel to a separate vessel.

 

Question 20:
The immune system of Hari is damaged by the attack of pathogen on his body.
(i) Name the disease he is suffering from.
(ii) Name the pathogen that has attacked his body.
(iii) Mention any three modes of transmission of this disease.
Answer:
(i) Hari is suffering from AIDS – Acquired Immuno – Deficiency Syndrome.
(ii) The pathogen that has caused AIDS is HIV – Human Immuno Deficiency Virus.
(iii) This virus is transmitted through following ways :
(a) Sexual contact with an infected person carrying AIDS virus.
(b) Transfusion of blood infected with HIV.
(c) Use of unsterilized needles, blades or razors.
(d) Transplacental transmission i.e., AIDS infected mother to the foetus developing in her womb.

 

Question 21:
Answer the following –
(i) Draw a labelled diagram of smooth muscle.
(ii) Differentiate between parenchyma and sclerenchyma.
(iii) Name the epithelium which has hair-like projection on outer surface of the cells.
(iv) Name a tissue that stores fat in the body.
Answer:
(i)

(ii) Distinction between parenchyma and sclerenchyma on the basis of their cells wall:

Parenchyma	Sclerenchyma
(a) A simple permanent plant tissue with thin walls. Cells are loosely packed with large intercellular spaces and form the packing tissue of all plant organs. (b) It is primarily involved in storage of food material. It also facilitates movement of gases. Parenchyma cells remain turgid and provide rigidity to softer parts.	(а) Sclerenchyma is the chief mechanical tissue of plants which comprises of highly thick walled cells with narrow lumen. The walls are thick on all sides due to presence of lignin. (b) It provides strength to the plants and enables them to bear various stresses. It also forms a protective covering around seeds and nuts.

 

SECTION – B

Question 22:
While determining the density of the material of a metallic sphere using a spring balance and measuring cylinder, a student noted the following readings :
(1) Mass of the sphere = 81 g.
(2) (i) Reading of water level in the cylinder without sphere = 54 mL.
(ii) Reading of water level in the cylinder with sphere = 63 mL.
On the basis of these observations calculate the density of the material of the sphere.
Answer:
Here mass of sphere M = 81 g and volume of sphere V = (63 – 54) mL = 9 mL
∴Density of the material of the sphere \(ρ={M\over V}={81g\over 9mL}=9g(mL)^{−1}\)

 

Question 23:
What happens in your experiment of verification of the laws of reflection of sound if pipe through which sound is heard is lifted vertically through a small distance ?
Answer:
If the pipe, through which sound is heard in our experiment, is slightly lifted vertically then the reflected sound beam does not remain in same plane as of incident sound beam and hence either a weakened sound of table clock is heard or no sound is heard at all.

 

Question 24:
Ramesh prepared the solution of chalk powder in water and Mahesh prepared the solution of milk in water. What type of solutions were prepared by them and whether the solutions are homogeneous or heterogeneous ?
Answer:
Solution prepared by Ramesh is suspension.
Solution prepared by Mahesh is colloidal solution.
Both the solutions are heterogeneous.

 

Question 25:
A horse magnet was moved over a mixture of iron filings and sulphur as shown in the figure.
(a) Tell whether the mixture of the two substances is homogeneous or heterogeneous.
(b) Which substance will be left on the watch glass after the experiment ?

Answer:
(a) The mixture of iron filings and sulphur is heterogeneous.
(b) Sulphur will be left on the watch glass after the experiment.

 

Question 26:
State any two specific features of Earthworm.
Answer:
Earthworm belongs to Phylum Annelida.
Its specific features are :
(i) Body is segmented i.e., shows metamerism.
(ii) Presence of clitellum.

 

Question 27:
Name two features which you would examine to categorise a plant into dicot or monocot.
Answer:
(i) Leaf : Are broad, with reticulate venation in dicots, and blade-like with parallel venation in monocots.
(ii) Flower: Flowers are pentamerous in dicots and trimerous in monocots.

 

Solved CBSE Sample Papers for Class 9 Science Set 2

Time Allowed: 3 hours

 

SECTION – A

Question 1:
A ball is thrown vertically upwards. What is its momentum at the highest point ?
Answer:
Zero because velocity of ball is zero at the highest point of its motion.

 

Question 2:
Degree Celsius and Kelvin are two units of measuring temperature. Which of these are SI and non-SI units ?
Answer:
Degree Celsius – Non-SI.
Kelvin – SI.

 

Question 3:
Why is crystallisation better than evaporation for the separation of mixtures ?
Answer:
In crystallisation, we obtain the components in the form of crystals which are the purest form of a substance and impurities are left in the mother liquor. Crystals are separated from the mother liquor. This is not possible in evaporation. Some solids may decompose on heating to dryness. Such solids cannot be purified by evaporation.

 

Question 4:
Write names of different kinds of connective tissue :

Answer:
(a) Blood
(b) Areolar
(c) Bone
(d) Cartilage.

 

Question 5:
Water hyacinth floats on water. Why ? How is this tissue different from that of succulent plants ?
Answer:
Water hyacinth being aquatic plant floats on water as its parenchyma has lots of air spaces (aerenchyma) which store air and help it in remain afloat.
Whereas in succulent plants it helps in storage of water.

 

Question 6:
In each of the adjoining diagram, a force F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Answer:
On observing the diagrams carefully we find as follows:
(i) In first (extreme left side) figure, the force is acting in a direction perpendicular to displacement, Hence, work done by the force is zero.
(ii) In second figure, force and displacement both are in same direction. So, the work done by the force is positive.
(iii) In third (extreme right side) figure, the force and displacement are in mutually opposite directions, hence, the work done by the force is negative.

 

Question 7:
Define power. Derive its SI unit. An electric bulb is rated 10 W. What does it mean? What is the energy consumed in joules if it is used for 5 minutes?
Answer:
The rate of doing work or the rate of transfer of energy is known as the power.
∴ \(Power ={ work \over time}{w\over t}\)
SI unit of power is a watt. Power is said to be 1 watt (1 W) if rate of doing work islJ \(s^{-1}\).
If an electric bulb is rated 10 W, it means that electric bulb supplied 10 J of energy per second.
If the bulb is used for a time t = 5 min = 5 × 60 s = 300 s, then total energy consumed by it E = P.t = 10 × 300 = 3000 J.

 

Question 8:
(a) What is reverberation of sound?
(b) Draw a diagram depicting low pitched sound and high pitched sound. What is the main difference between the two ?
Answer:
(a) Persistence of sound due to repeated reflection from walls and roof in an auditorium is called reverberation.
(b) The diagrams have been shown below :

High pitched sound has high frequency but low pitched sound has low frequency.

 

Question 9:
(a) When an athlete comes running from a distance, he is able to jump longer. Why ?
(b) The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in figure. How much force does the table exert on the ball to bring it to rest ?
Answer:
(a) He gains momentum which helps him in taking a longer jump.
(b) Here mass of ball m = 20 g = 0.02 kg, initial velocity of ball u = 20 m/s, final velocity v = 0 and time
t = 10 s

OR

(a) What is the relationship between force applied on a body and the resulting acceleration caused in it?
(b) A force produces an acceleration of 5 m s’2 when applied on a body of mass 2 kg. Find the magnitude of force. How much acceleration will the same force produce when applied to a body of mass 4 kg ?
Answer:
(a) Force (F) = Mass (m) x Acceleration (a).
(b) Here mass m = 2 kg and acceleration a = 5 m s-2
∴ Force F = ma = 2 × 5=10N
Now on applying same force on an object of mass m’ = 4 kg, the acceleration produced is given by

 

Question 10:
A substance A has high compressibility and can be easily liquefied. It can take the shape of any container. Predict the nature of the substance. Enlist four properties of this state of matter.
Answer:
The substance A is a gas because gases have large empty spaces between the particles. It has high compressibility and can be easily liquefied.
Properties of gases:
(i) Gases have no fixed shape and volume.
(ii) Gases can be compressed on applying pressure to give liquids.
(iii) Particles of a gas are moving in all possible directions with all possible speeds.
(iv) Gases show the property of diffusion.

 

Question 11:
List four properties of non-metals. Give two examples.
Answer:
(i) They show a variety of colours.
(ii) They are poor conductors of heat and electricity.
(iii) They are not malleable or ductile.
(iv) They are not lustrous or sonorous.
Sulphur and phosphorous are examples of non-metals.

 

Question 12:
Describe the structure of mitochondria with special emphasis on its membrane coverings. How can we relate the structure of the membrane with the function ?
Answer:
Mitochondria are rod shaped or sausage shaped cell organelles which are known as powerhouse of the cell. They contain enzymes necessary for the oxidation of food during the process of respiration and for release of energy in the form of ATP (Adenosine triphosphate). ATP is also known as the energy currency of the cell. This energy is used by cell to perform various functions such a mechanical work and biosynthesis of new chemical compounds.

Mitochondria to bounded by double membranes. The outer membrane is porous and the inner membrane is deeply folded. These folds are known as cristae and they provide a large surface area for ATP- generating chemical reactions. Mitochondria have their own DNA, ribosomes and enzymes to manufacture their own proteins. So they are also known as semiautonomous bodies.

 

Question 13:
How can poultry farming be integrated with crop production ? How improved poultry breeds are developed in poultry farming ?
Answer:
Poultry is rearing of domesticated fowl, ducks, geese, turkeys and pigeons for their meat and eggs.
Poultry feed consists of mashed cereals like wheat, maize, jowar, oil cakes, fish meat etc.
Therefore they should be kept along with such crops in the fields, thus benefitting the farmer.
The cross-breeding programmes between Indian and foreign poultry have developed improved poultry breeds. The new varieties have low maintenance requirements. The size of the egg laying bird is also reduced and it utilises more fibrous cheaper diets consisting of agricultural by products. They produce more eggs and broilers for meat. Hence, it is interesting to note that
poultry is India’s most efficient converter of low fibre food-stuff (which is unfit for human consumption) into highly nutritious animal protein food.

OR

What factors are responsible for storage losses in agricultural produces and how they can be controlled and prevented ?
Answer:
Following factors may be responsible for losses of grains during storage :
(i) Biotic factors : They include insects, rodents, bacteria, fungi and mites.
(ii) Abiotic factors : They include inappropriate moisture and temperature in the place of storage.
These factors cause poor quality, loss in weight, discolouration of produce, poor germinability and marketability of grains.
These factors can be controlled by proper treatment and by systematic management of ware housing.
Preventive measures include strict cleaning of produce before storage, proper drying of the produce first in sunlight and then in shade and fumigation using chemicals to kill pests.

 

Question 14:
Name the two types of diseases one caused by some external agents and other due to some internal disorder of the body.
Mention any two causative agents.
Answer:
Two types of diseases are :
By external agent – Acquired diseases like Jaundice, Tuberculosis, Polio etc.
By internal disorder – High blood pressure, Diabetes, Cancer etc.
Causative agents – Causes may be malnourishment or pathogen or congenital or stress induced etc.
For e.g., Malaria is caused by Plasmodium. Diabetes is caused due to insufficient insulin or insulin resistance.

 

Question 15:
Bhola and Rajni, who are studying in Class IX, were travelling in a train. Rajni observed a field with two crops growing simultaneously in a definite pattern. While Bhola was busy in playing with a video game. Rajni noticed that rows of bajra and lobia were grown in alternate rows. She asked her grandfather why bajra and lobia are grown together ?
(i) On what basis are the two crops selected in this pattern ?
(ii) How does this practice benefit the farmer ?
(iii) State any two values in Rajni’s behaviour here that differentiate her from Bhola.
Answer:
(i) This is inter-cropping.
The crops selected for such type of pattern should have different types of nutritional requirements. This will help the crops to utilise maximum nutrients from the soil.
(ii) This pattern prevents pests and diseases to spread in all plants of one crop thereby preventing losses.
(iii) Values that differentiate Rajni from Bhola are :
(a) Awareness
(b) Concern
(c) Inquisitive
(d) Observant etc.

 

Question 16:
(a) Draw a velocity-time graph for an object in uniformly accelerated motion. Show that the slope of the velocity-time graph gives the acceleration of the object.
(b) An aeroplane starts from rest with an acceleration of 3 m s-2 and takes a run for 35 s before taking off. What is the minimum length of the runway and with what velocity the plane took off ?
Answer:
(a) Velocity-time graph is shown in figure.

The Slope of the graph is given by \(BD\over AD\). However, from the
graph it is dear that BD = EA = v – u, and AD = OC = t.
∴ slope of velocity -time graph = \({BD\over AD} ={ v–u\over t}\)
But \(v–u\over t\) rate of change of velocity = acceleration ‘a’
So, we conclude that the slope of the velocity-time graph gives the acceleration of the object.
(b) Here u = 0, a = 3 m s’2 and t = 35 s
∴  final velocity v = u + at = 0 + 3 × 35 = 105 m \(s^{-1}\)
If minimum length of runway be s m, then
\(s = ut + {1\over2} ,at_2 =0 × 35 +{1\over2} × 3 × (35)_2 =1837.5\)

 

Question 17:
(i) State two factors on which the gravitational force between two objects depends.
(ii) Why is ‘G’ called as universal constant ?
(iii) What happens to the gravitational force between two objects if masses of both the objects are doubled and the distance between them is also doubled ?
(iv) What is the value of ‘G’ on moon ?
(v) What is the value of ‘g’ on moon ?
Answer:
(i) The gravitational force between two objects is (a) directly proportional to the product of their mases, and (b) inversely proportional to the square of the distance between them.
(ii) G is called a universal constant because its value never changes under any condition.
(iii) The force remains unchanged.

(iv) Value of G on moon is same as on earth having a value \( 6.673 × 10_{-11} N m^2 kg^{-2}\)
(v) Value of g on moon is 1/6 th of its value on the earth.

 

Question 18:
(a) What is meant by molecular formula ? State with example what informations can be derived from a molecular formula ?
(b) Write the names of the compounds represented by the following formulae :
(i)  \(Mg(NO_3)_2\)
(ii)  \(K_2SO_4\)
(iii)  \(Ca_3N_2\)
Answer:
(a) Molecular formula of a compound shows the constituent elements and the number of atoms of each element present in one molecule of the compound.
Take the example of \( H_2SO_4\). It is the molecular formula of sulphuric acid. It shows that
(i) Sulphuric acid molecule is constituted of hydrogen, sulphur and oxygen.
(ii) One molecule of sulphuric acid contains two atoms of hydrogen, one atom of sulphur and four atoms of oxygen.

OR

(a) State the basic difference between atoms and molecules.
(b) Why does not atomic mass of an element represent the actual mass of its atoms ?
(c) “The atomic mass of an element is in fraction.” What does this mean ?
Answer:
(a) Atom is the building block of all matter.
A molecule is a general a group of two or more atoms of the same or different elements
that are chemically bonded.
(b) The actual mass of an atom is extremely small. We represent the mass of an atom as
a relative mass compared to 1/12 of the mass of one carbon-12 atom.
(c) Many of the elements have isotopes with different masses. The atomic mass of an element is the average of the masses of these isotopes which comes out to be a fraction.

 

Question 19:
Give reasons for the following :
(a) Isotopes of an element are chemically similar.
(b) An atom is electrically neutral.
(c) Noble gases show least reactivity.
(d) Ions are more stable than atoms.
(e) Na+ has completely filled K and L shells.
Answer:
(a) Chemical properties of an element depend upon the atomic number. All the isotopes of an element have the same atomic number and therefore are chemically similar.
(b) An atom has the equal number of positively charged protons and negatively charged electrons. Therefore atom is electrically neutral.
(c) Noble gases contain 8 electrons in the outermost shell (2 in case of helium). This is the maximum capacity of the outermost shell. Hence they show least reactivity.
(d) Ions are more stable than atoms because the former have an octet structure.
(e) Electronic configuration of Na is
K         L           M
2          8           1
Na+ is formed from Na by removing the electron from the M shell. Therefore it has the completely filled K and L shells.

 

Question 20:
Give examples of organisms belonging to Kingdom Protista :
(i) Based on structure for locomotion.
(ii) Based on mode of nutrition,
Answer:
(i) Based on structure for locomotion :
(a) Pseudopodia – e.g., Amoeba.
(b) Flagella – e.g., Euglena.
(c) Cilia – e.g., Paramecium.
(ii) Based on mode of nutrition :
(a) Autotrophic —Euglena.
(b) Heterotrophic – Trypanosoma, Amoeba, etc.

 

Question 21:
(a) Mention the type of shelters which should be provided to cattles in dairy farming.
(b) Mention the preventive measures taken to control diseases of dairy animals.
(c) What are the food requirements of dairy animals ?
Answer:
(a) The shelter provided for the livestock depends on type of animals to be sheltered, number of animals and location of shelter.
Good animal shelter should be clean, dry, airy and well-ventilated. It should be spacious, have proper disposal of wastes, arrangement of clean drinking water and protection of animals from environment factors, predators and diseases.
(b) Cholera, diarrhoea and tuberculosis are a few examples of bacterial diseases of cattle. Viral disease are pox, rinderpest as well as foot and mouth disease.
Aspergillosis is an example of fungal disease in cattle.
Animal diseases can be prevented if animals are kept in clean, hygienic environment, if they are given regular bath, if their shelter prevent entry of germs and flies; if they are vaccinated at regular intervals.
(c) The cattle feed consists of :
(i) Roughage : It is a coarse and fibrous substance which has low nutrient content e.g., fodder, hay, straw and so on.
(ii) Concentrate : It is rich in nutrients with very little fibrous or cellulose matter. Concentrate is provided by grains, seeds and oil cakes.

 

SECTION – B

Question 22:
A string is stretched as shown and a ‘pulse’ is created along it. The stop watch is started, from its position A, when the pulse is in position X and is stopped, in its position B, when the ‘pulse’ has travelled back to its position ‘X’. Find the velocity of propagation of the pulse along the string.

Answer:
Distance covered by pulses – 2m + 2m – 4m Time taken t =45 s – (- 5 s) = 50 s
Speed of pulse  \( v = {s\over t}={4 \over 50 } {m \over s} =0.08m/s, or , 8 cm s^{-1}\)

 

Question 23:
While determining the density of a copper piece using a spring balance and a measuring cylinder, Seema carried out the following procedure :
(i) Noted the water level in the measuring cylinder without the copper piece.
(ii) Immersed the copper piece in the water.
(iii) Noted the water level in the measuring cylinder with the copper piece inside it.
(iv) Removed the copper piece from the water and immediately weighed it using a spring balance.
Which step in the procedure is wrong and why ? What should have been the correct step ?
Answer:
The step (iv) of the procedure is wrong because the copper piece just removed from the water is wet and on weighing it we shall not get the correct mass.
The copper piece should be weighed only when it is completely dry.

 

Question 24:
A student while carrying out the separation of a mixture of sand, common salt and camphor arranged the apparatus as shown in the figure but couldn’t collect camphor. What is the defect in his apparatus set up ? Name one more substance that can be purified by this method.

Answer:
The stem of the funnel is not closed with a cotton plug resulting in the escape of camphor.
Ammonium chloride can also be purified by this method.

 

Question 25:
On adding zinc to dilute sulphuric acid, zinc sulphate and hydrogen gas are formed.
(a) What is the colour of zinc sulphate solution ?
(b) How do we test hydrogen gas ?
Answer:
(a) Zinc sulphate solution is colourless.
(b) When a burning match stick is brought near the gas, it burns with a pop sound.

 

Question 26:
Spirogyra, Mosses and Ferns belong to which sub-kingdom ? Give two identifying features.
Answer:
Spirogyra is Thallophyta.
Mosses one Bryophyta.
Ferns are Pteriodophytes.
All these phyla belong to sub-kingdom Cryptogamae.
Features are:
(i) They do not produce external flowers or seeds.
(ii) Common means of reproduction is through spores.

 

Question 27:
Differentiate dicots from monocots on the basis of roots and leaves.
Answer:

Dicots	Monocots
Roots : (i) Dicots have tap root system.	(i) Monocots have fibrous roots,
Leaves: (ii) Leaves are broad with reticulate venation.	(ii) Leaves are thin and blade like with parallel venation.