(vi) Questions 27 to 41 are multiple choice questions carrying one mark each.
Question 1 ( 1.0 marks)
A substance ‘X’ takes the shape of the container in which it is kept. It has a fixed volume. What is the physical state of substance X’?
Solution:
Substance ‘X’ is a liquid.
Question 2 ( 1.0 marks)
Name any two cells with irregular shape.
Solution:
White blood cells and Amoeba have an irregular shape.
Question 3 ( 1.0 marks)
Under what situation can the acceleration in an object be zero even when several forces are acting on it?
Solution:
In case all the forces acting on it are balanced, the acceleration in the object will be zero.
Question 4 ( 1.0 marks)
What is meant by mass?
Solution:
The mass of an object is a measure of its inertia.
Question 5 ( 1.0 marks)
Give any two examples of crops grown during the period of June to October.
Solution:
Rice and maize (Kharif crops) are two of the crops grown during the period of June to October.
Question 6 ( 2.0 marks)
Why does water kept in an earthen pot remain cool?
Solution:
An earthen pot has minute pores throughout its surface through which water keeps coming out. This water absorbs heat from the surface of the pot and evaporates. Because of this heat loss, the water inside the pot remains cool.
Question 7 ( 2.0 marks)
How can cream be separated from milk?
Solution:
Cream can be separated from milk by the process of centrifugation. This process utilises the centrifugal force generated by a spinning motor to separate the molecules by size or density. During this process, the denser particles are forced to the bottom, while the lighter particles come on top. Cream collects in the upper layer of milk after centrifugation. This is because fat (cream) is the lightest component of milk.
Question 8 ( 2.0 marks)
Name and write the salient features of two components of protoplasm.
Solution:
Protoplasm consists of the cytoplasm and the nucleus.
Question 9 ( 2.0 marks)
We are able to bend our ears but cannot bend our legs other than at joints. Why?
Solution:
We are able to bend our ears but cannot bend our legs other than at the joints because the external part of our ears is made up of cartilage whereas leg consists of bone. Cartilage is a specialised connective tissue composed of widely spaced cells. These cells are embedded in a proteinaceous matrix, which is strong yet flexible in nature, hence enabling the ear to bend. Bones, on the other hand, have a hard matrix composed of calcium and phosphorus fibres.
Question 10 ( 2.0 marks)
(a). What are the units of speed and velocity?
(b). A car travels from point A to point B as shown in the figure in a time of 45 min. What is its velocity?
Solution:
(a). The units of speed and velocity are same i.e., m/s.
(b).
Thus, the velocity of the car is 20 km/h in the North-East direction.
Question 11 ( 2.0 marks)
What is the effect of force in the following cases?
(a). A fielder catches a cricket ball.
(b). Brakes are applied to a moving car.
(c). A soft rubber ball is squeezed.
(d). A football lying on the ground is kicked.
Solution:
(a). The speed of the ball becomes zero. Therefore, the applied force changes the state of motion of the ball.
(b). The speed of the car decreases. Therefore, the applied force changes velocity of the car.
(c). The shape of the ball changes. Therefore, the applied force deforms the ball.
(d). The speed of the football increases. Therefore, the applied force changes the state of motion of the ball.
Question 12 ( 2.0 marks)
According to the universal law of gravitation, between any two masses, there exists an attractive force. However, we do not experience this force in our lives with, say, a person sitting next to us. Why is it so?
Solution:
The force of attraction is given by
 , where M, m are the respective masses of the two bodies and r is the distance between them. If logical values for M and m (say, 60 kg each) and r as 0.5 m are put, the value of G is so small that the resulting force is negligible. Moreover, we experience gravitational force of attraction from every object around us. Thus, no net force acts on our body in any particular direction. Therefore, we do not feel any gravitational force of attraction.
Question 13 ( 2.0 marks)
In India, season determines the type of crop to be grown. Justify the statement.
Solution:
In India, different crops are grown in different seasons. This is because each crop requires a different climatic condition, temperature range, and photoperiod for its growth and to carry out its lifecycle. Therefore, some crops are grown in the rainy season (Kharif crops) while others are grown during winters (Rabi crops).
Question 14 ( 2.0 marks)
One of the varieties produces nutritionally-rich grains but only during the monsoon season. The other variety produces grains throughout year but they are poor in nutrition. Is it possible for the farmer to grow nutritionally-rich grains in all seasons? How?
Solution:
It is possible for the farmer to grow nutritionally-rich grains in all seasons by the process of hybridisation. Hybridisation is the method of crossing two organisms having different characteristics so as to bring different useful characteristics together into one organism. A hybrid plant produced from two varieties will have characteristics from both the parent plants. Therefore, by using hybridisation, the farmer will be able to produce nutritionally-rich grains, which can be cultivated throughout the year.
Question 15 ( 3.0 marks)
(a). In which of the following cases will the rate of evaporation be fastest?
i.
ii.
(b). When cold water is poured in a glass, small droplets get condensed on the surface of the glass. Why?
Solution:
(a).
i. Case I
The rate of evaporation increases as the surface area increases.
ii. Case III
Rate of evaporation is directly proportional to temperature. As temperature is maximum in case III, the rate of evaporation will also be maximum in this case.
(b). We know that air contains water vapour. When cold water is poured into a glass, the surface of the glass becomes cool. The water vapours present in the air come into contact with the surface, lose energy and get condensed into small droplets of water.
Question 16 ( 3.0 marks)
Write an activity to show that matter is made up of small particles.
Solution:
2-3 crystals of potassium permanganate (KMnO4) were taken and dissolved in a beaker containing 100 mL water. 10 mL of this solution was taken and added to another beaker containing 90 mL water. The procedure was repeated till the colour in the beaker became very light.
Figure: Serial dilution of potassium permanganate solution
We will observe that as soon as the crystals of potassium permanganate are dissolved in the first beaker, the water becomes violet-pink. When the solution is repeatedly diluted, the intensity of the colour gradually decreases.
Thus, we can conclude that 2-3 crystals of potassium permanganate consist of millions of particles that can be transferred from one beaker to another, causing the water present in each beaker to become pink in colour. Hence, we can say that particles of matter are very small and are not visible to the naked eye.
Question 17 ( 3.0 marks)
How will you show that the formation of iron sulphide is a chemical change? You are provided with iron fillings, sulphur powder and a magnet.
Solution:
Step 1: Take two china dishes and label them as I and II.
Step 2: Add 5 g of iron filings and 3 g of sulphur powder in each china dish. Mix the contents in both the dishes thoroughly.
Step 3: Heat the contents of dish II till it turns red-hot.
Step 4: Then, use a magnet to separate the iron filings from both the dishes. Compare the texture and colour of the materials obtained in both the dishes.
It will be observed that the material obtained in dish I is a mixture of two substances. The substances present in dish I are iron and sulphur. However, the substance obtained in dish II is a compound called iron sulphide. On heating iron and sulphur, ferrous sulphide (FeS) is formed. Ferrous sulphide (FeS) is a compound having completely different properties from that of iron and sulphur, which are its constituents. But, the composition of the compound (FeS) is same throughout.
Question 18 ( 3.0 marks)
Differentiate between cells composing unicellular and multicellular organisms on the basis of
(a). Their shape
(b). Their structure
(c). Their interdependence and division of labour in them
Solution:
(a). The cell of unicellular organisms such as Amoeba can change its shape according to the environment whereas cells in multicellular organisms exhibit definite shape. For example, neuron has a definite and distinct shape, which does not change.
(b). Cells of unicellular organisms can or cannot have a definite nucleus or membrane-bound organelles whereas cells of multicellular organisms always have a definite nucleus and membrane-bound organelles.
(c). A unicellular organism is independent and does not require help from other such cells for its growth and development. There is division of labour present within the cell whereas cells in multicellular organisms depend upon other cells present in the organism for their growth and development and show division of labour.
Question 19 ( 3.0 marks)
Is the location of dividing tissues significant in the overall plant development? Discuss with the help of a diagram.
Solution:
Meristematic tissue is composed of immature and continuously dividing cells. Meristem can be further classified based on its position in the plant in three classes:
Hence, it is evident that different types of dividing tissues present at different locations result in different growth patterns in a plant.
Question 20 ( 3.0 marks)
Study the distance−time graph and answer the following questions.
(a). Which part of the graph represents absence of motion?
(b). Which part of the graph represents the fastest motion?
(c). How can it be deduced from the graph if the motion is away from the origin or towards the origin?
Solution:
(a). The line graph between the time 0 s and 15 s represents absence of motion.
(b). The line graph between the time 15 s and 20 s represents the fastest motion.
(c). During 0 s to 15 s, there is no motion as there is no change in its distance from the origin.
During 15 s to 20 s, there is motion towards the origin as the distance reduces from 4 m to 2 m.
During 20 s to 35 s, there is motion away from origin as the distance increases from 2 m to 4 m.
Question 21 ( 3.0 marks)
A girl of mass 50 kg jumps onto a stationary skateboard of mass 5 kg with a velocity of 5 m/s. What is her velocity as the skateboard moves? Assume that there are no other external forces at work here.
Solution:
According to the law of conservation of energy:
Total initial momentum, Pi = Total final momentum, Pf
Pi = mg × vg + ms × vs (mg = mass of girl, vg = velocity of girl, ms = mass of skateboard, vs = velocity of skateboard)
Given that:
mg = 50 kg
vg = 5 m/s
ms = 5 kg
vs = 0 m/s
Pi = 50 × 5 = 250 kg m/s
Pf = (mg + ms) × v (where u = final velocity of girl and skateboard)
= 55 × v
Since Pf = Pi,
55v = 250
= 4.54 m/s
The velocity of the girl is 4.54 m/s.
Question 22 ( 3.0 marks)
Why is the weight of a man on the moon
 of his weight on the earth? It is given that the earth is 100 times heavier than the moon and the radius of the earth is 4 times that of the moon.
Solution:
Suppose there is an object of mass m. Let its weight on the moon be WM.
Weight of the object on moon,
 ... (i)
Where,
MM = Mass of the moon
RM = Radius of the moon
Suppose the weight of the same object on the earth is WE.
It is known that:
Mass of the earth (ME) = 100 × Mass of the moon (MM)
Radius of earth (RE) = 4 × Radius of the moon (RM)
∴ ME = 100MM and RE = 4 RM
Applying the universal law of gravitation:
Weight of the object on earth,
Or,
 … (ii)
Dividing (i) by (ii):
Or, Weight of the object on the moon
 (Weight of the object on the earth)
Therefore, a man weighs
 th of his weight on earth on the moon.
Question 23 ( 3.0 marks)
(a). Enumerate the beneficial aspects of improved varieties in poultry farming.
(b). Give examples of poultry breeds which are hybridised to bring about desirable results.
Solution:
(a). Some beneficial aspects of improved varieties in poultry farming are listed as follows:
(b). Hybridisation between an Indian breed such as Aseel and a foreign breed such as Leghorn has been done to bring about desirable traits in the improved variety.
Question 24 ( 4.0 marks)
(a). What is the significance of diffusion in biological processes? Explain with the help of examples. (In around 80 to 100 words)
(b). State the difference between diffusion and osmosis.
(c). What kind of membrane does diffusion take place through? Give example of any one such membrane.
Solution:
(a). Many biomolecules participate in the life processes. They travel through the living systems majorly through diffusion. In diffusion, the biomolecules travel across the concentration gradient i.e., from the region where they are less in concentration to the region of their higher concentration. Some of the examples stating the significance of diffusion in living world are given below:
(b).
(c). Diffusion does not essentially require a semi-permeable membrane, but can also occur through it. Plasma membrane is an example of a semi-permeable membrane.
Question 25 ( 4.0 marks)
(a). A person in a bus looks out and finds that a tree on the road appears to move away from him in a direction opposite to the direction the bus is travelling in. The bus is travelling with a speed of 90 km/h.
What is the speed with which the tree appears to move away?
(b). It is known that the tree is stationary and does not actually move away. How then is this apparent speed accounted for?
(c). What quantity does the odometer of a car show? Can it be used to calculate the velocity of the car?
Solution:
(a). The tree moves away with an apparent speed of 90 km/h in the opposite direction.
(b). The tree has an apparent displacement for an observer sitting in the bus. This happens because the person sitting in the bus has no movement with respect to the bus. Therefore, for the person, the bus is stationary and the objects outside the bus are moving. However, for a person standing outside the bus, the tree is stationary and the bus is moving. Thus, for a person sitting inside the bus, it is correct to say that the tree has a speed of 90 km/h with respect to the bus.
(c). The odometer of a car shows the distance travelled by the car. It may not be used to calculate the car’s velocity as velocity is the ratio of displacement and time. Displacement, being the shortest distance between origin and destination, may not be the same as the distance reading as shown in the odometer.
Question 26 ( 4.0 marks)
(a). What is gravitational force?
(b). What will be the value of g if the earth were to become three times as heavy and twice as large as it is now?
(c). What is the gravitational force of attraction between Hari and his friend, of masses 50 kg and 55 kg respectively, if they are sitting 1 m away from each other? (G = 6.673 × 10−11 Nm2kg−2)
Solution:
(a). The force of attraction between two objects by virtue of their masses is called gravitational force.
(b). Gravitational acceleration on earth surface,
Where,
G = Universal gravitational constant
M = Mass of the earth
R = Radius of the earth
Given that:
New mass of the earth, M1 = 3M
New radius, R1 = 2R
New value of gravitational acceleration on earth surface,
   
Therefore, the new value, g1, will be 0.75 g.
(c). Given that:
Mass of Hari, M1 = 50 kg
Mass of Hari’s friend, M2 = 55 kg
Distance between Hari and his friend, R = 1 m
G = 6.673 × 10−11 Nm2kg−2
Therefore, gravitational force of attraction between Hari and his friend,
 
= 1.84 × 10−7 N
Therefore, the force of attraction is only 1.84 × 10−7 N.
Question 27 ( 1.0 marks)
Use the following information to answer the next question.
What is the radius of the track?
A.
3.5 m
B.
7 m
C.
10.5 m
D.
14 m
Solution:
Speed of a cyclist, v = 2.2 m/s
Time taken for one revolution, t = 20 s
Radius of the circular track = r
Length of the track, l = 2πr
Speed is given by the relation,
Therefore, the radius of the track is 7 cm.
The correct answer is B.
Question 28 ( 1.0 marks)
Use the following information to answer the next question.
What should be the magnitude of the pull force so that acceleration of the system remains the same?
A.
8 N
B.
12 N
C.
24 N
D.
48 N
Solution:
Mass of the load, m = 2 kg
Force applied on the load, F = 8 N
Let a be the acceleration of the load in response to the applied force.
Acceleration is given as
F = ma
Mass of the box, m2 = 4 kg
It is tied to the load.
Total mass of the box and load system, M = 4 + 2 = 6 kg
New applied force is given as
Therefore, the required pull force should be 24 N.
The correct answer is C.
Question 29 ( 1.0 marks)
Which of the following objects has the maximum inertia about an axis that passes through the centre?
A.
Football
B.
Baseball
C.
Tennis ball
D.
Cricket ball
Solution:
The resistance offered by an object to change its state of motion is called the inertia of the object. It tends to keep the present state of motion of an object. Inertia is the content of a body. Inertia of a body is directly proportional to its mass. More the mass of a body, more is the inertia of the body and vice-versa.
The given table lists the mass of the given balls.
It can be observed that football is the heaviest among the given balls. Hence, football has the maximum inertia.
The correct answer is A.
Question 30 ( 1.0 marks)
Use the following information to answer the next question.
What is the acceleration due to gravity of the given planet?
A.
1.225 m/s2
B.
2.45 m/s2
C.
19.6 m/s2
D.
39.2 m/s2
Solution:
Mass of the earth = M
Radius of the earth = R
For gravitational constant G, acceleration due to gravity g is given by the relation,
Mass of the given planet,
Radius of the planet,
Acceleration due to gravity of the planet is given as
Acceleration due to gravity of earth, g = 9.8 m/s2
Therefore, acceleration due to gravity of the given planet is 1.225 m/s2.
The correct answer is A.
Question 31 ( 1.0 marks)
Use the following information to answer the next question.
With what velocity will the apple hit the ground?
A.
5.2 m/s
B.
9.8 m/s
C.
17.15 m/s
D.
24.12 m/s
Solution:
Distance covered by the apple = Height of the branch, h = 11 m
Acceleration of the apple (a) = Acceleration due to gravity (g) = 9.8 m/s2
A falling apple is a case of free fall. Hence, initial velocity of the apple, u = 0
Let v be the final velocity of the apple. It is given by the relation,
Therefore, the apple will hit the ground with velocity approximately 17.15 m/s.
The correct answer is C.
Question 32 ( 1.0 marks)
Which connective tissue connects muscles to bones?
A.
Tendon
B.
Areolar
C.
Cartilage
D.
Ligament
Solution:
Tendon is a dense regular connective tissue that connects muscles to bone. It is non-elastic in nature.
The correct answer is A.
Why alternative B is wrong:
Areolar connective tissue helps in supporting internal organs and in repairing the tissues of the skin and muscles.
Why alternative C is wrong:
Cartilage is a specialized connective tissue that provides support and flexibility to the various body parts.
Why alternative D is wrong:
A ligament is an elastic structure that connects two bones together.
Question 33 ( 1.0 marks)
Hari grow one type of crop in one season and other type of crop in next season. He performs this activity in a sequential manner. Which practice is followed by Hari?
A.
crop rotation
B.
inter-cropping
C.
crop protection
D.
mixed cropping
Solution:
Crop rotation is the method of growing two or more varieties of crop on the same land in sequential season. Rotation of a leguminous crop such as pea with wheat is an example of crop rotation.
The correct answer is A.
Why alternative B is wrong:
Inter-cropping is the method where few rows of a crop alternate with few rows of another crop.
Why alternative C is wrong:
Crop protection is the method of protecting crops from pests or insects.
Why alternative D is wrong:
Mixed cropping is the method of growing two or more varieties of crops simultaneously on the same field.
Question 34 ( 1.0 marks)
Use the following information to answer the next question.
The information in which alternative completes the given statement?
A.
B.
C.
D.
Solution:
Mites, bacteria, rodents, fungi, etc. are biotic factors that are responsible for the loss of grains.
The correct answer is A.
Why alternatives B, C and D are wrong:
Inappropriate temperature, moisture, lack of sunlight, etc. are abiotic factors that are responsible for loss of grains.
Question 35 ( 1.0 marks)
Use the following information to answer the next question.
The information in which alternative completes the given statement?
A.
B.
C.
D.
Solution:
Apiculture is the practice of bee keeping. Honey and wax are the two products obtained by apiculture.
The correct answer is A.
Why alternatives B, C and D are wrong:
Pollen and nectar are plant products and are not obtained from apiculture.
Question 36 ( 1.0 marks)
Use the following information to answer the next question.
Which of the following statements is correct regarding the given changes?
A.
Both I and II are physical changes.
B.
Both I and II are chemical changes.
C.
I is a physical change while II is a chemical change.
D.
I is a chemical change while II is a physical change.
Solution:
Physical changes are changes in which no new substance is formed. Only shape, size, colour, and state of the substance changes during the process. During boiling of water, no new substance is formed. Only the state of water changes from liquid to gaseous.
Chemical changes are changes in which at least one new substance is formed. Rusting of an iron gate is a chemical change as formation of a new substance, rust (iron oxide), takes place during the process.
Hence, I is a chemical change while II is a physical change.
The correct answer is D.
Question 37 ( 1.0 marks)
Cuboidal epithelial tissue is found in the inner lining of
A.
respiratory tract
B.
kidney tubules
C.
oesophagus
D.
mouth
Solution:
Cuboidal epithelium is found in the inner lining of kidney tubules. As the name suggests, it is made of cube-shaped cells.
The correct answer is B.
Why alternative A is wrong:
Ciliated columnar epithelium is found in the inner lining of respiratory tract.
Why alternatives C and D are wrong:
Simple squamous epithelium is found in the inner linings of mouth and oesophagus. It is made of thin, flat cells with delicate lining.
Question 38 ( 1.0 marks)
Which cell organelle provides energy in the form of ATP for vital activities of the living cell?
A.
Endoplasmic reticulum
B.
Mitochondrion
C.
Golgi apparatus
D.
Nucleus
Solution:
Mitochondria are the sites of cellular respiration. They provide energy in the form of ATP for vital activities of the living cells.
The correct answer is B.
Why alternative A is wrong:
Endoplasmic reticulum serves as a channel for the transport of materials (especially proteins) between various regions of the cytoplasm or between cytoplasm and the nucleus.
Why alternative C is wrong:
Golgi apparatus helps in storage, modification, and packaging of products in vesicles.
Why alternative D is wrong:
Nucleus is not the site for cellular respiration. It controls and coordinates the cellular activities and controls heredity.
Question 39 ( 1.0 marks)
Which of the following cell organelles contains chromosomes?
A.
Endoplasmic reticulum
B.
Golgi apparatus
C.
Lysosomes
D.
Nucleus
Solution:
Chromosomes are rod-like structures present inside the nucleus. Each chromosome is a complex of DNA and proteins.
The correct answer is D.
Why alternative A is wrong:
Endoplasmic reticulum does not contain chromosomes. It helps in the manufacture of proteins and lipids.
Why alternative B is wrong:
Golgi apparatus does not contain chromosomes. It helps in the storage and packaging of materials.
Why alternative C is wrong:
Lysosomes do not contain chromosomes. They help in the digestion of foreign materials.
Question 40 ( 1.0 marks)
Which of the following structures is associated with cellular reproduction?
A.
Nucleus
B.
Vacuole
C.
Lysosome
D.
Ribosome
Solution:
Cellular reproduction is the process by which a single cell divides to form two new cells. Nucleus plays an important role in cellular reproduction.
The correct answer is A.
Why alternative B is wrong:
Vacuole store materials such as amino acids, sugars, and some proteins.
Why alternative C is wrong:
Lysosome helps in the digestion of foreign materials.
Why alternative D is wrong:
Ribosomes play a vital role in protein synthesis.
Question 41 ( 1.0 marks)
Use the following information to answer the next question.
What is the distance travelled by the car before coming to rest?
A.
24 m
B.
30 m
C.
36 m
D.
42 m
Solution:
The distance travelled by the car is obtained by calculating the area under the velocity-time graph.
Distance travelled by the car before coming to rest = Area of triangle POQ
Hence, the car travels 36 metres before coming to rest.
The correct answer is C.
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are the biotic factors that lead to seed degradation.