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Chapter – 16 Water A Precious Resource | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter -16 Water A Precious Resource

Q.1. Mark ‘T’ if the statement is true and ‘F’ if it is false:
(a) The freshwater stored in the ground is much more than that present in the rivers and lakes of the world. (T/F)

(b) Water shortage is a problem faced only by people living in rural areas. (T/F

(c) Water from rivers is the only source for irrigation in the fields. (T/F)
(d) Rain is the ultimate source of water. (T/F)

Ans. (a) T (b) F (c) F (d) T

Q.2. Explain how groundwater is recharged.
Ans. The groundwater gets recharged through the process of infiltration. Infiltration means seeping of water from rivers and lakes into the empty spaces and cracks deep below the ground.

Q.3.There are ten tubewells in a lane of fifty houses. What could be the long-term impact on the water table?
Ans.The effect on the water table depends on the replenishment of underground water. Only five families will share a tubewell, the water used for daily domestic purposes will not effect the water table. But if there is shortage of rains, the water used by the family will not replenished and water table will fall down.

Q.4.You have been asked to maintain a garden. How will you minimise the use of water? „
Ans.To minimise the wastage of water we will use the drip irrigation which throws the water at the base of plants. We will check the leakages in the water pipes and arrange small pits for rainwater harvesting. The collected rainwater will be used later.

Q.5.Explain the factors responsible for the depletion of water table.
Ans.Various factors responsible for the depletion of water table are:
(i) Increased population: Demand of water has been increased by the increased population. As the number of humans increase, the consumption of water also increases.
(ii) Increasing industries: All industries need water. As the number of human population increase, the number of industries are also increased which definitely increases the consumption of water.

(iii) Lack of water conservation techniques: Main source of water on earth and for the underground water is rain. The water of the rain, if conserved can increase the ground water level. But this is not done due to lack of water conservative techniques.
(iv) Agricultural activities: India is a country which depends on agriculture. The land used for cultivation has increased. So, the consumption of water for agriculture has increased. Irregular rainfall has increased the consumption of groundwater. This has increased the depletion of groundwater.

Q.6.Fill in the blanks with the appropriate answers: .
(a) People obtain groundwater through and .
(b) Three forms of water are solid, and .
(c) The water bearing layer of the earth is .
(d) The process of water seepage into the ground is called .
Ans.(a) wells, hand pumps
(b) liquid, gas
(c) aquifer
(d) infiltration

Q.7.Which one of the following is not responsible for water shortage?
(i) Rapid growth of industries
(ii) Increasing population
(iii) Heavy rainfall
(iv) Mismanagement of water resources
Ans.(iii) heavy rainfall

Q.8.Choose the correct option. The total water
(i) in the lakes and rivers of the world remains constant.
(ii) under the ground remains constant.
(iii) in the seas’and oceans of the world remains constant.
(iv) of the world remains constant.
Ans.(iv) of the world remains constant.

Q.9.Make a sketch shoyving groundwater and water table. Label it.
Ans

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Chapter – 15 Light | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter -15 Light

Q.1.Fill in the blanks:
(a) An image that cannot be obtained on a screen is called ___________ .
(b) Image formed by a convex is __________ always virtual and smaller in size.

(c) An image formed by a __________ mirror is always of the same size as that of the object.

(d) An image which can be obtained on a screen is called a __________ image.
(e) An image formed by a concave __________ cannot be obtained on a screen.
Ans.(a) virtual image (b) mirror (c) plane (d) real (e) lens

Q.2.Mark ‘T’ if the statement is true and ‘F’ if it is false:
(a) We can obtain an enlarged and erect image by a convex mirror. (T/F)
(b) A concave lens always form a virtual image. (T/F)
(c) We can obtain a real, enlarged and inverted image by a concave mirror. (T/F)
(d) A real image cannot be obtained on a screen. (T/F)
(e) A concave mirror always form a real image. (T/F)
Ans. a) F (b) T (c) T (d) F (e) F

Q.3. Match the items given in Column I with one or more items of Column II

Ans.

Q.4.State the characteristics of the image formed by a plane mirror.
Ans.(i) Plane mirror forms an erect image.
(ii) It forms a virtual image.
(iii) Size of the image is same as that of the object.
(iv)Image is formed at the same distance behind the mirror as the object stands in front of it.
(v) Image formed is a laterally inverted image i.e., right hand side of the object seems to be the left hand side and vice-versa.

Q.5.Find out the letters of English alphabet or any other language known to you in which the image formed in a plane mirror appears exactly like the letter itself. Discuss your findings.
Ans.Letters like A, H, I, M, O, T, U ,V, W etc. appear same when seen through a plane mirror

Q.6.What is a virtual image? Give one situation where a virtual image is formed.
Ans.The image which cannot be taken on a screen is called virtual image. When some object is placed very close to the concave mirror we don’t get any image on the white screen placed behind the mirror. Such image is called virtual image.

Q.7. State two differences between a convex and a concave lens.

Q.8. Give one use each of a concave and a convex mirror.
Ans. Use of concave mirror:
Concave mirror is used by dentists to examine the teeth.
Use of convex mirror:
Convex mirror is used as side view mirror in vehicles

Q.9. Which type of mirror can form a real image?
Ans. Concave mirror can form a real image.

Q.10. Which type of lens forms always a virtual image?
Ans. Concave lens always forms a virtual image.

Choose the correct option in Questions 11-13:

Q.11. A virtual image larger than the object can be produced by a
(i) concave lens (ii) concave mirror (iii) convex mirror (iv) plane mirror

Ans. (ii) concave mirror

Q.12. David is observing his image in a plane mirror. Die distance between the mirror and his image is 4 m. If he moves 1 m towards the mirror, then the distance between David and his image will be
(i) 3 m (ii) 5 m (iii) 6 m (iv) 8 m

Ans. (iii) 6 m

Q.13. The rear view mirror of a car is a plane mirror. A driver is reversing his car at a speed of 2 m/s. The driver sees in his rear mew mirror the image of a truck parked behind his car. The speed at which the image of the truck appears to approach the driver will be
(i) 1 m/s (ii) 2 m/s (iii) 4 m/s {iv) 8 m/s

Ans. (ii) 4 m/s

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RD SHARMA SOLUTION CHAPTER-3 Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 3 Functions Exercise Ex. 3.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Express the function f: X ® R given by f (x) = x ³ + 1 as set of ordered pairs, where X = {-1, 0, 3, 9, 7}.Solution 18

Chapter 3 Functions Exercise Ex. 3.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

begin mathsize 12px style If space straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator 1 minus straight x end fraction comma space Show space that space straight f left square bracket straight f left curly bracket straight f left parenthesis straight x right parenthesis right curly bracket right square bracket equals straight x. end style

Solution 4

Question 5

Solution 5

Question 6

begin mathsize 11px style Find space colon left parenthesis straight i right parenthesis space straight f space left parenthesis 1 divided by 2 right parenthesis space space left parenthesis ii right parenthesis space straight f left parenthesis negative 2 right parenthesis space space left parenthesis iii right parenthesis space straight f left parenthesis 1 right parenthesis space left parenthesis iv right parenthesis space straight f left parenthesis square root of 3 right parenthesis space and space left parenthesis straight v right parenthesis space straight f left parenthesis square root of negative 3 end root right parenthesis. end style

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 10

Solution 10

Question 11

Solution 11

Chapter 3 Functions Exercise Ex. 3.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(viii)

find the domain and range of Solution 3(viii)

Question 3(vii)

Find domain and range of f (x) = -|x|Solution 3(vii)

As |x|is defined for all real numbers, its domain is R and range is only negative numbers because, |x| is always positive real number for all real numbers and -|x| is always negative real numbers.

Chapter 3 Functions Exercise Ex. 3.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 4(vi)

Solution 4(vi)

Question 4(vii)

Solution 4(vii)

Question 4(viii)

Solution 4(viii)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

f : R rightwards arrow R space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals 3 x minus 2
f : R rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals minus x plus 4
f : R minus open curly brackets 3 over 2 close curly brackets rightwards arrow R space d e f i n e d space b y space f over g left parenthesis x right parenthesis equals fraction numerator x plus 1 over denominator 2 x minus 3 end fraction

Question 9

Solution 9

f plus g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals square root of x plus x ; space
f minus g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals square root of x minus x ;
f g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f g right parenthesis left parenthesis x right parenthesis equals x to the power of 3 divided by 2 end exponent ;
f over g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space open parentheses f over g close parentheses left parenthesis x right parenthesis equals fraction numerator 1 over denominator square root of x end fraction ;

Question 10

Solution 10

left parenthesis f plus g right parenthesis : R rightwards arrow left square bracket 0 comma infinity right parenthesis space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals x squared plus 2 x plus 1 equals left parenthesis x plus 1 right parenthesis squared
left parenthesis f minus g right parenthesis : R rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals x squared minus 2 x minus 1
open parentheses f g close parentheses : R rightwards arrow R space d e f i n e d space b y space left parenthesis f g right parenthesis left parenthesis x right parenthesis equals 2 x cubed plus x squared
open parentheses f over g close parentheses : R rightwards arrow R space d e f i n e d space b y space open parentheses f over g close parentheses left parenthesis x right parenthesis equals fraction numerator x squared over denominator 2 x plus 1 end fraction

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Chapter – 14 Electric Current and its Effects | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter - 14 Electric Current and its Effects

Q.1.Draw in your notebook the symbols to represent the following components of electrical circuits: connecting wires, switch in the ‘OFF’ position, bulb, cell, switch in the ‘ON’ position and battery.
Ans.

Q.2.Draw the circuit diagram to represent the circuit shown in fig. 14.9.

Ans.

Q.3.Fig. 14.11 shows four cells fixed on a board. Draw lines to indicate how you will connect their terminals with wires to make a battery of four cells.

Ans.

Q.4.The bulb in the circuit shown in fig. 14.13 does not glow. Can you identify the problem? Make necessary changes in the circuit to make the bulb glow.

Ans.Problem in this circuit is the combination of two cells. In the circuit positive terminal of one cell should be connected with negative terminal of other to make the bulb glow

Q.5. Name any two effects of electric current.
Ans. Electric current has the following effect :
(i) Electric current can give rise to heating and lighting.
(ii) Electric current can convert a straight conductor into a temporary magnet.

Q.6.When the current is switched on through a wire, a compass needle kept nearby gets deflected from its north-south position. Explain.
Ans. When current is passed through the wire, it deflects the compass near it from its north-south position like a magnet. This is called magnetic effect of the current. As we know that needle of the compass is made up of a thin magnet. When this needle comes in contact with another magnet then the like poles of the magnet repel each other and opposite poles attract each other. So the deflection is seen in the needle. In this case the wire behaves like a magnet and causes deflection in needle of the compass.

Q.7.Will the compass needle show deflection when the switch in the circuit shown by fig. 14.15 is closed?

Ans. No, because there is no source of electric current in this circuit, i.e., there is no battery.

Q.8.Fill in the blanks:
(a) Longer line in the symbol for a cell represents its ___________terminal
(b) The combination of two or more cells is called a ____________.
(c) When current is switched ‘on’ in a room heater, it ___________.
(d) The safety device based on the heating effect of electric current is called a _____________.

Ans.(a) positive (b) battery (c) becomes red hot and emits heat (d) fuse.

Q.9.Mark ‘T’ if the statement is true and ‘F’ it is a false:
(a) To make a battery of two cells, the negative terminal of one cell is connected to the negative terminal of the other cell. [T/F]

(b) When the electric current through the fuse exceeds a certain limit , the fuse wire melts and breaks. [T/F]
(c) An electromagnet does not aatract a piece of iron. [T/F]
(d) An electric bell has an electromagnet. [T/F]
Ans. (a) F (b) T (c) F (d) T

Q.10. Do you think an electromagnet can be used for separating plastic bags from a garbage heap? Explain.
Ans. No, the plastic bags do not get attracted by the magnet, so they cannot be separated by an electromagnet. Plastic bags are not magnetic materials, only magnetic materials like iron can be attracted by the magnet.

Q.ll. An electrician is carrying out some repairs in your house. He wants to replace a fuse by a piece of wire. Would you agree? Give reasons for your response.
Ans. No, we would not agree to allow to replace the fuse by a wire. Wires in the fuses are of specific melting points. So wd should always use ISI marked fuses in our houses to prevent short circuits.

Q.12.Zubeda made an electric circuit using a cell holder shown in fig. 14.16, a switch and a bulb. When she put the switch in the ‘ON’ position, the bulb did not glow. Help Zubeda in identifying the possible defects in the circuit

Ans. It is important to put the cells in right series. The positive terminal of the first cell should be connected with negative terminal of the second cell. The switch should be closed properly and bulb should not be fused. If Zubeda will check these then the bulb will certainly glow.

Q.13. In the circuit shown in fig. 14.17.

Would any of the bulb glow when the switch is in the ‘OFF’ position?
What will be the order in which the bulbs A, B and C will glow when the switch is moved to the ‘ON’ position?

Ans. No bulb will glow.
All bulbs will glow simultaneously.

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RD SHARMA SOLUTION CHAPTER- 2 Relations I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 2 Relations Exercise Ex. 2.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 15(iii)

Solution 15(iii)

Question 15(iv)

Solution 15(iv)

Chapter 2 Relations Exercise Ex. 2.2

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Chapter 2 Relations Exercise Ex. 2.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

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Chapter – 13 Motion and Time | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter - 13 Motion and Time

Q.1.Classify the following as motion along a straight line, circular or oscillatory motion :
(i) Motion of your hands while running.
(ii) Motion of a horse pulling a cart on a straight road.
(iii) Motion of a child in a merry-go-round. •
(iv) Motion of a child on a see-saw.
(v) Motion of the hammer of an electric bell.
(vi) Motion of a train on a straight bridge.
Ans.(i) Oscillatory motion
(ii) Linear motion
(iii) Circular motion
(iv) Oscillatory motion
(v) Oscillatory motion
(vi) Linear motion

Q.2.Which of the following are not correct?
(i) The basic unit of time is second.
(ii) Every object moves with a constant speed.
(iii) Distances between two cities are measured in kilometers.
(iv) The time period of a given pendulum is not constant.
(v) The speed of a train is expressed in m/h
Ans. (ii), (iv), (v)

Q.3.A simple pendulum takes 32s to complete 20 oscillations, what is the time period of the pendulum?
Ans. Time taken to complete 20 oscillations = 32 s
Time taken to complete 1 oscillation = 32/20 s = 1.6 s
Time period of a pendulum is time taken by it to complete 1 oscillation.
Time period of pendulum is 1.6 second

Q.4.The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.
Ans.Distance = 240 km
Time taken = 4 hours
Speed =Distance covered/time taken = 240km/4 hammer
=60 km/h
Speed of train =60 km/h

Q.5.The odometer of a car reads 57321.0 km when the clock shows the time 08:30 AM. What is the distance moved by the car, if at 08:50 AM, the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time. Express the speed in km/h also.
Ans.Distance = 57336.0 km – 57321 km =15 km
Speed in km/min =15km/20 min = 3/4 km/min

Speed in km/hr =15 km/1/3 hr
=(15 x 3) km/hr
=45 km/hr.

Q.6.Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/min, calculate the distance between her house and the school.
Ans.Time taken =15 min
Speed = 2 m/min .
Distance = speed x time = 2 x 15 = 30 m
Distance between Salma’s school and her house is 30 m.

Q.7.Show the shape of the distance-time graph for the motion in the following cases :
(i) A car moving with a constant speed.
(ii) A car parked on a side road
Ans.

Q.8. Which of the following relations is correct?
(i) Speed = Distance Time
(ii) Speed = Distance/Time
(iii)Speed – Time/Distance
(iv)Speed = 1/Distance Time
Ans. (ii) Speed = Distance/Time is correct

Q.9.The basic unit of speed is:
(i) km/min (ii) m/min (iii) km/h (iv) m/s
Ans. (iv) m/s

Q.10.A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:
(i) 100 km (ii) 25 km (iii) 15 km (iv) 10 km
Ans.(ii) 25 km
Q.11.Suppose the two photographs, shown in fig. 13.1 and fig. 13.2 of NCERT had been taken at an interval of 10 seconds. If a distance of 100 metres is shown by 1 cm in these photographs, calculate the speed of the blue car.
Ans.0.1 cm/s or 10 m/s

Q.12.Fig 13.10 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

Ans. ‘A’ is moving faster.

Q.13.Which of the following distance-time graphs shows a truck moving with speed which is not constant?

Ans. (iii)

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Chapter – 12 Reproduction in Plants | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter - 12 Reproduction in Plants

Q.1.Fill in the blanks:
(a) Production of new individuals from the vegetative part of parent is called ___________.
(b) A flower may have either male or female reproductive parts. Such a flower is called _______.

(c) The transfer of pollen grains from the anther to the stigma of the same or of

another flower of the same kind is known as __________.
(d) The fusion of male and female gametes is termed as ____________ .
(e) Seed dispersal takes place by means of ________ and __________.

Ans. (a) vegetative reproduction (b) unisexual flower (c) pollination (d) fertilization (e) wind, water

Q.2. Describe the different methods of asexual reproduction. Give examples.
Ans. Different methods of asexual reproduction are:
(a) Binary Fission: This process takes place in unicellular organisms. Parent cell elongates and gets divided into two identical daughter cells. Each daughter cell grows into an independent adult.
(b) Endospore Formation: In this method the spore wall is formed around a bacterial cell to form an endospore. This endospore germinates to form an active bacterium under favourable conditions.
(c) Fragmentation: In this process, body of the organism breaks up into two parts. Then each part grows into a new filament thus forming two organisms from a single one.
(d) Spore Formation: The spores are tiny spherical unicellular structures protected by thick wall. The spores are stored in a hard outer covering and this is called sporangium. Under favourable conditions the hard cover breaks and spores spread for germination.
(e) Budding: In yeast, new organisms are produced by the bud formation from the parent organism. After growing to full size, the bud gets detached and forms a new independent individual.
(f) Vegetative propagation: When vegetative parts of a plant like stems, leaves and root etc., give rise to new ones, it is.called vegetative propagation.

Q.3. Explain what you understand by sexual reproduction.
Ans.Sexual reproduction means involvement of two parents in the process of reproduction. It is found mainly in higher plants where male gamete and female gamete fuse to form a zygote. These zygotes develop into individuals which are not identical. Offsprings inherit the characteristics of both the parents. In sexual reproduction both parents survive after the process of reproduction.

Q.4. State the main difference between asexual and sexual reproduction.
Ans.

Q.5.Sketch the reproductive parts of a flower.
Ans.

Q.6.Explain the difference between self-pollination and cross-pollination
Ans.

Q.7.How does the process of fertilization take place in flowers?
Ans.When the pollen grain reaches the stigma of a same species flower, it starts
growing out into the pollen tube of the stigma. This tube continues to grow inside the style till it reaches the ovule. Male cells are released into the ovule for the fertilization with the female egg cell and thus the zygote is formed. After this process of fertilization, the ovary develops into fruit and ovule into seeds.

Q.8.Describe the various ways by which seeds are dispersed.
Ans. Following are the ways in which the seeds are dispersed:
(i) Some light seeds like that of madar, which are hairy, dry and small are carried away by the wind to different places.
(ii) Spiny seeds and fruits like that of xanthium and urena, stick to the clothes of passers by and animals. These seeds are carried away by these agents to different places.
(iii) In some plants having heavy seeds like that of coconut, water acts as the dispersing agents.
(iv) Some seeds are dispersed with the fruit burst like in case of balsam and castor.

Q.9.Match items in Column I with those in Column II

Ans

Q.10. Tick (%/) the correct answer:
(a) The reproductive part of a plant is the
(i) leaf (ii) stem (iii) root (iv) flower
(b) The process of fusion of the male and the female gametes is called
(i) fertilisation (ii) pollination (iii) reproduction (iv) seed formation
c) Mature ovary forms the
(i) seed (ii) stamen (iii) pistil (iv) fruit
(d) A spore producing plant is
(i) rose (ii) bread mould (iii) potato (iv) ginger
(e) Bryophyllum can be reproduced by its
(i) stem (ii) leaves (iii) roots (iv) flower

Ans.(a) (iv) flower (b) (i) fertilisation (c) (iv) fruit (d) (ii) bread mould (e) (ii) leaves

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RD SHARMA SOLUTION CHAPTER- 1 Sets I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 1 Sets Exercise Ex. 1.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

If A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then insert the appropriate symbol or in each of the following blank spaces:

4…A
-4 …A
12 ….A
9 …A
0 …..A
-12 ….A

Solution 3

Chapter 1 Sets Exercise Ex. 1.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Chapter 1 Sets Exercise Ex. 1.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 1 Sets Exercise Ex. 1.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 4(vi)

Solution 4(vi)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Chapter 1 Sets Exercise Ex. 1.5

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 1 Sets Exercise Ex. 1.6

Question 2(i)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)Solution 2(i)

Question 2(ii)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)Solution 2(ii)

Question 2(iii)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A ∩ (B – C) = (A ∩ B) – (A ∩ C)Solution 2(iii)

Question 2(iv)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A – (B ∪ C) = (A – B) ∩ (A – C)Solution 2(iv)

Question 2(v)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A – (B ∩ C) = (A – B) ∪ (A – C)Solution 2(v)

Question 2(vi)

Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

A ∩ (B D C) = (A ∩ B) D (A ∩ C)Solution 2(vi)

Question 4(i)

For any two sets A and B, prove that

B ⊂ A ∪ BSolution 4(i)

Question 4(ii)

For any two sets A and B, prove that

A ∩ B ⊂ BSolution 4(ii)

Question 4(iii)

For any two sets A and B, prove that

A ⊂ B ⇒ A ∩ B = ASolution 4(iii)

Question 14(i)

Show that For any sets A and B,

A = (A ∩ B) ∩ (A – B)Solution 14(i)

Question 14(ii)

Show that For any sets A and B,

A ∪ (B – A) = A ∪ BSolution 14(ii)

Question 15

Each set X, contains 5 elements and each set Y, contains 2 elements and each element of S belongs to exactly 10 of the X’_r^s and to exactly 4 of Y’_r^s, then find the value of n.Solution 15

Question 1

Solution 1

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12(i)

Solution 12(i)

Question 12(ii)

Solution 12(ii)

Question 13

Solution 13

Chapter 1 Sets Exercise Ex. 1.7

Question 4(i)

For any two sets A and B, prove that

(A ∪ B) – B = A – BSolution 4(i)

Question 4(ii)

For any two sets A and B, prove that

A- (A ∩ B) = A – BSolution 4(ii)

Question 4(iii)

For any two sets A and B, prove that

A – (A – B) = A ∩ BSolution 4(iii)

Question 4(iv)

For any two sets A and B, prove that

A ∪ (B – A) = A ∪ BSolution 4(iv)

Question 4(v)

For any two sets A and B, prove that

(A – B) ∪ (A ∩ B) = ASolution 4(v)

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 3

Solution 3

Chapter 1 Sets Exercise Ex. 1.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

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Chapter – 11 Transportation in Animals and Plants | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter - 11 Transportation in Animals and Plants

Q.1.Match structures given in Column I with functions given in Column II.

Ans.

Q.2.-Fill in the blanks :
(i) The blood from the heart is transported to all parts of the body by the _____________
(ii) Haemoglobin is present in ___________ cells.
(iii) Arteries and veins are joined by a network of ___________ .
(iv) The rhythmic expansion and contraction of the heart is called ___________
(v) The main excretory product in human beings is ___________
(vi) Sweat contains water and .
(vii) Kidneys eliminate the waste materials in the liquid form called .
(viii) Water reaches great heights in the trees because of suction pull caused by .

Ans.(i) arteries (ii) red blood cells • (in) capillaries (iv) heartbeat (v) urea (vi) salts (vii) urine (viii) transpiration.

Q.3.Choose the correct options:
(a) In plants, water is transported through
(i) Xylem (ii) Phloem (iii) Stomata (iv) Root hair
(b) Water absorption through roots can be increased by keeping the plants
(i) In the shade (ii) in dim light (iii) under the fan (iv) covered with a polythene bag

Ans.(a) (i) Xylem {b) (iii) under the fan

Q.4.Why is transport of materials necessary in a plant or an animal? Explain.
Ans.Transport of materials is necessary for plants or animals because due to it the nutrients and oxygen are made available to all the parts of the body. If the transport of necessary nutrients and oxygen does not take place in the body, the body will not be able to survive.

Q.5.What will happen if there are no platelets in the blood?
Ans.The blood platelets are responsible for the clotting of the blood. When some injury occurs blood starts flowing. But it clots on its own. If there are no platelets, the blood will not be able to clot and keep on flowing. Huge loss of blood ultimately causes death.

Q.6.What are stomata? Give two junctions of stomata.
Ans.There are small openings on the lower surface of the leaves. These pores are called stomata. These openings are surrounded with guard cells.
Functions of stomata:
1. It helps in the transpiration of water, i.e., the loss of excess water from the plant.
2. Loss of water from the stomata creates an upward pull, i.e., suction pull which helps in absorption of water from the roots.
3. They help in exchange of gases.

Q.7.Does transpiration serve any useful function in plants? Explain.
Ans.(i) It helps to enhance the absorption of water and dissolved minerals by creation of a suction pull.
(ii) It helps in getting rid of the excess water.
(iii) It helps in transport of water and minerals to leaves and leaves use the water for photosynthesis. ‘
(iv)It produces a cooling effect on the plant

Q.8.What are the components of blood?
Ans.There are four components of blood:
(i) Plasma is a liquid part of blood which is yellowish in colour and contain 90% water. It contains food, enzymes, wastes and proteins etc
(ii) Red blood cells are disc shaped cells containing red coloured pigment called haemoglobin in it. Haemoglobin helps in transportation of oxygen.
(iii) White blood cells are the fighting cells which protect us against bacteria and foreign materials causing infections.
(iv) Platelets help in clotting of the blood

Q.9.Why is blood needed by all the parts of a body?
Ans. Blood is needed by all the parts because it contains the digested food and oxygen in it. It supplies the oxygen and digested food to various parts of the body to provide essential energy to them which helps the body to perform various functions

Q.10.What makes the blood look red?
Ans. A red pigment called haemoglobin gives the blood its red colour. This pigment helps in carrying oxygen to various parts of the body by blood

Q.11.Describe the function of the heart.
Ans. The right auricle and ventricle receive blood with carbon dioxide from all parts of the body. The collected blood is then pumped to the lungs for the purification. In lungs, the exchange of gases takes place and purified blood is sent back to left auricle. It pumps it to the left ventricle, which in turn pumps off the purified blood to all parts of body through arteries

Q.12.Why is it necessary to excrete waste products?
Ans. Certain waste and toxic products are formed during functioning of body cells. The waste products like urea etc. are toxic. When these toxic materials are not removed from the body, they get mixed with blood and can damage the cells of body. It is necessary to remove such poisonous waste materials from our body.

Q.13.Draw a diagram of the human excretory system and label the various parts.
Ans.

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Chapter – 10 Respiration in Animals and Plants | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter - 10 Respiration in Animals and Plants

Q1. Why does an athlete breathe faster and deeper than usual after finishing the race?
Answer:
During the run, the demand of energy is high but the supply of oxygen to produce energy is limited. Therefore, anaerobic respiration takes places in the muscle cells to fulfill the demand of energy. After finishing the race, an athlete breathe faster and deeper than usual so that more oxygen is supplied to the cells.

Q2. List the similarities and differences between aerobic and anaerobic respiration.
Answer:
Similarity:
(i) In both aerobic and anaerobic respiration, food is broken down to release energy.
(ii) Both takes place inside cells.
(iii) Both produces byproducts.

Differences:

Aerobic Respiration	Anaerobic Respiration
(i) It takes place in the presence of oxygen.	(i) It takes place in the absence of oxygen.
(ii) Energy is released in higher amount.	(ii) Energy is released in lesser amount.
(iii) Carbon dioxide and water are produced as byproducts.	(iii) Carbon dioxide and water are produced as byproducts.
(iv) It is a slow process.	(iv) It is a fast process.
(v) Examples: Animals and plants cells.	(iv) Examples: Human cells, yeast, Bacteria etc.

Q3. Why do we often sneeze when we inhale a lot of dust-laden air?
Answer:
We often sneeze when we inhale a lot of dust-laden air to expel out these foreign particles. These particles get past the hair in the nasal cavity and irritate the lining of the cavity which results in sneezing.

Q4. Take three test-tubes. Fill each of them with water. Label them A, B and C. Keep a snail in test-tube A, a water plant in test-tube B and in C, keep snail and plant both. Which test-tube would have the highest concentration of CO₂ ?
Answer:
Test-tube A will have the highest concentration of CO₂ because snail will take in oxygen and gives out CO₂.
In test-tubes B and C, the CO₂ will be utilized by the water plant for synthesizing food and hence there will be less concentration of CO₂ in these.

Q5. Tick the correct answer:
(a) In cockroaches, air enters the body through
(i) lungs
(ii) gills
(iii) spiracles
(iv) skin
Answer: (iii) spiracles
(b) During heavy exercise, we get cramps in the legs due to the accumulation of
(i) carbon dioxide
(ii) lactic acid
(iii) alcohol
(iv) water
Answer: (ii) lactic acid
(c) Normal range of breathing rate per minute in an average adult person at rest is:
(i) 9 – 12
(ii) 15 – 18
(iii) 21 – 24
(iv) 30 – 33
Answer: (ii) 15 – 18
(d) During exhalation, the ribs
(i) move outwards
(ii) move downwards
(iii) move upwards
(iv) do not move at all
Answer: (ii) move downwards

Q6. Match the items in Column I with those in Column II:

Column I	Column II
(a) Yeast	(i) Earthworm
(b) Diaphragm	(ii) Gills
(c) Skin	(iii) Alcohol
(d) Leaves	(iv) Chest cavity
(e) Fish	(v) Stomata
(f) Frog	(vi) Lungs and skin
–	(vii) Tracheae

Answer:

Column I	Column II
(a) Yeast	(iii) Alcohol
(b) Diaphragm	(iv) Chest cavity
(c) Skin	(i) Earthworm
(d) Leaves	(v) Stomata
(e) Fish	(ii) Gills
(f) Frog	(vi) Lungs and skin

Q7. Mark T if the statement is true and F if it is false:
(i) During heavy exercise the breathing rate of a person slows down. (T/ F)

(ii) Plants carry out photosynthesis only during the day and respiration only at night. (T/ F)

(iii) Frogs breathe through their skins as well as their lungs. (T/ F)
(iv) The fishes have lungs for respiration. (T/ F)
(v) The size of the chest cavity increases during inhalation. (T/ F)
Answer:
(i) F
(ii) F
(iii) T
(iv) F
(v) T

Q8. Given below is a square of letters in which are hidden different words related to respiration in organisms. These words may be present in any direction – upwards, downwards, or along the diagonals. Find the words for your respiratory system. Clues about those words are given below the square.

(i) The air tubes of insects
(ii) Skeletal structures surrounding chest cavity
(iii) Muscular floor of chest cavity
(iv) Tiny pores on the surface of leaf
(v) Small openings on the sides of the body of an insect
(vi) The respiratory organs of human beings
(vii) The openings through which we inhale
(viii) An anaerobic organism
(ix) An organism with tracheal system
Answer:

(i) The air tubes of insects → Trachea
(ii) Skeletal structures surrounding chest cavity → Ribs
(iii) Muscular floor of chest cavity → Diaphragm
(iv) Tiny pores on the surface of leaf → Stomata
(v) Small openings on the sides of the body of an insect → Spiracles
(vi) The respiratory organs of human beings → Lungs
(vii) The openings through which we inhale → Nostrils
(viii) An anaerobic organism → Yeast
(ix) An organism with tracheal system → Ant

Q9. The mountaineers carry oxygen with them because:
(a) At an altitude of more than 5 km there is no air.
(b) The amount of air available to a person is less than that available on the ground.
(c) The temperature of air is higher than that on the ground.
(d) The pressure of air is higher than that on the ground.
Answer:
The mountaineers carry oxygen with them because (b) The amount of air available to a person is less than that available on the ground.

Chapter – 16 Water A Precious Resource | Class 7th | NCERT Science Solutions | Edugrown

Chapter -16 Water A Precious Resource

Chapter – 15 Light | Class 7th | NCERT Science Solutions | Edugrown

Chapter -15 Light

RD SHARMA SOLUTION CHAPTER-3 Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 3 Functions Exercise Ex. 3.1

Chapter 3 Functions Exercise Ex. 3.2

Chapter 3 Functions Exercise Ex. 3.3

Chapter 3 Functions Exercise Ex. 3.4

Chapter – 14 Electric Current and its Effects | Class 7th | NCERT Science Solutions | Edugrown

Chapter - 14 Electric Current and its Effects

RD SHARMA SOLUTION CHAPTER- 2 Relations I CLASS 11TH MATHEMATICS-EDUGROWN

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Chapter 2 Relations Exercise Ex. 2.3

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Chapter - 13 Motion and Time

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Chapter 1 Sets Exercise Ex. 1.1

Chapter 1 Sets Exercise Ex. 1.2

Chapter 1 Sets Exercise Ex. 1.3

Chapter 1 Sets Exercise Ex. 1.4

Chapter 1 Sets Exercise Ex. 1.5

Chapter 1 Sets Exercise Ex. 1.6

Chapter 1 Sets Exercise Ex. 1.7

Chapter 1 Sets Exercise Ex. 1.8

Chapter – 11 Transportation in Animals and Plants | Class 7th | NCERT Science Solutions | Edugrown

Chapter - 11 Transportation in Animals and Plants

Chapter – 10 Respiration in Animals and Plants | Class 7th | NCERT Science Solutions | Edugrown

Chapter - 10 Respiration in Animals and Plants

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Chapter -16 Water A Precious Resource

Chapter -15 Light

Chapter 3 Functions Exercise Ex. 3.1

Chapter 3 Functions Exercise Ex. 3.2

Chapter 3 Functions Exercise Ex. 3.3

Chapter 3 Functions Exercise Ex. 3.4

Chapter - 14 Electric Current and its Effects

Chapter 2 Relations Exercise Ex. 2.1

Chapter 2 Relations Exercise Ex. 2.2

Chapter 2 Relations Exercise Ex. 2.3

Chapter - 13 Motion and Time

Chapter - 12 Reproduction in Plants

Chapter 1 Sets Exercise Ex. 1.1

Chapter 1 Sets Exercise Ex. 1.2

Chapter 1 Sets Exercise Ex. 1.3

Chapter 1 Sets Exercise Ex. 1.4

Chapter 1 Sets Exercise Ex. 1.5

Chapter 1 Sets Exercise Ex. 1.6

Chapter 1 Sets Exercise Ex. 1.7

Chapter 1 Sets Exercise Ex. 1.8

Chapter - 11 Transportation in Animals and Plants

Chapter - 10 Respiration in Animals and Plants

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