NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 8
Chapter Name Comparing Quantities
Exercise Ex 8.1
Number of Questions Solved 3
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

Question 1.
Find the ratio of:
(a) ₹ 5 to 50 paise
(b) 15 kg to 210 g
(c) 9 m to 27 cm
(d) 30 days to 36 hour

Solution:
(a) ₹ 5 : 50 paise
= ₹ 5 : 50 paise
= 5 × 100 paise : 50 paise
= 500 paise : 50 paise
= 10 : 1.

(b) 15 kg to 210g
= 15 kg : 210g
= 15 × 1000 g : 210g
= 15000 g : 210g = 500:7.

(c) 9 m to 27 cm
= 9m : 27 cm = 9 × 100 cm : 27 cm
= 900 cm : 27 cm
900 : 27 = 100 : 3.

(d) 30 days to 36 hours
= 30 days : 36 hours
30 × 24 hours : 36 hours
= 720 hours : 36 hours
= 720 : 36 = 20 : 1.

Question 2.
In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Solution:
Let x computers be needed for 24 students.
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1 1
Hence, 12 computers will be needed for
24 students.
Aliter:
In a computer lab,
Every 6 students need 3 computers.
∴ Every 1 student needs \(\frac { 3 }{ 6 } \) computers.
∴ Every 24 students need \(\frac { 3 }{ 6 } \) × 24 = 3 × 4 = 12 computers

Question 3.
Population of Rajasthan = 570 lakhs
and population of UP = 1660 lakhs.
Area of Rajasthan =3 lakh km2
and area of UP = 2 lakh km2.
(i) How many people are there per km2 in both these States?
(ii) Which State is less populated?
Solution:
(i) Number of people per sq. km. in Rajasthan state
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1 2
(ii) The Rajasthan State is less populated.

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NCERT Solutions for Class 7 Science Chapter 10 Respiration in Organisms

NCERT Solutions for Class 7 Science Chapter 10 Respiration in Organisms are part of NCERT Solutions for Class 7 Science. Here we have given NCERT Solutions for Class 7 Science Chapter 10 Respiration in Organisms.

Board CBSE
Textbook NCERT
Class Class 7
Subject Science
Chapter Chapter 10
Chapter Name Respiration in Organisms
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 7 Science Chapter 10 Respiration in Organisms

Question 1.
Why does an athlete breathe faster and deeper than usual after finishing the race?
Answer:
During running, the athlete uses a lot of energy. So, he/she needs more energy and breathes faster.

Question 2.
List the similarities and differences between aerobic and anaerobic respiration.
Answer:
Similarities:

Aerobic Respiration Anaerobic Respiration
1. It starts with the breakdown of a nutrient (glucose). 1. It also starts with the breakdown of a nutrient (glucose).
2. It yields byproducts. 2. It also yields byproducts.
3. It takes place in a cell. 3. It also takes place in a cell.
4. In this process energy is released. 4. In this process also energy is released.

 Differences:

Aerobic Respiration Anaerobic Respiration
1. It is the process of breakdown of glucose in the presence of oxygen. 1. It is the process of breakdown of glucose in the absence of oxygen.
2. Glucose is completely oxidized. 2. Glucose is incompletely oxidized.
3. The end products formed are CO2, H2O, and energy. 3. The end products formed are C02, ethyl alcohol (organic acid), and energy.
4. Energy released is more. (38 ATP molecules). 4. Energy released is less (2 ATP molecules).
5. It takes place in all higher organisms. 5. It takes place in lower organisms like yeast and the muscles of man.
6. The reactions take place in the cytoplasm and mitochondria. 6. The reactions take place only in the cytoplasm.

Question 3.
Why do we often sneeze when we inhale a lot of dust-laden air?
Answer:
The air around us has various types of unwanted particles, such as smoke, dust, pollens, etc. When we inhale, the particles get trapped in the hair present in our nasal cavity. However, sometimes these particles may get past the hair in thç nasal cavity. Then they irritate the lining of the cavity, as a result of which we sneeze. Sneezing expels these foreign particles from the inhaled air and dust-free, clean air enters our body.

Question 4.
Take three test-tubes. Fill 3/4th of each 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:
The exchange of gases in three test-tubes can be shown as in Fig.
NCERT Solutions for Class 7 Science Chapter 10 Respiration in Organisms Q.4
Fig. (A) In test tube A; CO2 is produced by the respiration of snail (No consumption of CO2).
(B) In test tube B; CO2 is produced by the respiration of the plant and a much higher amount of CO2 is consumed during photosynthesis.
(C) In test tube C; CO2 is produced by the respiration of snail and plant and CO2 is consumed during photosynthesis.
It is clear from the above observation that the highest concentration of CO2 will be in test tube A.

Question 5.
Tick the correct answer:
(a) In cockroaches, air enters the body through
(i) lungs
(ii) gills
(iii) spiracles
(iv) skin

(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

(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

(d) During exhalation, the ribs
(i) move outwards
(ii) move downwards
(iii) move upwards
(iv) do not move at all
Answer:
(a) (iii) spiracles
(b) (ii) lactic acid
(c) (ii) 15 – 18
(d) (ii) move downwards

Question 6.
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

 Question 7.
Mark ‘T’ if the statement is true and ‘F’ if it is false :
During heavy exercise, the breathing rate of a person slows down. (T/F)
Plants carry out photosynthesis only during the day and respiration only at night. (T/F)
Frogs breathe through their skins as well as their lungs. (T/F)
The fishes have lungs for respiration. (T/F)
The size of the chest cavity increases during inhalation. (T/F)
Answer:
(i) F
(ii) F
(iii) T
(iv) F
(v) T

Question 8.
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.
byjus class 7 science Chapter 10 Respiration in Organisms Q.8.1
(i) The air tubes of insects
(ii) Skeletal structures surrounding the chest cavity
(iii) Muscular floor of the chest cavity
(iv) Tiny pores on the surface of the 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 a tracheal system
Answer:
(i) TRACHEA
(ii) RIBS
(iii) DIAPHRAGM
(iv) STOMATA
(v) SPIRACLES
(vi) LUNGS
(vii) NOSTRILS
(viii) YEAST
(ix) ANT
These names are indicated by arrows and their serial number are indicated at starting point of the arrow.
NCERT Solutions for Class 7 Science Chapter 10 Respiration in Organisms Q.8.2

Question 9.
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 the air is higher than that on the ground.
(d) the pressure of air is higher than that on the ground.
Answer:
(b) The amount of air available to a person is less than that available on the ground.

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NCERT Solutions for Class 7 Science Chapter 15 Light

NCERT Solutions for Class 7 Science Chapter 15 Light are part of NCERT Solutions for Class 7 Science. Here we have given NCERT Solutions for Class 7 Science Chapter 15 Light.

Board CBSE
Textbook NCERT
Class Class 7
Subject Science
Chapter Chapter 15
Chapter Name Light
Number of Questions Solved 13
Category NCERT Solutions

NCERT Solutions for Class 7 Science Chapter 15 Light

Question 1.
Fill in the blanks:

  1. An image that cannot be obtained on a screen is called ………
  2. The image formed by a convex ……… is always virtual and smaller in size.
  3. An image formed by a ………. mirror is always of the same size as that of the object.
  4. An image which can be obtained on a screen is called a ………. image.
  5. An image formed by a concave ………. cannot be obtained on a screen.

Answer:

  1. virtual image
  2. mirror
  3. plane
  4. real
  5. lens

Question 2.
Mark “T” if the statement is true and ‘F’ if it is false:

  1. We can obtain an enlarged and erect image by a convex mirror. (T/F)
  2. A concave lens always forms a virtual image. (T/F)
  3. We can obtain a real, enlarged, and inverted image by a concave mirror. (T/F)
  4. A real image cannot be obtained on a screen. (T/F)
  5. A concave mirror always forms a real image. (T/F)

Answer:

  1. F
  2. T
  3. T
  4. F
  5. F

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

Column I Column II
(a) A plane mirror (i) Used as a magnifying glass.
(b) A convex mirror (ii) Can form images of objects spread over a large area.
(c) A convex lens (iii) Used by dentists to see an enlarged images of teeth.
(d) A concave mirror (iv) The image is always inverted and magnified.
(e) A concave lens (v) The image is erect and of the same size as the object.
(vi) The image is erect and smaller in size than the object.

Answer:

Column I Column II
(a) A plane mirror (v) The image is erect and of the same size as the object.
(b) A convex mirror (ii) Can form images of objects spread over a large area.
(c) A convex lens (i) Used as a magnifying glass.
(d) A concave mirror (iii) Used by dentists to see enlarged images of teeth.
(e) A concave lens (vi) The image is erect and smaller in size than the object.

Question 4.
State the characteristics of the image formed by a plane mirror.
Answer:
Image formation by a plane mirror. We are able to see images using a mirror. An image formed by a mirror (flat) has the following features:
NCERT Solutions for Class 7 Science Chapter 15 Light Q.4

In a plane mirror, the image is formed behind the mirror. It is erect, of the same size, and is at the same distance from the mirror as the object is in front of it.

Question 5.
Find out the letters of the 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.
Answer:
A, H, I, M, 0, T, U, V, W, X, Y.

Question 6.
What is a virtual image? Give one situation where a virtual image is formed.
Answer:
The image that cannot be formed on a screen is called a virtual image. The image formed by a plane mirror is virtual because the image cannot be obtained on a screen when placed either in front of the mirror or behind it.

Question 7.
State two differences between a convex and a concave lens.
Answer:

Convex lens Concave lens
1. It is thicker in the middle and thinner at the edges. 1. It is thinner in the middle and thicker at the edges.
2. Converges the light falling on it. 2. Diverges the light falling on it.
3. Can form virtual, erect, and magnified images. 3. Always forms erect, virtual, and smaller images.

Question 8.
Give one use each of a concave and a convex mirror.
Answer:

  1. Concave mirror: Used by dentists to see teeth.
  2. Convex mirror: Used in vehicles as a rearview mirror.

Question 9.
Which type of mirror can form a real image?
Answer:
A concave mirror can form a real image.
NCERT Solutions for Class 7 Science Chapter 15 Light Q.9
Fig. Convex mirror as a side-view mirror

Question 10.
Which type of lens forms always a virtual image?
Answer:
A concave lens.

Choose the correct option in questions 11-13:

Question 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
Answer:
(ii) Concave mirror

Question 12.
David is observing his image in a plane mirror. The 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
Answer:
(iii) 6 m

Question 13.
The rearview 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 rearview 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
Answer:
(iii) 4 m/s

We hope the NCERT Solutions for Class 7 Science Chapter 15 Light helps you. If you have any query regarding NCERT Solutions for Class 7 Science Chapter 15 Light, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects

NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects are part of NCERT Solutions for Class 7 Science. Here we have given NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects.

Board CBSE
Textbook NCERT
Class Class 7
Subject Science
Chapter Chapter 14
Chapter Name Electric Current and its Effects
Number of Questions Solved 13
Category NCERT Solutions

NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects

Question 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.
Answer:
NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.1.1
NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.1.2

Question 2.
Draw the circuit diagram to represent the circuit shown in Fig.
byjus class 7 science Chapter 14 Electric Current and its Effects Q.2.1
Answer:
NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.2.2

Question 3.
Figure 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.
NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.3.1
Answer:

NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.3.2

Question 4.
The bulb in the circuit shown in Fig. does not glow. Can you identify the problem? Make necessary changes in the circuit to make the bulb glow.
NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.4.1
Answer:
The problem in this circuit is the connection of two cells. Here both the positive terminals of the cells are connected to each other. This must be reversed for anyone cell to make the bulb glow i.e., the positive terminal of one cell should be connected with the negative terminal of the other.
byjus class 7 science Chapter 14 Electric Current and its Effects Q.4.2

Question 5.
Name any two effects of electric current.
Answer:

  • Heating effect.
  • Magnetic effect.

Question 6.
When the current is switched on through a wire, a compass needle kept nearby gets deflected from its north-south position. Explain.
Answer:
When an electric current passes through a wire, it behaves like a magnet. This is the magnetic effect of the electric current due to the attraction of the wire. So, the compass needle which is a magnet gets deflected.

Question 7.
Will the compass needle show deflection when the switch in the circuit shown by Fig. is closed?
NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.7
Answer:
No. Since, there is no cell so, no current will flow.

Question 8.
Fill in the blanks:

  1. A longer line in the symbol for a cell represents its ………… terminal.
  2. The combination of two or more cells is called a …………..
  3. When current is switched ‘on’ in a room heater, it ………….
  4. The safety device based on the heating effect of electric current is called a ……….

Answer:

  1. positive battery
  2. battery
  3. becomes red hot
  4. fuse

Question 9.
Mark ‘T’ if the statement is true and ‘F’ if it is false:

  1. 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)
  2. When the electric current, through the fuse exceeds a certain limit, the fuse wire melts and breaks. (T/F)
  3. An electromagnet does not attract a piece of iron. (T/F)
  4. An electric bell has an electromagnet. (T/F)

Answer:

  1. False
  2. True
  3. False
  4. True

Question 10.
Do you think an electromagnet can be used for separating plastic bags from a garbage heap? Explain.
Answer:
No. This is because electromagnets can only attract magnetic materials. The plastic bag is a non-magnetic material and will not be attracted by an electromagnet.

Question 11.
An electrician is carrying out some repairs in your house. He wants to replace a fuse with a piece of wire. Would you agree? Give reasons for your response.
Answer:
No, we would not agree to replace the fuse with a wire. The electric wire will not melt even if a high current flows through it. So, it will not prevent the damage done by high current.

Question 12.
Zubeda made an electric circuit using a cell holder shown in the following Fig., 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.
NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.12
Answer:
One of the reasons may be that the rubber band used in the cell holder may not be tight enough to keep the two cells in contact with each other. If the cells are not in proper contact with each other, then the circuit will not be complete, and the current will not flow through the circuit. Hence, the bulb will not glow. The other reason may be that the two cells are not connected properly. The negative terminal of one cell must be connected to the positive terminal of the other cell.

Question 13.
In the circuit shown in the following Fig.:
NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects Q.13

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

Answer:

  1. No bulb will glow
  2. All bulbs will glow at a time.

We hope the NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects help you. If you have any query regarding NCERT Solutions for Class 7 Science Chapter 14 Electric Current and its Effects, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 6
Chapter Name The Triangle and its Properties
Exercise Ex 6.1
Number of Questions Solved 3
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1

Question 1.
In ∆PQR, D is the mid-point of \(\overline { QR } \).
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1
Solution:
\(\overline { PM } \) is the altitude.
PD is the median.
No! QM ≠ MR.

Question 2.
Draw rough sketches for the following:

(a) In ∆ ABC, BE is a median.
(b) In ∆ PQR, PQ and PR are altitudes of the triangle.
(c) In ∆ XYZ, YL is an altitude in the exterior of the triangle.

Solution:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1 2
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1 3

Question 3.
Verify by drawing a diagram if the median and altitude of an isosceles triangle can be the same.
Solution:
AD is the median.
AL is the altitude.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1 4

Draw a line segment BC. By paper folding, locate the perpendicular bisector of BC. The folded crease meets BC at D, its mid-point.
Take any point A on this perpendicular bisector. Join AB and AC. The triangle thus obtained is an isosceles ∆ABC in which AB = AC.
Since D is the mid-point of BC, so AD is its median. Also, AD is the perpendicular bisector of BC. So, AD is the altitude of ∆ABC.
Thus, it is verified that the median and altitude of an isosceles triangle are the same.

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and it’s Properties Ex 6.1 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 12
Chapter Name Algebraic Expressions
Exercise Ex 12.1
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants, and arithmetic operations.
(i) Subtraction of z from y.
Solution:
y – z

(ii) One-half of the sum of numbers x and y.
Solution:
\(\frac{1}{2}\)  (x -y)

(iii) The number z multiplied by itself.
Solution:
z × z i.e., z2

(iv) One-fourth of the product of numbers p and q.
Solution:
\(\frac{1}{4}\) pq

(v) Numbers x and y both squared and added.
Solution:
x2 + y2

(vi) Number 5 added to three times the product of numbers m and n.
Solution:
3mn + 5

(vii) Product of numbers y and z subtracted from 10.
Solution:
10 – yz

(viii) Sum of numbers a and 6 subtracted from their product.
Solution:
ab – (a + b)

Question 2.
(i) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams :

(a) x – 3
(b) 1 + x + x2
(c) y – y3
(d) 5xy2 + 7x2y
(e) -ab + 2b2 -3a2

(ii) Identify terms and factors in the expressions given below :

(a) – 4x + 5
(b) – 4x + 5y
(c) 5y + 3y2
(d) xy + 2x2y2
(e) pq + q
(f) 1.2 ab -2.4 b + 3.6 a
(g) \(\frac { 3 }{ 4 } \) x + \(\frac { 1 }{ 4 } \)
(h) 0.1 p2 + 0.2 q2

Solution:
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 1
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 2
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 3
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 4
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 5

Question 3.
Identify the numerical coefficients of terms (other than constants) in the following expressions :

  1. 5 – 3t2
  2. 1 + t + t2 + t3
  3. x + 2xy + 3y
  4. 100m + 1000n
  5. -p2q2 + 7pq
  6. 1.2 a + 0.8 b
  7. 3.14 r2
  8. 2 (l + b)
  9. 0.1 y + 0.01 y2.

Solution:
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 6
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 7

Question 4.
(a) Identify terms which contain x and give the coefficient of x.

  1. y2x + y
  2. 13y2 – 8yx
  3. x + y + 2
  4. 5 + z + zx
  5. 1 + x + xy
  6. 12xy2 + 25
  7. 7x + xy2.

(b) Identify terms which contain y2 and give the coefficient of y2.

  1. 8 – xy2
  2. 5y2 + 7x
  3. 2x2y – 15xy2 + 7y2

Solution:
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 8
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 9

Question 5.
Classify into monomials, binomials and trinomials.

  1. 4y – 7z
  2. y2
  3. x + y – xy
  4. 100
  5. ab – a – b
  6. 5 – 3t
  7. 4p2q – 4pq2
  8. 7mn
  9. z2 – 3z + 8 a2 + b2
  10. z2 + z
  11. 1 + x+ x2

Solution:
We know that an algebraic expression containing only one term is called a monomial. So, the monomials are : (ii), (iv), and (viii).

We know that an algebraic expression containing two terms is called a binomial. So, the binomials are : (i), (vi), (vii), (x) and (xi).

We know that an algebraic expression containing three terms is called a trinomial. So, the trinomial are : (iii), (v), (ix) and (xii).

Question 6.
State whether a given pair of terms is of like or unlike terms :
(i) 1, 100
Solution:
Like

(ii) -7x, \(\frac{5}{2}\)x
Solution:
Like

(iii) – 29x, – 29y
Solution:
Unlike

(iv)14xy, 42yx
Solution:
Like

(v) 4m2p, 4mp2
Answer:
Unlike

(vi) 12xz, 12x2z2
Solution:
Unlike

(i) 4y – 7z.
This expression is a binomial because it contains two terms: 4y and – Iz.
(ii) y2.
This expression is a monomial because it contains only one term: y2
(iii) x + y – xy.
This expression is a trinomial because it contains three terms: x, y, and – xy.
(iv) 100.
This expression is a monomial because it contains only one term: 100
(v) ab – a – b.
This expression is a trinomial because it contains three terms: ab, -a, and -b
(vi) 5 – 3t.
This expression is a binomial because it contains two terms : 5 and – 31.
(vii) 4p2q – 4pq2.
This expression is a binomial because it contains two terms: 4p2q and – 4pq2.
(viii) 7mn.
This expression is a monomial because it contains only one term : 7mn.
(ix) z2 – 3z + 8.
This expression is a trinomial because it contains three terms : z2, – 3z and 8.
(x) a2 + b2.
This expression is a binomial because it contains two terms: a2 and b2.
(xi) z2 + z.
This expression is a binomial because it contains two terms : z2 and z.
(xii) 1 + x + x2.
This expression is a trinomial because it contains three terms: 1, x, and x2.

Question 6.
State whether a given pair of terms is of like or unlike terms :
(i) 1, 100
Solution:
Like

(ii) -7x, \(\frac{5}{2}\)x
Solution:
Like

(iii) – 29x, – 29y
Solution:
Unlike

(iv)14xy, 42yx
Solution:
Like

(v) 4m2p, 4mp2
Answer:
Unlike

(vi) 12xz, 12x2z2
Solution:
Unlike

Question 7.
Identify like terms in the following :
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 10
Solution:
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 11

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RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B

RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B.

Other Exercises

Question 1.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6cm
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 1
(ii) At B, draw an arc of the radius 5cm.
(iii) At C, draw another arc of the radius 5.4cm which intersects the first arc at A
(iv) Join AB and AC
(v) With centre B and C and radius more than half of BC, draw arcs intersecting each other at L and M.
(vi) Join LM which intersects BC at Q and produce it to P.
Then PQ is perpendicular bisector of side BC.

Question 2.
Solution:
Steps of construction :
(i) Draw a line segment QR = 6cm
(ii) With centre Q and radius 4.4 cm draw an arc.
(iii) With centre R and radius 5.3 cm, draw another arc intersecting the first arc at P.
(iv) Join PQ and PR.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 2
(v) With centre P and a suitable radius, draw an arc meeting PR at E and PQ at F.
(vi) With centres E and F, with same radius, draw two arcs intersecting each other at G.
(vii) Join PG and produce it to meet QR at S. Then PS is the bisector of ∠P.

Question 3.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.2 cm.
(ii) With centres B and C radius 6.2 cm, draw arcs intersecting each other at A.
(iii) Join AB and AC.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 3
∆ABC is the required equilateral triangle.
On measuring, each angle is equal to 60°.

Question 4.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5.3 cm.
(ii) With centre B and C, and radius 4.8 cm, draw arcs intersecting each other at A
(iii) Join AB and AC, Then ∆ABC is the required triangle.
(iv) Now, with centre A and a suitable radius draw an arc intersecting BC at L and M.
(v) Then with centre L and M, draw two arcs intersecting eachother at E.
(vi) Join AE intersecting BC at D. Then AD is perpendicular to BC.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 4
On measuring ∠B and ∠C, each is equal to 55°.

Question 5.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8cm.
(ii) At A draw a ray AX making an angle of 60°.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 5
(iii) Cut off AC = 5 cm from AX
(iv) Join CB.
Then ∆ABC is the required triangle.

Question 6.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.3 cm.
(ii) At C, draw a ray CY making an angle equal to 45°.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 6
(iii) With centre C, and radius 6cm, draw an arc intersecting CY at A.
(iv) Join AB.
Then ∆ABC is the required triangle.

Question 7.
Solution:
(i) Draw a line segment AB = 5.2cm.
(ii) At A, draw a ray AX making an angle equal to 120°.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 7
(iii) From AX cut off AC = 5.2cm.
(iv) Join BC.
Then ∆ABC is the required triangle.
(v) With centre A and some suitable radius draw an arc intersecting BC at L and M.
(vi) With centres L and M, draw two arcs intersecting each other at E.
(vii) Join AE intersecting BC at D Then AD is the perpendicular to BC.

Question 8.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.2 cm.
(ii) At B draw a ray BX making an angle of 60°.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 8
(iii) At C draw another ray CY making an angle of 45° which intersect the ray BX at A. .
Then ∆ABC is the required triangle.

Question 9.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5.8 cm.
(ii) At B draw a ray BX making an angle of 30°
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 9
(iii) At C draw another ray CY making an angle of 30° intersecting the BX at A
Then ∆ABC is the required triangle.
On measuring AB and AC, AB = 3.5cm and AC = 3.5cm.
AB = AC
∆ABC is an isosceles triangle.

Question 10.
Solution:
Steps of construction :
In ∆ABC, ∠A = 45° and ∠C = 75°
But ∠A + ∠B + ∠C = 180°
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 10
⇒ 45° + ZB + 75° = 180°
⇒ ∠B = 180° – 45° – 75°
⇒ ∠B = 180° – 120° = 60°
(i) Draw a line segment AB = 7cm.
(ii) At A, draw a ray AX making an angle of 45°.
(iii) At B, draw another ray BY making an angle of 60° which intersects AX at C
Then ∆ABC is the required triangle.

Question 11.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.8 cm
(ii) At C, draw a ray CX making an angle of 90°.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 11
(iii) With centre B, an radius 6.3cm draw an arc intersecting CX at A.
(iv) Join AB.
Then ∆ABC is the required triangle.

Question 12.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.5cm
(ii) At B, draw a ray BX making an angle of 90°.
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 12
(iii) With centre C and radius 6cm draw an arc intersecting BX at A
(iv) Join AC.
Then ∆ABC is the required triangle.

Question 13.
Solution:
One acute angle = 30°, then
second acute angle will be = 90° – 30° = 60° (Sum of acute angles = 90°)
Steps of construction :
(i) Draw a line segment BC = 6cm.
(ii) At B draw a ray BX making an angle of 30°.
(iii) At C draw another ray CY making an angle of 60° which intersects BX at A
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B 13
Then ∆ABC is the required triangle.

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NCERT Solutions for Class 7 English An Alien Hand Chapter 8 The Bear Story

NCERT Solutions for Class 7 English An Alien Hand Chapter 8 The Bear Story are part of NCERT Solutions for Class 7 English. Here we have given NCERT Solutions for Class 7 English An Alien Hand Chapter 8 The Bear Story.

Board CBSE
Textbook NCERT
Class Class 7
Subject English An Alien Hand
Chapter Chapter 8
Chapter Name The Bear Story
Number of Questions Solved 8
Category NCERT Solutions

NCERT Solutions for Class 7 English An Alien Hand Chapter 8 The Bear Story

EXERCISES
(Page 57)

Answer the following questions :
Question 1.
Where did the lady find the bear cub ? How did she bring it up ?
Answer:
The lady found the bear cub in the dense forest near her house. She brought it up on the bottle with the help of the cook.

Question 2.
The bear grew up but “he was a most amiable bear”. Give three examples to prove this. (Imp.)
Answer:
The bear was a most amiable bear. He did not dream of harming anybody, man or beast. He looked amicably at the cattle grazing in the field but caused them no harm. Sometimes he went into the stable with his mistress. The three horses there did not mind it in the least. The children used to ride his back. More than once they had been found asleep between his two paws.

Question 3.
What did the bear eat ? There were two things he was not allowed to do. What were they ?
Answer:
The bear ate the same food as the dogs and often out of the same plate. He ate bread, porridge, potatoes, cabbages and turnips. Like all bears he was a vegetarian and he liked fruit the most. He was not allowed to climb the apple tree and eat them. He was also not allowed to touch the beehive.

Question 4.
When was the bear tied up with a chain ? Why ?
Answer:
Generally the bear was tied up with a chain only at night. But on Sundays the lady visited her sister who lived on the other side of the mountain lake. It was not safe to go with the bear through the forest. So on Sundays the bear was chained all day.

Question 5.
What happened one Sunday when the lady was going to her sister’s house ? What did the lady do ? What was the bear’s reaction ? (Imp.)
Answer:
One Sunday the lady was going to her sister’s house. She heard the cracking of tree branches in the forest. She looked back and was very angry to see the bear following her. She told him in her hardest tone to go back. The bear did not seem to listen. Then the lady saw that the bear had even lost his new collar. She hit the bear with her parasol so hard that it broke into two. The bear stopped and opened its mouth several times as if wanting to say something. Then the bear turned round and went back. However, it stopped now and then to look back at the lady. Ultimately the lady lost sight of him.

Question 6.
Why was the bear looking sorry for himself in the evening ? Why did the cook get angry with her mistress ? (V. Imp.)
Answer:
The bear was looking sorry for himself in the evening. The poor creature had been on chain all day. He did not like to be chained. He was waiting for the lady to come and release him. However, when the lady came, she began to scold the bear. She also threatened him with punishment. The cook heard it and rushed out from the kitchen. She got angry with her mistress because she loved the bear much. She knew that the bear had been as gentle as an angel. So she asked her mistress to bless him instead of scolding him.

EXERCISES
[Page 57)

Discuss the following topics in groups.
Question 1.
Most people keep dogs and cats as pets. Can you think of some unusual pets that people keep ?
Answer:
Not very long ago, many Indians kept many wild animals as pets. Among these were included not only deers and bears but even wolves and tigers. There are many snake-charmers in our country who keep snakes as pets. Some foreigners and even Indians in the South have crocodiles as their pets.

Question 2.
The second bear did not attack the lady because he was afraid of her. Do you agree ? (Imp.)
Answer:
It seems that it was the lady’s confidence which frightened the bear. She thought that it was her pet bear. She dealt with him as one would deal with one’s dog or cat. Her fearlessness obviously frightened the bear.

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B

RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data (Mean, Median and Mode) Ex 21B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21B.

Other Exercises

Question 1.
Solution:
(i) Arranging in ascending order.
2, 2, 3, 5, 7, 9, 9, 10, 11
Here number of terms = 9 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 9 + 1 }{ 2 }\) th term
= 5th term = 7
Hence median = 7
(ii) Arranging in ascending order,
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, number of terms (n) = 9 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 9 + 1 }{ 2 }\) th term
= 5th term = 16
(iii) Arranging in ascending order,
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25 Here number of terms (n) = 11 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 11 + 1 }{ 2 }\) th term
= 6th term = 16

Question 2.
Solution:
(i) Arranging in ascending order,
9, 10, 17, 19, 21, 22, 32, 35
Here, number of terms = 8 which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 1
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 2

Question 3.
Solution:
First 15 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.
Here, number of terms (n) = 15 which is odd
Median = \(\frac { n + 1 }{ 2 }\) = \(\frac { 15 + 1 }{ 2 }\) th term
= 8th term = 15

Question 4.
Solution:
First 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Here, number of terms = 10 which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 3

Question 5.
Solution:
First 50 whole numbers
0, 1, 2, 3, 4, …, 49
Here, number of terms = 50, which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 4

Question 6.
Solution:
Arranging in ascending order,
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40 .
Here, number of terms = 15 which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 15 + 1 }{ 2 }\) = \(\frac { 16 }{ 2 }\) th
= 8th term = 23

Question 7.
Solution:
Arranging is ascending order,
31, 34, 36, 37, 40, 43, 46, 50, 52, 53
Here, number of terms = 10, which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 5

Question 8.
Solution:
Preparing the cumulative frequency table
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 6
Here, number of terms (N) = 41, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term
= \(\frac { 41 + 1 }{ 2 }\) = \(\frac { 42 }{ 2 }\) th = 21 th term = 50kg (value of 20 to 28 = 50)
Hence median = 50kg

Question 9.
Solution:
Arranging in order and preparing the cumulative frequency table.
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 7
Here, number of terms (N) = 37 which is odd.
Median = \(\frac { n + 1 }{ 2 }\) th term
= \(\frac { 37 + 1 }{ 2 }\) = \(\frac { 38 }{ 2 }\) th term
= 19th term = 22 (Value of 18 to 21 = 22)
Hence median = 22

Question 10.
Solution:
Arranging in order and then preparing its cumulative frequency table :
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 8
Here, number of terms (N) = 50, which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 9
Hence median = 154.5 cm

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NCERT Solutions for Class 7 Science Chapter 7 Weather, Climate and Adaptations of Animals to Climate

NCERT Solutions for Class 7 Science Chapter 7 Weather, Climate and Adaptations of Animals to Climate are part of NCERT Solutions for Class 7 Science. Here we have given NCERT Solutions for Class 7 Science Chapter 7 Weather, Climate and Adaptations of Animals to Climate.

Board CBSE
Textbook NCERT
Class Class 7
Subject Science
Chapter Chapter 7
Chapter Name Weather, Climate and Adaptations of Animals to Climate
Number of Questions Solved 12
Category NCERT Solutions

NCERT Solutions for Class 7 Science Chapter 7 Weather, Climate and Adaptations of Animals to Climate

Question 1.
Name the elements that determine the weather of a place.
Answer:
Temperature, rainfall, humidity.

Question 2.
When are the maximum and minimum temperatures likely to occur during the day?
Answer:
The maximum temperature of the day generally occurs in the afternoon while the minimum temperature occurs generally in the early morning.

Question 3.
Fill in the blanks:
(i) The average weather taken over a long time is called …….
(ii) A place receives very little rainfall and the temperature is high throughout the year, the climate of that place will be …… and ………
(iii) The two regions of the earth with extreme climatic conditions are ……… and ……..
Answer:
(i) climate
(ii) dry, very hot
(iii) tropical, polar

Question 4.
Indicate the type of climate of the following areas:
(а) Jammu and Kashmir : ……….
(b) Kerala : ………
(c) Rajasthan : ………..
(d) Northeast India : ………
Answer:
(a) Moderately hot and moderately wet (very cold in winter).
(b) Very hot and wet.
(c) Hot and dry.
(d) Wet.

Question 5.
Which of the two changes frequently, weather or climate?
Answer:
It is the weather that changes frequently because it is such a complex phenomenon that it can vary over a very short period of time.

Question 6.
Following are some of the characteristics of animals:
(i) Diets heavy on fruits
(ii) White fur
(iii) Need to migrate
(iv) Loud voice
(v) Sticky pads on feet
(vi) Layer of fat under skin
(vii) Wide and large paws
(viii) Bright colours
(ix) Strong tails
(x) Long and large beak
Each characteristic indicates whether it is an adaptation for tropical rainforests or polar regions. Do you think that some of these characteristics can be adapted for both regions?
Answer:

Characteristics Adaptation for Tropical Rainforest/Polar Regions
(i) Diets heavy on fruits Tropical rainforest
(ii) White fur Polar regions
(iii) Need to migrate Polar regions
(iv) Loud voice Tropical rainforest
(v) Sticky pads on feet Tropical rainforest
(vi) Layer of fat under the skin Polar regions
(vii) Wide and large paws Polar regions
(viii) Bright colours Tropical rainforest
(ix) Strong tails Tropical rainforest
(x) Long and large beak Tropical rainforest

 Question 7.
The tropical rainforest has a large population of animals. Explain why it is so.
Answer:
Because of continuous warmth and rain, this region supports wide variety of plants and animals.

Question 8.
Explain, with examples, why we find animals of certain kinds living in particular climatic conditions.
Answer:
Animals are adapted to survive in the conditions in which they live. Features and habits that help animals to adapt to their surroundings are a result of the process of evolution. Since to survive in a particular type of climate all the animals must have certain adaptive features, therefore we find animals of a certain kind living in a particular climatic conditions. For example, animals in the polar region are adapted to the extremely cold climate by having some special characteristics as in polar bear such as white fur, strong sense of smell, a layer of fat under the skin, wide and large paws for swimming and walking etc.

Question 9.
How does elephant living in the tropical rainforest adapt themselves?
Answer:
The elephant has adapted to the conditions of rainforests in many remarkable ways. It uses the trunk as a nose because of which it has a strong sense of smell. The trunk is also used for picking up food. Moreover, its tusks are modified teeth. These can tear the bark of trees that elephant loves to eat. So, the elephant is able to handle the competition for food rather well. The large ears of the elephant help it to hear even very soft sounds. They also help the elephant to keep cool in the hot and humid climate of the rainforest.

Choose the correct option which answers the following question:
Question 10.
A carnivore with stripes on its body moves very fast while catching its prey. It is likely to be found in
(i) polar regions
(ii) deserts
(iii) oceans
(iv) tropical rainforests
Answer:
(iv) tropical rainforests

Question 11.
Which features adapt polar bears to live in an extremely cold climate?
(i) A white fur, fat below the skin, keen sense of smell.
(ii) Thin skin, large eyes, white fur.
(iii) A long tail, strong claws, white large paws.
(iv) White body, paws for swimming, gills for respiration.
Answer:
(i) A white fur, fat below the skin, keen sense of smell.

Question 12.
Which option best describes a tropical region?
(i) Hot and humid
(ii) Moderate temperature, heavy rainfall
(iii) Cold and humid
(iv) Hot and dry
Answer:
(i) Hot and humid

We hope the NCERT Solutions for Class 7 Science Chapter 7 Weather, Climate and Adaptations of Animals to Climate help you. If you have any query regarding NCERT Solutions for Class 7 Science Chapter 7 Weather, Climate and Adaptations of Animals to Climate, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Science Chapter 6 Physical and Chemical Changes

NCERT Solutions for Class 7 Science Chapter 6 Physical and Chemical Changes are part of NCERT Solutions for Class 7 Science. Here we have given NCERT Solutions for Class 7 Science Chapter 6 Physical and Chemical Changes.

Board CBSE
Textbook NCERT
Class Class 7
Subject Science
Chapter Chapter 6
Chapter Name Physical and Chemical Changes
Number of Questions Solved 12
Category NCERT Solutions

NCERT Solutions for Class 7 Science Chapter 6 Physical and Chemical Changes

Question 1.
Classify the changes involved in the following processes as physical or chemical changes:

  1. Photosynthesis
  2. Dissolving sugar in water
  3. Burning of coal
  4. Melting of wax
  5. Beating aluminium to make aluminium foil
  6. Digestion of food

Answer:

  1. Chemical change
  2. Physical change
  3. Chemical change
  4. Physical change
  5. Physical change
  6. Chemical change

Question 2.
State whether the following statements are true or false. In case a statement is false, write the correct statement in your notebook:
(a) Cutting a log of wood into pieces is a chemical change. (True/False)
(b) Formation of manure from leaves is a physical change. (True/False)
(c) Iron pipes coaled with zinc do not get rusted easily. (True/False)
(d) Iron and rust are the same substances. (True/False)
(e) Condensation of steam is not a chemical change. (True/False)
Answer:
(a) False
Correct statement: Cutting a log of wood into pieces is a physical change, ft, False
(b) True
Correct statement: The formation of manure from leaves is a chemical change.
(c) True
(d) False
Correct statement: Iron and rust are two different substances.
(e) True

Question 3.
Fill in the blanks in the following statements:

  1. When carbon dioxide is passed through lime water, it turns milky due to the formation of …….
  2. The chemical name of baking soda is ………
  3. Two methods by which rusting of iron can be prevented are ……. and ……….
  4. Changes in which only ………. properties of a substance change are called physical changes.
  5. Changes in which new substances are formed are called ……. changes.

Answer:

  1. calcium carbonate
  2. sodium hydrogen carbonate
  3. painting or greasing, galvanization
  4. physical
  5. chemical

Question 4.
When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. What type of change is it? Explain.
Answer:
It is a chemical change. Here, a new substance carbon dioxide gas is formed.

Question 5.
When a candle burns, both physical and chemical changes take place. Identify these changes. Give another example of a familiar process in which both the chemical and physical changes take place.
Answer:
When candle bums, both physical and chemical changes occur as given below:
(i) Physical change: Melting of wax, vapourisation of melted wax.
(ii) Chemical change: Burning of the vapours of wax to give carbon dioxide, heat, and light.
The burning of LPG in our kitchen is another such example in which a physical change occurs when LPG comes out of the cylinder and is converted to hum a liquid state to a gaseous state and a chemical change occurs when this gas burns in air.

Question 6.
How would you show that the setting of curd is a chemical change?
Answer:
When some sour substance is added to milk or kept overnight, it turns into curd. The curd in no way can be converted into milk i.e. irreversible change. Curd is a different substance than milk. So, the formation of curd is a chemical change.

Question 7.
Explain why burning of wood and cutting it into small pieces are considered as two different types of changes.
Answer:
Burning of wood is a chemical change because in burning new substances are formed as given below:
Wood + Oxygen → Coal + Carbon dioxide + Heat + Light.
While cutting it into small pieces is physical change because by cutting we can only reduce the size of the log of wood and no change in its chemical properties occurs and no new substance is formed.

Question 8.
Describe how crystals of copper sulphate are prepared.
Answer:
A cup of water is taken in a beaker and a few drops of dilute sulphuric acid are added into it. The water is heated. When it starts boiling, copper sulphate powder is added slowly while stirring continuously. Copper sulphate powder is added continuously till no more powder can be dissolved. The solution is filtered and allowed to cool down. Crystals of copper sulphate slowly form at the bottom of the beaker.
NCERT Solutions for Class 7 Science Chapter 6 Physical and Chemical Changes Q 8.
Fig. Crystals of copper sulphate

Question 9.
Explain how the painting of an iron gate prevents it from rusting.
Answer:
We know that the process of rusting requires exposure to both oxygen and water (or moisture). Painting prevents the surface of the iron gate from coming in contact with oxygen and moisture thus prevents it from rusting.

Question 10.
Explain why rusting of iron objects is faster in coastal areas than in deserts.
Answer:
In coastal areas, there is more moisture in the air due to the presence of the sea. But in the desert, there is a scarcity of water and hence the air is almost dry there. Both air and moisture are necessary conditions for rusting. So, rusting is faster in coastal areas than in deserts.

Question 11.
The gas we use in the kitchen is called liquified petroleum gas (LPG). In the cylinder, it exists as a liquid. When it comes out from the cylinder it becomes a gas (Change – A) then it burns (Change – B). The following statements pertain to these changes. Choose the correct one.
(i) Process – A is a chemical change.
(ii) Process – B is a chemical change.
(iii) Both processes A and B are chemical changes.
(iv) None of these processes is a chemical change.
Answer:
(ii) Process – B is a chemical change.

Question 12.
Anaerobic bacteria digest animal waste and produce biogas (Change – A). The biogas is then burnt as fuel (Change – B). The following statements pertain to these changes. Choose the correct one.
(i) Process – A is a chemical change.
(ii) Process – B is a chemical change.
(iii) Both processes A and B are chemical changes.
(iv) None of these processes is a chemical change.
Answer:
(iii) Both processes A and B are chemical changes.

We hope the NCERT Solutions for Class 7 Science Chapter 6 Physical and Chemical Changes help you. If you have any query regarding NCERT Solutions for Class 7 Science Chapter 6 Physical and Chemical Changes, drop a comment below and we will get back to you at the earliest.