Term 1 · Class 7 Science · Chapter 1

Samacheer Class 7 Science - Measurement

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Chapter-wise textbook exercise answers for Measurement with validation-aware solutions.

Answers marked verified were checked during generation against the chapter context and source question text.
Sections in this chapter
I. Choose the appropriate answer: 5II. Fill in the blanks: 1III. State whether the following statements are true or false. 5IV. Match the items in column – I to the items in column – II : 2V. Arrange the following in correct sequence : 2VI. Use the analogy to fill in the blank: 2VII. Assertion and reason type questions: 3VIII. Give very short answer: 7IX. Give Short Answer. 5X. Answer In detail. 2XI. Questions based on Higher er Thinking skills: 1XII. Numerical problems: 5XIII. Cross word puzzle: 1Activity -1 1Activity – 2 1Activity – 3 1
Your Progress - Chapter 10% complete
1I. Choose the appropriate answer:5 questions
Q.1Which of the following is a derived unit?v
  1. A. mass
  2. B. time
  3. C. area
  4. D. length
Solution

(c) area

Answer:

(c) area

Q.2Which of the following is correct?v
  1. A. 1L=lcc
  2. B. 1L= l0cc
  3. C. 1L= l00cc
  4. D. 1L= l000cc
Solution

(d) 1L = 1000cc

Answer:

(d) 1L = 1000cc

Q.3SI unit of density isv
  1. A. kg/m 2
  2. B. kg/m 3
  3. C. kg/m
  4. D. g/m 3
Solution

(b) kg/m 3

Answer:

(b) kg/m 3

Q.4Two spheres have equal mass and volume in the ratio 2:1. The ratio of their density isv
  1. A. 1:2
  2. B. 2:1
  3. C. 4:1
  4. D. 1:4
Solution

(b) 2:1

Answer:

(b) 2:1

Q.5Light year is the unit ofv
  1. A. Distance
  2. B. time
  3. C. density
  4. D. both length and time
Solution

(a) Distance

Answer:

(a) Distance

2II. Fill in the blanks:1 questions
Q.G1Volume of irregularly shaped objects are measured using the law of ___________ One cubic metre is equal to ___________ cubic centimetre. Density of mercury is ___________ One astronomical unit is equal to ___________ The area of a leaf can be measured using a ___________v
Solution

Archimedes
10,00,000 or 106 6
13,600 kg/m 3
1.496×10 11 m
graph sheet

Answer:

Archimedes
10,00,000 or 106 6
13,600 kg/m 3
1.496×10 11 m
graph sheet

3III. State whether the following statements are true or false.5 questions
Q.1The region covered by the boundary of the plane figure is called its volume.v
Solution

(False) Correct statement: The region covered by the boundary of plane figure is called its area.

Answer:

(False) Correct statement: The region covered by the boundary of plane figure is called its area.

Q.2Volume of liquids can be found using measuring containers.v
Solution

True

Answer:

True

Q.3Water is denser than kerosene.v
Solution

True

Answer:

True

Q.4A ball of iron floats in mercury.v
Solution

True

Answer:

True

Q.5A substance which contains less number of molecules per unit volume is said to be denser.v
Solution

False. Correct statement: A substance which contains more number of molecules per unit volume is said to be denser.

Answer:

False. Correct statement: A substance which contains more number of molecules per unit volume is said to be denser.

4IV. Match the items in column – I to the items in column – II :2 questions
Q.1Column -1 Column – II i. Areav
  1. (a) light year ii. Distance
  2. (b) m 3 iii. Density
  3. (c) m 2 iv. Volume
  4. (d) kg V. Mass
  5. (e) kg/ m 3
Solution

i
ii
iii
iv
v

Answer:

i
ii
iii
iv
v

Q.2Column -1 Column – II i. Areav
  1. (a) g / cm 3 ii. Length
  2. (b) measuring jar iii. Density
  3. (c) amount of a substance iv. Volume
  4. (d) rope V. Mass
  5. (e) plane figures
Solution

i
ii
iii
iv
v

Answer:

i
ii
iii
iv
v

5V. Arrange the following in correct sequence :2 questions
Q.11L, 100 cc, 10 L, 10 ccv
Solution

10 cc, 100 cc, 1L, 10L

Answer:

10 cc, 100 cc, 1L, 10L

Q.2Copper, Aluminium, Gold, Ironv
Solution

Aluminium, Iron, Copper, Gold

Answer:

Aluminium, Iron, Copper, Gold

6VI. Use the analogy to fill in the blank:2 questions
Q.2Liquid : Litre :: Solid : _________v
Solution

cm 3

Answer:

cm 3

Q.3Water: Kerosene :: ______ : Aluminiumv
Solution

Iron

Answer:

Iron

7VII. Assertion and reason type questions:3 questions
Q.1Assertion (A) : Volume of a stone is found using a measuring cylinder. Reason (R) : Stone is an irregularly shaped object.v
Solution

(a) If both assertion and reason are true and reason is the correct explanation of assertion

Answer:

(a) If both assertion and reason are true and reason is the correct explanation of assertion

Q.2Assertion (A) : Wood floats in water. Reason (R) : Water is a transparent liquid.v
Solution

(b) If both assertion and reason are true, but reason is not the correct explanation of assertion
Correct explanation: Density of water is more than the density of wood.

Answer:

(b) If both assertion and reason are true, but reason is not the correct explanation of assertion
Correct explanation: Density of water is more than the density of wood.

Q.3Assertion (A) : Iron ball sinks in water. Reason (R) : Water is denser than iron.v
Solution

(b) If both assertion and reason are true, but reason is not the correct explanation of assertion
Correct explanation : Density of iron is more than that of water.

Answer:

(b) If both assertion and reason are true, but reason is not the correct explanation of assertion
Correct explanation : Density of iron is more than that of water.

8VIII. Give very short answer:7 questions
Q.1Name some of the derived quantities.v
Solution

Area, volume, density.

Answer:

Area, volume, density.

Q.2Give the value of one light year.v
Solution

One light year = 9.46 x 10 15 m

Answer:

One light year = 9.46 x 10 15 m

Q.3Write down the formula used to find the volume of a cylinder.v
Solution

Volume of a cylinder = πr 2 h

Answer:

Volume of a cylinder = πr 2 h

Q.4Give the formula to find the density of objects.v
Solution

Samacheer Kalvi Guru 7th Science Term 1 Chapter 1 Measurement

Answer:

Samacheer Kalvi Guru 7th Science Term 1 Chapter 1 Measurement

Q.5Name the liquid in which an iron ball sinks.v
Solution

Iron ball sinks in water. The density of an iron ball is more than that of water so it sinks in water.

Answer:

Iron ball sinks in water. The density of an iron ball is more than that of water so it sinks in water.

Q.6Name the unit used to measure the distance between celestial objects.v
Solution

Astronomical unit and light year are the units used to measure the distance between celestial objects.

Answer:

Astronomical unit and light year are the units used to measure the distance between celestial objects.

Q.7What is the density of gold?v
Solution

Density of gold is 19,300 kg/m 3

Answer:

Density of gold is 19,300 kg/m 3

9IX. Give Short Answer.5 questions
Q.1What are derived quantities?v
Solution

The physical quantities which can be obtained by multiplying, dividing or by mathematically combining the fundamental quantities are known as derived quantities.
(or)
The physical quantities which are expressed is terms of fundamental quantities are called

Answer:

The physical quantities which can be obtained by multiplying, dividing or by mathematically combining the fundamental quantities are known as derived quantities.
(or)
The physical quantities which are expressed is terms of fundamental quantities are called

Q.2Distinguish between the volume of liquid and capacity of a container.v
Solution

S.No
Volume of liquid
Capacity of a container
1.
Volume is the amount of space taken up by a liquid
Capacity is the measure of an objects ability to hold a substance like solid, liquid or gas
2.
It is measured in cubic units.
It is measured in litres, gallons, pounds, etc.
3.
It is calculated by multiplying the length, width and height of an object.
It’s measurement is cc or ml.

Answer:

S.No
Volume of liquid
Capacity of a container
1.
Volume is the amount of space taken up by a liquid
Capacity is the measure of an objects ability to hold a substance like solid, liquid or gas
2.
It is measured in cubic units.
It is measured in litres, gallons, pounds, etc.
3.
It is calculated by multiplying the length, width and height of an object.
It’s measurement is cc or ml.

Q.3Define the density of objects.v
Solution

Density of a substance is defined as the mass of the substance contained in unit volume
Measurement 7th Standard Science Term 1 Chapter 1 Measurement

Answer:

Density of a substance is defined as the mass of the substance contained in unit volume
Measurement 7th Standard Science Term 1 Chapter 1 Measurement

Q.4What is one light year?v
Solution

One light year is the distance travelled by light in vacuum during the period of one year.
1 Light year = 9.46 x 10 15 m.

Answer:

One light year is the distance travelled by light in vacuum during the period of one year.
1 Light year = 9.46 x 10 15 m.

Q.5Define one astronomical unit?v
Solution

One astronomical unit is defined as the average distance between the earth and the sun.
1 AU = 1.496 5 10 6 km = 1.496 × 10 11 m.

Answer:

One astronomical unit is defined as the average distance between the earth and the sun.
1 AU = 1.496 5 10 6 km = 1.496 × 10 11 m.

10X. Answer In detail.2 questions
Q.1Describe the graphical method to find the area of an irregularly shaped plane figure.v
Solution

To find the area of an irregularly shaped plane figure, we have to use graph paper.
Place a piece of paper with an irregular shape on a graph paper and draw its outline.
To find the area enclosed by the outline, count the number of squares inside it (M).
You will find that some squares lie partially inside the outline.
Count a square only if half (p) or more of it (N) lies inside the outline.
Finally count the number of squares, that are less than half. Let it be
For the shape in figure we have the following:
M = 50
N = 7
P = 4
Q = 4
Now, the approximate area of the can be calculated using the following formula.
Measurement Lesson 7th Standard Term 1 Chapter 1 Measurement
Area of the leaf = M+(\(\frac { 3 }{ 4 }\)) N + (\(\frac { 1 }{ 2 }\)) P+\(\frac { 1 }{ 4 }\) Qsq.cm
= 52 + 5.25 = 58.25 sq.mm = 0.5825 sq.cm

Answer:

To find the area of an irregularly shaped plane figure, we have to use graph paper.
Place a piece of paper with an irregular shape on a graph paper and draw its outline.
To find the area enclosed by the outline, count the number of squares inside it (M).
You will find that some squares lie partially inside the outline.
Count a square only if half (p) or more of it (N) lies inside the outline.
Finally count the number of squares, that are less than half. Let it be
For the shape in figure we have the following:
M = 50
N = 7
P = 4
Q = 4
Now, the approximate area of the can be calculated using the following formula.
Measurement Lesson 7th Standard Term 1 Chapter 1 Measurement
Area of the leaf = M+(\(\frac { 3 }{ 4 }\)) N + (\(\frac { 1 }{ 2 }\)) P+\(\frac { 1 }{ 4 }\) Qsq.cm
= 52 + 5.25 = 58.25 sq.mm = 0.5825 sq.cm

Q.2How will you determine the density of a stone using a measuring jar?v
Solution

Determination of density of a stone using a measuring cylinder.
7th Science Measurement Term 1 Chapter 1 Measurement
In order to determine the density of a solid, we must know the mass and volume of the stone.
The mass of the stone is determined by a physical balance very accurately. Let it be ‘m’ grams.
In order to find the volume, take a measuring cylinder and pour in it some water.
Record the volume of water from the graduations marked on measuring cylinder. Let it be 40 cm 3 .
Now tie the given stone to a fine thread and lower it gently in the measuring cylinder, such that it is completely immersed in water.
Record the new level of water. Let it be 60 cm 3
∴Volume of the solid = (60-40) cm 3
= 20 cm 3 = V cm 3 (assume)
Knowing the mass and the volume of the stone, the density can be calculate by the formula:
Samacheer Kalvi 7th Science Solutions Term 1 Chapter 1 Measurement

Answer:

Determination of density of a stone using a measuring cylinder.
7th Science Measurement Term 1 Chapter 1 Measurement
In order to determine the density of a solid, we must know the mass and volume of the stone.
The mass of the stone is determined by a physical balance very accurately. Let it be ‘m’ grams.
In order to find the volume, take a measuring cylinder and pour in it some water.
Record the volume of water from the graduations marked on measuring cylinder. Let it be 40 cm 3 .
Now tie the given stone to a fine thread and lower it gently in the measuring cylinder, such that it is completely immersed in water.
Record the new level of water. Let it be 60 cm 3
∴Volume of the solid = (60-40) cm 3
= 20 cm 3 = V cm 3 (assume)
Knowing the mass and the volume of the stone, the density can be calculate by the formula:
Samacheer Kalvi 7th Science Solutions Term 1 Chapter 1 Measurement

11XI. Questions based on Higher er Thinking skills:1 questions
Q.1There are three spheres A, B, C as shown below :v
Solution

7th Standard Science Measurement Term 1 Chapter 1 Measurement
Sphere A and B are made of the same material. Sphere C is made of a different material. Spheres A and C have equal radii. The radius of sphere B is half that of A. Density of A is double that of C.

Answer:

7th Standard Science Measurement Term 1 Chapter 1 Measurement
Sphere A and B are made of the same material. Sphere C is made of a different material. Spheres A and C have equal radii. The radius of sphere B is half that of A. Density of A is double that of C.

12XII. Numerical problems:5 questions
Q.1A circular disc has a radius 10 cm. Find the area of the disc in m2. (Use n = 3.14)v
Solution

Given radius = 10 cm = 0.1m
π= 3.14
Area of a circular disc A = ?
Formula : Area of a circle A = πr 2
= 3.14 × 0.1 × 0.1
A = 0.0314 m 2

Answer:

Given radius = 10 cm = 0.1m
π= 3.14
Area of a circular disc A = ?
Formula : Area of a circle A = πr 2
= 3.14 × 0.1 × 0.1
A = 0.0314 m 2

Q.2The dimension of a school playground is 800 m x 500 m. Find the area of the ground.v
Solution

Given :The dimension of a school
Playground = l x b = 800 m x 500 m
Formula :Area of the ground A = l x b
= 800 x 500 = 4,00,000
A = 4,00,000 m 2

Answer:

Given :The dimension of a school
Playground = l x b = 800 m x 500 m
Formula :Area of the ground A = l x b
= 800 x 500 = 4,00,000
A = 4,00,000 m 2

Q.3Two spheres of same size are made from copper and iron respectively. Find the ratio between their masses. Density of copper 8,900 kg/m and iron 7,800 kg/m 3v
Solution

Given : Density Copper Dc = 8900 kg/m 3
Density of Iron D 1 = 7800 kg/m 3
Volume of Copper sphere = Volume of Iron sphere
To find : Ratio of Masses of Copper (M C ) and Iron (M I )
Mass = Density x Volume
M C = D C x V, M 1 = D 1 x V
M C = 8900 V, M 1 = 7,800 V
M C = M 1
8900 V : 7800 V
= 1.14: 1

Answer:

Given : Density Copper Dc = 8900 kg/m 3
Density of Iron D 1 = 7800 kg/m 3
Volume of Copper sphere = Volume of Iron sphere
To find : Ratio of Masses of Copper (M C ) and Iron (M I )
Mass = Density x Volume
M C = D C x V, M 1 = D 1 x V
M C = 8900 V, M 1 = 7,800 V
M C = M 1
8900 V : 7800 V
= 1.14: 1

Q.4A liquid having a mass of 250 g fills a space of lOOOcc. Find the density of the liquid.v
Solution

Given : Mass of a liquid M = 250 g
Volume V = l000cc
Density of the liquid D = ?
Measurement Lesson Question Answer Samacheer Kalvi 7th Science Solutions Term 1
Density of the liquid = 0.25 g/cc

Answer:

Given : Mass of a liquid M = 250 g
Volume V = l000cc
Density of the liquid D = ?
Measurement Lesson Question Answer Samacheer Kalvi 7th Science Solutions Term 1
Density of the liquid = 0.25 g/cc

Q.5A sphere of radius 1cm is made from silver. If the mass of the sphere is 33 g, find the density of silver (Take π = 3.14)v
Solution

Given : radius of a sphere r = 1cm
Volume of the sphere V = ?
Mass of the sphere M = 33 g
Density of silver D = ?
Samacheer Guru 7th Science Term 1 Chapter 1 Measurement
Density of silver sphere = 7.889 g/cc.

Answer:

Given : radius of a sphere r = 1cm
Volume of the sphere V = ?
Mass of the sphere M = 33 g
Density of silver D = ?
Samacheer Guru 7th Science Term 1 Chapter 1 Measurement
Density of silver sphere = 7.889 g/cc.

13XIII. Cross word puzzle:1 questions
Q.G2Clues – Across 1. SI unit of temperature 2. A derived quantity 3. Mass per unit volume 4. Maximum volume of liquid a container can hold Clues – Downv
  1. A. A derived quantity
  2. B. SI unit of volume
  3. C. A liquid denser than iron
  4. D. A unit of length used to measure very long distances
Solution

7th Standard Measurement Lesson Term 1 Chapter 1 Measurement Samacheer Kalvi
Clues – Across
1. KELVIN
2. VOLUME
3. DENSITY
4. CAPACITY
Clues – Down
a. VELOCITY
b. CUBIC METRE
c. MERCURY
d. LIGHT YEAR

Answer:

7th Standard Measurement Lesson Term 1 Chapter 1 Measurement Samacheer Kalvi
Clues – Across
1. KELVIN
2. VOLUME
3. DENSITY
4. CAPACITY
Clues – Down
a. VELOCITY
b. CUBIC METRE
c. MERCURY
d. LIGHT YEAR

14Activity -11 questions
Q.G3Take a leaf from any one of trees in your neighborhood. Place the leaf on a graph sheet and draw the outline of the leaf with a pencil. Remove the leaf. You can see the outline of the leaf on the graph sheet. Now, count the number of whole squares enclosed within the outline of the leaf. Take it to be M. Then, count the number of squares that are more than half. Take it as N. Next, count the number of squares which are half of a whole square. Note it to be P. Finally, count the number of squares that are less than half. Let it be Q. M = _______;N = _______; P = _______; Q = _______ Now, the approximate area of the leaf can be calculated using the following formula: Approximate area of the leaf = M +(\(\frac { 3 }{ 4 }\)) N+(\(\frac { 1 }{ 2 }\)) P+(\(\frac { 1 }{ 4 }\)) Q square cm Area of the leaf =________. This formula can be used to calculate the area of any irregularly shaped plane figures.v
Solution

M = 50
N = 7
P = 4
Q = 4
Samacheer Kalvi 7th Science Term 1 Chapter 1 Measurement

Answer:

M = 50
N = 7
P = 4
Q = 4
Samacheer Kalvi 7th Science Term 1 Chapter 1 Measurement

15Activity – 21 questions
Q.G4Draw the following regularly shaped figures on a graph sheet and find their area by the graphical method. Also, find their area using appropriate formula. Compare the results obtained in two methods by tabulating them.v
  1. A. A rectangle whose length is 12 cm and breadth is 4 cm.
  2. B. A square whose side is 6 cm.
  3. C. A circle whose radius is 7 cm.
  4. D. A triangle whose base is 6 cm and height is 8 cm.
Solution

Samacheer Kalvi 7th Science Solution Term 1 Chapter 1 Measurement

Answer:

Samacheer Kalvi 7th Science Solution Term 1 Chapter 1 Measurement

16Activity – 31 questions
Q.G5Take a measuring cylinder and pour some water into it (Do not fill the cylinder completely). Note down the volume of water from the readings of the measuring cylinder. Take it as V . Now take a small stone and tie it with a thread. Immerse the stone inside the water by holding the thread. This has to be done such that the stone does not touch the walls of the measuring cylinder. Now, the level of water has raised. Note down the volume of water and take it to be V . The volume of the stone is equal to the raise in the volume of water. V1= _______ V2=_______ Volume of stone = v2 – v1 =_______.v
Solution

V 1 = 30 cc, V 2 = 40 cc; Volume of stone =v 2 – v 1 = 40cc – 30cc = 10cc

Answer:

V 1 = 30 cc, V 2 = 40 cc; Volume of stone =v 2 – v 1 = 40cc – 30cc = 10cc