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.

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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
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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
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
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
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
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
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
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.6The region covered by the boundary of the plane figure is called its volume.v
Answer:

This statement is false. The region covered by the boundary of a plane figure is called its area, not volume. Area is a two-dimensional measurement that represents the space enclosed within the perimeter of a flat shape such as a square, rectangle, triangle, or circle. Volume, on the other hand, is a three-dimensional measurement that represents the amount of space occupied by a solid object. It is important to distinguish between these two concepts as they measure different properties and use different units. Area is measured in square units like square centimeters or square meters, while volume is measured in cubic units like cubic centimeters or cubic meters.

Q.7Volume of liquids can be found using measuring containers.v
Answer:

This statement is true. The volume of liquids can be found using measuring containers such as measuring cylinders, graduated cylinders, beakers, or measuring jugs. These containers have markings or graduations on their sides that indicate the volume of liquid they contain. By pouring the liquid into the measuring container and reading the level at which the liquid reaches, we can accurately determine the volume of the liquid. This is a practical and commonly used method in laboratories and kitchens for measuring liquid volumes.

Q.8Water is denser than kerosene.v
Answer:

This statement is true. Water is denser than kerosene. Density is defined as the mass of a substance per unit volume. Water has a density of approximately 1 gram per cubic centimeter, while kerosene has a lower density of approximately 0.8 grams per cubic centimeter. Because water is denser than kerosene, if the two liquids are mixed together, kerosene will float on top of water and they will form separate layers due to their different densities.

Q.9A ball of iron floats in mercury.v
Answer:

This statement is true. A ball of iron floats in mercury. Mercury is a liquid metal with a very high density of approximately 13.6 grams per cubic centimeter, which is much greater than the density of iron at approximately 7.8 grams per cubic centimeter. Since mercury is denser than iron, an iron ball will float on the surface of mercury rather than sink. This demonstrates the principle that objects float in liquids that are denser than themselves.

Q.10A substance which contains less number of molecules per unit volume is said to be denser.v
Answer:

This statement is false. The correct statement is that a substance which contains more number of molecules per unit volume is said to be denser. Density is a measure of how closely packed the molecules or atoms of a substance are within a given volume. When a substance has more molecules packed into the same unit volume, it means the molecules are closer together, resulting in greater density. Conversely, a substance with fewer molecules per unit volume has lower density because its molecules are more spread out. For example, iron has a higher density than wood because iron atoms are packed more tightly together in the same volume compared to wood molecules. This relationship between molecular packing and density is fundamental to understanding why some objects sink while others float in different liquids.

4IV. Match the items in column – I to the items in column – II :2 questions
Q.11Column -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
Answer:

i
ii
iii
iv
v

Q.12Column -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
Answer:

i
ii
iii
iv
v

5V. Arrange the following in correct sequence :2 questions
Q.131L, 100 cc, 10 L, 10 ccv
Answer:

The correct order from smallest to largest is: 10 cc, 100 cc, 1 L, 10 L. To arrange these volumes in ascending order, we need to convert them to the same unit. Since 1 liter equals 1000 cubic centimeters, we have 10 cc, 100 cc, 1000 cc, and 10000 cc respectively. Therefore, the ascending order is 10 cc, 100 cc, 1 L, 10 L.

Q.14Copper, Aluminium, Gold, Ironv
Answer:

The correct order of metals from least dense to most dense is: Aluminium, Iron, Copper, Gold. Aluminium has the lowest density at approximately 2.7 grams per cubic centimeter, followed by Iron at approximately 7.8 grams per cubic centimeter, then Copper at approximately 8.9 grams per cubic centimeter, and finally Gold which is the densest at approximately 19.3 grams per cubic centimeter. This ordering reflects how tightly the atoms of each metal are packed together.

6VI. Use the analogy to fill in the blank:2 questions
Q.15Liquid : Litre :: Solid : _________v
Answer:

cm 3

Q.16Water: Kerosene :: ______ : Aluminiumv
Answer:

Iron

7VII. Assertion and reason type questions:3 questions
Q.17Assertion (A) : Volume of a stone is found using a measuring cylinder. Reason (R) : Stone is an irregularly shaped object.v
Answer:

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

Q.18Assertion (A) : Wood floats in water. Reason (R) : Water is a transparent liquid.v
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.19Assertion (A) : Iron ball sinks in water. Reason (R) : Water is denser than iron.v
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.20Name some of the derived quantities.v
Answer:

Some of the derived quantities are area, volume, and density. Derived quantities are physical quantities that are obtained by combining fundamental quantities through mathematical operations such as multiplication, division, or other combinations. Area is derived from length multiplied by length, volume is derived from length multiplied by length multiplied by length, and density is derived by dividing mass by volume. Other examples of derived quantities include speed, acceleration, force, and pressure.

Q.21Give the value of one light year.v
Answer:

One light-year is a unit of distance used in astronomy to measure very large distances, such as those between stars and galaxies. It is defined as the distance that light travels in a vacuum in one Julian year (365.25 days). The value of one light-year is approximately 9.46 x 10^15 meters. This immense distance highlights the vastness of space and is crucial for understanding cosmic scales.

Q.22Write down the formula used to find the volume of a cylinder.v
Answer:

The formula used to find the volume of a cylinder is Volume = πr^2h, where 'π' (pi) is a mathematical constant approximately equal to 3.14159, 'r' is the radius of the circular base of the cylinder, and 'h' is the height of the cylinder. This formula calculates the amount of three-dimensional space occupied by the cylinder.

Q.23Give the formula to find the density of objects.v
Answer:

The formula used to find the density of an object is Density = Mass / Volume. Density is defined as the mass of a substance per unit of volume. It is a fundamental property of matter and helps to distinguish between different materials. The standard SI unit for density is kilograms per cubic meter (kg/m³).

Q.24Name the liquid in which an iron ball sinks.v
Answer:

An iron ball sinks in water. The density of an iron ball is greater than the density of water, which is why it sinks. Iron has a density of approximately 7.8 grams per cubic centimeter, while water has a density of 1 gram per cubic centimeter. Since the iron ball is denser than water, the gravitational force acting on it is greater than the buoyant force exerted by the water, causing the iron ball to sink to the bottom of the container.

Q.25Name the unit used to measure the distance between celestial objects.v
Answer:

The units used to measure the distance between celestial objects are the astronomical unit and the light year. An astronomical unit, abbreviated as AU, is defined as the average distance between the Earth and the Sun, which is approximately 150 million kilometers. A light year is the distance that light travels in one year through the vacuum of space, which is approximately 9.46 trillion kilometers. These units are used because the distances between celestial objects are extremely large, and using standard units like kilometers would result in unwieldy numbers that are difficult to work with.

Q.26What is the density of gold?v
Answer:

The density of gold is approximately 19,300 kg/m³. This high density indicates that gold is a very heavy metal for its size, meaning a small volume of gold contains a large amount of mass. This property is one of the reasons gold is valued and has been used for coinage and jewelry throughout history.

9IX. Give Short Answer.5 questions
Q.27What are derived quantities?v
Answer:

Derived quantities are physical quantities that can be obtained by multiplying, dividing, or mathematically combining fundamental quantities. They are expressed in terms of fundamental quantities, which are the basic quantities that cannot be derived from other quantities. Examples of fundamental quantities include length, mass, and time. Derived quantities, on the other hand, are quantities that depend on fundamental quantities and are expressed using combinations of them. For instance, area is a derived quantity obtained by multiplying length by length, volume is obtained by multiplying length by length by length, speed is obtained by dividing distance by time, and density is obtained by dividing mass by volume. Derived quantities are essential in physics and science because they allow us to describe and measure complex physical phenomena in terms of simpler, more fundamental properties.

Q.28Distinguish between the volume of liquid and capacity of a container.v
Answer:

Volume of liquid and capacity of a container are related but distinct concepts. Volume of liquid refers to the amount of space occupied by a liquid and is measured in cubic units such as cubic centimeters or cubic meters. It is calculated based on the dimensions of the liquid itself. Capacity of a container, on the other hand, refers to the maximum amount of substance that a container can hold, whether that substance is a solid, liquid, or gas. Capacity is typically measured in liters, milliliters, gallons, or other volume-based units. While volume is measured in cubic units, capacity is often expressed in liters or milliliters, which are equivalent to cubic centimeters. The key difference is that volume describes the space taken up by the liquid itself, whereas capacity describes the ability or potential of a container to hold a substance. For example, a measuring cylinder might have a capacity of 100 milliliters, and if it contains 50 milliliters of water, the volume of that water is 50 milliliters.

Q.29Define the density of objects.v
Answer:

Density of a substance is defined as the mass of the substance contained in unit volume. It is a derived physical quantity that tells us how much mass is packed into a given volume. Density is calculated by dividing the total mass of an object by its total volume. Objects with higher density are heavier for the same volume, while objects with lower density are lighter for the same volume. For example, iron has a higher density than wood, which is why iron sinks in water while wood floats. The SI unit of density is kilogram per cubic metre (kg/m³).

Q.30What is one light year?v
Answer:

One light year is the distance travelled by light in a vacuum during the period of one year. Since the speed of light in a vacuum is constant (approximately 299,792 kilometers per second), the distance covered in a year is enormous. The value of one light year is 9.46 x 10^15 meters. This unit is crucial for measuring vast interstellar and intergalactic distances.

Q.31Define one astronomical unit?v
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.32Describe the graphical method to find the area of an irregularly shaped plane figure.v
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.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.33How will you determine the density of a stone using a measuring jar?v
Answer:

Determination of density of a stone using a measuring cylinder.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:

11XI. Questions based on Higher er Thinking skills:1 questions
12XII. Numerical problems:5 questions
Q.34A circular disc has a radius 10 cm. Find the area of the disc in m2. (Use n = 3.14)v
Answer:

To find the area of the circular disc, we use the formula for the area of a circle, A = πr^2. Given the radius (r) is 10 cm, we first convert it to meters to maintain consistency with the required unit for the area. Since 1 meter equals 100 centimeters, 10 cm is equal to 0.1 meters. Using the given value of π = 3.14, we can substitute these values into the formula. So, A = 3.14 × (0.1 m) × (0.1 m). This calculation results in A = 3.14 × 0.01 m^2, which gives an area of 0.0314 m^2. Therefore, the area of the circular disc is 0.0314 square meters.

Q.35The dimension of a school playground is 800 m x 500 m. Find the area of the ground.v
Answer:

To find the area of the school playground, we use the formula for the area of a rectangle, which is length (l) multiplied by breadth (b). The dimensions of the school playground are given as 800 m by 500 m. So, the length (l) is 800 m and the breadth (b) is 500 m. Applying the formula, Area (A) = l × b = 800 m × 500 m. Multiplying these values, we get A = 400,000 m^2. Therefore, the total area of the school playground is 400,000 square meters. This large area indicates a significant space for various activities.

Q.36Two 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
Answer:

Given the densities of copper (Dc) as 8,900 kg/m³ and iron (Dᵢ) as 7,800 kg/m³. Since the two spheres are of the same size, their volumes (V) are equal. The formula for mass is Mass = Density × Volume. Therefore, the mass of the copper sphere (M<0xE2><0x82><0x9C>) is M<0xE2><0x82><0x9C> = Dc × V = 8900V, and the mass of the iron sphere (Mᵢ) is Mᵢ = Dᵢ × V = 7800V. The ratio between their masses is M<0xE2><0x82><0x9C> : Mᵢ = 8900V : 7800V. Simplifying this ratio by dividing both sides by V, we get 8900 : 7800, which further simplifies to 89 : 78. This ratio is approximately 1.14:1, meaning the copper sphere is about 1.14 times more massive than the iron sphere of the same size.

Q.37A liquid having a mass of 250 g fills a space of lOOOcc. Find the density of the liquid.v
Answer:

Given : Mass of a liquid M = 250 g
Volume V = l000cc
Density of the liquid D = ?Density of the liquid = 0.25 g/cc

Q.38A 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
Answer:

Given the radius of the silver sphere (r) is 1 cm and its mass (M) is 33 g. We are asked to take π = 3.14. First, we calculate the volume of the sphere using the formula V = (4/3)πr³. Substituting the values, V = (4/3) × 3.14 × (1 cm)³ = (4/3) × 3.14 × 1 cm³ ≈ 4.187 cm³. Now, we find the density of silver using the formula Density (D) = Mass / Volume. So, D = 33 g / 4.187 cm³ ≈ 7.88 g/cc. Therefore, the density of the silver is approximately 7.88 grams per cubic centimeter.

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
Answer:

Clues – Across: 1. The SI unit of temperature is KELVIN, which is one of the seven base units in the International System of Units. 2. VOLUME is a derived quantity, as it is obtained by multiplying three fundamental length measurements. 3. DENSITY is defined as mass per unit volume, indicating how much mass is contained in a given space. 4. CAPACITY refers to the maximum volume of liquid a container can hold, representing its internal volume. Clues – Down: a. VELOCITY is a vector quantity that describes both the speed and direction of an object's motion. b. CUBIC METRE is the SI unit of volume, representing the space occupied by a cube with sides of one meter. c. MERCURY is a liquid metal often used in thermometers due to its uniform expansion with temperature changes. d. LIGHT YEAR is a unit of distance, representing the distance light travels in one year, used for astronomical measurements.

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
Answer:

M = 50
N = 7
P = 4
Q = 4

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.
Answer:

To find the area of regularly shaped figures like squares, rectangles, or triangles using the graphical method, first draw the figure accurately on a graph sheet. Then, count the number of complete squares inside the figure. Next, count the number of incomplete squares that are more than half-filled and divide this number by two. Finally, add the number of complete squares and the number of half-filled squares to get the total area in square units. For comparison, calculate the area using the standard formula for that shape (e.g., length x width for a rectangle, 1/2 x base x height for a triangle). Tabulate both results to compare their accuracy. The graphical method provides an approximate area, while the formula gives the precise area.

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
Answer:

This experiment demonstrates how to measure the volume of an irregularly shaped object, like a stone, using the water displacement method. First, we pour some water into a measuring cylinder and note the initial volume, V1. For instance, if V1 = 30 cc, this is our baseline. Next, we carefully immerse the stone into the water, ensuring it is fully submerged but does not touch the sides or bottom of the cylinder, which could affect the accuracy of the reading. The water level will rise due to the stone displacing an equivalent volume of water. We then note the new, final volume, V2. If V2 = 40 cc, the difference between the final and initial volumes gives us the volume of the stone. Therefore, the volume of the stone = V2 – V1 = 40 cc – 30 cc = 10 cc. This method is effective for finding the volume of solids that do not dissolve in water.