Physical Quantities and Measurement

One litre is equal to:

1. 1 cm³
2. 1 m³
3. 10^-3 cm³
4. 10^-3 m³

Answer

10^-3 m³

Reason — 1 litre = 10^-3 m³

A metallic piece displaces water of volume 15 mL. The volume of piece is:

1. 15 cm³
2. 15 m³
3. 15 x 10³ cm^-3
4. 15 x 10³ cm³

Answer

15 cm³

Reason — Volume of body is equal to volume of liquid displaced = 15 mL.

Since

1 mL = 1 cm³,

So

15 mL = 15 cm³

Hence, the volume of piece is 15 cm³.

A piece of paper of dimensions 1.5 m x 20 cm has area:

1. 30 m²
2. 300 cm²
3. 0.3 m²
4. 3000 m³

Answer

0.3 m²

Reason — 20 cm = 0.2 m

Area of paper = 1.5 m x 0.2 m = 0.3 m²

Hence, the area of the paper is 0.3 m².

The correct relation is:

1. d = M x V
2. M = d x V
3. V = d x M
4. d = M + V

Answer

M = d x V

Reason — Mass = Volume x density

The density of alcohol is 0.8 g cm^-3. In S.I. unit, it will be:

1. 0.8 kg m^-3
2. 0.0008 kg m^-3
3. 800 kg m^-3
4. 8 x 10³ kg m^-3

Answer

800 kg m^-3

Reason — As

1 g cm^-3 = 1000 kg m^-3

Then

0.8 g cm^-3 = 0.8 x 1000 = 800 kg m^-3

The density of aluminium is 2.7 g cm^-3 and of brass is 8.4 g cm^-3. For the same mass, the volume of:

1. both will be same
2. aluminium will be less than that of brass
3. aluminium will be more than that of brass
4. nothing can be said.

Answer

aluminium will be more than that of brass

Reason — For a given mass of a substance, volume is inversely proportional to density.

As density of aluminium is less than that of brass, so volume of aluminium will be more than that of brass.

A block of wood of density 0.8 g cm^-3 has a volume of 60 cm³. The mass of block will be:

1. 60.8 g
2. 75 g
3. 48 g
4. 0.013 g

Answer

48 g

Reason — Given,

• Density of the block = 0.8 g cm^-3
• Volume of the block = 60 cm³

Now, mass of the block is given by,

Mass = density x volume
= 0.8 x 60
= 48 g

So, mass of block is 48 g.

The correct relation for speed is:

1. $\text {Speed} = \text {distance} \times \text {time} \\[1em]$
2. $\text {Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em]$
3. $\text {Speed} = \dfrac{\text{time}}{\text{distance}} \\[1em]$
4. $\text {Speed} = \dfrac{\text{1}}{\text{distance x time}}$

Answer

$\text {Speed} = \dfrac{\text{distance}}{\text{time}}$

Reason — Speed of a body = $\dfrac{\text{distance travelled by the body}}{\text{time of travel}}$

A boy travels a distance 150 m in 1 minute. His speed is:

1. 150 m s^-1
2. 2.5 m s^-1
3. 25 m s^-1
4. 9 m s^-1

Answer

2.5 m s^-1

Reason — Given,

• Distance travelled by the boy = 150 m
• Time taken = 1 minute = 60 sec

Speed of the boy is given by,

$\text{Speed} = \dfrac{\text{Distance travelled by the boy}}{\text{Time taken}} \\[1em] = \dfrac{\text{150}}{\text{60}} \\[1em] = \dfrac{5}{2} \\[1em] = 2.5 \text { m s}^{-1}$

Hence, the speed of the boy is 2.5 m s^-1.

The density of a substance ............... with the increase in the temperature.

1. increases
2. decreases
3. remains same
4. none of the above

Answer

decreases

Reason — When a substance is heated, its particles move faster and spread farther apart, causing the substance to expand so its volume increases.

Since density is defined as mass ÷ volume and the mass remains the same while the volume increases, the density decreases.

Hence, the density of a substance decreases with the increase in the temperature.

Assertion (A): The density of water decreases when it is cooled from 4 °C to 0 °C.

Reason (R): Volume of water increases in cooling from 4 °C to 0 °C.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true and R is not the correct explanation of A
3. Assertion is false but reason is true
4. Assertion is true but reason is false

Answer

Both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because when water is cooled from 4 °C to 0 °C, its density decreases instead of increasing as water shows anomalous expansion in this temperature range.

Reason (R) is true because as water cools from 4 °C to 0 °C, its volume increases even though the temperature is decreasing.

Since

$\text {Density} = \dfrac{\text {Mass}}{\text {Volume}}$

So, an increase in volume (with mass constant) leads to a decrease in density.

Therefore, both A and R are true and R is the correct explanation of A.

Fill in the blanks:

(a) 1 m³ = ............... cm³.

(b) The volume of an irregular solid is determined by the method of ............... .

(c) Volume of a cube = ............... .

(d) The area of an irregular lamina is measured by using a ............... .

(e) Equal masses of different substances have different ............... .

(f) The S.I. unit of density is ............... .

(g) 1 g cm^-3 = ............... kg m^-3.

(h) 36 km h^-1 = ............... m s ^-1.

(i) Speed of a vehicle at a particular instant is shown by ............... .

Answer

(a) 1 m³ = 10⁶ cm³.

(b) The volume of an irregular solid is determined by the method of displacement of liquid.

(c) Volume of a cube = (one side)³.

(d) The area of an irregular lamina is measured by using a graph paper.

(e) Equal masses of different substances have different volumes.

(f) The S.I. unit of density is kg m^-3.

(g) 1 g cm^-3 = 1000 kg m^-3.

(h) 36 km h^-1 = 10 m s ^-1.

(i) Speed of a vehicle at a particular instant is shown by speedometer.

Write true or false for each statement:

(a) The S.I. unit of volume is litre.

(b) A measuring beaker of capacity 200 mL can measure only the volume of 200 mL of a liquid.

(c) cm² is a smaller unit of area than m².

(d) Equal volumes of two different substances have equal masses.

(e) The S.I. unit of density is g cm^-3.

(f) 1 g cm^-3 = 1000 kg m^-3.

(g) The density of water is maximum at 4 °C.

(h) The speed 5 m s^-1 is less than 25 km h^-1.

(i) The S.I. unit of speed is m s^-1.

Answer

(a) False
Correct Statement — The S.I. unit of volume is cubic metre (m³).

(b) True

(c) True

(d) False
Correct Statement — Equal volumes of two different substances have different masses.

(e) False
Correct Statement — The S.I. unit of density is kg m^-3.

(f) True

(g) True

(h) True

(i) True

Match the following:

Answer

Define the term volume of an object.

Answer

The space occupied by an object is called its volume.

State and define the S.I. unit of volume.

Answer

The S.I. unit of volume is cubic metre (m³).

One cubic metre is the volume of a cube with each side 1 metre long.

State two smaller units of volume. How are they related to the S.I. unit?

Answer

The two smaller units of volume are cubic centimetre (cm³) and cubic decimetre (dm³).

1 m³ = 1 m x 1 m x 1 m
= 100 cm x 100 cm x 100 cm
= 1000000 cm³
= 10⁶ cm³

So, 1 m³ = 10⁶ cm³

1 m³ = 1 m x 1 m x 1 m
= 10 dm x 10 dm x 10 dm
= 1000 dm³
= 10³ dm³

So, 1 m³ = 10³ dm³

How will you determine the volume of a cuboid? Write the formula you will use.

Answer

The volume of a cuboid is determined by finding its three dimensions i.e., length, breadth and height.

The formula for volume of cuboid is length x breadth x height

You are required to take out 200 mL of milk from a bucket full of milk. How will you do it?

Answer

A measuring beaker of 200 mL is used to measure 200 mL of milk from a bucket full of milk.

1. Take a measuring beaker of 200 mL, wash and dry it.
2. Immerse the beaker inside the bucket full of milk.
3. Fill the beaker completely with the milk.
4. Take out the beaker from bucket gently so that no milk splashes out.
5. Pour the milk from measuring beaker into another empty vessel.

Define the term density of a substance.

Answer

The density of a substance is defined as the mass of a unit volume of that substance.

State the S.I. and C.G.S. units of density. How are they related?

Answer

The S.I. unit of density is kg m^-3 and its C.G.S. unit is g cm^-3.

1 g cm^-3 = 1000 kg m^-3.

'The density of brass is 8.4 g cm^-3'. What do you mean by the statement?

Answer

The statement means one cubic centimeter volume of brass has mass of 8.4 g.

Arrange the following substances in order of their increasing density:

1. iron
2. cork
3. brass
4. water
5. mercury.

Answer

Cork < Water < Iron < Brass < Mercury.

How does the density of water change when:

(a) it is heated from 0 °C to 4 °C,

(b) it is heated from 4 °C to 10 °C ?

Answer

(a) Water when heated from 0°C to 4°C, it contracts so density of water increases and becomes maximum at 4°C.

(b) When water is heated from 4°C to 10°C it expands, so density decreases.

Write the density of water at 4 °C.

Answer

The density of water at 4°C is 1 g cm^-3 or 1000 kg m^-3.

Explain the meaning of the term speed.

Answer

Speed of the body is distance travelled by the body in unit time. It tells us about the motion of the body i.e. how fast or slow it is moving.

Write the S.I. unit of speed.

Answer

The S.I. unit of speed is metre per second (m s^-1).

A car travels with a speed 12 m s^-1, while a scooter travels with a speed 36 km h^-1. Which of the two travels faster?

Answer

Given,

• Speed of car = 12 m s^-1
• Speed of scooter = 36 km h^-1

1 km h^-1 = $\dfrac{\text{5}}{\text{18}}$ m s^-1

36 km h^-1 = 36 x $\dfrac{\text{5}}{\text{18}}$ = 10 m s^-1

⟹ Speed of scooter = 10 m s^-1

Since

Speed of scooter (10 m s^-1) < Speed of car (12 m s^-1)

So, car travels faster than scooter

Name two devices which are used to measure the volume of an object. Draw their neat diagrams.

Answer

The two devices which are used to measure the volume of an object are:

1. Measuring cylinder

2. Measuring beaker

How can you determine the volume of an irregular solid (say a piece of brass)? Describe in steps with neat diagrams.

Answer

The volume of an irregular solid can be measured by using a measuring cylinder by the method of displacement of liquid.

1. Place a measuring cylinder on a flat horizontal surface.
2. Fill it partially with water.
3. Note the reading of water level very carefully. Let it be V₁.
4. Now tie the piece of brass with a thread and dip it completely into water.
5. The level of water rises. Note the reading of new water level V₂. The difference in the two levels of water gives the volume of the piece of brass.

Describe the method in steps to find the area of an irregular lamina using a graph paper.

Answer

To find the area of an irregular lamina using a graph paper.

1. Place the lamina over a graph paper.
2. Draw its boundary line on the graph paper with a pencil.
3. Remove the lamina.
4. Count and note the number of complete squares, half squares and more than half squares within the boundary line. Ignore the squares which are less than half.

The sum of the areas of the number of complete squares and the number of incomplete squares gives the approximate area of the irregular object.

The length, breadth and height of a water tank are 5 m, 2.5 m and 1.25 m respectively. Calculate the capacity of the water tank in:

(a) m³

(b) litre

Answer

Given,

• Length = 5 m
• Breadth = 2.5 m
• Height = 1.25 m

(a) Volume of water tank = Length x Breadth x Height
= 5 m x 2.5 m x 1.25 m
= 15.625 m³

So, capacity of water tank is 15.625 m³

(b) As

1 m³ = 1000 L

15.625 m³ = 15.625 x 1000 = 15625 L

So, capacity of water tank in litres is 15,625 L

A solid silver piece is immersed in water contained in a measuring cylinder. The level of water rises from 50 mL to 62 mL. Find the volume of silver piece.

Answer

Given,

• Initial level of water V₁ = 50 mL
• Final level of water V₂ = 62 mL

Volume of silver piece = V₂ - V₁
= 62 - 50
= 12 mL
= 12 cm³

So, volume of silver piece is 12 cm³.

Find the volume of a liquid present in a dish of dimensions 10 cm x 10 cm x 5 cm.

Answer

Given,

• Dimensions of the dish = 10 cm x 10 cm x 5 cm

Volume of dish = Length x Breadth x Height
= 10 cm x 10 cm x 5 cm
= 500 cm³

Now,

Volume of liquid = Volume of dish
= 500 cm³

Since

1 cm³ = 1 mL

Then

500 cm³ = 500 mL

So, volume of a liquid present in the dish is 500 mL.

A rectangular field is of length 60 m and breadth 35 m. Find the area of the field.

Answer

Given,

• Length = 60 m
• Breadth = 35 m

Area of the field = Length x Breadth
= 60 m x 35 m
= 2100 m².

So, area of the field is 2100 m².

Find the approximate area of an irregular lamina of which boundary line is drawn on the graph paper shown in figure below.

Answer

From the figure,

• Number of complete squares = 11
• Number of more than half squares = 8
• Number of half squares = 2

So,

Total number of squares = 11 + 8 + 2 = 21

Since

Area of the square = 1 cm x 1 cm = 1 cm²

Then,

Area of 21 squares = 21 x 1 = 21 cm²

So, approximate area of lamina is 21 cm².

A piece of brass of volume 30 cm³ has a mass of 252 g. Find the density of brass in:

(i) g cm^-3

(ii) kg m^-3

Answer

Given,

• Mass of brass = 252 g
• Volume of brass = 30 cm³
• Density = ?

(i) Density of brass is given by, $\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] = \dfrac{\text{252}}{\text{30}} \\[1em] = 8.4 \text{g cm}^{-3}$

So, density of brass is 8.4 g cm^-3.

(ii) Since

1 g cm^-3 = 1000 kg m^-3

Then

8.4 g cm^-3 = 8.4 x 1000 = 8400 kg m^-3

So, density of brass is 8400 kg m^-3.

The mass of an iron ball is 312 g. The density of iron is 7.8 g cm^-3. Find the volume of the ball.

Answer

Given,

• Mass of the ball = 312 g
• Density of the ball = 7.8 g cm^-3

Density of the ball is given by,

$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] \Rightarrow \text{Volume} = \dfrac{\text{Mass}}{\text{Density}} \\[1em] = \dfrac{\text{312}}{7.8} \\[1em] = 40 \text{ cm}^{3}$

So, volume of the ball is 40 cm³.

A cork has a volume 25 cm³. The density of cork is 0.25 g cm^-3. Find the mass of the cork.

Answer

Given,

• Volume of the cork = 25 cm³
• Density of the cork = 0.25 g cm^-3

Density of the cork is given by,

$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] \text{Mass} = \text{Density} \times \text{Volume} \\[1em] = 0.25 \times 25 \\[1em] = 6.25 \text{ g}$

So, mass of the cork is 6.25 g.

The mass of 5 litre of water is 5 kg. Find the density of water in g cm^-3.

Answer

Given,

• Mass of water = 5 kg
• Volume of water = 5 litre

Density of water is given by,

$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] = \dfrac{\text{5}}{\text{5}} \\[1em] = 1 \text{ kg litre}^{-1}$

As

1 kg = 1000 g and 1 litre = 1000 cm³

Then

1 kg litre^-1 = $\dfrac{\text{1000}}{\text{1000}}$ = 1 g cm^-3

So, density of water is 1 g cm^-3.

A cubical tank of side 1 m is filled with 800 kg of a liquid. Find:

(i) the volume of tank,

(ii) the density of liquid in kg m^-3.

Answer

• Length of side of the cubical tank = 1 m;
• Mass of the liquid = 800 kg

(i) Volume of cubical tank = (side)³
= (1 m)³
= 1 m³

So, volume of the tank = 1 m³

(ii) Density of water is given by, $\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] = \dfrac{\text{800}}{\text{1}} \\[1em] = 800 \text{ kg m}^{3}$

So, density of the liquid is 800 kg m^-3.

A block of iron has dimensions 2 m x 0.5 m x 0.25 m. The density of iron is 7.8 g cm^-3. Find the mass of block.

Answer

Given,

• Dimensions of the block = 2 m x 0.5 m x 0.25 m
• Density of iron = 7.8 g cm^-3

Since

1 g cm^-3 = 1000 kg m^-3

Then

7.8 g cm^-3 = 7.8 x 1000 = 7800 kg m^-3

Now,

Volume of iron = Length x Breadth x Height
= 2 m x 0.5 m x 0.25 m
= 0.25 m³

Now, density of water is given by,

$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] \Rightarrow \text{Mass} = \text{Density} \times \text{Volume} \\[1em] = 7800 \times 0.25 \\[1em] = 1950 \text{ kg}$

So, the mass of iron is 1950 kg.

The mass of a lead piece is 115 g. When it is immersed into a measuring cylinder, the water level rises from 20 mL mark to 30 mL mark. Find:

(i) the volume of the lead piece,

(ii) the density of the lead in kg m^-3.

Answer

Given,

• Mass of the lead piece = 115 g

• Initial volume (V₁) = 20 mL

• Final volume (V₂) = 30 mL

(i) Volume of lead piece = V₂ - V₁
= 30 - 20
= 10 mL

Since

1 mL = 1 cm³

Then

10 mL = 10 cm³

So, volume of the lead piece is 10 cm³.

(ii) Density of the lead piece is given by, $\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] = \dfrac{\text{115}}{\text{10}} \\[1em] = 11.5 \text{ g cm}^{3}$

As

1 g cm^-3 = 1000 kg m^-3

Then

11.5 g cm^-3 = 1000 x 11.5 = 11500 kg m^-3

So, density of lead is 11500 kg m^-3.

The density of copper is 8.9 g cm^-3. What will be its density in kg m^-3?

Answer

Given,

• Density of copper = 8.9 g cm^-3

As

1 g cm^-3 = 1000 kg m^-3

Then

8.9 g cm^-3 = 8.9 x 1000 = 8900 kg m^-3

So, density of copper = 8900 kg m^-3

A car travels a distance of 15 km in 20 minute. Find the speed of the car in:

(i) km h^-1

(ii) m s^-1

Answer

(i) Given,

• Distance travelled by the car = 15 km;
• Time taken = 20 min

As

60 min = 1 hr

Then

20 min = $\dfrac{\text{20}}{\text{60}}$ = $\dfrac{\text{1}}{\text{3}}$ h

Now, speed of the car is given by,

$\text{Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em] = \dfrac{15}{\dfrac{1}{3}} \\[1em] = 15 \times 3 \\[1em] = 45 \text{ km h}^{-1}$

So, speed of the car is 45 km h^-1.

(ii) Here,

• Distance = 15 km

And

1 km = 1000 m

So

15 km = 15 x 1000 = 15000 m

Also

• Time = 20 min

Since

1 min = 60 sec

So

20 min = 20 x 60 = 1200 sec

Speed of the car is given by,

$\text{Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em] = \dfrac{\text{15000}}{\text{1200}} \\[1em] = 12.5 \text{ m s}^{-1}$

So, speed of the car is 12.5 m s^-1.

How long a train will take to travel a distance of 200 km with a speed of 60 km h^-1?

Answer

Given,

• Distance travelled by the train = 200 km
• Speed of the train = 60 km h^-1

Speed of the train is given by,

$\text{Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em] \Rightarrow \text{time} = \dfrac{\text{distance}}{\text{Speed}} \\[1em] = \dfrac{\text{200}}{\text{60}} \\[1em] = 3\text{ h } 20\text{ min}$

So, time taken by train will be 3 hrs 20 minutes.

A boy travels with a speed of 10 m s^-1 for 30 minute. How much distance does he travel?

Answer

Given,

• Speed of the boy = 10 m s^-1
• Time taken to cover the distance = 30 minutes

Since

1 min = 60 sec
⟹ 30 min = 60 x 30
= 1800 sec

Speed of the boy is given by,

$\text{Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em] \Rightarrow \text{distance} = \text{Speed} \times \text{time} \\[1em] = 1800 \times 10 \\[1em] = 18000 \text{ m}$

So, distance travelled by the boy is 18000 m or 18 km.

Express 36 km h^-1 in m s^-1.

Answer

As

1 km h^-1 = $\dfrac{\text{5}}{\text{18}}$ m s^-1

Then

36 km h^-1 = 36 x $\dfrac{\text{5}}{\text{18}}$ = 10 m s^-1

So, 36 km h^-1 = 10 m s^-1

Express 15 m s^-1 in km h^-1.

Answer

As

1 m s^-1 = 3.6 km h^-1

Then

15 m s^-1 = 3.6 x 15 = 54 km h^-1

So, 15 m s^-1 = 54 km h^-1

Read the clues across and clues downwards and fill up the blank squares.

Across :

1. The device is used to measure the atmospheric pressure.
2. The area of a regular object can be found by measuring its
3. The S.I. unit of volume is ............... metre.

Down :

2. The ............... of a substance does not change with change in its shape or size.
3. The volume of ............... lamina is measured by using a graph paper.
4. The space occupied by an object

Answer

The solved crossword puzzle is given below:

1 kg of salt occupies more space than 1 kg of copper. Give reason.

Answer

1 kg of salt occupies more space than 1 kg of copper because the density of salt is less than that of copper. Therefore, for the same mass, salt occupies more volume, while copper, being denser, occupies less volume.

Gunjan went with her father to a wood shop to buy blocks for a school project. The shopkeeper showed them several wooden cubes, each of size 5 cm x 5 cm x 5 cm. When Gunjan picked them up one by one, she noticed that some cubes felt lighter and some heavier even though they were exactly the same size. To check their weights, she placed two cubes on a digital weighing machine. One cube had a mass of 150 g and the other had a mass of 250 g. Her father smiled and told her that her careful observation was helping him choose the right material.

Answer the following :

(i) Since both blocks have the same dimensions but different masses, what can Gunjan conclude about their material ?

(ii) Calculate the density of the heavier block of mass 250 g.

(iii) Which property of matter explains why two objects of the same size can feel heavier or lighter?

(iv) If there is another block with the same density as the heavier block but double its volume, what happens to its mass ?

Answer

(i) Since both cubes have the same dimensions and therefore the same volume, but different masses, Gunjan can conclude that they are made of different materials and have different densities.

(ii) Given,

• Dimensions of each cube = 5 cm x 5 cm x 5 cm
• Mass = 250 g

Volume of cube = Length x Breadth x Height

= 5 × 5 × 5

= 125 cm³

$\text {Density} = \dfrac{\text {Mass}}{\text {Volume}} \\[1em] = \dfrac{250}{125} \\[1em] = 2 \text { g cm}^{-3}$

So, the density of the heavier block is 2 g cm^-3.

(iii) The property is density. It explains why objects of the same size can have different masses and feel heavier or lighter.

(iv) If the density remains the same, then mass is directly proportional to volume. Therefore, if the volume is doubled, the mass will also double.

So, the mass will become 500 g.