Physical Quantities and Measurement

10^-3 m³

Reason — 1 litre = 10^-3 m³

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.

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.

M = d x V

Reason — Mass = Volume x density

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

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.

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.

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

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

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,

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

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

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.

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

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

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.

(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.

(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

The space occupied by an object is called its volume.

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.

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

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

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.

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

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

Cork < Water < Iron < Brass < Mercury.

(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.

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

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.

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

Given,

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

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

36 km h^-1 = 36 x $\frac{\text{5}}{\text{18}}$ = 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

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

1. Measuring cylinder

2. Measuring beaker

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.

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.

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

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.

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.

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.

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.

Given,

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

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

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.

Given,

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

Density of the ball is given by,

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

So, volume of the ball is 40 cm³.

Given,

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

Density of the cork is given by,

$$\begin{aligned}\text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1em] \text{Mass} = \text{Density} \times \text{Volume} \\[1em] = 0.25 \times 25 \\[1em] = 6.25 \text{ g}\end{aligned}$$

So, mass of the cork is 6.25 g.

Given,

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

Density of water is given by,

$$\begin{aligned}\text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1em] = \frac{\text{5}}{\text{5}} \\[1em] = 1 \text{ kg litre}^{-1}\end{aligned}$$

As

1 kg = 1000 g and 1 litre = 1000 cm³

Then

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

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

• 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, $$\begin{aligned}\text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1em] = \frac{\text{800}}{\text{1}} \\[1em] = 800 \text{ kg m}^{3}\end{aligned}$$

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

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,

$$\begin{aligned}\text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1em] \Rightarrow \text{Mass} = \text{Density} \times \text{Volume} \\[1em] = 7800 \times 0.25 \\[1em] = 1950 \text{ kg}\end{aligned}$$

So, the mass of iron is 1950 kg.

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, $$\begin{aligned}\text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1em] = \frac{\text{115}}{\text{10}} \\[1em] = 11.5 \text{ g cm}^{3}\end{aligned}$$

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.

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

(i) Given,

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

As

60 min = 1 hr

Then

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

Now, speed of the car is given by,

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

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,

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

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

Given,

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

Speed of the train is given by,

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

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

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,

$$\begin{aligned}\text{Speed} = \frac{\text{distance}}{\text{time}} \\[1em] \Rightarrow \text{distance} = \text{Speed} \times \text{time} \\[1em] = 1800 \times 10 \\[1em] = 18000 \text{ m}\end{aligned}$$

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

As

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

Then

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

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

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

The solved crossword puzzle is given below:

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.

(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

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

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.