Physical Quantities and Measurement

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 in m³ = 15.625 m³.

b. 1 m³ = 1000 L
15.625 m³ = 15.625 x 1000 = 15625 L

So, capacity of water tank in litres = 15,625 L.

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 = 12 cm³.

Volume of dish = 10 cm x 10 cm x 5 cm = 500 cm³.

Volume of liquid is equal to volume of dish = 500 cm³.

1 cm³ = 1 mL
500 cm³ = 500 mL

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

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 = 2100 m².

Number of complete squares = 11
Number of more than half squares = 8
Number of half squares = 2
Total number of squares = 11 + 8 + 2 = 21
Area of the square = 1 cm x 1 cm = 1 cm²
Area of 21 squares = 21 x 1 = 21 cm²

So, approximate area of lamina = 21 cm².

Mass = 252 g
Volume = 30 cm^-3
Density = ?

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

So, Density = 8.4 g cm^-3.

(ii) 1 g cm^-3 = 1000 kg m^-3
8.4 g cm^-3 = 8.4 x 1000 = 8400 kg m^-3

So, Density = 8400 kg m^-3.

Mass = 312 g
Density = 7.8 g cm^-3
Volume = ?

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

So, volume of the ball is 40 cm³.

Volume = 25 cm³
Density = 0.25 g cm^-3
Mass = ?

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

So, mass of the cork is 6.25 g.

Mass = 5 kg
Volume = 5 litre
Density = ?

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

Density = 1 kg litre^-1

1 kg = 1000 g; 1 litre = 1000 cm³

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

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

Side = 1 m; mass = 800 kg

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

So, volume of tank = 1 m³.

(ii) $\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] \text{Density} = \dfrac{\text{800}}{\text{1}} = 800 \text{kg m}^{3}$

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

Density of iron = 7.8 g cm^-3
1 g cm^-3 = 1000 kg m^-3
7.8 g cm^-3 = 7.8 x 1000 = 7800 kg m^-3

Volume of iron = 2 m x 0.5 m x 0.25 m = 0.25 m³

Mass = ?

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

So, mass of iron = 1950 kg.

(i) Mass = 115 g;
Initial volume V₁ = 20 mL;
Final volume V₂ = 30 mL;
Volume of lead piece = V₂ - V₁ = 30 - 20 = 10 mL

1 mL = 1 cm³
10 mL = 10 cm³

So, volume of the lead piece = 10 cm³.

(ii) Density = ? $\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] \text{Density} = \dfrac{\text{115}}{\text{10}} = 11.5 \text{g cm}^{3}$

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

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

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

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

(i) distance = 15 km; time = 20 min

60 min = 1 hr

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

$\text{Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em] = \dfrac{15}{\dfrac{1}{3}} \\[1em] = 15 \times 3 \\[1em] = 45 \text{km h}^{-1}$ So, speed of car = 45 km h^-1.

(ii) distance = 15 km
1 km = 1000 m
15 km = 15 x 1000 = 15000 m

time = 20 min
1 min = 60 sec
20 min = 20 x 60 = 1200 sec

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

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

distance = 200 km; speed = 60 km h^-1

time = ?

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

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

Speed = 10 m s^-1
time = 30 minutes
1 min = 60 sec
30 min = 60 x 30 = 1800 sec

distance = ?

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

So, distance travelled = 18000 m or 18 km.

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

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

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

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

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

(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

(a) 10⁶

(b) displacement of liquid

(c) (one side)³

(d) graph paper

(e) volume

(f) kg m^-3

(g) 1000

(h) 10

(i) speed v

10^-3 m³

Reason — 1 litre = 10^-3 m³

15 cm³

Reason — Volume of body is equal to volume of liquid displaced = 15 mL.
1 mL = 1 cm³, so 15 mL = 15 cm³.

0.3 m²

Reason — 20 cm = 0.2 m
Area of paper = 1.5 m x 0.2 m = 0.3 m²

M = d x V

Reason — Mass = Volume x density

800 kg m^-3

Reason — 1 g cm^-3 = 1000 kg m^-3
So 0.8 g cm^-3 = 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 — Mass = density x volume
= 0.8 x 60
= 48 g
So, mass of block is 48 g.

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

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

2.5 m s^-1

Reason — time = 1 min = 60 sec, distance = 150 m $\text{Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em] \text{Speed} = \dfrac{\text{150}}{\text{60}}$ Speed = 2.5 m s^-1

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), 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

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.

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.

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 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).

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

So speed of scooter = 10 m s^-1

So car travels faster than scooter; since speed of car is more than that of scooter.