Selina Concise Class 7 Physics Solution Chapter No. 1- ‘Physical Quantities and Measurements’ For ICSE Board Students.
A) Objective Questions:
l.) Write true or false for each statement :
(a) The S.I. unit of volume is litre.
Answer: False SI unit of volume is m³.
(b) A measuring beaker of capacity 200 mL can measure only the volume of 200 mL of a liquid.
Answer: True, Athens marking of scale on measuring beaker is at 200 mL only. Thus we can measure 200 mL of liquid using a measuring beaker of capacity 200 mL.
(c) Cm² is a smaller unit of area than m².
Answer: True, because cm is 100 times less than m.
1 m = 100 cm.
(d) Equal volumes of two different substances have equal masses.
Answer: False, equal volume of two different substances may be unequal masses because their density may be different.
(e)The S.I. unit of density is g cm-3.
Answer: False, The SI unit of density is kg/m³.
(f) 1 g cm-3 = 1000 kg m-3.
Answer: True,
As we know that,
1 g = 10-3 kg and
1 cm³ = (10-2)-3
1 cm³ = 10-6
Thus,
1 g/cm³ = 10-3 / 10-6
1 g/cm³ = 1000 g cm-3
(g) The density of water is maximum at 4⁰C.
Answer: True, the density of water is maximum at 4⁰ C.
(h) The speed 5 m/s is less than 25 km/h.
Answer: True.
We have to convert 5 m/s in km/h.
As we know that,
1 m/s = 18/5 km/h.
Thus,
5 m/s = 5 × 18/5 km/h
5 m/s = 18 km/h.
Therefore 5 m/s is less than 25 km/h.
(i) The S.I. unit of speed is m s-1.
Answer: True.
As we know that,
Speed = distance / time.
Speed = m/s.
2.) Fill in the blanks:
(a) 1 m³ = ——-cm³.
Answer: 1 m³ = 10⁶ cm³.
As we know that,
1 m = 10² cm,
1 m³ = (10²)³ cm³.
1 m³ = 10⁶ cm³
(b) The volume of an irregular solid is determined by the method of ———–.
Answer: The volume of an irregular solid is determined by the method of displacement of liquid.
(c) Volume of a cube = ———.
Answer: Volume of a cube = side³.
Cube has three equal sides and volume has 3 dimensional geometry.
(d) The area of an irregular lamina is measured by using a ——–.
Answer: The area of an irregular lamina is measured by using a graph paper.
(e) Mass = density x ……..
Answer: Mass = density × volume.
As we know that,
Density = mass/volume.
There fore,
Mass = density × volume.
f) The S.I. unit of density is —–.
Answer: The S.I. unit of density I kg/m³.
As we know that,
Density = mass / volume.
But, SI unit mass is kg and volume is m³.
Therefore,
The S.I. unit of density I kg/m³.
(g) 1g/cm³= ——- kg/m³.
Answer: 1 g/cm³= 1000 kg/m³.
As we know that,
1 g = 10-3 kg and
1 cm = 10-2 m
1 cm³ = (10 -2)³ m³
1 cm³ = 10-6 m³
Thus,
1 g /cm³ = 10-3 /10-6 kg /cm³
1 g / cm³ = 1000 kg/cm³.
(h) 36 km/h = ——– m/s.
Answer: 36 km/h = 10 m/s.
As we know,
1 km = 1000 m and
1 hour = 3600 seconds.
Thus,
1 km/h = 1000/3600 m/s.
1 km/h = 5/18 m/s.
So,
36 km/h = 36 × 5/18 m/s.
36 km/h = 10 m/s.
(i) Distance travelled D = —— x time.
Answer: Distance travelled D = speed Vx time t.
As we know,
Speed = distance /time
Distance = speed ×time.
3.) Match the following :
Column A Column B
(a) Volume of a liquid (i) kg /m³
(b) Area of a leaf (ii) m³
(c) S.I. unit of volume (iii) graph paper
(d) S.I. unit of density (iv) m/s
(e) S.I. unit of speed (v) measuring cylinder
Answer:
Column A Column B
(a) Volume of a liquid (v) Measuring cylinder
(b) Area of a leaf (iii) graph paper
(c) S.I. unit of volume (ii) m³
(d) S.I. unit of density (i) kg/m³
(e) S.I. unit of speed (iv) m/s
4.) Select the correct alternative :
(a) One litre is equal to :
(i) 1 cm³
(ii) 1 m³
(iii) 10-3cm³
(iv) 10-3 m³
Answer:(iv) 10-3 m³
(b) A metallic piece displaces water of volume 15 mL. The volume of piece is :
(i) 15 cm³
(ii) 15 m³
(iii) 15 x 10³ cm-3
(iv) 15 x 10³ cm³
Answer: i) 15 cm³.
If a metallic piece displaces water of volume 15 mL then the volume of piece is 15 cm³.
(c) A piece of paper of dimensions 1.5 m x 20 cm has area :
(i) 30 m²
(ii) 300 cm²
(iii) 0.3 m²
(iv) 3000 m³
Answer: (iii) 0.3 m².
Given, first side = 1.5 m,
Second side = 20 cm = 0.20 m,
As we know,
Area = side × side
Area = 1.5 × 0.20
Area = 0.3 m².
Thus the area of that piece of paper is 0.3 m².
(d) The correct relation is :
(i) d = M × V
(ii) M = d x V
(iii) V=dxM
(iv) d=M+V
Answer: (ii) M = d × V
As we know that,
Density = mass / volume.
Therefore,
Mass = density × volume.
(e) The density of alcohol is 0.8 g cm-3. In S.I. unit, it will be
(i) 0.8 kg m-3
(ii) 0.008 kg m-3
(iii) 800 kg m-3
(iv) 8 x 10³ kg m-3
Answer: (iii) 800 kg m-3
As we know that,
1 g cm-3 = 1000 kg m-3
Thus,
0.8 g cm-3 = 0.8 × 1000 kg m-3
0.8 g cm-3 = 800 kg m-3
(f) The density of aluminum is 2.7 g cm-3 and of brass is 8.4 g cm-3. For the same mass, the volume of :
(i) both will be same
(ii) aluminum will be less than that of brass (iii) aluminum will be more than that of brass (iv) nothing can be said.
Answer:
(iii) aluminum will be more than that of brass.
Given,
density of aluminum = 2.7 g cm-3
And density of brass = 8.4 g cm-3.
As we know that,
Density = mass/volume.
But mass of the object is constant.
If the mass of object is constant then density is inversely proportional to volume and vice versa.
The density aluminum is less than brass thus its volume will be greater than brass.
(g) A block of wood of density 0.8 g cm-3 has a volume of 60 cm³. The mass of block will be :
(i) 608 g
(ii) 75 g
(iii) 48 g
(iv) 0.013 g.
Answer: (iii) 48 g
Given,
Density of block of wood = 0.8g cm-3
Volume of block of wood = 60 cm³.
As we know that,
Density = mass / volume
Mass = density × volume
Mass = 0.8 × 60
Mass = 48 g.
Thus the mass of that block of wood is 48 g.
(h) The correct relation for speed is :
(i) Speed = distance x time
(ii) Speed = distance/time
(iii) Speed =time/ distance
(iv) Speed =1/( distance x time)
Answer: (ii) Speed = distance/time
(i) A boy travels a distance 150 m in 1 minute. His speed is :
(i) 150 m/s
(ii) 2.5 m/s
(iii) 25 m/s
(iv) 9 m/s.
Answer: (ii) 2.5 m/s
Given,
Distance = 150 m,
Time = 1 minute = 60 seconds.
As we know that,
Speed = Distance / time
Speed = 150 / 60
Speed = 2.5 m/s
The speed of that boy is 2.5 m/s.
B.) Short/Long Answer Questions :
l.) Define the term volume of an object.
Answer: Every object occupied a space. That occupied space is called as volume of that object.
2.) State and define the S.I. unit of volume.
Answer: SI unit of volume is m³ (cubic meter).
Definition : If length of each side of a cube is one meter then the volume of that cube is one cubic meter.
3.) State two smaller units of volume. How are they related to the S.I. unit ?
Answer: Two smaller units of volume are cubic centimeter and litre.
Relationship between cubic meter and cubic centimeter.
As we know that,
1 m = 10² cm
Thus,
Taking cube on both side, we get
1 m³ = (10²)³ cm³
1 m³ = 10⁶ cm³.
One cubit meter = 10⁶ cubic centimeter.
Relationship between cubic meter and Liter.
1 m³ = 10³ litre.
4.) How will you determine the volume of a cuboid ? Write the formula you will use.
Answer:
- As we know that, cuboid is 3 dimensional geometry shape.
- It has three sides along three different axis.
- Length of each side of cuboid is different.
- We have tomultiply three different side to getting volume of a cuboid.
- Let three side of cuboid are a, b and c.
As discussed above,
Volume of cuboid = a × b × c.
5.) Name two devices which are used to measure the volume of an object. Draw their neat diagrams.
Answer: Measuring cylinder and beakers are the two devices which used to measure the volume of an object.
Fig: Measuring cylinder
6.) How can you determine the volume of an irregular solid (say a piece of brass) ? Describe in steps with neat diagrams.
Answer: We can determine the volume of irregular solid using the concept of density. As we know that, when an object immersed into the liquid then it displaced the level of liquid upto the density of that substance.
To understand this, we perform an experiment.
Apparatus: Measuring cylinder, stone, string, water etc.
Diagram:
Procedure:
- Fill the water upto a certain limit.
- Note down the reading of water level on beaker and say V¹.
- Tie the stone with the help of string.
- Immersed it into the water.
- Note down the final level of water in the Measuring cylinder and say it V².
- As we know that, Change in level of liquid is equal to volume of the object immersed in it.
- Volume of object = V² – V¹.
In this way, we can calculate the volume of object.
7.) You are required to take out 200 mL of milk from a bucket full of milk. How will you do it?
Answer: We can use beaker of 200 mL for Measuring the 200 mLmilk from a bucket of milk.
- Sink empty beaker of value 200 mL into a bucket of milk.
- Fill the milk into beaker upto reading 200 mL.
- Take it out from the bucket.
In this way , we measure 200 mL milk from the bucket of milk.
8.) Describe the method in steps to find the area of an irregular lamina using a graph paper.
Answer: We can measure the area of irregular lamina using graph paper.
- Put the lamina on graph paper.
- Draw outline of lamina on graph paper As shown in figure below.
- Count the Total no of complete squares and incomplete squares (greater than half shaded).
- Suppose, total number of complete squares = a m² and
Total number of incompletes squares = b m².
Area of lamina =( a + b) m²
In this way, we can calculate the area of irregular lamina.
9.) Define the term density of a substance.
Answer: Density is the property of material associated with mass and volume of that substance.
Definition: Density of the substance is defined as the ratio of mass to unit volume of that substance.
10.) State the S.I. and C.GS. units of density. How are they related ?
Answer: Density is the ratio of mass and volume.
Density = mass/volume.
SI unit of density is kg/m³.
CGS unit of density is g/cm³.
The relationship between SI unit and CGS unit of density:
As we know that,
1 kg = 10 g and
1 m = 10² cm.
Therefore,
1 Kg/m³ = 10³/(10²)³
1 kg/m³ = 10³/10⁶
1 kg/m³ = 10-3 g/cm³
11.) ‘The density of brass is 8.4 g cm-3. What do you mean by the statement ?
Answer: The density of brass is 8.5 g/cm³. The meaning of this statement is-
- The mass of that brass is 8.4 g.
- The volume of that 8.4 g mass is 1 mL or 1 cm³.
- In simple way, the mass of unit volume in ml of brass is 8.4 g.
12.) Arrange the following substances in order of their increasing density :
(a) iron
(b) cork
(c) brass
(d) water
(e) mercury
Answer: The density of a substance is depend on the ration of mass to volume. The order of the above
Cork < water <Iron < Brass <mercury.
13.) How does the density of water change when :
(a) it is heated from O⁰ C to 4⁰ C,
(b) it is heated from 4⁰ C to 10⁰ C ?
Answer: As we studied that, when temperature increases then volume also increases and density decreases. But water does notobey this rule. The variation of density of water with temperature is as-
a) From 0⁰ C to 4⁰ C-
Water contract from 1⁰ C to 4⁰ C. Thus the volume decreases and density increases.
The density of water increases when we heat it from 0⁰ C to 4⁰ C.
b) From O⁰ C to 4⁰ C-
Water expand from 4⁰ C to 10⁰ C. Thus the volume of water increases and density decreases.
The density of water decreases when we heat it from 4⁰ C to 10⁰ C.
14.) Write the density of water at 4⁰ C.
Answer: The density of water is maximum at 4⁰ C. It is 1 g/cm³ or 1000 kg/m³.
15.) Explain the meaning of the term speed.
Answer: Speed gives information about how the fast object changes it’s position .
It is the rate of distance and time.
We can calculate speed using the formula below.
Speed = mass/volume.
The SI unit of speed is m/s.
16.) Write the S.I. unit of speed
Answer: As we know that,
Speed = distance /time.
SI unit of distance is meter and time is second.
Thus SI unit of speed is m/s.
17.) A car travels with a speed 12 m /s, while a scooter travels with a speed 36 km/h. Which of the two travels faster ?
Answer: Given,
Speed of car A = 12 m/s, ——1
Speed of car B = 36 km/h.
As we know,
1 km= 10000 m and 1 hour = 3600 seconds.
Thus,
1 km/h = 1000/3600 m/s.
1km/h = 5/18 m/s.
Therefore,
36 km/h = 36 × (5/18) m/s.
36 km/h = 10 m/s.
The speed of car B = 10 m/s. ——-2
From 1 and 2, we conclude that car A moves faster than car B.
C.) Numericals:
1.) The length, breadth and height of a water tank are 5 m, 2.5 m and 1.25 m Calculate the capacity of the water tank in (a) m³ (b) liter.
Answer: Given,
Length = 5 m, width = 2.5 m and height = 1.25 m.
From above data, we conclude that the shape of water tank must be cuboid.
As we know that,
Volume of cuboid = length × width × height
Volume of cuboid = 5 × 2.5 × 1.25
Volume of cuboid = 15.625 m³
The volume of water tank is 15.625 m³.
Now we convert this volume in liter.
As we know,
1 m³ = 1000 liter.
Therefore,
15.625 m³ = 15.625 × 1000 liter
15.625 m³ = 15625 liter.
The volume of water tank is 15625 liter.
2.) 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:
As we know that,
When a solid immersed into water then it displaced the level of water upto the volume of that solid.
The level of water rises from 50 mL to 62 mL.
The rise of level = 62 -50 = 12 mL.
Now we concert this into cm³.
As we know, 1 mL = 1 cm³
Thus the rise in water level = 12 cm³
Therefore the volume of that solid is also 12 cm³.
3.) Find the volume of a liquid present in a dish of dimensions 10 cm x 10 cm x 5 cm.
Answer: Given,
Length = 10 m, width = 10 cm and height = 5 cm.
From above data, we conclude that the shape of dish must be cuboid.
As we know that,
Volume of cuboid = length × width × height
Volume of cuboid = 10 × 10 × 5
Volume of cuboid = 500 cm³
As we know,
1 cm³ = 1 mL.
Thus the volume of liquid is also 500 ml.
4.) A rectangular field is of length 60 m and breadth 35 m. Find the area of the field.
Answer: Given,
Length of field = 60 m, breadth of field = 35 m.
The shape of field is rectangular.
We know that,
Area of rectangle = length × breadth
Area of rectangle = 60 × 35
Area of rectangle = 2100 m²
The area of that rectangular field is 2100 m².
5.) Find the approximate area of an irregular lamina of which boundary line is drawn on the graph paper shown in Fig. 1.16. below.
Answer: we can calculate the area of irregular lamina using graphical method.
- Count the shaded complete square.
Complete shaded squares = a = 10 cm².
- Now calculate incomplete shaded squares( more than half shaded)
Incomplete shaded squares = b = 11 cm²
- Ignore less than half shaded squares.
- Now add ‘a’ and ‘b’.
Total area = a + b = 10 + 11
Total area = 21 cm².
The area of irregular lamina is 21 cm².
6.) A piece of brass of volume 30 cm³ has a mass of 252 g. find the density of brass in (i) g/cm³. (ii) kg/m³
Answer:
Given, volume of piece of brass = 30 cm³,
Mass of piece of brass = 252 g.
As we know,
Density = mass/ volume
Density = 252/30
Density = 8.4 g/cm³.
The density of piece of brass = 8.4 g/cm³.
As we know that,
1 g/cm³ = 1000 kg/m³.
Therefore ,
8.4 g/cm³ = 8.4 × 1000 kg/m³.
8.4 g/cm³ = 8400 kg/m³
The density of piece of brass is 8.4 g/cm³ or 8400 kg/m³.
7.) The mass of an iron ball is 312 g. The density of iron is 7.8 g/cm³. find the volume of the ball.
Answer: Given,
Mass of ball = 312 g, density of iron ball = 7.8 g/cm³.
As we know that,
Density = mass/volume
Volume = mass / density
Volume = 312/7.8
Volume = 40 cm³
The volume of that iron ball is 40 cm³.
8.) A cork has a volume 25 cm³. The density of cork is 0.25 g/cm³. find the mass of the cork.
Answer: Given,
Volume of cork = 25 cm³,
Density of cork = 0.25 g/cm³.
As we know,
Density = mass/volume.
Mass = density × volume.
Mass = 0.25 × 25
Mass = 6.25 g.
The mass of that cork is 6.25 g.
9.) The mass of 5 litre of water is 5 kg. Find the density of water in g/cm³
Answer: Given,
Mass = 5 kg = 5 × 10³ g,
Volume = 5 liter = 5 × 10³ cm³.
As we know that
Density = mass/volume.
Density = (5 × 10³)/(5 × 10³)
Density = 1 g/cm³.
The density of that mass is 1 g/cm³.
10.) 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/m³
Answer: Given, side of cube of a tank = 1 m.
Mass of cube of tank = 800 kg.
Now we find the volume of cubical tank.
Volume of a cube = l³
Volume of a cube = 1³
Volume of a cube = 1 m³.
Now we find the density of that cubical tank.
As we know,
Density = mass/volume.
Density = 800/1
Density= 800 kg/m³.
The density of that cubical tank is 800 kg/m³.
11.) A block of iron has dimensions 2 m x0.5 m x 0.25 m. The density of iron is 7.8 g/cm³. Find the mass of block.
Answer: Given,
Length of block = 2 m,
Breadth = 0.5 m, height = 0.25 m.
Density of iron block = 7.8 g/cm³.
We know that,
1 g /cm³ = 10³ kg/m³.
Therefore,
density= 7.8 g/cm³ = 7.8 × 10³ kg/m³.
Firstly we calculate the volume of that iron block.
As we know that,
Volume of cuboid = length × breadth × height
Volume of a cuboid = 2 × 0.5 × 0.25
Volume of a cuboid = 0.25 m³
Now we find mass of that iron block.
We know that,
Density = mass/volume
Mass = density × volume.
Mass = 7.8 × 10³ × 0.25
Mass = 1.95 × 10³ kg
Mass = 1950 kg
The mass of that iron block is 1950 kg.
12.) 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/m³.
Answer: Given,
Mass of lead piece = 115 g = 0.115 kg
We know that, when an object immersed into water then it displaced the level of water upto the volume of that object.
Thus,
volume of the lead piece = difference in level of water.
Volume of water level = 30 mL- 20 mL
Volume of water level = 10 mL.
The volume of lead piece is 10 mL or 10 cm³. (1 ml = 1 cm³).
We have to convert this volume into m³.
10 cm³ = 10 × 10-6 m³
Now we calculate the density of lead piece.
We know that
Density = mass /volume
Density = 0.115/(10× 10-6)
Density =0.115 × 10⁵
Density = 11500 kg/m³
The volume of that lead piece is 10 cm³ and density is 11500 kg/m³.
13.) The density of copper is 8.9 g/cm³. What will be its density in kg/m³ ?
Answer:
Density of copper = 8.9 g/cm³.
We know that,
1 g/cm³ = 10³ kg/m³.
Thus,
8.9 g/cm³ = 8.9 × 10³ kg/m³
8.9 g/cm³ = 8900 kg/m³.
14) A car travels a distance of 15 km in 20 minute. find the speed of the car in (i) km/h , (ii) m/s.
Answer: Given,
Distance = 15 km,
Time=20 minutes = 20/60 hours
Time = 1/3 hours.
As,
Speed = distance/time
Speed = 15/(1/3)
Speed = 45 km/h.
Now we concert this speed onto m/s.
As we know,
1 km= 10000 m and 1 hour = 3600 seconds.
Thus,
1 km/h = 1000/3600 m/s.
1km/h = 5/18 m/s.
Therefore,
45 km/h = 45 × (5/18) m/s.
45 km/h = 12.5 m/s.
Thus, the speed of that car is 45 km/h or 12.5 m/s.
15.) How long a train will take to travel a distance of 200 km with a speed of 60 km /h?
Answer: Given,
Speed = 60 km/h.
Distance = 200 km.
As we know,
Speed = distance / time
Time = distance/ speed
Time = 200/60
Time = 10/3 hours.
Time = (10/3)× 60
Time = 200 minutes or 3 hours 20 minutes.
Time required for train to cover 200 km at the rate of 60 km/h is 3 hours 20 minutes.
16.) A boy travels with a speed of 10 m/s for 30 minute. How much distance does he travel ?
Answer: Given,
Speed = 10 m/s,
Time = 30 minutes = 1800 seconds.
As,
Speed = distance /time
Distance = speed × time
Distance = 10 × 1800
Distance = 18000 m.
The boy covers 18000 m or 18 km in 30 minutes.
17) Express 36 km/h in m/s.
Answer: Given, Speed = 36 km/h.
As we know,
1 km= 10000 m and 1 hour = 3600 seconds.
Thus,
1 km/h = 1000/3600 m/s.
1km/h = 5/18 m/s.
Therefore,
36 km/h = 36 × (5/18) m/s.
36 km/h = 10 m/s.
18) Express 15 m/s in km /h.
Answer: Given, Speed = 15m/s.
As we know,
1 km= 10000 m and 1 hour = 3600 seconds.
Thus,
1 km/hour = 1000/ 3600 m/s.
1 km/hour = 5/18 m/s.
18 /5 km/hour = 1 m/s.
1 m/s = 18/5 km/h
Thus,
15 m/s = 15 × 18/5 km/h
15 m/s = 54 km/h