Fun with magnets

1. How will you do this?

Question a.
Determine whether a material is magnetic or non-magnetic.
Answer:

1. To determine whether the material is magnetic or non-magnetic, a magnet is moved over it.
2. If the material sticks to the magnet, it is called magnetic material.
3. If the material does not stick to the magnet, it is non-magnetic.

Question b.
Explain that a magnet has a certain magnetic field.
Answer:

1. The space around a magnet in which the magnetic force is active is called the magnetic field.
2. Place a white paper on a drawing board and place a bar magnet in the middle of the paper.
3. Spread the iron filings on the sheet and gently tap the sheet.
4. The iron filings arrange around the magnet in definite curved lines forming a symmetric pattern.
5. The lines are closer to each other near the poles and less crowded in the middle region around the magnet.
6. Beyond a particular region, the iron filings, will not get attracted.
7. The region where iron filings are attracted is the magnetic field of the magnets.

Question c.
Find the north pole of a magnet.
Answer:

1. Take a bar magnet. Tie a thread to the centre of a bar magnet and hang it from a stand.
2. Note the direction in which the magnet settles and turn it around again.
3. Allow it to settle and note the direction.
4. The end of the magnet that points to the north is called the north pole, while the end that points to the south is called the south pole.
5. The north pole is indicated by ‘N’ and the south pole by ‘S’.

2. Which magnet will you use?

Question a.
Iron is to be separated from a trash.
Answer:

1. Sharp and heavy iron scrap material is attached to a big disc.
2. The disc is a magnet and all scrap is attracted to it.
3. It is not possible to create, store such a big size magnet. Therefore magnetism is induced in the disc with the help of electricity.

An electromagnet is used which is attached to a crane for loading and unloading, transporting scrap and loose iron material from a trash.

Question b.
You are lost in a forest.
Answer:

1. If we are lost in a forest, we should take help of a mariner’s compass which will help us to find the directions while travelling through unknown regions.
2. If mariner’s compass is not available, a bar magnet when suspended in the centre will rest in north-south direction.

Question c.
A window shutter opens and shuts continuously in the wind.
Answer:
A bar magnet can be attached to the window pane so that the window will be closed tight during strong winds also.

3. Fill in the blanks with appropriate word.

Question a.
If a bar magnet is hung by a thread tied at its centre, its north pole becomes steady in the direction of the …………… pole of the earth. (south, north, east, west)
Answer:
North

Question b.
If a bar magnet is cut into equal pieces by cutting it at right angles to its axis at two pieces …………… bar magnets are formed, and a total of …………… poles are formed. (6,3,2)
Answer:
3, 6

Question c.
There is a repulsion between the …………… poles of a magnet and attraction between its ……………. poles. (opposite, like.)
Answer:
like, opposite

Question d.
When magnetic material is taken close to a magnet, the material acquires …………… . (permanent magnetism, induced magnetism, temporary, magnet keeper)
Answer:
induced magnetism

Question e.
If a magnet attracts a piece of metal, that piece must be made of ………… .(any other metal but iron, magnetic material or iron, non-magnetic material, electromagnets)
Answer:
magnetic material or iron

Question f.
A magnet remains steady in a ………….. direction. (east-west, north-south,)
Answer:
north-south

4. Write the answers in your words.

Question a.
How is an electromagnet made?
Answer:
1. To make an electromagnet we need the following apparatus; An iron nail of 10 cm length, 1 metre long insulated copper wire, a battery cell, pins.
2. Wind the copper wire around the nail as shown in the figure. Connect the free ends of the wire to the two terminals of a cell through a plug key.
3. Close the key to complete the circuit.
4. Bring small pins near the tip of the nail and observe.
5. When the circuit is completed, the iron pins are attracted by the nail and hence, they stick to the nail.

6. When the circuit is broken, the pins fall off.
7. A magnet is prepared by passing an electric current through an insulated wire wound around the iron nail. This is an electromagnet.
8. When the current is allowed to pass, the nail becomes a magnet and attracts pin / pins stick to it.
9. When the current is put off the nail does not behave as a magnet and therefore, pins fall off.
10. The magnetism is temporary in the case of an electromagnet.

Question b.
Write the properties of a magnet.
Answer:
Magnet possess following properties/ characteristics.

1. Magnet always settles in the north-south direction.
2. The magnetic force is concentrated at the two ends or poles of a magnet.
3. If a magnet is divided into two parts, two independent magnets are formed. It means that the two poles of a magnet cannot be separated from each other.
4. A magnetic material acquires magnetism when placed near a magnet. This magnetism is called induced magnetism.
5. There is repulsion between like poles of a magnet, while there is attraction between the opposite poles.

Question c.
What are the practical uses of a magnet?
Answer:

1. Magnets are materials to which objects made from iron, nickel, cobalt are attracted. But man explored magnets and its properties and made his life comfortable.
2. Permanent Magnets: are used in caps of pin holders, doors of fridges, doors of cupboards etc.
3. Temporary magnets: Electromagnets are used in electric bells, circuit of various machines, ATM card swipe machines, MRI- Magnetic Resonance Image, loudspeakers, electric cranes, microphones, Mariner’s compasses, etc.

Activity

Question 1.
Collect information regarding how the various magnets used in our day-to-day tasks are produced.

Question 2.
Collect information about the magnetism of the earth.

Fill in the blanks.

Question 1.
Iron objects ………….. to a magnet.
Answer:
stick

Question 2.
A magnet is used in ………….. and ………….. .
Answer:
gadgets and machines

Question 3.
The materials that stick to a magnet are called ………….. materials.
Answer:
magnetic

Question 4.
Materials that do not stick to a magnet are called ………….. materials.
Answer:
non-magnetic

Question 5.
When a magnet attracts an object, that object is ………….. due to the magnetic force.
Answer:
displaced

Question 6.
Magnetism is a form of ………….. .
Answer:
energy

Question 7.
A magnet always settles in the ………….. direction.
Answer:
north-south

Question 8.
The north pole is indicated by’ …………… and the south pole by ‘……………’.
Answer:
‘N’-‘S’

Question 9.
The end of the magnet that points to the north is called the ………….. .
Answer:
Northpole

Question 10.
The end of the magnet that points to the south is called the ………….. .
Answer:
South pole

Question 11.
The magnetic force is concentrated at the two ends or ………….. of a magnet.
Answer:
poles

Question 12.
If a magnet is divided into two parts, two ………….. magnets are formed.
Answer:
independent

Question 13.
It means that the two poles of a magnet cannot be ………….. from each other.
Answer:
separated

Question 14.
A magnetic material acquires magnetism when placed near a
Answer:
magnet

Question 15.
Iron filling stick to the iron bar when the ………….. is near it.
Answer:
magnet

Question 16.
There is ………….. between like poles of a magnet.
Answer:
repulsion

Question 17.
There is ………….. between the opposite poles of a magnet.
Answer:
attraction

Question 18.
Magnetic objects ………….. magnetism.
Answer:
induce

Question 19.
Material ………….. is a mixture of aluminium, nickel and cobalt.
Answer:
Alnico

Question 20.
………….. magnets are made from a mixture of nickel, cobalt and iron.
Answer:
Permanent

Question 21.
The bar of soft or pure iron which protects a magnet is called ………….. .
Answer:
magnet keeper

Question 22.
Magnetism gets ………….. when a magnet is heated, thrown, knocked about or broken into pieces.
Answer:
destroyed

Question 23.
Electromagnetic energy is used in our ………….. life.
Answer:
day-to-day

Question 24.
The metals iron, cobalt, nickel are ………….. materials.
Answer:
magnetic

Question 25.
………….. is a natural magnet.
Answer:
Magnetite

Match the columns.

Question a.

Column ‘A’	Column ‘B’
1. Iron, nickel, cobalt	(a) Electromagnet
2. Door bell magnet	(b) Permanent magnet
3. Nickel, cobalt, aluminium	(c) Magnetic metal
4. Cupboard magnet	(d) Mariner’s compass
5. Lodestone	(e) Alnico

Answer:

Column ‘A’	Column ‘B’
1. Iron, nickel, cobalt	(c) Magnetic metal
2. Door bell magnet	(a) Electromagnet
3. Nickel, cobalt, aluminium	(e) Alnico
4. Cupboard magnet	(b) Permanent magnet
5. Lodestone	(d) Mariner’s compass

State whether true or false. If false, correct the statement.

Question 1.
Material alnico is a mixture of aluminium, nickel and iron.
Answer:
False: Material alnico is a mixture ofaluminium, nickel and cobal.

Question 2.
Magnetism of electromagnet is permanent.
Answer:
False: Magnetism of electromagnet is temporary.

Question 3.
The bar of soft or pure iron protects the magnet.
Answer:
True

Question 4.
Like poles attract each other and unlike poles repel each other.
Answer:
False: Like poles repel each other and unlike poles attract each other.

Question 5.
The magnetic force is concentrated at the centre of the magnet.
Answer:
False: The magnetic force is concentrated at the poles of the magnet

Question 6.
Magnetism is a kind of energy.
Answer:
True

Question 7.
Mariner’s compass is used for finding directions while travelling.
Answer:
True

Question 8.
Cobalt is a magnetic material.
Answer:
True

Question 9.
The north pole is indicated by ‘S’ and the south pole is indicated by ‘N’.
Answer:
False: The north pole is indicated by Wand the south pole is indicated by ‘S’.

Question 10.
Electromagnetism is used in many places in our day-to-day life.
Answer:
True

Answer the following questions in one sentence.

Question 1.
What is a magnet?
Answer:
The material to which objects made from iron, nickel, cobalt get attracted is called as magnet.

Question 2.
What is magnetism?
Answer:
The property of a material to which objects made from iron, nickel, cobalt get attracted is called as magnetism.

Question 3.
What are magnetic materials?
Answer:
Materials that stick to a magnet are called magnetic materials, e.g. cobalt, nickel, iron.

Question 4.
What are non-magnetic materials?
Answer:
Materials that do not stick to a magnet are called non-magnetic material, e.g. plastic, rubber, glass etc.

Question 5.
What are lodestones?
Answer:
Lodestones are leading stones which are used for finding the directions while travelling through unknown regions.

Question 6.
How is magnetism a kind of energy?
Answer:
Work is done by magnetic force. Thus, magnetism is a kind of energy.

Question 7.
What is an electromagnet?
Answer:
When magnetism is produced in the iron due to the electric current, it is called an electromagnet.

Question 8.
How are permanent magnets made?
Answer:
Permanent magnets are made from a mixture of nickel, cobalt and iron.

Question 9.
List the instruments where electromagnets are used.
Answer:
Electromagnets are used in doorbells, cranes, loudspeakers, voltameters, TVs, antennas, radios etc.

Question 10.
How is magnetism destroyed?
Answer:
When magnets are heated, thrown, knocked about or broken into pieces, magnetism gets destroyed.

Question 11.
What is a magnet keeper?
Answer:
A magnet keeper is a bar of soft or pure iron which protects a magnet. It is a piece of soft iron placed in the box in which a magnet is kept.

Question 12.
Magnets exist in variety of shapes.
Answer:
Today, magnets are used in many machines, gadgets and devices. They are all man-made. Hence, they can have a variety of shapes depending upon their use.

Answer the following briefly.

Question 1.
What are leading stones?
Answer:

1. It was known quite long ago to the people in China and Europe that a piece of magnetite, hung freely always settled in the north-south direction.
2. These rocks then came to be used for finding the directions while travelling through unknown regions.
3. That is why they are called leading stones or Lodestones.

Question 2.
What has led to the invention of the Mariner’s compass?
Answer:
Leading stones have led to the invention of the mariner’s compass.

Question 3.
List the different shapes of magnets.
Answer:

1. Magnets have a variety of shapes depending on their uses.
2. They are bar magnets, disc magnets, horseshoe magnets, ring shaped magnets, cylindrical magnets, and small button magnets.

Question 4.
What are permanent magnets?
Answer:

1. Magnets which do not lose their magnetism easily are called permanent magnets or Magnets which are made up of magnetic substances are permanent magnets.
2. e.g. Magnets fixed in a pin holder, magnets of a door of a cupboard are permanent magnets.
3. Permanent magnets are made from a mixture of
   • Nickel, cobalt, iron
   • Aluminium, nickel, cobalt – alnico

Give scientific reasons.

Question 1.
Why is it important to place a magnet keeper in a box along with magnets?
Answer:
Magnetism gets destroyed when a magnet is heated, thrown, knocked about or broken into pieces. A magnet keeper which is a bar of soft or pure iron protects a magnet.

Question 2.
Cranes with magnets are used.
Answer:
When a magnet attracts an object, that object is displaced due to the magnetic force. In factories, ports, garbage depots, large objects are lifted and shifted from place to place using cames. Hence cranes are fitted with magnets.

Can you tell?

Question 1.
Pins in a pin holder do not fall? While we are shutting the door of a fridge, we find that it closes automatically from certain distance and does not open unless pulled again.
Answer:
Magnet is fitted in the cap of a pin holder and in the door of a fridge. Iron objects stick to the magnet.

Question 2.
Take a magnet from the laboratory and bring it near various objects in your use. Which of them stick to the magnet? What material is each of them made of? Observe these things carefully. Classify the objects into two groups: those which stick to the magnet, those which do not.
Comb, table, cupboard – iron, spoon, scissors, pen, pencil, eraser, books, mobile, laptops, glass bangles, hair pin, cupboard handle, chair, steel lunch box, magnetic stickers, toys, gold ring.
Answer:

Stick to the magnet	Doesn’t stick to the magnet
Iron cupboard, spoon, scissors, hairpin, steel lunch box, magnetic stickers	Comb, table, pen, pencil, eraser, books, glass bangles, chair, mobile, laptops, cupboard handle, toys, gold ring

Question 3.
Take a mixture of sand, pieces of paper, sawdust, iron filings and pins in a saucer and pass a magnet around the mixture. What do you see?
Answer:
When magnet is moved over a mixture of sand, pieces of paper, sawdust, iron filings and pins, pins and iron filings will cling to the magnet. Sand, sawdust and pieces of paper will remain behind.

Question 4.
How is a Mariner’s Compass used?
Answer:

1. A Mariner’s Compass is a magnetic needle used in navigation to show direction by deflections.
2. It is a direction-finding instrument used in navigation.
3. It is placed on the maps, grounds, decks as it will point to the magnetic north pole.
4. It has two or more magnets permanently attached to a compass card which moves freely on a pivot.
5. The needle fixed on the compass bowl indicates the ship’s heading position.

Question 5.
Find out where the magnet given are used?
Answer:

Magnets	Uses
Horseshoe magnet	used in electric bell
Circular magnet	used in loudspeaker.
Magnetic needle	used in Mariner’s Compass.
Disc magnets	used in toys
Bar magnets	used in cupboard doors
Button magnet	supporting side rails or blockouts
Square magnet	Industries
Arc magnet	Electric motors and generators.
Cylindrical magnet	used in medicine, used in treatment of scoliosis patients.

Question 6.
Identify the different types of magnets as shown in the picture below.
Answer:
a. Circular magnet
b. Cylindrical magnet
c. Horseshoe magnet
d. Bar magnet

Std. 7 _Perimeter & Area

Question 1:

If the length and breadth of a rectangle are doubled, how many times the perimeter of the old rectangle will that of the new rectangle be?

ANSWER:

Let the length of old rectangle = l
Let the breadth of old rectangle = b
Perimeter of the old rectangle = 2(length + breadth) = 2(l + b)
When the length and the breadth of the rectangle are doubled, then
length of the new rectangle = 2l
breadth of the new rectangle = 2b
∴ Perimeter of new rectangle = 2(length + breadth)
= 2(2l + 2b)
= 2 × 2(l + b)
= 2 × perimeter of the old rectangle    [∵ perimeter of old rectangle = 2(l + b)]
Hence, the perimeter of the new rectangle will become two times of the perimeter of the old rectangle.

Page No 80:

Question 2:

If the side of a square is tripled, how many times the perimeter of the first square will that of the new square be?

ANSWER:

Let the length of each side of the old square = s
Then, perimeter of the old square = 4 × side = 4 × s = 4s
When the length of each side of the square is tripled, then
length of each side of the new square = 3s
∴ perimeter of the new square = 4 × side 
= 4 × 3s
= 3 × 4s
= 3 × perimeter of the old square    [∵ perimeter of the old square = 4s]
Hence, the perimeter of the new square will become three times of the perimeter of the old square. 

Page No 80:

Question 3:

Given alongside is the diagram of a playground. It shows the length of its sides. Find the perimeter of the playground.

ANSWER:

It is known that, perimeter of a polygon is equal to sum of the lengths of all the sides of the polygon.
Let us mark the vertices of the given polygon as A, B, C, D, E and F.
The given figure can be broken into rectangle and square by drawing a line CG || BA.

Now, ABCG is a rectangle and GDEF is a square.
Since, opposite sides of rectangle are equal, then
GC = AB = 10 m and AG = BC = 15 m
Since length of each sides of square is equal, then
GD = DE = EF = FG = 15 m
Perimeter of ABCDEFG = AB + BC + CD + DE + EF + FG + GA
= 10 + 15 + 5 + 15 + 15 + 15 + 15
= 90 m

Page No 80:

Question 4:

As shown in the figure, four napkins all of the same size were made from a square piece of cloth of length 1 m. What length of lace will be required to trim all four sides of all the napkins ?

ANSWER:

Length of each side of a square piece of cloth = 1 m
Now, four napkins all of same size were made from this square piece of cloth.
∴ Length of each side of a napkin, s = Length of each side of a square piece of cloth ÷ 2
= 1 ÷ 2
= 0.5 m
Length of lace needed to trim all the four sides of a napkin = perimeter of a napkin
= 4 × s
= 4 × 0.5
= 2 m
∴ Length of lace needed to trim all the four sides of four napkins = 4 × length of lace needed for a napkin
= 4 × 2
= 8 m

Page No 82:

Question 1:

If the side of a square is 12 cm, find its area.

ANSWER:

Length of each side of square, s = 12 cm
∴ Area of square = s²
= (12)²
= 12 × 12
= 144 cm²
Thus, the area of the square is 144 cm².

Page No 82:

Question 2:

If the length of a rectangle is 15 cm and breadth is 5 cm, find its area.

ANSWER:

Length of the rectangle, l = 15 cm
Breadth of the rectangle, b = 5 cm
∴ Area of rectangle = l × b
= 15 × 5
= 75 cm²
Hence, the area of rectangle is 75 cm².

Page No 82:

Question 3:

The area of a rectangle is 102 sqcm. If its length is 17 cm, what is its perimeter ?

ANSWER:

Area of rectangle = 102 cm²
Length of the rectangle, l = 17 cm
∴ Breadth of the rectangle, b = area of rectangle÷ length of rectangle  (∵ area of rectangle = length × breadth)

⇒b = 102÷17 = 6 cm
Perimeter of rectangle = 2(length + breadth)
= 2(17 + 6)
= 2 × 23
= 46 cm

Page No 82:

Question 4:

If the side of a square is tripled, how many times will its area be as compared to the area of the original square ?

ANSWER:

Let the length of each side of original square = s
Area of original square = side × side = s²
When the side of the square is tripled, then
length of each side of new square = 3s
Area of new square = side × side
= 3s × 3s
= 9s²
= 9 × s²
= 9 × area of original square    [∵ area of original square = s²]
Hence, the area of the new square will become nine times the area of the original square.

Page No 84:

Question 1:

A page of a calendar is 45 cm long and 26 cm wide. What is its area ?

ANSWER:

Length of the calendar, l = 45 cm
Breadth of the calendar, b = 26 cm
∴ Area of calendar = l × b
= 45 × 26
= 1170 cm²

Page No 84:

Question 2:

What is the area of a triangle with base 4.8 cm and height 3.6 cm ?

ANSWER:

We have, base of a triangle = 4.8 cm
Height of a triangle = 3.6 cm
∴ Area of a triangle = 12$\frac{1}{2}$ × base × height
= 12$\frac{1}{2}$ × 4.8 × 3.6
= 8.64 cm²

Page No 84:

Question 3:

What is the value of a rectangular plot of land 75.5 m long and 30.5 m broad at the rate of 1000 rupees per square metre ?

ANSWER:

Length of the rectangular plot of land, l = 75.5 m
Breadth of the rectangular plot of land, b = 30.5 m
∴ Area of rectangular plot of land = l × b
= 75.5 × 30.5
= 2302.75 m²
Cost of 1 m² of rectangular plot of land = 1000 rupees
∴ Cost of 2302.75 m² of rectangular plot of land = 1000 × 2302.75
= 2302750 rupees
Hence, the value of the rectangular plot of land is Rs 23,02,750.

Page No 84:

Question 4:

A rectangular hall is 12 m long and 6 m broad. Its flooring is to be made of square tiles of side 30 cm. How many tiles will fit in the entire hall ? How many would be required if tiles of side 15 cm were used?

ANSWER:

Length of the rectangular hall, l = 12 m
Breadth of the rectangular hall, b = 6 m
Area of the floor of the rectangular hall = l × b
= 12 × 6
= 72 m²
Length of each side of square tile used in the first case = 30 cm = 30÷ 100= 0.3 m    (∵ 100 cm = 1 m)
Area of each tile = side × side
= 0.3 × 0.3
= 0.09 m²
∴ Required number of square tiles each of side 30 cm = area of floor ÷ area of each tile 
= 72÷ 0.09
= 800
Length of each side of square tile used in the second case = 15 cm = 15÷ 100= 0.15 m   
 Area of each tile = side × side
= 0.15 × 0.15
= 0.0225 m²
∴ Required number of square tiles each of side 15 cm = area of floor÷ area of each tile
= 72÷ 0.0225
= 3200

Page No 84:

Question 5:

Find the perimeter and area of a garden with measures as shown in the figure alongside.

ANSWER:

Let us mark the vertices of the given polygon as A, B, C, D, E, F, G, H, I, J, K and L.
Perimeter of garden ABCDEFGHIJKL = sum of all the sides of polygon = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL
= 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13
= 143 m

The given figure of the garden can be broken into five squares as follows.



Now, length of each side of each of the five squares is equal to 13 m.
Area of each square = side × side
= 13 × 13
= 169 m²
∴ Area of the five squares = 5 × area of each square
= 5 × 169
= 845 m²
Hence, the area of the given garden is 845 m².

Page No 86:

Question 1:

Find the total surface area of cubes having the following sides.

(i) 3 cm (ii) 5 cm  (iii) 7.2 m (iv) 6.8 m (v) 5.5 m

ANSWER:

(i) Length of each side of cube, l = 3 cm
∴ Total surface area of the cube = 6 × l²
= 6 × (3)²
= 6 × 9
= 54 cm²

(ii) Length of each side of cube, l = 5 cm
∴ Total surface area of the cube = 6 × l²
= 6 × (5)²
= 6 × 25
= 150 cm²

(iii) Length of each side of cube, l = 7.2 m
∴ Total surface area of the cube = 6 × l²
= 6 × (7.2)²
= 6 × 51.84
= 311.04 m²

(iv) Length of each side of cube, l = 6.8 m
∴ Total surface area of the cube = 6 × l²
= 6 × (6.8)²
= 6 × 46.24
= 277.44 m²

(v) Length of each side of cube, l = 5.5 m
∴ Total surface area of the cube = 6 × l²
= 6 × (5.5)²
= 6 × 30.25
= 181.50 m²

Page No 86:

Question 2:

Find the total surface area of the cuboids of length, breadth and height as given below:

(i) 12 cm, 10 cm, 5 cm  (ii) 5 cm, 3.5 cm, 1.4 cm

(iii) 2.5 cm, 2 m, 2.4 m  (iv) 8 m, 5 m, 3.5 m

ANSWER:

(i) Length of the cuboid, l = 12 cm
Breadth of the cuboid, b = 10 cm
Height of the cuboid, h = 5 cm
∴ Total surface area of the cuboid = 2(l × b + b × h + h × l)
= 2(12 × 10 + 10 × 5 + 5 × 12)
= 2(120 + 50 + 60)
= 2 × 230
= 460 cm²

(ii) Length of the cuboid, l = 5 cm
Breadth of the cuboid, b = 3.5 cm
Height of the cuboid, h = 1.4 cm
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(5 × 3.5 + 3.5 × 1.4 + 1.4 × 5)
= 2(17.5 + 4.9 + 7.0)
= 2 × 29.4
= 58.8 cm²

(iii) Length of the cuboid, l = 2.5 cm = 2.5100$\frac{2.5}{100}$ m = 0.025 m    [∵ 100 cm = 1 m]
Breadth of the cuboid, b = 2 m
Height of the cuboid, h = 2.4 m
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(0.025 × 2 + 2 × 2.4 + 2.4 × 0.025)
= 2(0.05 + 4.8 + 0.06)
= 2 × 4.91
= 9.82 m²

(iv)  Length of the cuboid, l = 8 m
Breadth of the cuboid, b = 5 m
Height of the cuboid, h = 3.5 m
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(8 × 5 + 5 × 3.5 + 3.5 × 8)
= 2(40 + 17.5 + 28)
= 2 × 85.5
= 171 m²

Page No 86:

Question 3:

A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height. Its outer sides are to be covered exactly with craft paper. How much paper will be required to do so ?

ANSWER:

Length of the matchbox, l = 4 cm
Breadth of the matchbox, b = 2.5 cm
Height of the matchbox, h = 1.5 cm
∴ Surface area of the box = 2(l × b + b × h + h × l)
= 2(4 × 2.5 + 2.5 × 1.5 + 1.5 × 4)
= 2(10 + 3.75 + 6)
= 2 × 19.75
= 39.5 cm² 
Hence, 39.5 cm² of the craft paper will be needed to cover the matchbox.

Page No 86:

Question 4:

An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box ? The inside and outside surface of the box is to be painted with rust proof paint. At a rate of 150 rupees per sqm, how much will it cost to paint the box?

ANSWER:

Length of the open box, l = 1.5 m
Breadth of the open box, b = 1 m
Height of the open box, h = 1 m
Surface area of open box = Total surface area of the box − Area of the top
= 2(l × b + b × h + h × l) − l × b
= 2(1.5 × 1 + 1 × 1 + 1 × 1.5) − 1.5 × 1
= 2(1.5 + 1 + 1.5) − 1.5
= 2 × 4 − 1.5
= 8 − 1.5
= 6.5 m²
Hence, 6.5 m² of the sheet will be needed to make the open box.
Now, it is given that the inside and the outside surface of the open box is to be painted with rust proof paint.
∴ Total area of the box to be painted = 2 × Surface area of open box
= 2 × 6.5
= 13 m²
Now, cost of painting of 1 m² of area = 150 rupees
∴ Cost of painting of 13 m² of area = 13 × 150 = 1950 rupees
Hence, it will cost 1950 rupees to paint the open box from inside and outside with rust proof paint.

LIGHT (Std.6)

Question 1:

Choose an appropriate word and fill in the blanks.
(a) A ........... is a natural source of light.
(b) A ......... is an artificial source of light.
(c) When light passes through a prism, it gets separated into ....... colours.
(d) The image obtained in the pinhole camera is ......... .
(e) A shadow is formed when an ........... object comes in the way of light.
(f) When a ........... object comes in the way of light, light passes ......... it.
(options : seven, star, through, transparent, opaque, colors, shape, erect, inverted, luminous, candle)

ANSWER:

(a) A star is a natural source of light.
(b) A candle is an artificial source of light.
(c) When light passes through a prism, it gets separated into seven colours.
(d) The image obtained in the pinhole camera is inverted.
(e) A shadow is formed when an opaque object comes in the way of light.
(f) When a transparent object comes in the way of light, light passes through it.

Page No 104:

Question 2:

Write whether of the following objects are luminous or non-luminous.

Object	Luminous/Non-luminous
A book
A burning candle
A wax cloth
A pencil
A pen
A light bulb
A tyre
A torch

ANSWER:

Object	Luminous/Non-luminous
A book	Non-luminous
A burning candle	Luminous
A wax cloth	Non-luminous
A pencil	Non-luminous
A pen	Non-luminous
A light bulb	Luminous
A tyre	Non-luminous
A torch	Luminous

Page No 104:

Question 3:

Match the following.

Group A	Group B
(a) Mirror	(1) Non-luminous
(b) Firefly	(2) Inverted image
(c) Pinhole camera	(3) Reflection
(d) Moon	(4) Luminous

ANSWER:

Group A	Group B
(a) Mirror	(3) Reflection
(b) Firefly	(4) Luminous
(c) Pinhole camera	(2) Inverted image
(d) Moon	(1) Non-luminous

Page No 104:

Question 4:

Write the answers to the following.
(a) What things are necessary for the formation of a shadow ?
(b) When can an object be seen?
(c) What is a shadow?

ANSWER:

(a) The things necessary for the formation of a shadow are:

• A source of light
• An opaque object whose shadow has to be formed
• A screen on which shadow is to be formed

(b) An object is seen when the light rays reflected from its surface reaches our eyes.

(c) Shadow is a dark patch obtained on a screen where the light from a light source is blocked by an opaque object.

SOUND

Question 1: Fill in the blanks with the proper words. 

(a) The propagation of sound does not occur through a ........... . 

(b) Noise pollution is a .......... issue. 

(c) The sound which is disagreeable to the ears is called..... .

 (d) Noise has adverse effects on our .......... .

ANSWER: (a) The propagation of sound does not occur through a vacuum.

(b) Noise pollution is a social issue.

(c) The sound which is disagreeable to the ears is called noise.

 (d) Noise has adverse effects on our health.

 Page No 96: 

Question 2: 

What should we do ?

 (a) The silencer of a motorcycle is broken. (b) A factory in the surroundings is producing continuous loud noise.

 ANSWER: 

a) If the silencer of a motorcycle is broken, then take the motorcycle to a mechanic and get it reinstalled or repaired.

 (b) If a factory in the surroundings is producing continuous loud noise, then we can lodge a complaint against the factory to a local government body citing the troubles faced by the residents around and the danger it is causing to the environment.

 Page No 96:

 Question 3:

 Write the answers in your own words.

 (a) What is meant by vibration ?

 (b) Explain with the help of practical examples how sound is propagated through solids.

 (c) What is meant by noise pollution? 

(d) What measures will you take to control noise pollution ? 

 ANSWER: (a) The to and fro motion of a body about its mean position is known as vibration.

 (b) Sound is propagated in the form of waves in solids. It can be explained clearly from the below examples: Place your ear on one end of a long table. Ask your friend to tap the table from the other end. You can hear the produced sound. Sound reaches your ear after travelling through the table. This indicates that sound can travel through solids and in the form of waves. ​Stethoscope is another example of sound travelling through solids. Doctors use stethoscopes to listen to our heartbeat.

 (c) Noise pollution is the excess of unwanted sound in the environment. It can adversely affect human and other living beings health.

 (d) Following measures can be taken to control noise pollution: To control noise pollution, we must control its source. Hence, silencers must be installed in vehicles such as motorcycles, cars, trucks, buses, and other noise producing machines. We should watch television and listen to music at a low volume. Also, use of loudspeakers as well as horns of buses and trucks should be minimized. Regular maintenance of automobiles should be done so that noise produced by them can be kept under check. All industrial work should be done away from residential areas. More trees should be planted in residential areas as they help in reducing noise.

Direct Proportion

Question 1:

If 7 kg onions cost 140 rupees, how much must we pay for 12 kg onions?

ANSWER:

Let us suppose the cost of 12 kg onions is x rupees.
The number of onions and their cost vary in direct proportion.
∴7140=12x⇒x=12×1407$$\begin{gathered} \therefore \frac{7}{140}=\frac{12}{x} \\ \Rightarrow x=\frac{12\times 140}{7} \end{gathered}$$
= 240 rupees
Hence, the cost of 12 kg onions is 240 rupees.

Page No 64:

Question 2:

If 600 rupees buy 15 bunches of feed, how many will 1280 rupees buy?

ANSWER:

Let us suppose x bunches of feed can be bought in 1280 rupees.
The number of bunches of feed and their cost vary in direct proportion.
∴15600=x1280⇒x=15×1280600$$\begin{gathered} \therefore \frac{15}{600}=\frac{x}{1280} \\ \Rightarrow x=\frac{15\times 1280}{600} \end{gathered}$$
= 32
Hence, 32 bunches of feed can be bought in 1280 rupees.

Page No 64:

Question 3:

For 9 cows, 13 kg 500 g of food supplement are required every day. In the same proportion, how much will be needed for 12 cows?

ANSWER:

Let us suppose x kg of food supplement required for 12 cows.
The quantity of food supplement and the number of cows vary in direct proportion.
∴913.5=12x⇒x=12×13.59$$\begin{gathered} \therefore \frac{9}{13.5}=\frac{12}{x} \\ \Rightarrow x=\frac{12\times 13.5}{9} \end{gathered}$$
= 18 kg
Hence, 18 kg of food supplement required for 12 cows.

Page No 64:

Question 4:

The cost of 12 quintals of soyabean is 36,000 rupees. How much will 8 quintals cost?

ANSWER:

Let us suppose the cost of 8 quintals of soyabean is x rupees.
The quantity of soyabeans and their cost vary in direct proportion. 
∴123600=8x⇒x=8×360012$$\begin{gathered} \therefore \frac{12}{3600}=\frac{8}{x} \\ \Rightarrow x=\frac{8\times 3600}{12} \end{gathered}$$
= 24000 rupees
Hence, the cost of 8 quintals of soyabean is 24000 rupees.

Page No 64:

Question 5:

Two mobiles cost 16,000 rupees. How much money will be required to buy 13 such mobiles ?

ANSWER:

Let us suppose the cost of 13 mobiles is x rupees.
The number of mobiles and their cost vary in direct proportion.
∴216000=13x⇒x=13×160002$$\begin{gathered} \therefore \frac{2}{16000}=\frac{13}{x} \\ \Rightarrow x=\frac{13\times 16000}{2} \end{gathered}$$
= 104000 rupees
Hence, the cost of 13 mobiles is 104000 rupees.

Page No 66:

Question 1:

Five workers take 12 days to weed a field. How many days would 6 workers take ? How many would 15 take ?

ANSWER:

Let us suppose 6 workers will take x days to weed a field. 
As the number of workers increases, the number of days decreases.
So, the number of workers and number of days are in inverse proportion. 
∴ 5 × 12 = 6 × x
⇒x=606$\Rightarrow x=\frac{60}{6}$
⇒ x = 10 days
Let us suppose 15 workers will take y days to weed a field. 
∴ 5 × 12 = 15 × y
⇒y=6015$\Rightarrow y=\frac{60}{15}$
⇒ y = 4 days
Hence, 6 workers will take 10 days, while 15 workers will take 4 days to weed a field.

Page No 66:

Question 2:

Mohanrao took 10 days to finish a book, reading 40 pages every day. How many pages must he read in a day to finish it in 8 days?

ANSWER:

Let us suppose Mohanrao will have to read x pages every day to finish the book in 8 days.
As the number of days decreases, the number of pages increases.
So, the number of days and number of pages are in inverse proportion. 
∴ 10 × 40 = 8 × x
⇒x=400/8$\Rightarrow x=\frac{\mathrm{400/}}{8}$
⇒ x = 50 pages
Hence, Mohanrao will have to read 50 pages every day to finish the book in 8 days

Page No 66:

Question 3:

Mary cycles at 6 km per hour. How long will she take to reach her Aunt’s house which is 12 km away ? If she cycles at a speed of 4 km/hr, how long would she take ?

ANSWER:

Given:
Case -1:
Speed = 6 km/hr
Distance = 12 km
∴Time=DistanceSpeed=126$$\begin{gathered} \therefore \mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}} \\ =\frac{12}{6} \end{gathered}$$
= 2 hours
Case -2:
​Speed = 4 km/hr
Distance = 12 km
∴Time=DistanceSpeed=124$$\begin{gathered} \therefore \mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}} \\ =\frac{12}{4} \end{gathered}$$
= 3 hours
Hence, if the speed of cycle is 6 km/hr then, Marry will take 2 hours and if the speed of cycle is 4 km/hr then, she will take 3 hours to reach her Aunt’s house.

Page No 66:

Question 4:

The stock of grain in a government warehouse lasts 30 days for 4000 people. How many days will it last for 6000 people ?

ANSWER:

Let us suppose the stock of grain in a government warehouse lasts x days for 6000 people.
As the number of people increases, the number of days decreases.
So, the number of days and number of people are in inverse proportion. 
∴ 30 × 4000 = 6000 × x
⇒x=1200006000$\Rightarrow x=\frac{120000}{6000}$
⇒ x = 20 days
Hence, the stock of grain in a government warehouse lasts 20 days for 6000 people.

PARTNERSHIP 

Page No 68:

Question 1:

Suresh and Ramesh together invested 144000 rupees in the ratio 4:5 and bought a plot of land. After some years they sold it at a profit of 20%. What is the profit each of them got?

ANSWER:

The proportion of Suresh's and Ramesh’s investment is 4:5.
The profit is shared in the same proportion as the investment, hence, proportion of profit is 4:5.
Now, profit = 20100×144000$\frac{20}{100}\times 144000$
= 28800 rupees
Therefore, the profit of Suresh and Ramesh  is given by
Suresh's profit = 49×28800$\frac{4}{9}\times 28800$
= 12800 rupees
Ramesh's profit = 59×28800$\frac{5}{9}\times 28800$
= 16000 rupees
Hence, Suresh and Ramesh got the profit of 12800 and 16000 rupees respectively.

Page No 68:

Question 2:

Virat and Samrat together invested 50000 and 120000 rupees to start a business. They suffered a loss of 20%. How much loss did each of them incur ?

ANSWER:

The proportion of Virat's and Samrat’s investment is given by
50000:120000 = 5:12
The loss is shared in the same proportion as the investment, hence, proportion of profit is 5:12.
Now, Loss = 20100×(50000+120000)$\frac{20}{100}\times \left(50000+120000\right)$
20100×(170000)$\frac{20}{100}\times \left(170000\right)$
= 34000 rupees
Therefore, the loss incurred by Virat and Samrat is given by
Virat's loss = 517×34000$\frac{5}{17}\times 34000$
= 10000 rupees
Samrat's loss = 1217×34000$\frac{12}{17}\times 34000$
= 24000 rupees
Hence, Virat and Samrat incurred the loss of ​10000 and 24000 rupees respectively.

Page No 68:

Question 3:

Shweta, Piyush and Nachiket together invested 80000 rupees and started a business of selling sheets and towels from Solapur. Shweta’s share of the capital was 30000 rupees and Piyush’s 12000. At the end of the year they had made a profit of 24%. What was Nachiket’s investment and what was his share of the profit?

ANSWER:

Nachiket’s investement = Total investment − (Shweta’s investment + Piyush’s investment)
= 80000 − (30000 + 12000)
= 80000 − 42000
= 38000 rupees
The proportion of Shweta's, Piyush's and Nachiket's investment is given by
30000:12000:38000 = 15:6:19
The profit is shared in the same proportion as the investment, hence, proportion of profit is 15:6:19.
Now, Profit = 24100×(80000)$\frac{24}{100}\times \left(80000\right)$
= 19200 rupees
Therefore, Nachiket’s share of the profit is given by
= 1940×19200$\frac{19}{40}\times 19200$
= 9120 rupees
Hence, Nachiket’s investment and his share of the profit are ​38000 and 9120 rupees respectively.

Page No 68:

Question 4:

A and B shared a profit of 24500 rupees in the proportion 3:7. Each of them gave 2% of his share of the profit to the Soldiers’ Welfare Fund. What was the actual amount given to the Fund by each of them?

ANSWER:

Amount of share to the Soldiers’ Welfare Fund = 2% of 24500
= 490 rupees
The profit is shared in the proportion 3:7.
Therefore, A’s share to the Fund is given by
= 310×490$\frac{3}{10}\times 490$
= 147 rupees
Therefore, B’s share to the Fund is given by
= ​710×490$\frac{7}{10}\times 490$
= 343 rupees
Hence, A's and B's share to the fund are 147 and 343 rupees respectively.

Page No 68:

Question 5:

Jaya, Seema, Nikhil and Neelesh put in altogether 360000 rupees to form a partnership, with their investments being in the proportion 3 : 4 : 7 : 6. What was Jaya’s actual share in the capital ? They made a profit of 12%. How much profit did Nikhil make ?

ANSWER:

Total investment = 360000 rupees
Total profit = 12100×360000$\frac{12}{100}\times 360000$
= 43200 rupees
The profit is shared in the same proportion as the investment, hence, proportion of profit is 3:4:7:6.
Jaya's share is given by
320×360000$\frac{3}{20}\times 360000$
= 54000 rupees
Nikhil's share in the profit is given by
720×43200$\frac{7}{20}\times 43200$
= 15120 rupees
Hence, Jaya's share and Nikhil's profit are 54000 and 15120 rupees respectively.

Circle

Question 1:

Complete the table below

Sr. No.	Radius (r)	Diameter (d)	Circumference (c)
(i)	7 cm	....................	.................
(ii)	............	28 cm	..............
(iii)	..............	...............	616 cm
(iv)	...........	.............	72.6 cm

ANSWER:

(i) Radius, r = 7 cm
 Diameter, d = 2r = 2 × 7 = 14 cm
∴ Circumference, c = πd 
= 22/7 × 14
= 22 × 2
= 44 cm

(ii) Diameter, d = 28 cm
Radius, r = d2$\frac{d}{2}$ = 282$\frac{28}{2}$ = 14 cm
∴ Circumference, c = 2πr
= 2 × 22/7 × 14
= 88 cm

(iii) Circumference, c = 616 cm
Now, c = 2πr     (where 'r' is the radius)
⇒616 = 2 × 22/7× r
⇒r = 616 x 7/44
⇒r = 98
So, radius = 98 cm
Diameter, d = 2r = 2 × 98 = 196 cm

(iv) Circumference, c = 72.6 cm
Now, c = 2πr     (where 'r' is the radius)
⇒72.6 = 2 × 22/7 × r
⇒r = 72.6 × 7/44
⇒r = 11.55
So, radius = 11.55 cm
Diameter, d = 2r = 2 × 11.55 = 23.1 cm

The complete table is shown below.
 

Sr. No.	Radius (r)	Diameter (d)	Circumference (c)
(i)	7 cm	14 cm	44 cm
(ii)	14 cm	28 cm	88 cm
(iii)	98 cm	196 cm	616 cm
(iv)	11.55 cm	23.1 cm	72.6 cm

Page No 77:

Question 2:

If the circumference of a circle is 176 cm, find its radius.

ANSWER:

Circumference, c = 176 cm
Now, c = 2πr    (where 'r' is the radius of circle)
⇒176 = 2 × 22/7 × r
⇒r = 176 / 2 x 3.14
⇒r = 28
∴ Radius of the circle = 28 cm

Page No 77:

Question 3:

The radius of a circular garden is 56 m. What would it cost to put a 4-round fence around this garden at a rate of 40 rupees per metre ?

ANSWER:

Radius of the circular garden, r = 56 m
Circumference of the circular garden, c = 2πr
= 2 × 22/7 × 56
= 352 m
∴ Length of the wire needed for one round of fencing = c = 352 m
Cost of one round of fencing = length of wire × cost per metre
= 352 × 40
= 14080 rupees
Cost of four round of fencing = 4 × 14080 = 56320 rupees

Page No 77:

Question 4:

The wheel of a bullock cart has a diameter of 1.4 m. How many rotations will the wheel complete as the cart travels 1.1 km ?

ANSWER:

Diameter of the wheel, d = 1.4 m
Circumference, c = πd
= 22/7 × 1.4
= 4.4 m
When the wheel completes 1 rotation, it covers a distance that is equal to its circumference.
So, number of rotations taken by the wheel to cover 4.4 m = 1
Now, the wheel covered a total distance of 1.1 km.
We know that, 1 km = 1000 m
∴ 1.1 km = 1.1 × 1000 m = 1100 m
∴ Total number of rotations taken by wheel = total distancecircumference$\frac{\mathrm{total} \mathrm{distance}}{\mathrm{circumference}}$ 
= 11004.4$\frac{1100}{4.4}$
= 1100044$\frac{11000}{44}$
= 250
Hence, the wheel completes 250 rotations to cover a distance of 1.1 km.

Page No 79:

Question 1:

Choose the correct option.

If arc AXB and arc AYB are corresponding arcs and m(arc AXB) = 120^° then m(arc AYB) = 12 eryyt4$\boxed{12 eryyt4}$
 

(i) 140^°   (ii) 60^°   (iii) 240^°  (iv) 160^°

 

ANSWER:

Consider that arc AXB is the minor arc and arc AYB is the corresponding major arc.
It is known that, measure of major arc = 360^∘ − measure of the corresponding minor arc.
We have, m(arc AXB) = 120^°.
So, m(arc AYB) = 360^∘ − m(arc AXB) = 360^∘ − 120^∘ = 240^∘
Hence, the correct answer is option (iii).

Page No 79:

Question 2:

Some arcs are shown in the circle with centre ‘O’. Write the names of the minor arcs, major arcs and semicircular arcs from among them.

ANSWER:

Minor arc : An arc of a circle having measure less than 180^∘.
Major arc : An arc of a circle having measure greater than 180^∘.
Semicircular arc : An arc of a circle having measure equal to 180^∘.

Names of minor arcs : 
(i) arc PXQ
(ii) arc PR
(iii) arc RY
(iv) arc XP
(v) arc XQ
(vi) arc QY

Names of major arcs :
(i) arc PYQ
(ii) arc PQR
(iii) arc RQY
(iv) arc XQP
(v) arc QRX

Names of semicircular arcs :
(i) arc QPR
(ii) arc QYR

Page No 79:

Question 3:

In a circle with centre O, the measure of a minor arc is 110^°. What is the measure of the major arc PYQ?

ANSWER:

Suppose PQ is the minor arc and then m(arc PQ) = 110^∘.
We know that, measure of major arc = 360^∘ − measure of corresponding minor arc.
∴ m(arc PYQ) = 360^∘ − m(arc PQ) 
=  360^∘ − 110^∘
= 250^∘
Hence, the measure of major arc PYQ is 250^∘.