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MP Board Class 7th Science Solutions Chapter 14 Electric Current and its Effects

Electric Current and its Effects Intex Questions

Question 1.
Paheli and Boojho wonder whether the batteries used in tractors, trucks and inverters are also made from cells. Then why is it called a battery? Can you help them to find the answer to this question?
Answer:
The battery used in tractors, trucks and inverters is a combination of several cells. The cells are not dry cells. These are the several sets of plates and each set of plates acts like a cell.

Question 2.
Paheli and Boojho saw a magic trick sometime back. The magician placed an iron box on a stand. He then called Boojho and asked him to lift the box. Boojho could easily lift the box. Now the magician made a show of moving his stick around the box while muttering some thing. He again asked Boojho to lift the box. This time Boojho could not even move it. The magician again muttered some thing and now Boojho could lift the box? The audience, including Paheli and Boojho, were very impressed with the show and felt that the magician had some supernatural powers. However, after reading this chapter Paheli is wondering if the trick was indeed some magic or some science was involved? Can you guess what science might be involved?
Answer:
The magician arranged an electromagnet below the box. The electromagnet could be turned ‘ON’ and ‘OFF’ as when the magician signals his assistance. When the electromagnet is turned ‘ON’ it attracts the iron box and hence it could not be lifted.

Electric Current and its Effects Text Book Exercises

Question 1.
Draw in your notebook the symbols to represent the following components of electrical circuits: connecting wires, switch in the ‘OFF’ position, bulb, cell, switch in the ‘ON’ position, and battery?
Answer:

Question 2.
Draw the circuit diagram to represent the circuit shown in fig?

Answer:

Question 3.
Fig. shows four cells fixed on a board. Draw lines to indicate how you will connect their terminals with wires to make a battery of four cells?

Answer:

Question 4.
The bulb in the circuit shown in figure, does not glow. Can you identify the problem? Make necessary changes in the circuit to make the bulb glow?

Answer:
The problem with the circuit is that both the negative terminals are connected to the bulb. The corrected diagram is shown below:

Question 5.
Name any two effects of electric current?
Answer:
The two effects of electric current are heating effect and magnetic effect.

Question 6.
When the current is switched on through a wire, a compass needle kept nearby gets deflected from its north-south position. Explain?
Answer:
When electric current passes through a wire, it behaves like a magnet. This is the magnetic effect of the electric current due to attraction of the wire. So, the compass needle which is a magnet, gets deflected.

Question 7.
Will the compass needle show deflection when the switch in the circuit shown by fig is closed?

Answer:
No, because there is no cell so no current will flow.

Question 8.
Fill in the blanks:

1. Longer line in the symbol for a cell represents its ………………………… terminal.
2. The combination of two or more cells is called a …………………………..
3. When current is switched ‘on’ in a room heater, it ………………………..
4. The safety device based on the heating effect of electric current is called a ………………………

Answer:

1. Positive
2. Battery
3. Get heated
4. Fuse.

Question 9.
Mark T if the statement is true and F if it is false:

1. To make a battery of two cells, the negative terminal of one cell is connected to the negative terminal of the other cell.
2. When the electric current through the fuse exceeds a certain limit, the fuse wire melts and breaks.
3. An electromagnet does not attract a piece of iron.
4. An electric bell has an electromagnet.

Answer:

1. False
2. True
3. False
4. True

Question 10.
Do you think an electromagnet can be used for separating plastic bags from a garbage heap? Explain?
Answer:
No, the plastic bags do not get attracted by the magnet, so they cannot be separated by an electromagnet. The plastic bags are not magnetic materials, only magnetic materials like iron can be attracted by magnet.

Question 11.
An electrician is carrying out some repairs in your house. He wants to replace a fuse by a piece of wire. Would you agree? Give reasons for your response?
Answer:
No, because this piece of wire will not melt even if high current flows through it. So, it will not prevent the damage done by high current.

Question 12.
Zubeda made an electric circuit using a cell holder shown in figure a switch and a bulb. When she put the switch in the ‘ON’ position, the bulb did not glow. Help Zubeda in identifying the possible defects in the circuit?

Answer:
It is important to put the cells in right series. The positive terminal of the cell should be connected with negative terminal of the second cell. The switch should be closed properly and bulb should not be fused. If Zubeda will check these then the bulb will certainly glow. Also, in the circuit, there is no any bulb and switch, the connecting wires are not connected with bulb and switch.

Question 13.
In the circuit shown in Fig?
Answer:

1. Would any of the bulb glow when the switch is in the ‘OFF’ position?
2. What will be the order in which the bulbs A, B and C will glow when the switch is moved to the ‘ON’ position?

Answer:

1. No, bulb will glow.
2. The bulb A will glow first, follow by bulb B and then bulb C, because bulb A comes first in the path of electric current flowing from positive terminal towards the negative terminal of the battery.

Extended Learning – Activities and Projects

Question 1.
Set up the circuit shown in Fig. again. Move the key to ‘ON’ position and watch carefully in which direction the compass needle gets deflected. Switch ‘OFF’ the current. Now keeping rest of the circuit intact, reverse the connections at the terminal of the cell. Again switch ‘on’ the current. Note the direction in which the needle gets deflected. Think of an explanation?

Answer:
When the connection of cell is reversed, the compass needle moves in opposite direction. This shows that the polarity of the magnet depends on the direction of current.

Question 2.
Make four electro – magnets with 20, 40, 60 and 80 turns. Connect them one by one to a battery of 2 cells. Bring the electromagnet near a box of pins. Count the number of pins attracted by it. Compare the strengths of the electromagnets?
Answer:
The strength increases with number of turns in the coil.

Question 3.
Using an electromagnet, you can make a working model of a railway signal as shown in Fig?
Answer:
Do with the help of your subject teacher.

Question 4.
Visit an electric shop. Request a mechanic to show you the various types of fuses and MCB and to explain how they work?

Answer:
Go and visit the nearby shop with your friends.

Electric Current and its Effects Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (a)
The electric circuit which is broken at some point is called?
(a) closed
(b) open
(c) short
(d) None of these.
Answer:
(b) open

Question (b)
The amount of heat produced in a wire depends on its?
(a) material
(b) length
(c) thickness
(d) all of these.
Answer:
(d) all of these.

Question (c)
The direction of flow of conventional current is taken as
(a) -ve to +ve
(b) +ve to -ve
(c) both of these
(d) None of these.
Answer:
(b) +ve to -ve

Question 2.
Fill in the blanks:

1. A continuous flow of electric charge is called ………………………….
2. In a dry cell energy is converted into ……………………… energy.
3. A combination of two or more cells is called a ………………………
4. The positive terminal of one cell is connected to the ……………………. terminal of the next cell.
5. When the switch of a circuit is in the OFF position, the circuit is ………………………..
6. In the bulb, thin wire is called …………………………….
7. Current flowing in our household appliances is ………………………….
8. A ………………………. is a safety device which prevents damages to electrical circuits and possible fires.
9. The switches which automatically turn off when current in a circuit exceeds the safe limit are called ……………………….

Answer:

1. Current
2. Chemical, electrical
3. Battery
4. Negative
5. Incomplete
6. Filament
7. A.C.
8. Fuse
9. MCB

Question 3.
Which of the following statements are true (T) or false(F):

1. Switch is used to regulate the current.
2. An electric cell has three terminals.
3. Two or more than two cells are interconnected to make a battery.
4. The coil of wire in a electric iron is called an element.
5. Never use just any wire or strip of metal in place of a fuse.
6. Heating effect of electric current is used in a heater.
7. An electric bell is an application of an electromagnet.
8. A cell or a battery are the sources of electric current in an electric circuit.

Answer:

1. True
2. False
3. True
4. True
5. True
6. True
7. True
8. True.

Electric Current and its Effects Very Short Answer Type Questions

Question 1.
What is meant by electric current?
Answer:
The flow of negative charges i.e., electrons is called electric current.

Question 2.
Name the various sources of electric current?
Answer:
Cells and storage batteries.

Question 3.
What do you mean by electrically neutral substances?
Answer:
Those substances which have an equal quantity of positive and negative charge.

Question 4.
What are the two kinds of electric charges?
Answer:
The two kinds of electric charges are positive and negative.

Question 5.
Which terminal of a cell is represented by longer line?
Answer:
Positive.

Question 6.
Where should be the key placed in a circuit?
Answer:
The key or switch can be placed anywhere in the circuit.

Question 7.
What is an open electric circuit?
Answer:
An electric circuit having the switch in off position is called open electric circuit.

Question 8.
What is a closed electric circuit?
Answer:
An electric circuit having the switch in on position is called a closed circuit or a complete circuit.

Question 9.
What is the significance of using symbols in an electric circuit?
Answer:
Using symbols is significant as if we have to show the various batteries, bulbs and switches as they actually look, it would be very cumber some and take a lot of time.

Question 10.
What is a filament?
Answer:
In the bulb there is a thin wire, called the filament, which glows when an electric current passes through it.

Question 11.
What is the full form of CFL?
Answer:
Compact Fluoroscent Lamp.

Question 12.
What is the full form of MCB?
Answer:
Miniature Circuit Breaker.

Question 13.
What are different hazards of electricity?
Answer:

1. ‘Electric shock’
2. Tire due to short circuiting’.

Question 14.
List the appliances around you that depend on electricity for their operation. List the appliances that do not use electrical energy?
Answer:
Appliances that depend on electricity : Electric iron, Television, Geysers, Kitchen blender. Appliances that do not use electricity: Solar cells, Sewing Machine, Vehicles.

Question 15.
What energy is converted to electrical energy in a cell?
Answer:
Chemical energy is converted to electrical energy.

Question 16.
Name the three components of the cable used in a house circuit?
Answer:
The cable used in a house circuit has three cores of different colours. The red wire is the live wire. The black is the neutral wire and the green is the earth wire.

Question 17.
What kind of energy transformation takes place in an electric heater?
Answer:
In electric heater, electric energy is converted into heat energy.

Question 18.
Why is alternating current of same voltage more dangerous than direct current?
Answer:
The alternating current of same voltage is more dangerous than direct current because the direction and magnitude of AC changes alternatively and it may produce electric shock.

Question 19.
What happens when fuse melts away?
Answer:
When fuse melts away, the circuit is broken and the flow of current stops.

Question 20.
What is the main purpose of earthing an electrical appliance?
Answer:
To prevent electric shock.

Question 21.
Why is mica used in elements of an Electric iron?
Answer:
Mica is a insulator.

Question 22.
List the appliances in your house that do not have ground wires?
Answer:
Electric grinder, electric heater, electric oven, electric toaster.

Question 23.
Who discovered magnetic effect of electric current?
Answer:
Hans Christian Oersted.

Question 24.
What are the various effects shown by electricity?
Answer:

1. Heating Effect
2. Chemical Effect
3. Magnetic Effect.

Question 25.
What is electrolysis?
Answer:
The phenomenon of breaking up of a chemical compound under the action of electric current is called electrolysis.

Question 26.
What causes heating effect in electric appliances?
Answer:
Heating effect is produced by heating element i.e., mainly ‘Nichrome’ which is an alloy of nickel and chromium.

Question 27.
Why is ‘Nichrome’ most commonly used as heating element?
Answer:
Since ‘Nichrome’ has a very high melting point, it does not melt with the flow of current.

Question 28.
Name few processes which are based on chemical effect of electricity?
Answer:
Electroplating, Refining of metals and electrolysis of water.

Question 29.
What is meant by magnetic effect of current?
Answer:
A current carrying wire behaves like a magnet.

Electric Current and its Effects Short Answer Type Questions

Question 1.
Can electrically neutral substances be made to acquire charge?
Answer:
Yes, the delicate balance between the charges can be disturbed. Thus electrically neutral substances can be made to acquire charge. For example, if we rub a glass rod with silk we brush off some negative charges from the rod. The rod now acquires a net positive charge because it has more positive charges than negative.

Question 2.
What happens when a charged electroscope is connected into a uncharged electroscope by a wire? What is that called?
Answer:
When a charged electroscope is joined to a uncharged electroscope by a wire, charges flow from the charged to the uncharged electroscope through the wire till they are equalised. This flow of charges form an Electric Current.

Question 3.
What are the differences between a simple cell and a storage cell?
Answer:
A simple cell:

1. Once the chemicals are used up the cell stope supplying current.
2. A small amount of current can be drawn.

A storage cell:

1. The used up chemicals can be restored to their original state by reversing the reaction.
2. A large current can be drawn.

Question 4.
What should be the characteristics of an ideal heating element and that of an ideal safety fuse?
Answer:

1. Firstly the heating element should have a very high melting point.
2. Secondly, it should be a conductor.
3. Thirdly it should have high resistance.
4. The ideal safety fuse should be firstly it should have low melting point.
5. Secondly, it should be a conductor.
6. Thirdly, it have low resistance.

Question 5.
Make a list of materials around you which conduct electricity and a list of those that do not.
Answer:
Conductors of electricity are:
All metals, acid, base and salt solutions, aluminium, iron, copper, nickel. Do not conduct electricity are: Paper, rubber, wood, nylon, polythene, bakelite.

Question 6.
What is the significance of the earth wire?
Answer:
The metallic body of an electric appliance is connected to the earth through this earth wire. Sometimes due to faulty wire, the insulation breaks and the live w;re comes into contact with the body of appliance. If the appliance is earthed the charge will quickly flow to the earth, thus protecting us from electric shock.

Question 7.
How does fuse act as a safety device?
Answer:
A fuse is a porcelain holder having a short and thin piece of wire in it called fuse wire which easily melts on heating. Sometimes the insulation of a wire is broken i.e., live wire touches the neutral wire there is a short circuit. When there is a short circuit the fuse wire melts due to heating produced by the high current. The circuit is then broken and the current stops flowing. So damages are prevented.

Question 8.
On which effect of current does the fuse wire in electric circuits depend? Write two characteristics of a good quality fuse wire?
Answer:
Whenever a current passes through a material, a part or whole of its energy is converted into heat. The part which heats up is called heating elements. An alloy of nickel and chromium called Nichrome is one of the most heating elements.
Two characteristics of a good quality fuse wire are as follows:

1. It should have low melting point.
2. Is should have high specific resistance.

Question 9.
Write four safety measures to be taken while working with live electric lines?
Answer:

1. Do not touch the switch in the main switch board.
2. Do not cut the cable of an appliance with a pair of scissors when the appliance is switched on.
3. Do not touch any electric poles in the street, especially during the rainy season.
4. Do not touch any appliance (when it is switched on) with naked hands. Use rubber gloves..

Question 10.
Draw a battery of four cells?

Question 11.
How does electromagnetic induction find its use or application in generating electricity?
Answer:
The act of moving a magnet quickly in or cut of a coil of wire, can cause a current to flow in the wire. This phenomenon is called electromagnetic induction. This finds its application in generators where the coil is connected to a turbine. That rotates by the energy of falling water and moves the coil and causes current to flow, generators used in power stations have coil kept stationary and the magnet itself is rotated.

Question 12.
What forms the basic principle in electromagnets?
Answer:
Producing electromagnets is based on the principle that when current is passed through a coil it behaves like a magnet i.e., achieves north and south pole. Also if we reverse the direction of current in the coil the two poles exchange places. In electromagnets we insert a piece of iron in such a coil, its domain will get aligned in a direction normal to the face of the coil and the piece of iron gets strongly magnatised and act as electromagnets.

Electric Current and its Effects Long Answer Type Questions

Question 1.
Find the various ways in which two batteries and two bulbs can be connected in a working circuit?
Answer:
Two batteries and two bulbs can be connected in working circuit in two ways:
1. In series as shown in figure.

2. In parallel as shown in figure.

Question 2.
Give the symbols used for: open key; closed key, dry cell, a wire point, a battery and a bulb?
Answer:
The symbols are:

Question 3.
How can conductivity of a substance be tested?
Answer:
We can perform an experiment to test whether the given material conducts electricity or not. Make a circuit with wire, battery, bulb and connect them. If the bulb lights up, we can be sure that current is flowing in the circuit. Break the circuit. Now Fi8- TestinS whether a maternal close the circuit by inserting conducts electricity. different materials in between as (A) Match Stick (B) Pencil (C) Water (D) Eraser (E) Paper cup. Now see whether the bulb lights up. If it lights up, the material used for insertion in a conductor and if the bulb does not light up, the material is an insulator.

Question 4.
Define Miniature Circuit Breaker (MCB) with diagram?
Answer:
These day Miniature Circuit Breakers (MCBs) are increasingly being used in place of fuses. These are switches which automatically turn off when current in a circuit exceeds the safe limit. You turn them join and the circuit is once again complete. Look for ISI mark on MCBs also.

Question 5.
Define the working of an electric bell with circuit?
Answer:
We are quite familiar with an electric bell. It has an electromagnet in it. Fig. shows the circuit of an electric bell. It consists of a coil of wire wound on an iron piece. The coil acts as an electromagnet. An iron strip with a hammer at one end is kept close to the electromagnet.

There is a contact screw near the iron strip. When the iron strip is in contact with the screw, the current flows through the coil which becomes an electromagnet. It, then, pulls the iron strip. In the process, the hammer at the end of the strip strikes the gong of the bell to produce a sound. However, when the electromagnet pulls the iron strip, it also breaks the circuit. The current through the coil stops flowing.

The coil is no longer an electromagnet. It no longer attracts the iron strip. The iron strip comes back to its original position and touches the contact screw again. This completes the circuit. The current flows in the coil and the hammer strikes the gong again. This process is repeated in quick succession. The hammer strikes the gong every time the circuit is completed. This is how the bell rings.

MP Board Class 7th Science Solutions

MP Board Class 9th Science Solutions Chapter 8 Motion

Motion Intext Questions

Motion Intext Questions Page No. 100

Question 1.
An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:
Yes, if an object start from a point and finally, reaches the same point. Then, it has distance but displacement is zero.

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Answer:

Suppose farmer starts from A of the square field.
Now, he covered 10 m in 40 sec.
Now, 2 minutes 20 seconds =140 sec.
So, he cover in 140 seconds = \(\frac { 10 }{ 40 }\) × 140 m
= \(\frac { 140 }{ 4 }\)m
= 35 m.
Now, in square field = 35m
AB + BC + CD + DE (\(\frac { 1 }{ 2 }\)AD)
So, he will reach at E.
Then, his displacement is AE = 5 m.

Question 3.
Which of the following is true for displacement?

1. It cannot be zero.
2. Its magnitude is greater than the distance travelled by the object.

Answer:

1. False
2. False.

Motion Intext Questions Page No. 102

Question 1.
Distinguish between speed and velocity.
Answer:

Question 2.
Under what condition (s) is magnitude of average velocity of an object equal to its average speed?
Answer:
When a body is moving in a straight line in a particular direction.

Question 3.
What does the odometer of an automobile measure?
Answer:
It measures the distance travelled by the vehicle.

Question 4.
What does the path of an object look like when it is in uniform motion?
Answer:
Straight line.

Question 5.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is 3 × 10⁸ m/s.
Answer:
Time taken by signal to reach ground station = 5 minutes
= 5 × 60 sec.
= 300 seconds.
Speed of the signal = 3 × 10⁸ m/s.
∴ Distance between spaceship and ground station = speed × time
= 3 × 10⁸ × 300
= 9 × 10¹⁰ m.

Motion Intext Questions Page No. 103

Question 1.
When will you say a body is in

1. Uniform acceleration?
2. Non – uniform acceleration?

Answer:

1. A body is in uniform acceleration if its velocity changes uniformly with equal intervals of time. Example: Freely falling object.
2. A body is in non – uniform acceleration if its velocity changes non – uniformly with equal intervals of time. Example: A bus running in a city.

Question 2.
A bus decreases its speed from 80 km/h^-1 to 60 km/h^-1 in 5 s. Find the acceleration of the bus.
Answer:
Initial velocity of the bus, u = 80 km/h
= \(\frac { 80\times 100 }{ 60\times 60 } \) m/s
u = \(\frac { 800 }{ 36} \) m/s
Final velocity of the bus, v = 60 km/h
= \(\frac { 60\times 100 }{ 60\times 60 } \) m/s
v = \(\frac { 600 }{ 36 } \) m/s
Time taken by bus to change its speeds, t = 5 seconds
∴ Acceleration, a = \(\frac { v-u }{ t } \)
= \(\frac { 600 }{ 36 } \) – \(\frac { 800 }{ 36 } \) ÷ 5
= \(\frac { 600-800 }{ 36 } \) ÷ 5
= \(\frac { 200 }{ 36 } \) × \(\frac { 1 }{ 5 } \) m/s²
= \(\frac { -10 }{ 9 } \) m/s²
a = -1.11 m/s².

Question 3.
A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h^-1 in 10 minutes. Find its acceleration.
Answer:
Initial velocity, u = 0
Final velocity, v = 40 km/h
= \(\frac { 40\times 1000 }{ 60\times 60 } \) m/s
= \(\frac { 400 }{ 36 } \) m/s
= \(\frac { 200 }{ 18 } \) m/s.
Time taken, t = 10 minutes
= 10 × 10 seconds = 100 seconds
a = \(\frac { v-u }{ t } \)
= \(\frac { 200 }{ 18 } \) – 0 ÷ 100
= \(\frac { 200 }{ 18 } \) ÷ 100
= \(\frac { 200 }{ 18 } \) × \(\frac { 1 }{ 100 } \) m/s²
= \(\frac { 1 }{ 9 } \) m/s²
a = 0.018 m/s².

Motion Intext Questions Page No. 107

Question 1.
What is the nature of the distance – time graphs for uniform and non – uniform motion of an object?
Answer:
For uniform motion, it is a straight line.

For non – uniform motion, it is a curved line.

Question 2.
What can you say about the motion of an object whose distance – time graph is a straight line parallel to the time axis?
Answer:
It shows the body at rest.

Question 3.
What can you say about the motion of an object if its speed – time graph is a straight line parallel to the time axis?
Answer:
It means body has constant speed or uniform motion with zero acceleration.

Question 4.
What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
Distance travelled.

Motion Intext Questions Page No. 109 – 110

Question 1.
A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled.
Answer:
Initial velocity, u = 0
Acceleration, a = 0.1 m/s²
Time, t = 2 min = 2 × 60 seconds
= 120 seconds
(a) Velocity, v = ?
We know, v = u + at
v = 0 + 0.1 × 120 m/s
= 12 m/s.

(b) Distance, s = ?
We Know, s = ut + \(\frac { 1 }{ 2 }\) at²
= 0 × 120 + \(\frac { 1 }{ 2 }\) × 0.1 × 120 × 120
= \(\frac { 1 }{ 2 }\) × \(\frac { 0.1 }{ 10 }\) × 120 × 120 m
= 720 m.

Question 2.
A train is travelling at a speed of 90 km/h^-1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s². Find how far the train wilt go before it is brought to rest.
Answer:
Initial velocity, u = 90 km/h
= \(\frac { 90\times 5 }{ 18 } \) m/s
= 25 m/s
Final velocity, v = 0
Acceleration, a = -0.5 m/s².
Distance, s = ?
We know,
v² = u² + 2as
0² = (25)² + 2 × (-0.5) × s
0 = 625 – 1 × s
= 625 – 1 × s
= 625 – s
∴ s = 625 m.
∴ Distance travelled by the train = 625 m.

Question 3.
A trolley while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity in 3 s after the start?
Answer:
Acceleration, a = 2 cm/s²
Time, t = 3 sec.
Initial velocity, u = 0
Final velocity, v = ?
Now, v= u + at
v = 0 + 2 × 3
= 6 cm/s.

Question 4.
A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?
Answer:
Acceleration, a = 4 m/s²
Time, t = 10 sec,
Initial velocity, u = 0
Distance, s = ?
Now, s = ut + \(\frac { 1 }{ 2 }\) at²
s = 0 × 10 + \(\frac { 1 }{ 2 }\) × 4 × (10)²
s = \(\frac { 1 }{ 2 }\) × 4 × 100 m
s = 200 m.
So, the distance covered is 200 m.

Question 5.
A stone is thrown in a vertically upward direction with a velocity of 5 m/s^-1. If the acceleration of the stone during its motion is 10 m/s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer:
Initial velocity, u = 5 m/s
Final velocity, v = 0
Acceleration, a = -10 m/s² (∴ It is from opposite direction)
Height attained, s = ?
Time taken, t = ?
Now, V² = u² + 2 as
(0)² = (5)² + 2 × (-10) × s
0 = 25 – 20s
⇒ 20s = 25
∴ s = \(\frac { 25 }{ 20 }\) = \(\frac { 5 }{ 4 }\) = 1025m
Now, v = u + at
0 = 5 + (-10) × t
0 = 5 – 10 × t
-5 = -10 × t
∴ t = \(\frac { -5 }{ -10 }\) = \(\frac { 1 }{ 2 }\) = 0.5 sec
So, height attained by the stone is 1.25 m and time taken to reach there is 0.5 sec.

Motion NCERT Textbook Exercises

Question 1.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer:

Diameter of the track = 200 m
So, radius = \(\frac { 200 }{ 2 }\) m
= 100m
Circumference = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 100 m
= \(\frac { 4400 }{ 7 }\) m.
Now, athlete covers 200 m in 40 s.
So, rounds covered in 40 seconds = 1
Rounds covered in 1 sec = \(\frac { 1 }{ 40 }\)
Rounds covered in 140 sec = \(\frac { 1 }{ 40 }\) × 140
= 3.5 rounds
So, distance covered = 3.5 × circumference
= \(\frac { 3.5 }{ 10 }\) × \(\frac { 4400 }{ 7 }\) m
= 2200 m.
And, displacement after 3.5 rounds
= Diameter of track
= 200 m.

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer:

(a) A to B,
Time taken, t = 2 min, 30 sec
= 150 sec.

= \(\frac { 300 }{ 150 }\) m/s = 2 m/s

= \(\frac { 300 }{ 150 }\) m/s = 2 m/s.

(b) A to C,
Time taken = Time from A to B + Time from B to C
= 2 min 30 sec + 1 min
= 150 sec + 60 sec
= 210 sec.
Total distance covered (AB + BC) = (300 + 100) m
= 400 m
Displacement = AC = (300 – 100) m
= 200 m

= \(\frac { 400 }{ 210 }\) m/s = 1.9 m/s

= \(\frac { 200 }{ 210 }\) m/s = 0.95 m/s.

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20 kmh^-1. On his return trip along the same route, there is less traffic and the average speed is 30 kmh^-1. What is the average speed for Abdul’s trip?
Answer:
Let the distance between Abdul’s home and school = x km.
Lets time taken from home to school = t₁ hr.
and time taken from school to home = t₂ hr.
Now,

∴ t₁ = \(\frac { x }{ 20 }\) hr.
t₂ = \(\frac { x }{ 30 }\) hr.
Now,

∴ Average Speed = \(\frac { 120 }{ 5 }\) 24 km/h.

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^-2 for 8.0 s. How far does the boat travel during this time?
Answer:
Initial velocity, u = 0
Acceleration, a = 3.0 m/s²
= 3 m/s²
Time, t = 8 sec
Now,
s = ut + \(\frac { 1 }{ 2 }\) at²
s = 0 × 8 + \(\frac { 1 }{ 2 }\) × 3 × (8)²
= 0 + \(\frac { 1 }{ 2 }\) × 3 × 64
s = 96 m.
∴ Distance Travelled by Boat is 96 m.

Question 5.
A driver of a car travelling at 52 kmh^-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 34 kmh^-1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer:
We have,
Speed of first driver = 52 km/h
= \(\frac { 52\times 1000 }{ 60\times 60 } \) m/s = 14.4 m/s
Speed of second driver = 3 km/h
= \(\frac { 3\times 1000 }{ 60\times 60 } \) m/s = 0.83 m/s
Now, distance covered by first driver = Area of ΔAOB
= \(\frac { 1 }{ 2 }\) × OA × OB
= \(\frac { 1 }{ 2 }\) × 14.4 × 5 = 36.1 m
Distance covered by second driver = Area of ΔMON
= \(\frac { 1 }{ 2 }\) × OM × ON
= \(\frac { 1 }{ 2 }\) × O.83 × 10 = 4.16 m
So, first driver covers more distance.

Question 6.
Figure below shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

1. Which of the three is travelling the fastest?
2. Are all three ever at the same point on the road?
3. How far has C travelled when B passes A?
4. How far has B travelled by the time it passes C?

Answer:

1. B is travelling the fastest.
2. No.
3. C was at around 9 km from origin when B passes A.
4. B travelled around 5.5 km when it passes C.

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^-2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Distance, s = 20 m
Initial velocity, u = 0
Acceleration, a = 10 m/s²
Now, s = ut +\(\frac { 1 }{ 2 }\) at²
20 = 0 × t + \(\frac { 1 }{ 2 }\) × 10 × t²
20 = 5 × t²
∴ t² = \(\frac { 20 }{ 5 }\) = 4
∴ t = 2 sec.
Also, v = u + at
v = 0 + 10 × 2
y = 20 m/s.
So, it will strike the ground after 2 seconds with the velocity of 20 m/s.

Question 8.
The speed – time graph for a car is shown in figure below:

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer:
(a) The distance travelled by car in first 4 seconds
= Area of triangular like region OAB
= \(\frac { 1 }{ 2 }\) × AB × OB = \(\frac { 1 }{ 2 }\) × 6.9 × 4
= 12 m (approximately).

(b) Part of the graph after 5.5 seconds shows the uniform motion.

Question 9.
State which of the following situations are possible and give an example for each of these:

1. an object with a constant acceleration but with zero velocity.
2. an object moving in a certain direction with an acceleration in the perpendicular direction. Both are possible.

Answer:

1. Freely falling.
2. Motion of an object in a circular path.

Question 10.
An artificial, satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer:
Radius of the orbit = 42250 km = 42250 × 1000 m.
Time taken to complete one revolution = 24 hours.
= 24 × 60 × 60 sec
Distance covered by satellite to complete 1 revolution = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 42250 × 1000 m

Motion Additional Questions

Motion Multiple Choice Questions

Question 1.
Deceleration of a body is expressed in _______ .
(a) m
(b) ms^-1
(c) ms^-2
(d) -ms^-2
Answer:
(c) ms^-2.

Question 2. A car goes from town A to another town B with a speed of 40 km/h and returns back to the town A with a speed of 60 km/h. The average speed of the car during the complete journey is _______ .
(a) 48 km/h
(b) 50 km/h
(c) Zero
(d) None of these.
Answer:
(a) 48 km/h

Question 3.
A ball is thrown vertically upwards. It rises to a height of 50 m and comes back to the thrower, _______ .
(a) The total distance covered by the ball is zero.
(b) The net displacement of the ball is zero.
(c) The displacement is 100 m.
(d) None of the above.
Answer:
(b) The net displacement of the ball is zero.

Question 4.
The SI unit of velocity is _______ .
(a) ms^-1
(b) ms^-2
(c) ms^-3
(d) Nm^-1.
Answer:
(a) ms^-1

Question 5.
In 12 minutes a car whose speed is 35 km/h travels a distance of _______ .
(a) 7 km
(b) 3.5 km
(c) 14 km
(d) 28 km.
Answer:
(a) 7 km

Question 6.
A 50 m long train passes over a 250 m long bridge at a velocity of 60 km/h. How long will it take to pass completely over the bridge?
(a) 18 s
(b) 20 s
(c) 24 s
(d) none of these.
Answer:
(a) 18 s

Question 7.
The initial velocity of a train which is stopped in 20 s by applying brakes (retardation due to brakes being 1.5 ms^-2) is _______ .
(a) 30 ms^-1
(b) 30 cm^-1
(c) 20 cm^-1
(d) 24 ms^-1.
Answer:
(a) 30 ms^-1

Question 8.
A wooden slab starting from rest, slides down a 10 m long inclined plane with an acceleration of 5 ms². What would be its speed at the bottom of the inclined plane?
(a) 10 ms^-1
(b) 12 ms^-1
(c) 10 cm s^-1
(d) 12 cm s^-1.
Answer:
(a) 10 ms^-1

Question 9.
m/s² is the SI unit of _______ .
(a) Distance
(b) Displacement
(c) Velocity
(d) Acceleration.
Answer:
(d) Acceleration.

Question 10.
A car increases its speed from 20 km/h to 50 km/h in 10 seconds. Its acceleration is _______ .
(a) 30 m/s²
(b) 3 m/s²
(c) 18 m/s²
(d) 0.83 m/s².
Answer:
(d) 0.83 m/s².

Motion Very Short Answer type Questions

Question 1.
Define uniform motion.
Answer:
When an object covers equal distances in equal intervals of time: it is said to be in uniform motion.

Question 2.
Define speed.
Answer:
It is defined as the distance travelled by an object in unit time. Its unit is m/s.

Question 3.
Define average speed.
Answer:
The total distance travelled by an object divided by the total time taken.

Question 4.
Define velocity.
Answer:
Velocity is the speed of an object moving in definite direction.

Question 5.
Define acceleration.
Answer:
Change in the velocity of an object per unit time.
a = \(\frac { v-u }{ t } \), S.I. unit is m/s².

Motion Short Answer type Questions

Question 1.
While plotting a distance – time graph, why do we plot time on x – axis.
Answer:
While plotting a distance – time graph, we should plot time on x – axis as, all independent quantities are plotted on x – axis.

Question 2.
What is odometer?
Answer:
Odometer is a device fitted in the automobiles to show the distance travelled.

Question 3.
Draw a distance – time graph that represents uniform speed.
Answer:

Question 4.
With the help of distance – time graph show that the object is stationary.
Answer:

Question 5.
What does the area under velocity – time graph represent?
Answer:
The area under the velocity – time graph represents the displacement.

Question 6.
A body travels a distance from A to B, its physical quantity is measured to be -15 m/s. Is it speed or velocity? Give reason for the answer.
Answer:
The physical quantity is -15 m/s. It represents velocity because sign of speed cannot be a negative. Negative sign indicates the opposite direction.

Question 7.
What type of velocity – time graph will you get for a particle moving with a constant acceleration?
Answer:
For a particle moving with a constant acceleration velocity – time graph will be a straight line inclined to x – axis.

Question 8.
Define uniform circular motion.
Answer:
When a body moves in a circular path with uniform speed, its motion is called uniform circular motion.

Question 9.
What is deceleration?
Answer:
When the speed of an object decreases, the object is said to be decelerating. In this case, the direction of acceleration is opposite to the velocity of the object.

Question 10.
What conclusion do you draw about acceleration of the particle in motion from given velocity – time graphs?

Answer:
(a) No acceleration.
(b) Uniform acceleration.
(c) Non – uniform acceleration.

Question 11.
With the help of graph, show the uniform acceleration and uniform retardation of a body.
Answer:

• OA – Shows uniform acceleration.
• CB – Shows uniform retardation.

Question 12.
Give difference between acceleration and velocity.
Answer:
table

Motion Long Answer Type Questions

Question 1.
How can we get speed from distance – time graph?
Answer:
Let us assume that an object moves with a uniform speed. The distance – time graph will be a straight line.
Let d₁ be the distance covered in time t₁.
Let d₂ be the distance covered in time t₂.
∴ d₂ – d₁ will be the distance covered in t₂ – t₂.

But, here \(\frac { BC }{ AC }\) = slope of the graph.
Conclusion:
To find the speed from distance – time graph, we can get its slope which is equal to the speed of the object.

Question 2.
Represent a graph that shows the following:
(a) Uniform speed
(b) Non – uniform speed
(c) Stationary object.
Answer:
Time is taken on x – axis as it is independent quantity and all dependent quantities are taken along y – axis.

Question 3.
(a) What is motion?
(b) State types of motion.
(c) Derive the unit for acceleration.
Answer:
(a) Motion: An object is said to be in motion when its position changes with time. We describe the location of an object by specifying a reference point. Motion is relative. The total path covered by an object is said to be the distance travelled by it.

(b) There are two types of motion: Uniform motion and non – uniform motion

• Uniform motion: When an object covers equal distances in equal intervals of time, it is said to be in uniform motion.
• Non – uniform motion: Motion where object cover unequal distances in equal intervals of time.

(c) Acceleration, ‘a’ is change in velocity per unit time.

Question 4.
(a) What is uniform circular motion?
(b) An athlete runs on a circular track, whose radius is 50 m with a constant speed. It takes 50 seconds to reach point B from starting point A. Find:

1. the distance covered.
2. the displacement.
3. the speed.

Answer:
(a) When a body moves in a circular path with uniform speed, its motion is called uniform circular motion.

(b) Radius = 50 m
Time = 50 s
The distance covered by an athlete A to B i.e., semicircle of the track
∴ Circumference = 2πr
circumference half of = πr.

Question 5.
Derive the equation for velocity – time relation (v = u + at) by graphical method.
Answer:

Let, the initial velocity of a body be, u = OA = CD.
The final velocity of a body be, v = OE = CB
Time, t = OC = AD
AD is parallel to OC
BC = BD + DC
= BD + OA
v = BD + u
∴ BD =v – u …….. (1)
In velocity – time graph, slope gives acceleration.
∴ a = \(\frac { BD }{ AD }\) = \(\frac { BD }{ OC }\) = OC = t
we get, a = \(\frac { BD }{ t }\)
∴ BD = at …….. (2).

Question 6.
A car accelerates uniformly from 20 km/h to 35 km/h in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Answer:
u =20 km/h = 5.5 m/s
v =35 km/h = 9.72 m/s
t = 5 s
a = \(\frac { v-u }{ t } \)
= \(\frac { 9.7-5.5 }{ 5 } \)
a = \(\frac { 4.2 }{ 5 } \)
= 0.84 m/s².

(ii) The distance covered by the car s = ut + \(\frac { 1 }{ 2 }\) at²
= 5.5 × 5 + \(\frac { 1 }{ 2 }\)(0.84)(5)²
= 27.5 + \(\frac { 1 }{ 2 }\) × 0.84 × 25
= 27.5 + 10.5
s = 38 m.

Question 7.
Draw graph to show the following:
(a) Uniform acceleration.
(b) Non – uniform acceleration.
(c) Uniform motion.
Answer:

Uniform acceleration – A body travels equal distances in equal intervals of time.

Non – uniform acceleration – A body travels unequal distances in equal intervals of time.

Uniform motion or zero acceleration A body moves with constant speed in same line.

Question 8.
How can you get the distance travelled by an object from its speed – time graph?
Answer:

Suppose an object is moving with a same speed v, the distance covered by this object when it was at (point A) t₁ to t₂ is:
AB = t₂ – t₁
AD = BC = v

= AD . AB
∴ s = Area of rectangle ABCD
∴ To find the distance travelled by a body from its speed – time graph, we need to find the area enclosed by the graph.

Motion Higher Order Thinking Skills (HOTS)

Question 1.
A blacksmith strikes a nail with a hammer of mass 500 g moving with a velocity of 20 ms^-1. The hammer comes to rest in 0.02 s after striking the nail. Calculate the force exerted by the nail on the hammer.
Answer:
m = 500 g = \(\frac { 1 }{ 2 }\) kg
u = 20 ms^-1 , v = 0
t = 0.02 s, F = ?

Therefore, force exerted by nail on the hammer = 500 N.

Motion Value Based Question

Question 1.
A jeep was moving at a speed of 120 km/h on an highway. All of a sudden tyre of the jeep got punctured and burst The driver of the jeep did not apply sudden brakes but he slowed down its speed and finally stopped it at the road side. All the passengers in the jeep were shocked but happy with the driver’s decison.

1. What would have happened if the driver applied sudden brakes?
2. On applying brakes in which direction would the passengers fall?
3. What value of driver is reflected in this act?

Answer:

1. If the driver would have applied sudden brakes, then all the passengers would get hurt and even other vehicles moving on the highway would have collided with this jeep from behind.
2. On applying brake, the passengers would fall in the forward direction.
3. The driver showed the presence of mind, concern and responsible behaviour and quick decision taking capability.

MP Board Class 9th Science Solutions

MP Board Class 7th Science Solutions Chapter 3 Fibre to Fabric

Fibre to Fabric Intext Questions

Question 1.
Boojho is wondering why it hurts when someone pulls his hair but not when he goes for a haircut?
Answer:
When someone pulls his hair, it hurts because it’s roots is connected to the skin which has sensation. But, during haircut the tip of the hair is cut which is dead and does not have any sensation. Thus, hair cut does not hurt.

Question 2.
Boojho is wondering why a cotton garment cannot keep us as warm in winter as a woollen sweater does?
Answer:
Cotton cloths are thin and does not trap air. Wool is thicker than cotton and have spaces in which air traps. Air is a poor conductor of heat and so it prevent heat coming out of our body. Thus, wool gives better protection than cotton.

Question 3.
Paheli wants to know if the cotton thread and silk thread are spun and woven in the same manner?
Answer:
No.

Fibre to Fabric Text book Exercises

Question 1.
You must be familiar with the following nursery rhymes?

1. ‘Baa baa black sheep, have you any wool.’
2. ‘Mary had little lamb, whose fleece was white as snow.’

Answer the following:

1. Which parts of the black sheep have wool?
2. What is meant by the white, fleece of the lamb?

Answer:

1. Basically the abdomen and back of the sheep have wool.
2. The white fleece refers to the white hair of lamb that is used to make wool.

Question 2.
The silkworm is (a) a caterpillar, (b) a larva. Choose the correct option.
(i) (a)
(ii) (b)
(iii) both (a) and (b)
(iv) neither (a) nor (b).
Answer:
(iii) both (a) and (b).

Question 3.
Which of the following does not yield wool?
(i) Yak
(ii) Camel
(iii) Goat
(iv) Woolly dog.
Answer:
(iv) Woolly dog.

Question 4.
What is meant by the following terms?

1. Rearing
2. Shearing
3. Sericulture

Answer:
1. Rearing:
The bringing up and looking after the sheep is called rearing.

2. Shearing:
The fleece of the sheep along with a thin layer of skin is removed from its body. This process is called shearing.

3. Sericulture:
The rearing of silkworms for obtaining silk is called sericulture.

Question 5.
Given below is a sequence of steps in the processing of wool. Which are the missing steps? Add them.
Shearing, …………… sorting, ………….., ………….
Answer:
Scouring, drying, dying, spinning, weaving.

Question 6.
Make sketches of the two stages in the life history of the silk moth which are directly related to the production of silk?
Answer:

Question 7.
Out of the following, which are the two terms related to silk production? Sericulture, floriculture, moriculture, apiculture and silviculture.
Hint:

1. Silk production involves cultivation of mulberry leaves and rearing silkworms.
2. Scientific name of mulberry is Morus alba.

Answer:

1. Sericulture
2. Moriculture.

Question 8.
Match the words of Column I with those given in Column II:

Answer:

(i) (e)
(ii) (c)
(iii) (b)
(iv) (a)

Question 9.
Given below is a crossword puzzle based on this lesson. Use hints to fill in the blank spaces with letters that complete the words?

Answer:

Extended Learning – Activities and Projects:

Question 1.
Paheli wants to know the maximum length of continuous silk thread that can be obtained from a cocoon. Find out for her?
Answer:
980 m.

Question 2.
Boojho wants to know why caterpillars need to shed their skin when they grow bigger but we humans do not. Do you have any idea?
Answer:
During its growing stage, the caterpillar eats own shed skin. It has no other food option.

Question 3.
Boojho wants to know why caterpillars should not be collected with bare hands. Can you help him?
Answer:
Because caterpillar’s skin can cause alergy.

Question 4.
Paheli wanted to buy a silk frock and went to the market with her mother. There they found that the artificial (synthetic) silk was much cheaper and wanted to know why. Do you know why? Find out.
Answer:
The cost of production for artificial silk is very low. The raw materials used for artificial silk are available at very low cost and are available in huge quantity. Also, the production time is very less. Thus artificial silk is cheaper.

Question 5.
Someone told Paheli that an animal called ‘Vicuna’ also gives wool. Can you tell her where this animal is found? Look for this in a dictionary or an encyclopaedia?
Answer:
Vicuna is found in Spain.

Question 6.
When handloom and textile exhibitions are held, certain stalls display real moths of various varieties of silk and their life Stories. Try and visit these stalls with elders or teachers and these moths and stages of their life history.
Answer:
Do yourself.

Question 7.
Look for eggs of any moth or butterfly in your garden or park or any other place full of plants. They look like tiny specks (dots) laid in a cluster on the leaves. Pull out the leaves containing eggs and place them in a cardboard box. Take some leaves of the same plant or another plant of the same variety, chop them and put them in the box.

Eggs will hatch into caterpillars, which are busy eating day and night. Add leaves everyday for them to feed upon. Sometimes you may be able to collect the caterpillars. But be careful. Use a paper napkin or a paper to hold a caterpillar. Observe everyday. Note the:

1. Number of days taken for eggs to hatch.
2. Number of days taken to reach cocoon stage and
3. To complete life cycle. Record your observations in your notebook.

Answer:
Do yourself.

Fibre to Fabric Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (i)
Silk is a –
(a) Rough fibre
(b) Way fibre
(c) Staple fi bre
(d) Filament fibre.
Answer:
(d) Filament fibre.

Question (ii)
Which country is the largest producer of silk –
(a) India
(b) Pakistan
(c) China
(d) None of these.
Answer:
(c) China

Question (iii)
Which country is the largest producer of wool –
(a) India
(b) Australia
(c) Pakistan
(d) China.
Answer:
(b) Australia

Question (iv)
Mooga is the strongest variety of –
(a) Silk
(b) Wool
(c) Cotton
(d) None of these.
Answer:
(a) Silk

Question (v)
An International trade mark for new and pure wool is –
(a) ISI
(b) ISO
(c) Woolmark
(d) Agmark.
Answer:
(c) Woolmark

Question (vi)
Virgin wool is obtained from –
(a) Adult sheep
(b) Dead sleep
(c) Lamb
(d) Carnal.
Answer:
(a) Adult sheep

Question 2.
Fill in the blanks:

1. Silk is a natural ………….. fibre.
2. Silk fibres come from …………….. of the silk moth.
3. The fine hair provide the fibres for making ……………
4. Yak wool is common in Tibet and ………….
5. Angora wool is obtained from …………. goats.
6. Sheep are herbivores and prefer grass and …………….
7. Sericulture of silk worms is a very old …………….. in India.
8. …………… leads the world in silk production.
9. ……………… is a dried perspiration found is the raw wool.
10. The science of raising silk worms so as to obtain silk cocoons is …………..

Answer:

1. Protein
2. Cocoons
3. Wool
4. Ladakh
5. Angora
6. Leaves
7. Occupation
8. China
9. Suint
10. Sericulture

Question 3.
Which of the following statements are true (T) or false (F):

1. Silk worm feeds an mulberry leaves.
2. A yarn is not a long continuous thread.
3. The wool and cotton are staple fibres.
4. Silk is a good conductor of heat.
5. The high grade silk is obtained from the filaments of damaged cocoons.
6. A female silk moth lays 100 of eggs at a time.
7. Sheeps are reared is many parts of our country for wool.
8. Lohi and Nali are the breeds of sheep.
9. Nowadays scouring is done by machines.
10. Sheep hair is sheared off from the body, scoured, sorted, dried, dyed, spun and woven to yield wool.

Answer:

1. True (T)
2. False (F)
3. True (T)
4. False (F)
5. False (F)
6. True (T)
7. True (T)
8. True (T)
9. True (T)
10. True (T)

Question 4.
Match the items in Column A with Column B:

Answer:

(i) (b)
(ii) (d)
(iii) (f)
(iv) (e)
(v) (c)
(vi) (a)

Fibre to Fabric Very Short Answer Type Questions

Question 1.
Write the name of some wool yielding animals.
Answer:
Sheep, goat, angora goat, camel, yak, lamb, etc.

Question 2.
Feel the hair on your body and arms and these on your head. Do you find any difference? Which are seems coarse and which are is soft?
Answer:
Yes, hairs in arms are soft.

Question 3.
In which area a yak found in plenty?
Answer:
Tibet and Ladakh.

Question 4.
What are the sources of obtaining fibres?
Answer:
Animals and plants are the sources of obtaining fibres.

Question 5.
What do you meant by shearing?
Answer:
The fleece of the sheep along with a thin layer of skin is removed from its body. This process is known as shearing.

Question 6.
Which animal provides pashmina shawls?
Answer:
Kashmiri goats.

Question 7.
Define wool?
Answer:
Wool is the common name of applied to soft curly fibres obtained chiefly from the fleece of domesticated sheep.

Question 8.
In which season, sleep sheared?
Answer:
Summer season.

Question 9.
What are the uses of wool?
Answer:
Wool is used for making fabrics, blankets, shawls, carpets, felt and upholstery.

Question 10.
What are the properties of wool fibre?
Answer:
Wool fibre is considerably resilient, has high tensile strength, light weight and is heat insulator.

Question 11.
What are silkworms?
Answer:
The female silk moth lays eggs from which hatch larvae are called silkworms of caterpillars.

Question 12.
What is the most common silk moth?
Answer:
Mulberry silk moth.

Question 13.
Name the strongest variety of silk.
Answer:
Mooga.

Question 14.
Name the different varieties of processed silk.
Answer:
Gepe, Tram, Organize, Trown singles are different varieties of processed silk.

Question 15.
What is reeling of silk?
Answer:
A pile of cocoons is used for obtaining silk fibres. This process is called reeling the silk.

Question 16.
Name some Indian breeds of sheep?
Answer:
Lohi, Rampur bushair, Nali, Bakharwal, Marwari and Patanwadi.

Question 17.
Where are sheep reared in India?
Answer:
Sub – Himalyan region because this area has a cooler climate.

Question 18.
Name one disease caused to sheep.
Answer:
Anthrax.

Fibre to Fabric Very Short Answer Type Questions

Question 1.
Name the animals that yield wool?
Answer:
The fleece of sheep is not the only source of wool, though wool commonly available in the market is sheep wool. Yak wool is common in Tibet and Ladakh. Angora wool is obtained from angora goats, found in hilly regions such as Jammu and Kashmir. Wool is also obtained from goat hair. The under fur of Kashmiri goat is soft. The fur (hair) on the body of camels is also used as wool. Uama and Alpaca, found in South Africa, also yield wool.

Question 2.
What is raw silk?
Answer:
After brushing, filaments from four to eight cocoons are joined and twisted. They are then combined with a number of other similarly twisted filaments, to make a thread that is wound on a reel. The thread is called raw silk.

Question 3.
How is shearing of .wool done? Describe.
Answer:
The fleece of the sheep along with a thin layers of skin is removed from its body .This process is called shearing. Machines similar to those used by barbers are used to shave off hair. Usually, hair are removed during the hot weather. This enables sheep to survive without their protective coat of hair. The hair provide woolen fibers. Woolen fibres are then processed to obtain woollen yarn. Shearing does not hurt the sheep just as it does not hurt when you get a hair cut or your father shaves his beard.

Question 4.
Name the some breeds of sheep reared in our country. Also indicate the quality and texture of the fibre, obtained from them?
Answer:
Some Indian breeds of sheep are given in following table:

Question 5.
Why do the wool fibres have greater bulk as compared to other fibres?
Answer:
The scales and crimps is the wool fibre make it possible to spin and felt the fleece. They help the individual wool fibres grab each other so that they stay together. Because of the crimp, wool fabrics have a greater bulk than other textiles. They can retain lot of air and hence wool fabrics are good insulators of heat. The heat insulation also works both ways.

Question 6.
How is silk processed?
Answer:
A pile of cocoons is used for obtaining silk fibres. This process is called reeling of silk. The cocoons are kept under the sun or boiled or exposed to steam. The silk fibres separate out. The process of taking out threads from the cocoon for use as silk is called reeling the silk. Reeling is done in special machines, which unwind the threads or fibres of silk from the cocoon. Silk fibres are then spun into silk threads, which are woven into silk cloth by weavers.

Question 7.
Write a short note on'”discovery of silk”?
Answer:
The exact time of discovery of silk is perhaps unknown. According to an old Chinese legend, the empress Si – lung – Chi was asked by the emperor Huang – ti to find the cause of the damaged leaves of mulberry trees growing their garden. The empress found white worms eating up mulberry leaves. She also noticed that they were spinning shiny cocoons around them. Accidentally a cocoon dropped into her cup of tea and a tangle of delicate threads separated from the cocoon.

Silk industry began in China and was kept a closely guarded secret for hundreds of years. Latter on, traders and travellers introduced silk to other countries. The route they travelled is still called the ‘silk route’.

Fibre to Fabric Long Answer Type Questions

Question 1.
How wool is produced from fibres?
Ans. For obtaining wool. Sheep are reared. Their hair is cut and processed into wool. Sheep are herbivores and prefer, grass and leaves. Apart from grazing sheep, rearers also feed them on a mixture of pulses, corn, jowar, oil cakes (material left after taking out oil from seeds) and minerals. In winter, sheep are kept indoors and fed on leaves, grain and dry fodder.

Sheep are reared in many parts of our country for wool. The quality and texture of the fibres obtained from different sheeps are different. Once the teared sheep have developed a thick growth of hair, hair is shaved off for getting wool.

Question 2.
Define with diagram the process of scouring?
Answer:

Question 3.
Define the process of sorting and rolling of fibres into wool?
Answer:
After scouring sorting is done. The hairy skin is sent to a factory where hair of different textures are separated or sorted. The small fluffy fibres, called burrs, are picked out from the hair. These are the same burrs which sometimes appear on your sweaters. The fibres are scoured again and dried. This is the wool ready to be drawn into fibres.

The fibres can be dyed in various colours, as the natural fleece of sheep and goats is black, brown or white. The fibres are straightened, combed and rolled into yarn (Fig.). The longer fibres are made into wool for sweaters and the shorter fibres are spun and woven into woollen cloth.

Question 4.
How we can obtain silk from cocoon? Explain.
Answer:
For obtaining silk, moths are reared and their cocoons are collected to get silk threads. A female silk moth lays hundreds of eggs at a time. The eggs are stored carefully on strips of cloth or paper and sold to silk worm farmers. The farmers keep eggs under hygienic conditions and under suitable conditions of temperature and humidity.

The eggs are warmed to a suitable temperature for the larvae to hatch from eggs. This is done when mulberry trees bear a fresh crop of leaves. The larvae, called caterpillars or silkworms, eat day and night and increase in size enormously.

The worms are kept in clean bamboo trays along with freshly chopped mulberry leaves. After 25 to 30 days, the caterpillars stop eating and move to a tiny chamber of bamboo in the tray to spin cocoons. Small racks or twigs may be provided in the trays to which cocoons get attached. The caterpillar or silkworm spins the cocoon inside which develops the silk moth.

MP Board Class 7th Science Solutions

MP Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

Question 1.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all Ijbe angles of the quadrilateral.

Solution:
∠A = 3x, ∠B = 5x, ∠C = 9x, ∠D = 13x
In quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
(∴ Sum of all the angles of ♢ is equal to 360°)

3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac{360^{\circ}}{30^{\circ}}\)
Let angle in ratio be x then angles are x= 12°
∠A = 3 x 12° = 36°
∠B = 5 x 12° = 60°
∠C = 9 x 12° = 108°
∠D = 13 x 12° = 156°

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Given
ABCD is a parallelogram in which
AC =DB
To prove:
ABCD is a rectangle.
Proof
In ∆DAB and ∆CBA
DB = CA (given)
AB = BA (common)
AD = BC (∴ opposites sides of ∥^gm are equal)
∆DAB = ∆CBA (by SSS)
and so ∠DAB = ∠CBA
AD∥BC and AB is the transversal (by CPCT)
∴ ∠A + ∠B = 180° (CIA’s)
⇒ ∠A + ∠A = 180° (∴ ∠A = ∠B)
∴ ∠A = 90°
∠A = ∠C = 90°
and ∠B = ∠D = 90°
In ∥^gm ABCD, all the angles are right angles.
ABCD is a rectangle.

Question 3.
Show that if the diagonals ofa quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Given
OA = OC, OB = OD and ∠AOD = 90°
To prove:
ABCD is a rhombus.
Proof:
In ∆AOD and ∆COB

OA = OC (given)
OD = OB (given)
∠1 = ∠2 (V.O.A.’s)
∴ ∆AOD = ∆COB (by SAS)
and so AD = CB (by CPC 7)
∠3 = ∠4 (by CPCT)
∠3 and ∠4 are A.I.A.’s and are equal
∴ AD ∥ PC (proved)
AD = BC
∴ ABCD is a parallelogram
In ∆AOD and ∆COD,
OA = OC (given)
OD = OD (common)
∠1 = ∠5 = 90°
(∴ ∠1 + ∠5 = 180° ⇒ 90° + ∠5 = 180° ∴ ∠5 = 90°)
∴ ∠AOD = ∠COD (by SAS)
and so AD = CD (by CPCT)
In ∥^gm APCD, all the sides are equal.
ABCD is rhombus.

Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Given
ABCD is a square.
To prove:

1. AC = BD
2. OA = OC and OB = OD
3. ∠AOD = 90°

Proof:
In ∆DAB and ACBA,
DA = CB (given)
AB = BA (common)
∠A = ∠B (each 90°)
∴ ∆DAB = ∆CBA (by SAS)
and so BD =AC (by CPCT)

2. In ∆AOD and ∆COB, AD = CB(given)

∠4 = ∠5 (A.I.A.’s)
∠6 = ∠7 (A.I.A. ’s)
∆AOD = ∆COB (by ASA)
and so OA=OC , (byCPCT)
OD = OB (byCPCT)

3. In ∆AOD and ∆COD,
AO = CO (proved)
OD = OD (common)
AD = CD (given)
∆AOD = ∆COD (by SSS)
and so ∠1 = ∠3 (byCPCT)
∠1 + ∠3 = 180° (LPA’S)
⇒ ∠1 + ∠1 = 180° (∠1 = ∠3)
⇒ 2∠1 = 180°
∴ ∠1 = \(\frac{180^{\circ}}{2}\)

Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:

Give
ABCD is in which
AC = BD
OA = OC
and OB = OD
∠AOD = 90°
To prove
ABCD is a square
Proof:
In ∆AOD and ∆COB,
OA = OC (given)
OD = OB (given)
∠7 = ∠8 (V.CXA.’s)
∴ ∆AOD = ∠COB
and so AD = BC
and ∠3 = ∠1 (byCPCT)
∠3 and ∠1 are A.I.A.’s and are equal
∴ AD ∥ BC
Similarly, AB ∥ CD
∴ ABCD is a parallelogram.
In ∆AOD and ∆COD,
OA = OC (given)
OD = OD (common)
∠7 = ∠9 (each 90°)
∆AOD = ∠COD (by SAS)
AD = CD (byCPCT)
In ∥^gmABCD, adjacent sidesAD = CD
∴ ABCD is a rhombus
In ∆DAB and ∆CBA,
DA = CB (proved)
AB = BA (common)
DB = CA (given)
∆DAB = ∆CBA (bySSS)
∠A = ∠B (by CPCT)
∠A + ∠B = 180° (CIA’s)
2∠A = 180°
∠A =90°
ABCD is a square.

Question 6.
Diagonals AC of a parallelogram ABCD bisects ∠A (see Fig). Show that

1. it bisects ∠C also,
2. ABCD is a rhombus.

Solution:
Given
ABCD is a parallelogram in which ∠1 = ∠2
To prove:

1. ∠3 = ∠4
2. ABCD is a rhombus.

Proof
1. ∠1 = ∠4 (A.I.A.’s) ….(i)
(∴ AD ∥ SC and AC is the transversal)
∠2 = ∠3 (A.I.A.’s) …(ii)
(∴ AB ∥ DC and AC is the transversal)
∠1 = ∠2 (given) …(iii)

From (i), (ii) and (iii), we get
∠4 = ∠3
(ii) From (ii) and (iii), we get,
∠1 = ∠3
In ∆ADC,
∠1 = ∠3
∴ AD = DC (In a A, sides opposites to equal angles are equal) and so ABCD is a rhombus
(∴ In a ∥^gm, if adjacent sides are equal then it is a rhombus)

Question 7.
ABCD is a rhombus, show that diagonal AC biusects ∠A as well as ∠C and diagonal BD biusects ∠B as well as ∠D.
Solution:
Given
ABCD is a rhombus.
To prove
∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8.
Proof:
In ∆ADC and ∆ABC.

AD = AB (Adj. sides of a rhombus)
DC = BC (Adj. sides of a rhombus)
AC = AC (common)
∴ ∆ADC = ∆ABC (by SSS)
so ∠1 = ∠2 (by CPCT)
and ∠3 = ∠4 (by CPCT)
∴ AC bisects ∠A and ∠C
Similarly, ∠5 = ∠6 and ∠7 = ∠8
BD bisects ∠B and ∠D.

Question 8.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that

1. ABCD is a square
2. diagonal BD bisects ∠B as well as ∠D.

Solution:
Given
ABCD is a rectangle.
∠1 = ∠2 and ∠3 = ∠4
To prove:

1. ABCD is a square
2. ∠5 = ∠6 and ∠7 = ∠8

Proof:
1. ∠A = ∠C
(∵ Rectangle is ∥^gm and in a ∥^gm opp. angles are equal.)

\(\frac{1}{2}\)∠A = \(\frac{1}{2}\)∠C
∠A = ∠C
∠2 = ∠4
In ∆ABC, ∠2 = ∠4
AB = BC
(In a A, sides opp. to equal angles are always equal)
ABCD is a rectangle in which adjacent sides are equal.
∴ ABCD is a square.

2. In ∆ABD,
AB = AD (∴ ABCD is a square)
∴ ∠5 = ∠7 (∴ In a A, angles opp. to equal sides are equal) ….(1)
AB ∥ DC and BD is the transversal
∴ ∠6 = ∠7 …(2)
AD ∥ BC and BD is the transversal.
∴ ∠5 = ∠8 …(3)
From (1) and (3), we get
∠7 = ∠8
From (1) and (2), we get
∠5 = ∠6
Diagonal BD bisects ∠B as well as ∠D.

Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig.). Show that

1. ∆APD = ∆CQB
2. AP = CQ
3. ∆AQB = ∆CPD
4. AQ = CP A
5. APCQ is a parallelogram

Given
ABCD is a parallelogram.
∴ AD = BC, AB= DC and DP = BQ
To prove:

1. ∆APD = ∆CQB
2. AP = CQ
3. ∆AQB = ∆CPD
4. AQ = CP
5. APCQ is a parallelogram

Proof:
In ∆APD and ∆CQB
PD = QB (given)
AD = CB (given)
∠2 = ∠1 (AIA’s)
∆APD = ∆CQB (by SAS)
and so AP = CQ (by CPCT)

In ∆AQB and ∆CPD,
AB = CD (given)
∠3 = ∠4 (AIA’s)
BQ = DP (given)
∆AQB = ∆CPD (by SAS)
and so AQ = CP (by CPCT)
In quadrilaterals AQCP,
AP = CQ
AQ = CP
AQCP is a parallelogram.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD (see Fig.). Show that

1. ∆APB = ∆CQD
2. AP = CQ

Solution:
Given
ABCD is a ∥^gm in which Ap ⊥ BD and CQ ⊥ BD.
To prove:

1. ∆APB = ∆CQD
2. AP = CQ

Proof:
In ∆APB and ∆CQD,
∠P = ∠Q (each 90°)
∠1 = ∠2 (AIA’s)
AB = CD (given)
∆APB ≅ ∆CQD (byAAS)
and so AP = CQ (by CPCT)

Question 11.
In ∆ABC and ∆DEF, AB = DE, AB ∥ DE, BC = EF and BC ∥ EE. Vertices A, B and C are joined to vertices, D,E and F respectively (see Fig.). Show that

1. quadrilateral ABED is a parallelogram
2. quadrilateral BEFC is a parallelogram
3. AD ∥ CF and AD = CF
4. quadrilateral ACFD is a parallelogram
5. AC = DF
6. ∆ABC = ∆DEF.

Solution:
Given
AB = DE and AB ∥ DE
BC = EF and BC ∥ EF
To prove

1. ABED is a ∥^gm
2. BEFC is a ∥^gm
3. AD ∥ CF and AD – CF
4. ACFD is a ∥^gm
5. AC =DF
6. ∆ABC ≅ ∆DEF

Proof:
1. AB = DE and AB ∥ DE (given)
ABED is ∥^gm
and so AD ∥ BE and AD = BE …(1)

2. BC = EF and BC ∥ FC (given)
BEFC is a ∥^gm
and so BE ∥ CF and BE = CF …..(2)

3. From (1) and (2), we get
AD ∥ CF and AD = CF

4. AD ∥ CF and AD = CF (proved)
ACFD is a ∥^gm

5. and so AC = DF
(In a parallelogram, opp. sides are equal)

6. In ∆ABC and ∆DEF,
AB = DE (given)
BC = EF (given)
AC = DF (proved)
∆ABC = ∆DEF (by SSS)

Question 12.
ABCD is a trapezium in which AB ∥ CD and AD = BC (see Fig.). Show that:

1. ∠A = ∠B
2. ∠C = ∠D
3. ∆ABC ≅ ∆BAD
4. Diagonal AC = diagonal BF)

[Hint: Extend AB and draw line through C parallel to DA intersecting AB produced at E.]
Solution:
Given
AB ∥ CD, AD = BC
To prove:

1. ∠A = ∠B
2. ∠C = ∠D
3. ∆ABC ≅ ∆BAD
4. diagonal AC = diagonal BD

Construction:
Draw a line CE ∥ DA which intersect AB produced at E.
Proof:
1. In quadrilateral ADCE,
AD ∥ EC (by const)
and AE ∥ DC ( AB ∥ DC)
∴ ADCE is a parallelogram
and so AD = EC (opp. sides of a ∥ are equal)…(i)
AD = BC (given) ….(ii)
From (i) and (ii), we get
BC = EC
In ∆BCE BC = EC (proved)
∠4 = ∠3
(∴ In a ∆, angles opp. to equal sides are equal)
∠2 + ∠3 = 180° (LPA’s) …(iii)
∠1 + ∠4 = 180° (CIA’s) …(iv)
From (iii) and (iv), we get
∠2 + ∠3 = ∠1 + ∠4
∠2 = ∠1 (∠3 = ∠4)
i.e., ∠A = ∠B

2. ∠3 = ∠5 (AIA’s) …(v)
∠6 = ∠4
(∴ ADCE is a ∥^gm and in a ∥^gm opp. angles are equal) …(vi)
∠4 = ∠3 (proved) …(vii)
From (v), (vi) and (vii), we get
∠5 = ∠6
i.e., ∠C = ∠D

3. In ∆ABC and ∆BAD,
AB = BA (common)
BC = AD (given)
∠2 = ∠1 (proved)
∆ABC ≅ ∆BAD (by SAS)
and so AC =BD (by CPCT)

4. diagonal AC = diagonal BD (proved)

Mid Point Theorem:
The line segment joining the mid-points of the sides of a triangle is parallel to the third side and equal to half of it.
Given.
ABC is a A in which D and E are the mid-points of sides AB and AC respectively.
To prove.
DE ∥ BC and DE = \(\frac{1}{2}\) BC
Construction:
Extend DE uptoFsuch that DE = EF. Join CF.
Proof:
In ∆AED and ∆CEE
AE = CE (E is the mid – point of AC)
∠AED = ∠CEF (VOA’s)
DE = FE (By constriction)
∆AED = ∆CEF , (By SAS)
and so ∠DAE = ∠FCE (By CPCT)
AD = CF (By CPCT)
∠DAE and ∠FCE are alternate interior angles and are equal.

AD ∥ FC
⇒ DB ∥ FC
Now, AD = DB and AD = FC
DB = FC
In BCFD, DB ∥ FC and DF = BC
BCFD is a ∥^gm
and so DF ∥ BC and DF = BC
⇒ DF ∥ BC and 2DF = BC
DE ∥ BC and DE = \(\frac{1}{2}\) BC

Converse of mid point theorem:
The line drawn through the mid – point of one side of a triangle and parallel to another side, bisects the third side.
Given
ABC is a A in which D is the mid-point of AB and DE ∥ BC.
To Prove:
E is the mid – point of AC.
Construction:
Mark a point F on AC and join DF.
Proof:
Let E be not the mid – point of AC. Let us assume that F be the mid – point of AC.
Then by mid-point theorem

DF ∥ BC
DE ∥ BC (Given)
From (i) and (ii), we get
DE ∥ DF
But lines DE and DF are intersecting lines, intersecting at D. This is a contradiction. So our supposition is wrong. Hence E is the mid – point of AC.

MP Board Class 9th Maths Solutions

MP Board Class 7th Science Solutions Chapter 13 Motion and Time

Motion and Time Intex Questions

Question 1.
Paheli wondered how time was measured when pendulum clocks were not available?
Answer:
Many times measuring devices were used in different parts of the world before the pendulum clocks became popular. Water clocks, sand clocks and sundials are some examples of such devices.

Question 2.
Boojho wants to know whether there is any device that measures the speed?
Answer:
Yes, speedometer.

Activities

Activity – 1
Table gives some common examples of motions. Identify the type of motion in each case?
Answer:
Some examples of different types of motion

Activity – 2
Draw a straight line on the ground with chalk powder or lime and ask one of your friends to stand 1 to 2 m away from it. Let your friend gently roll a ball along the ground in a direction perpendicular to the line. Note the time at the moment the ball crosses the line and also when it comes to rest (Fig.) How much time does the ball take to come to rest? Measure the distance between the point at which the ball crosses the line and the point where it comes to rest. You can use a scale or a measuring tape. Let different groups repeat the activity. Record the measurements in Table. In each case calculate the speed of the ball?

Answer:

Activity – 3
The unit of speed in the given table is in km/h. Change this speed in to m/s?
Answer:
Speed of some animals.

Motion and Time Text Book Exercises

Question 1.
Classify the following as motion along a straight line, circular or oscillatory motion:

1. Motion of your hands while running.
2. Motion of a horse pulling a cart on a straight road.
3. Motion of a child in a merry – go – round.
4. Motion of a child on a see – saw.
5. Motion of the hammer of an electric bell.
6. Motion of a train on a straight bridge.

Answer:

1. Oscillatory
2. Straight line
3. Circular
4. Oscillatory
5. Oscillatory
6. Straight line.

Question 2.
Which of the following are not correct?

1. The basic unit of time is second.
2. Every object moves with a constant speed.
3. Distances between two cities are measured in kilometres.
4. The time period of a given pendulum is not constant.
5. The speed of a train is expressed in m/h.

Answer:

2. Every object moves with a constant speed.
5. The speed of a train is expressed in m/h.

Question 3.
A simple pendulum takes 32 s to complete 20 oscillations?What is the time period of the pendulum?
Answer:
Given, time taken to complete 20 oscillations = 32 seconds
∴Time taken to complete 1 oscillation = \(\frac{32}{20}\)s = 1.6 s
Thus, the time period of pendulum is 1.6 s

Question 4.
The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train?
Answer:
Given, distance between two stations = 240 km and time taken to cover this distance = 4 hour
We know that,

\(\frac{240}{4}\) = 60 km/hour.
Thus, the speed of the train is 60 km/hour.

Question 5.
The odometer of a car reads 57321.0 km when the clock shows the time 08.30 AM. What is the distance moved by the car, if at 08:50 AM, the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time. Express the speed in km/h also?
Answer:
Given, reading of odometer at time 08:30 AM = 57321.0 km and reading of odometer at time 08:50 AM = 57336.0 km.
∴Distance covered by car = (57336.0 – 57321.0) km = 15 km and Time taken = 8.50 – 8.30 = 20 minutes
We know that,

Question 6.
The odometer of a car reads 57321.0 km when the clock shows the time 08.30 AM. What is the distance moved by the car, if at 08:50 AM, the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time. Express the speed in km/h also?
Answer:
Given, reading of odometer at time 08:30 AM = 57321.0 km and reading of odometer at time 08:50 AM = 57336.0 km
∴Distance covered by car = (57336.0 – 57321.0) km = 15 km and Time taken = 8.50 – 8.30 = 20 minutes We know that,

Thus, the distance between Salma’s school from her house is 1800 m or 1.8 km.

Question 7.
Show the shape of the distance – time graph for the motion in the following cases:

1. A car moving with a constant speed.
2. A car parked on a side road.

Answer:

Question 8.
Which of the following relations is correct?

Answer:

Question 9.
The basic unit of speed is:
(a) 100km.
(b) m/min
(c) km/h
(d) m/s.
Answer:
(d) m/s.

Question 10.
A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:
(a) 100 km
(b) 25 km
(c) 15 km
(d) 10 km.
Answer:
(b) 25 km.
Distance travelled in first 15 minutes = Speed x Time = 40 km/h x 15 minutes.
= 40 km/h × \(\frac{15}{60}\) hour
= 10 km
Distance travelled in last 15 minutes = Speed x Time = 60 km/h x 15 minutes
= 60 km/h × 15 minutes
= 60 km/h × \(\frac{15}{60}\) hour
= 15 km.
Total distance = (10 + 15) km = 25 km.

Question 11.
Suppose the two photographs, shown in Fig. (a) and Fig. (b) had been taken at an interval of 10 seconds. If a distance of 100 meters is shown by 1 cm in these photographs, calculate the speed of the blue car?


Answer:
Speed = \(\frac{100 km}{10 h}\) = 10 km/h.

Question 12.
Fig. shows the distance – time graph for the motion of two vehicles A and B. Which one of them is moving faster?

Answer:
Car A is moving faster.

Question 13.
Which of the following distance – time graphs shows a truck moving with speed which is not constant?

Answer:
(iii)

Motion and Time Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative :

Question (a)
The time taken for one oscillation is called its
(a) Pitch
(b) Frequency
(c) Vibration
(d) Time period.
Answer:
(d) Time period.

Question (b)
1 minute has?
(a) 30 second
(b) 60 second
(c) 100 second
(d) None of these.
Answer:
(b) 60 second

Question (c)
The smallest time interval that can be measured with commonly available clocks and watches is?
(a) 1 second
(b) 1 minute
(c) 1 hour
(d) None of these.
Answer:
(c) 1 hour

Question (d)
The ages of stars and planet are often expressed in –
(a) billions of years
(b) days
(c) years
(d) None of these.
Answer:
(a) billions of years

Question (e)
The distance moved by a vehicle can be measured by –
(a) speedometer
(b) odometer
(c) regulator
(d) None of these.
Answer:
(b) odometer

Question (f)
The distance – time graph of a body moving with uniform speed is –
(a) a curve
(b) a straight line
(c) line parallel to x – axis
(d) None of these.
Answer:
(b) a straight line

Question 2.
Fill in the blanks:

1. We shall use the term speed for average ……………………….
2. An object moving along a straight line with a constant speed is said to be in …………………………
3. The to and fro motion of a simple pendulum is an example of a ……………….. or ……………………. motion.
4. The basic unit of time is a ……………………….
5. The basic unit of speed is ………………………..
6. Periodic events are used for the measurement of ……………………….
7. The distance – time graph for the motion of an object moving with a constant speed is a ……………………
8. The speedometer records the speed in ……………………..
9. Odometer is used to measure ………………………
10. The distance moved by an object is a unit time is called its ………………………….

Answer:

1. Speed
2. Uniform speed
3. Periodic, oscillatory
4. Second
5. m/s
6. Time
7. Straight line
8. km/h
9. Distance
10. Speed.

Question 3.
Which of the following statements are true (T) or false(F).

1. The distance moved by objects in a given interval of time can help us to decide which one is faster or slower.
2. We shall not use the term speed for average speed.
3. The to and fro motion of a simple pendulum is an example of a periodic motion.
4. Janter Manter in Delhi and Jaipur have big Sundials.
5. In non – uniform motion, the speed of an object changes.
6. Clocks or watches are perhaps the most common time measuring devices.
7. Different units of time are used depending on the need.

Answer:

1. True
2. False
3. True
4. True
5. True
6. True
7. True.

Motion and Time Very Short Answer Type Questions

Question 1.
Define motion.
Answer:
Motion is the change in position of a body with respect to line and its surroundings.

Question 2.
What type of motion do the vehicles on a straight road perform?
Answer:
The vehicles perform rectilinear motion on the road.

Question 3.
What type of motion does the wheel of a bicycle perform?
Answer:
The wheel of a bicycle perform rotatory or circular motion.

Question 4.
What is uniform motion?
Answer:
Motion of a body along a straight line covering equal distances in equal intervals of time is called uniform motion.

Question 5.
Give two example of non – uniform motion?
Answer:

1. Motion of the train when its driver applies breaks.
2. Motion of the cricket ball when the bowler throws the ball and the ball is hit by the player.

Question 6.
Give two example of periodic motion?
Answer:

1. Rotation of earth on its axis.
2. Moon moves around the earth.

Question 7.
What is oscillatory motion?
Answer:
The to and fro motion is called periodic or an oscillatory motion.

Question 8.
An object is hung from a spring is pulled down and left? What type of motion does the object perform?
Answer:
The object perform oscillatory motion which is also a periodic motion.

Question 9.
Write the formula for speed?
Answer:

Question 10.
What is the unit of speed?
Answer:
Unit of speed is meter per second.

Question 11.
A snail covers a distance of 1 meter in 10 sec. What is the speed of the snail?
Answer:
image
= 0.1 ms^-1
= 0.1 m per sec.

Question 12.
What happens to the speed of your bicycle when you pedal faster?
Answer:
On pedalling fast the speed of bicycle increase.

Question 13.
“Speed of a car is 8 km/h”. What does it mean?
Answer:
When we say that a car is moving with a speed of 80 kilometer per hour, it implies that it will cover a distance of 80 kilometre in one hour.

Question 14.
Define a day?
Answer:
The time between one sunrise and the next is called a day.

Question 15.
What is month ?
Answer:
A month is measured from one new moon to the next.

Question 16.
What is year?
Answer:
A year is fixed as the time taken by the earth to complete one revolution of the sun.

Question 17.
What is bob?
Answer:
The metallic ball is called bob of the pendulum.

Question 18.
What is the use of stop – watch?
Answer:
Stop – watch is used to measure exact time in case of atheletic events as it can be stopped and started any moment.

Question 19.
When you travel in bus or car then, generally, what kind of motion it is?
Answer:
Linear motion.

Question 20.
There is 100 metres line on which two sprinters have to run, when they run, what kind of motion it will be?
Answer:
Linear motion.

Question 21.
See the picture and tell whose speed is more:


Answer:
Speed of tiger is more than the speed of horse.

Motion and Time Short Answer Type Questions

Question 1.
Define periodic motion?
Answer:
Periodic motion:
The motion of a body which is repeated at regular intervals of time is called periodic motion. In periodic motion, a body covers equal distances in equal intervals of time; e.g. revolution of earth round the sun, the motion of moon round the earth, the motion of a swing, the motion of a pendulum, etc.

Question 2.
Define non – periodic motion?
Answer:
Non – periodic motion:
The motion of a body which is not repeated at regular intervals of time is called non – periodic motion. In non – periodic motion, a body does not cover equal distances in equal intervals of time; e.g. an athelete running on a track, a moving car on busy road, etc.

Question 3.
Define oscillatory motion?
Answer:
Oscillatory motion:
A body is said to be in oscillatory motion when it moves to and fro about a fixed point; e.g. the pendulum of a wall clock (Fig.), a swing, a wire of sitar when plucked, striking of drum, etc. Mostly the hanging objects show oscillatory motion.

Question 4.
How can we make a pendulum of air one? What is the time – period of a pendulum?
Answer:
We can make a pendulum by suspending a metal ball with a cotton thread. The other end of the thread can be tied to same support. The time taken by a pendulum to complete its to and from movement i.e., one oscillation is called the time period of the pendulum.

Question 5.
One student reaches school in 1 hour by travelling a distance of 8 km from his house by a cycle and the other student reaches school in one and half hours by travelling a distance of 12 kms from his house by cycle. Tell whose speed was slow?
Answer:

From above calculations it is clear that the speed of both the students is same.

Question 6.
What are quartz clocks? Give a diagram of digital clock?
Answer:
Now a days most clocks or watches have an electric circuit with one or more cells. These clocks are called quartz clocks. The time measured by quartz clocks is much more accurate than that by the clocks available earlier.

Question 7.
Mark x – axis, y – axis, origin on a graph paper?
Answer:

Motion and Time Long Answer Type Questions

Question 1.
Write briefly about the development of pendulum clock?
Answer:
There is an interesting story about the discovery that the time period of a given pendulum is constant. You might have heard the name of famous scientist Galileo Galilie (A.D. 1564 – 1642). It is said that once Galileo was sitting in a church. He noticed that a lamp suspended from the ceiling with a chain was moving slowly from one side to the other.

He was surprised to find that his pulse beat the same number of times during the interval in which the lamp completed one oscillation. Galileo experimented with various pendulums to verify his observation. He found that a pendulum of a given length takes always the same time to complete one oscillation. This observation led to the development of pendulum clocks. Winding clocks and wristwatches were refinements of the pendulum clocks.

Question 2.
Discuss in details about the different range to measure time interval?
Or
Discuss the different units for measuring time interval?
Answer:
The smallest time interval that can be measured with commonly available clocks and watches is one second. However, now special clocks are available that can measure time intervals smaller than a second. Some of these clocks can measure time intervals as small as one millionth or even one billionth of a second. We might have heard the terms like microsecond and nanosecond. One microsecond is one millionth of a second. A nanosecond is one billionth of a second.

Clocks that measure such small time intervals are used for scientific research. The time measuring devices used in sports can measure time intervals that are one tenth or one hundredth of a second. On the other hand, times of historical everts are stated in terms of centuries or millenniums. The ages of stars and planet are often expressed in billions of years.

Question 3.
Draw a bar graph to show the runs scored by a team in each over?

Answer:

Question 4.
Draw a pie chart to show the composition of air?
Answer:

Question 5.
Write down the steps to make a distance – time graph?
Answer:
We can make the graph by following steps:
Step 1.
Draw two perpendicular lines to represent the two axes and mark them as OX and OY as shown in figure.

Steps 2.
Decide the quantity to be shown along the x – axis and that to be shown along the y – axis. In this case we show the time along the x – axis and the distance along the y – axis.

Steps 3.
Choose a scale to represent the distance and another to represent the time on the graph. For the motion of the car scales could be
Time : 1 min = 1 cm
Distance : 1 km = 1 cm

Step 4.
Mark values for the time and the distance on the respective axes according to the scale chosen. For the motion of the cat let mark the time 1 min, 2 min,..,on the origin x – axis from the origin Similarly, mark the distance 1 km. 2 km… on the y – axis.

Steps 5.
Now you have to mark the points on the graph paper to represent each set of values for distance and time.

Step 6.
Join all the points on the graph as shown in figure. It is a straight line. This is the distance-time graph for the motion of the car.
Note:
If the distance – time graph is a straight line, it indicates that the object is moving with a constant speed. However, if the speed of the object keeps changing, the graph can be of any shape.

MP Board Class 7th Science Solutions

MP Board Class 7th Science Solutions Chapter 2 Nutrition in Animals

Nutrition in Animals Intext Questions

Question 1.
Paheli wants to know how food moves in the opposite direction during vomiting?
Answer:
The food pipe runs along the neck and the chest. Food is pushed down by movement of the wall of the food pipe. Actually this movement takes place throughout the alimentary canal and pushes the food downwards. At times the food is not accepted by our stomach and is vomited out.

Question 2.
Paheli wants to know why these animals cannot chew food properly at the time they take it in?
Answer:
The ruminants mainly feed on grass and bush which primarily contain cellulose or roughage. The grass is rich in cellulose, so lot of chewing and saliva are need. Thus, the ruminants need to chew the grass for long time. If they chew for long while eating, they will get less time to eat.

Question 3.
Boojho wants to know why we cannot digest cellulose like the cattle do?
Answer:
Ruminants have a large sac – like structure between the small intestine and large intestine. The cellulose of the food is digested here by the action of certain bacteria which are not present in humans.

Activities
Activity 1
What is the type of food and mode of feeding of the following animals? Write down your observations in the given Table. You may find the list of modes of feeding given below the Table helpful.
Table 1:
Various modes of feeding

(Scraping, chewing, brewing, capturing and swallowing, sucking etc.)
Answer:

Activity 2
Wash your hands Look into the mirror and count your teeth. Use your index finger to feel the teeth. How many kinds of teeth could you find? Take a piece of an apple or bread and eat it. Which teeth do you use for biting and cutting and which ones for piercing and tearing? Also find out the ones that are used for chewing and grinding?
Record your observations in Table 2.

Answer:
Four.

1. Incisors are used for biting and cutting.
2. Canines are used for piercing and tearing.
3. Premolars and molars are used for chewing and grinding.

Nutrition in Plants Text book Exercises

Question 1.
Fill in the blanks:

1. The main steps of digestion in humans are …………, …………, …………, …………, and …………
2. The largest gland in the human body is …………
3. The stomach releases hydrochloric acid and ………… juices which act on food.
4. The inner wall of the small intestine has many finger – like outgrowths called …………
5. Amoeba digests its food in the …………

Answer:

1. Ingestion, digestion, absorption, assimilation and egestion
2. Liver
3. Digestive
4. Villi
5. Food vacuole.

Question 2.
Mark ‘T’ if the statement is true and ‘F’ if it is false:

1. Digestion of starch starts in the stomach. (T/F)
2. The tongue helps in mixing food with saliva. (T/F)
3. The gall bladder temporarily stores bile. (T/F)
4. The ruminants bring back swallowed grass into their mouth and chew it for some time. (T/F)

Answer:

1. False (F)
2. True (T)
3. True (T)
4. True (T)

Question 3.
Tick (S) mark the correct answer in each of the following:

Question (a)
Fat is completely digested in the –
(i) Stomach
(ii) Mouth
(iii) Small intestine
(iv) Large intestine
Answer:
(iii) Small intestine

Question (b)
Water from the undigested food is absorbed mainly in the –
(i) Stomach
(ii) Food pipe
(iii) Small intestine
(iv) Large intestine
Answer:
(iv) Large intestine

Question 4.
Match the items of Column I with those given in Column II:

Answer:

(i) (b)
(ii) (c)
(iii) (a).

Question 5.
What are villi? What is their location and function?
Answer:
The inner walls of the small intestine have thousands of finger – like outgrowths, these are called villi. They are situated in the small intestine.
Functions of Villi:

1. The villi increase the surface area for absorption of the digested food.
2. Each villus has a network of thin and small blood vessels close to its surface.
3. The surface of the villi absorbs the digested food materials.

The absorbed substances are transported via the blood vessels to different organs of the body where they are used to build complex substances such as the proteins required by the body.

Question 6.
Where is the bile produced? Which component of the food does it digest?
Answer:
The liver secretes bile juice that is stored in a sea called gall bladder. Bile juice digests fat.

Question 7.
Name the types of carbohydrate that can be digested by ruminants but not by humans. Give the reason also?
Answer:
The grass is rich in cellulose, a type of carbohydrate. Many animals including humans cannot digest cellulose. The cellulose can be digested by ruminants but not by humans because they have a large sac – like structure between the small intestine and large intestine. The cellulose of the food is digested here by the action of certain bacteria which are not present in humans.

Question 8.
Why do we get instant energy from glucose?
Answer:
We get instant energy from glucose because glucose is the simplest form of carbohydrate which can be broken down easily to give energy.

Question 9.
Which part of the digestive canal is involved in:

1. Absorption of food ……………….
2. Chewing of food
3. ……………….
4. killing of bacteria …………………..
5. Complete digestion of food ………………….
6. Formation of faeces ……………….

Answer:

1. Small intestine
2. Buccal cavity
3. Stomach
4. Small intestine
5. Large intestine.

Question 10.
Write one similarity and one difference between the nutrition in Amoeba and human beings?
Answer:
Similarity:
Both amoeba and human use the process of digestion involves ingestion, digestion, absorption, assimilation and egestion.

Difference:
Human needs to chew food whereas in amoeba, there is no chewing.

Question 11.
Match the items of Column I with suitable items in Column II

Answer:

(a) (iii)
(b) (iv)
(c) (i)
(d) (vii)
(e) (v)
(f) (vi).

Question 12.
Label Fig of the digestive system

Answer:

Question 13.
Can we survive only on raw, leafy vegetables/grass? Discuss.
Answer:
No, because to live healthy life we need a complete balance of all nutrients. Raw green vegetables may have cellulose which can not be digested by us. Thus, only green leafy vegetbles will not solve the purpose.

Extended Learning Activities and Project

Question 1.
Visit a doctor and find out:

1. Under what conditions does a patient need to be on a drip of glucose?
2. Till when does a patient need to be given glucose?
3. How does glucose help the patient recover?

Write the answers in your notebook.
Answer:

1. A patient is given glucose. When he/she is unable to eat food, unable to digest food.
2. Till the crisis is over.
3. Glucose chip has the following positive points to recover patient:

1. It need not be digested
2. It need not be eaten
3. It gives instant energy.

Question 2.
Find out what vitamins are and get the following information:

1. Why are vitamins necessary in the diet?
2. Which fruits or vegetables should be eaten regularly to get vitamins?

Answer:

1. Vitamins help to protect against diseases. They also help m proper functioning of body and brain.
2. All green leafy vegetables, fruits, milk, egg, butter, etc. provide vitamins.

Question 3.
Collect data from your friends, neighbours and classmates to know more about milk teeth?
Tabulate your data. One way of doing it is given below:

Find out from at least twenty children and find the average age. at which children lose the milk teeth. You may take help of your friends.
Answer:
Do with the help of your subject teacher.

Nutrition in Plants Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (i)
Which one of the following is not an excretory organ –
(a) Oesophagus
(b) Skin
(c) Lever
(d) Kidney.
Answer:
(a) Oesophagus

Question (ii)
How many times heart of an healthy person beats in one minute –
(a) 70
(b) 71
(c) 72
(d) 73.
Answer:
(c) 72

Question (iii)
Which organ produces bile –
(a) Liver
(b) Stomach
(c) Gall bladder
(d) Pancreas.
Answer:
(a) Liver

Question (iv)
The undigested food is eliminated through –
(a) Anus
(b) Lungs
(c) Caecum
(d) Colon.
Answer:
(a) Anus

Question (v)
The teeth which help in biting the food are –
(a) Incisors
(b) Canines
(c) Premolars
(d) Molars.
Answer:
(a) Incisors

Question (vi)
The number of molors present in a adult are –
(a) 2
(b) 4
(c) 6
(d) None of the above
Answer:
(c) 6

Question (vii)
The number of premolors present in a adult are –
(a) 2
(b) 4
(c) 6
(d) None of the above
Answer:
(b) 4

Question 2.
Fill in the blanks:

1. The breakdown of complex components of food into smaller substances is called …………….
2. The process of taking food into the body is called …………….
3. Our mouth has the salivary glands which secrete …………….
4. The acids gradually damage the teeth, this is called tooth …………….
5. The digestive juices break down the ……………. into simpler substances.
6. The bile plays an important role in the digestion of …………….
7. The grass is rich in …………….

Answer:

1. Digestion
2. Ingestion
3. Saliva
4. Decay
5. Proteins
6. fats
7. Cellulose.

Question 3.
Which of the following statements are True (T) or False (F):

1. Animal nutrition includes nutrient requirement, mode of intake of food and its utilisation in the body.
2. The human digestive system consists of the alimentary canal and secretory glands.
3. The modes of feeding do not vary in different organisms.
4. The absorbed substances are transported to different parts of the body.
5. The undigested and unabsorbed residues are expelled out of the body as faces through the anus.
6. Amoeba ingests its food with the help of its false feet or pseudopodia.
7. Gastric juice is secreted by the small intestine.
8. An adult has 22 permanent teeth.
9. Mouth cavity contains salivary glands and teeth only.
10. We can taste food with our tongue.

Answer:

1. True (T)
2. True (T)
3. False (F)
4. True (T)
5. True (T)
6. True (T)
7. False (F)
8. False (F)
9. False (F)
10. True (T).

Question 4.
Match the items in Column A with Column B.

Answer:

(i) (b)
(ii) (d)
(iii) (a)
(iv) (c).

Nutrition in Plants  Very Short Answer Type Questions

Question 1.
Name the largest gland of the human body?
Answer:
Liver.

Question 2.
Which chemical reaction takes place in internal respiration?
Answer:
Glucose + Oxygen → Carbon – di – oxide + Water + Energy.

Question 3.
Give an example of an oxy respiration?
Answer:
In human beings.

Question 4.
Which organs of plants participate in respiration?
Answer:
There is no special organ in plants for breathing.

Question 5.
Write the names of various stages of digestion?
Answer:
The digestion process is completed in following stages:

1. Ingestion
2. Digestion
3. Absorption
4. Assimilation
5. Egestion

Question 6.
Write the name of secretion secreted in human stomach?
Answer:
Gastric juce and hydrochloric acid is secreted in human stomach.

Question 7.
What is the meaning of enzyme?
Answer:
The complex molecules of ingested food are decomposed into smaller molecules. This process takes place in presence of a special substance called as enzyme. Enzyme increases the spread of reaction.

Question 8.
In which part of food canal water is absorbed?
Answer:
Large intestine.

Question 9.
Write the digestive organs given below in proper sequence?
Large intestine → stomach mouth cavity → small intestine → anus oesophagus.
Answer:
Mouth cavity Oesophagus → Stomach → Small intestine → Large intestine →Anus.

Question 10.
Which are the excretory organs in humans?
Answer:
The excretory organs in human are kidney, skin, sweat and lever.

Question 11.
Which is the excretory organ in paramoecium?
Answer:
The contracitile is the excretory organ in paramoecium.

Question 12.
Explain the digestion of food in stomach?
Answer:
The food halts here for longest time. Gastric juice and hydrochloric acid is secreted h£re. Proteins are digested here and food becomes acidic in nature due to presence of hydrochloric acid.

Question 13.
What is called excretion?
Answer:
Excretion is the process of the removal of the waste product from the body of a plant or an animal. The different organisms eliminate their waste in different ways.

Question 14.
What is common between our nose, gills of a fish and stomata of leaves?
Answer:
All these organs are the respiratory organs. The exchange of gases (oxygen and carbon dioxide) takes place through them. Thus they have similarities.

Question 15.
Name five herbivores and five carnivores?
Answer:
Herbivores:
Cow goat, buffalow, horse and deer.

Carnivores:
Lion, frog, wolf, snake, peacock.

Question 16.
What is the function of saliva secreted by salivary glands?
Answer:
The functions of saliva are:

1. It makes the food soft and easy to swallow.
2. It also helps in digestion to small extent.

Question 17.
Name the four types of teeth?
Answer:
Incisor, canine, premolar and molar.

Question 18.
Which teeth do you use for biting and cutting?
Answer:
Inscisors are use for biting and cutting.

Question 19.
What is cud ?
Answer:
The partially digested food stored in rumen is called cud.

Question 20.
Write any two functions of villi?
Answer:

1. Villi provides space for food.
2. Digestive juice is secreted out from the digestive glands present in the villi in the embedded form.

Question 21.
What is anylase?
Answer:
Saliva contains an enzyme called anylase. Amylase acts on start and changes it into a sugar called maltose.

Question 22.
Define stomach?
Answer:
Stomach is a thick walled bag Its shape is like a flattened U and it is the widest part of the alimentary canal. It received food from the food pipe at one end and opens into the small intestine at the other.

Question 23.
Name all the parts of alimentary canal?
Answer:
The buccal cavity, oesophagus, stomach, small intestine, large intestine and anus.

Nutrition in Plants Short Answer Type Questions

Question 1.
What do you understand by respiration?
Answer:
All the living organism penorm a number of vital activities. The energy is obtained by the oxidation of food or respiration. When the living organisms are completely at the stage of rest, even then they require some minimum amount of energy for the maintenance of cells and tissues. Thus, respiration is one of the most important process for the living organisms.

Question 2.
Write names of organs in sequence through which food passes starting from mouth?
Answer:
First of all food is taken in mouth with help of hands. Food passes further through various organs of food canal. The sequence is as following:
Mouth cavity → Pharynx – Oesophagus → stomach → Large intestine ←Duodenum Ileum

Question 3.
What things in our mouth help in physical and chemical digestion?
Answer:
The teeth and tongue help on physical digestion. The teeth help in cutting, crushing and chewing the food. The tongue helps in mixing the saliva with food. The enzymes and saliva secreted by salivary glands help in chemical digestion.

Question 4.
What makes the food move in digestive organ?
Answer:
The walls of the digestive organs are made of involuntary muscles. When the food enters into the food pope, the inner walls start contracting and expanding. These movements are called peristatic movements. This activity sends the food into digestive organs.

Question 5.
How will you prove that we exhale CO₂ gas during respiration?
Answer:
Pass the exhaled air given out by us into lime water. It will turn milky in colour. We know that CO₂ gas turn lime water milky. This confirms that we exhale CO₂ gas in respiration.

Question 6.
What organ helps amoeba to move from one place to another?
Answer:
In amoeba pseudopodia are present which are locomotory organs. By means of these pseudopodia, amoeba move from one place to another. A protuberance is developed in the direction in which amoeba has to move. Cytoplasm reaches in it, due to which pseudopodium is formed.

Question 7.
What is digestion?
Answer:
Digestion is a chemical process. It involves the conversion of the complex substances of the food into simpler water soluble substances by the action of the enzymes given out by the digestive glands. The body can utilise this food easily. The body absorbs the digested food and utilises it for getting energy.

Question 8.
Define:
1. Incisors
2. Canines.
Answer:
1. Incisors:
These are the flat front teeth. They have a sharp straight edge that help us to cut food and hence they are also called cutting teeth. There are four incisors in each jaw.

2. Canines:
These are the pointed teeth present on either side of the incisors. They help us to tear the food and hence they are called tearing teeth. There are two canines in each jaw.

Question 9.
Give a labelled diagram to show the location of taste buds on the tongue?
Answer:

Question 10.
Write down some functions of the tongue?
Answer:
We use our tongue for taking. Besides, it mixes saliva with the food during chewing and helps in swallowing food. We also taste food with our tongue. It has so many taste buds that detect different tastes of food

Question 11.
Draw a diagram to show the movement of the food in the alimentary canal?
Answer:

Question 12.
What is meant by diarrhoea? Explain.
Answer:
Sometime you may have experienced the need to pass watery stool frequently. This condition is known as diarrhoea. It may be caused by an infection, food poisoning or indigestion. It is very common in India, particularly among children. Under severe conditions it can be fatal. This is because of the excessive loss of water and salts from the body. Diarrhoea should not be neglected. Even before a doctor is consulted the patient should be given plenty of boiled and cooled water with a pinch of salt and sugar dissolved in it. This is called Oral Rehydration Solution (ORS).

Nutrition in Plants  Long Answer Type Questions

Question 1.
Explain ingestion of food in Amoeba with diagram?
Answer:
The organs which ingests food in pseudopodium. The food is caught by encircling it with Pseudopodia. When it is totally encircled then it is taken inside cell.

Question 2.
Describe the respiratory system of human beings?
Answer:
The various respiratory organs alongwith their functions are:
Air enters through the nostrils in the nose. The air then enters the nasal cavity which leads to larynx or voice box. From the larynx the air passes into the wind pipe or trachea. The trachea bifurcates into two branches. These are called bronchi. Bronchi further divide into smaller tubes called bronchioles. Later terminate in small thin wall air sacs called alveoli. The thin walls of the alveoli are provided with a network of blood capillaries. It is here that blood absorbs oxygen from alveoli and carbon dioxide from the blood enters the alveoli. From here the carbon dioxide is thrown out from the body. This process is called exhalation.

Question 3.
Describe in details the process of digestion in human beings?
Answer:
The various digestive organs and the process of digestion are:
1. Mouth, Teeth and Tongue:
The process of digestion beings with mouth. Food is taken inside the mouth, chewed into smaller pieces by the teeth. Mouth has few salivary glands which secrete saliva. The tongue helps in mixing saliva with food and swallowing the food.

2. Oesophagus:
Then the food enters into food pipe which is also called oesophagus. The wall of the oesophagus undergo peristaltic movements. This sends the food into the stomach.

3. Stomach:
Stomach is a bag like structure. Here the digestion of food starts. Here food is thoroughly mixed with the gastric juice secreted by gastric gland present in the stomach.

4. Small Intestine:
The process of digestion continues into small intestine. The upper part of the small intestine is called duodenum. In duodenum food is mixed with bile juice made by the liver and pancreatic juice from pancreas. The digested food is absorbed in the lower part of the small intestine by tiny projections called villi.

5. Large Intestine:
The undigested food now moves to the large intestine. Here water is absorbed and solid residue left behind called faeces.

6. Rectum and Anus:
The faeces then passer to rectum and expelled from the body through the anus.

Question 4.
Draw a diagram to show the arrangement of teeth in mouth?
Answer:

Question 5.
What is tooth decay? Explain.
Answer:
Normally bacteria are present in our mouth but they are not harmful to us. However, if we do not dean our teeth and mouth after eating, many harmful bacteria also begin to live and grow in it. These bacteria break down the sugars present from the leftover food and release acids. The acids gradually damage the teeth (Fig.). This is called tooth decay. If it is not treated in time, it causes severe toothache and in extreme cases results in tooth loss. Chocolates, sweets, cold drinks and other sugar products are the major culprits of tooth decay.

Therefore, one should clean the teeth with a brush or dantun and dental floss (a special strong thread which is moved between two teeth to take out trapped food particles) at least twice a day and rinse the mouth after every meal. Also, one should not put dirty fingers or any unwashed object in the mouth.

Question 6.
Explain excretion in human?
Answer:
Excretion is done by following organs:
1. Kidneys:
These excrete urea, salts, water etc., through urine.

2. Skin:
Skin possesses sweet glands. These glands absorb urea, salts and a large quantity of water from the blood capillaries near by and excrete them through their pores on the skin in the form of sweet.

3. Liver:
Liver is not a direct excretory organ. Liver cells convert more toxic ammonia into less toxic urea and uric acid and then pass them to the kidneys through the blood for excretion.

Liver also censer toxic substances. It remove bile salts in bile juice. These bile salts are formed due to break down of heamoglobin. If bile salts are not removed. They cause jaundice. Bile salts are removed through intestine.

4. Lungs:
It excretes gases like carbon dioxide, ammonia and water vapours.

5. Large Intestine:
It excretes undigested solid food waste (faeces).

MP Board Class 7th Science Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Given
ABC is a right angle A.
To prove:
AC > AB and AC > BC

Proof:
In ∆ABC
∠B = 90°
∴ ∠A + ∠C = 90° (by ASP)
and so ∠B > ∠A and ∠B > ∠C
AC > BC and AC > AB
(In a A, sides opposite to large angle are always longer).

Question 2.
In Fig. given below, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution:
Given
∠PRC < ∠QCB To prove AC > AB

Proof:
In ∆ ABC
Exterior angle is equal to sum of two opposite angles
∠PBC = ∠1 + ∠3 and
∠QCB = ∠1 + ∠2
∠QCB > ∠PBC (given)
⇒ ∠1 + ∠2 > ∠1 + ∠3
⇒ ∠2 > ∠3
∴ AC > AB
(∴ In a A, side opposite to larger angle is always longer).

Question 3.
In Fig. below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Given
∠B < ∠A i.e., ∠A > ∠B
∠C < ∠D i.e., ∠D> ∠C
To prove: AD > BC
i.e., BC > AD
Proof:
In ∆OCD
∠D > ∠C
OC > OD
(∴ In a ∆, sides opposite to larger angle are always longer) …(1)
In ∆OBA
∠A > AB
OB > OA
(∴ In a A, sides oppositedo larger angle are always longer) …(2)
Adding (1) and (2), we get
OC + OB > OD + OA
BC > AD

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. below). Show that ∠A > ∠C and ∠B > ∠D.
Solution:
Given
AB is the smallest siBe and CD is the longest side
To prove:

1. ∠A > ∠C and
2. ∠B > ∠D

Construction:
Join AC

1. Proof:
In, ∆ABC
BC > AB (∴ AB is the smallest side) ∠1 > ∠3
(∴ In a ∆ angle opposite to longer side is always larger) …..(1)
In ∆ACD
CD > AD (∴ CD is the largest side) ∠2 > ∠4
(∴ In a ∆ angles opposite to longer side are always larger) …(2)
Adding (1) and (2), we get
∠1 + ∠2 > ∠3 + ∠4
∠A > ∠C

2. To prove: AB > AD
Construction:
Join BD
Proof:
In ∆ABD
AD > AB (AB is the smallest side)
∠5 > ∠7 …(3)
In ∆BCD
CD > BC (CD is the longest side)

∠6 > ∠8 …(4)
Adding (3) and (4), we get
∠5 + ∠6 > ∠7 + ∠8
∠B > ∠D

Question 5.
In Fig. below, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Sol.
Given:
PR > PQ
∠1 = ∠2
To prove:
∠PSR > ∠PSQ
Proof:
In ∆PSQ
∠PSR = ∠1 + ∠Q (EAP)
In ∆PSR
∠PSQ = ∠2 + ∠R (EAP)
= ∠1 + ∠R (∠1 = ∠2)
In ∆PQR
PR > PQ (given)
∠Q > ∠R
(∴ In a ∆, angle opposite to longer side is always larger)
Adding ∠1 on both sides
∠Q + ∠l > ∠R + ∠1
∴ ∠PSR > ∠PSQ
(∴ ∠PSR = ∠1 + ∠Q and ∠PSQ = ∠1 + ∠R)

Question 6.
Show that of all the line segments drawn from a give point not on it, the perpendicular line segment is the shortest.
Solution:
Given
Let us consider the ∆PMN such that ∠M = 90°
Since, ∠M + ∠N + ∠P = 180° [Sum of angles of a triangle]

∠M = 90° [PM ⊥ l]
∠N < ∠M
PM < PN …..(1)
Similarly PM < PN₁ …..(2)
PM < PN₂ …..(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l.
Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. below). If AD is extended to intersect BC at P, show that

1. ∆ABD = ∆ACD
2. ∆ABP = ∆ACP
3. AP bisects ∠A as well as ∠D.
4. AP is the perpendicular bisector of BC.

Solution:
Given
AB = AC
DB = DC
To prove:

1. ∆ABD = ∆ACD
2. ∆ABP = ∆ACP
3. ∠1 = ∠2 and ∠5 = ∠6
4. ∠3 = ∠4 = 90° and BP = PC

Proof
1. In ∆ABD and ∆ACD
AB = AC (given)
BD = CD (given)
AD =AD (common)
∆ABD = ∆ACD (by SSS)
and so ∠1 = ∠2 (by CPCT)

2. In ∆ABP and ∆ACP
AB = AC (given)
∠1 = ∠2 (proved)
AP = AP (common)
∆ABP ≅ ∆ACP (by SAS)
and so ∠3 = ∠4
and BP = CP (by CPCT)

∠1 = ∠2 AP bisects ∠A

3. In ∆DBP and ∆DCP
BP = CP (proved)
∠3 = ∠4 (proved)
DP = DP (common)

∆DBP ≅ ∆DCP (by SAS)
and so ∠5 = ∠6 (by CPCT)
∴ AP bisects ∠D.

4. BP = CP (proved)
∠3 = ∠4 (proved)
∠3 + ∠4 = 180° (LPA’s)
∠3 + ∠3 = 180° (∠3 = ∠4)
2∠3 = 180°
∠3 = 90°
∠3 = ∠4 (each 90°)
and therefore, AP is the perpendicular bisector of BC

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

1. AD bisects BC
2. AD bisects ∠A.

Solution:
Given
AB = AC and AD ⊥ BC
To prove
∠1 = ∠2 and
BD = CD
Proof:
In ∆ABD and ∆ACD
AB = AC
AD = AD
∠3 = ∠4
∆ABD = ∆ADC
and so BD = CD and ∠1 = ∠2

Question 3.
Two sides AS and BC and median AM on one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see-Fig. below). Show that:

1. ∆ABM ≅ ∆PQN
2. ∆ABC ≅ ∆PQR

Solution:
Given
AB = PQ
BC = QR
AM = PN
To prove:

1. ∆ABM ≅ ∆PQN
2. ∆ABC ≅ ∆PQR

Proof:
BC = QR (given)
\(\frac{1}{2}\) BC = \(\frac{1}{2}\)QR
BM = QN

1. In ∆ABM and ∆PQN
AB = PQ (given)
BM = QN (proved)
AM = PN (given)
∆ABM ≅ ∆PQN (by SSS)
and so ∠ABC = ∠PQR (by CPCT)

2. In ∆ABC and ∆PQR
AB = PQ (given)
BC = QR (given)
∠B = ∠Q (proved)
∆ABC ≅ ∆PQR by (SAS)

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that triangle ABC is isosceles.

Solution:
Given
∠E = ∠F
BE = CF
To prove
AB = AC
Proof:
In ∆FBC and ∆ECB
BE = CF (given)
∠F = ∠E (each 90°)
BC = CB (common)
∴ ∆FBC = ∆ECB (by RHS)
and so ∠B = ∠C (by CPCT)
In ∆ABC, ∠B = ∠C
AB = AC
(sides opposite to equal angles of a A are equal)
and so ABC is isosceles.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
Given
AB = AC
∠APB = ∠APC = 90°
To prove:
∠B = ∠C

Proof:
In ∆ABP and ∆ACP
AP = AP (common)
∠APB = ∠APC (each 90°)
AB = AC (given)
∴ ∆ABP = ∆ACP (by RHS)
and ∠B = ∠C (by CPCT)

Theorem 7.6.
If two angles of a triangle are equal, then the sides opposite to them are also equal.
Given
In ∆ABC, ∠C = ∠B

To prove: AB = AC
Construction:
Draw the bisector of ∠A and let it meet BC an D
Proof:
In ∆s ABD and ACD, we have ∠B = ∠C (Given)
∠BAD = ∠CAD (Construction)
AD = AD (Common)
∴ ∆ABD ≅ ∆ACD (AAS Cong. Criterion)
Hence, AB = AC (CPCT)

Theorem 7.7.
If two sides of a triangle are unequal, the longest side has greater angle opposite to it.
Given:
In ∆ABC; AC > AB
To prove: ∠ABC > ∠ACB.

Construction:
Mark point D on AC such that AB = AD. Join BD.
Proof:
In ∆ABD,
AB = AD
∴ ∠1 = ∠2 (Const. As opp. equal sides) ….(1)
But ∠2 is an exterior angle of ABCD.
∠2 > ∠ACB (Exterior Angle Theorem) …(2)
From (1) and (2), we have
∠1 > ∠ACB (Const.)
But ∠ABC > ∠1
∴ ∠ABC > ∠ACB

Theorem 7.8. (Converse of Theorem 7.7)
In a triangle the greater angle has the longer side opposite to it.
Given:
In ∆ABC, ∠ABC > ∠ACB
To prove: AC > AB

Proof:
For ∆ABC, there are only three possibilities of which exactly one must be true.

1. AC = AB
2. AC < AB (iii) AC > AB.

Case 1:
If AC = AB, then ∠ABC = ∠ACB, which is contrary to what is given.
AB ≠ AC

Case 2:
If AC < AB, the longer side has the greater angle opposite to it.
∴ ∠ACB > ∠ABC.
This is also contrary to what is given.

Case 3:
We are left with the only possibility, namely AC > AB which is true.
AC > AB.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 9 Force and Laws of Motion

Force and Laws of Motion Intext Questions

Force and Laws of Motion Intext Questions Page No. 118

Question 1.
Which of the following has more inertia:

1. a rubber ball and a stone of the same size?
2. a bicycle and a train?
3. a five – rupees coin and a one – rupee coin?

Answer:
As we know, inertia is the calculated value for the mass of the body. It is proportional to mass of the body:

1. Inertia of the stone is greater than that of a rubber ball as mass of a stone is more than the mass of a rubber ball for the same size.
2. Inertia of the train is greater than that of the bicycle. As mass of a train is more than the mass of a bicycle.
3. Mass of a five rupee coin is more than that of a one – rupee coin. Hence, inertia of the five – rupee coin is greater than that of the one – rupee coin.

Question 2.
In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kick it towards a player of his own team”. Also, identify the agent supplying the force in each case.
Answer:
Four times:

• First, when a football player kicks to another player. Agent supplying the force: First case – First player. Second when that player kicks the football to the goalkeeper. Agent supplying the force. Second case – Second player.
• Third when the goalkeeper stops the football. Agent supplying the force: Third case – Goalkeeper.
• Fourth when the goalkeeper kicks the football towards a player of his own team. Agent supplying the force: Fourth case – Goalkeeper.

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
When we shake any tree’s branches vigorously some leaves of that tree get detached because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
Due to inertia of motion, we fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest.

1. Case I: Since the driver applies brakes and bus comes to rest. But, the passenger tries to maintain its inertia of motion. As a result, a forward force is exerted on him.
2. Case II: The passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus.

Force and Laws of Motion Intext Questions Page No. 126

Question 1.
If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
According to Newton’s third law of motion, a force is exerted by the Earth on the horse in the forward direction while horse pushes the ground in the backward direction. As a result, the cart moves forward.

Question 2.
Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer:
According to Newton’s third law of motion, a reaction force is exerted over fireman by the ejecting water in the backward direction when a fireman holds a hose, which is ejecting large amounts of water at a high velocity. As a result of the backward force, the stability of the fireman get affected. Hence, it is difficult for him to remain stable while holding the hose.

Question 3.
From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35 ms^-1. Calculate the initial recoil velocity of the rifle.
Answer:
Given,
Mass of the rifle, m₁ = 4kg
Mass of the bullet, m₂ = 50g = 0.05kg
Recoil velocity of the rifle = v₁
Bullet is fired with an initial velocity, v₂= 35 m/s
Condition:
Initially, the rifle is at rest.
Thus, its initial velocity, v = 0
Total initial momentum of the rifle and bullet system = (m₁+ m₂)v = 0
Total momentum of the rifle and bullet system after firing = m₁v₁ + m₂v₂
= 0.05 × 35 = 4v₁ + 1.75
As per law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing
⇒ 4v₁ + 1.75 = 0
v₁ = –\(\frac { 1.75 }{ 4 }\) = -0.4375 m/s
So, the rifle recoils backwards with a velocity of 0.4375 m/s because value is negative.

Question 4.
Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2 ms^-1 and 1  ms^-1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^-1. Determine the velocity of the second object.
Answer:
Given,
m₁ = 100g = 0.1kg
m₂ = 200g = 0.2kg
Velocity of m₁ before collision, v₁ = 2 m/s
Velocity of m₂ before collision, v₂ = 1 m/s
Velocity of m₁ after collision, v₃ = 1.67 m/s
Velocity of m₂ after collision = v₄
As per the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
Hence,
m₁v₁ + m₂v₂ = m₁v₃+ m₂v₄
Putting values
2(0.1) + 1(0.2) = 1.67(0.1) + v₄(0.2)
0.4 = 0.167 + 0.2v₄
v₄ = 1.165 m/s
Velocity of the second object = 1.165 m/s.

Force and Laws of Motion NCERT Textbook Exercises

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non –  zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Yes, an object may travel with a non – zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the upthrust and the viscosity of air. The net force on the drop is zero.

Question 2.
When a carpet is beaten with a stick, dust comes out of it Explain.
Answer:
When we beat the carpet with a stick, it comes into motion. But the dust particles continue to be at rest due to inertia and get detached from the carpet.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
Due to sudden jerks or due to the bus taking sharp turn on the road, the luggage may fall down from the roof because of its tendency to continue moving in the original direction. To avoid this, the luggage is tied with a rope on the roof.

Question 4.
A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would come to rest.
Answer:
(c) there is a force on the ball opposing the motion.

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)
Answer:
Here, u = 0, s = 400m, t = 20 s, a = ?, F = ?.
m = 7 tonnes
= 7 × 1000kg
= 7000kg
⇒ s = ut + \(\frac { 1 }{ 2 }\)at²
400 = (0 + 20) + \(\frac { 1 }{ 2 }\)a(20)²
a = \(\frac { 400\times 2 }{ { 20 }^{ 2 } } \)
∴ a = 2 m/s²
Force,
F = ma
= 7000 × 2 = 14,000 N.

Question 6.
A stone of 1kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice?
Answer:
m = 1kg, u = 20 m/s, s = 50m, v = 0, F = ? a = ?.
⇒ v² – u² = 2as
(0)² – (20)² = 2a (50)
∴ – 400 = 100a
= \(\frac { 400 }{ 100 }\) – 4 m/s²
Force of friction, F = m × a
= 1kg × -4 m/s² = -4 N

Question 7.
A 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Answer:
m = 8,000 + 5 × 2,000 = 18,000kg
(a) The net accelerating force,
F = Engine force – friction force
= 40,000 – 5,000 = 35,000 N.

(b) The acceleration of the train,
a = \(\frac { F }{ m }\) = \(\frac { 35,000 }{ 18,000 }\) = \(\frac { 35 }{ 18 }\) = 1.94 ms^-2

(c) The force of wagon 1 on wagon 2
= The net accelerating force – mass of wagon × acceleration
= 35,000 – 2,000 × \(\frac { 35 }{ 18 }\)
= 35,000 – 3888.8 = 31,111.2 N.

Question 8.
An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is stopped with a negative acceleration of 1.7 ms^-2?
Answer:
Here, mass = 1500kg
a = -1.7 ms²
F = m × a
= 1500 × (-1.7)
= -2550 N
The force between the vehicle and the road is 2,550 is, m a direction opposite to the direction of the vehicle.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v? Choose correct option.
(a) (mv)²
(b) mv²
(c) \(\frac { 1 }{ 2 }\) × mv²
(d) mv.
Answer:
(d) mv.

Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
The cabinet will move with constant velocity only when the net force on it is zero.
Force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.

Question 11.
Two objects, each of mass 1.5kg, ate moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms^-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer:
Here, m₁ = m₂ = 1.5kg,
u₁ = 2.5 ms^-1, u₂ = – 2.5 ms^-1
Let v be the velocity of the combined object after the collision.
By conservation of momentum,
Total momenta after collision = Total momenta before collision
= (m₁ + m₂) v = m₁u₁ + m₂u₂
= (1.5 + 1.5) v = 1.5 × 2.5 + 1.5 × (-2.5)
= 3.0 v = 0
or
⇒ v = 0 ms^-1

Question 12.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
Action and reaction always act on different bodies, so they cannot cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so, the truck does not move.

Question 13.
A hockey ball of mass 200g travelling at 10 ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Here, m = 200g = 0.2kg,
u = 10 ms^-1,
v = -5 ms^-1
change in momentum = m (v – u)
= 0.2 (- 5 – 10) = -3kg ms^-1.

Question 14.
A bullet of mass 10g travelling horizontally with a velocity of 150 ms^-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
Here m = 10g = 0.01kg,
u = 150 ms^-1,
v = 0, t = 0.03 s
a = \(\frac { v-u\quad }{ t } \) = \(\frac { 0-150\quad }{ 0.03 } \) = -5,000 ms^-1
The distance of penetration of the bullet into the block,
s = ut + \(\frac { 1 }{ 2 }\)at²
150 × 0.03 + \(\frac { 1 }{ 2 }\) × (-5,000) × (0.03)
= 4.5 – 2.25 = 2.25
The magnitude of the force exerted by the wooden block on the bullet
= ma = 0.01 × 5,000 = 50 N.

Question 15.
An object of mass 1kg travelling in a straight line with a velocity of 10 ms^-1 collides with, and sticks to, a stationary wooden block of mass 5kg. Then, they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer:
Here, m₁ = 1kg, u₁ = 10 ms^-1, m₂ = 5kg, u₂ = 0
Let v be the velocity of the combined object after the collision.
Total momentum just before the impact
= m₁u₁ + m₂u₂  = 1 × 10 + 5 × 0 = 10kg ms^-1
Total momentum just after the impact = (m₁ + m₂)v = (1 + 5)v
= 6v kg ms^-1 by conservation of momentum,
6v = 10
= v = \(\frac { 10 }{ 6 }\) = \(\frac { 5 }{ 3 }\) ms^-1
∴ Total momentum just after the impact
= 6 × \(\frac { 5 }{ 3 }\) = 10 ms^-1

Question 16.
An object of mass 100kg is accelerated uniformly from a velocity of 5 ms^-1 to 8 ms^-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Here, m = 100kg,
u = 5 ms^-1,
v = 8 ms^-1,
t = 6 s
Initial momentum, P₁ = mu = 100 × 5 = 500kg ms^-1
Final momentum, P₂ = mu = 100 × 8 = 800kg ms^-1
The magnitude of the force exerted on the object.
F = \(\frac { { P }_{ 2 }-{ P }_{ 1 } }{ t } \) = \(\frac { 800-500 }{ 6 }\) = \(\frac { 300 }{ 6 }\) = 50 N.

Question 17.
Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer:
Both, the motorcar and insect experience the equal force and hence, a same change in their momentum. So, we agree with Rahul. But due to smaller mass or inertia, the insect dies.

Question 18.
How much momentum will a dumb – bell of mass 10kg transfer to the floor if it falls from a height of 80cm? Take its downward acceleration to be 10 ms^-2.
Answer:
Here, m = 10kg, u = 0,
s = 80cm = 0.80m,
a = 10 m/s^-2
Let v be the velocity gained by the dumb-bell as it reaches the floor.
As v² – u² = 2as
v² – 0² = 2 × 10 × 0.80 = 16
or
v = 4 ms^-1
Momentum transferred by the dumb-bell to the floor
p = mv = 10 × 4 = 40kg ms^-1

Force and Laws of Motion Additional Questions

Force and Laws of Motion Multiple Choice Questions

Question 1.
Which of the following statements is not correct for an object moving along a straight path in an accelerated motion?
(a) Its speed keeps changing.
(b) Its velocity always changes.
(c) It always goes away from the earth.
(d) A force is always acting on it.
Answer:
(c) It always goes away from the earth.

Question 2.
According to the third law of motion, action and reaction ________ .
(a) Always act on the same body.
(b) Always act on different bodies in opposite directions.
(c) Have same magnitude and directions.
(d) Act on either body at normal to each other.
Answer:
(b) Always act on different bodies in opposite directions.

Question 3.
A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to ________ .
(a) Exert larger force on the ball.
(b) Reduce the force exerted by the ball on hands.
(c) Increase the rate of change of momentum.
(d) Decrease the rate of change of momentum.
Answer:
(b) Reduce the force exerted by the ball on hands.

Question 4.
The inertia of an object tends to cause the object ________ .
(a) To increase its speed.
(b) To decrease its speed.
(c) To resist any change in its state of motion.
(d) To decelerate due to friction.
Answer:
(c) To resist any change in its state of motion.

Question 5.
A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is ________ .
(a) Accelerated
(b) Uniform
(c) Retarded
(d) Along circular tracks.
Answer:
(a) Accelerated

Question 6.
An object of mass 2kg is sliding with a constant velocity of 4 ms^-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is ________ .
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N.
Answer:
(b) 0 N

Question 7.
Rocket works on the principle of conservation of ________ .
(a) Mass
(b) Energy
(c) Momentum
(d) Velocity.
Answer:
(c) Momentum

Question 8.
A water tanker filled up to \(\frac { 2 }{ 3 }\) of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would ________ .
(a) Move backward
(b) Move forward
(c) Be unaffected
(d) Rise upwards.
Answer:
(b) Move forward

Question 9.
Which of the following represents example(s) of potential energy?
(a) A moving car
(b) A moving fan
(c) A book resting on the table
(d) Both (a) and (c).
Answer:
(d) Both (a) and (c).

Question 10.
Unit of force is ________ .
(a) Ampere
(b) Volt
(c) Joule
(d) Hertz.
Answer:
(c) Joule

Question 11.
Product of mass and acceleration of a body is called ________ .
(a) Acceleration
(b) Work
(c) Power
(d) Energy.
Answer:
(b) Work

Question 12.
Which of the following is correct about energy?
(a) Energy is not required to do work.
(b) Work can be expressed as Force × Displacement.
(c) Unit of power is joule.
(d) Power is the amount of work done per unit time.
Answer:
(c) Unit of power is joule.

Question 13.
An object of mass 3kg is falling from the height of 1m. The kinetic energy of the body will be when it touches the ground ________ .
(a) 29.4 N
(b) 29.4 J
(c) 30 N
(d) 15 J
Answer:
(b) 29.4 J

Question 14.
Two objects with masses 1kg and 9kg, and equal momentum. Calculate the ratios of their kinetic energies ________ .
(a) 3 : 1
(b) 9 : 1
(c) 1 : 1
(d) 1 : 2
Answer:
(b) 9 : 1

Question 15.
Considering air resistance negligible, the sum of potential and kinetic energies of the free falling body would be ________ .
(a) zero
(b) infinite
(c) would decrease
(d) remains fixed.
Answer:
(d) remains fixed.

Force and Laws of Motion Very Short Answer Type Questions

Question 1.
What do we call to the product of mass and velocity of an object?
Answer:
Momentum.

Question 2.
Define inertia.
Answer:
The property by which an object tends to remain in the state of rest or of uniform motion unless acted upon by some force is called inertia.

Question 3.
Which property has S.I. unit kilogram metres per second i.e., 1kg m/s?
Answer:
Momentum.

Question 4.
Give an example of scalar quantity.
Answer:
Mass.

Question 5.
Give an example of vector quantity.
Answer:
Momentum.

Question 6.
Calculate the total momentum of the bullet and the gun before firing.
Answer:
For both, it would be zero because both of them are at rest.

Question 7.
Which force slows down a moving bicycle when we try to stop?
Answer:
The force of friction.

Question 8.
Which kind of force of gravity work when an object is under free fall?
Answer:
Unbalanced force.

Question 9.
Which property of an object resist a change in their state of rest or motion?
Answer:
Inertia.

Question 10.
Which law of Newton is also known as Galileo’s law of inertia?
Answer:
First law.

Question 11.
Is force a vector quantity?
Answer:
Yes.

Question 12.
Which force of motion opposes motion of an object?
Answer:
Force of friction.

Question 3.
When action and reaction forces act on two different bodies, what kind of magnitude they have?
Answer:
Action and reaction forces act on two different bodies but they are equal in magnitude.

Question 14.
When gun moves in the backward direction, which kind of velocity is generated?
Answer:
Recoil.

Question 15.
Which factor of body is dependent on its mass?
Answer:
Inertia of a body depends on its mass.

Force and Laws of Motion Short Answer Type Questions

Question 1.
There are three solids made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have highest inertia?
Answer:
Steel has highest inertia because it has greatest density and greatest mass, therefore, it has highest inertia.

Question 2.
Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer.
Answer:
If the breaks are applied suddenly then, the balls will start rolling in the direction in which the train was moving. Due to the application of the brakes, the train comes to rest but due to inertia the balls try to remain in motion, therefore, they begin to roll. Direction and speed of all balls will not be same because the masses of the balls are not the same, therefore, the inertial forces are not same on both the balls. Thus, the balls will move with different speeds.

Question 3.
Two identical bullets are fired, one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?
Answer:
According to law of conservation of momentum or explanation by Newton’s laws of motion, light rifle will hurt the shoulder more.

Question 4.
A horse continues to apply a force in order to move a cart with a constant speed. Explain why?
Answer:
The force applied by the horse balances the force of friction

Question 5.
Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed v again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain.
Answer:
Law of conservation of momentum is applicable to isolated system (no external force is applied). In this case, the change in velocity is due to the gravitational force of earth.

Question 6.
In which of the following conditions work done will be equal to zero?
Answer:
In the absence of any one of the two conditions given below, work done will be equal to zero, that is work is not considered to be executed:

1. Force should act on the object.
2. Object must be displaced.

Question 7.
Define energy and explain its forms.
Answer:

1. Energy: Energy is the capacity of doing work. More the power, more will be energy and vice – versa. For example, a motorcycle has more energy than a bicycle.
2. Forms of energy: There are many forms of energy, such as kinetic energy, potential energy, mechanical energy, chemical energy, electrical energy etc.

Force and Laws of Motion Long Answer Type Questions

Question 1.
Give the formulation of work. In which conditions work can occur?
Answer:
Work = Force × Displacement
or W = F × s
where, W is work ‘F’ is force and ‘s’ is displacement.
If force, F = 0
Therefore, work done, W = 0, s = 0
If displacement, s = 0
Therefore, Work done, W = F × 0 = 0
It proves that, there are two conditions for work to occur or be done:

1. Force should act on the object.
2. Object must be displaced.

Question 2.
Give the conditions when work done become positive and negative.
Answer:
When force is applied in the direction of displacement, the work done is considered as positive.
i.e., W = F × s
When force is applied in opposite direction of displacement, the work done is considered as negative.
i.e., W = -F × s = -Fs.

Question 3.
Explain positive and negative work.
Answer:
1. Positive work:
If a force displaces the object in its direction, then the work done is positive.
Here,
W = Fd
Example:
Motion of ball falling towards ground where displacement of ball is in the direction of force of gravity.

2. Negative work. If the force and the displacement are in opposite directions, then the work is said to be negative.
Here,
W = -Fd.
Example:
If a ball is thrown in upward direction but the force due to earth’s gravity is in the downward direction.

Question 4.
A cyclist moving along a circular path of radius 63m completes three rounds in 3minutes.
1. The total distance covered by him during this time.
2. Net displacement of cyclist.
3. The speed of the cyclist
Answer:
1. Total distance covered
s = 2πr × t
s = 2πr × 3
= 2 × \(\frac { 22 }{ 7 }\) × 63 × 3 = 1188m

2. Displacement = Zero

3. Speed = \(\frac { Distance }{ Time }\)
= \(\frac { 1188 }{ 180 }\)
= 6.6 m/s.

Force and Laws of Motion Higher Order Thinking Skills (HOTS)

Question 1.
As per Newton’s third law, every force is accompanied by equal and opposite force. How then can anything move?
Answer:
According to the Newton’s third law, action and reaction are two equal and opposite forces but they act on different bodies. This make the motion of a body possible.

Question 2.
The passengers travelling in a bus fall ahead when a speeding bus stops suddenly. Why?
Answer:
When the speeding bus stops suddenly lower part of the body, a long with the bus comes to rest while the upper part tends to remain in motion due to inertia of motion. That is why passengers fall ahead.

Question 3.
A player always runs some distance before taking a jump. Why?
Answer:
A player always runs for some distance before taking a jump because inertia of motion helps him to take a longer jump.

Force and Laws of Motion Value Based Question

Question 1.
Sushil saw his karate expert breaking a slate. He tried to break the slate but Sushil’s friend stopped him from doing so and told him that it would hurt, one needs lot of practice in doing such activity.

1. How can a karate expert break the slate without any injury to his hand?
2. What is Newton’s third law of motion?
3. What value of Sushil’s friend is seen in the above case?

Answer:

1. A karate expert Sushil applies the blow with large velocity in a very short interval of time on the slate, therefore large force is exerted on the slate and it breaks.
2. To every action, there is an equal and opposite reaction, both act on different bodies.
3. Sushil’s friend showed the value of being responsible and caring friend.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

1. OB = OC
2. AO bisect ∠A.

Solution:
Given
AB = AC
∠1 = ∠2, ∠3 = ∠4
To prove:

1. OB = OC
2. ∠5 = ∠6

Proof:
In ∆ABC,
AB = AC
∠B = ∠C
\(\frac{1}{2}\) ∠B = \(\frac{1}{2}\) ∠C
∠1 = ∠3 or ∠2 = ∠4
In ∆OBC
∠2 = ∠4
and so OB = OC
(In a A, sides opposite to equal angles are equal)
In ∆ABO and ∆ACO
BO = CO (proved)
∠1 = ∠3 (proved)
AB = AC (given)
∆ABO = ∆ACO (by SAS)
and so ∠5 = ∠6 (by CPCT)

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see Fig. below). Show that AABC is an isosceles triangle in which AB =AC.

Solution:
Given
∠ABC = ∠ADC
To prove:
AB = AC
Proof:
In A ABD and A ACD
BD = CD (given)
∠ADB = ∠ADC (given each 90°)
AD = AD (common)
∴ ∆ABD = ∆ACD (BySAS)
and so AB = AC (by CPCT)

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. below). Show that these altitudes are equal.

Solution:
Given
AB = AC
∠E = ∠F (each 90°)
To prove: BE = CF
Proof:
In ∆ABE and ∆ACE
∠A = ∠A (common)
∠E = ∠F (each 90°)
AB = AC (given)
∆ABE = ∆ACE (byAAS)
and so BE = CF (by CPCT)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. below). Show that:

1. ∆ABE ≅ ∆ACF
2. AB = AC,

i. e., ABC is an isosceles triangle.

Solution:
Given
BE = CF
∠E = ∠F (each 90°)
To prove:

1. ∆ABE = ∆ACE
2. AB = AC

Proof:
In ∆ABE and ∆ACF
∠A = ∠A (common)
BE = CF (given)
∠E = ∠F (each 90°)
∆ABE = ∆ACF (by AAS)
and so AB = AC (by CPCT)

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see Fig. below). Show that ∠ABD = ∠ACD.

Solution:
Given
AB = AC
BD = CD
To prove
∠ABD = ∠ACD
Construction: Join AD
Proof:
In ∆ABD and ∆ACD
AD = AD (common)
AB = AC (given)
BD = CD (given)
∆ABD = ∆ACD (by SSS)
and so ∠ABD = ∠ACD (by CPCT)

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. below). Show that ∠BCD is a right angle.
Solution:
Given: AB = AC
AD = AB
To show: ∠BCD = 90°
i.e., ∠2 + ∠3 = 90°

Proof:
AB = AC …..(1)
AB = AD …..(2)
From (1) and (2), we get
AC = AD
In ∆ABC
AB = AC
∠1 = ∠2
(In a A, angles opposite to equal sides are always equal) …p) …(3)
In ∆ACD
AC = AD
∠3 = ∠4
(In a A, angles opposite to equal sides are always equal) …(4)
In ∆BCD
∠1 + ∠2 + ∠3 + ∠4 = 180° (ASP)
∠2 + ∠2 + ∠3 + ∠3 = 180°
(∴ ∠1 = ∠2, ∠3 = ∠4)
2 (∠2 + ∠3) = 180°
(∠2 + ∠3) = 90°
∠BCD = 90°

Question 7.
ABC is a right angled triangle in which ∠A – 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆BAC
AB =AC
∠B = ∠C = x
∠A + ∠B + ∠C= 180°
∠B + ∠C = 180° – 90°
∠B + ∠C = 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\)

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Given
ABC is an equilateral ∆
i. e., AB = BC = AC
To prove
∠A = ∠B = ∠C = 60°
Proof:
In ∆BAC
AB = AC
∠B = ∠C
(In a A, angles opposite to equal sides are always equal) ……(1)
AC = BC
∠A = ∠B
(In a A, angles opposite to equal sides are always equal) …..(2)
From (1) and (2), we get,

∠A = ∠B = ∠C = x (say)
∠A + ∠B + ∠C = 180° (ASP)
⇒ x + x + x = 180°
⇒ 3x = 180°
⇒ x = \(\frac{180^{\circ}}{3}\) = 60°
∴ ∠A = ∠B = ∠C = 60

Theorem 7.4.
SSS (Side-Side-Side) Congruence Theorem:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given
In ∆s ABC and DEF we have,

AB =DE
BC = EE
and AC = DF
To prove:
∆ABC = ∆DEF
Construction:
Suppose BC is the longest side.
Draw EF such that EE = AB and FEG = ∠CBA.
Join GF and DG.

Proof:
In ∆s ABC and GEE, we have
AB = GE (Const.)
∠ABC = ∠GEF (Const.)
and BC = EF (Given)
∴ ∆ABC = ∆GEF (SAS Cong. Axiom)
∠A = ∠G (CPCT) …..(1)
and AC = GF (CPCT) …..(2)
Now AB = EG (Const.)
AB = DE (Given)
∴ DE = EG ……(3)
Similarly, DF = GF ……(4)
In ∆EDG
DE = EG (Proved above)
∴ ∠A = ∠2 (∠s opp. equal side) …..(5)
In ∆DFG
FD = FG (Proved above)
∴ ∠3 = ∠4 (∠s ppp. equal side) …..(6)
∴ ∠1 + ∠3 – ∠2 + ∠4 [From (5) and (6)]
i. e. ∠D = ∠G …..(7)
But ∠G = ∠A [From (1)]
∴ ∠A = ∠D …..(8)
In ∆s ABC and DEF,
AB – DE (Given)
AC = DF (Given)
∠A = ∠D [From (8)]
∆ABC ≅ ∆DEF (SAS Cong. Axiom)

Theorem 7.5.
RHS (Right Angle Hypotenuse Side) Congruence Theorem:
Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.
Given
In ∆s ABC and DEF,
∠B = ∠E = 90°
AC = DF
BC = EF.

∆ABC ≅ ∆DEF
Construction:
Produce DE to M so that
EM = AB, Join ME.
Proof:
In ∆s ABC and MEF
AB = ME (Const.)
BC = EE (Given)
∠B = ∠MEF (each 90°)
∴ ∆ABC = ∆MEF (SAS Cong. Axiom)
Hence ∠A = ∠M (CPCT) …(1)
AC = MF (CPCT) …(2)
Also AC =DF (Given)
∴ DF = MF
∴ ∠D = ∠M (∠s opp. equal side of ADFM) …(3)
From (1) and (3), we have
∠A = ∠D …..(4)
Now, in ∆s ABC and DEF, we have
∠A = ∠D [From (4)]
∠B = ∠E (Given)
∴ ∠C = ∠F …..(5)
Again, in ∆s ABC and DEF, we have
BC = EF (Given)
AC = DF (Given)
∠C = ∠F [From (5)]
∴ ∆ABC = ∆DEF (SAS Cong. Axiom)

MP Board Class 9th Maths Solutions