Goyal Brothers Prakashan Class 9 ICSE Physics Solutions

A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 1

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 1

Unit I
Exercise 1

Question 1.
There is a positively charged sphere A and negatively charged sphere B, such that they are brought in electrical contact by a copper wire. Answer the following questions :

(a) Which sphere is at higher potential before electrical contact on the basis of convention?
(b) Which sphere is at lower potential before electrical contact on the basis of convention?
(c) In which direction conventional current flows?
(d) In which direction electronic current flows?
(e) What is potential of the spheres after electrical contact?

Answer:
Sphere A is positively charged and sphere B is negatively charged.
Both the sphere A and B are brought in electrical contact by a copper wire.

(a) On the basis of convention, positively charged sphere A is at higher potential before electrical contact.
(b) On the basis of convention, negatively charged sphere B is lower potential before electrical contact.
(c) Conventional current flows from sphere A to sphere B i. e. from a body at higher potential to the body at lower potential.
(d) Electronic current flower from sphere B to sphere A i.e. from a body at lower potential to the body at higher potential.
(e) After electrical contact, both the spheres will be at same potential.

Question 2.

(a) What do you understand by the term electric potential?
(b) Define electric potential in terms of energy spent.
(c) State the unit of electric potential and define it.

Answer:
(a) Electric potential : The amount of work done in moving a unit positive charge from infinity to a given point in an electric field is called electric potential.
(b) Electric potential : The amount of energy spent in moving a unit positive charge from infinity to a given point in an electric field is called electric potential.
(c) Volt is the SI unit of electric potential.
One volt : When one coulomb of charge is brought from infinity to a given point in an electric field, such that work done is one joule, then electric potential is said to be one volt.
OR
Electric potential is said to be one volt if one Joule of work is done in moving one coulomb of charge from infinity to a given point in an electric field.

Question 3.

(a) What do you understand by the term quantity of electric charge?
(b) State SI unit of electric charge and define it.
(c) How many electrons constitute one unit electric charge in SI system?

Answer:

(a) Quantity of electric charge : The number of charge (electrons) which drift from lower to higher potential is called quantity of charge.
(b) SI unit of electric charge is coulomb (C).
One coulomb : The quantity of electric charge which will deposit 0.00118 g of silver on the cathode, when passed through silver nitrate is called one coulomb.
(c) 6.25 × 10¹⁸ electrons constitute one unit (IC) electric charge in SI system.

Question 4.

(a) What do you understand by the term electric current?
(b) State and define the SI unit of electric current.
(c) State the relation between electric current; number of electrons moving in a circuit and time in seconds.

Answer:
(a) Electric current : The rate of flow of electric charge in an electric circuit is called electric current.
(b) Ampere (A) is the SI unit of electric current.
One ampere : When one coulomb charge flows through an electric circuit in one second, then the electric current flowing the circuit is said to be ampere.
(c) If Q is the charge (in coulombs) flowing through conductor in time t (in seconds) such that current I flows through the conductor then
Rate of flow of charge = Q/t
We know rate of flow of charge = I = Electric current.
⇒ I = Q/t

Question 5.
How electric current flows in (i) solids, (ii) liquids?
Answer:
(i) Flow of electric current in solids : In solids, the positive charges are associated with atomic nuclei. As the nuclei are firmly packed and closely held by inter-atomic forces, therefore, positive charges cannot drift.
On the other hand, negative charges (electrons) are not held firmly. Thus, when a potential difference, however small, is applied they start drifting from lower to higher potential.
The continuous drift of electrons, through the body of a solid conductor constitutes the current.
(ii) Flow of electric current in liquids : Within a liquid no electrons move. However, when a negatively charged and a positively charged electrodes are placed in a liquid, it sets up an electric field.
Under the influence of the electric field, the positively charged ions migrate towards the negatively charged electrode and vice versa.
At the cathode the positively charged ions gain electrons. At the anode the negatively charged ions lose same number of electrons.
Thus, in a way number of electrons given by the cathode is equal to the number of electrons accepted by the anode. The sum up, we can say that simultaneous movement and discharge of positive and negative ions in the opposite directions constitutes the current in the liquids.

Question 6.

(a) Define the term potential difference.
(b) How is potential difference related to work done and quantity of charge?

Answer:
(a) Potential difference : The amount of work done in moving a unit positive charge from one point to another point in an electric field is called potential difference.
(b) If Q = Charge moving from one point to another in an electric field.
W = Work done to move the charge Q from one point to another.
V = Potential difference between two points.
Then work done in moving Q units of charge = W
Work done in moving one unit of charge = W/Q
But work done in moving one unit of charge = Potential difference = V
⇒ V = W/Q

Practice Problems 1

Question 1.
A charge of 5000 C flows through an electric circuit in 2 hours and 30 minutes. Calculate the magnitude of current in circuit.
Answer:

Question 2.
A charge of 8860 C flows through an electric circuit in 2 min and 40 s. Calculate the magnitude of current in circuit.
Answer:

Practice Problems 2

Question 1.
A battery can supply a charge of 25 × 10⁴ C. If the current is drawn from battery at the rate of 2.5 A, calculate the time in which battery will discharge completely.
Answer:

Question 2.
A dry cell can supply a charge of 800 C. If continuous current of 8.0 mA is drawn, calculate the time in which cell will discharge completely.
Answer:

Practice Problems 3

Question 1.
Calculate the total number of electrons flowing through a circuit in 20 mins and 40 s, if a current of 40 μA flows through the circuit.
[1 e^– = 1.6 × 10^-19 C]
Answer:

Question 2.
4 × 10²⁰ electrons flow through a circuit in 10 hours. Calculate magnitude of current. [1 e- = 1.6 × 10^-19 C]
Answer:

Practice Problems 4

Question 1.
What is the electrical potential at a point in an electric field when 24 J of work is done in moving a charge of 96 C from infinity?
Answer:

Question 2.
A charge of 75 C is brought from infinity to a given point in an electric field, when amount of work done is 3.75 J. Calculate the electrical potential at that point.
Answer:

Practice Problems 5

Question 1.
A work of 25 J and 30 J is done when 5 C charge is moved first to point A and then to point B from infinity. Calculate the potential difference between points A and B.
Answer:

Question 2.
A charge of 25 C is moved from infinity to points A and B in an electric field when the work done to do so is 10 J and 10.5 J respectively. Calculate the potential difference between the points A and B.
Answer:

Unit II
Exercise 2

(A) Objective Questions

Multiple Choice Questions.
Select the correct option :

1. SI unit potential difference is :
(a) coulomb
(b) kelvin
(c) volt
(d) ampere
Ans. (c) volt

2. Current in a circuit flows :
(a) in a direction from high potential to low potential
(b) in a direction from low potential to high potential
(c) in a direction of flow of electron
(d) in any direction
Ans. (a) in a direction from high potential to low potential

3. In a metallic conductor, electric current is thought to be due to movement of :
(a) ions
(b) amperes
(c) electrons
(d) protons
Ans. (c) electrons

4. Assuming that the charges of an electron is 1.6 × 10^-19 coulombs, the number of electrons passing through a section of wire per sec, when the wire carries a current of one ampere is :
(a) 0.625 × 10¹⁹
(b) 1.6 × 10^-19
(c) 1.6 × 10¹⁹
(d) 0.627 × 10^-17
Ans. (a) 0.625 × 10¹⁹

5. Which of the following is best conductor of electricity?
(a) copper
(b) gold
(c) platinum
(d) silver
Ans. (d) silver

(B) Subjective Questions

Question 1.
What do you understand by the term electric cell?
Answer:
Electric cell is an arrangement which maintains constant potential difference between conductors.
A cell basically consists of two conducting rods, which are called electrodes, immersed in a solution, which is called the electrolyte.

Question 2.
Draw a neat and labelled diagram of simple voltaic cell showing clearly the direction of flow of conventional current and direction of flow of electrons.
Answer:
Simple voltaic cell was invented by Alessandro Volta in year 1800. It was the first device which could creat a constant potential difference between two plates with the help of chemical energy.

Question 3.
Briefly describe the theory of simple voltaic cell.
Answer:
Theory of simple voltaic cell: Amongst the zinc and copper plates, zinc is more electro-positive (ionisation potential – 0.76 V) as compared to copper
(ionisation potential + 0.34 V) in electrochemical series. The dilute sulphuric acid is used as an electrolyte in the ionised state.

When zinc plate comes in contact with (H⁺) hydrogen ions, it being more electro-positive, ionises to form zinc ions and free electrons.

The free electrons so formed, take the passage of least resistance, and hence, move out in the external circuit through the copper wires. The zinc ions, however, enter in the dilute sulphuric acid. Since Zn²⁺ ions are positively charged, they repel H⁺ ions in the acid solution, with the result that H⁺ ions start crowding at the copper plate.
The copper plate in turn loses its free electrons to H⁺ ions, which . form nascent hydrogen.

The nascent hydrogen atoms so formed unite to form molecular hydrogen.

From the above explanation, it is clear that free electrons actually drift from zinc to copper in external circuit, and hence, current should flow from zinc to copper.
However, we still continue saying the electric charge flows from copper to zinc and this current is called conventional current, whereas the actual flow of electrons is from zinc to copper which constitutes electronic current.
The emf between the zinc copper plate in the external circuit is 0.34 V – (-0.75 V)= 1.10 V.

Question 4.
What do you understand by the following terms?

1. electric circuit
2. closed electric circuit
3. open electric circuit.

Answer:

1. Electric circuit : The path along which electric current flows is known as electric circuit.
2. Closed electric circuit : W’hen the path of an electric circuit starting from one terminal of the cell, ends at the other terminal of cell, without any break, then such a circuit is called closed circuit.
3. Open electric circuit : When the path of an electric circuit, starting from one terminal of the cell, is broken at some point, then such a circuit is called open electric circuit.

Question 5.
State two conditions necessary for a circuit, such that electric current flows through it.
Answer:
For the flow of electric current through a circuit, following are the necessary conditions :

1. Electric circuit must be closed or complete.
2. Every part of the circuit is a conductor.

Question 6.
Draw a neat diagram showing

1. closed electric circuit
2. open electric circuit.

Answer:

1. Closed circuit
   
2. Open circuit
   

Question 7.
Name four electric conductors and four electric insulators.
Answer:

• Conductors : Silver, copper, aluminium and iron.
• Insulators : Plastic, nylon, dry wood and rubber.

Question 8.

(a) What do you understand by the term electric resistance?
(b) Why does the filament of an electric bulb in an electric circuit get white hot, but not the connecting wires?

Answer:

(a) Electric resistance : The obstruction offered to the passage of electric current by a material is called resistance of the material.
(b) Filament of an electric bulb is made up of tungsten having high resistance. Due to its high resistance, on passing electric current through it, a electrical energy changes into heat energy. So much heat is produced that filament of bulb becomes white hot and gives light.
Resistance of connecting wires is very low and hence the connecting wires do not get heated.

Question 9.
Is it correct to say that a resistance wire is an insulator or a bad conductor? Explain your answer.
Answer:
It is not correct to say that a resistance wire is an insulator or bad conductor. From resistance, it is implied that a gien material will conduct electricity, but will also offer obstruction to the passage of electric current.

Question 10.

(a) What do you understand by the term series circuit?
(b) State two characteristics of resistances in the series circuit.
(c) Draw a diagram showing two bulbs connected in series to a dry cell.

Answer:
(a) Series circuit : When a number of resistances are connected in an electrical circuit in such a way that positive of one resistance acts as the negative of the other resistance, then resistances are said to be in series.
OR
A number of resistors are said to be in series if these are joined end to end and same current flows through each one of them when a potential difference is applied across the combination.

(b) Characteristics of resistances in series :

1. The sum total of resistances in series increases with increase in number of resistors.
2. The potential difference remaining constant, the current in series circuit decreases with the increase in number of resistors in series.
3. All the elements in series circuit work simultaneously. If the circuit is broken anywhere between the elements, none of the elements work.

(c) Two bulbs glowing dimly

Question 11.

(a) What do you understand by the term parallel circuit?
(b) State two characteristics of resistance in the parallel circuit.
(c) Draw a diagram showing two bulbs connected in parallel to a dry cell.

Answer:
(a) Parallel circuit : When a number of resistances (bulbs) are connected in an electrical circuit in such a way that all of them are connected to common positive and common negative terminal of a cell, then the resistance (bulbs) are said to be connected in parallel.
OR
A number of resistors are said to be connected in parallel if one end of each resistor is connected to one point and other end of each resistor is connected to another point so that the potential difference across each resistor is same.

(b) Characteristics of resistances in Parallel :

1. The sum total of resistances in parallel decreases with the increase in number of resistors.
2. The current flowing in any resistor in parallel will be inversely proportional to resistance i.e., more the resistance, less the current.
3. Each resistor in parallel functions independently with respect to the other resistors in parallel.

(c)

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 1 are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.

A New Approach to ICSE Physics Part 1 Class 9 Solutions Sound

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Sound

Exercise

(A) Objective Questions

Multiple Choice Questions.
Select the correct option :

1. In case of longitudinal waves, the particles of the medium vibrate :
(a) in the direction of wave propagation
(b) opposite to the direction of wave propagation
(c) at right angles to the direction of wave propagation
(c) none of the above
Ans. (a) in the direction of wave propagation

2. A longitudinal wave consists of :
(a) crest and trough in the medium
(b) compression and rarefaction in the medium
(c) both (a) and (b)
(d) neither (a) nor (b)
Ans. (b) compression and rarefaction in the medium

3. The longitudinal waves can propagate only in .
(a) solids
(b) liquids
(c) gases
(d) all of these
Ans. (d) all of these

4. A part of the longitudinal wave in which particles of the medium are closer than the normal particles is called :
(a) rarefaction
(b) crest
(c) trough
(d) compression
Ans. (d) compression

5. A part of longitudinal wave in which particles of the medium are farther away than the normal particles is called :
(a) rarefaction
(b) trough
(c) compression
(d) crest
Ans. (a) rarefaction

6. In the region of compression or rarefaction, in a longitudinal wave, the physical quantity which does not change is :
(a) pressure
(b) mass
(c) density
(d) volume
Ans. (b) mass

7. The wavelength is the linear distance between the :
(a) two consecutive compressions
(b) two consecutive rarefactions
(c) one compression and one rarefaction
(d) both (a) and (b)
Ans. (d) both (a) and (b)

8. The number of oscillations passing through a point in unit time is called :
(a) vibration
(b) frequency
(c) wavelength
(d) amplitude
Ans. (b) frequency .

9. The SI unit of frequency is
(a) hertz
(b) gauss
(c) decibel
(d) none of these
Ans. (a) hertz

10. If the frequency of a wave is 25 Hz. the total number of compressions and rarefactions passing through a point in 1 second is :
(a) 25
(b) 50
(c) 100
(d) none of these
Ans. (b) 50

11. Which of the following is an elastic wave?
(a) light wave
(b) ratio wave
(c) sound wave
(d) microwave
Ans. (c) sound wave

(B) Subjective Questions

Question 1.

(a) What do you understand by the term sound energy?
(b) State three conditions necessary for hearing sound.

Answer:
(a) Sound energy : “It is a form of energy that produces the sensation of hearing in our ears.” Sound is produced when a body vibrates.
(b) Necessary conditions for hearing sound :

1. There must be a vibrating body, capable of transferring its energy to its surroundings.
2. These must be a material medium to pick the energy and then propagate it in forward direction.
3. There must be a receiver, so as to receive the sound vibrations and then transmit them to the brain for final interpretation, such as human ear.

Question 2.
Describe briefly an experiment to prove that vibrating bodies produce sound.
Answer:
Experiment : Take a tuning fork. It is U-shaped fork made of steel provided with a handle.

Strike the tuning fork with a rubber hammer and hold it close to ear. A sound is heard. Now take a freely suspended pith ball and touch one end of the tuning fork (which is already hit with a rubber hammer) to it. It is observed that pith ball repeatedly flies outward. This experiment too proves that sound is produced by a vibrating body.

Question 3.

(a) What do you understand by the term infrasonic vibrations?
(b) What do you understand by the term sonic vibrations? State the range of sonic vibrations for the human ear.

Answer:
(a) Infrasonic vibrations : Those vibrations whose frequency is less than 20 hertz are known as infrasonic vibrations
(b) Sonic vibrations : Sonic vibrations are also known as audio vibrations.
Those vibrations whose frequency is from 20 hertz to 20000 hertz are known as sonic vibrations.
The range of sonic vibrations is from 20 vibrations per second to 20000 vibrations per second.

Question 4.

(a) What do you understand by the term ultrasonic vibrations?
(b) Name three animals which can hear ultrasonic vibrations.

Answer:

(a) Ultrasonic vibrations : Those vibrations, whose frequency is more than 20000 hertz and are not perceived by human ear, are known as ultrasonic vibrations.
(b) Dogs, bats and dolphins can hear ultrasonic vibrations.

Question 5.
How do bats locate their prey during flight?
Answer:
Bats produce ultra sound which returns after striking an obstacle in their way. By hearing the reflected round, bats can judge the distance and direction of obstacle/prey in their way and hence bats can catch their prey during flight.

Question 6.
What is Galton’s whistle? To what use is it put?
Answer:
A special whistle which can produce ultra sound, not heard by humans, is called Galton’s whistle.
It is used to train dogs because they can hear ultrasounds upto a frequency of 40000 hertz.

Question 7.
State four practical uses of ultrasonic vibrations.
Answer:
Uses of ultrasonic vibrations :

1. These are used for dissipating fogs on the runways at the airports.
2. These are used in the ultrasound scanning of internal organs of human body.
3. These are used for making dish washing machines. In these machines, water and detergents are vibrated with ultlrasonics vibrator. The vibrating particles of the dissolved detergent rub against the plates and clean them.
4. These are used in SONAR (Sound navigation and ranging) to detect and find the distance of objects under water.

Question 8.
Describe an experiment to prove that material medium is necessary for the propagation of sound.
Answer:
A material medium is necessary for the propagation of sound. It can be proved with the help of following experiment :
An electric circuit consisting of a cell, a switch and an electric bell is arranged inside a bell-jar, which stands on the platform of an evacuating pump.
The switch of the bell is pressed to close the electric circuit. Sound is heard when there is air within the bell-jar. Air is now gradually pumped out of the bell-jar. The intensity of sound goes on decreasing, and no sound is heard when the air is completely removed from the bell-jar. It is because, the air which acts as a medium for the propagation of sound energy is removed.

From above experiment, it is clear that sound can not be heard in the absence of air i.e. a material medium is necessary for the propagation of sound.

Question 9.
Why do astronauts talk to each other through radio telephone in space?
Answer:
Sound can not travel through vacuum. There is no material medium in the space for propagation of sound. Hence astronauts talk to each other through radio telephone in space. In radio telephone, message of one astronauts reach to the other with the help of radio waves, which can travel through vacuum.

Question 10.
Define the terms :

1. wavelength
2. amplitude
3. frequency.

Answer:
(i) Wavlength : The linear distance between the two consecutive particles of a vibrating medium in the same phase is called its wavelength. It is denoted by Greek letter lambda (λ).
OR
The distance travelled by the wave in one time period of vibration of the particles of medium is called its wavelength.
OR
In a longitudinal wave, distance between the two consecutive compressions or rare fractions is called wavelength.
(ii) Amplitude : The maximum displacement of a vibrating particles about its mean position is called amplitude.
(iii) Frequency : It is denoted by letter (ƒ). The number of complete vibrations executed by a vibrating particle of a medium about its mean position in one second is called its frequency.
OR
It may be defined as the number of waves passing through one particular point in one second.

Question 11.
State four differences between the sound wave and the light wave.
Answer:
Differences between the sound wave and light wave :
Light waves :

1. They are produced from the electrons in an excited state.
2. They travel at a very high speed of 3 x 108 ms-1 in air.
3. They can travel through vacuum.
4. Their velocity does not change with the change in temperature, humidity, etc.
5. They can produce the sensation of vision.

Sound waves :

1. They are produced due to vibrations of various objects in a material medium.
2. They travel in air at a very low speed of 332 ms-1 at 20°C.
3. They cannot travel through vacuum and always require some material medium for propagation.
4. Their velocity changes with the change in velocity, humidity, etc.
5. They can produce the sensation of hearing.

Question 12.
What is meant by the term wave motion?
Answer:
Wave motion : The transference of energy when the particles of a medium, move about their mean position is called wave motion.

Question 13.
State the relation between the wavelength and the frequency.
Answer:
Consider a wave is propagating through a medium,
Let,ƒ = Frequency of wave
λ = Wavelength of wave
T = Time period of the wave.
In the time ‘T’, distance covered by wave = λ
In the time 1 second, distance covered by wave = λ/T
But distance covered by wave in one second = v = Velocity of the wave

Which is the required relation between wavelength and frequency.

Question 14.
What kinds of the waves are produced in solids, liquids and gases?
Answer:
Elastic waves or material waves are produced in solids, liquids and gases.

Question 15.
The sound of an explosion on the surface of lake is heard by a boatman 100 m away and a diver 100 m below the point of explosion.

1. Of the two persons mentioned (boatman and diver), who would hear the sound first?
2. Give reason for your answer in (i).
3. If the sound takes ‘t’ seconds to reach the boatman, approximately how mcuh time it will take to reach the diver?

Answer:

1. Diver would hear the sound first.
2. Because velocity of sound in water (1450 ms^-1) is more than the velocity of sound in air (332 ms^-1).
3. Time taken by sound to reach the diver would be less than t, where t is the time taken by sound to reach the boatman.

Note : Time taken by sound to reach the diver would be 4.36 times less than the time taken by sound to reach the boatman.

Question 16.
What is approximate value of speed of sound in iron as compared to that in air? Illustrate your answer with a simple experiment.
Answer:
Speed of sound in iron is approximately sixteen times the speed of sound in air.
[Speed of sound in iron = 5100 ms^-1; Speed of sound in air=332 ms^-1]
Experiment : If we put our ears to rails, we can hear the sound of the train through metal. But at the same time, when we stand near by the railway track, we are not able to hear the sound.
This occurs because when we put our ears to rails, the sound travels through iron. But in standing position, sound travels through air and due to smaller speed of sound in air as compared to iron, we are not able to hear the sound.

Question 17.
How does a bat avoid obstacles in its way when in flight?
Answer:
Bats make a series of twittering sound, so high pitched that human ear can not hear. These sound waves strike against the obstacles less in their path of flight and send back echoes to the bat’s ear. The echoes tell the bats, how they must turn in the air to avoid colliding with the obstacles or with one another.

Question 18.
A continuous disturbance is created on the surface of water in a ripple tank with a small piece of cork floating on it. Describe the motion of the cork. What does the motion of the cork tell about the disturbance?
Answer:
Cork will move up and down about the mean position at the same position along horizontal.
It tells as that only the disturbance/energy is transferred from one particle to the other but particles of the medium do not move from one position to the other.

Question 19.
Draw a displacement-time graph for water wave.
Answer:
Displacement-time graph for water wave :

Practice Problems 1

Question 1.
200 waves pass through a point in one second. Calculate the time period of wave.
Answer:
Waves passing through a point in 1 second = 200
We know, frequency (ƒ) = Number of waves passing through point in 1 second.
∴ ƒ = 200Hz

Question 2.
A bat emits an ultrasonic sound of frequency 0.25 MHz. Calculate the time in which one vibration is completed.
Answer:
Frequency of ultrasound = ƒ = 0.25 MHz
ƒ = 0.25 × 10⁶ Hz

Practice Problems 2

Question 1.
The sonic boom of an aircraft has a time period of 0.00005s. Calculate the frequency of sound produced.
Answer:

Question 2.
An electromagnetic wave has a time period of 4 × 10^-8 s. Calculate its frequency in MHz.
Answer:
Time period = T = 4 × 10^-8 s
Frequency of sound = ƒ = ?

Practice Problems 3

Question 1.
An ultraviolet radiation has a wavelength of 300Å. If the velocity of electromagnetic wave is 3 × 10⁸ ms^-1. Calculate

1. frequency
2. time period.

Answer:

Question 2.
The wavelength of the vibrations produced on the surface of water is 2 cm. If the wave velocity is 16 ms^-1, calculate

1. no. of waves produced in one second
2. time required to produce one wave.

Answer:

Practice Problems 4

Question 1.
A continuous progressive transverse wave of frequency 8 Hz moves across the surface of a ripple tank

(a) With reference to the frequency, describe the movement of water on the surface
(b) If the wavelength of transverse wave is 32 mm, calculate the speed with which wave travels across the surface of water.

Answer:

Question 2.
A thin metal plate is placed against the teeth of cog wheel. Cog wheel is rotated at a speed of 120 rotations per minute and has 160 teeth. Calculate :

1. frequency of node produced.
2. speed of sound, if wavelength is 1.05 m.
3. what will be the effect when speed of cog wheel is doubled?

Answer:
Frequency for rotation of wheel = 120 rotations per minute

Practice Problems 5

Question 1.
A sound wave of wavelength 1/3 m has a frequency 996 Hz. Keeping the medium same, if frequency changes to 1328 Hz. Calculate

1. velcoity of sound
2. new wavelength.

Answer:

Question 2.
Two tuning forks A and B of frequencies 256 Hz and 192 Hz respectively are vibrated in air. If the wavelength of A is 1.25 m, calculate the wavelength produced by B.
Answer:

Practice Problems 6

Question 1.
The distance between one crest and one trough of a sea wave is 4.5 m. If the waves are produced at the rate of 240/min, calculate

1. time period
2. wave velocity.

Answer:
Distance between one crest and one trough of a sea wave = 4.5 m

Question 2.
The distance between three consecutive crests of wave is 60 cm. If the waves are produced at the rate of 180/ min, calculate

1. wavelength
2. time period
3. wave velocity.

Answer:

Practice Problems 7

Question 1.
The diagram given below shows a displacement distance graph of a wave. If the velocity of wave is 160 ms-1, calculate

1. wavelength
2. frequency
3. amplitude.

Answer:

Question 2.
From diagram given below calculate

1. velocity of P and Q
2. frequency of P, when frequency of Q is 512 Hz.

Assume that both wave are travelling in same medium.

Answer:

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given A New Approach to ICSE Physics Part 1 Class 9 Solutions Sound are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.

A New Approach to ICSE Physics Part 1 Class 9 Solutions Light

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Light.

Unit I
Exercise 1

Question 1.
(a) What do you understand by the following terms?

1. Light
2. Diffused light

(b) By giving one example and one use explain or define

1. Regular reflection
2. Irregular reflection.

Answer:
(a)
(i) Light : Light is a form of energy which produces in us sensation of seeing.
(ii) Diffused light : Light obtained after reflection from rough surface is known as diffused light.
It is a soft light with neither the intensity nor the glare of direct light. It is scattered and comes from all directions. It does not cause harsh shadows.
(b)
(i) Regular reflection : The phenomenon due to which a parallel beam of light travelling through a certain medium, on striking some polished surface, bounces off from it, as parallel beam, in some other direction is called regular reflection.
For example : Reflection taking place from the objects like looking glass, still water, oil, highly polished metals is regular reflection.
Regular reflection is useful in the formation of images. We can see our face in a mirror only due to regular reflection.
(ii) Irregular reflection or Diffused reflection : The phenomenon due to which a parallel beam of light, travelling through some medium, gets reflected in various possible directions, on striking some rough surface is called irregular reflection.
For example : Reflection taking place from ground, walls, trees, suspended particles in air is irregular reflection.
Use : It helps in the general illumination of places and helps us to see things around us.

Question 2.
By drawing a neat diagram define the following :

1. Mirror
2. Incident ray
3. Reflected ray
4. Angle of incidence
5. Angle of reflection
6. Normal

Answer:

1. Mirror is a highly polished and smooth surface which reflects almost the entire light falling on it. A plane mirror is made by silvering one side of a glass plate as shown in figure.
   
2. Incident ray : The light ray striking a reflecting surface is called the incident ray.
3. Reflected ray : The light ray obtained after reflection from the surface, in the same medium in which the incident ray is travelling, is called the reflected ray.
4. Angle of incidence : The angle which the incident ray makes with the normal at the point of incidence is called the angle of incidence. It is denoted by the letter “i”.
5. Angle of reflection : The angle which the reflected ray makes with the normal at point of incidence, is called the angle of reflection. It is denoted by the letter “r”.
   
6. Normal : The perpendicular drawn at the point of incidence, to the surface of mirror is called normal.

Question 3.
State the laws of reflection.
Answer:
Laws of reflection :

1. The incident ray, the reflected ray and the normal ray at the point of incidence, lie in the same plane.
2. The angle of incidence i is equal to the angle of reflection r i.e. ∠i = ∠r.

Question 4.
A ray of light strikes a plane mirror, such that angle with the mirror is 20°. What is value of angle of reflection? What is the angle between the incident ray and the reflected ray?
Answer:
∵ Light ray makes an angle of 20° with the mirror
∴ ∠ABM = 20°
Angle of incidence = ∠i = 90° – 20°
∠i = 70°
∵ ∠i = r ∴ ∠r = 70°

Angle between incident ray and reflected ray = ∠i + ∠r = 70°+ 70°= 140°

Question 5.
Prove experimentally that images are formed as far behind in a plane mirror as the object is in front of it.
Answer:
Consider an object‘O’ situated in front of a plane mirror MM,. A ray of light which starts from point ‘O’ perpendicularly, is reflected back along the same path (See figure).
However, another ray which moves along OB is reflected along BC, obeying the laws of reflection, such that BN is the normal. Produce OA and CB backward, such that they meet at point I. Then ‘I’ is image of ‘O’.

Thus, in particular OA = IA.

Question 6.
Prove geometrically that when plane mirror turns through a certain angle, the reflected ray turns through twice the angle.
Answer:
Consider a ray of light AB, incident on plane mirror in position MM’, such that BC is the reflected ray and BN is the normal.
∠ABM = ∠CBN = ∠i
∠ABC = 2 ∠i …(i)
Let the mirror be rotated through an angle ‘0’ about point B, such that M₁M₁ is the new position of the mirror and BN₁ is the new position of normal. As the position of the incident ray remains same, therefore new angle of the incidence is ∠ABN₁ whose magnitude is (i + θ). Let BD be the reflected ray, such that ∠DNB₁ is the new angle of reflection.

Thus, for a given incident ray, if the plane mirror is rotated through a certain angle, then the reflected ray rotates through twice the angle.

Question 7.

(a) What do you understand by the term lateral inversion?
(b) A printed card has letters PHYSICS. By drawing the diagram show the appearance of the letters. (No ray diagram is required).

Answer:
(a) Lateral Inversion. The phenomenon due to which the image of an object turns through an angle of 180° through verticle axis rather the horizontal axis, such that right side of image appears as left or vice-versa is called Lateral Inversion. During lateral inversion the left side of object appears as right side of image and vice-versa. In a way the image turns through the angle of 180° about vertical axis.

Question 8.

(a) State the mirror formula for the formation of total number of images formed in two plane mirrors, held at an angle.
(b) Calculate the number of images formed in two plane mirrors, when they are held at the angle of (i) 72° (ii) 36°.

Answer:
(a) If θ = Angle of inclination between two mirrors
n = number of images formed

Numeral one is subtracted because of the loss of one image due to overlapping of the images.
(b) (i) When θ = 72°

Question 9.
Draw a neat two ray diagram for the formation of images in two plane mirrors, when mirrors are

1. at right angles to each other
2. facing each other.

Answer:
(a) When two mirrors are inclined at right angles
‘O’ is an object placed in between two mirrors XY and
XZ, inclined at an angle of 90°. (See figure)

Taking normal incidence, I₁ and I₂ are the images formed in the plane mirror XY and XZ respectively as far behind the mirrors, as point ‘O’ is in front of them.
However, image I₁ acts as a virtual object for image mirror XZ₁ and forms an image I₃. Similarly, image I₂ acts as a virtual object for the image mirror XY₁ and forms the image I₄. The images I₃ and I₄ overlap to form a very bright image. Thus, on the whole three images are seen. In order to draw two-ray diagrams, from the position FE of the eye, draw two rays meeting at I₃,I₄ such that these ray intersect the mirror XZ at D and C.
Now draw two rays from point I₁ to join C and D intersecting mirror XY at A and B. Join O with A and B.
Similarly, in order to show image I₂, draw two rays from I₂ to the position of eye FE, such that they intersect at H and G Join H and G to ‘O’ so as to form incident beam.
(b) When two mirrors are parallel to each other

Consider two plane mirrors XY and PQ facing each other and ‘A’ as an object situated anywhere between them (See figure).
It is clear that mirror XY forms its image in mirror PQ and vice versa. These images by themsleves will act as image mirror or virtual mirrors.
Let us consider the normal incidence towards the mirror XY for the object A. First of all an image X₁ is formed as far behind, as the object is in front of it. This image X₁ will fall in front of mirror PQ, and hence, forms an image X₂. The image X₂ falls in front of mirror XY and hence, forms an image X₃. Thus, it continues and infinite images can be formed.
Similarly, taking normal incidence for mirror PQ image P₁,P₂,P₃ etc. are formed.
In order to draw ray diagram, from point A, draw a divergent beam, meeting mirror PQ at points 1 and 2. With P₁ as reference point, draw rays 1, 3 and 2, 4 meeting mirror XY. With P₃ as reference point draw rays, such that they enter the eye.

Question 10.
Why are infinite images not seen when two plane mirrors are facing each other?
Answer:
Infinite images are not seen when two plane mirrors are facing each other because :

1. After every successive reflection, some amount of light energy is absorbed. Thus luminosity of image goes on decreasing, till they are no longer visible.
2. As the distance of images from the eye goes on increasing, it is unable to resolve far off images.

Question 11.

(a) State four characteristics of the image formed in a plane mirror.
(b) State three ways in which the image formed in a plane mirror differs from the image formed in a pin hole camera.

Answer:
(a) Characteristics of Image formed by a Plane Mirror.

1. Image is of the same size as that of object.
2. Image is laterally inverted.
3. It is upright.
4. It is virtual.
5. Image formed is as far behind the mirror as the object infront of the mirror.

(b) Image formed by plane mirror :

1. Image formed by plane mirror is virtual.
2. It can not be obtained on the screen.
3. Size of image is same as that of the object.
4. Image appears to be inverted only on the vertical axis.

Image formed by Pinhole camera :

1. Image formed by pinhole camera is real.
2. It can be obtained on the screen.
3. Size of the image is smaller than the object.
4. Image appears to be inverted on both the vertical axis and horizontal axis.

Question 12.
(a) What should be the minimum size of a plane mirror, so that a person 182 cm high can see himself completely?
(b) A boy stands 4 m away from plane mirror. If the boy moves 1/2 m towards mirror, what is now the distance between the boy and his image? Give a reason for your answer.
Answer:
(a) Size (height) of a person = 182 cm
We know to see the full length image of a person, we need a plane mirror of size as half of the size of the person.
So, size of plane mirror required = Height of person/2 = 182/2 = 91 cm
(b) Distance of boy from the plane mirror = u = 4 m When boy moves 1/2 m towards the mirror.
Then, distance between boy and mirror = 4 – 1/2 = 7/2 m
We know, in a plane mirror, the image formed is as behind the mirror as the object is in front of it.
So, distance between the mirror and image of boy = 7/2 m
Distance between the boy and his image
= Distance of boy from mirror + Distance of image of boy from mirror

Question 13.
State four uses of a plane mirror.
Answer:
Four uses of plane mirror are :

1. Plane mirrors are used in construction of reflecting periscope.
2.  They are used as looking glass.
3. They are used in solar cookers for reflecting the rays of the sun into the cooker.
4. They are used for signalling purpose.

Question 14.

(a) Draw a neat diagram of reflecting periscope.
(b) State two advantages and two disadvantages of the reflecting periscope.

Answer:
(a)

(b) Advantages of Reflecting Periscope :

1. It is used to see above the head of crowds.
2. It is used by soldiers in trench warfare.

Disadvantages of Reflecting Periscope :

1. The final images is not brightly illuminated as light energy is absorbed due to two successive reflections.
2. Any deposition of moisture of dust on the mirror reduces the reflection almost to nil, and hence, the periscope cannot be used in places where there is a lot of dust or moisture.

Question 15.
What must be the minimum length of a plane mirror in which a person can see himself full length? Draw a diagram to justify your answer. Does the distance of person from the mirror affect the above answer?
Answer:
Minimum Height Of Plane Mirror Required For A Person To See Full Length Consider a person AB, such that A represents the highest point on his head, and B the lowest point on the foot such that E is the fixed eye level (see figure). The person will be able to see every part of his body if the can see points A and B. Let MN be the minimum length of mirror fixed on the wall, such that the rays AM and BN, after reflection, reach the eye of person, thereby forming an image A₁B₁ when produced backward.

In ∆AEA₁, CM is parallel to AE and C is the mid-point of AA₁. M is the mid-point of A₁E.
Similarly, in ∆BEB₁, ND is parallel to BE and D is the mid-point of BB₁
∴ N in the mid-point of B₁E.
Now in ∆A₁B₁E, M is mid-point of A₁E and N is mid-point of B₁E.
∴ MN is paralle and half of A₁B₁
But, A₁B₁ = AB
So, MN = 1/2 AB.
Thus, in order to see full length, a person requires a plane mirror which is half his own height. This relation is true for any distance of object from plane mirror.

Question 16.
An insect is sitting in front of a plane mirror at a distance of one metre from it.

1. Where is the image of insect formed?
2. What is the distance between insect and its image?
3. State any two characteristics of image formed in a plane mirror.

Answer:
(i) Distance of insect from plane mirror = u = 1 m
We know in case of plane mirror, distance of object from plane mirror is equal to distance of image of the object from the plane mirror.
∴Distance of the image of insect from plane mirror = v = u – 1 m
(ii) Distance between insect and its image = u + v = 1 + 1 = 2 m
(iii) Characteristics of the image formed by a plane mirror :

(a) Image formed is virtual.
(b) Image formed is erect.
(c) Image formed is of same size as that of the object.

Question 17.
(i) Draw a diagram to show reflection of a ray of light using plane mirror. In the diagram label the incident ray, the reflected ray, the normal, the angle of incidence and angle of reflection. (ii) State the laws of reflection.
Answer:

1. See Q. No. (2) of Exercise (1).
2. See Q. No. (3) of Exercise (1).

Question 18.
(i) Parallel rays are incident :

1. on regular surface and
2. on irregular surface. In what respect do reflected rays in (1) differ from those of (2)?

(ii) Write down four characteristics of image formed in a plane mirror.
Answer:
(i) Parallel rays of light after reflection from a regular surface . goes in a particular direction while after reflection from irregular surface go in different directions.
(ii) Characteristics of Image formed by a Plane Mirror.

(a) Image is of the same size as that of object.
(b) Image is laterally inverted.
(c) It is upright.
(d) It is virtual.
(e) Image formed is as far behind the mirror as the object in front of the mirror.

Question 19.
How many images will be formed when an object is placed between two parallel plane mirrors with their reflecting surfaces facing each other? Why do more distant images appear fainter?
Answer:
Angle between two mirrors facing each other = θ = 0°

So, infinite number of images are formed when an object is placed between two parallel plane mirrors with their reflecting surfaces facing each other.
More distant images appear fainter when two plane mirror with their reflecting surfaces facing each other because,
After every successive reflection, some amount of light energy is absorbed. Thus luminosity of images goes on decreasing and hence they appear fainter.

Question 20.

(a) Write down the letters of the word ‘POLEX’ as seen in a plane mirror, held parallel to the plane of this paper.
(b) Distinguish between real and virtual image.

Answer:
(a)

(b)
Real image :

1. Can be taken on the screen.
2. It is inverted.
3. It is formed when light rays after reflection (or refraction) actually meet.

Virtual image :

1. Cannot be taken on screen.
2. It is erect.
3. It is formed when rays of light after reflection (or refraction) appear to meet.

Question 21.

(a) Describe the principle of simple periscope through an outline ray diagram. Give one of its uses.
(b) Draw diagrams to show difference between regular and irregular reflection.

Answer:
(a) Simple periscope is based upon the principle of reflection. It consists of a cardboard or wooden tube, bent twice at right angles and is provided with two openings as shown in figure. Two plane mirrors are fixed at the bends of the tube at an angle 45° to the framework, such that the mirrors face each other. The tube is completely blackened from inside to avoid any reflection from its sides.

The parallel rays coming from an object at a higher plane, strike the plane mirror at an angle of 45° and hence are reflected through an angle of 45°.
These reflected rays strike the second mirror at an angle of 45° and hence are further reflected through an angle of 45°. These reflected rays on reaching the eye form the image on retina.
Use of periscope : It is used by soldiers in trench warfare.
(b)

Question 22.
An object is placed 2 cm from a plane mirror. If the object is moved by 1 cm towards the mirror, what will be the distance between the object and its new image?
Answer:
Distance of the object from plane mirror = 2 cm
If object is moved by 1 cm towards the mirror, then distance between the object and plane mirror = u = 1 cm
We know in case of plane mirror,
Distance of object from plane mirror = Distance of the image of the object from plane mirror
⇒ v = u = 1 cm
∴ Distance between its object and new image = u + v = 1 + 1 = 2 cm

Unit II

Practice Problem 1

Question 1.
An object 3 cm high produces a real image 4.5 cm high, when placed at a distance of 20 cm from a concave mirror. Calculate :

1. the position of image
2. focal length of the concave mirror.

Answer:
Size of object = h_o = 3 cm
Size of image = h₁ = -4.5 cm [Image formed is real]
Distance of the object from the concave mirror = u = – 20 cm
Distance of the image from the concave mirror = v = ?
Focal length of the concave mirror = f = ?

Practice Problem 2

Question 1.
An object 1.5 cm high when placed in front of a concave . mirror, produces a virtual image 3 cm high. If the object is placed at a distance of 6 cm from the pole of the mirror, calculate :

1. the position of the image
2. the focal length of the mirror.

Answer:

Question 2.
A converging mirror, forms a three times magnified virtual image when an object is placed at a distance of 8 cm from it. Calculate :

1. the position of the image
2. the focal length of the mirror.

Answer:
Let size of the object = h_o = x
Size of image = h₁= +3 x (Virtual image)
Distance of the object from the mirror = u = -8 cm
Distance of the image from the mirror = v = ?
Focal length of the mirror = f = ?

Practice Problems 3

Question 1.
An object 5 cm high forms a virtual image of 1.25 cm high, when placed in front of a convex mirror at a distance of 24 cm. Calculate :

1. the position of the image
2. the focal length of the convex mirror.

Answer:
Size of the object = h₀ = 5 cm
Size of the image = h₁ = +1.25 cm
Distance of the object from mirror = u = – 24 cm
Distance of the image from the mirror = v = ?
Focal length of the concave mirror = f = ?

Question 2.
An object forms a virtual image which is 1/8th of the size of the object. If the object is placed at a distance of 40 cm from the convex mirror, calculate :

1. the position of the image
2. the focal length of the convex mirror.

Answer:

Exercise 2

(A) Objective Questions

Multiple Choice Questions.
Select the correct option:

1. A concave mirror is made by cutting a portion of a hollow glass sphere of radius 30 cm. The focal length of the concave mirror is :
(a) 24 cm
(b) 12 cm
(c) 15 cm
(d) 60 cm
Ans. (c) 15 cm
Explanation :
Radius of curvature = R = 30 cm

2. A mirror forms a virtual image (diminished) of an object, whatever be the positiion of object :
(a) it must be a concave mirror
(b) it must be a convex mirror
(c) it must be a plane mirror
(d) it may be (b) or (c) or both
Ans. (b) it must be a convex mirror

3. A ray of light is incident on a concave mirror. If it is parallel to principal axis, the reflected ray will :
(a) pass through its principal focus
(b) pass through its centre of curvature
(c) pass through its pole
(d) retraces its path
Ans. (a) pass through its principal focus

4. If an incident ray passes through the centre of curvature of a spherical mirror, the reflected ray will :
(a) pass through its pole
(b) retraces its path
(c) pass through its focus
(d) be parallel to principal axis
Ans. (b) retraces its path

5. In case of concave mirror, the minimum distance between an object and its real image is :
(a) f
(b) 2f
(c) 4f
(d) zero
Ans. (d) zero

6. Looking into a mirror one finds her image diminished, the mirror is :
(a) concave
(b) convex
(c) cylindrical
(d) parabolic
Ans. (b) convex

7. Which mirror is used in periscope?
(a) Convex mirror
(b) Concave mirror
(c) Plane mirror
(d) Parabolic mirror
Ans. (c) Plane mirror

(B) Subjective Questions

Question 1.
Define the following terms :

1. spherical mirror
2. convex mirror
3. concave mirror

Answer:

1. Spherical mirror : “A mirror which is made from a part of a hollow sphere is called Spherical Mirror.
2. Convex mirror : “A mirror made by silvering the inner surface such that reflection takes place from the bulging surface” is called Convex Mirror.
   Centre of curvature is towards the silvered surface.
3. Concave mirror : “A mirror made by silvering the outer or the bulging surface such that the reflection takes place from the concave surface.” Centre of curvature is towards the reflecting surface.

Question 2.
Define the following terms in relation to concave mirror.

1. Pole
2. Centre of curvature
3. Principal axis
4. Principal focus
5. Focal length
6. Radius of curvature
7. Aperture

Answer:

1. Pole : Pole “is the mid-point of the mirror”.
2. Centre of curvature : The centre of hollow sphere of which the mirror forms a part, is called centre of curvature.
3. Principal axis : An imaginary line passing through the pole and the centre of curvature of a spherical mirror is called principal axis
4. Principal focus : It is a point on the principal axis, where a beam of light, parallel to principal axis, after reflection actually meet.
5. Focal length : The linear distance between the pole and the principal focus is called focal length.
6. Radius of curvature : The linear distance between the pole and the centre of curvature is called radius of curvature.
7. Aperture : The diameter of a spherical mirror is called its aperture.

Question 3.

(a) Define the term principal focus in case of convex mirror. Draw a convex mirror and show its principal focus and focal length clearly.
(b) What is the relation between focal length and radius of curvature of a concave mirror?

Answer:
(a) Focal length is the distance between pole (P) and focus (F)

(F) Focus is a point on principal axis where rays of light appear to meet.
(b) Focal length of a spherical mirror is equal to half of the radius of curvature of spherical mirror.

Question 4.

(a) What do you understand by the term real image?
(b) What type of mirror is used to obtain a real image?
(c) Does the mirror named by your form real image for all locations? Give reason for your answer.
(d) Is real image always inverted?

Answer:

(a) Real image : When rays of light after reflection or refraction actually meet at some other point” the image is real.
(b) Concave mirror.
(c) No, this mirror does not give real image of the object that lies between principal focus and pole.
(d) Yes. Real image is always ihverted.

Question 5.
Copy the figure. By taking two rays from point A, show the formation of image. State four characteristics of the image.

Answer:

Characteristics of the image :

1. Image formed is real.
2. Image formed is inverted.
3. Image formed is enlarged.
4. Image is formed beyond centre of curvature in front of concave mirror.

Question 6.
Draw a neat two ray diagram to illustrate how a concave mirror is used as a shaving mirror.
Answer:
We know that when an object is placed between P and F of a concave mirror, it forms a virtual and enlarged image. Thus, by using concave mirror we can have a proper shave, as the tiny hairs are clearly visible.

Question 7.
Copy the figure. By taking two rays from point A, show the formation of image. State four characteristics of the image.

Answer:

Characteristics of image :

1. Image formed is virutal.
2. Image formed is erect.
3. Image formed is diminished.
4. Image is always formed between pole and principal focus, behind the convex mirror.

Question 8.
Why do automobile drivers prefer convex mirror as a rear vew mirror? Illustrate your answer.
Answer:
A convex mirror always forms a small and upright image between pole and focus. That means in small area of mirror driver can see all the traffic coming from behind.

Question 9.
Give two uses of

1. convex mirror
2. concave mirror.

Answer:
(i) Two uses of convex mirror are :

(a) It is used as a rear view mirror in automobile to see the traffic behind.
(b) It is used as reflector for street light bulbs.

(ii) Two uses of concave mirror are :

(a) Concave mirror is used as A Reflector in head lights of cars and in search light. The source of light (bulb) is placed at the principal focus and the reflector forms parallel beam of light.
(b) For doctors to examine throat, ear, nose and eyes, light is focused with the help of concave mirror.

Question 10.
You are provided a convex mirror, a concave mirror and a plane mirror. How will you distinguish between them, without touching or using any other apparatus?
Answer:
We can identify convex mirror, concave mirror and a plane mirror by looking at them one by one.

1. If the size of the image of an object is of the same size as that of the object, then that mirror is a plane mirror.
2. If the size of image of an object increases as the object is brought closer to the mirror and size of the image of the object decreases when the object is taken away from the mirror, then that mirror is a concave mirror.
3. If the size of the iamge of an object remains diminished, either the object is moved away from the mirror or moved towards the mirror, then that mirror is a convex mirror.

Question 11.
Compare the characteristics of an image formed by a convex mirror and a concave mirror, when object is beyond centre of curvature, but not at infinity.
Answer:
Characteristics of the image formed by concave mirror when object is beyond centre of curvature, but not at infinity are :

1. Image formed is real.
2. Image formed is inverted.
3. Image formed is diminished.
4. Image is formed between centre of curvature (C) and principal focus (F), in front of the concave mirror.

Characteristics of the image formed by convex mirror when object is beyond centre of curvature, but not at infinity are :

1. Image formed is virutal.
2. Image formed is erect.
3. Image formed is diminished.
4. Image is always formed between pole and principal focus and behind the convex mirror.

Question 12.

1. Why does a driver use a convex mirror as a rear view mirror?
2. Illustrate your answer with the help of ray diagram.

Answer:
See Q. No. 8 of Exercise (2).

Question 13.

1. What is a real image?
2. What type of mirror is used to obtain a real image-of an object?
3. Does the mirror named by you above give real images for all locations of object?

Answer:

1. Real image : When the rays of light diverging from a point, after reflection of refraction, actually converge at some point, then that point is the real image of the object.
2. Concave mirror is used to obtain the real image of an object.
3. Concave mirror can not give real image for all the locations of object.

Question 14.
In the figure is shown a concave mirror. A is a point on the principal axis. If an object O is kept at A, image is formed on A itself. Copy the diagram. Draw the image in the diagram. Is the image real or virtual?

Measure the distance PA and write it in the diagram. What is the distance PA called?
Mark a point B on the principal axis, at which, if a point source of light is kept, the rays travel parallel to principal axis after reflection from M. What is point B called?
Answer:
As the object is placed at A and its image is also formed at A, so object must be at centre of curvature (C).

PA is called radius of curvature and on measuring PA = 4.8 cm. Point B marked on principal axis is called principal focus.

Question 15.
An object OA is placed on the principal axis of a concave mirror as shown in the figure. Copy and complete the diagram to show the formation of image.

Answer:

Question 16.
Copy the figure and complete it, by drawing two rays to show the formation of the image of the object AB. State the size, position and nature of image formed.

Answer:

• Nature : Image formed is virtual and erect.
• Position : Image is formed between principal focus (F) and pole.
• Size : Image formed is diminished.

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given A New Approach to ICSE Physics Part 1 Class 9 Solutions Light are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.

A New Approach to ICSE Physics Part 1 Class 9 Solutions Energy Flow and Practices for Conservation of Resources

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Energy Flow and Practices for Conservation of Resources.

Exercise

Question 1.
Name two renewable and the two non-renewable sources of energy.
Answer:

• Renewable sources : Solar energy, wind energy, hydro-energy, geo-thermal energy and nuclear energy.
• Non-renewable sources : Coal, petroleum and natural gas.

Question 2.
Why should we conserve the non-renewable sources of energy?
Answer:
We should conserve the non-renewable sources of energy because if these sources once finished, cannot be regenerated over a reasonable period of time.

Question 3.
State the law of conservation of energy.
Answer:
Law of conservation of energy : According to law of conservation of energy— “energy is neither created nor destroyed in a system and sum total of energy remains same.”

Question 4.
State the main difference between the renewable and non-renewable sources of energy.
Answer:
Renewable sources :

1. These are the sources from which energy can be obtained continuously over a long period of time.
2. They are non-conventional sources.
3. These are the natural sources which will never get exhausted.
4. These sources can be regenerated.
5. Examples : Solar energy, wind energy, hydro-energy, geo-thermal energy and nuclear energy.

Non-renewable sources :

1. These are the sources from which energy can not be continuously obtained over a long period of time.
2. They are conventional sources.
3. These are natural sources which ould soon depletes.
4. These sources cannot be regenerated.
5. Examples : Coal, petroleum and natural gas.

Question 5.
Explain why is it no wise to regard wood as a renewable sources of energy.
Answer:
Use of wood is renewable source, but is not advised to use, because saplings take a long time to become a tree.

Question 6.
Why are fossil fuels called non-renewable sources of energy?
Answer:
Fossil fuels (coal, petrol, diesel etc.) are exhaustible i.e. if these are once finished can not be regenerated over a reasonable period of time. That is why fossil fuels are called non-renewable sources of energy.

Question 7.
State two limitations of solar energy.
Answer:
Limitations of solar energy :

1. This energy can not be used at night.
2. This cannot be used where solar rays are less in power
3. Solar energy produces only d.c. electricity which cannot be used for household purposes.

Question 8.
State four traditional uses of solar energy.
Answer:
Uses of solar energy :

1. Solar energy is used in solar cookers for cooking of food.
2. Solar energy is used by plants to make their food by photosynthesis.
3. Solar energy is used to produce electricity with the help of solar cells.
4. Solar energy is used for providing electricity to electronic calculators and wrist watches.

Question 9.
State three ways by which you can enhance the collection of solar energy.
Answer:
Following are the three ways by which we can enhance the collection of solar energy :
(i) Use of black painted surface : Black surfaces are good absorbers as well as good radiators of radiant heat. Thus the surfaces of the objects which are to collect radiant heat are painted black, so that their temperature rises rapidly. But after some time, an equilibrium state is reached when the solar energy absorbed by a blackened object is equal to radiant heat radiating out from its surface. To overcome this difficulty, some means should be provided to that the rate of radiant heat from its surface is reduced or stopped altogether.
(ii) Use of insulated box with glass cover : The objects painted black, should be placed in well insulated box provided with glass cover. The insulated box will prevent heat losses due to conduction, convection and radiation. The glass cover will allow the solar energy to pass through it as it is transparent to radiant heat of smaller wavelength. However, when the object with blackened surface gets hot within the box and radiates out radiant heat, the glass will not allow these radiations to pass through, as they are of longer wavelength. Thus, the heat radiations are trapped within the insulated box, thereby raising the temeperature of objects with blackened surface.
(iii) Use of reflectors : As mentioned earlier, the amount of solar energy per square metre per second is too small to do any useful work. Thus, if solar energy is directly allowed to enter a solar heating device, it does not raise its temperature sufficiently. However, if solar energy is collected from a sufficiently large area and then reflected into the small area of a solar heating device, its temperature rises sufficiently to do some useful work. It is here that reflectors are useful for solar heating devices.

Figure shows the use of plane mirror as a reflector. In the diagram it is clearly shown that a part of solar energy enters directly into solar heating device and a part is reflected into it by the plane mirror. As the solar heating device receives more energy per unit time, its temperature rises quickly.

Question 10.
State two advantages of box type solar cooker.
Answer:
Advantage of box type solar cooker :

1. It is used for cooking food.
2. It costs nothing for cooking the food.
3. No loss of nutritions value of ood.
4. It causes no pollution.

Question 11.
What is the range of temperature which can be achieved in it?
Answer:
The temperature within solar cooker rises to about 140°C.

Question 12.

(a) What are semiconductors? Name two semi conductors.
(b) What are doped semiconductors? Name two materials used for doping the semiconductors. What is the advantage of doping the semiconductors?

Answer:

(a) Semiconductors : Those materials whose conductivity is less than conductors but more than insulators are known as semiconductors.
Semiconductors are neither good conductors nor insulators of electricity. Germanium (Ge) and Silicon (Si) are the two semiconductors.
(b) Doping : The process of adding impurities to semi conductors is called doping.
Materials used for doping are Boron and phosphorous.
Advantages of doping the semiconductors : Semi conductors are doped to increase the conductivity of the semiconductor.

Question 13.
Name a device which directly converts solar energy to electric energy.
Answer:
Solar cell.

Question 14.
What is a solar panel? To what uses are the solar panels put?
Answer:
Solar Panel : The group of solar cells connected in specific pattern to produce desired potential difference and magnitude of current is called Solar Panel.

Question 15.
What is biogas? Which component of biogas is used as fuel? Describe a fixed dome type biogas plant with the help of a labelled diagram.
Answer:
Biogas : Biogas is a mixture of gases formed when slurry of animal dung and water is allowed to ferment in the absence of oxygen (or air). Biogas is a mixture of methane, carbon dioxide, hydrogen and traces of hydrogen sulphide along with water vapours.
Its chief component methane is used as fuel.
Fixed dome type biogas plant :
It consists of follows parts :
Digester tank : It is an underground tank. Its base and walls are made of bricks and cement. Its roof is made in the form of dome either with cement or bricks or with concrete. The dome collects the biogas. It is provided with an outlet for biogas. The outflow of the gas is controlled by gas valve.
Slurry mixing tank : On the left hand side of the digester is constructed a brick lined tank above the ground level. The bottom of this tank connects the digester through a brick-lined channel near the base.
The fresh slurry of animal dung and water in equal proportions is poured into the slurry mixing tank. This slurry slowly flows into the digester.

Spent slurry tank : On the right hand side of the digester is constructed spent slurry tank, below the ground level. The base of this tank is in line with the lowest point of the dome of the digester. The base of this tank is connected to the base of the digester through a brick-lined channel.
Working of fixed dome type biogas plant : Everyday the slurry of animal dung and water in equal proportions is poured into the mixing tank. This slurry flows into the digester. In the digester anaerobic fermentation starts with the release of biogas. The biogas formed collects in the dome.
As more and more slurry flows into the digester everyday, the level of slurry rises up. It takes about 30 to 60 days for the slurry to fill the digester depending upon the temperature within the digester. During this time the biogas collects in the dome under pressure.
When the pressure of biogas exceeds certain limit, the biogas forces the spent slurry into overlfow tank. The biogas is withdrawn from the dome through outlet pipe. Its flow is regulated by a gas valve. The spent slurry is periodically removed and is allowed to dry. It is then used as manure.
Once the biogas plant gets operative, it is a continuous process. Everyday fresh slurry is added in the mixing tank and spent slurry is removed from the overflow tank.

Question 16.
Name an agent which decomposes animal dung into biogas.
Answer:
Anaerobic bacteria decomposes animal dung into biogas.

Question 17.
Write any two uses of biogas.
Answer:
Uses of biogas :

1. It is used for cooking.
2. Its calorific value is very high.
3. It is an eco-friendly source of energy.

Question 18.

(a) Name the kind of energy possessed by the wind.
(b) State two advantages and two limitations of wind as a source of energy.

Answer:
(a) Kinetic energy is possessed by wind.
(b) Advantages of wind energy :

1. It does not cause any kind of pollution.
2. It is a renewable source.

Limitations of wind energy :

1. The wind farms can be establishes only at places when wind blows around the year steadily.
2. A large area of land is needed to establish a wind farm.

Question 19.
With the help of a diagram explain how wind energy is converted into electric energy.
Answer:
Windmill : It is a machine which converts kinetic energy of wind into electrical energy.
A windmill consists of two or three blades like an electric fan which rotates about an axle mounted on a pole.
The shaft of the windmill is connected to the shaft of armature of a generator. When the fast moving wind rotates the blades of windmill, its shaft rotates the armature inside the generator. The rotating armature produces electric current. Thus, the kinetic energy of wind is converted into electric energy.

Question 20.

(a) What do you understand by the term hydroelectric power?
(b) Explain the energy changes taking place in a hydroelectric dam?

Answer:
(a) Hydroelectric power : Hydroelectric power is the power derived from energy of falling water or fast running water. A hydroelectric power plant is an arrangement in which the kinetic energy of flowing water is transformed into electric energy and the electric energy so generated is called hydroelectric energy.
(b) Energy changes taking place in a hydroelectric dam :

1. As the water flows into reservoir from the catchment area, the kinetic energy of flowing water changes to potential energy.
2. As the water is released through control valve, the potential energy of water changes to kinetic energy.
3. As the water flowing at a high speed strikes the blades of the turbine, its kinetic energy changes to rotational kinetic energy of the turbine.
4. As the turbine is coupled to the shaft of generator, the rotational kinetic energy of turbine changes to the rotational kinetic energy of armature.
5. As the coils of armature rotate in magnetic field, the rotational kinetic energy of armature changes to the electric energy in its coils.

Question 21.
State two advantages and two limitations of Hydel Power.
Answer:
Advantages of Hydel Power :

1. It does not produce any environmental pollution.
2. It is renewable source of energy.
3. Dam help us in irrigation and they control floods.

Limitations of Hydel Power :

1. The flowing water is not available every where.
2. The ecological balance in the downstream areas of rivers gets disturbed.

Question 22.
Give four reasons, why the use of cowdung in biogas preferred to burning of cowdung cakes.
Answer:
Use of cow-dung in biogas is preferred to burning of cow-dung cakes because :

1. Biogas does not produce any smoke and hence causes no pollution.
2. Biogas leaves behind no ash on burning.
3. Calorific value of biogas is very high as compared to cow-dung cakes.
4. Biogas can be used for driving engines of tube wells while cow-dung cakes can not be used for this purpose.

Question 23.
Why is charcoal considered better fuel than wood?
Answer:
Charcoal is considered a better fuel than wood because of the
following reasons :

1. Calorific value of charcoal is higher than the calorific value of wood.
2. It produces no smoke.
3. It burns easily without any flame.

Question 24.
Name two forms of sea energy.
Answer:
Two forms of sea energy are :

1. Tidal energy
2. Wave energy

Question 25.
By which name heat energy in the interior of earth is known?
Answer:
Geothermal energy.

Question 26.
Name the major constituent of natural gas.
Answer:
Methane is the major constituent of natural gas.

Question 27.
Name the nuclear process that is responsible for

1. explosion of an atom bomb
2. release of energy by sun.

Answer:

1. Nuclear fission is responsible for explosion of atom bomb.
2. Nuclear fusion is responsible for release of energy by the sun.

Question 28.

1. Define nuclear fission
2. Nuclear fusion.

Answer:

1. Nuclear fission : The process by which a heavy unstable nucleus (such as w-235) is broken into two medium-weight nuclei by the bombardment of a slow neutron, so as to liberate more neutrons and tremendous amount of energy is called nuclear fission.
2. Nuclear fusion : The process in which two or more light nuclei combine together to form a heavy nucleus along with the release of very large amount of energy is called nuclear fusion.

Question 29.
Describe condition for

1. nuclear fission
2. nuclear fusion.

Answer:

1. Bombardment of slow neutrons on heavy unstable nucleus (such as U-235).
2. A very high temperature of the order of millions of degree Calsius is required for nuclear fusion.

Question 30.
Explain the difference between nuclear fission and nuclear fusion.
Answer:
Difference between nuclear fission and nuclear fusion are :

Nuclear fission :

1. In a nuclear fission, a heavy nucleus splits up into lighter nuclei.
2. Harmful radiations are emitted in nuclear fission
3. It starts when slow neutrons bombards the heavy nucleus like ranium-235
4. It causes much pollution.
5. A large amount of energy is released in nuclear fission.
6. It can be controlled.

Nuclera fusion :

1. In a nuclear fusion, two or more light nuclei combine to form a heavy nucleus.
2. No harmful radiations are emitted in nuclear fusion.
3. It starts when light nuclei are heated at very high temperautre.
4. It causes no pollution.
5. Energy released in nuclear fusion is much more than nuclear fission.
6. It can not be controlled.

Question 31.
Explain how geothermal energy is used in generating electricity.
Answer:
Geothermal energy : The natural occurring thermal energy found within rock formations of the earth and the fluids hold within those formations is known as geothermal energy.
The places under the surface of the earth, where the hot magma collects at fairly less depths, are called hot spots of the earth. The hot spots are the source of the geothermal energy which is used to generate electricity as follows :
(i) In some geological regions of the earth, the underground water comes in contact with hot spots and changes into steam. The steam so formed gets trapped between the underground rocks and gets compressed to very high pressure. This steam is extracted from the ground by sinking pipes from the surface of the earth. The steam coming up at high pressure is used in running turbines connected to the generator. This in turn produces electric energy.

(ii) In some place, the steam formed under the surface of the earth does not get trapped. Instead it forces its way up through the cracks in the rocks along with hot water and rushes out from the surface of the earth to form natural geysers. The heat energy brought up by the natural geysers is used to generate electricity as stated above.
(iii) In some geological regions of the earth, there may be hot spots, but the underground water does not come in contact with them. In such regions, two holes are made in the Earth’s crust. Through one hole is pumped in cold water. The cold water on coming in contact with hot spot changes to super heated steam, which emerges out from the other hole. The steam is then made to run a turbine coupled with generator to produce electricity.

Question 32.
What are hot spots? How can you extract energy from a hot spot, if it does not come in contact with underground water?
Answer:
Hot spots : The places under the surface of earth, where hot magma collects at fairly less depths, are called hot spots of the earth.
If hot spot does not come in contact with underground water then two holes are made in the earth’s crust. Through one hole is pumped in cold water. The cold water on coming in contact with the hot spot changes to super heated steam, which emerges out from the other hole. The steam is then made to run a turbine coupled with generator to produce electricity.

Question 33.
Explain, why nuclear fusion is not being used to meet day to day energy needs.
Answer:
Nuclear fusion is not being used to meet day to day energy needs because extremely high temperature and pressure is required to initiate the nuclear fusion. Such an extremely high temperature and pressure can not be achieved in daily life in a controlled way.

Question 34.
What is nuclear waste? What are hazard of nuclear waste to living beings? How is nuclear waste disposed off?
Answer:
Nuclear waste : Nuclear waste is the material that nuclear fuel becomes after it is used in a reactor.
Hazards of nuclear waste :
(i) Nuclear waste can cause environmental contamination which can affect the health of millions of people.
(ii) Nuclear waste is the source of radiations which can lead to the following disorders:

(a) Radiations can cause genetic disorders.
(b) These can cause leukemia.
(c) These can destroy the immunity and hence may lead to death.
(d) Long exposure to radiations can cause cancer, blindness etc.

Question 35.
What are environmental consequences of using fossil fuels?
Answer:
Environmental consequences of using fossil fuels are :

1. Burning of fossil fuel causes air pollution.
2. Burning of fossil fuel can cause acid rain which corrode the buildings, monuments and reduces the fertility of the soil.
3. Smog is caused by burning of fossil fuels which block the sun’s radiations and hence hamper the process of photosynthesis.
4. Burning of fossil fuels release green house gases, which in turn be the source of global warming.

Question 36.
Energy from various sources is considered to have been derived from the Sun. Do you agree? Justify your answer by giving two examples.
Answer:
Yes, sun is the ultimate source of energy directly or indirectly, all the forms of energy derived from solar energy.
(i) Non-renewable sources of energy : Fossil fuels like coal, petroleum and natural gas are formed due to burial of large plants and ancient creatures whose ultimate source of energy is sun.
(ii) Renewable sources of energy : They are indirectly derived from solar energy such as :

(a) Energy from flowing water : In lakes, rivers, seas etc., evaporates due to solar energy. They bring rainfall and snow fall.
(b) Wind energy : Wind energy arises due to uneven heating of the earth’s surface by the sun rays at two different adjoining places. Due to this, wind possesses kinetic energy.
(c) Bio energy : Plants in the process of photosynthesis convert the solar energy into food (chemical energy). This food is consumed by animals.
(d) Wave energy : The waves are generated by strong winds (due to solar energy) blowing across the sea.
(e) Ocean thermal energy : Sun is responsible for the temperature difference between the water at the surface and water at depth in seas and oceans.

(iii) Solar devices : They derive their energy directly from solar energy and convert it into other usable forms of energy.

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given A New Approach to ICSE Physics Part 1 Class 9 Solutions Energy Flow and Practices for Conservation of Resources are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.