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Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves

November 9, 2023May 28, 2022 by Veerendra

Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 8 Propagation of Sound Waves. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 8 Propagation of Sound Waves

Exercise 8(A)

Solution 1S.

Sound is caused due to vibrations of a body.

Solution 2S.

Sound is a form of energy that produces the sensation of hearing in our ears. Sound is produced by a vibrating body.

Solution 3S.

Vibrating

Solution 4S.

Experiment: A tuning fork is taken and its one arm is struck on a rubber pad and it is brought near a tennis ball suspended by a thread as shown in figure.
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It is noticed that as the arm of the vibrating fork is brought close to the ball, it jumps back and forth and sound of the vibrating tuning fork is heard. When its arm stop vibrating, the ball becomes stationary and no sound is heard.

Solution 5S.

Experiment to demonstrate that a material medium is necessary for the propagation of sound:
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An electric bell is suspended inside an airtight glass bell jar. The bell jar is connected to a vacuum pump as shown in figure. As the circuit of electric bell is completed by pressing the key, the hammer of the electric bell begins to strike the gong repeatedly due to which sound is heard.

Keeping the key pressed, air is gradually withdrawn from jar by starting the vacuum pump. It is noticed that the loudness of sound goes on decreasing as the air is taken out from the bell jar and finally no sound is heard when all the air from the jar has been drawn out. The hammer of the electric bell is still seen striking the gong repeatedly which means that sound is still produced but it is not heard.

When the jar is filled with air, the vibrations produced by the gong are carried by the air to the walls of jar which in turn set the air outside the jar in vibration and sound is heard by us but in absence of air, sound produced by bell could not travel to the wall of the jar and thus no sound is heard. It proves that material medium is necessary for the propagation of sound waves.

Solution 6S.

We cannot hear each other on moon’s surface because there is no air on moon and for sound to be heard, a material medium is necessary.

Solution 7S.

Requisites of the medium for propagation of sound:

The medium must be elastic.
The medium must have inertia.
The medium should be frictionless.

Solution 8S.

Take a vertical metal strip with its lower end fixed and upper end being free to vibrate as shown in fig (a).

As the strip is moved to right from a to b as shown in Fig (b), the air in that layer is compressed (compression is formed at C). The particles of this layer compress the layer next to it, which then compresses the next layer and so on. Thus, the disturbance moves forward in form of compression without the particles themselves being displaced from their mean positions.

As the metal strip returns from b to a as shown in Fig (c) after pushing the particles in front, the compression C moves forward and particles of air near the strip return to their normal positions.

When the strip moves from a to c as shown in Fig (d), it pushes back the layer of air near it towards left and thus produces a low pressure space on its right side i.e. layers of air get rarefied. This region is called rarefaction (rarefaction is formed at R).

When the strip returns from C to its mean position A in Fig (e), the rarefaction R travels forward and air near the strip return to their normal positions.

Thus, one complete to and fro motion of the strip forms one compression and one rarefaction, which together form one wave. This wave through which sound travels in air is called longitudinal wave.
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Solution 9S.

the disturbance

Solution 10S.

Sound travels in a medium in form of longitudinal and transverse waves.

Solution 11S.

A type of wave motion in which the particle displacement is parallel to the direction of wave propagation is called a longitudinal wave. It can be produced in solids, liquids as well as gases.

Solution 12S.

A type of wave motion in which the particle displacement is perpendicular to the direction of wave propagation is called a transverse wave. It can be produced in solids and on the surface of liquids.

Solution 13S.

A longitudinal wave propagates by means of compressions and rarefactions.

When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called a compression (C), as shown in Fig. This compression starts to move away from the vibrating object. When the vibrating object moves backwards, it creates a region of low pressure called rarefaction (R), as shown in Fig.
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Compressions are the regions of high density where the particles of the medium come very close to each other and rarefactions are the regions of low density where the particles of the medium move away from each other.

Solution 14S.

A crest is a point on the transverse wave where the displacement of the medium is at a maximum.
A point on the transverse wave is a trough if the displacement of the medium at that point is at a minimum.
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Solution 15S.

Experiment to show that in a wave motion, only energy is transferred, but particles of the medium do not move:

If we drop a piece of stone in the still water of pond, we hear a sound of stone striking the water surface. Actually a disturbance is produced in water at the point where the stone strikes it. This disturbance spreads in all directions radially outwards in form of circular waves on the surface of water.

If we place a piece of cork on water surface at some distance away from the point where the stone strikes it, we notice that cork does not move ahead, but it vibrates up and down, while the wave moves ahead. The reason is that particles of water (or medium) start vibrating up and down at the point where the stone strikes. These particles then transfer their energy to the neighboring particles and they themselves come back to their mean positions. Thus only energy is transferred but the particles of the medium do not move.

Solution 16S.

The maximum displacement of the particle of medium on either side of its mean position is called the amplitude of wave.
Its SI unit is metre.

Solution 17S.

The number of vibrations made by the particle of the medium in one second is called the frequency of the wave. It can also be defined as the number of waves passing through a point in one second.
Its SI unit is hertz (Hz).

Solution 18S.

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Solution 19S.

The distance travelled by a wave in one second is called its wave velocity.
Its SI unit is metre per second (ms^-1).

Solution 20S.

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Solution 21S.

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Solution 22S.

Let the velocity of a wave be V, time period T, frequency ν and wavelength λ. By the definition of wavelength,
Wavelength = Distance travelled by the wave in one time period i.e., in T second
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Solution 23S.

The speed of sound in a medium depends upon its elasticity and density.

Solution 24S.

V_g < V_l < V_s

Solution 25S.

(i) Speed of light in air = 3 x 10⁸ m s^-1 (ii) Speed of sound in air = 330 m s^-1.

Solution 26S.

1 : 4 : 15

Solution 27S.

(i) No, sound cannot travel in vacuum as it requires a material medium for its propagation.
(ii) Speed of sound is maximum in solids, less in liquids and least in gases.

Solution 28S.

This happens because the light travels much faster than sound.

Solution 29S.

Sound travels in iron faster than in air so first the sound travelled in iron rail is heard and then the sound travelled in air is heard.

Solution 30S.

(i) The diver would hear the sound first.
(ii) This is because sound travels faster in water than in air.
(iii) It would take 0.25t to reach the diver because sound travels almost four times faster in water.

Solution 31S.

(i) Frequency of sound has no effect on the speed of sound.
(ii) Speed of sound increases with the increase in the temperature of sound.
(iii) Pressure of sound has no effect on the speed of sound.
(iv) Speed of sound increases with the increase in presence of moisture in air.

Solution 32S.

(i) Speed of sound does not change with a change in amplitude.
(ii) Speed of sound does not change with a change in wavelength.

Solution 33S.

Speed of sound is more in humid air because in presence of moisture, the density of air decreases and sound travels with greater speed.
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Solution 34S.

The speed of sound increases by 0.61 m s^-1 for each 1°C rise in temperature.

Solution 35S.

The simple experiment that a person can do to calculate the speed of sound in air is that a person stands at a known distance (d meter) from the cliff and fires a pistol and simultaneously start the stopwatch. He stops the stopwatch as soon as he hears an echo. The distance travelled by the sound during the time (t) seconds is 2d. So, speed of sound = distance travelled / time taken = 2d/t

The approximation made is that speed of sound remains same for the time when the experiment is taking place.

Solution 36S.

(a) Vacuum, medium (b) do not move, moves (c) rarefaction (d) trough.

Solution 1M.

Sound needs medium, but light does not need medium for its propagation.

Solution 2M.

Longitudinal wave

Solution 3M.

330 m s^-1

Solution 4M.

3 x 10⁸ m s^-1

Solution 1N.

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Solution 2N.

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Solution 3N.

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Solution 4N.

Wave velocity = 0.3 m/s
Frequency = 20 Hz
Separation between two consecutive compressions is the wavelength of a wave.
We know that,
Wave velocity = Frequency x Wavelength
Or, wavelength = Wave velocity / frequency
Or, λ = 0.3 / 20 = 1.5 x 10^-2 m

Solution 5N.

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Solution 6N.

Distance between the two observers = 1650 m
Speed of sound = 330 m/s
Time in which B hears the sound = Distance / speed = 1650/330 = 5s
Thus, B will hear the sound 5s after the gun is shot.

Solution 7N.

Speed of sound in air (V) = 330 m/s
Time in which thunder is heard after lighting is seen (t) = 5s
Thus, distance between flash and observer = V x t = (330 x 5) = 1650 m

Solution 8N.

Speed of sound in air (V) = 340 m/s
Time in which sound of fire is heard after flash is seen (t) = 2.5s
Thus, distance between flash and observer = V x t = (340 x 2.5) = 850 m

Solution 9N.

Time taken by the observer to hear the sound of the first tank A= 3.5s
Time taken by the observer to hear the sound of the second tank B = 2s
Time taken by the tank B to hear the sound of tank A= (3.5 – 2)s = 1.5s
Distance between the two tanks = 510m
Speed = 510/1.5=340m/s

Solution 10N.

(a) Length of iron rail (D) = 3.3 km = 3300 m
Speed of sound in iron (V) = 5280 m/s
Time taken by sound to travel in iron rod (t) = D/V
Or, t = (3300 / 5280) s = 0.625 s

(b) Length of iron rail (D) = 3.3 km = 3300 m
Speed of sound in air (V) = 330 m/s
Time taken by sound to travel in iron rod (t) = D/V
Or, t = (3300/330) s = 10 s

Solution 11N.

(i) Distance travelled (D) = 1700
Speed of sound in air (V) = 340 m/s
Time taken (t) = D/V = (1700 / 340) s = 5 s

(ii) Distance travelled (D) = 1700
Speed of sound in water (V’) = 1360 m/s
Time taken (t) = D/V = (1700 / 1360) s = 1.25 s

Exercise 8(B)

Solution 1S.

The range of frequency within which the sound can be heard by a human being is called the audible range of frequency.

Solution 2S.

The audible range of frequency for humans is 20 Hz to 20 kHz.

Solution 3S.

Human ears are most sensitive for the range 2000 Hz to 3000 Hz.

Solution 4S.

Ultrasonic has higher frequency.

Solution 5S.

(a) 20 Hz, 20 kHz (b) above 20 kHz (c) below 20 Hz (d) ultrasonic (e) infrasonic.

Solution 6S.

(a) Infrasonic (b) Audible (c) Audible (d) Ultrasonic.

Solution 7S.

No, we cannot hear the sound produced due to vibrations of a seconds pendulum because the frequency of sound produced due to vibrations of seconds pendulum is 0.5 Hz which is infrasonic.

Solution 8S.

Sounds of frequency above 20 kHz are called ultrasound.

Solution 9S.

The approximate speed of ultrasound in air is 330 m/s.

Solution 10S.

Two properties of ultrasound which make it useful to us are:

High energy contents
High directivity

Solution 11S.

Bats locate the obstacles and prey in their path by producing and hearing the ultrasound. They emit an ultrasound which returns after striking an obstacle in their way. By hearing the reflected sound and from the time interval (when they produce ultrasound and they receive them back), they can judge the direction and the distance of the obstacle in their way.

Solution 12S.

Two applications of ultrasound:

Ultrasound is used for drilling holes or making cuts of desired shape in materials like glass.
Ultrasound is used in surgery to remove cataract and in kidneys to break the small stones into fine grains.

Solution 1M.

1000 Hz

Solution 2M.

High power and good directivity

Solution 3M.

Ultrasound

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Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation

November 9, 2023May 28, 2022 by Veerendra

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation

ICSE Solutions Selina ICSE Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation

Exercise 1(A)

Solution 1S.

Measurement is the process of comparing a given physical quantity with a known standard quantity of the same nature.

Solution 2S.

Unit is a quantity of constant magnitude which is used to measure the magnitudes of other quantities of the same manner.

Solution 3S.

The three requirements for selecting a unit of a physical quantity are

It should be possible to define the unit without ambiguity.
The unit should be reproducible.
The value of units should not change with space and time.

Solution 4S.

Definitions of three fundamental quantities:

S.I. unit of length (m): A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy made of 90% platinum and 10% iridium) rod kept at 0°C in the International Bureau of Weights and Measures at serves near Paris.
S.I. unit of mass (kg): In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at International Bureau of Weights and Measures at serves near Paris.
S.I. unit of time (s): A second is defined as 1/86400th part of a mean solar day, i.e.

Solution 5S.

Three systems of unit and their fundamental units:

C.G.S. system (or French system): In this system, the unit of length is centimeter (cm), unit of mass is gramme (g) and unit of time is second (s).
F.P.S. system (or British system): In this system, the unit of length is foot (ft), unit of mass is pound (lb) and unit of time is second (s).
M.K.S. system (or metric system): In this system, the unit of length is metre (m), unit of mass is kilogramme (kg) and unit of time is second (s).

Solution 6S.

A fundamental (or basic) unit is that which is independent of any other unit or which can neither be changed nor can be related to any other fundamental unit.

Solution 7S.

Fundamental quantities, units and symbols in S.I. system are

Quantity	Unit	Symbol
Length	metre	m
Mass	kilogramme	kg
Time	second	s
Temperature	kelvin	K
Luminous intensity	candela	cd
Electric current	ampere	A
Amount of substance	mole	mol
Angle	radian	rd
Solid angle	steradian	st-rd

Solution 8S.

The units of quantities other than those measured in fundamental units can be obtained in terms of the fundamental units, and thus the units so obtained are called derived units.

Example:
Speed = Distance/time
Hence, the unit of speed = fundamental unit of distance/fundamental unit of time
Or, the unit of speed = metre/second or ms^-1.
As the unit of speed is derived from the fundamental units of distance and time, it is a derived unit.

Solution 9S.

A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy with 90% platinum and 10% iridium) rod kept at 0^o C in the International Bureau of Weights and Measures at serves near Paris.

Solution 10S.

Astronomical unit (A.U.) and kilometer (km) are units of length which are bigger than a metre.
1 km = 1000 m
1 A.U. = 1.496 × 10¹¹ m

Solution 11S.

Centimeter (cm) and millimeter (mm) are units of length smaller than a metre.
1 cm = 10^-2 m
1 mm = 10^-3 m

Solution 12S.

1 nm = 10 Å

Solution 13S.

Three convenient units of length and their relation with the S.I. unit of length:

1 Angstrom (Å) = 10^-10 m
1 kilometre (km) = 10³ m
1 light year (ly) = 9.46 × 10¹⁵ m

Solution 14S.

S.I. unit of mass is ‘kilogramme’.
In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at the International Bureau of Weights and Measures at serves near Paris.

Solution 15S.

(a) 1 light year = 9.46 × 10¹⁵ m
(b) 1 m = 10¹⁰ Å
(c) 1 m = 10⁶ µ (micron)
(d) 1 micron = 10⁴ Å
(e) 1 fermi = 10^-15 m

Solution 16S.

The units ‘gramme’ (g) and ‘milligramme’ (mg) are two units of mass smaller than ‘kilogramme’.
1 g = 10^-3 kg
1 mg = 10^-6 kg

Solution 17S.

The units ‘quintal’ and ‘metric tonne’ are two units of mass bigger than ‘kilogramme’.
1 quintal = 100 kg
1 metric tonne = 1000 kg

Solution 18S.

(a) 1 g = 10^-3 kg
(b) 1 mg = 10^-6 kg
(c) 1 quintal = 100 kg
(d) 1 a.m.u (or u) = 1.66 x 10^-27 kg

Solution 19S.

The S.I. unit of time is second (s).
A second is defined as 1/86400th part of a mean solar day, i.e.

Solution 20S.

The units ‘minute’ (min) and ‘year’ (yr) are two units of time bigger than second(s).
1 min = 60 s
1 yr = 3.1536 × 10⁷ s

Solution 21S.

A leap year is the year in which the month of February has 29 days.

Solution 22S.

Yes, the given statement is true.

Solution 23S.

One lunar month is the time in which the moon completes one revolution around the earth. A lunar month is made of nearly 4 weeks.

Solution 24S.

(a) 1 nanosecond = 10^-9 s
(b) 1 µs = 10^-6 s
(c) 1 mean solar day = 86400 s
(d) 1 year = 3.15 × 10⁷ s

Solution 25S.

(a) Mass (b) Distance (or length) (c) Time (d) Length

Solution 26S.

(a) ms^-1 (b) kg ms^-2 (c) kg m²s^-2 (d) kg m^-1s^-2

Solution 27S.

(a) kg ms^-2 (b) kg m²s^-3(c) kg m²s^-2 (d) kg m^-1s^-2

Solution 28S.

(a) Area (b) Force (c) Energy
(d) Pressure (f) Power

Solution 1M.

Second

Solution 2M.

Litre

Solution 3M.

Leap year

Solution 4M.

0.1 nm

Solution 5M.

Length

Solution 1N.

Wavelength of light of particular colour = 5800 Å

(a) (i) 1 Å = 10^-1 nm
5800 Å = 5800 × 10^-1 nm
= 580 nm

(ii) 1 Å = 10^-10 m
5800 Å = 5800 × 10^-10 m
= 5.8 × 10^-7 m

(b) The order of its magnitude in metre is 10^-6 m because the numerical value of 5.8 is more than 3.2.

Solution 2N.

Size of a bacteria = 1 µ
Since 1 µ = 10^-6 m
Number of the particle = Total length/size of
one bacteria
= 1 m/10^-6 m
= 10⁶

Solution 3N.

Distance of galaxy = 5.6 × 10²⁵ m
Speed of light = 3 × 10⁸ m/s

(a) Time taken by light = Distance travelled/speed of light
= (5.6 × 10²⁵/ 3 × 10⁸) s
= 1.87 × 10¹⁷ s

(b) Order of magnitude = 10⁰ × 10¹⁷ s = 10¹⁷ s
(This is because the numerical value of 1.87 is less than the numerical value 3.2)

Solution 4N.

Wavelength of light = 589 nm
= 589 × 10^-9 m
= 5.89 × 10^-7 m
Order of magnitude = 10¹ × 10^-7 m
= 10^-6 m
(This is because the numerical value of 5.89 is more than the numerical value 3.2)

Solution 5N.

Mass of an oxygen atom = 16.00 u
Now, 1 u = 1.66 × 10^-27 kg
Hence, mass of oxygen in kg = 16 × 1.66 × 10^-27 kg
= 26.56 × 10^-27 kg
Because the numerical value of 26.56 is greater than the numerical value of 3.2, the order of magnitude of mass of oxygen in kg
= 10¹ × 10^-27 kg
= 10^-26 kg

Solution 6N.

Time taken by light to reach from the Sun to the Earth = 8 min = 480 s.
Speed of light = 3 × 10⁸ m/s
Distance from the Sun to the Earth = Speed × time
= 3 × 10⁸ × 480 m
= 1440 × 10⁸ m
= 1440 × 10⁸ × 10^-3 km
= 1440 × 10⁵ km
= 1.44 × 10⁸ km
Because the numerical value of 1.44 is less than the numerical value of 3.2, the order of magnitude of distance from the Sun to the Earth in km = 10⁰ × 10⁸ km
= 10⁸ km

Solution 7N.

The statement ‘the distance of a star from the Earth is 8.33 light minutes’ means that the light from the star takes 8.33 minutes to reach Earth.

Exercise 1(B)

Solution 1S.

The least count of an instrument is the smallest measurement that can be taken accurately with it. For example, if an ammeter has 5 divisions between the marks 0 and 1A, then its least count is 1/5 = 0.2 A or it can measure current up to the value 0.2 accurately.

Solution 2S.

Total length of the scale = 1 m = 100 cm
No. of divisions = 100
Length of each division = Total length/total no. of divisions
= 100 cm/100
= 1 cm
Thus, this scale can measure with an accuracy of 1 cm.
To increase the accuracy, the total number of divisions on the scale must be increased.

Solution 3S.

The least count of a metre rule is 1 cm.
The length cannot be expressed as 2.60 cm because a metre scale measures length correctly only up to one decimal place of a centimeter.

Solution 4S.

The least count of vernier callipers is equal to the difference between the values of one main scale division and one vernier scale division.
Let n divisions on vernier callipers be of length equal to that of (n – 1) divisions on the main scale and the value of 1 main scale division be x. Then,
Value of n divisions on vernier = (n – 1) x
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Solution 5S.

Vernier constant is equal to the difference between the values of one main scale division and one vernier scale division. It is the least count of vernier callipers.

Solution 6S.

A vernier calipers is said to be free from zero error, if the zero mark of the vernier scale coincides with the zero mark of the main scale.

Solution 7S.

Due to mechanical errors, sometimes the zero mark of the vernier scale does not coincide with the zero mark of the main scale, the vernier callipers is said to have zero error.

It is determined by measuring the distance between the zero mark of the main scale and the zero mark of the vernier scale.
The zero error is of two kinds

Positive zero error
Negative zero error

1. Positive zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the right of the zero mark of the main scale, the error is said to be positive.

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To find this error, we note the division of the vernier scale, which coincides with any division of the main scale. The number of this vernier division when multiplied by the least count of the vernier callipers, gives the zero error.

For example, for the scales shown, the least count is 0.01 cm and the 6^th division of the vernier scale coincides with a main scale division.

Zero error = +6 × L.C. = +6 × 0.01 cm
= +0.06 cm

2. Negative zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the left of the zero mark of the main scale, then the error is said to be negative.

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To find this error, we note the division of the vernier scale coinciding with any division of the main scale. The number of this vernier division is subtracted from the total number of divisions on the vernier scale and then the difference is multiplied by the least count.

For example, for the scales shown, the least count is 0.01 cm and the sixth division of the vernier scale coincides with a certain division of the main scale. The total number of divisions on vernier callipers is 10.

Zero error = – (10 – 6) × L.C.
= – 4 × 0.01 cm = – 0.04 cm

Correction:
To get correct measurement with vernier callipers having a zero error, the zero error with its proper sign is always subtracted from the observed reading.
Correct reading = Observed reading – zero error (with sign)

Solution 8S.

Selina Concise Physics Class 9 ICSE Soluti

Solution 9S.

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Main parts and their functions:

Main scale: It is used to measure length correct up to 1 mm.
Vernier scale: It helps to measure length correct up to 0.1 mm.
Outside jaws: It helps to measure length of a rod, diameter of a sphere, external diameter of a hollow cylinder.
Inside jaws: It helps to measure the internal diameter of a hollow cylinder or pipe.
Strip: It helps to measure the depth of a beaker or a bottle.

Solution 10S.

Three uses of vernier callipers are

Measuring the internal diameter of a tube or a cylinder.
Measuring the length of an object.
Measuring the depth of a beaker or a bottle.

Solution 11S.

Two scales of vernier calipers are

Main scale
Vernier scale

The main scale is graduated to read up to 1 mm and on vernier scale, the length of 10 divisions is equal to the length of 9 divisions on the main scale.
Value of 1 division on the main scale= 1 mm
Total no. of divisions on the vernier scale = 10
Thus, L.C. = 1 mm /10 = 0.1 mm = 0.01 cm.
Hence, a vernier callipers can measure length correct up to 0.01 cm.

Solution 12S.

Measuring the length of a small rod using vernier calipers:
The rod whose length is to be measured is placed in between the fixed end and the vernier scale as shown in the figure.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 6
In this position, the zero mark of the vernier scale is ahead of 1.2 cm mark on main scale. Thus the actual length of the rod is 1.2 cm plus the length ab (i.e., the length between the 1.2 cm mark on the main scale and 0 mark on vernier scale).
To measure the length ab, we note the pth division of the vernier scale, which coincides with any division of main scale.
Now, ab + length of p divisions on vernier scale = length of p divisions on main scale
Alternatively, ab = length of p divisions on the main scale – length of p divisions on the vernier scale.
= p (length of 1 division on main scale – length of 1 division on vernier scale)
= p × L.C.
Therefore, total reading = main scale reading + vernier scale reading
= 1.2 cm + (p × L.C.)

Solution 13S.

(a) Outside jaws
(b) Inside jaws
(c) Strip
(d) Outer jaws

Solution 14S.

(i) Pitch: The pitch of a screw gauge is the distance moved by the screw along its axis in one complete rotation.
(ii) Least count (L.C.) of a screw gauge: The L.C. of a screw gauge is the distance moved by it in rotating the circular scale by one division.
Thus, L.C. = Pitch of the screw gauge/total no. of divisions on its circular scale.
If a screw moves by 1 mm in one rotation and it has 100 divisions on its circular scale, then pitch of screw = 1 mm.
Thus, L.C. = 1 mm / 100 = 0.01 mm = 0.001 cm

Solution 15S.

The least count of a screw gauge can be increased by decreasing the pitch and increasing the total number of divisions on the circular scale.

Solution 16S.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 7

Main parts and their functions:

Ratchet:It advances the screw by turning it until the object is gently held between the stud and spindle of screw.
Sleeve:It marks the main scale and base line.
Thimble:It marks the circular scale.
Main scale:It helps to read the length correct up to 1 mm.
Circular scale:It helps to read length correct up to 0.01 mm.

Solution 17S.

A screw gauge is used for measuring diameter of circular objects mostly wires with an accuracy of 0.001 cm.

Solution 18S.

Ratchet helps to advance the screw by turning it until the object is gently held between the stud and spindle of the screw.

Solution 19S.

Due to mechanical errors, sometimes when the anvil and spindle end are brought in contact, the zero mark of the circular scale does not coincide with the base line of main scale. It is either above or below the base line of the main scale, in which case the screw gauge is said to have a zero error. It can be both positive and negative.
It is accounted by subtracting the zero error (with sign) from the observed reading in order to get the correct reading.
Correct reading = Observed reading – zero error (with sign)

Solution 20S.

Diagram of a screw gauge with L.C. 0.001 cm and zero error +0.007 cm.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 8

Solution 21S.

Backlash error: If by reversing the direction of rotation of the thimble, the tip of the screw does not start moving in the opposite direction immediately but remains stationary for a part of rotation; it is called backlash error.

Reason: It happens due to wear and tear of the screw threads.
To avoid the backlash error, while taking the measurements the screw should be rotated in one direction only. If the direction of rotation of the screw needs to be changed, then it should be stopped for a while and then rotated in the reverse direction.

Solution 22S.

Measurement of diameter of wire with a screw gauge:

The wire whose thickness is to be determined is placed between the anvil and spindle end, the thimble is rotated until the wire is firmly held between the anvil and spindle. The rachet is provided to avoid excessive pressure on the wire. It prevents the spindle from further movement. The thickness of the wire could be determined from the reading as shown in the figure below.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 9
The pitch of the screw = 1 mm
L.C. of screw gauge = 0.01 mm
Main scale reading = 2.5 mm
46th division of circular scale coincides with the base line.
Therefore, circular scale reading = 46 × 0.01 = 0.46 mm
Total reading = Main scale reading + circular scale reading
= (2.5 + 0.46) mm
= 2.96 mm

Solution 23S.

(a) Screw gauge
(b) Screw gauge
(c) Vernier calipers
(d) Screw gauge

Solution 24S.

Screw gauge measures a length to a high accuracy.

Solution 25S.

(a) Vernier callipers (b) Metre scale (c) Screw gauge.

Solution 1M.

The least count of a vernier calipers is 0.01 cm

Solution 2M.

0.002 cm

Solution 3M.

A screw gauge

Solution 1N.

Range of the stop watch = 5s
Total number of divisions = 10
L.C. = 5/10 = 0.5 s

Solution 2N.

Value of 1 m.s.d. = 1 mm
10 vernier divisions = 9 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on vernier scale
= 1 mm/10
= 0.1 mm or 0.01 cm

Solution 3N.

There are 20 divisions in 1 cm on the main scale.
Therefore, the value of 1 m.s.d. (x) = 1/20 cm = 0.05 cm
No. of divisions on the vernier scale (n) = 25
Hence, the L.C. of the microscope = x/n = (0.05 / 25) cm
= 0.002 cm

Solution 4N.

Thickness of the pencil (observed reading) = 1.4 mm
Zero error = + 0.02 cm = + 0.2 mm
Correct reading = observed reading – zero error (with sign)
= 1.4 mm – 0.2 mm
= 1.2 mm

Solution 5N.

(i) Value of 1 m.s.d. =1 mm
10 vernier divisions = 9 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on the vernier scale
= 1 mm/10
= 0.1 mm or 0.01 cm

(ii) On bringing the jaws together, the zero of the vernier scale is ahead of zero of the main scale, the error is positive.
3rd vernier division coincides with a main scale division.
Total no. of vernier divisions = 10
Zero error = +3 × L.C.
= +3 × 0.01 cm
= +0.03 cm

Solution 6N.

(i) Value of 1 m.s.d = 1 mm = 0.1 cm
20 vernier divisions = 19 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on the vernier scale
= 1mm/20
= (0.1/20) cm
= 0.005 cm

(ii) Main scale reading = 35 mm = 3.5 cm
Since 4th division of the main scale coincides with the main scale, i.e. p = 4.
Therefore, the vernier scale reading = 4 × 0.005 cm = 0.02 cm
Total reading = Main scale reading + vernier scale reading
= (3.5 + 0.02) cm
= 3.52 cm
Radius of the cylinder = Diameter (Total reading) / 2
= (3.52/2) cm
= 1.76 cm

Solution 7N.

(a) L.C. of vernier callipers = 0.01 cm
Main scale reading = 1.8 cm
Since 4th division of the main scale coincides with the main scale, i.e. p = 4.
Therefore, the vernier scale reading = 4 × 0.01 cm = 0.04 cm
Total reading = Main scale reading + vernier scale reading
= (1.8 + 0.04) cm
= 1.84 cm

(b) Observed reading = 1.84 cm
Zero error = -0.02 cm
Correct reading = Observed reading – Zero error (with sign)
= [1.84 – (-0.02)] cm
= 1.86 cm

Solution 8N.

L.C. of vernier callipers = 0.01 cm
In the shown scale,
Main scale reading = 3.3 mm
6th vernier division coincides with an m.s.d.
Therefore, vernier scale reading = 6 × 0.01 cm = 0.06 cm
Total reading = m.s.r. + v.s.r.
= 3.3 + 0.06
= 3.36 cm

Solution 9N.

Pitch of a screw gauge = 0.5 mm
No. of divisions on the circular scale = 100
L.C. = (0.5/100) mm
= 0.005 mm or 0.0005 cm

Solution 10N.

No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm

(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm

Solution 11N.

Pitch of the screw gauge = 1mm
No. of divisions on the circular scale = 100

(i) L.C. = Pitch/No. of divisions on the circular head
= (1/100) mm
= 0.01 mm or 0.001 cm

(ii) Main scale reading = 2mm = 0.2 cm
No. of division of circular head in line with the base line (p) = 45
Circular scale reading = (p) × L.C.
= 45 x 0.001 cm
= 0.045 cm
Total reading = M.s.r. + circular scale reading
= (0.2 + 0.045) cm
= 0.245 cm

Solution 12N.

(i) L.C. of screw gauge = 0.01 mm or 0.001 cm
Main scale reading = 1 mm or 0.1 cm
No. of division of circular head in line with the base line (p) = 27
Circular scale reading = (p) × L.C.
= 27 × 0.001 cm
= 0.027 cm
Diameter (Total reading) = M.s.r. + circular scale reading
= (0.1 + 0.027) cm
= 0.127 cm

(ii) Zero error = 0.005 cm
Correct reading = Observed reading – zero error (with sign)
= [0.127 – (+0.005)] cm
= 0.122 cm

Solution 13N.

No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm

(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm

(iii) Because the zero of the circular scale lies below the base line, when the flat end of the screw is in contact with the stud, the error is positive.
No. of circular division coinciding with m.s.d. = 4
Zero error = + (4 × L.C.)
= + (4 × 0.01) mm
= + 0.04 mm

Solution 14N.

No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/1 = 1 mm.

(ii) L.C. = Pitch/No. of divisions on the circular head
= (1/50) mm
= 0.02 mm

(iii) Main scale reading = 4 mm
No. of circular division coinciding with m.s.d. (p) = 47
Circular scale reading = p × L.C.
= (47 × 0.02) mm
= 0.94 mm
Diameter (Total reading) = M.s.r. + circular scale reading
= (4 + 0.94) mm
= 4.94 mm

Solution 15N.

Pitch of the screw gauge = 0.5 mm
L.C. of the screw gauge = 0.001 mm
No. of divisions on circular scale = Pitch / L.C.
= 0.5 / 0.001
= 500

Exercise 1(C)

Solution 1S.

A simple pendulum is a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string.

No, the pendulum used in pendulum clock is not a simple pendulum because the simple pendulum is an ideal case. We cannot have a heavy mass having the size of a point and string having no mass.

Solution 2S.

(i) Oscillation: One complete to and fro motion of the pendulum is called one oscillation.
(ii) Amplitude: The maximum displacement of the bob from its mean position on either side is called the amplitude of oscillation. It is measured in metres (m).
(iii) Frequency: It is the number of oscillations made in one second. Its unit is hertz (Hz).
(iv)Time period: This is the time taken to complete one oscillation. It is measured in second (s).

Solution 3S.

Simple Pendulum:
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 10

Solution 4S.

Two factors on which the time period of a simple pendulum depends are

Length of pendulum (l)
Acceleration due to gravity (g)

Solution 5S.

Two factors on which the time period of a simple pendulum does not depend are

Material of the bob
Amplitude

Solution 6S.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 12
Therefore, the time period is doubled.

(b) If the acceleration due to gravity is reduced to one-fourth,
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 13
Therefore, the time period is doubled.

Solution 7S.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 14

Solution 8S.

Measurement of time period of a simple pendulum:

To measure the time period of a simple pendulum, the bob is slightly displaced from its mean position and is then released. This gives a to and fro motion about the mean position to the pendulum.
The time ‘t’ for 20 complete oscillations is measured with the help of a stop watch.
Time period ‘T’ can be found by dividing ‘t’ by 20.

To find the time period, the time for the number of oscillations more than 1 is noted because the least count of stop watch is either 1 s or 0.5 s. It cannot record the time period in fractions such as 1.2 or 1.4 and so on.

Solution 9S.

The time period of a simple pendulum is directly proportional to the square root of its effective length.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 15
From this graph, the value of acceleration due to gravity (g) can be calculated as follows.
The slope of the straight line can be found by taking two points P and Q on the straight line and drawing normals from these points on the X- and Y-axis, respectively. Then, the value of T2 is to be noted at a and b, the value of l at c and d. Then,
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 16
where g is the acceleration due to gravity at that place. Thus, g can be determined at a place from these measurements using the following relation:

Solution 10S.

The ratio of their time periods would be 1:1 because the time period does not depend on the weight of the bob.

Solution 11S.

Pendulum A will take more time (twice) in a given time because the time period of oscillation is directly proportional to the square root of the length of the pendulum. Therefore, the pendulum B will have a greater time period and thus making lesser oscillations.

Solution 12S.

(a) The time period of oscillations is directly proportional to the square root of the length of the pendulum.
(b) The time period of oscillations of simple pendulum does not depend on the mass of the bob.
(c) The time period of oscillations of simple pendulum does not depend on the amplitude of oscillations.
(d) The time period of oscillations of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

Solution 13S.

A pendulum with the time period of oscillation equal to two seconds is known as a seconds pendulum.

Solution 14S.

The frequency of oscillation of a seconds’ pendulum is 0.5 s^-1. It does not depend on the amplitude of oscillation.

Solution 1M.

Half

Solution 2M.

2 s

Solution 3M.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 18

Solution 1N.

(a) Frequency = Oscillations per second
= (40/60) s^-1 = 0.67 s^-1

(b) Time period = 1/frequency
= (1/0.67) s
= 1.5 s

Solution 2N.

Time period = 2 s
Frequency = 1/time period
= (½)s^-1 = 0.5 s^-1Such a pendulum is called the seconds’ pendulum.

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 19

Solution 4N.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 20

Solution 5N.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 21

Solution 6N.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 22

Solution 7N.

Let T₁and T₂ be the time periods of the two pendulums of lengths l₁ and l₂, respectively.
Then, we know that the time period is directly proportional to the square root of the length of the pendulum.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 23

Solution 8N.

Time period = Time taken to complete 1 oscillation
= (4 × 0.2) s
= 0.8 s

Solution 9N.

Time period of a seconds’ pendulum = 2 s
Time taken to complete half oscillation, i.e. from one extreme to the other extreme = 1 s.

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Selina Concise Biology Class 9 ICSE Solutions Five Kingdom Classification

November 9, 2023May 28, 2022 by Veerendra

Selina Concise Biology Class 9 ICSE Solutions Five Kingdom Classification

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Biology Chapter 8 Five Kingdom Classification. You can download the Selina Concise Biology ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Biology for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Biology Chapter 8 Five Kingdom Classification

Exercise 1

Solution A.

(d) vertebrates and invertebrates
(d) Bat, Parrot, Oyster – Vertebrates
(c) Rana tigrina
(d) Mule is neither a donkey nor a horse.

Solution B.1.

Carolus Linnaeus had introduced the binomial system of naming living beings.

Solution B.2.

The two characters common to dog, humans, squirrel, bat, camel and monkey are:
(c ) external ears
(d) give birth to young ones

Solution B.3.

Column I	Column II
1.Pine	(iv) Gymnosperm (v) Plantae
2. Earthworm	(vi) Animalia
3. Bread mould	(i) Fungi
4. Amoeba	(vii) Protista
5. Moss	(v) Plantae (viii) Bryophyta
6.Bacteria	(ii) Monera (iii) Prokaryote

Solution C.1.

Phylum → Class → Order → Family → Genus → Species

Solution C.2.

Man	Domestic cat	Peepal tree
Homo sapiens	Felis domesticus	Ficus religiosa

Solution C.3.

In science, people from different countries with different languages have to read about each others research. So, it was necessary to eliminate any possible confusion created by local names. Scientific names are based on certain rules which are universal. They are unique and can be used to identify an organism anywhere around the world. That is why, scientific names of living beings are considered better than their common names.

Solution C.4.

According to the ‘Two-Kingdom Classification’, proposed by Carolus Linnaeus in 1758, living organisms were classified into two broad kingdoms, Plants and Animals.
The drawbacks in classifying organisms under the old two kingdom classification are:

Bacteria were kept in Kingdom Plantae. These organisms have no chlorophyll and do not carry out photosynthesis. Bacteria do not have a definite nucleus nor a nuclear membrane nor chromosomes.
Fungi were kept in Kingdom Plantae. Bread mould is a multicellular fungi. However, it does not possess roots, stem and leaves, lacks chlorophyll and does bear any flowers, fruits and seeds like plants.

Solution C.5.

Species means an organism of a particular kind whose members can interbreed among themselves to produce fertile young ones.
All humans on the earth today may differ widely in their facial features, colour, height, etc. Yet, they belong to a single species Homo sapiens because they can interbreed among themselves and produce a normal offpsring.

Solution C.6.

Ficus religiosa (Peepal)
Zea mays (Maize)
Bombyx mori (Silkmoth)

Solution C.7.

The five kingdoms according to the new classification are:

Kingdom Monera
Kingdom Protista
Kingdom Fungi
Kingdom Plantae
Kingdom Animalia

Solution C.8.

(a) Animals with a backbone: Vertebrata
(b) Animals with a hairy skin: Mammalia
(c ) Animals with three pairs of legs: Insecta
(d) Animals with feathers: Aves

Solution C.9.

Invertebrate animals:

Housefly
Silverfish
Jellyfish
Sponge

Solution C.10.

(a) Protozoa and Metazoa

Protozoa	Metazoa
Unicellular organisms	Multi-cellular organisms

(b) Vertebrate and Invertebrate

Vertebrates	Invertebrates
Have their unique backbone with the spinal cord	Do not have a backbone

Insecta	Arachnida
Have three pairs of legs	Have four pairs of legs

(d) Flatworm and Roundworm

Flatworm	Roundworm
Dorso-ventrally flattened	Cylindrical in shape and are tapered at both ends

Solution C.11.

(i) Amoeba – Nucleus, tentacle, food vacuole
(ii) Hydra – Invertebrata, Cnidaria, Crustacea
(iii) Fish – Gills, paired fins, ear drum
(iv) Earthworm – Invertebrata, Annelida, Insecta
(v) Grasshopper – Wings, trachea, proboscis
(vi) Butterfly – Insecta, Invertebrata, Mollusca
(vii) Whale – Gills, mammary glands, fat under the skin
(viii) Pigeon – Feathers, wings, hair
(ix) Monkey – External ear, sweat glands, lateral line
(x) Bat – Aves, Mammalia, Chordata

Solution C.12.

Cold-blooded animals	Warm-blooded animals
These animals cannot maintain their body temperature. Their body temperature is regulated by the external environment.	Animals whose body temperature is kept relatively constant by internal mechanisms.
E.g. Insects, Amphibians	E.g. Birds, Mammals

Solution C.13.

Class Amphibia: Tree frog
Class Reptilia: Cobra
Class Aves: Duck

Solution D.1.

(a) Insects and Birds

Similarity	Difference
Have wings	Insects	Birds
Have wings	Invertebrates	Vertebrates

(b) Whales and Fishes

Similarity	Difference
Aquatic	Whales	Fishes
Aquatic	Have lungs for breathing	Have gills for breathing

Similarity	Difference
Do not have any limbs	Snakes	Earthworm
Do not have any limbs	Vertebrates	Invertebrates

(d) Bat and Pigeon

Similarity	Difference
Breathe through lungs	Bat	Pigeon
Breathe through lungs	Have external ears	Have internal ears

(e) Cuttlefish and Dogfish

Similarity	Difference
Marine animals	Cuttlefish	Dogfish
Marine animals	Invertebrates	Vertebrates

(f) Wall lizard and Frog

Similarity	Difference
Cold-blooded animals	Wall lizard	Frog
Cold-blooded animals	Completely adapted to life on land	Live partly on land and partly in water

Solution D.2.

Column I	Column II
Annelida	Earthworm
Porifera	Sponge
Mollusca	Octopus
Reptilia	Snake
Pisces	Trout
Mammal	Rabbit

Amphibia from Column I and Pigeon from column II are left out. They do not match as Pigeon comes under Class Aves and not Class Amphibia.

Solution E.

No, it is not a relative of Kangaroo as Kangaroo is a mammal and Tyrannosaurus is a reptile.
Characteristics of Tyrannosaurus:

Body has scales which may be horny
Lays eggs which have a leathery shell
One of the dinosaurs which moved on the earth. These ruled the earth. Some were vegetarian, others were non-vegetarian. Archaeopteryx is a link between reptiles and birds. Its fossils have been found, which vanished from the earth due to ice-age.
Had three-chambered heart as ventricles are partially divided
Cold-blooded animals

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Selina Concise Biology Class 9 ICSE Solutions Waste Generation and Management

November 9, 2023May 28, 2022 by Veerendra

Selina Concise Biology Class 9 ICSE Solutions Waste Generation and Management

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Biology Chapter 9 Chapter 19 Waste Generation and Management. You can download the Selina Concise Biology ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Biology for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Biology Chapter 19 Waste Generation and Management

Exercise 1

Solution A.

(a) Old newspapers
(a) plastics

Solution B.1.

(a) Sludge
(b) Scrubber, Plate type electrostatic precipitator

Solution B.2.

(a) True
(b) True
(c) False. Electricity is one of the secondary needs of human society.

Solution B.3.

Column I	Column II
1. Cow dung	(iii) Manure (i) Sugarcane
2. Bagasse	(i) Sugarcane
3. Old newspapers	(ii) Raddiwalas (i) Sugarcane

Solution C.1.

DEGRADABLE WASTE	NON-DEGRADABLE WASTE
The waste that can be decomposed by micro-organisms is called degradable waste.	The waste that cannot be decomposed by micro-organisms is called non-degradable waste.
E.g. Vegetable peel, paper, cloth etc.	E.g. Plastic

Solution C.2.

(a) Broken glass utensils are a kind of non-degradable waste as they cannot be decomposed and broken down by living micro-organisms. These need to be disposed off in deep dug out pits so that they do not cause any harm. However, sometimes broken glass in large quantities can be used in glass industries after melting them.

(b) Landfills are coming up fast near large cities due to huge population growth and lack of using efficient and safe disposal method. As construction activities are very active, new colonies are being raised and construction units have become very active. These produce large quantities of waste stones, pebbles, broken bricks and wood waste. Mostly, these are used as landfills. One can see heaps of landfills in the suburbs.

(c) Municipal sewage is first separated into degradable and non-degradable wastes because degradable wastes can be broken down into non-toxic waste in septic tanks, while non-degradable waste needs to be dumped or buried at safe places to avoid any hazardous effects.

Solution D.1.

Electronic waste is the waste generated by discarded electrical appliances.

Items which come under the category of electronic waste:

Fluorescent tubes
Medicalinstruments
Toys
Lead acid batteries
Mobile phones
Radios

Solution D.2.

Common wastes produced in mining operation:

Coal, Dust, Iron, Copper, Zinc, etc.

Reusing of wastes produced in mining operation:

During operation of getting minerals, a large quantity of waste material is produced. This waste material is called mine tailing. It is the left over. The mine tailing can be mixed with materials to produce tiles and masonry cement.

Solution D.3.

The rotting and conversion of organic waste into manure is known as composting. The product formed after composting is called compost.

Method of preparing compost:

A trench of about 5 m long, 1.5 m wide and 1.5 m deep is dug.
A 30 cm thick layer of well-mixed refuse and waste is spread on it. This layer is completely wetted with a watery mixture of cow dung and some mud.
A second layer of mixed refuse is spread over the first layer, till the heap rises about half a metre.
The set-up is kept undisturbed for about 3 months, during which water is sprinkled over it at regular intervals.
A trench is then dug, its material is taken out and rearranged in conical heaps and covered with a layer of soil.
After 50-60 days, compost is ready for use in cultivation fields or in garden flower beds, etc.

Solution D.4.

Usefulness of incineration of wastes: