Selina ICSE Solutions for Class 9 Physics

Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves

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Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves

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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Keeping the key pressed, air is gradually withdrawn from jar by starting the vacuum pump.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:

  1. The medium must be elastic.
  2. The medium must have inertia.
  3. 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.

Vg < Vl < Vs

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:

  1. High energy contents
  2. 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:

  1. Ultrasound is used for drilling holes or making cuts of desired shape in materials like glass.
  2. 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

More Resources for Selina Concise Class 9 ICSE Solutions

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation

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Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation

ICSE SolutionsSelina 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.

Download Formulae Handbook For ICSE Class 9 and 10

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

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

Solution 4S.

Definitions of three fundamental quantities:

  1. 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.
  2. 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.
  3. 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:

  1. 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).
  2. 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).
  3. 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

QuantityUnitSymbol
Lengthmetrem
Masskilogrammekg
Timeseconds
TemperaturekelvinK
Luminous intensitycandelacd
Electric currentampereA
Amount of substancemolemol
Angleradianrd
Solid anglesteradianst-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 0o 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. 1 Angstrom (Å) = 10-10 m
  2. 1 kilometre (km) = 103 m
  3. 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
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 1

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

  1. Positive zero error
  2. 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 6th 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.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 24

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

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

Solution 11S.

Two scales of vernier calipers are

  1. Main scale
  2. 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.

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Main parts and their functions:

  1. Ratchet:It advances the screw by turning it until the object is gently held between the stud and spindle of screw.
  2. Sleeve:It marks the main scale and base line.
  3. Thimble:It marks the circular scale.
  4. Main scale:It helps to read the length correct up to 1 mm.
  5. 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.
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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:
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Solution 4S.

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

  1. Length of pendulum (l)
  2. Acceleration due to gravity (g)
    Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 11

Solution 5S.

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

  1. Material of the bob
  2. Amplitude

Solution 6S.

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Therefore, the time period is doubled.

(b) If the acceleration due to gravity is reduced to one-fourth,
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Therefore, the time period is doubled.

Solution 7S.

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Solution 8S.

Measurement of time period of a simple pendulum:

  1. 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.
  2. The time ‘t’ for 20 complete oscillations is measured with the help of a stop watch.
  3. 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.
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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,
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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:
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 17

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.

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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-1
Such a pendulum is called the seconds’ pendulum.

Solution 3N.

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Solution 4N.

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Solution 5N.

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Solution 6N.

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Solution 7N.

Let T1 and 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.
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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 Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes’ Principle and Floatation

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Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes’ Principle and Floatation

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Selina ICSE Solutions for Class 9 Physics Chapter 5 Upthrust in Fluids, Archimedes’ Principle and Floatation

Exercise 5(A)

Solution 1S.

When a body is partially or wholly immersed in a liquid, an upward force acts on it. This upward force is known as an upthrust.

Upthrust can be demonstrated by the following experiment:

Take an empty can and close its mouth with an airtight stopper. Put it in a tub filled with water. It floats with a large part of it above the surface of water and only a small part of it below the surface of water. Push the can into the water. You can feel an upward force and you find it difficult to push the can further into water. It is noticed that as the can is pushed more and more into the water, more and more force is needed to push the can further into water, until it is completely immersed. When the can is fully inside the water, a definite force is still needed to keep it at rest in that position. Again, if the can is released in this position, it is noticed that the can bounces back to the surface and starts floating again.

Solution 2S.

Buoyant force on a body due to a liquid acts upwards at the centre of buoyancy.

Solution 3S.

The property of a liquid to exert an upward force on a body immersed in it is called buoyancy.

Solution 4S.

The upward force exerted on a body by the fluid in which it is submerged is called the upthrust. Its S.I. unit is ‘newton’.

Solution 5S.

A liquid contained in a vessel exerts pressure at all points and in all directions. The pressure at a point in a liquid is the same in all directions – upwards, downwards and sideways. It increases with the depth inside the liquid.
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When a body is immersed in a liquid, the thrusts acting on the side walls of the body are neutralized as they are equal in magnitude and opposite in direction. However, the magnitudes of pressure on the upper and lower faces are not equal. The difference in pressure on the upper and lower faces cause a net upward force (= pressure x area) or upthrust on the body.
It acts at the centre of buoyancy.

Solution 6S.

Upthrust due to water on block when fully submerged is more than its weight. Density of water is more than the density of cork; hence, upthrust due to water on the block of cork when fully submerged in water is more than its weight.

Solution 7S.

A piece of wood if left under water comes to the surface of water because the upthrust on body due to its submerged part is equal to its own weight.

Solution 8S.

Experiment to show that a body immersed in a liquid appears lighter:
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Take a solid body and suspend it by a thin thread from the hook of a spring balance as shown in the above figure (a). Note its weight. Above figure (a) shows the weight as 0.67 N.
Then, take a can filled with water. Immerse the solid gently into the water while hanging from the hook of the spring balance as shown in figure (b). Note its weight. Above figure (b) shows the weight as 0.40 N.
The reading in this case (b) shall be less than the reading in the case (a), which proves that a body immersed in a liquid appears to be lighter.

Solution 9S.

The readings in the spring balance decreases.
As the cylinder is immersed in the jar of water, an upward force acts on it, which is in opposition to the weight component of the cylinder. Hence the cylinder appears to be lighter.

Solution 10S.

A body shall weigh more in vacuum because in vacuum, i.e. in absence of air, no upthrust will act on the body.

Solution 11S.

Upthrust on a body depends on the following factors:

  1. Volume of the body submerged in the liquid or fluid.
  2. Density of liquid or fluid in which the body is submerged.

Solution 12S.

Larger the volume of body submerged in liquid, greater is the upthrust acting on it.

Solution 13S.

A stone falls faster.
Because the volume of stone is less than the volume of bunch of feathers of the same mass, the upthrust due to air on stone is less than that on the bunch of feathers, and hence, the stone falls faster.
However, in vacuum, both shall fall together because there will be no upthrust.

Solution 14S.

F2 > F1; Sea water is denser than river water; therefore, the upthrust due to sea water will be greater than that due to river water at the same level. This shall make the body to appear lighter in the sea water.

Solution 15S.

Observation: Volume of a block of wood immersed in glycerine is smaller as compared to the volume of block immersed in water.
Explanation: Density of glycerine is more than that of water. Hence, glycerine exerts more upthrust on the block of wood than water, causing it to float in glycerine with a smaller volume.

Solution 16S.

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Solution 17S.

Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 3

Solution 18S.

(a) Both have equal volumes.
(b) Bounce back to the surface.
(c) More than

Solution 19S.

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Consider a cylindrical body PQRS of cross-sectional area A immersed in a liquid of density ρ as shown in the figure above. Let the upper surface PQ of the body is at a depth h1 while its lower surface RS is at depth h2 below the free surface of liquid.

At depth h1, the pressure on the upper surface PQ,
P1 = h1 ρg.

Therefore, the downward thrust on the upper surface PQ,
F1 = Pressure x Area = h1 ρgA ……………….(i)

At depth h2, pressure on the lower surface RS,
P2 = h2 ρg

Therefore, the upward thrust on the lower surface RS,
F2 = Pressure x Area = h2 ρgA …………………(ii)

The horizontal thrust at various points on the vertical sides of body get balanced because the liquid pressure is the same at all points at the same depth.

From the above equations (i) and (ii), it is clear that F2 > F1 because h2 > h1 and therefore, body will experience a net upward force.

Resultant upward thrust or buoyant force on the body,

FB = F2 – F1
 = h2 ρgA – h1 ρgA
= A (h2 – h1) ρg

However, A (h2 – h1) = V, the volume of the body is submerged in a liquid.
Therefore, upthrust FB = V ρg.

Now, V g = Volume of solid immersed x Density of liquid x Acceleration due to gravity
= Volume of liquid displaced x Density of liquid x Acceleration due to gravity
= Mass of liquid displaced x Acceleration due to gravity
= Weight of the liquid displaced by the submerged part of the body

Thus, Upthrust FB = weight of the liquid displaced by the submerged part of the body…..(iii)

Now, let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight.

Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid is collected in the measuring cylinder.Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 5

When the water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.

From the diagram, it is clear that
Loss in weight (Weight in air – Weight in water) = Volume of water displaced.
Or, Loss in weight = Volume of water displaced x 1 gcm-3 [Because the density of water = 1 gcm-3]
Or, Loss in weight = Weight of water displaced ……………(iv)

From equations (iii) and (iv),
Loss in weight = Upthrust or buoyant force

Solution 20S.

Since the spheres have the same radius, both will have an equal volume inside water, and hence, the upthrust acted by water on both the spheres will be the same.
Hence, the required ratio of upthrust acting on two spheres is 1:1.

Solution 21S.

Sphere of iron will sink.

Density of iron is more than the density of water, so the weight of iron sphere will be more than the upthrust due to water in it; thus, it causes the iron sphere to sink.

Density of wood is less than the density of water, so the weight of sphere of wood shall be less than the upthrust due to water in it. So, the sphere of wood will float with a volume submerged inside water which is balanced by the upthrust due to water.

Solution 22S.

The bodies of average density greater than that of the liquid sink in it. While the bodies of average density equal to or smaller than that of liquid float on it.

Solution 23S.

(i) The body will float if ρ ≤ ρL
(ii) The body will sink if ρ > ρL

Solution 24S.

It is easier to lift a heavy stone under water than in air because in water, it experiences an upward buoyant force which balances the actual weight of the stone acting downwards. Thus, due to upthrust there is an apparent loss in the weight of the heavy stone, which makes it lighter in water, and hence easy to lift.

Solution 25S.

Archimedes’ principle states that when a body is immersed partially or completely in a liquid, it experiences an upthrust, which is equal to the weight of liquid displaced by it.

Solution 26S.

Let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight (Fig a).
Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid gets collected in the measuring cylinder.
Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 6

When water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.
From diagram, it is clear that
Loss in weight (Weight in air – weight in water) = 300 gf – 200 gf = 100 gf
Volume of water displaced = Volume of solid = 100 cm3
Because density of water = 1 gcm-3 
Weight of water displaced = 100 gf = Upthrust or loss in weight
This verifies Archimedes’ principle.

Solution 1M.

Turpentine

Solution 2M.

N

Solution 3M.

ρ > ρL

 

Solution 1N.

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Solution 2N.

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Solution 3N.

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Solution 4N.

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Solution 5N.

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Solution 6N.

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Solution 7N.

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Solution 8N.

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Solution 9N.

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Solution 10N.

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Exercise 5(B)

Solution 1S.

The density of a substance is its mass per unit volume.

Solution 2S.

(i) The C.G.S. unit of density is gcm-3.
(ii) The S.I. unit of density is kgm-3.

Solution 3S.

1 gcm-3 = 1000 kgm-3

Solution 4S.

It means the mass of 1 m-3 of iron is 7800 kg.

Solution 5S.

Density of water at 4°C in S.I. units is 1000 kgm-3.

Solution 6S.

(i) Mass of a metallic body remains unchanged with increase in temperature.
(ii) Volume of metallic body increases with an increase in temperature.
(iii) Density (= Mass/volume) of a metallic body decreases with an increase in temperature.

Solution 7S.

On heating from 0°C, the density of water increases up to 4°C and then decreases beyond 4°C.

Solution 8S.

(i) Volume, (ii) kg m-3, (iii) 1000 and (iv) 1000

Solution 9S.

The relative density of a substance is the ratio of density of that substance to the density of water at 4°C.

Solution 10S.

Relative density is the ratio of two similar quantities; thus, it has no unit.

Solution 11S.

Density of a substance is the ratio of its mass to its volume but R.D. of a substance is the ratio of density of that substance to the density of water at 4°C.

Solution 12S.

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Steps:

  1. With the help of a physical balance, find the weight, W1 of the given solid.
  2. Immerse the solid completely in a beaker filled with water such that it does not touch the walls and bottom of beaker, and find the weight W2 of solid in water.

Observations:

Loss in weight of solid when immersed in water = (W1 – W2) gf
R.D. = Weight of solid in air/Loss of weight of solid in water
R.D. = W1/(W1 – W2).
If the solid is soluble in water, then instead of water, take a liquid in which the solid is insoluble and it sinks in the liquid.
Then, R.D. = (Weight of solid in air/Loss of weight of solid in liquid) x R.D. of the liquid

Solution 13S.

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Solution 14S.

Experimental determination of R.D. of a solid lighter than water (such as cork):

  1. Take a sinker (i.e. a piece of metal or stone).
  2. Place a beaker nearly two-third filled with water on a wooden bridge kept over the left pan of a physical balance.
  3. Suspend the sinker with a thread from the hook of the left pan of balance so that it is completely immersed in water (as shown in the figure below). Find the weight W1 of the sinker in water.
  4. Tie the given solid (say, a cork) in the middle of a thread, and then measure the weight W2 of a solid in the air along with the sinker in water.
  5. Tie the cork with the sinker and immerse both of them completely in water of beaker and measure the weight W3 of the solid and sinker both in water.
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Observations:

Weight of the sinker in water = W1gf
Weight of the sinker in water and cork in air = W2gf
Weight of sinker and cork together in water = W3gf

Calculations:

Weight of cork in air = (W2 – W1) gf
Weight of cork in water = (W3 – W1) gf
Loss in weight of the cork in water = Weight of cork in air  Weight of cork in water.
= [(W2 – W1) – (W3 – W1)] gf
= (W2 – W3) gf
R.D. of cork = Weight of cork in air/Loss of weight of cork in water
Or, R.D. of cork = (W2 – W1)/(W2 – W3).

Solution 15S.

The weight of the sinker and cork combined, in water will be less than the weight of the sinker alone in water because the upthrust due to water on cork (when completely immersed) is more than the weight of cork itself.

 

Solution 1M.

Water

Solution 2M.

No unit.

Solution 3M.

1 g cm-3 

Solution 1N.

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Solution 2N.

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Solution 3N.

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Solution 4N.

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Solution 5N.

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Solution 6N.

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Solution 7N.

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Solution 8N.

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Solution 9N.

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Solution 10N.

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Solution 11N.

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Solution 12N.

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Solution 13N.

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Solution 14N.

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Solution 15N.

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Solution 16N.

a. The mass of stone is 15.1 g. Hence, its weight in air will be Wa = 15.1 gf

b. When stone is immersed in water its weight becomes 9.7 gf. So, the upthrust on the stone is 15.1 – 9.7 = 5.4 gf, Since the density of water is 1 g cm-3, the volume of stone is 5.4 cm3.

 

c. Weight of stone in liquid is Wl = 10.9 gf
Weight of stone in water is Ww = 9.7 gf
Therefore, the relative density of stone is

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Exercise 5(C)

Solution 1S.

According to the principle of floatation, the weight of a floating body is equal to the weight of the liquid displaced by its submerged part.

Solution 2S.

(i) Two forces acting on the body are as listed below:
(a) Weight of the body (downwards)
(b) Upthrust of the liquid (upwards)
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(ii) If the weight of the body is greater than the upthrust acting on it, the body will sink
If the weight of the body is equal to or less than the upthrust acting on it, the body will float.

(iii) (a) The net force acting on the body when it sinks is body’s own weight.
(b) The net force acting on the body when it floats is the upthrust due to the liquid.

Solution 3S.

The reading on the spring balance will be zero because wood floats on water and while floating the apparent weight = 0.

Solution 4S.

(a) The ball will float because the density of ball (i.e. iron) is less than the density of mercury.
(b) While floating, the apparent weight = 0.

Solution 5S.

The body will float if its density is less than or equal to the density of the liquid ρS ≤ ρL.
The body will sink if its density is greater than the density of the liquid ρS > ρL.

Solution 6S.

Density of iron is less than the density of mercury; hence, an iron nail floats in mercury and density of iron is more than the density of water; hence, an iron nail sinks in water.

Solution 7S.

(i) Weight of the floating body is equal to the upthrust.
(ii) While floating, the apparent weight is zero.

Solution 8S.

When the body is partially immersed, its centre of buoyancy will be below the centre of gravity of the block.
When the body is completely immersed, its centre of buoyancy will coincide the centre of gravity.

Solution 9S.

The upthrust on the body by each liquid is the same and equal to the weight of the body.
However, upthrust = Volume submerged × ρL × g,
For the liquid C, since the volume submerged is least so the density ρ3 must be maximum.

Solution 10S.

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The forces acting are as listed below:

  1. Weight of the body acting downwards.
  2. Upthrust due to water acting upwards.

Weight of water displaced by the floating body = Weight of the body

Solution 11S.

Centre of buoyancy: It is the point through which the resultant of the buoyancy forces on a submerged body act; it coincides with the centre of gravity of the displaced liquid, if the body is completely immersed.
For a floating body with its part submerged in the liquid, the centre of buoyancy is at the centre of gravity of the submerged part of the body and it lies vertically below the centre of gravity of the entire body.

Solution 12S.

Observation : The balloon will sink.
Explanation : As air is pumped out from jar, the density of air in jar decreases, so the upthrust on balloon decreases. As weight of balloon exceeds the upthrust on it, it sinks.

Solution 13S.

(a) It will float with some part outside water.
Reason : On adding some salt to water, the density of water increases, so upthrust on a block of wood increases, and hence, the block rises up till the weight of salty water displaced by the submerged part of block becomes equal to the weight of the block.

(b) The block will sink.
Reason: On heating, the density of water decreases, so upthrust on the block decreases and the weight of block exceeds upthrust due to which it sinks.

Solution 14S.

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Solution 15S.

Density of brine is more than the density of water. Hence, the upthrust exerted by brine is more than the upthrust exerted by water on ice. Therefore, floating ice is less submerged in brine.

Solution 16S.

(i) 1:1; The weight of the water displaced by the man in sea and river will be same and will be equal to his own weight.
(ii) He finds it easier to swim in the sea because the density of sea water is more than the density of river water. So his weight is balanced in sea water with a part of his body submerged in the water.

Solution 17S.

An iron nail sinks in water because density of iron is more than the density of water, so the weight of the nail is more than the upthrust of water on it.

On the other hand, ships are also made of iron, but they do not sink. This is because the ship is hollow and the empty space in it contains air, which makes its average density less than that of water. Therefore, even with a small portion of ship submerged in water, the weight of water displaced by the submerged part of ship becomes equal to the total weight of ship and it floats.

Solution 18S.

Due to the hollow and empty space in the ship, the average density of a ship is less than the density of water.

Solution 19S.

When a floating piece of ice melts into water, it contracts by the volume equal to the volume of ice pieces above the water surface while floating on it. Hence, the level of water does not change when ice floating on it melts.

Solution 20S.

Forces acting on the body are listed below:

  1. Weight of the body vertically downwards.
  2. Upthrust of water on body vertically upwards.
  3. Tension in thread vertically downwards.

Solution 21S.

A ship submerges more as it sails from sea water to river water.
Density of river water is less than the density of sea water. Hence, according to the law of floatation, to balance the weight of the ship, a greater volume of water is required to be displaced in river water of lower density.

Solution 22S.

(a) Icebergs are dangerous for ships as they may collide with them. Icebergs being lighter than water, float on water with a major part of their surfaces laying under the water surface and only a small part lies outside water. Thus, it becomes difficult for the driver of the ship to estimate the size of the iceberg.

(b) Density of a strong salt solution is more than the density of fresh water. Hence, the salt solution exerts a greater upthrust on the egg which balances the weight of the egg, so the egg floats in a strong salt solution but sinks in fresh water.

(c) Density of hydrogen is much less than the density of carbon dioxide. When a balloon is filled with hydrogen, the weight of the air displaced by an inflated balloon (i.e. upthrust) becomes more than the weight of a gas filled balloon, and hence, it rises. In case of a balloon filled with carbon dioxide, weight of the balloon becomes more than the upthrust of the air, and hence, it sinks to the floor.

(d) As a ship in harbor is unloaded, its weight decreases. As a result, it displaces less water, and the ship’s hull rises in water till the weight of the water displaced balances the weight of the unloaded ship.

(e) The reason is that the density of air decreases with altitude. Therefore, as the balloon gradually goes up, the weight of the displaced air (i.e. uphrust) decreases. It keeps on rising as long as the upthrust exceeds its weight. When upthrust becomes equal to its weight, it stops rising.

(f) Density of river water is less than the density of sea water. Hence, according to the law of floatation, to balance the weight of the ship, a great volume of water is required to be displaced in river water having a comparitively lower density.

Solution 1M.

W= FB

Solution 2M.

Zero

Solution 3M.

ρ1 > ρ2

Solution 1N.

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Solution 2N.

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Solution 3N.

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Solution 4N.

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Solution 5N.

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Solution 6N.

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Solution 7N.

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Solution 8N.

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Solution 9N.

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Selina Concise Physics Class 9 ICSE Solutions Heat and Energy

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Selina ICSE Solutions for Class 9 Physics Chapter 6 Heat and Energy

Exercise 6(A)

Solution 1S.

Heat is the energy of random motion of molecules constituting the body.
Its S.I. unit is ‘joule’.

Solution 2S.

Heat will flow from a hot body (body at a higher temperature) to a cold body (body at a lower temperature).

Solution 3S.

S.I. unit of heat is ‘joule’.
1 joule = 0.24 cal

Solution 4S.

Temperature is the parameter which tells the thermal state of a body (i.e. the degree of hotness or coldness).
The S.I. unit of temperature is ‘kelvin’.

Solution 5S.

On touching a piece of ice, heat flows from our hand (hot body) to the ice (cold body), and hence, it appears cold.

Solution 6S.

Heat is a form of energy obtained due to the random motion of molecules in a substance but temperature is a quantity which decided the direction of flow of heat when two bodies at different temperature are placed in contact. Two quantities having the same amount of heat may differ in temperature.

Solution 7S.

The expansion of a substance on heating is called thermal expansion.

Solution 8S.

Brass and iron expand on heating.

Solution 9S.

Water contracts on heating from 0°C to 4°C.

Silver iodide contracts on heating from 80°C to 141°C.

Solution 10S.

The expansion of water when it is cooled from 4°C to 0°C is known as the anomalous expansion of water.

Solution 11S.

Density of water is maximum at 4°C. Its value is 1000 kgm-3.

Solution 12S.

When a given mass of water is heated from 0°C to 4°C, it contracts, i.e. its volume decreases.
On heating from 4°C to 10°C, it expands, i.e. its volume increases.

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Solution 13S.

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Solution 14S.

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Solution 15S.

Hope’s experiment to demonstrate that water has maximum density at 4°C:

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Hope’s apparatus consists of a tall metallic cylinder provided with two side openings P and Q, P near the top and Q near the bottom, fitted with thermometers T1 and T2 in them. The central part of the cylinder is surrounded with a cylindrical trough containing a freezing mixture of ice and salt. The cylinder is fitted with pure water at room temperature.

Observations: (i) Initially, both thermometers T1 and T2 are at the same temperature.

(ii) First, the temperature recorded by the lower thermometer T2 starts decreasing and finally it becomes steady at 4°C, while the temperature recorded in the upper thermometer T1 remains almost unchanged during this time.

(iii) Then, the temperature recorded by the lower thermometer T2 remains constant at 4°C and upper thermometer T1 records a continuous fall in temperature up to 0°C and then it becomes steady.

Thus, finally, the temperature recorded by the upper thermometer is 0°C and that by lower thermometer is 4°C.

As the freezing mixture cools water in the central portion of the cylinder, water contracts and its density increases, consequently it sinks to the bottom, thereby causing the reading of the lower thermometer T2 to fall rapidly. The reading of the upper thermometer T1 does not change as the temperature of water in the upper part does not change. This continues till the entire water below the central portion reaches 4°C. On cooling further below 4°C, due to anomalous expansion, water of the central portion expands, so its density decreases and hence it rises up. As a result, reading of the upper thermometer T1 falls rapidly to 0°C and water freezes to form ice at 0°C near the top. This proves that water has maximum density at 4°C.

This anomalous expansion of water helps in preserving the aquatic life during the very cold weather. In winters, when the temperature falls, the top layer of water in a pond contracts, becomes denser and sinks to the bottom. A circulation is thus set up until the entire water in the pond reaches its maximum density at 4°C. If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, even though the upper layers are frozen, the water near the bottom is at 4°C and the fishes can survive in it easily.

Solution 16S.

(i) Water just in contact with ice is at 0°C.
(ii) Water at the bottom of the pond is at 4°C.

Solution 17S.

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Solution 18S.

(a) On winter nights, as the atmospheric pressure starts falling below 4°C, water in the pipe lines expand and exert a great pressure on the pipes, causing them to burst.

(b) In winters, when temperature falls, the surface of water in the tank contracts, becomes denser and sinks to the bottom. A circulation is thus set up until the entire water in the tank reaches its maximum density at 4°C. If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, water in a tank starts freezing from the top and not from the bottom.

(c) The anomalous expansion of water helps preserve aquatic life during very cold weather. When temperature falls, the top layer of water in a pond contracts becomes denser and sinks to the bottom. A circulation is thus set up until water in the pond reaches its maximum density at 4°C. If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, even though the upper layer are frozen, the water near the bottom is at 4°C and the fishes can survive in it easily.

(d) On heating water above 4°C, the density of water decreases. As a result, the upthrust acting due to water on hollow glass sphere also decreases, which causes it to sink.

(e) Inside the freezer, when the temperature of water falls below 4°C, the water in the bottle starts expanding. If the bottle is completely filled and tightly closed, there is no space for water to expand, and hence, the bottle may burst.

Solution 1M.

Calorie is the unit heat.

Solution 2M.

1 calorie = 4.186 J
Therefore,
1 J = 1/4.186 = 0.24 cal

Solution 3M.

The SI unit of temperature is kelvin (K).

Solution 4M.

Water shows anomalous behavior between 0°C and 4°C. Hence, when it is cooled it expands.

Solution 5M.

Water shows anomalous behavior between 0°C and 4°C. It has lowest volume at 4°C. Hence, its density will be maximum at 4°C.

Exercise 6(B)

Solution 1S.

A unit composed of biotic components (i.e. producers, consumers and decomposers) and abiotic components (i.e. light, heat, rain, and humidity, inorganic and organic substances) is called an ecosystem.

Solution 2S.

The source of energy for all ecosystems is the Sun.

Solution 3S.

Green plants absorb most of the energy falling on them and by the process of photosynthesis they produce food for the consumers. Plants, being primary producers are of great importance in the ecosystem. They also maintain the balance of oxygen and carbon dioxide on earth.

Solution 4S.

Producers like plants and some bacteria are capable of producing its own food using the energy of sun but consumers are not capable of producing their own food. They depend on producers for food.

Solution 5S.

The role of a decomposer is to break down dead organisms and then feed on them. The nutrients created by the dead organisms are returned to the soil to be later used by the producers. Once these deceased organisms are returned to the soil, they are used as food by bacteria and fungi by transforming the complex organic materials into simpler nutrients. The simpler products can then be used by producers to restart the cycle. These decomposers play an important role in every ecosystem.

Solution 6S.

A food chain shows the feeding relationship between different living things in a particular environment or habitat. Often, a plant will begin a food chain because it can make its own food using energy from the Sun. In addition, a food chain represents a series of events in which food and energy are transferred from one organism in an ecosystem to another. Food chains show how energy is passed from the sun to producers, from producers to consumers, and from consumers to decomposers.

Solution 7S.

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Solution 8S.

Ecosystems maintain themselves by cycling energy and nutrients obtained from external sources. At the first trophic level, primary producers (plants, algae, and some bacteria) use solar energy to produce organic plant material through photosynthesis. Herbivores-animals that feed solely on plants-make up the second trophic level. Predators that eat herbivores comprise the third trophic level; if larger predators are present, they represent still higher trophic levels. Decomposers, which include bacteria, fungi etc. break down wastes and dead organisms and return nutrients to the soil.
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On average about 10 percent of net energy production at one trophic level is passed on to the next level. Processes that reduce the energy transferred between trophic levels include respiration, growth and reproduction, defecation, and non-predatory death.

The low rate of energy transfer between trophic levels makes decomposers generally more important than producers in terms of energy flow. Decomposers process large amounts of organic material and return nutrients to the ecosystem in inorganic forms, which are then taken up again by primary producers.

Solution 9S.

The laws of thermodynamics govern the energy flow in the ecosystem.
According to the first law of thermodynamic, the energy can be transformed from one form to the other form, but it can neither be created nor destroyed.
According to the second law of thermodynamics, when energy is put to work, a part of it is always converted in un-useful form such as heat mainly due to friction and radiation.

Solution 10S.

The energy flow in ecosystem is linear i.e., it moves in a fixed direction. The solar energy is absorbed by plants and a part of it is converted into food. These plants (or primary producers) are then eaten by the primary consumers, which are consumed by secondary consumers and the secondary by tertiary consumers. This cycle is unidirectional. The dead and decomposed are fed by decomposers, which return the nutrients to the soil. At the end, the energy reaches the degraded state. It does not return to the sun to make the process cyclic, thus energy flow is linear.

Solution 11S.

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Solution 12S.

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Solution 1M.

Photosynthesis

Solution 2M.

Sun

Solution 3M.

Producers

Solution 4M.

Consumer

Exercise 6(C)

Solution 1S.

  1. A source of energy should be safe and convenient to use.
  2. A source of energy should be economical and easy to store and transport.

Solution 2S.

The two groups in which various sources of energy are classified are renewable or non-conventional sources of energy and non-renewable or conventional sources of energy.
These sources are classified on the basis of their availability and utility.

Solution 3S.

Renewable: The natural sources providing us energy continuously are called renewable sources of energy.
Non-renewable: The sources of energy which have accumulated in nature over a very long period of time and cannot be quickly replaced when exhausted are called non-renewable sources of energy.

Difference:

Renewable sourcesNon-renewable sources
They can be utilised continuously.They cannot be utilised once exhausted.
Examples: Sun, WindExample: Coal, Petroleum

Solution 4S.

Renewable: Wood, Water and Wind
Non-renewable: Coal, Diesel and Oil

Solution 5S.

Wood is obtained from trees. Hence, trees need to be cut down for wood to be used as a fuel.
Also, burning wood releases a lot of smoke which pollutes the atmosphere.

Solution 6S.

Renewable:

  1. Sun
  2. Wind
  3. Flowing water
  4. Tides
  5. Nuclear fuel

Non-renewable:

  1. Coal
  2. Petroleum
  3. Natural gas

Solution 7S.

  1. Tidal energy: The energy possessed by rising and falling water in tides is known as tidal energy.Dams are constructed across a narrow opening to the sea to harness tidal energy and produce electricity. However, it is not a major source of energy as the rise and fall of seawater during tides is not enough to generate electricity on a large scale.
  2. Ocean energy: Water in the oceans possesses energy in two forms:
    1. Ocean thermal energy- The energy available due to the difference in temperature of water at the surface and at deeper levels of ocean is called the ocean thermal energy. This energy is harnessed for producing electricity by a device called ocean thermal energy conversion power plant (OCTEC power plant).
    2. Oceanic waves energy- The kinetic energy possessed by fast moving oceanic (or sea) waves is called oceanic waves energy. Though models have been made to generate electricity from oceanic waves, but so far it has not been put to practical use.
  3. Geo thermal energy: The heat energy possessed by the rocks inside the Earth is called geothermal energy.The hot rocks present at the hot spots deep inside the Earth, heat the underground water and turn it into steam. This steam is compressed at high pressure between the rocks. Holes are drilled deep into the Earth up to the hot spots to extract the steam through pipes, which is utilized to rotate the turbines connected to the armature of an electric generator to produce electricity.

Solution 8S.

Sun is the main source of energy on Earth.

Solution 9S.

The energy obtained from Sun is called solar energy.

A solar power plant is a device in which heat energy of sun is used to generate electricity. It consists of a large number of concave reflectors, at the focus of which there are black painted water pipes. The reflectors concentrate the heat energy of the sun rays on the pipes due to which water inside the pipes starts boiling and produces steam. The steam thus produced is used to rotate a steam turbine which drives a generator producing electricity.

Solution 10S.

A solar cell is an electrical device that converts light energy directly into electricity with the help of photovoltaic effect. Solar cells are usually made from semiconductors like silicon and gallium with some impurity added to it. When sunlight is made incident on a solar cell, a potential difference is produced between its surface, due to which a current flows in the circuit connected between the opposite faces of the semiconductor.

Two uses of solar cells are as listed below:

  1. They do not require maintenance and last over a long period of time at zero running cost.
  2. They are very useful for remote, inaccessible and isolated places where electric power lines cannot be laid.Solar cell produces d.c. (direct current).

One disadvantage of solar cell is listed below:

  1. The initial cost of a solar panel is sufficiently high.

Solution 11S.

Advantages of using solar panels:

  1. They do not cause any pollution in the environment.
  2. Running cost of solar panel is almost zero.
  3. They last over a long period of time.
  4. They do not require any maintenance.
  5. They are suitable for remote and inaccessible places where electricity power lines cannot be laid.

Disadvantages of using solar panels:

  1. The initial cost of a solar panel is sufficiently high.
  2. The efficiency of conversion of solar energy to electricity is low.
  3. A solar panel produces d.c. electricity which cannot be directly used for many household purposes.

Solution 12S.

The kinetic energy of the moving large masses of air is called the wind energy. Wind energy is used in a wind generator to produce electricity by making use of wind mill to drive a wind generator.

At present in India, more than 1025 MW electric power is generated using wind energy.

Solution 13S.

Advantages of using wind energy:

  1. It does not cause any kind of pollution.
  2. It is an everlasting source.

Limitations of using wind energy:

  1. The establishment of a wind farm is expensive.
  2. A large area of land is needed for the establishment of a wind farm.

Solution 14S.

The kinetic energy possessed by flowing water is called the water or hydro energy.
Principle of a hydroelectric power plant is that the water flowing in high altitude rivers is collected in a high dam (or reservoir). The water from the dam is then allowed to fall on a water turbine which is located near the bottom of the dam. The shaft of the turbine is connected to the armature of an electric generator or dynamo.

At present only 23% of the total electricity is generated by the hydro energy.

Solution 15S.

Advantages of producing the hydro electricity:

  1. It does not produce any environmental pollution.
  2. It is a renewable source of energy.

Disadvantages of producing hydroelectricity:

  1. Due to the construction of dams over the rivers, plants and animals of that place get destroyed or killed.
  2. The ecological balance in the downstream areas of rivers gets disturbed.

Solution 16S.

When a heavy nucleus is bombarded with slow neutrons, it splits into two nearly equal light nuclei with a release of tremendous amount of energy. In this process of nuclear fission, the total sum of masses of products is less than the total sum of masses of reactants. This lost mass gets converted into energy. The energy so released is called nuclear energy.

Principle: The heat energy released due to the controlled chain reaction of nuclear fission of uranium-235 in a nuclear reactor is absorbed by the coolant which then passes through the coils of a heat exchanger containing water. The water in heat exchanger gets heated and converts into steam. The steam is used to rotate the turbine which in turn rotates the armature of a generator in a magnetic field and thus produces electricity.

Solution 17S.

At present only about 3% of the total electrical power generated in India is obtained from the nuclear power plants.
Tarapur in Maharahtra and Narora in Uttar Pradesh are the places where electricity is produced using nuclear energy.

Solution 18S.

Advantages of using nuclear energy:

  1. A very small amount of nuclear fuel can produce a tremendous amount of energy.
  2. Once the nuclear fuel is loaded into nuclear power plant, it continues to release energy for several years.

Disadvantages of using nuclear energy:

  1. It is not a clean source of energy because very harmful nuclear radiations are produced in the process.
  2. The waste causes environmental pollution.

Solution 19S.

i. Light energy into electrical energy
ii. Mechanical energy into electrical energy.
iii. Mechanical energy into electrical energy.
iv. Nuclear energy (or heat energy) into electrical energy.

Solution 20S.

Four ways for the judicious use of energy are:

  1. The fossil fuels such as coal, petroleum, natural gas should be used only for the limited purposes when there is no other alternative source of energy available.
  2. The wastage of energy should be avoided.
  3. Efforts must be made to make use of energy for community or group purposes.
  4. The cutting of trees must be banned and more and more new trees must be roped to grow.

Solution 21S.

The gradual decrease of useful energy due to friction etc. is called the degradation of energy.
Examples:

  1. When we cook food over fire, the major part of heat energy from the fuel is radiated out in the atmosphere. This radiated energy is of no use to us.
  2. When electrical appliances are run by electricity, the major part of electrical energy is wasted in the form of heat energy.

Solution 22S.

The conversion of part of energy into a non-useful form of energy is called degradation of energy.

Solution 1M.

The ultimate source of energy is the Sun.

Solution 2M.

Renewable source of energy is the Sun.

Exercise 6(D)

Solution 1S.

Greenhouse effect is the process of warming of planet’s surface and its lower atmosphere by absorbtion of infrared radiations of longer wavelength emitted out from the surface of planet.

Solution 2S.

Carbon-di-oxide, water vapour and methane are greenhouse gases.

Solution 3S.

Visible light rays and short infrared radiation pass through the atmosphere of earth.

Solution 4S.

Infrared radiations of long wavelength are absorbed by the green house gases.

Solution 5S.

The concentration of carbon-di-oxide content’s of earth’s atmosphere has increased due to industrial growth, combustion of fossil fuels and clearing of forests.

Solution 6S.

In absence of green house gases, the average temperature on earth would be -18°C.

Solution 7S.

The greenhouses gases have an average warming effect on Earth’s surface of about 15.5°C (or 60°F ).

Solution 8S.

Global warming means the increase in average effective temperature near the earth’s surface due to an increase in the amount of green house gases in its atmosphere.

Solution 9S.

With activities industrialization, deforestation, excess burning of fossil fuel, the concentration of green house gases has increased on earth’s atmosphere. This increase in the amount of greenhouse gases present in atmosphere has caused the rise in atmospheric temperature.

Solution 10S.

The increase in green house gases due to activities like industrialization, deforestation, natural gas exploration, burning of biomass, natural gas exploration, more use of gadgets like refrigerators has caused the increase of green house effect.

Solution 11S.

At the poles, due to increase in temperature, the snow and ice will melt which will cause flood in coastal countries. The icebergs of dark land and oceans will melt, so the dark land and oceans will become uncovered and will absorb more heat radiations coming from sun, increasing the green house effect further.

Solution 12S.

Due to global warming, the snow and ice around the poles will melt and cause flood in coastal countries.

Solution 13S.

Due to melting of polar ice and glaciers, there will be rise in sea level on coastal wet lands. It would raise worldwise sea level, thereby, many big cities in the coastal areas will be covered by sea water.

Solution 14S.

Global warming will cause drastic changes in the patterns of wind, rainfall etc. Thus it will result in low agricultural yield.

Solution 15S.

  1. Use of renewable sources of energy to generate electricity in place of generating electricity from the fossil fuels based power plants.
  2. Controlling population through family planning, welfare reforms and the empowerment of women.

Solution 16S.

The tax calculated on the basis of carbon emission from industry, number of employee hour and turnover of the factory is called carbon tax.
This tax shall be paid by industries. This will encourage the industries to use the energy efficient techniques.

Solution 1M.

Carbon dioxide

Solution 2M.

Increase in temperature

Solution 3M.

Without green house effect, the average temperature of Earth’s surface would have been -18 °C.

Solution 4M.

The global warming has resulted in the increase in sea levels.

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Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension

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Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension

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Selina ICSE Solutions for Class 9 Physics Chapter 2 Motion in One Dimension

Exercise 2(A)

Solution 1S.

ScalarVector
They are expressed only by their magnitudes.They are expressed by magnitude as well as direction.
They can be added, subtracted, multiplied or divided by simple arithmetic methods.They can be added, subtracted or multiplied following a different algebra.
They are symbolically written by English letter.They are symbolically written by their English letter with an arrow on top of the letter.
Example: mass, speedExample: force, velocity

Solution 2S.

a) Pressure is a scalar quantity.
b) Momentum is a vector quantity.
c) Weight is a vector quantity.
d) Force is a vector quantity.
e) Energy is a scalar quantity.
f) Speed is a scalar quantity.

Solution 3S.

A body is said to be at rest if it does not change its position with respect to its immediate surroundings.

Solution 4S.

A body is said to be in motion if it changes its position with respect to its immediate surroundings.

Solution 5S.

When a body moves along a straight line path, its motion is said to be one-dimensional motion.

Solution 6S.

The shortest distance from the initial to the final position of the body is called the magnitude of displacement. It is in the direction from the initial position to the final position.
Its SI unit is metre (m).

Solution 7S.

Distance is a scalar quantity, while displacement is a vector quantity. The magnitude of displacement is either equal to or less than the distance. The distance is the length of path travelled by the body so it is always positive, but the displacement is the shortest length in direction from initial to the final position so it can be positive or negative depending on its direction. The displacement can be zero even if the distance is not zero.

Solution 8S.

Yes, displacement can be zero even if the distance is not zero.

For example, when a body is thrown vertically upwards from a point A on the ground, after sometime it comes back to the same point A. Then, the displacement is zero, but the distance travelled by the body is not zero (it is 2h; h is the maximum height attained by the body).

Solution 9S.

The magnitude of displacement is equal to distance if the motion of the body is one-dimensional.

Solution 10S.

The velocity of a body is the distance travelled per second by the body in a specified direction.
Its SI unit is metre/second (m/s).

Solution 11S.

The speed of a body is the rate of change of distance with time.
Its SI unit is metre/second (m/s).

Solution 12S.

Speed is a scalar quantity, while velocity is a vector quantity. The speed is always positive-it is the magnitude of velocity, but the velocity is given a positive or negative sign depending upon its direction of motion. The average velocity can be zero but the average speed is never zero.

Solution 13S.

Velocity gives the direction of motion of the body.

Solution 14S.

Instantaneous velocity is equal to average velocity if the body is in uniform motion.

Solution 15S.

If a body travels equal distances in equal intervals of time along a particular direction, then the body is said to be moving with a uniform velocity. However, if a body travels unequal distances in a particular direction in equal intervals of time or it moves equal distances in equal intervals of time but its direction of motion does not remain same, then the velocity of the body is said to be variable (or non-uniform).

Solution 16S.

Average speed is the ratio of the total distance travelled by the body to the total time of journey, it is never zero. If the velocity of a body moving in a particular direction changes with time, then the ratio of displacement to the time taken in entire journey is called its average velocity. Average velocity of a body can be zero even if its average speed is not zero.

Solution 17S.

The motion of a body in a circular path with uniform speed has a variable velocity because in the circular path, the direction of motion of the body continuously changes with time.
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 1

Solution 18S.

If a body starts its motion from a point and comes back to the same point after a certain time, then the displacement is zero, average velocity is also zero, but the total distance travelled is not zero, and therefore, the average speed in not zero.

Solution 19S.

Acceleration is the rate of change of velocity with time.
Its SI unit is metre/second2 (m/s2).

Solution 20S.

Acceleration is the increase in velocity per second, while retardation is the decrease in velocity per second. Thus, retardation is negative acceleration. In general, acceleration is taken positive, while retardation is taken negative.

Solution 21S.

The acceleration is said to be uniform when equal changes in velocity take place in equal intervals of time, but if the change in velocity is not the same in the same intervals of time, the acceleration is said to be variable.

Solution 22S.

Retardation is the decrease in velocity per second.
Its SI unit is metre/second2 (m/s2).

Solution 23S.

Velocity determines the direction of motion.

Solution 24S.

(a) Example of uniform velocity: A body, once started, moves on a frictionless surface with uniform velocity.
(b) Example of variable velocity: A ball dropped from some height is an example of variable velocity.
(c) Example of variable acceleration: The motion of a vehicle on a crowded road is with variable acceleration.
(d) Example of uniform retardation: If a car moving with a velocity ‘v’ is brought to rest by applying brakes, then such a motion is an example of uniform retardation.

Solution 25S.

Initially as the drops are equidistant, we can say that the car is moving with a constant speed but later as the distance between the drops starts decreasing, we can say that the car slows down.

Solution 26S.

When a body falls freely under gravity, the acceleration produced in the body due to the Earth’s gravitational acceleration is called the acceleration due to gravity (g). The average value of g is 9.8 m/s2.

Solution 27S.

No. The value of ‘g’ varies from place to place. It is maximum at poles and minimum at the Equator on the surface of the Earth.

Solution 28S.

In vacuum, both will reach the ground simultaneously because acceleration due to gravity is same (=g) on both objects.

Solution 1M.

Velocity is a vector quantity. The others are all scalar quantities.

Solution 2M.

m s-1

Solution 3M.

m s-2

Solution 4M.

The displacement is zero.

Solution 5M.

5 m s-1

Solution 1N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 2

Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 3

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 4

Solution 4N.

18 km h-1 < 10 m s-1 < 1 km min-1

Solution 5N.

Total time taken = 3 hours
Speed of the train = 65 km/hr
Distance travelled = speed x time
= 65 x 3 = 195 km

Solution 6N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 5

Solution 7N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 6
(ii) Average velocity of the train is zero because the train stops at the same point from where it starts, i.e. the displacement is zero.

Solution 8N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 7

Solution 9N.

Here, final velocity = 10 m/s
Initial velocity = 0 m/s
Time taken = 2s
Acceleration = (Final Velocity – Initial Velocity)/time
= (10/2) m/s2
 = 5 m/s2

Solution 10N.

Here, final velocity = 180 m/s
Initial velocity = 0 m/s
Time taken = 0.05 h or 180 s
Acceleration = (Final Velocity – Initial Velocity)/time
= (180-0)/180 m/s2
 = 1 m/s2

Solution 11N.

Here, final velocity = 20 m/s
Initial velocity = 50 m/s
Time taken = 3 s
Acceleration = (Final Velocity – Initial Velocity)/time
= (20 – 50)/3 m/s2
 = -10 m/s2
Negative sign here indicates that the velocity decreases with time, so retardation is 10 m/s2.

Solution 12N.

Here, final velocity = 18 km/h or 5 m/s
Initial velocity = 0 km/h
Time taken = 2 s
Acceleration = (Final Velocity – Initial Velocity)/time
= (5 – 0) / 2 m/s2
 = 2.5 m/s2

Solution 13N.

Acceleration = Increase in velocity/time taken
Therefore, increase in velocity = Acceleration × time taken
= (5 × 2) m/s
= 10 m/s

Solution 14N.

Initial velocity of the car, u = 20 m/s
Retardation = 2 m/s2
Given time, t = 5 s
Let ‘v’ be the final velocity.
We know that, Acceleration = Rate of change of velocity /time
= (Final velocity – Initial velocity)/time
Or, -2 = (v – 20) / 5
Or, -10 = v – 20
Or, v = -20 + 10 m/s
Or, v = -10 m/s
Negative sign indicates that the velocity is decreasing.

Solution 15N.

Initial velocity of the bicycle, u = 5 m/s
Acceleration = 2 m/s2
Given time, t = 5 s
Let ‘v’ be the final velocity.
We know that, acceleration = Rate of change of velocity/time
= (Final velocity – Initial velocity)/time
Or 2 = (v – 5)/5
Or, 10 = (v – 5)
Or, v = -5 – 10
Or, v = -15 m/s
Negative sign indicates that the velocity is decreasing.

Solution 16N.

Initial velocity of the bicycle, u = 18 km/hr
Time taken, t = 5 s
Final velocity, v = 0 m/s (As the car comes to rest)
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 8
(iii) Let ‘V’ be the speed of the car after 2 s of applying the brakes.
Then, Acceleration = (V – 5)/ 2
Or, -1 = (V – 5)/2
Or, V = -2 + 5
Or, V = 3 m/s2

Exercise 2(B)

Solution 1S.

For the motion with uniform velocity, distance is directly proportional to time.

Solution 2S.

From displacement-time graph, the nature of motion (or state of rest) can be understood. The slope of this graph gives the value of velocity of the body at any instant of time, using which the velocity-time graph can also be drawn.

Solution 3S.

(a) Slope of a displacement-time graph represents velocity.
(b) The displacement-time graph can never be parallel to the displacement axis because such a line would mean that the distance covered by the body in a certain direction increases without any increase in time, which is not possible.

Solution 4S.

(a) There is no motion, the body is at rest.
(b) It depicts that the body is moving away from the starting point with uniform velocity.
(c) It depicts that the body is moving towards the starting point with uniform velocity.
(d) It depicts that the body is moving with variable velocity.

Solution 5S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 9

Solution 6S.

(i) The slope of the velocity-time graph gives the value of acceleration.
(ii) The total distance travelled by a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (without any sign).
(iii) The displacement of a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (with proper signs).

Solution 7S.

Vehicle A is moving with a faster speed because the slope of line A is more than the slope of line B.

Solution 8S.

(a) Fig. 4.33 (a) represents uniformly accelerated motion. For example, the motion of a freely falling object.
(b) Fig. 4.33 (b) represents motion with variable retardation. For example, a car approaching its destination.

Solution 9S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 10
In this graph, initial velocity = u
Velocity at time t = v
Let acceleration be ‘a’
Time = t
Then, distance travelled by the body in t s = area between the v-t graph and X-axis
Or distance travelled by the body in t s = area of the trapezium OABD
= (1/2) × (sum of parallel sides) × (perpendicular distance between them)
= (1/2) × (u + v) × (t)
= (u + v)t /2

Solution 10S.

The slope of the velocity-time graph represents acceleration.

Solution 11S.

Car B has greater acceleration because the slope of line B is more than the slope of line A.

Solution 12S.

For body A: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration for body A is zero.
For body B: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration for body B is also zero.
For body C: The slope of the graph is decreasing with time. Hence, the acceleration is negative.
For body D: The slope of the graph is increasing with time. Hence, the acceleration is positive.

Solution 13S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 11
Velocity-time for a body moving with uniform velocity and uniform acceleration.

Solution 14S.

Retardation is calculated by finding the negative slope.

Solution 15S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 12
The area enclosed between the straight line and time axis for each interval of time gives the value of change in speed in that interval of time.

Solution 16S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 13

Solution 17S.

For motion under uniform acceleration, such as the motion of a freely falling body, distance is directly proportion to the square of the time.

Solution 18S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 14

The value of acceleration due to gravity (g) can be obtained by doubling the slope of the S – t2 graph for a freely falling body.

Solution 1M.

B
Acceleration is uniform.

Solution 2M.

The solution is C.
The speed-time graph is a straight line parallel to the time axis.

Solution 3M.

D
A straight line inclined to the time axis.

Solution 1N.

Velocity of body at t = 1s is 2 m/s
Velocity of body at t = 2s is 4 m/s
Velocity of body at t = 3s is 6 m/s
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 15

Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 16
Displacement-time graph
From the part AB of the graph,
Average velocity = (Displacement at B – Displacement at A)/Time taken
= (30 – 20) m/( 6 – 4) s
= (10/2) m/s
= 5 m/s
(b) (i) From the graph, the displacement of car at 2.5 s is 12.5 m.
(ii) From the graph, the displacement of car at 4.5 s is 22.5 m.

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 17

Solution 4N.

(a) (i) Velocity from 0 to 5 s = Displacement /time
= (3/5) m/s
= 0.6 m/s

(ii) Velocity from 5 s to 7 s = Displacement /time
= (0/2) m/s
= 0 m/s.

(iii) Velocity from 7 s to 9 s = Displacement /time
= (7 – 3)/(9 – 7) m/s
= (4/2) m/s
= 2 m/s

(b) From, 5 s to 9 s, displacement = 7m – 3m = 4m.
Time elapsed between 5 s to 9 s = 4 s
Average velocity = Displacement/time
= (4/4) m/s
= 1 m/s

Solution 5N.

(i) Displacement in first 4s = 10 m
Therefore, the average velocity = Displacement/time
= (10/4) m/s
= 2.5 m/s

(ii) Initial position = 0 m
Final position at the end of 10 s = -10m
Displacement = Final position – Initial position
= (-10) m – 0
= -10 m

(iii) At 7 s and 13 s, the cyclist reaches his starting point.

Solution 6N.

(i) Initially, the car B was 40 km ahead of car A.
(ii) Straight line depicts that cars A and B are moving with uniform velocities.
For car A
Displacement at t = 1 h is 40 m
Velocity = Displacement /time
= (40/1) km/h
= 40 km/h

For car B
Displacement at t = 4 h is (120 – 40) km, i.e. 80 km
Velocity = Displacement /time
= (80/4) km/h
= 20 km/h

(iii) Car A catches car B in 2 hours.
(iv) After starting, car A will catch car B at 80 km.

Solution 7N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 18

Solution 8N.

Velocity of the body at t = 1 s is 1 m/s.
Displacement of the body at t = 1 s is velocity × time = (1) × (1) m or 1 m.

Velocity of the body at t = 2s is 2 m/s.
Displacement of the body at t = 1 s is velocity × time = (2) × (2) m or 4 m.

Velocity of the body at t = 3 s is 3 m/s.
Displacement of the body at t = 3 s is velocity × time = (3) × (3) m or 9 m

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 19

Solution 9N.

(i) Distance travelled in any part of the graph can be determined by finding the area enclosed by the graph in that part with the time axis.

(ii) Distance travelled in part BC = Area of the rectangle tBC2t = base × height.
= (2t – t) × vo
 = vot

Distance travelled in part AB = Area of the triangle ABt
= (1/2) × base × height
= (1/2) × t × vo
 = (1/2) vo t
Therefore, distance travelled in part BC:distance travelled in part AB :: 2:1.

(iii) (a) BC shows motion with uniform velocity.
(b) AB shows motion with uniform acceleration.
(c) CD shows motion with uniform retardation.

(iv) (a) The magnitude of acceleration is lower as the slope of line AB is less than that of line CD.
(b) Slope of line AB = vo/t
Slope of line CD = vo/0.5t
Slope of line AB/Slope of line CD = (vo /t)/(vo /0.5t)
Slope of line AB:Slope of line CD :: 1:2.

Solution 10N.

(i) Acceleration in the part AB = Slope of AB
= tan (∠BAD)
= (30/4) ms-2
 = 7.5 ms-2

Acceleration in the part BC = 0 ms-2
Acceleration in the part CD = slope of CD = -tan (∠CDA)
= -(30/2) ms-2
 = -15 ms-2

(ii) Displacement of part AB = Area of ΔAB4 = (1/2) (4) (30)
= 60 m
Displacement of part BC = Area of rectangle 4BC8
= (30) × (4) = 120 m
Displacement of part CD = Area of ΔC8D = (1/2) (2) (30)
= 30 m

(iii) Total displacement = Displacement of part AB + Displacement of part BC + Displacement of part CD
= 60 + 120 + 30 = 210 m

Solution 11N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 20
Distance travelled in first 6 s = velocity × time
= 10 m/s × 6
= 60 m/s

Distance travelled in next 6 s = velocity × time
= 10 m/s × 6
= 60 m/s

Total distance travelled in 12 s = (60 + 60) m = 120 m
Total displacement = 0, as the ball returns its starting point.

Solution 12N.

(i) From 0 to 4 seconds, the motion is uniformly accelerated and from 4 to 6 seconds, the motion is uniformly retarded.

(ii) Displacement of the particle at 6 s = (1/2) (6) (2) = 6 m

(iii) The particle does not change its direction of motion.

(iv) Distance travelled by the particle from 0 to 4s (D1) = (1/2) (4) (2) = 4 m
Distance travelled by the particle from 4 to 6s (D2) = (1/2) (2) (2) = 2 m
D1:D2:: 4:2
D1:D2:: 2:1

(v) Acceleration from 0 to 4 s = (2/4) ms-2 or 0.5 ms-2
Retardation from 4 s to 6 s = (2/2) ms-2 or 1 ms-2.

Exercise 2(C)

Solution 1S.

Three equations of a uniformly accelerated motion are
v = u + at
s = ut + (1/2)at2
v2 = u2 + 2as

Solution 2S.

Derivation of equations of motion

First equation of motion:
Consider a particle moving along a straight line with uniform acceleration ‘a’. At t = 0, let the particle be at A and u be its initial velocity, and at t = t, let v be its final velocity.
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 21
Acceleration = Change in velocity/Time
a = (v – u)/t
at = v – u
v = u+ at … First equation of motion.

Second equation of motion:
Average velocity = Total distance traveled/Total time taken Average velocity = s/t …(1)
Average velocity can be written as (u+v)/2 Average velocity = (u+v)/2 …(2)
From equations (1) and (2) s/t = (u+v)/2 …(3)
The first equation of motion is v = u + at.
Substituting the value of v in equation (3), we get
s/t = (u + u + at)/2 s = (2u + at) t/2 = 2ut + at2/2 = 2ut/2 + at2/2
s = ut + (1/2) at2 …Second equation of motion.

Third equation of motion:
The first equation of motion is v = u + at. v – u = at … (1)
Average velocity = s/t …(2)
Average velocity =(u+v)/2 …(3)
From equation (2) and equation (3) we get,
(u + v)/2 = s/t …(4)
Multiplying eq (1) and eq (4) we get,
(v – u)(v + u) = at × (2s/t) (v – u)(v + u) = 2as
[We make the use of the identity a2 – b2 = (a + b) (a – b)]
v2 – u2 = 2as …Third equation of motion.

Solution 3S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 22

Solution 1M.

v = u + at

Solution 2M.

5 km

Solution 1N.

Initial velocity u = 0
Acceleration a = 2 m/s2
Time t = 2 s
Let ‘S’ be the distance covered.
Using the second equation of motion,
S = ut + (1/2) at2
S = 0 + (1/2) (2) (2) 2
S = 4 m

Solution 2N.

Initial velocity u = 10 m/s
Acceleration a = 5 m/s2
Time t = 5s
Let ‘S’ be the distance covered.
Using the second equation of motion,
S = ut + (1/2) at2
S = (10)(5) + (1/2) (5) (5) 2
S = 50 + 62.5
S = 112.5 m

Solution 3N.

Acceleration = Change in velocity/time taken
In the first two seconds,
Acceleration = [(33.6 – 30)/2] km/h2
 = 1.8 km/h2
 = 0.5 m/s2 …(i)

In the next two seconds,
Acceleration = [(37.2 – 33.6)/2] km/h2
 = 1.8 km/h2
 = 0.5 m/s2…(ii)
From (i) and (ii), we can say that the acceleration is uniform.

Solution 4N.

Initial velocity u = 0 m/s
Acceleration a = 2 m/s2
Time t = 5 s

(i) Let ‘v’ be the final velocity.
Then, (v – u)/5 = 2
v = 10 m/s

(ii) Let ‘s’ be the distance travelled.
Using the third equation of motion,
v2 – u2 = 2as
We get,
(10) 2 – (0) 2 = 2(2) (s)
Thus, s = (100/4) m = 25 m

Solution 5N.

Initial velocity u = 20 m/s
Final velocity v = 0
Distance travelled s = 10 cm = 0.1 m
Let acceleration be ‘a’.
Using the third equation of motion,
v2 – u2 = 2as
We get,
(0) 2 – (20) 2 = 2(a) (0.1)
a = -(400/0.2) m/s2
a = -2000 m/s2
Thus, retardation = 2000 m/s2

Solution 6N.

Initial velocity u = 20 m/s
Final velocity v = 0
Time taken t = 5 s
Let acceleration be ‘a’.
Using the first equation of motion,
v = u + at
0 = 20 + 5a
a = -4 m/s2
Thus, retardation = 4 m/s2

Solution 7N.

Let ‘s’ be the distance between stations A and B.
(i) Average speed = Total distance/total time taken
Here, total distance = s + s = 2s
Total time taken = Time taken to travel from A to B + Time taken to travel from B to A.
= [(s/ 60) + (s/ 80)] s
= [ 140 s / 4800] s

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 23

(ii) Average velocity = Displacement/total time taken
Because the train starts and ends at the same station, the displacement is zero. Thus the average velocity is zero.

Solution 8N.

Initial velocity u = 90 km/h = 25 m/s
Final velocity v = 0 m/s
Acceleration a = -0.5 m/s2

(i) Let ‘V’ be the velocity after time t = 10 s
Using the first equation of motion,
v = u + at
We get,
V = 25 + (-0.5) (10) m/s
V = 25 – 5 = 20 m/s

(ii) Let t’ be the time taken by the train to come to rest.
Using the first equation of motion,
v = u + at
We get,
t’ = [(0 – 25)/ (-0.5)] s
t’ = 50 s

Solution 9N.

Distance travelled s = 100 m
Average velocity V = 20 m/s
Final velocity v = 25 m/s

(i) Let u be the initial velocity.
Average velocity = (Initial velocity + Final velocity)/2
V = (u + v)/2
20 = (u + 25)/2
u = 40 – 25 = 15 m/s

(ii) Let ‘a’ be the acceleration of the car.
Using the third equation of motion,
v2 – u2 = 2as
We get,
(25) 2 – (15) 2 = 2 (a) (100)
625 – 225 = 200 a
a = (400/200) m/s2 = 2 m/s2

Solution 10N.

Final velocity v = 0
Acceleration = -25 cm/s2 or -0.25 m/s2
Time taken t = 20 s

(i) Let ‘u’ be the initial velocity.
Using the first equation of motion,
v = u + at
We get,
u = v – at
u = 0 – (-0.25)(20) = 5 m/s

(ii) Let ‘s’ be the distance travelled.
Using the third equation of motion,
v2 – u2 = 2as
We get,
(0) 2 – (5)2 = 2 (-0.25) (s)
s = (25/0.5) = 50 m.

Solution 11N.

Initial velocity u = 0 m/s
Distance travelled s = 270 m
Time taken to travel s distance = 3 s
Let ‘a’ be the uniform acceleration.
Using the second equation of motion,
S = ut + (1/2) at2
We get,
270 = 0 + (1/2) a (3)2
270 = 9a/2
a = 60 m/s2

Let v be the velocity of the body 10 s after the start.
Using the first equation of motion, we get
v = u + at
v = 0 + (60)(10) = 600 m/s

Solution 12N.

Let the constant acceleration with which the body moves be ‘a’.
Given, the body travels distance S1 = 3 m in time t1 = 1 s.
Same body travels distance S2 = 8 m in time t2 = 2 s.

(i) Let ‘u’ be the initial velocity.
Using the second equation of motion,
S = ut + (1/2) at2
Substituting the value for S1 and S2, we get
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 24

(ii) Putting u = 2 m/s in the equation
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 25

Solution 13N.

Initial velocity u = 25 m/s
Final velocity v = 0

(i) Before the brakes are applied, let S be the distance travelled.
Distance = Speed × time
S = (25) × (5) m
S = 125 m

(ii) Acceleration = (Final velocity – Initial velocity)/Time taken
= [(0 – 25)/15] ms-2
= (-5/2) ms-2
 = -2.5 ms-2
Therefore, retardation = 2.5 ms-2

(iii) After applying brakes, the time taken to come to stop = 10 s
Let S’ be the distance travelled after applying the brakes.
Initial velocity u = 25 m/s
Final velocity v = 0
Using the third equation of motion,
v2 – u2 = 2as
We get,
(0) 2 – (25)2 = 2 (-2.5) (S’)
625 = 5(S’)
S’ = 125 m

Solution 14N.

Given, the initial velocity u = 75 km/s
Final velocity v = 120 km/s
Time taken = 6 s

(i) Acceleration = (Final velocity – Initial velocity)/time taken
= [(120 – 75)/6] kms-2
= (45/6) kms-2
 = 7.5 kms-2

(ii) Distance travelled by the aircraft in the first 10 s = Distance travelled in the first 6 s + Distance travelled in the next 4 s.
Distance travelled in the first 6s (S1) = ut + (1/2) at2
(S1) = ut + (1/2) at2
(S1) = (75)(6) + (1/2) (7.5)(6)2
(S1) = 450 + 135
(S1) = 585 km

Distance travelled in the next 4 s (S2) = Speed × time
Speed at the end of 6 s is 120 km/s.
(S2) = (120) (4)
(S2) = 480 km
Thus, the distance travelled by the aircraft in the first 10 s = (S1) + (S2) = 585 + 480 = 1065 km.

Solution 15N.

(i) For the first 10 s, initial velocity u = 0
Acceleration a = 2 m/s2
Time taken t = 10 s
Let ‘v’ be the maximum velocity reached.
Using the first equation of motion
v = u + at
We get
V = (0) + (2) (10) = 20 m/s

(ii) For the last 50 s: Final velocity = 0 m/s, initial velocity = 20 m/s.
Acceleration = (Final velocity – Initial velocity)/time
= (0 – 20)/50 = -0.4 m/s2
Retardation = 0.4 m/s2

(iii) Total distance travelled = Distance travelled in the first 10 s + Distance travelled in 200 s + Distance travelled in last 50 s
Distance travelled in first 10s (s1) = ut + (1/2) at2
S1= (0) + (1/2) (2) (10)2
S1= 100 m
Distance travelled in 200s (s2) = speed × time
S2 = (20) (200) = 4000 m

Distance travelled in last 50s (s3) = ut + (1/2) at2
Here, u = 20 m/s, t = 50 s and a = -0.4 m/s2
S3= (20)(50) + (1/2) (-0.4) (50)2
S3= 1000 – 500
S3= 500 m
Therefore, total distance travelled = S1 + S2 + S3 = 100 + 4000 + 500 = 4600 m

(iv) Average velocity = Total distance travelled/total time taken
= (4600/260) m/s
= 17.69 m/s

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