Month: June 2022

JAC Board Class 9th Science Solutions Chapter 12 Sound

JAC Class 9th Science Sound InText Questions and Answers

Poge 162

Question 1.
How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
When an object vibrates, it sets the particles of the medium around it in vibration. The particles in the medium in contact with the vibrating object are displaced from their equilibrium position. It then exerts a force on the adjacent particles. After displacing the adjacent particle, the first particle of the medium comes back to its original position. This process continues in the medium till the sound reaches our ear.

Page 163

Question 1.
Explain how sound is produced by your school bell.
Answer:
When the bell continues to vibrate forward and backward, it creates a series of compressions and rarefactions resulting in the production of sound.

Question 2.
Why are sound waves called mechanical waves?
Answer:
Sound waves need material medium to propagate, therefore, they are called mechanical waves. Sound waves can propagate through a medium only because of the interaction of the particles present in that medium.

Question 3.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
No, because a sound wave needs a medium through which it can propagate. Since there is no atmosphere on the moon, no material medium is available and we cannot hear any sound on the moon.

Page 166

Question 1.
Which wave property determines:
(a) loudness
(b) pitch?
Answer:
(a) Amplitude
(b) Frequency

Question 2.
Guess which sound has a higher pitch: guitar or car horn?
Answer:
Guitar has a higher pitch than car horn, because sound produced by the strings of guitar has a higher frequency than that of a car horn. The higher the frequency, the higher is the pitch.

Page 166

Question 1.
What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:

1. Wavelength: The distance between two consecutive compressions or two consecutive rarefactions is known as wavelength. Its SI unit is metre (m).
2. Frequency: The number of complete oscillations per second is known as the frequency of a sound wave. It is measured in hertz (Hz).
3. Time period: The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the wave.
4. Amplitude: The maximum height reached by the crest or the maximum depth reached by the trough of a
   sound wave is called its amplitude.

Question 2.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
Speed, wavelength and frequency of a sound wave are related by the following equation:
Speed (v) = Wavelength (λ) x Frequency (v)
V = λ × v

Question 3.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:
Frequency of the sound wave, v = 220 Hz
Speed of the sound wave, v = 440 ms^-1
For a sound wave,
Speed = Wavelength × Frequency
v = λ × v
∴ λ = \(\frac{v}{v}\) = \(\frac{440}{220}\) = 2
Hence, the wavelength of the sound wave is 2 m.

Question 4.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Answer:
The time interval between two successive compressions is equal to the time period of the wave. This time period is reciprocal of the frequency of the wave and is given by the relation:
T = \( \frac{1}{\text { Frequency }}\) = \(\frac{1}{500}\) = 0.002 s

Page 166

Question 1.
Distinguish between loudness and intensity of sound.
Answer:
Intensity of a sound wave is defined as the amount of sound energy passing through a unit area per second. Loudness is a measure of the response of the ear to the sound. The loudness of a sound is defined by its amplitude.

Page 167

Question 1.
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
The speed of sound depends on the nature of the medium. Sound travels the fastest in solids. Its speed decreases in liquids and it is the slowest in gases. Therefore, for a given temperature, sound travels fastest in iron.

Page 168

Question 1.
An echo is heard in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms^-1?
Answer:
Speed of sound, v = 342 ms^-1
Echo returns in time, t = 3s
Distance travelled by sound
= v × t = 342 × 3 = 1026 m
In the given time interval, sound has to travel a distance that is twice the distance between the reflecting surface and the source.
Hence, the distance of the reflecting surface from the source
= \(\frac{1026}{2}\) m = 513 m.

Page 169

Question 1.
Why are the ceilings of concert halls curved?
Answer:
The ceilings of concert halls are curved so that the sound, after reflection (from the walls), spreads uniformly in all directions.

Page 170

Question 1.
What is the audible range of the average human ear?
Answer:
The audible range of an average human ear is 20 Hz to 20,000 Hz.

Question 2.
What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
Answer:
(a) Infrasound has frequencies less than 20 Hz.
(b) Ultrasound has frequencies more than 20,000 Hz.

Page 172

Question 1.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Answer:
Time taken by the sonar pulse to return, t = 1.02s
Speed of sound in salt water, v = 1531 ms^-1
Distance travelled by the sonar pulse
= Speed of sound × Time taken
= 1.02 × 1531 = 1561.62 m
Distance travelled by the sonar pulse during its transmission and reception in water
= 2 × actual distance = 2d
Actual distance of the cliff from the submarine, d
Distance travelled by the sonar pulse
= \(\frac{Distance travelled by the sonar pulse2}{2}\) = \( \frac{1561.62}{2}\) = 780.81 m

JAC Class 9th Science Sound Textbook Questions and Answers

Question 1.
What is sound and how is it produced?
Answer:
Sound is a form of energy which gives the sensation of hearing. It is produced by the vibrations caused in the medium by vibrating objects.

Question 2.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:
When a vibrating body moves forward, it creates a region of high pressure in its vicinity. This region of high pressure is known as compression. When it moves backward, it creates a region of low pres – sure in its vicinity. This region is known as rarefaction. As the body continues to move forward and backward, it produces a series of compressions and rarefactions. This is shown in the figure below.

Question 3.
Cite an experiment to show’ that sound needs a material medium for its propagation.
Answer:
Take an electric bell and an air tight glass bell jar connected to a vacuum pump. Suspend the bell inside the jar, and press the switch of the bell. You will be able to hear the bell ring. Now pump out the air from the glass jar. The sound of the bell will become progressively fainter and after some time, the sound will not be heard. This is so because almost all air has been pumped out. This shows that sound needs a material medium to travel.

Question 4.
Why is sound wave called a longitudinal wave?
Answer:
Sound wave is called a longitudinal wave because it is produced by compressions and rarefactions in the air. The air particles vibrate parallel to the direction of propagation of sound.

Question 5.
Which characteristics of the sound help you to identify your friend by his voice while sitting with others in a dark room?
Answer:
The quality or timber of sound enables us to identify our friend by his voice.

Question 6.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer:
The speed of sound (344 m/s) is less than the speed of light (3 × 108 m/s). Sound of thunder takes more time to reach the earth as compared to light. Hence, a flash is seen before we hear a thunder.

Question 7.
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms^-1.
Answer:
For a sound wave,
Speed = Wavelength × Frequency = λ × v
Speed of sound in air = 344 m/s (Given)
(a) For, v = 20 Hz
λ₁ = \(\frac{v}{v}\) = \(\frac{344}{20}\)
= 17.2 m

(b) For, v = 20000 Hz
λ₂ = \(\frac{v}{v}\) = \(\frac{344}{20,000}\)
= 0.0172 m
Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

Question 8.
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Answer:
Velocity of sound in air = 346 m/s
Velocity of sound wave in aluminium = 6420 m/s
Let the length of the rod be l
Time taken for sound wave in air,
t₁ =\(\frac{l}{Velocity in air}\)
Velocity in air Time taken for sound wave in aluminium,
t₂ = \(\frac{l}{Velocity in aluminium}\)
Therefore, = \(\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}\)
= \(\frac{Velocity in aluminium}{Velocity in air}\) = \(\frac{6420}{ 346}\)
= 18.55

Question 9.
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Frequency = 100 Hz (given)
This means that the source of sound vibrates 100 times in one second. Therefore, number of vibrations in 1 minute, i.e., in 60 seconds = 100 × 60 = 6000 times.

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:
Sound follows the same laws of reflection as light does. The incident sound wave and the reflected sound wave make equal angles with the normal to the surface at the point of incidence. Also, the incident sound wave, the reflected sound wave and the normal to the point of incidence, all lie in the same plane.

Question 11.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
Answer:
An echo is heard when the time for the reflected sound is heard after 0.1s.
Time taken = \(\frac{Total distance}{Velocity}\)
On a hotter day, the velocity of sound is more. If the time taken by echo is less than 0.1s, it will not be heard.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
Two practical applications of reflection of sound waves are:

1. Reflection of sound is used to measure the distance and speed of underwater objects. This technique is known as SONAR.
2. Working of a stethoscope is also based on reflection of sound. In a stethoscope, the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflections.

Question 13.
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g =10 ms^-2 and speed of sound = 340 ms^-1.
Answer:
Height of the tower, s = 500m
Velocity of sound, v = 340 ms^-1.
Acceleration due to gravity, g = 10 ms^-2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Let the time taken by the stone to fall to the base of the tower be t₁
According to the second equation of motion:
s= ut₁ + \(\frac{1}{2} \mathrm{gt}_{1}^{2}\)
500 = (0 × t₁) + \( \left(\frac{1}{2} \times 10 \times t_{1}^{2}\right)\)
\( \mathrm{t}_{1}^{2}\) = 100
t₁ = 10s
Now, time taken by the sound to reach the top from the base of the tower, t₂ = \(\frac{500}{340}\) = 1.47s
Therefore, the splash is heard at the top after time, t.
Where, t = t₁ + t₂ = 10 + 1.47 = 11.47s.

Question 14.
A sound wave travels at a speed of 339 ms^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer:
Speed of sound, v = 339 ms^-1
Wavelength of sound,
λ = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
v = λ × v
∴ v = \(\frac{v}{λ}\) = \(\frac{ 339 }{0.015}\) = 22600 Hz
The frequency range of audible sound for humans is between 20 Hz to 20,000 Hz. Since the frequency of the given sound is more than 20,000 Hz, it is not audible.

Question 15.
What is reverberation? How can it be reduced?
Answer:
The repeated or multiple reflections of sound in a large enclosed space is known as reverberation. The reverberation can be reduced by covering the ceiling and walls of the enclosed space with sound absorbing materials, such as fibre board, loose woollens, etc.

Question 16.
What is loudness of sound? What factors does it depend on?
Answer:
Loudness is a measure of sound energy reaching the ear per second. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Question 17.
Explain how bats use ultrasound to catch a prey.
Answer:
Bats produce high – pitched ultrasonic squeaks. These high – pitched squeaks are reflected by objects and their preys and returned to the bats’ ears. This allows the bat to know the distance and direction of their prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Objects to be cleaned are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasonic waves detaches the dirt from the objects.

Question 19.
Explain the working and application of a sonar.
Answer:
SONAR is an acronym for sound Navigation And Ranging. It is an acoustic device used to measure the depth, direction and speed of underwater objects, such as submarines and ship wrecks, with the help of ultrasounds. It is also used to measure the depth of seas and oceans.

A beam of ultrasonic sound is produced and transmitted by the transducer (a device that produces ultrasonic sound) of the SONAR, which travels through sea water.
The echo produced by the reflection of this ultrasonic sound is detected and recorded by the detector, which is converted into electrical signals. The distance (d) of the underwater object is calculated from the time (t) taken by the echo to return with speed (v) which is given by 2d = v × t.
This method of measuring distance is also known as ‘echo – ranging’.

Question 20.
A sonar device on a submarine sends out a signal and receives an echo 5s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625m.
Time taken to hear the echo, t = 5s
Distance of the object from the submarine, d = 3625m
Total distance travelled by the sonar waves during the transmission and reception in water = 2d
Velocity of sound in water,
v = \(\frac{2 \mathrm{~d}}{t}\) = \(\frac{2 \times 3625}{5}\) = 1450ms^-1

Question 21.
Explain how7 defects in a metal block can be detected using ultrasound.
Answer:
Defects in metal blocks do not allow ultrasound waves to pass through them and they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. The defective part of the metal block does not allow ultrasound to pass through it. As a result, it will not be detected by the detector. Hence, defects in metal blocks can be detected using ultrasound.

Question 22.
Explain how the human ear works.
Answer:
The human ear consists of three parts – outer ear, middle ear and inner ear.

1. Outer ear: This is also called ‘pinna’. It collects the sound from the surroundings and directs it towards auditory canal.
2. Middle ear: The sound reaches the end of the auditory canal where there is a thin membrane called eardrum or tympanic membrane. The sound waves set this membrane in vibration. These vibrations are amplified by three small bones. Malleus, Incus and Stapes in the middle ear.
3. Inner ear: These vibrations reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound.
   

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 11 Work and Energy

JAC Class 9th Science Work and Energy InText Questions and Answers

Page 148

Question 1.
A force of 7N acts on an object. The displacement is, say 8m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer:
When a force F acts on an object to displace it through a distance s in its direction, the work done, W on the body by the force is given by:
Work done = Force × Displacement
W = F × s
Where, F = 7N, and s = 8m Therefore, work done,
W = 7 × 8 = 56 Nm = 56J

Page 149

Question 1.
When do we say that work is done?
Answer:
Work is said to be done whenever the following two conditions are satisfied:

1. A force acts on the body.
2. There is a displacement in the body caused by the applied force along the direction of the force.

Question 2.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
When a force ‘F’ displaces a body through a distance ‘s’ in the direction of the applied force, the work done, ‘W’ on the body is given by the expression:
Work done = Force × Displacement W = F × s

Question 3.
Define 1J of work.
Answer:
1J is the amount of work done by a force of 1N on an object that displaces it through a distance of 1m in the direction of the applied force.

Question 4.
A pair of bullocks exerts a force of 140N on the plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?
Answer:
Work done by the bullocks is given by the expression:
Work done = Force × Displacement
W=F × s
Where,
Applied force, F = 140N
Displacement, s = 15 m
W= 140 × 15 = 2100J
Hence, 2100 J of work is done in ploughing the length of the field.

Page 152

Question 1.
What is the kinetic energy of an object?
Answer:
The energy possessed by a body by the virtue of its motion is called kinetic energy. It has various applications:
(a) Kinetic energy of flowing water is used to generate electricity.
(b) Kinetic energy of hammer is used in driving a nail into a log of wood.
(c) Kinetic energy of air is used to run windmills, etc.

Question 2.
Write an expression for the kinetic energy of an object.
Answer:
If a body of mass ‘m’ is moving with a velocity ‘v’ its kinetic energy K.E. is given by the expression,
K.E. = \(\frac{1}{2}\) mv²
Its SI unit is joule (J).

Question 3.
The kinetic energy of an object of mass, m moving with a velocity of 5 m s^-1 is 25J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
K.E. of the object = 25J
Velocity of the object, v = 5 m/s
∴ K.E. = \(\frac{1}{2}\) mv²
25 = \(\frac{1}{2}\)m(5)²
m = 2 kg

(a) If velocity is doubled, v = 10m/s, m = 2 kg
K.E. = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\)(2) × (10)² = 100J.
K.E will become four times.

(b) If velocity is increased 3 times,
v = 15 m/s, m = 2 kg
K.E = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) × 2 × (15)² = 225J.
K.E will become nine times.

Page 156

Question 1.
What is power?
Answer:
Power is the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by the expression,
Power = \(\frac{Work}{Time }\) = \(\frac{w}{t}\)
It is expressed in watt (W).
watt = \(\frac{Joule}{Second}\)

Question 2.
Define 1 watt of power.
Answer:
A body is said to have a power of 1 watt if it does 1 joule work in 1 s, i.e.,
1W =\(\frac{1 \mathrm{~J}}{1 \mathrm{~s}}\)

Question 3.
A lamp consumes 1000J of electrical energy in 10s. What is its power?
Answer:
Power = \(\frac{Work done}{Time }\)
Work done = Energy consumed by the lamp = 1000J
Time = 10s
Power = \(\frac{1000}{10}\) = 100J s^-1 = 100W

Question 4.
Define average power.
Answer:
The average power is defined as the total work done or total energy consumed divided by the total time taken.
Average power = \(\frac{Total work done}{Total time taken }\)

JAC Class 9th Science Work and Energy Textbook Questions and Answers

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
1. Suma is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A windmill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Foodgrains are getting dried in the sun.
7. A sailboat is moving due to wind energy.
Answer:
Work is done whenever the following two conditions are satisfied:

• A force acts on the body.
• There is a displacement of the body by the application of force.

1. While swimming, Suma applies a force to push the water backwards. Therefore, Suma swims in the forward direction under the influence of the forward reaction of water. Here, the force causes a displacement. Hence, work is done by Suma while swimming.
2. While carrying a load, the donkey has to apply a force in the upward direction. But, displacement of the load is in the forward direction. Since, displacement is perpendicular to force, the work done is zero.
3. A windmill works against the gravitational force to lift water. Hence, work is done by the windmill in lifting water from the well.
4. In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero.
5. An engine applies force to pull the train. This allows the train to move in the direction of force. Therefore, there is a displacement of the train in the direction of force. Hence, work is done by the engine on the train.
6. Foodgrains do not move in the presence of solar energy. Hence, the work done is zero when the foodgrains are getting dried in the sun.
7. Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

Question 2.
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Work done by the force of gravity on the object is zero. Force of gravity acts in the vertically downward direction and the distance covered by the object is in the horizontal direction. As there is no  displacement in the direction of force, hence the work done is zero.

Question 3.
A battery lights a bulb Describe the energy changes involved in the process.
Answer:
When a bulb is connected to a battery, the chemical energy of the battery is transformed into electrical energy. When the bulb receives this electrical energy, it converts it into heat and light energy. Hence, the transformation of energy in the given situation can be shown as:
Chemical Energy → Electrical Energy → Heat Energy + Light Energy

Question 4.
Certain force acting on a 20 kg mass changes its velocity from 5 ms^-1 to 2 ms^-1. Calculate the work force.
Answer:
Here,
m = 20 kg,
u = 5 m^-1,
v = 2 m^-1
Work done = Change in K.E.
W = \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mu²
= \(\frac{1}{2}\) × 20 × (2² – 5²)
= 10 × (4 – 25)
= – 10 × 21 = – 210J
The negative sign indicates that work is done against the applied force.

Question 5.
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Therefore, work done by gravity is given by the expression,
W = mgh
Where, vertical displacement, h = 0
∴ W = mg × 0 = 0
Hence, the work done by gravity on the body is zero.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
No. The process does not violate the law of conservation of energy. This is because when a body falls from a height, its potential energy progressively changes into kinetic energy. The decrease in potential energy is equal to the increase in kinetic energy of the body. During this process, the total mechanical energy of the body remains conserved. Therefore, the law of conservation of energy is not violated.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:
While riding a bicycle, the muscular energy of the rider gets transformed into heat energy and kinetic energy of the bicycle. Heat energy heats the rider’s body. Kinetic energy provides velocity to the bicycle. The transformation can be shown as: Muscular Energy → Kinetic Energy + Heat Energy During this transformation, the total energy remains conserved.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
When we push a huge rock, there is no transfer of muscular energy to the stationary rock. Also, there is no loss of energy because muscular energy is transferred into heat energy which causes our body to become hot.

Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer:
1 unit of energy is equal to 1 kilowatt hour (kW h).
1 unit = 1 kW h
1 kWh = 3.6 × 10⁶J
Therefore, 250 units of energy
= 250 × 3.6 × 10⁶ = 9 × 10⁸J

Question 10.
An object of mass 40 kg is raised to a height of 5m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half – way down.
Answer:
Gravitational potential energy is given by the expression,
P_E= mgh Where,
h = Vertical displacement = 5 m
m = Mass of the object = 40 kg
g = Acceleration due to gravity = 9.8 ms^-2
P_E = 40 × 5 × 9.8= 1960J.
At half – way down, the potential energy of the object will be \(\frac{1960}{2}\) = 980J.
At this point, the object has an equal amount of potential and kinetic energy. This is due to the law of conservation of energy. Hence, half – way down, the kinetic energy of the object will also be 980J.

Question 11.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:
Zero. When the satellite moves around the earth, the force of gravity acts on it along the radius of its orbit, while its direction of motion is along the tangent to the orbit at any point. Thus, the force acts perpendicular to the displacement. Hence, the work done on the satellite is zero.

Question 12.
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer:
Yes. For a uniformly moving object. Suppose an object is moving with a constant velocity. The net force acting on it is zero. But, there is a displacement along the motion of the object. Hence, there can be a displacement without a force.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
When a person holds a bundle of hay over his head, there is no displacement in the bundle of hay. Although, force of gravity is acting on the bundle, the person is not applying any force on it. Hence, in the absence of force and displacement, work done by the person on the bundle is zero.

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Ans.
Energy consumed by an electric heater can be obtained with the help of the expression,
P= \(\frac{\mathrm{W}}{\mathrm{t}}\)
Where, power rating of the heater,
P= 1500 W
Time for which the heater has operated, t = 10h
Work done = Energy consumed by the heater
Therefore, energy consumed
= P × t = 1500 W × 10 h
= \( \frac{15000}{1000}\) = 15kW h

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Answer:
The law of conservation of energy states that energy can neither be created nor destroyed. It can only be converted from one form to another. Consider the case of an oscillating pendulum. When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height ‘h’ above the mean level ‘P’. At this point, the kinetic energy of the bob changes completely into potential energy.

The kinetic energy becomes zero, and the bob possesses only potential energy. As it moves towards point P, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates. The bob does not oscillate forever. It comes to rest because air resistance resists its motion.

The pendulum loses its kinetic energy to overcome this friction and stops after some time. The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system remains conserved.

Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer:
Work done on the object = Change in K.E. of the object.
W = \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) m(0)² = \(\frac{1}{2}\) mv²
To bring the object to rest, \(\frac{1}{2}\) mv² amount of work is required to be done on the object.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Kinetic energy, KE = \(\frac{1}{2}\) mv²
Where,
mass of the car, m = 1500 kg and velocity of the car,
v = 60 km/h = \(\frac{60 \times 5}{18}\) ms^-1
KE = \(\frac{1}{2}\) × 1500 × \(\left(\frac{60 \times 5}{18}\right)^{2}\)
= 20.8 × 10⁴ J
Hence, 20.8 × 10⁴ J of work is required to stop the car.

Question 18.
In each of the following, a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the . force is negative, positive or zero.

Answer:

1. In this case, the direction of force acting on the block is perpendicular to the displacement. Therefore, work done by force on the block will be zero.
2. In this case, the direction of force acting on the block is in the direction of displacement. Therefore, work done by force on the block will be positive.
3. In this case, the direction of force acting on the block is opposite to the direction of displacement. Therefore, work done by the force on the block will be negative.

Question 19.
Soni says that the acceleration of an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
Acceleration in an object could be zero even when several forces are acting on it. This happens when all the forces cancel out each other, i.e., the net force acting on the object is zero. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

Question 20.
Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Answer:
Here, P = 500 W = \(\frac{500}{1000}\) kW= 0.5 kW and t =10 h
Energy consumed by four devices
= 4P × t = 4 × 0.5 kW × 10h = 20 kW h

Question 21.
A freely falling object eventually stops reaching the ground. What happens to its kinetic energy?
Answer:
When a body is falling freely, the energy remains conserved. Potential energy is transformed into kinetic energy. But when the body reaches the ground, its kinetic energy is converted into sound energy and also into a little heat energy (due to friction between ground and object). A part of the energy is also transferred to the ground. Hence, the total energy of the system remains conserved.

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 10 Gravitation

JAC Class 9th Science Gravitation InText Questions and Answers

Page 134

Question 1.
State the universal law of gravitation.
Answer:
The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. For two objects of masses m₁ and m₂ separated by a distance r, the force (F) of attraction acting between them is given by the universal law of gravitation as:
F = \(=\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{R}^{2}}\)
where, G is the universal gravitational constant given by:
G = 6.67 × 10^-2 Nm^-2 kg^-2.

Question 2.
Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer:
Let M_E be the mass of the earth and m be the mass of an object on its surface.
If R is the radius of the earth, then according to the universal law of gravitation, the gravitational force (F) acting between the earth and the object is given by the relation:
F = \(=\frac{G M_{e} m}{R^{2}}\)

Page 136

Question 1.
What do you mean by free fall?
Answer:
Gravity of earth attracts every object towards its centre. When an object is dropped from a certain height, it begins to fall towards earth’s surface under the influence of gravitational force. Such a motion of object is called free fall.

Question 2.
What do you mean by acceleration due to gravity?
Answer:
When an object falls freely towards the surface of the earth from a certain height, its velocity changes. This change in velocity produces acceleration in the object which is known as acceleration due to gravity, denoted by ‘g’. The value of acceleration due to gravity is 9.8 m/s².

Page 138

Question 1.
What are the differences between the mass of an object and its weight?
Answer:

Mass	Weight
1. Mass is the quantity of matter contained in a body.	1. Weight is the force of gravity acting on a body.
2. It is the measure of inertia of the body.	2. It is the measure of gravity.
3. Mass is a constant quantity.	3. Weight is not a constant quantity. It is different at different places.
4. It only has magnitude.	4. It has magnitude as well as direction.
5. Its SI unit is kilogram (kg)	5. Its SI unit is the same as the SI unit of force, i.e., newton (N).

Question 2.
Why is the weight of an object on the moon l/6th its weight on the earth?
Answer:
The mass of moon is 1/100 times and its radius 1/4 times that of the earth. As a result, the gravitational attraction on the moon is about one sixth when compared to the earth. Hence, the weight of an object on the moon is l/6th of its weight on the earth.

Page 141

Question 1.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller the surface area, the larger will be the pressure on the surface. In case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

Question 2.
What do you mean by buoyancy?
Answer:
The upward force exerted by a liquid on an object immersed in it is known as buoyancy.

Question 3.
Why does an object float or sink when placed on the surface of water?
Answer:
The density of the object and water decides the floating or sinking of the object in water. The density of water is 1g/cm³. An object sinks in water if its density is greater than that of water. An object floats in water if its density is less than that of water.

Page 142

Question 1.
You find your mass to be 42kg on a weighing machine. Is your mass more or less than 42kg?
Answer:
The weighing machine actually measures the mass of the body. Hence, the mass reading of 42kg given by a weighing machine is same as the actual mass of body. As mass is the quantity of inertia, it remains the same.

Question 2.
You have a bag of cotton and an iron bar, each indicating a mass of 100kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?
Answer:
The cotton bag is heavier than the iron bar. The cotton bag experiences larger upthrust of air than the iron bar because it has a larger volume and displaces more air. Hence, in the given situation, cotton is heavier in reality.

JAC Class 9th Science Gravitation Textbook Questions and Answers

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
According to the universal law of gravitation, the gravitational force of attraction between any two objects of masses say,
M and m, is proportional to the product of their masses and inversely proportional to the square of distance r between them. So, force F is given by
F = \(\mathrm{G} \frac{\mathrm{M} \times \mathrm{m}}{\mathrm{r}^{2}}\)
Now, when the distance r is reduced to half, force between the two masses becomes
F = \(\mathrm{G} \frac{\mathrm{M} \times \mathrm{m}}{\left(\frac{\mathrm{r}}{2}\right)^{2}}\) or F = 4F
Hence, if the distance is reduced to half, the gravitational force becomes four times larger than the previous value.

Question 2.
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Answer:
All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistance). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

Question 3.
What is the magnitude of the gravitational force between the earth and a 1kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10²⁴ m).
Answer:
Given that,
Mass of the body, m = 1kg
Mass of the earth, M = 6 × 10²⁴ kg
Radius of the earth, R = 6.4 × 10⁶ m
Now, magnitude of the gravitational force (F) between the earth and the body can be given as,
F = \(\mathrm{G} \frac{\mathrm{M} \times \mathrm{m}}{\mathrm{r}^{2}}\) = \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^{6}\right)^{2}}\)
= \(\frac{6.67 \times 6 \times 10}{6.4 \times 6.4}\) = 9.8N (approx)

Question 4.
The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
Answer:
According to the universal law of gravitation, two objects attract each other with equal forces, but in opposite directions. The earth attracts the moon with a force which is equal to the force with which the moon attracts the earth.

Question 5.
If the moon attracts the earth, why does the earth not move towards the moon?
Answer:
The earth and the moon experience equal gravitational forces from each other. However, the mass of the earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the earth. For this reason, the earth does not move towards the moon.

Question 6.
What happens to the force between two objects, if
1. the mass of one object is doubled?
2. the distance between the objects is doubled and tripled?
3. the masses of both objects are doubled?
Answer:

1. If the mass of one object is doubled, the force between two objects will be doubled (increases).
2. If the distance between the objects is doubled, the force between the two objects will be one – fourth and if the distance will be tripled, the force will be one – ninth (1/9).
3. If the masses of both objects are doubled, the force will be 4 times.

Question 7.
What is the importance of universal law of gravitation?
Answer:
Universal law of gravitation is important because it tells us about:

1. the force that is responsible for binding us to the earth.
2. the motion of moon around the earth.
3. the motion of planets around the sun.
4. the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both the sun and the moon on the earth.

Question 8.
What is the acceleration of free fall?
Answer:
Acceleration of free fall is the acceleration produced when a body falls under the. influence of the force of gravitation of the earth alone. It is denoted by g and its value on the surface of the earth is 9.8 ms^-2.

Question 9.
What do we call the gravitational force between the earth and an object?
Answer:
The gravitational force between the earth and an object is called force due to gravity. It is measured as the weight of an object

Question 10.
Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? Hint: The value of g is greater at the poles than at the equator.
Answer:
Weight of a body on the earth is given by W = mg
Where, m = mass of the body, g = acceleration due to gravity.
The value of g is greater at poles than at the equator. Therefore, gold at the equator weighs less than at the poles. ‘Hence, his friend will not agree with the weight of the gold bought.

Question 11.
Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer:
The sheet of paper will experience a larger air resistance due to its larger surface area than that of its ball form. Hence, the sheet falls slower than its ball form.

Question 12.
Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton of a 10 kg object on the moon and on the earth?
Answer:
Mass of the object = 10 kg
Weight of the object on the earth = W = m × g
Weight of the object on the moon = 1/6th the weight of an object on the earth Also, acceleration due to gravity, g = 9.8 m/s²
Therefore, weight of a 10 kg object on the earth = 10 × 9.8 = 98 N And, weight of the same object on the moon = \(\frac{1}{6}\) × 98 = 16.3 N.

Question 13.
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:
(a) the maximum height to which it rises.
(b) the total time it takes to return to the surface of the earth.
Answer:
(a) According to the equation of motion under gravity:
v² – u² = 2gs
Where,
u = initial velocity of the ball
v = final velocity of the ball
s = height achieved by the ball
g = acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0, u = 49 m/s
During upward motion, g = – 9.8 ms^-2
Let h be the maximum height attained by the ball. Hence,
0 – (49)^-2 = 2 × 9.8 × h
h – \(\frac{49 \times 49}{2 \times 9.8}\) = 122.5m

(b) Let Ttfe the time taken by the ball to reach the height 122.5 m, then according to the equation of motion,
v = u + gt
We get
0 = 49 – (9.8 × t)
∴ 9. 8t = 49 t =\(\frac{49}{9.8}\) = 5s
But, time of ascent = time of descent. Therefore, total time taken by the ball to return = 5 + 5 = 10s.

Question 14.
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer:
According to the equation of motion under gravity: v² – u² = 2gs Where,
u = initial velocity of the stone = 0
v = final velocity of the stone
s = height of the stone = 19.6 mg
g = acceleration due to gravity = 9.8 ms^-2
v^-2 – 0^-2 = 2 × 9.8 × 19.6
v^-2 = 2 × 9.8 × 19.6 = (19.6)^-2
v = 19.6 ms^-1
Hence, the velocity of the stone just before touching the ground is 19.6 ms^-1

Question 15.
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s^-2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:
According to the equation of motion under gravity: v² – u² = 2gs
Where, u = initial velocity of the stone = 40 m/s
v = final velocity of the stone = 0
s = height of the stone
g = acceleration due to gravity = -10 ms^-2
Let h be the maximum height attained by the stone.
Therefore, 0 – (40)² = 2 × (-10) h
⇒ \(\frac{40 \times 40}{20}\) = 80m
Therefore, total distance covered by the stone during its upward and downward journey
= 80 + 80= 160 m
Net displacement of the stone during its upward and downward journey = 80 + (- 80) = 0m

Question 16.
Calculate the force of gravitation between the earth and the sun, given that the mass of the earth = 6 × 10² kg and of the sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.
Answer:
According to the question,
M = mass of the sun = 2 × 10³⁰ kg
m = mass of the earth = 6 × 10²⁴ kg
R = average distance between the earth and the sun = 1.5 × 10¹¹ m
From universal law of gravitation, F = \(G \frac{M \times m}{R^{2}}\)
Therefore, putting all the given values in above equation we get;
F = 6.67 × 10^-11× \(\frac{\left(6 \times 10^{24}\right) \times\left(2 \times 10^{30}\right)}{\left(1.5 \times 10^{11}\right)^{2}}\)
= 3.56 × 10²² N

Question 17.
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:
Let ‘t’ be the point at which two stones meet and let ‘h’ be their height from the ground.
It is given that the height of the tower = 100m.
Now, first consider the stone which falls from the top of the tower. So, distance covered by this stone in time t can be calculated using the second equation of motion.
Height covered by the falling stone = s₁
s₁ = ut + \(\frac{1}{2}\) gt²
s₁ = (0 × t) + \(\frac{1}{2}\) gt²
s₁ = \(\frac{1}{2}\) gt² …………(1)
The distance covered by the stone thrown upward = s₂
g = – 10 m/s
u = 25 m/s 1
s₂ = ut + \(\frac{1}{2}\) gt²
s₂ = 25t + \(\frac{1}{2}\) (-g)t²
s₂ = 25t – \(\frac{1}{2}\) gt² ………..(2)
Total height given= 100m
s₁ + s₁ = 100 m
from equations (1) and (2)
25t + (\(\frac{1}{2}\) gt² – \(\frac{1}{2}\) gt²) = 100m
25t = 100 m
t = \(\frac{100}{25}\) = 4 seconds …(3)
Putting the value of equation (3) in equation (1), we get
\(\frac{1}{2}\) × 9.8 × 4² = 78.4 m 2
The two stones will meet after 4 seconds when the falling stone has covered a height of 78.4 m and meet the other stone after 4 s at a height = (100 – 78.4)
= 21.6 m from the ground.

Question 18.
A ball thrown up vertically returns to the thrower after 6s. Find:
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4s.
Answer:
(a) Time of ascent is equal to the time of descent. The ball takes a total of 6s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g = -9.8 ms^-2
Using, equation of motion, v = u + gt
We get, 0 = u + (-9.8 × 3),
u = 9.8 × 3 = 29.4 ms^-1
Hence, the ball was thrown upwards with a velocity of 29.4 ms^-1

(b) Let the maximum height attained by the ball be ‘h’.
Initial velocity during the upward journey, u = 29.4 ms^-1
Final velocity, v = 0
Acceleration due to gravity, g = -9.8 ms^-2
From the equation of motion,
s = ut +\(\frac{1}{2}\) at²
h = (29.4 × 3) + (\(\frac{1}{2}\) × (-9.8) × 3²)
= 44.1 m

(c) Ball attains the maximum height after 3s. After attaining this height, it will start falling downwards. In this case, initial velocity, u = 0
Position of the ball after 4s of the throw is given by the distance travelled by it during its downward journey in 4s – 3s = 1s.
Equation of motion, s = ut + \(\frac{1}{2}\) gt²
will give,
s = (0 × t) + (\(\frac{1}{2}\) × 9.8 × 1²) = 4.9 m
Total height = 44.1 m
This means that the ball is 39.2 m, i.e., (44.1 m – 4.9 m) above the ground after 4 seconds.

Question 19.
In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
The buoyant force on an object immersed in a liquid acts upwards, i.e., opposite to the direction of the force exerted by the object.

Question 20.
Why does a block of plastic released under water come up to the surface of water?
Answer:
For an object immersed in water, two forces act on it:

1. gravitational force, which tends to pull the object in downward direction,
2. buoyant force that pushes the object in upward direction. In this case, buoyant force is greater than the gravitational pull on the plastic block.
3. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water.

Question 21.
The volume of 50g of a substance is 20 cm³. If the density of water is 1g cm^-3 will the substance float or sink?
Answer:
If the density of an object is more than the density of a liquid, it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, it floats on the surface of the liquid.
Here, density of the substance = \(\frac{Mass of the substance}{Volume of the substanc}\) = \(\frac{50}{20}\) = 2.5g cm^-3
The density of the substance is more than the density of water (1g cm^-3). Hence, the substance will sink in water.

Question 22.
The volume of a 500g sealed packet is 350 cm^-3. Will the packet float or sink in water if the density of water is 1g cm^-3? What will be the mass of the water displaced by this packet?
Answer:
The density of the 500g sealed packet
= \(\frac{Mass of the packet }{Volume of the packet }\)
= \(\frac{500}{350}\) = 1.428g cm^-3
The density of the substance is more than the density of water (1g cm^-3). Hence, it will sink in water. The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.

JAC Class 9 Science Solutions

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

JAC Class 9th Science Force and Laws of Motion InText Questions and Answers

Page 118

Question 1.
Which of the following has more inertia:
(a) a rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five – rupees coin and a one – rupee coin?
Answer:
(a) A stone
(b) A train
(c) A five – rupee coin
As the mass of an object is a measure of its inertia, objects with more mass have more inertia.

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

Agent supplying the force	Change in velocity of ball
(a) First player kicks a football.	(a) Velocity changes from ‘zero’ to ‘u’.
(b) Second player kicks the football towards the goal.	(b) Velocity changes again by change in direction.
(c) The goalkeeper collects the football.	(c) Velocity becomes zero.
(d) Goalkeeper kicks it towards a player of his team.	(d) Change in velocity takes place.

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
Before shaking the branches, leaves are at rest. When branches are shaken, they come in motion while the leaves tend to remain at rest due to inertia of rest. As a result, leaves get detached from the branches and fall down.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
1. When a moving bus brakes to a stop: When the bus is moving, our body is also in motion. But due to sudden brakes, the lower part of our body comes to rest as soon as the bus stops. But the upper part of our body continues to be in motion and hence we fall in forward direction due to inertia of motion.

2. When the bus accelerates from rest: When the bus is stationary, our body is at rest but when the bus accelerates, the lower part of our body, being in contact with the floor of the bus, comes in motion but the upper part of our body remains at rest due to inertia of rest. Hence we fall in backward direction.

Page 126

Question 1.
If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
The horse pulls the cart with a force (action) in the forward direction. The cart also pulls the horse with an equal force (reaction) in the backward direction. The two forces get balanced While pulling the cart, the horse also pushes the ground with its feet in the backward direction, the reaction of the earth makes it move in the forward direction along with the cart.

Question 2.
Explain, why it is difficult for a fireman to hold a hose, which ejects a large amount of water at a high velocity.
Answer:
The water that is ejected out from the hose in the forward direction, comes out with a large momentum and an equal amount of momentum is developed in the hose in the opposite direction and hence the hose is pushed backward It hence becomes difficult for a fireman to hold a hose which experiences this large momentum.

Question 3.
From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35m/s. Calculate the initial recoil velocity of the rifle.
Answer:
(m₁) Mass of rifle = 4kg
(m₂) Mass of bullet = 50g = 0.05kg
(v₂) Velocity of bullet = 35m/s
(v₁) Recoil velocity of rifle = v₁
According to the law of conservation of momentum,
momentum of rifle = momentum of bullet
m₁v₁ = v₂m₂
4 kg × v₁ = 0.05 × 35m/s
v₁ = \(\frac{0.05 \times 35}{4}\) = \(\frac{1.75}{4}\) = 0.4375 m/s.

Question 4.
Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2m/s and 1m/s respectively. They collide and after the collision, the first object moves at a velocity of 1.67m/s. Determine the velocity of the second object.
Answer:
m₁ = 100g = 0.1kg
m₂= 200g = 0.2kg
u₁ = 2m/s
u₂ = 1m/s
After collision
v₁ = 1.67m/s
v2=?
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.1 × 2) + (0.2 × 1)
= (0.1 × 1.67) + (0.2 × v₂)
0. 2 + 0.2 = 0.167 + 0.2 v₂
0. 4 = 0.167 + 0.2 v₂
\(\frac{0.4-0.167}{0.2}\) = v₂
\(\frac{0.233}{0.2}\) = 1.165 m/s
The velocity of the second object is 1.165 m/s.

JAC Class 9th Science Force and Laws of Motion Textbook Questions and Answers

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non – zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
When an object experiences a net zero external unbalanced force, in accordance with the second law of motion, its acceleration is zero. If the object was initially in a state of motion, then in accordance with the first law of motion, the object will continue to move in the same direction with the same speed It means that the object may be travelling with a non – zero velocity but the magnitude as well as direction of velocity must remain unchanged or constant throughout.

Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
The carpet with dust is in state of rest. When it is beaten with a stick the carpet is set in motion, but the dust particles remain at rest. Due to inertia of rest, the dust particles retain their position of rest and fall down due to gravity.

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

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

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes.
(Hint: 1 tonne = 1000 kg).
Answer:
u = 0m/s
m = 7 tonnes = 7 × 1000kg = 7000kg
s = 400m, t = 20s,
a = ? F = ?
s = ut +\(\frac{1}{2}\) at²
400 = (0 × 20) + \(\frac{1}{2}\) a (20)²
⇒ \(\frac{400 \times 2}{(20)^{2}}\) = a
or, a = 2m/s²
Force (F) = ma = 7000 × 2 = 14000N

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

Question 7.
A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate.
(a) the net accelerating force and
(b) the acceleration of the train.
Answer:
(a) The net accelerating force = Force exerted by the engine – friction force = 40000 N – 5000 N = 35000 N

(b) The acceleration of the train (a) = ?
F = 35000 N
Mass of 5 wagons pulled by the engine = 5 × 2000 = 10000kg
F = ma
35000 = 10000 × a
a = 35000/10000 = 3.5 m/s²

Question 8.
An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7ms²?
Answer:
Mass = 1500kg
a = -1.7 m/s²
F = ?
F = m × a = 1500 × (- 1.7) = – 2550 N
The force between the vehicle and road is – 2550 N.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v?
Answer:
(a) (mv)²
(b) v²
(c) \(\frac{1}{2}\)mv²
(d) mv

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

Question 11.
Two objects, each of mass 1.5kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5ms¹ before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer:
Mass of the objects, m₁ = m₂ = 1.5kg
Velocity of first object, v₁ = 2.5 m/s
Velocity of second object, v₂ = – 2.5 m/s
Momentum before collision = m₁v₁ + m₂v₂
= (1.5 × 2.5)+ (1.5 × (- 2.5)) = 0
Mass of combined object = m₁ + m₂ = 1.5 + 1.5 = 3.0kg
After collision, v = ?
According to the law of conservation of momentum,
Momentum after collision = Momentum before collision
mv = 0
or v = 0 ms^-1
(since mass cannot be zero)

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

Question 13.
A hockey ball of mass 200 g travelling at 10 ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1 Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Mass of ball, m = 200g = 0.2kg
Initial speed of ball, u = 10m/s
Final speed of ball, v = – 5 m/s
Initial momentum of the ball
= mu = 0.2kg × 10m/s = 2kg m/s
Final momentum of the ball
= mv = 0.2kg × (- 5 m/s) = – 1kg m/s
Hence, change in momentum
= difference in the momentum = (-1) – 2 = -3kg m/s

Question 14.
A bullet of mass 10g travelling horizontally with a velocity of 150ms^-1 strikes a stationary wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
m = 10g = 0.01kg, u = 150ms^-1,
v = 0, t = 0.03s
a = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) = \(\frac{0-150}{0.03}\) = – 5000ms^-2,
The distance of penetration of the bullet into the block,
s = ut +\(\frac{1}{2}\) at²
= (150 × 0.03) + \(\frac{1}{2}\) × (- 5000) × (0.03)²
= 4.5 – 2.25 = 2.25 m
The magnitude of the force exerted by the wooden block on the bullet
= ma = 0.01 × 5000 = 50N.

Question 15.
An object of mass 1kg travelling in a straight line with a velocity of 10ms^-1 collides with and sticks to a stationary wooden block of mass 5kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer:
m₁= 1 kg
v₁ = 10 m/s
Mass of wooden block = 5 kg
m₂ = 5kg + 1kg (combined object)
= 6kg
Velocity of combined object = v₂ = ?
Momentum before impact (p₁) = m₁v₁ = 1 × 10 = 10 kg m/s
∴ Momentum before impact = Momentum after impact
m₁v₁ = m₂v₂
10Kg m/s = 6v₂
\(\frac{10}{6}\) = v₂ or v₂ = 1.76 m/s

Question 16.
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms^-1 to 8 m s^-1 in 6s. Calculate the initial and final momentum of the object. Also, find out the magnitude of the force exerted on the object.
Answer:
Given:
m = 100 kg,
u = 5 m/s,
v = 8 m/s,
t = 6s
p₁ = ?
P₂ = ?
F = ?
Initial momentum, p₁ = mu = 100 × 5 = 500 kg m/s
Final momentum, p₁ = mv = 100 × 8 = 800 kg m/s
Force exerted on the object, F = ma
= 100 \(\left(\frac{v-u}{t}\right)\) = 100 \(\left(\frac{8-5}{6}\right)\)
= 100 × \(\frac{3}{6}\) = 50 N

Question 17.
Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect and as a result, the insect died Rahui while putting on entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer:
Rahul gave the correct explanation that both the motorcar and the insect experienced the same force and a change in their momentum. As per the law of conservation of momentum, when two bodies collide:
Initial momentum before collision = Final momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Equal force is exerted on both the bodies but because the mass of insect is very small, it will suffer a greater change in velocity.

Question 18.
How much momentum will a dumb bell of mass 10kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10ms².
Answer:
Here, m = 10kg, u = 0,
s = 80 cm = 0.80 m, a = 10 m/s^-2
Let v be the velocity gained by the dumb bell as it reaches the floor.
As v² – u² = 2as
v² – 0² = 2 × 10 × 0.80 = 16
v = 4 ms^-1
Momentum transferred by the dumb bell to the floor:
p = mv = 10 × 4 = 40kg ms^-1

Page 130

Question A1.
The following is the distance – time table of an object in motion:

Time in seconds	Distance in metres
0	0
1	1
2	8
3	27
4	64
5	125
6	216
7	343

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b) What do you infer about the forces acting on the object?
Answer:

Time (s)	Distance (m)	Vloctiy, v = \(\frac{S_{2}-S_{1}}{t_{1}-t_{1}}\left(\mathrm{~s}^{-1}\right)\)	Acceleration, a = \(\frac{v_{2}-v_{1}}{t_{2}-t_{1}}\left(m^{-2}\right)\)
0	0	0	0
1	1	\(\frac{1-0}{1-0}=1 \mathrm{~ms}^{-1}\)	\(\frac{1-0}{1-0}=1 \mathrm{~ms}^{-2}\)
2	18	\(\frac{8-1}{2-1}=7 \mathrm{~ms}^{-1}\)	\(\frac{7-1}{2-1}=6 \mathrm{~ms}^{-2}\)
3	27	\(\frac{27-8}{3-2}=19 \mathrm{~ms}^{-1}\)	\(\frac{19-7}{3-2}=12 \mathrm{~ms}^{-2}\)
4	64	\(\frac{64-27}{4-3}=37 \mathrm{~ms}^{-1}\)	\(\frac{37-19}{4-3}=18 \mathrm{~ms}^{-2}\)
5	125	\(\frac{125-64}{5-4}=61 \mathrm{~ms}^{-1}\)	\(\frac{61-37}{5-4}=24 \mathrm{~ms}^{-2}\)
6	216	\(\frac{216-125}{6-5}=91 \mathrm{~ms}^{-1}\)	\(\frac{91-61}{6-5}=30 \mathrm{~ms}^{-2}\)
7	343	\(\frac{343-216}{7-6}=127 \mathrm{~ms}^{-1}\)	\(\frac{127-91}{7-5}=36 \mathrm{~ms}^{-2}\)

(a) There is an unequal change of distance covered in equal intervals of time. Thus, the given object is having a non – uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.
(b) The object is in an accelerated condition. According to Newton’s second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. So, we can say that unbalanced forces are acting on the object.

Question A2.
Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level roa(d) The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s 2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort).
Answer:
Mass of the motorcar = 1200kg.
Only two persons manage to push the car with uniform velocity. Hence, the acceleration acquired by the car is given by the third person alone.
Acceleration produced by the car when it is pushed by the third person is, ‘a’
= 0.2 m/s².
Let the force applied by the third person be F.
From Newton’s second law of motion, Force = Mass × Acceleration => F = 1200 × 0.2 = 240 N.
Thus, the third person applies a force of magnitude 240 N. Hence, each person applies a force of 240 N to push the motorcar.

Question A3.
A hammer of mass 500 g, moving at 50ms¹, strikes a nail. The nail stops the hammer in a very short time of 0.01s. What is the force of the nail on the hammer?
Answer:
Mass of the hammer,
m = 500g = 0.5kg.
Initial velocity of the hammer,
u = 50m/s.
Time taken by the nail to stop the hammer, t = 0.01s.
Velocity of the hammer, v = O (since the hammer finally comes to rest). From
Newton’s second law of motion,
Force, F = \(\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\) = \(\frac{0.5(0-50)}{0.01}\)
= – 2500N
The hammer strikes the nail with a force of -2500N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e.. + 2500 N.

Question A4.
A motorcar of mass 1200kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18km/h in 4s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Answer:
Mass of the motorcar, m = 1200kg.
Initial velocity of the motorcar, u = 90km/h = 25m/s.
Final velocity of the motorcar,
v = 18 km/h = 5m/s.
Time taken, t = 4s.
According to the first equation of motion:
v = u + at
=> 5 = 25 + a(4) or, a = -5m/s²
Negative sign indicates that it is a retarding motion, i.e., velocity is decreasing.
Change in momentum = mv – mu
= m(v – u)= 1200 (5 – 25)
= – 24000 kg ms^-1
Force = Mass × Acceleration
= 1200 × (- 5) = – 6000 N

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 8 Motion

JAC Class 9th Science Motion InText Questions and Answers

Page 100

Question 1.
An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:
Yes, an object can have zero displacement even after it has moved through a distance. We know that distance is just the length of path an object has covered irrespective of its direction or position with reference to a certain point, whereas the shortest distance measured from the initial to the final position of an object is known as displacement. For example, an object starts from point A and after covering a distance of say 50 metres, reaches at point B. Hereafter, it again moves back to point A.

Here the distance covered by an object is = AB + BA = 50 m + 50 m = 100 m
whereas displacement of object is = AB – BA = 50 m – 50 m = 0 m.
As initial position of object is same as that of its final position, hence its displacement, which is distance measured from the initial to the final position, is zero.

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer:
Given, length of each side = 10 m
Distance covered in 1 lap = Perimeter of ABCD = 4 × 10 = 40 m
Time taken by farmer to cover 1 lap = 40s
Speed of farmer = \(\frac{40}{40 s}\) = 1m/s
Distance covered by farmer in 2 minutes 20s = Speed × Time = 1 × 140 s = 140 m
Number of laps covered = \(\frac{140}{40}\)
= 3.5 laps
After 140s, the farmer will be at position C (i.e., 3 and \(\frac{1}{2}\) laps)
Displacement = AC = (AB² + BC²)^1/2
= (100 + 100)^1/2
= 10√2 = 10 × 1.414 = 14.14m

Question 3.
Which of the following is true for displacement?
1. It cannot be zero.
2. Its magnitude is greater than the distance travelled by the object.
Answer:
Both of the statements are not true as:

1. Displacement can be zero.
2. Its magnitude is either less or equal to the distance travelled by the object

Page 102

Question 1.
Distinguish between speed and velocity.
Answer:

Speed	Velocity
(a) The distance travelled by a moving body per unit time is called its speed.	(a) The distance travelled by a moving body, in a particular direction, per unit time is called its velocity.
(b) It is a scalar quantity.	(b) It is a vector quantity.
(c) It can be changed only by changing the distance travelled by a body in a particular time.	(c) It can be changed by changing the object’s speed, direction of motion or both.

Question 2.
Under what conditions is the magnitude of average velocity of an object equal to its average speed?
Answer:
The magnitude of average velocity of an object is equal to its average speed only when it is moving in a straight line without turning back.

Question 3.
What does the odometer of an automobile measure?
Answer:
Odometer of an automobile measures the distance covered by the automobile.

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

Question 5.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3^1/2 ,10⁸ms^-1
Answer:
Time taken = 5 minutes = 5 × 60s = 300 seconds
Speed of signal u = 3 × 10⁸ m/s
Distance = ?
Speed = Distance / Time
∴ Distance = speed × time
Distance = 3 × 10⁸ × 300 = 9 × 10¹⁰ m

Page 103

Question 1.
When will you say a body is in
(a) uniform acceleration?
(b) non – uniform acceleration?
Answer:
(a) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time. Acceleration due to gravity is an example of uniform acceleration.

(b) A body is said to be in non – uniform acceleration if the rate of change of its velocity is not constant, Acceleration of a train running from one station to another is a nonuniform acceleration.

Question 2.
A bus decreases its speed from 80 km h^-1 to 60 km h^-1 in 5s. Find the acceleration of the bus.
Answer:
Initial speed of the bus (u) = 80 km h^-1
\(\frac{80 \times 1000 \mathrm{~ms}^{-1}}{60 \times 60}\) = \(\frac{800 \mathrm{~ms}^{-1}}{36}\)
Final speed of the bus (v) = 60 km h^-1
= \(\frac{60 \times 1000 \mathrm{~ms}^{-1}}{60 \times 60}\) = \(\frac{600 \mathrm{~ms}^{-1}}{36}\)
Time in transition (t) = 5s
The acceleration of the bus (a)
\(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) = \(\frac{\left(\frac{800}{36}-\frac{600}{36}\right)}{5}\) = \(\frac{\left(\frac{-200}{36}\right)}{5}\)
\(\frac{5.55 \mathrm{~ms}^{-1}}{5 \mathrm{~s}}\) = 1.11 ms^-1

Question 3.
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^-1 in 10 minutes. Find its acceleration.
Answer:
Initial speed of the train (u)
= 0 ms^-1
Final speed of the train (v) = 40 km h^-1
= \(\frac{40 \times 1000}{60 \times 60} \mathrm{~ms}^{-1}\) = \(\frac{400}{36} \mathbf{m s}^{-1}\)
Time in transition (t) = 10 minutes = 10 × 60s = 600s
The acceleration of the train (a) = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) = \(\frac{\left(\frac{400-0}{36}\right)}{600}\) = \(\frac{11.11}{600}\) ms^-2
= 0.0185 ms^-2

Page-107

Question 1.
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer:
(a) For uniform motion, the distance time graph is a straight line inclined with the time-axis.
(b) For non – uniform motion, the distance – time graph is a curved line.

Question 2.
What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:
Motion of an object whose distance – time graph is a straight line parallel to the time axis is not moving at all and is in the state of rest.

Question 3.
What can you say about the motion of an object if its speed – time graph is a straight line parallel to the time axis?
Answer:
The motion of an object, if its speed – time graph is a straight line parallel to the time axis, indicates that the object is moving with a uniform speed.

Question 4.
What is the quantity which is measured by the area occupied below the velocity – time graph?
Answer:
Displacement covered by the body in the given time interval.

Page 110

Question 1.
A bus starting from rest moves with a uniform acceleration of 0.1 ms^-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled.
Answer:
Here,
u = 0, a = 0.1 ms^-2, t = 2 min = 120s
(a) v = u + at = 0 + (0.1 × 120) = 12 ms^-1
(b) s = ut + \(\frac{1}{2}\) at²
= (0 × 120) + (\(\frac{1}{2}\) × 0.1 × (120)²)
= 720m.

Question 2.
A train is travelling at a speed of 90 km h^-1. Brakes are applied so as to produce a uniform acceleration of -0.5 ms^-2. Find how far the train will go before it is brought to rest.
Answer:
Here, u = 90 km h^-1 = 90 x \(\frac{5}{18}\) ms^-1 = 25 ms^-1
a = O.5 ms^-2
v= O
As v² – u² = 2as
O² – (25)² = 2 × (-0.5) × s
or s = \(\frac{-(25 \times 25)}{2 \times-0.5}\) = \(\frac{-625}{-1}\) = 265m.

Question 3.
A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3s after the start?
Answer:
Initial velocity of trolley (u) = 0 cm s^-1
Acceleration (a) = 2 cm s^-2
Time (t) = 3s
We know that final velocity (v) = u + at = 0 + (2 × 3) cm s^-1
∴ The velocity of train after 3 seconds = 6 cm s^-1

Question 4.
A racing car has a uniform acceleration of 4ms^-2. What distance will it cover in 10 s after start?
Answer:
Initial velocity of the car (u) = 0 ms^-1
Acceleration (a) = 4ms^-2
Time (t) = 10s
We know distance (s) = ut + \(\frac{1}{2}\) at²
∴ Distance covered by car in 10 s
=0 × 10 + \(\frac{1}{2}\) × 4 × 10²
= 0 + (\(\left(\frac{1}{2} \times 4 \times 10 \times 10\right)\)m
= \(\frac{1}{2}\) x 400 m = 200m.

Question 5.
A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer:
Given, initial velocity of the stone (u) = 5ms^-1, final velocity (v) = 0ms^-1
Downward or negative acceleration (a) = – 10ms^-2
We know that 2as = v² – u²
∴Height attained by the stone (s)
\(\frac{(0)^{2}-(5)^{2}}{2(-10)}\) = \(\frac{-25}{-20}\) = 1.25m
Also we know that final velocity (v) = u + at
Also, Time (t) = \(\frac{v-u}{a}\)
Time (t) taken by the stone to attain the height (s) = \(\frac{0-5}{-10}\) s = 0.5s.

JAC Class 9th Science Motion Textbook Questions and Answers

Question 1.
An athlete completes one round of a circular track of diameter 200 m in 40s. What will be the distance covered and the displacement at the end of 2 minutes 20s?
Answer:
Diameter of circular track (D) = 200m
Radius of circular track (r) = \(\frac{200}{2}\) =100m
Time taken by the athlete for one round (t) = 40s
Distance covered by athlete in one round
(s) = 2π r = 2 x \(\frac{22}{7}\) x 100m
Speed of theathlete (v) = \(=\frac{\text { Dis tan ce }}{\text { Time }}\)
\(\frac{2 \times 2200}{7 \times 40}\) = \(\frac{4400}{7 \times 40}\) ms^-1

∴ Distance covered in 2 minutes 20 seconds (s) or 140s = Speed (s) × Time (t) 4400
= \(\vec{a}\) × 4 = 2200 m
Number of rounds in 40s = 1 round
Number of rounds in 140s = \(\frac{140}{40}\) = 3.5
After taking start from position X, the athlete will be at position Y after 3.5 rounds as shown in the figure.
∴ Hence, displacement of the athlete with respect to initial position at X = XY
= Diameter of circular track = 200m.

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging: (a) from A to B and (b) from A to C?
Answer:
(a) From A to B
Time for A → B = 2 min 30 s = (2 x 60) + 30 = 150s Average speed
\(\frac{Total distance}{Time taken}\) = \(\frac{300}{150}\) = 2m/s
Average velocity
\(\frac{\text { Displacement }}{\text { Time taken }}\) = \(\frac{300}{150}\) = 2m/s

(b) From A to C Total distance
= (A to B) + (B to C)
= (300 + 100) = 400m
Time taken= 150 + 60 = 210s
Average speed
\(\frac{Total distance}{Time taken}\)= \(\frac{400}{210}\) = 1.9m/s
Total displacement = AC
= 300 – 100 = 200m
Average velocity
\(\frac{Displacement}{Time taken}\) = \(\frac{200}{210}\) = 0.95 m/s.

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20 km h^-1. On his return trip along the same route, there is less traffic and the average speed is 40 km h^-1. What is the average speed for Abdul’s trip?
Answer:
Average speed = \(\frac{Total distance}{Time}\)
Suppose the distance from Abdul’s home to school = x km
While driving to school:
Average speed = \(\frac{Distance}{Time}\)
20 km/h = \(\frac{x}{t_{1}}\)…………..(1)
While on his return trip

Average speed =\(\frac{Distance}{Time}\)
40 km/h = \(\frac{x}{t_{2}}\) ……………(2)
Total distance = x + x = 2x
Total time (t₁ + 1₂)
\(\frac{x}{20}\) = \(\frac{x}{40}\) = \(\frac{2 x+x}{40}\) = \(\frac{3 x}{40}\)
(∴from eq. (1) and (2))
Average speed = \(\frac{2 x}{\left(\frac{3 x}{40}\right)}\) = \(\frac{2 x \times 40}{3 x}\)
= 26.67 km/h.

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^-2 for 8.0s. How far does the boat travel during this time?
Answer:
Given initial velocity of the motorboat, u = 0 m s^-1
Acceleration of the motorboat, a = 3.0 m s^-2
Time under consideration, t = 8.0 s
We know that, Distance s = ut + \(\frac{1}{2}\) at²
∴ The distance travelled by the
motorboat = (0 × 8) + \(\frac{1}{2}\) × 3.0 × 8²
= \(\frac{1}{2}\) × 3 × 8 × 8m = 96m

Question 5.
A driver of a car travelling at 52 km h^-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3 km h^-1 in another car applies his brakes slowly and stops in 10s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Answer:
Initial speed of the first driver
\(=\frac{52 \times 1000}{60 \times 60}\) =14.44 m/s
Initial speed of the second driver
\(\frac{3 \times 1000}{60 \times 60}\) = 0.833 m/s
Distance covered by the first driver
\(\frac{1}{2}\) × 14.44 × 5 = 36.5 m
Distance covered by the second driver
\(\frac{1}{2}\) × 0.8333 × 10
= 4.165m
Hence, distance travelled by the first car is more.

Question 6.
Figure shows the distance – time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?

Answer:
(a) B is travelling the fastest.

(b) No. Because three lines do not meet at any point.

(c) When B passes A, C is at a distance of approximately 8 km from the origin.

(d) By the time B passes C, it has travelled 5.5 km.

Question 7.
A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10ms^-2 with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Let us assume that the final velocity with which the ball will strike the ground is V and the time it takes to strike the ground is ‘t’
Initial velocity of the ball, u = 0ms^-1
Distance or height of fall, s = 20m
Downward acceleration, a = 10ms^-2
As we know, 2as = v² – u²
v² = 2as + u² = (2 × 10 × 20) + 0 = 400ms^-1
∴ v = 20ms^-1
Final velocity of the ball, v = 20 ms^-1
∴ Time taken by the ball to strike,
t =\(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\) = \(\frac{20-0}{10}\) = \(\frac{20}{10}\) = 2 seconds.

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

(a) Find how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer:
(a) In the given graph, 56 full squares and 12 half squares come under the slope for the time of 4 seconds.
Total number of squares = 56 + \(\frac{12}{2}\) = 62 squares The total area of the squares will give the distance travelled by the car in 4 seconds.
On the time axis, 5 squares
= 2s ∴ 1 square = \(\frac{2}{5}\)s
On the speed axis, 3 squares
= 2m/s ∴ 1 square = \(\frac{2}{3}\) m/s^-1
∴ area of 1 square = \(\frac{2}{3}\) s x \(\frac{2}{5}\)ms^-1
= \(\frac{4}{15}\)
∴ area of 62 squares
\(=\frac{4}{15}\) m × 62 = \(\frac{248}{15}\) m = 16.53 m
Therefore, car travels 16.53m in first 4 seconds. The shaded area under speed – time graph represents the distance which the car will travel in the first 4 seconds.

(b) The straight line part of graph, from point A to point B, represents uniform motion.

Question 9.
State which of the following situations are possible and give an example for each of these:
1. An obj ect with a constant acceleration but with zero velocity.
2. An object moving with an acceleration but with uniform speed.
Answer:

1. Possible. Object under free fall due to gravity.
2. Possible. Object moving in circular path.

Question 10.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer:
Let us assume an artificial satellite that is moving in a circular orbit of radius 42250 km and covers a distance ‘s’ as it revolves around earth with speed ‘v’ in given time ‘t’ of 24 hours.
Radius of circular orbit, r = 42250 × 1000m
Time taken by artificial satellite, t = 24 hours = 24 × 60 × 60s
Distance covered by satellite, s = circumference of circular orbit = 2πr
Speed of satellite, v = \(\frac{2 \pi \mathrm{r}}{\mathrm{t}}\)
= \(\frac{\left[2 \times \frac{22}{7} \times 42250 \times 1000\right]}{24 \times 60 \times 60}\)
= \(\frac{2 \times 22 \times 42250 \times 1000}{7 \times 24 \times 60 \times 60}\)ms^-1
= 3073.74 ms^-1

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter Chapter 7 Diversity in Living Organisms

JAC Class 9th Science Diversity in Living Organisms InText Questions and Answers

Page 80

Question 1.
Why do we classify organisms?
Answer:
We classify organisms for easier and convenient study of the variety of life forms.

Question 2.
Give three examples of the range of variations that you see in life forms around you.
Answer:
Examples of range of variations observed in daily life are:

1. Variety of living organisms in terms of size ranges from microscopic bacteria to tall trees upto 100.
2. The colour, shape and size of snakes are completely different from those of lizards.
3. The life span of different organisms is also quite varied, e.g., a crow lives for only 15 years, whereas a parrot lives for about 140 years.

Page 82

Question 1.
Which, do you think, is a more basic characteristic for classifying organisms?
(a) the place where they live.
(b) the kind of cells they are made o(f) Why?
Answer:
Classification based on the kind of cells they are made of is more basic as there can be wide variations in organisms living in a given place. Hence it cannot be a characteristic for classifying organisms.

Question 2.
What is the primary characteristic on which the first division of organisms is made?
Answer:
Whether organism is a eukaryotic cell, i.e., has membrane – bound cell organelles, or is a prokaryotic cell, i.e., does not have membrane – bound cell organelles.

Question 3.
On what bases are plants and animals put into different categories?
Answer:
Plants and animals are put into different categories on the bases of whether the organisms produce their own food through the process of photosynthesis (plants) or organisms get food from outside (animals). relatively recently. There is a possibility that advanced or younger organisms have undergone complexity in body structure during evolution to compete and survive.

Page 83

Question 1.
Which organisms are called primitive and how are they different from the so – called advanced organisms?
Answer:
Organisms with simple cellular structure and no division of labour are called primitive. Advanced organisms like mammals, have millions of cells and have different organs and organ systems for different biological functions.

Question 2.
Will advanced organisms be the same as complex organisms? Why?
Answer:
Yes, advanced organisms will be the same as complex organisms. This is because the advanced organisms have acquired their particular body designs relatively recently. There is a possibility that advanced or younger organisms have undergone complexity in body structure during evolution to compete and survive.

Page 85

Question 1.
What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?
Answer:
It is the presence or absence of a well defined nucleus. Monera has no nuclear membrane, while Protista shows well defined nucleus.

Question 2.
In which kingdom will you place an organism which is single-celled, eukaryotic and photosynthetic?
Answer:
Kingdom Protista.

Question 3.
In the hierarchy of classification, which group will have the smallest number of organisms with a maximum number of characteristics in common and which will have the largest number of organisms?
Answer:
Organisms belonging to a species will have the smallest number of organisms with a maximum number of characteristics in common. Kingdom will have the largest number of organisms.

Page 88

Question 1.
Which division among plants has the simplest organisms?
Answer:
Thallophyta or algae.

Question 2.
How are pteridophytes different from the phanerogams?
Answer:
Pteridophytes have naked embryo and inconspicuous reproductive organs whereas phanerogams have well differentiated reproductive organs and covered embryo.

Question 3.
How do gymnosperms and angiosperms differ from each other?
Answer:
In gymnosperms, reproductive organs are present in cones. The plants bear naked seeds. In angiosperms, reproductive organs are flowers. The seeds are enclosed within fruit.

Question 1.
How do poriferan animals differ from coelenterate animals?
Answer:

Poriferans	Coelenterates
1. Poriferan animals have many small pores, called ostia, all over the body and a large opening at the top.	1. Coelenterate animals lack ostia and have only one opening.
2. They have canal system for circulating water throughout the body.	2. They do not have water canal system in the body.
3. External skeleton is present.	3. Skeleton is absent.
4. Their body is less differentiated.	4. Their body is more differentiated.
5. Tentacles are absent.	5. Tentacles are present.

Question 2.
How do annelid animals differ from arthropods?
Answer:

Arthropods	Annelids
1. Exoskeleton pres – ent.	1. No skeleton
2. Body is segmented into head, thorax and abdomen.	2. Body is segmented into rings.
3. Sexes are usually separate.	3. Sexes may be united (hermaphrodites) or separate.

Question 3.

What are the differences between amphibians and reptiles?
Answer:

Amphibians	Reptiles
1. Adapted to live in water and land, can breathe by skin in water.	1. Can live in water but need to come to surface to breathe oxygen.
2. Skin is moist and soft.	2. Skin is dry and has scales.
3. Respiration is either through gills or lungs.	3. Respiration is through lungs.
4. Move by hopping	4. Move by crawling.
5. Development is indirect through tadpole stage.	5. Development is direct with no intermediate stage in life cycle.

Question 4.
What are the differences between animals belonging to the Aves group and those in the Mammalia group?
Answer:

Aves	Annelids
1. Aves have beak.	1. Mammalia do not have beak.
2. Their body is cov – ered with feathers. with hair. Feathers are absent.	2. Their body is covered
3. Forelimbs are modi	3. Forelimbs are not modified into wings flight. as in birds. However, forelimbs may be modified for various purposes.

 

Aves	Mammalia
1. They lay eggs	1. Most of the mammals produce young ones.
2. They do not have glands to produce mammary glands to produce milk for milk for young ones.	2.They have mammary glands to produce milk for milk for young ones.

JAC Class 9th Science Diversity in Living Organisms Textbook Questions and Answers

Question 1.
What are the advantages of classifying organisms?
Answer:
Following are the advantages of classification:

1. Classification helps us identify the living organisms easily.
2. It makes the study of a wide variety of organisms easy and systematic.
3. It makes the study of similarities and dissimilarities among organisms easy.
4. It enables us to understand how complex organisms evolved over time.
5. Classification helps us understand the inter – relationships among different groups.
6. It forms the basis of other branches of bio – sciences like bio-geography, environmental biology, ecology, etc.
7.  It also provides a systematic way to identify known and unknown organisms.
8. Classification systems are accepted internationally. This aids communication between scientists.

Question 2.
How would you choose between two characteristics to be used for developing a hierarchy in classification?
Answer:
The characteristic which is dependent on the previous one and would decide the variety in the next level should be chosen for developing a hierarchy in classification.

1. Presence of vertebral column in human beings can be considered under vertebrata.
2. Presence of four limbs makes them the members of tetrapoda.
3. Presence of mammary glands classifies them under mammalia.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:
The bases for grouping organisms into five kingdoms are as follows:

1. Cellular structure: The two major divisions based on the cellular structure within living things are prokaryotes and eukaryotes.
2. Number of cells: It divides organisms into unicellular and multicellular.
3. Mode and source of nutrition: Organisms are divided, based on the mode of nutrition, into autotrophis and heterotrophic.
4. Presence or absence of cell wall: It divides organisms into plants and animals.
5. Phylogenetic relationship and body organisation: Based on body organisation and evolution, organisms are divided into simple and complex organisms.

Question 4.
What are the major divisions in kingdom Plantae? What is the basis for these divisions?
Answer:
Divisions in kingdom Plantae are:
1. Thallophyta,
2. Bryophyta,
3. Pteridophyta,
4. Gymnosperms,
5. Angiosperms.
Bases for classification of kingdom Plantae into these divisions are:

1. Whether the plant body has well differentiated and distinct components or not.
2. Whether the plant body has special tissues for the transport of water and other substances within it or not.
3. Whether the plant bears the seeds or not.
4. Whether the seeds are enclosed within fruits or not.

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?
Answer:
The criteria for deciding divisions in plants are the presence or absence of seeds and flowers, differentiation of body parts, presence or absence of specialised vascular tissues and nature of the seed The criteria for subdivisions among animals are the presence or absence of notochord and coelom, position of nerve cord, gill slits, body segmentation, habitat and oviparity or viviparity.

Question 6.
Explain how animals in Vertebrata are classified into further subgroups.
Answer:
The subphylum Vertebrata has been classified into two divisions, viz. Agnatha and Gnathostomata, on the basis of presence or absence of jaws and paired appendages. Agnatha does not have jaws and paired appendages while Gnathostomata bear jaws and paired appendages. These division have further been classified into Pisces, Amphibia, Reptilia, Aves and Mammalia.
1. The major characteristics used to classify these groups are as follows:

• The kind of exoskeleton or endoskeleton,
• The kind of respiratory organs,
• The method of reproduction and giving birth to young ones.

2. The classes with their characteristics are as follows:

1. Exoskeleton of scales: endoskeleton of cartilage or bones; breathing through gills Pisces (fishes).
2. Breathing through gills only in larva: skin slimy – Amphibia.
3. Exoskeleton of scales: laying eggs outside the water – Reptilia.
4. Exoskeleton of feathers: lay eggs; flight possible – Aves (Birds).
5. Exoskeleton of hair: external ears give birth to young ones – Mammalia.

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 6 Tissues

JAC Class 9th Science Tissues InText Questions and Answers

Page 69

Question 1.
What is a tissue?
Answer:
A group of cells that are similar in structure and work together to achieve a particular function is called a tissue.

Question 2.
What is the utility of tissues in multicellular organisms?
Answer:
In multicellular organisms, different types of tissues perform different functions. Since a particular group of cells carries out only a particular function, these cells do it very efficiently. So, multicellular organisms possess a definite division of labour.

Page 74

Question 1.
Name types of simple tissues.
Answer:
The types of simple tissues are parenchyma, collenchyma and sclerenchyma.

Question 2.
Where is apical meristem found?
Answer:
Apical meristem is found at the tip of root or shoot of the plant.

Question 3.
Which tissue makes up the husk of coconut?
Answer:
The husk of coconut is made of sclerenchyma tissue.

Question 4.
What are the constituents of phloem?
Answer:
Phloem is made up of four types of elements sieve tubes, companion cells, phloem fibres and phloem parenchyma.

Page 77

Question 1.
Name the tissue responsible for movement in our body.
Answer:
The combination of both muscular tissue and nervous tissue is responsible for movement in our body.

Question 2.
What does a neuron look like?
A neuron consists of a cell body with a nucleus and cytoplasm. It has two important extensions known as the axon and dendrites. An axon is a long thread – like extension of nerve cells that transmits impulses away from the cell body. Dendrites, on the other hand, are thread-like extensions of cell body that receive nerve impulses. Thus, the axon transmits impulses away from the cell body, whereas the dendrite receives nerve impulses. This coordinated function helps in transmitting impulses very quickly.

Question 3.
Give three features of cardiac muscles.
Answer:
Three features of cardiac muscles are as follows:

1. Cardiac muscles are involuntary muscles that contract rapidly but do not get fatigued.
2. The cells of cardiac muscles are cylindrical, branched and uninucleate.
3. They control the contraction and relaxation of the heart.

Question 4.
What are the functions of areolar tissues?
Answer:
Areolar tissue helps in supporting internal organs. It helps in repairing the tissues of the skin and muscles.

JAC Class 9th Science Tissues Textbook Questions and Answers

Question 1.
Define the term ‘tissue’.
Answer:
Group of cells that are similar in structure and perform same function is called a tissue.

Question 2.
How many types of elements together make up the xylem tissue? Name them.
Answer:
Tracheids, vessels, xylem parenchyma, and xylem fibres.

Question 3.
How are simple tissues different from complex tissues in plants?
Answer:
Simple tissues are made up of one type of cells which coordinate to perform a common function. Complex tissues are made up of more than one type of cells. All these coordinate to perform a common function.

Question 4.
Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell wall.
Answer:

Parenchyma	Collenchyma	Sclerenchyma
Cell walls are relatively thin, and the cells in parenchyma tissues are loosely packed.	The cell wall is irregularly thickened at the corners and there is very little space between the cells.	The cell walls are uniformly thickened and there is no intercellular space.
The cell wall in this tissue is made up of cellulose.	Pectin and hemi – cellulose are the major constituents of the cell wall.	An additional layer of the cell wall composed mainly of lignin is found.

Question 5.
What are the functions of the stomata?
Answer:
The outermost layer of the cell is called epidermis and is very porous. These pores are called stomata The stomata help in transpiration and exchange of gases.

Question 6.
Diagrammatically show the difference between the three types of muscle fibres.
Answer:

1. Striated muscles
(a) They are connected to bones (skeletal muscles).

(b) They are voluntary muscles.

(c) The cells are long, cylindrical with many nuclei and are unbranched.

2. Smooth muscles
(a) They are found in alimentary canal and lungs.
(b) They are involuntary muscles.

(c) They are spindle in shape and have single nucleus.

3. Cardiac muscles
(a) They are found in heart.
(b) They are involuntary in action.

(c) They are branched and have one nucleus.

Question 7.
What is the specific function of cardiac muscle?
Answer:
The specific function of cardiac muscle is to contract and relax rhythmically throughout life. Their rhythmic contraction and relaxation helps in pumping action of heart.

Question 8.
Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.
Answer:

Striated muscles	Unstriated muscles	Cardiacmuscles
Cells are long, cylindrical, non – tapering and are unbranched.	Cells are long with tapering ends and are unbranched.	Cells are non – tapering and cylindrical in shape and are branched.
In hands, legs and skeletal muscles.	The wall of stomach,intestine, ureter, bronchi, etc.	In the heart.
Light and dark bands are present.	Absent	Present but less prominent.

Question 9.
Draw a labelled diagram of a neuron.
Answer:

Question 10.
Name the following:
(a) Tissue that forms the inner lining of our mouth
(b) Tissue that connects muscle to bone in humans
(c) Tissue that transports food in plants
(d) Tissue that stores fat in our body
(e) Connective tissue with a fluid matrix
(f) Tissue present in the brain
Answer:

(a) Tissue that forms the inner lining of our mouth	Epithelial tissue (squamous epithelium)
(b) Tissue that connects muscle to bone in humans	Tendons
(c) Tissue that transports food in plants	Phloem
(d) Tissue that stores fat in our body	Adipose tissue
(e) Connective tissue with a fluid matrix	Blood
(f) Tissue present in the brain	Nervous tissue

Question 11.
Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:
Skin – Striated squamous epithelial tissue Bark of tree – Simple permanent tissues Bone – Connective tissue Lining of kidney tubule – Cuboidal epithelial tissue

Question 12.
Name the regions in which parenchyma tissue is present.
Answer:
Parenchyma is found in cortex and pith of root and stem. When it contains chlorophyll, it is called chlorenchyma, found in green leaves.

Question 13.
What is the role of epidermis in plants?
Answer:
Epidermis is present on the outer surface of the entire plant body. The cells of the epidermal tissues form a continuous layer without any intercellular space. It performs the following important functions:

• It is a protective tissue of the plant body.
• It protects the plant against mechanical injury.
• It allows exchange of gases through the stomata.

Question 14.
How does the cork act as a protective tissue?
Answer:
Cork acts as a protective tissue because its cells are dead and compactly arranged without intercellular spaces. They have deposition of suberin on the walls that makes them impervious to gases and water.

Question 15.

Answer:

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 5 The Fundamental Unit of Life

JAC Class 9th Science The Fundamental Unit of Life InText Questions and Answers

Page 59

Question 1.
Who discovered cells and how?
Answer:
Robert Hooke discovered cells in 1665. He observed the cells in thin slices of cork. They appeared like small compartments when viewed under the microscope.

Question 2.
Why is the cell called structural and functional unit of life?
Answer:
A cell is capable of carrying out all the life functions such as nutrition, excretion, respiration, etc Hence, a cell is called the functional unit of life. Additionally, the cell is the smallest unit of life and all the living beings are made up of cells. Therefore, a cell is called the structural unit of life.

Page 61

Question 1.
How do substances like CO, and water move in and out of the cell? Discuss.
Answer:
Substances move in and out of the cell because of diffusion. Diffusion is the random movement of particles in order to attain concentration equilibrium. The movement of water molecules through a semi – permeable membrane is called osmosis. It is important to note that plasma membrane is a semi – permeable membrane. Water always moves from its higher concentration to its lower concentration.

Question 2.
Why is the plasma membrane called a selectively permeable membrane?
Answer:
Plasma membrane allows passage to some selected substances. Hence, it is called a selectively permeable or semi- permeable membrane.

Page 63

Question 1.
Fill in the gaps in the following table illustrating difference between prokaryotic and eukaryotic cells.

Prokaryotic cell	Eukaryotic cell
(a) Size: generally small (1 – 10µm)	(a) Size: generally large (5 – 100µm)
(b) Nuclear region ……………. ………………… …………………and known as …. …………………	(b) Nuclear region: well – defined and surrounded by a nuclear membrane.
(c) Chromosome: single	(c) More than one chromosome.
(d) Membrane bound cell organelles absent.	(d) …………… ………………. ……………….

Answer:

• Prokaryotic cell (b): Nuclear region is poorly defined due to absence of a nuclear membrane and is known as nucleoid.
• Eukaryotic cell (d): Membrane bound cell organelles are present.

Page 65

Question 1.
Can you name the two organelles we have studied that contain their own genetic material?
Answer:
Mitochondria and chloroplast contain their own genetic material.

Question 2:
If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen?
Answer:
Various parts of a cell are responsible for various functions. They work in tandem
to continue life in the cell. In case, the organisation of a cell is destroyed due to some physical or chemical influence, the cell will die.

Question 3.
Why are lysosomes known as suicide bags?
Answer:
Lysosomes contain digestive enzymes. In case of a rupture in lysosomes, the enzymes are released in the cytoplasm and end up digesting the contents of the cell. This results in cell death. Due to this, lysosomes are also known as suicide bags of cells.

Question 4.
Where are proteins synthesised inside the cell?
Answer:
The proteins are synthesised in the ribosomes present on RER (rough endoplasmic reticulum). They are also known as protein factories.

JAC Class 9th Science The Fundamental Unit of Life Textbook Questions and Answers

Question 1.
Make a comparison and write the ways in which plant cells are different from animal cells.
Answer:

Plant cell	Animal cell
(a) Plant cells have cell wall.	(a) Animal cells don’t have a cell wall.
(b) They contain chloroplast.	(b) They don’t have chloroplasts.
(c) They do not have centrioles.	(c) Centriole is present in them.
(d) Vacuole is large and present in centre of the cell.	(d) Vacuoles are numerous and small.
(e) Nucleus is present at the periphery of the plant cell.	(e) Nucleus is present in the centre of the animal cell.

Question 2.
How is a prokaryotic cell different from a eukaryotic cell?
Answer:

Prokaryotic cell	Eukaryotic cell
(a) Cell size is generally small (1 – 10µm).	(a) Cell is , generally large (5 – 100µm).
(b) Nuclear region, called nucleoid, is not surrounded by a nuclear membrane.	(b) Nuclear material is surrounded by a nuclear membrane.
(c) Nucleolus is absent.	(c) Nucleolus is present.
(d) Membrane bound cell organelles are absent.	(d) Cell organelles bounded by membrane are present.
(e) Cell division by fission or budding (no mitosis).	(e) Cell division by mitosis or meiosis.

Question 3.
What would happen if the plasma membrane ruptures or breaks down?
Answer:
Plasma membrane provides a container for the cell organelles and cytoplasm. Moreover, plasma membrane also protects the contents of a cell from external environment. In case the plasma membrane ruptures or breaks down, the cell contents would be exposed to the external environment. This would prove lethal for the cell and the cell would cease to exist.

Question 4.
What would happen to the life of a cell if there was no Golgi apparatus?
Answer:
If Golgi apparatus is not present, the packaging and transport of materials would cease to happen. Various substances would not be transformed into useful materials of the cell. Plasma membrane will also be affected The secretory activities of the cell would also cease to occur. Hence, the cell will eventually die off.

Question 5.
Which organelle is known as the powerhouse of the cell? Why?
Answer:
Mitochondrion is known as the powerhouse of the cell. The reason for this is the fact that cellular respiration takes place in mitochondria and the energy released after that, gets stored in mitochondria in the form of ATP. As ATP instantly provides energy, they are called energy currency of the cell.

Question 6.
Where do the lipids and proteins constituting the cell membrane get synthesised?
Answer:
Lipids are synthesised in the endoplasmic reticulum (smooth ER). Protein is synthesised in ribosomes which are usually present on the rough ER.

Question 7.
How does Amoeba obtain its food?
Answer:
Amoeba obtains its food through a process called phagocytosis. The cell membrane of amoeba is projected into numerous finger – like outgrowths called pseudopodia Amoeba surrounds a food particle by pseudopodia and makes a food vacuole after engulfing the food.

Question 8.
What is osmosis?
Answer:
Osmosis is the process of movement of water molecules from a region of higher water concentration, through a semi – permeable membrane, to a region of lower water concentration.

Question 9.
Carry out the following osmosis experiment:
Take four peeled potato halves and scoop each one out to make potato cups, one of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now,
(a) Keep cup A empty.
(b) Put one teaspoon sugar in cup B.
(c) Put one teaspoon salt in cup C.
(d) Put one teaspoon sugar in the boiled potato cup D.
Keep these for two hours. Then, observe the four potato cups and answer the following:
1. Explain why water gathers in the hollowed portion of B and C.
2. Why is potato A necessary for this experiment?
3. Explain why water does not gather in the hollowed out portions of A and D.
Answer:

1. Water gathers in the hollowed portion of potato B and C because: Living plasma membrane of potato cells act as semi – permeable membrane. There is higher concentration of water in trough than the hollowed portions of B and C. So water, by the process of osmosis, moves into the hollowed portions of potato cups B and C.
2. Potato cup A is kept empty to act as control set – up.
3.  potato is necessary because
   • (a) As the potato cup A is empty, water does not gather in hollowed out portions of A.
   • (b) In the potato cup D, the potato cell membrane lose semi – permeability due to boiling. So, no water movement occurs from the trough into the potato cup D.

Question 10.
Which type of cell division is required for growth and repair of body and which type is involved in formation of gametes?
Answer:
Growth and repair of body – Mitosis. Formation of gametes – Meiosis.

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 4 Structure of the Atom

JAC Class 9th Science Structure of the Atom InText Questions and Answers

Page 47

Question 1.
What are canal rays?
Answer:
Canal rays are radiations which are positively charged. They were the key in the discovery of proton, another positively charged sub – atomic particle.

Question 2.
If an atom contains one electron and one proton, will it carry any charge or not?
Answer:
Since an electron is a negatively charged particle and the proton, a positively charged one, the net charge becomes neutral as both particles neutralise each

Page 48

Question 1.
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Answer:
According to Thomson’s model of an atom:

1. an atom consists of a positively charged sphere in which the negatively charged electrons are embedded.
2. the number of protons and electrons are equal in an atom, thereby, neutralising their charge keeping the overall system neutral.

Question 2.
On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom?
Answer:
As per Rutherford’s model of atom, the positively charged protons are the ones that are present in the nucleus of an atom.

Question 3.
Draw a sketch of Bohr’s model of an atom with three shells.
Answer:
Bohr’s model of an atom with three shells:

Question 4.
What do you think would be the observation if the α – particle scattering experiment is carried out using a foil of a metal other than gold?
Answer:
When any other metal foil is used instead of gold, the observation would remain the same. This is because the structure of an atom, when considered individually, remains the same.

Page 49

Question 1.
Name the three subatomic particles of an atom.
Answer:
An atom consists of three subatomic particles:

1. Protons : Positively charged
2. Electrons : Negatively charged
3. Neutrons : Neutral in nature (no charge)

Question 2.
Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?
Answer:
Atomic mass of He = 4u
Atomic mass = number of (protons + neutrons)
4 = 2 + number of neutrons Number of neutrons = 4 – 2
= 2 Helium atom has 2 neutrons.

Page 50

Question 1.
Write the distribution of electrons in carbon and sodium atoms.
Answer:
1. Carbon atom:
Atomic number = 6
Number of protons = 6 = Number of electrons
Distribution of electrons = K – 2, L – 4.

2. Sodium atom:
Atomic number = 11
Number of protons = 11 = Number of electrons
Distribution = K – 2, L – 8, M – 1.

Question 2.
If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Answer:
Number of electrons K shell can hold = 2 Number of electrons L shell can hold = 8
Hence, when both the shells are full, the total number of electrons present = 2 + 8 ⇒ 10 electrons.

Page 52

Question 1.
How will you find the valency of chlorine, sulphur and magnesium?
Answer:
Valency is the combining capacity of the atom of an element.
1. Chlorine: Atomic number = 17
Number of protons = Number of electrons = 17
Distribution: K – 2, L – 8, M – 7 Chlorine needs 1 electron to complete its outermost orbit shell. Its valency is – 1 (gains 1 electron).

2. Sulphur: Atomic number = 16 Number of protons = Number of electrons = 16
Distribution: K – 2, L – 8, M – 6 Sulphur needs 2 electrons to complete its outermost shell. Its valency is – 2 (gains 2 electrons).

3. Magnesium: Atomic number = 12
Number of protons = Number of electrons = 12
Distribution: K – 2, L – 8, M – 2 Magnesium needs to donate 2 electrons from its outermost shell to become stable. Its valency is + 2 (donate 2 electrons).

Question 2.
If the number of electrons in an atom is 8 and number of protons is also 8, then
(a) what is the atomic number of the atom?
(b) what is the charge on the atom?
Answer:
Number of electrons = 8, Number of protons = 8
(a) Atomic number of the atom = Number of protons = 8

(b) As the number of electrons is equal to the number of protons on the atom, their charges neutralise each other. Therefore, the atom does not possess any charge.

Question 3.
With the help of the table 4.1, find out the mass number of oxygen and sulphur atom.
Answer:

1. Oxygen: Number of protons = 8 Number of neutrons = 8 Atomic number = 8
   Mass number = Number of (protons + neutrons) = 8 + 8 = 16u.
2. Sulphur: Number of protons = 16 Number of neutrons = 16 Atomic number = 16
   Mass number = Number of (protons + neutrons) = 16 + 16 = 32u.

Page 53

Question 1.
For the symbols H, D and T, tabulate three subatomic particles found in each of them.
Answer:
The symbols H, D and T, tabulate three subatomic particles:

Element	H (Protium) (11H)	D (Deute – rium) (2H)	T (Tritium) (31H)
Number of protons	1	1	1
Number of electrons	1	1	1
Number of neutrons	Nill	1	2

Question 2.
Write the electronic configuration of any one pair of isotopes and isobars.
Answer:
(a) Isotopes : Isotopes are atoms which have the same number of protons but the number of neutrons differs. This leads to the variation in mass number too.
Example :
The simplest example is the carbon molecule which exists as \({ }_{6} \mathrm{C}^{12}\) and \({ }_{6} \mathrm{C}^{14}\) but when their electronic configuration is noticed, both have K – 2, L – 4.

(b) Isobars : Isobars are the atoms having the same mass number but differ in the atomic numbers. Electronic configuration of an isobar pair is as follows:
Example :
⁴⁰Ca₂₀: K – 2, L – 8, M – 8, N – 2
⁴⁰Ar₁₈ : K – 2, L – 8, M – 8

JAC Class 9th Science Atoms and Molecules Textbook Questions and Answers

Question 1.
Compare the properties of electrons, protons and neutrons.
Answer:
The properties of electrons, protons and neutrons:

Electrons	Protons	Neutrons
Negatively charged	Positively charged	No charge
Present outside the nucleus	Present within the nucleus	Present inside the nucleus of an atom
Negligible mass	1 a.m.u.	1 a.m.u.
Get attracted towards positive electrode	Get attracted towards negative electrode	Do not get attracted to any charged particle.

Question 2.
What are the limitations of J.J. Thomson’s model of an atom?
Answer:
According to J.J. Thomson’s model of an atom, the electrons are embedded all over in the positively charged sphere. But experiments done by the other scientists show that protons are present only in the centre of the atom and electrons are distributed around it.

Question 3.
What are the limitations of Rutherford’s model of atom?
Charged bodies, when move in circular motion, emit radiations. Thus, electrons revolving round the nucleus, as suggested by Rutherford, will lose energy and will come closer and closer to the nucleus and will finally merge into the nucleus. This means that atoms are quite unstable which is not true.

Question 4.
Describe Bohr’s model of atom.
Answer:
(a) The nucleus of an atom is present in the centre.
(b) Negatively charged electrons revolve around this nucleus.
(c) Discrete orbits of electrons are present inside the atom.
(d) While revolving in the orbit, the electrons do not radiate energy.
(e) These discrete orbits are represented as K, L, M, N orbits or denoted by

Question 5.
Compare all the proposed models of an atom given in this chapter.
Answer:

Thomson’s atomic model	Rutherford’s atomic model	Bohr’s atomic model
Sphere of positive charge	Sphere of positive charge in centre is called nucleus. All mass of an atom resides in the nucleus.	Positive charge in centre is called nucleus.
Electrons are spread randomly all over in the sphere.	Electrons revolve around the nucleus in well defined orbits.	Electrons revolve in discrete orbits and do not radiate energy.
Positive charge = negative charge.	Size of nucleus is very small as compared to the size of atom.	The orbits were termed as energy shells
Atom is electricity – neutral.	Rutherford’s atomic model	labelled as K, L, M, N or n = 1,2, 3, 4.

Question 6.
Summarise the rules for writing of distribution of electrons in various shells for the first 18 elements.
Answer:
(a) Generally, the maximum number of electrons that can be accommodated in a shell is given by the formula: 2n², where n = 1, 2, 3 … .

(b) Maximum number of electrons in different shells are:
K shell (n = 1), 2n² = 2(1)² = 2
L shell (n = 2), 2n² = 2(2)² = 8
M shell (n = 3), 2n² = 2(3)² = 18
N shell (n = 4), 2n² = 2(4)² = 32.

(c) The electrons are not taken in unless the inner shells are completely filled.

Question 7
Define valency by taking examples of silicon and oxygen.
Answer:
Valency is the combining capacity of an atom. Take the examples of silicon and oxygen:

Oxygen	Silicon
Atomic Number : 8	Atomic Number : 14
Electronic Config : K – 2, L – 6	Electronic Config : K – 2, L – 8, M – 4
Valence electrons : 6	Valence electrons : 4
Valency : 8 – 6 = 2	Valency : 8 – 4 = 4

In the atoms of oxygen, the valence electrons are 6.
To fill the orbit, 2 electrons are required In the atom of silicon, the valence electrons are 4.
To fill this orbit 4 electrons are required Hence, the  combining capacity of oxygen is 2 and of silicon is 4, i.e., valency of oxygen = 2 and valency of silicon = 4.

Question 8.
Explain with examples.
(a) Atomic number
(b) Mass number
(c) Isotopes
(d) Isobars
Give any two uses of isotopes.
Answer:
(a) Atomic number : The atomic number of an element is the total number of protons present in the atom of that element. For example, nitrogen has 7 protons in its atom. Thus, the atomic number of nitrogen is 7.

(b) Mass number : The mass number of an element is the. sum of the number of protons and neutrons present in the atom of that element. For example, the atom of boron has 5 protons and 6 neutrons. So, the mass number of boron is 5 + 6 =11.

(c) Isotopes : These are atoms of the same element having the same atomic number, but different mass numbers. For example, chlorine has two isotopes with atomic number 17 but mass numbers 35 and 37 represented by \({ }_{17}^{35} \mathrm{Cl}\) \({ }_{17}^{37} \mathrm{Cl}\).

(d) Isobars : These are atoms having the same mass number but different atomic number, i.e, isobars are atoms of different elements having the same mass number. For example,
Ca has atomic number 20 and Ar has atomic number 18 but both of them have mass number 40 represented by \({ }_{20}^{40} \mathrm{Ca}\) and \({ }_{18}^{40} \mathrm{Ar}\) respectively.

Two uses of isotopes:

1. An isotope of uranium is used as a fuel in nuclear reactors.
2. An isotope of cobalt is used in the treatment of cancer.

Question 9.
Na⁺ has completely filled K and L shells. Explain.
Answer:
The atomic number of sodium is 11. So, neutral sodium atom has 11 electrons and its electronic configuration is 2, 8, 1. But Na⁺ has 10 electrons. Out of 10, K – shell contains 2 and L – shell has 8 electrons. Thus, Na⁺ has completely filled K and L shells.

Question 10.
If bromine atom is available in the 79 form of, say, two isotopes \({ }_{35}^{79} \mathrm{Br}\) (49.7%) and \({ }_{35}^{81} \mathrm{Br}\) (50.3%), calculate the average atomic mass of bromine atom.
Answer:
The atomic masses of two isotopic atoms are 79 (49.7%) and 81 (50.3%).
Thus, total mass = (79 × \(\frac{49.7}{100} \) ) + (81 × \(\frac{50.3}{100} \)) = 39.263 + 40.743 = 80.006u.

Question 11.
The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes \({ }_{8}^{16} \mathrm{X}\) and \({ }_{8}^{18} \mathbf{X}\) in the sample?
Answer:
It is given that the average atomic mass of the sample of element X is 16.2 u. Let the % of isotope \({ }_{8}^{16} \mathrm{X}\) be y%. Thus, the % of isotopes \({ }_{8}^{18} \mathbf{X}\) will be (100 – y) %. Therefore,
16 × \(\frac{\mathrm{y}}{100}\) + \(\frac{18 \times(100-y)}{100}\) = 16.2
\(\frac{16 y}{100}\) + \(\frac{18(100-y)}{100}\) = 16.2
\(\frac{16 y+1800-18 y}{100}\) = 16.2
1800 – 2y = 1620 or 2y = 1800 – 1620 = y – 90
Therefore, the % of isotope \({ }_{8}^{16} \mathrm{X}\) is 90%.
And, the % of the isotope \({ }_{8}^{18} \mathbf{X}\) is (100 – 90) % = 10%.

Question 12.
If Z = 3, what would be the valency of the element? Also, name the element.
Answer:
Z = atomic number = 3 (given) Electronic configuration = K – 2, L – 1 Thus, valency = 1 The element with atomic number 3 is lithium.

Question 13.
The composition of the nuclei of two atomic species X and Y are given as under

	X	Y
Protons	6	6
Neutrons	6	8

Give the mass number of X and Y. What is the relation between the two species?
Answer:
Mass number of X = Protons + Neutrons = 6 + 6 = 12
Mass number of Y = Protons + Neutrons = 6 + 8 = 14
Since the atomic numbers of both the species are the same, they are the same element. Also, since they have different number of neutrons, their mass number is different and they are the isotopes.

Question 14.
For the following statements, write T for true and F for false.
(a) J. J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1/2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer:
(a) False
(b) False
(c) True
(d) False
Put tick against correct choice and cross against wrong choice in questions 15, 16 and 17.

Question 15.
Rutherford’s alpha – particle scattering experiment was responsible for the discovery of:
(a) Atomic nucleus
(b) Proton
(c) Electron
(d) Neutron
Answer:
(a) Atomic nucleus.

Question 16.
Isotopes of an element have:
(a) the same physical properties
(b) different number of neutrons
(c) different number of protons
(d) different atomic number
Answer:
(b) different number of neutrons.

Question 17.
Number of valence electrons in Cl^– ion are:
(a) 16
(b) 8
(c) 17
(d) 18
Answer:
(b) 8.

Question 18:
Which one of the following is a correct electronic configuration of sodium?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1
Answer:
(d) 2, 8, 1.

Question 19.
Complete the following table.

Atomic number	Mass number	Number of neutrons
9	–	10
16	32	–
–	24	–
–	2	–
–	1	0
Number of protons	Number of electrons	Name of the atomic species
–	–	–
12	–	Sulphur
1	–	–

Answer:

Atomic number	Mass number	Number of neutrons
9	19	10
16	32	16
12	24	12
1	2	1
1	1	0
Number of protons	Number of electrons	Name of the atomic species
9	9	Fluorine
16	16	Sulphur
12	12	Magnesium
1	1	Hydrogen
1	0	Deuterium
1	1	Hydrogen
1	0	Protium

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 3 Atoms and Molecules

JAC Class 9th Science Atoms and Molecules InText Questions and Answers

Page 32

Question 1.
In a reaction, 5.3g of sodium carbonate reacted with 6g of ethanoic acid. The products were 2.2g of carbon dioxide, 0.9g water and 8.2g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. Sodium carbonate + Acetic acid → Sodium acetate + Carbon dioxide + Water
Answer:

This shows that during a chemical reaction, mass of reactants mass of products. Hence the given observation are in agreement with the law of conservation of mass.

Question 2.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3g of hydrogen gas?
Answer:
Ratio of H : O by mass in water is:
Hydrogen : Oxygen → H₂O ⇒ 1 : 8 = 3 : x or x = 24g
24 g of oxygen gas would be rcquircd to react completely with 3g of hydrogen.

Question 3.
Which postulate of Dalton’s atomic theory is the result of the Law of consersatlon of mass?
Answer:
Awms are indivisible particles, which can neither be created nor destroyed in a chemical rcacuon.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
“The relative number and kind of atoms are constant in a given compound”

Page 35

Question 1.
Define the atonik mass unit.
Answer:
A unit of mass used in express atomic and molecular weights, equal to one twelfth (1/12th.) of the mass of an stom of carbon – 12 The relative atomic masses of all elements have been found with respect to an atom of carbon – 12.

Question 2.
Why Is it not possible to see an atom with naked eyes?
Answer:
The size of an atom is very small, Further, atoms of most elements do not exist independently. The radius of an atom is of the order of 10^-10m.

Page 39

Question 1.
Write down the formulae of
(a) sodium oxide
(b) aluminium chloride
(c) sodium sulphide
(d) magnesium hydroxide
Answer:
(a) Sodium oxide — Na₂O
(b) Aluminium chloride — AlCl₃
(c) Sodium sulphide — Na₂S
(d) Magnesium hydroxide — Mg(OH)₂

Question 2.
Write down the names of compounds represented by the following formulae:
(a) Al₂(SO₄)₃
(b) CaCl₂
(c) K₂SO₄
(d) KNO₃
(e) CaCO₃
Answer:
(a) Al₂(SO₄)₃ : Aluminium sulphate
(b) CaCl₂ : Calcium chloride
(c) K₂SO₂ : Potassium sulphate
(d) KNO₃ : Potassium nitrate
(e) CaCO₂ : Calcium carbonate

Question 3.
What Is meant by the term chemical formula?
Answer:
The chemical formula of a compound is the symbolic representation of its composition. It gives the number and kinds of atoms which are chemically united in a given compound. For example, chemical formula of sodium chloride is NaCl.

Question 4.
How many atoms are present in a
(a) H₂S molecule and
(b) \(\mathrm{PO}_{4}^{3-}\) ion?
Answer:
(a) H₂S → 3 atoms are present: 2 atoms of hydrogen and 1atom of sulphur.
(b) \(\mathrm{PO}_{4}^{3-}\) → 5 atoms are present: 1 atom of phosphorus and 4 atoms of oxygen.

Page 40

Question 1.
Calculate the molecular masses of H₂, O₂ C₁₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
Answer:
The molecular masses are:
H₂ → 1 + 1 = 2u
O₂ → 16 + 16 = 32u
Cl₂ → 35.5 + 35.5 = 71u
CO₂ → p 12 + 32 = 44u
CH₄ → 12 + 4 = 16u
C₂H₆ → (12 × 2) + (1 × 6) = 30u
C₂H₄ → (12 × 2) + (1 × 4) = 28u
NH₃ → (1 × 14) + (1 × 3) = 17u
CH₃OH → (1 × 12) + (1 × 3) + (1 × 16)(1 × 1) = 32u.

Question 2.
Calculate the formula unit of masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u and O = 16u.
Answer:
The formula Unit mass of:
(a) ZnO = 65u + 16u = 81u
(b) Na₂O = (23u × 2) + 16u = 46u + 16u = 62u
(c) K₂CO₃ = (39u × 2) + 12u + (16u × 3)
= 78u + 12u + 48u
= 138u.

Page 42

Question 1.
If 1 mole of carbon atoms weigh 12grams, what Is the mass (in grams) of 1 atom of carbon?
Answer:
1 mole of carbon atoms = 6.022 × 10²³ atoms
Now, 12/6.022 × 10²³ atoms of carbon weigh = 12g
One atom of carbon weighs = \(\frac{12}{6.023}\) × 10²³
= 1.99 × 10^-23g.

Question 2.
Which has more number of atoms 100 grams of sodium or loo grams of iron (given atomic mass of Na = 23u, Fe 56u)?
Answer:
23 gram atomic unit or 23g sodium (1 mole) = 6.022 × 10²³ atoms
100 gram atomic unit or 100g sodium = \(\frac{6.022 \times 10^{23} \times 100}{23}\)
= 2.617 × 10²⁴ atoms Again 56 gram atomic unit or 56 g iron (1 mole) 6.022 × 10²³
100 gram atomic unit or 100 g iron = \(\frac{6.022 \times 10^{23} \times 100}{56}\)
= 1. 075 × 10²⁴ atoms Thus, 100 g of sodium has more atoms than 100g of iron

JAC Class 9th Science Atoms and Molecules Textbook Questions and Answers

Question 1.
A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:
Percentage (%) of boron in the sample
\(\frac{0.096}{0.24}\) × 100 = 40%
Percentage (%) of oxygen in the sample
\(\frac{0.144}{0.24}\) × 100 = 60%
The sample of compound contains 40% boron and 60% oxygen by weight.

Question 2.
When 3.0g of carbon is burnt in 8.00 g oxygen, 11.00g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon Is burnt In 50.00g of oxygen? Which law of chemical combination will govern your answer?
Answer:
When 3.0g of carbon is burnt in 8.00g oxygen, 11.00g carbon dioxide is produced. It means all of carbon and oxygen are used up and carbon and oxygen are combined in the ratio of 3 : 8 to form carbon dioxide. Thus when there is 3g carbon and 50.0g oxygen, then also only 8g oxygen will be used and 11.0g carbon dioxide will be formed. The remaining oxygen is not used up. This indicates law of defmite proportions which Say that in compounds, the combining elements are present in definite proportions by mass.

Question 3.
What are polyatomic ions? Give examples.
Answer:
The ions which contain more than one atom (same kind or may be of different kind) and behave as a single unit are called polyatomic ions. For example:

1. Ammonium ion NH is a compound ion which is made up of two types of atoms joined together, viz., nitrogen and hydrogen.
2. Carbonate ion CO is a compound ion which is made up of two types of atoms joined together, viz., carbon and oxygen.

Question 4.
Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oside
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:
(a) Magnesium chloride : MgCl
(b) Calcium oxide : CaO
(e) Copper nitrate : Cu(NO₃)₂
(d) Aluminium chloride : AlCl₃
(e) Calcium carbonate: CaCO₃

Question 5.
Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer:
(a) Quick lime : Calcium and oxygen
(b) Hydrogen bromide : Hydrogen and bromine
(c) Baking powder : Sodium, hydrogen. carbon and oxygen
(d) Potassium sulphate : potassium. sulphur and oxygen

Question 6.
Calculate the molar mass of the following substances.
(a) Ethvne, C₂H₂
(b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31u)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃
Answer:
(a) Ethyne, C₂H₂ = (2 × 12) + (2 × 1) = 26g
(b) Sulphur molecule, S₈ = 8 × 32 = 256g
(c) Phosphorus molecules, P₄ = 4 × 31 = 124g
(d) Hydrochloric acid, HCl = (1 × 1) + (1 × 35.5) = 36.5g
(e) Nitric acid, HNO₃ = (1 × 1) + (1 × 14) + (3 × 16) = 63g

Question 7.
What is the muss of
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (atomic mass of aluminium 27u)?
(c) 10 moles of sodium sulphite (Na₂SO₃)?
Answer:
(a) 1 mole of nitrogen atoms 14u = 14g
(b) 4 moles of aluminium atoms = 4 × 27 = 108u = 108g
(c) 1 mole of sodium sulphite = (2 × 23) + (1 × 32) + (3 × 16) = 126u = 126g
10 moles of sodium sulphite = 10 × 126 = 1260g.

Question 8.
Convert Into mole.
(a) 12g of oxygen gas
(b) 20g of water
(c) 22 g of carbon dioxide
Answer:
(a) Given mass of oxygen gas = 12g
Molar mass of oxygen gas (O₂) = 32g
Mole of oxygen gas = \(\frac{12}{32}\)
= 0.375 mole.

(b) Given mass of water = 20g
Molar mass of water(H₂O) (2 × 1) + 16= 18g
Mole of water = \(\frac{20}{18}\)= 1.11 mole.

(c) Given mass of carbon dioxide = 22g
Molar mass of carbon dioxide (CO₂) (1 × 12) + (2 × 16) = 12 + 32 = 44g
Mole of carbon dioxide = \(\frac{22}{44}\)
= 0.5 mole.

Question 9.
What is the mass of
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer:
(a) Mole of oxygen atoms = 0.2 mole
Molar mass of oxygen atoms = 16g
Mass of oxygen atoms 16 × 0.2 = 3.2g.

(b) Mole of water molecule 0.5 mole
Molar mass of water molecules = (2 × 1) + 16 = 18g
Mass of H₂O = 18 × 0.5 = 9g.

Question 10.
Calculate the number of molecules of sulphur (S₈) present In 16 g of solid sulphur.
Answer:
Molar mass of sulphur (S₈) = 256 g = 6.022 × 10₂₃ molecules
Given mass of sulphur = 16g
Molecules of sulphur = \(\frac{16 \times 6.022 \times 10^{23}}{256}\)
\(=\frac{96.35 \times 10^{23}}{256}\) = 0.376 × 10²³ molecules.

Question 11.
Calculate the number of aluminium ions present In 0.051 g of aluminium oxide.
Hint: The mass of an ion is the same as that of an atom of the saine element Atomic mass of Al = 27u
Answer:
1 mole of aluminium oxide, Al₂O₃ = (2 × 27) + (3 × 16) = 102u = 102g
102 g of Al₂O₃ has = 6.022 × 10₂₃ Al₂O₃ molecules
0.051 g of Al₂O₃has = \( \frac{6.022 \times 10^{23} \times 0.051}{102}\)
= 3.01 × 10₂₀ molecules
1 molecule of Al₂O₃ gives = 2 Al³⁺ ions
Hence, 0.051g Al₂O₃ gives = 2 × 3.01 × 10²⁰ Al³⁺ ions
= 6.022 × 10²⁰ aluminium ions.

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 2 Is Matter Around Us Pure

JAC Class 9th Science Is Matter Around Us Pure InText Questions and Answers

Page 15

Question 1.
What is meant by a pure substance?
Answer:
A pure substance consists of a single type of particles and it cannot be separated into other kinds of matter by any separation process.

Question 2.
List the points of difference between ho-mogeneous and heterogeneous mixtures.
Answer:
Differences between homogeneous and heterogeneous mixtures:

Homogeneous mixture	Heterogeneous mixture
It has no visible boundary or boundaries of separation between its constituents.	It has visible boundary boundaries of separation between its constituents.
It has a uniform composition.	It does not have a uniform composition.
They form solutions.	They form suspensions colloids.
The particle size is very small.	The particle size is larger.
For example, sugar + water → sugar solution.	For example, sugar + sand.

Page 18

Question 1.
Differentiate between homogeneous and heterogeneous mixtures with examples.
Answer:
See answer 2 above.

Question 2.
How are sol, solution and suspension different from each other?
Answer:
Comparison among properties of true solution, colloidal solution and suspension:

Property	True solution	Colloidal solution	Suspension
Appearance	Hetero geneous and transparent.	Hetero geneous and translucent.	Hetero geneous and opaque.
Particle size	lnm (10-9 m)	lnm – 1000 nm	1000 lnm (10-6m)
Visibility	Particles are not visible even with a powerful microscope.	Particles can be seen with a high power microscope.	Particles can be seen with naked eyes.
Stability	Stable	Stable	Unstable
Diffusion	Diffuse rapidly	Diffuse slowly	Do not diffuse
Filterability	Passes through filter paper, e.g., sodium chloride dissolved in water.	Passes through filter paper, e.g., blood.	Can be separated by filter paper, e.g., mud water.

Question 3.
To make a saturated solution, 36g of sodium chloride is dissolved in lOOg of water at 293K. Find its concentration at this temperature.
Answer:
Mass of solute (sodium chloride) = 36g
Mass of solvent (water) = 100 g
Mass of solution = Mass of solute + Mass of solvent = 36g + 100g =136g
\(\frac{ Mass of solute}{Mass of solution}\) × 100 ⇒ \(\frac{36 \times 100}{136}\)
= 26.47%

Page 24

Question 1.
How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?
Answer:
Set up the apparatus as shown in the figure. Take the given mixture in a distillation flask. Heat the mixture slowly, keeping a close watch at the thermometer. At a certain point, temperature becomes constant. Petrol vaporises first as it has lower boiling point. It condenses in the condenser and is collected from the condenser outlet. Stop heating when the temperature further starts rising. Kerosene is left behind in the distillation flask.

Question 2.
Name the technique to separate:
(a) Butter from curd
(b) Salt from sea – water
(c) Camphor from salt
Answer:
(a) Butter from curd – Centrifugation
(b)Salt from sea – water – Evaporation
(c) Camphor from salt – Subiimation

Question 3.
What type of mixtures are separated by the technique of crystallisation?
Answer:
Crystallisation technique is used to purify a solid with some impurities in it. For example, purification of salt obtained from sea – water.

Page 25

Question 1.
Classify the following as chemical or physical changes: cutting of trees, melting of butter in a pan, rusting of almirah, boiling of water to form steam, passing of electric current, through water and the water breaking down into hydrogen and oxygen gases. making of fruit salad with raw fruits, dissolving common salt in water, burning of paper and wood
Answer:

1. Physical change:
   • Cutting of trees
   • Melting of butter in a pan
   • Boiling of water to form steam
   • Dissolving common salt in water
   • Making a fruit salad with raw fruits
2. Chemical change:
   • Rusting of almirah
   • Passing of electric current through water and the water breaking down into hydrogen and oxygen gases
   • Burning of paper and wood

Question 2.
Try segregating the things around you as pure substances or mixtures.
Answer:

1. Pure substances: Water, sugar, gold, copper wire, salt, ice, etc.
2. Mixtures: Steel, plastic, paper, talc, milk, air, ink, soda water, lemonade, bread, etc.

JAC Class 9th Science Is Matter Around Us Pure Textbook Questions and Answers

Question 1.
Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.
Answer:
(a) Crystallisation/Evaporation

(b) Sublimation

(c) Filtration

(d) Chromatography

(e) Centrifugation

(f) Separating funnel

(g) Filtration

(h) Magnetic separation

(i) Winnowing

(j) Decantation/Sedimentation

Question 2.
Write the steps you would use for making tea Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Answer:
Take a cup of water in a kettle as solvent and heat it. When the solvent boils, add sugar in it which is the solute. Heat it till entire sugar dissolves. Water and sugar form a solution. Then, add some tea leaves in this solution. Boil the contents, add milk which is also soluble in this mixture, and boil again. Filter the tea with the help of a strainer. The tea collected in cup is the filtrate and the tea leaves collected in the strainer is residue.

Question 3.
Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

	Temperature in K Solubility
Substance Dissolved	283	293	313	333	353
Potassium nitrate	21	32	62	106	167
Sodium chloride	36	36	36	37	37
Potassium chloride	35	35	40	46	54
Ammonium chloride	24	37	41	55	66

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
(b) Pragya makes a saturated solution of potassium chloride in water at 353K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at 293K. Which salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Answer:
(a) Mass of potassium nitrate (KNO₃) needed to produce a saturated solution of KNO₃ in 100 grams of
water at 313 K = 62 g
Mass of KNO₃ needed in 50 g of
water at 313 K = \(\frac{62.0 \times 50}{100}\)
= 31.0g

(b) As solution cools, potassium chloride gets crystallised. This is because the solubility of a solid decreases with decrease in temperature.

(c) At 293K, solubility of KNO₃ is 32, NaCl is 36, KCL is 35 andNH₄Cl is 37. Ammonium chloride has the highest solubility at 293 K temperature.

(d) As the temperature increases, solubility increases and vice – versa.

Question 4.
Explain the following giving examples.
(a) saturated solution
(b) pure substance
(c) colloid
(d) suspension
Answer:
(a) Saturated solution: A solution in which no more solute can be dissolved at a given temperature is called a saturated solution.

(b) Pure substance: A pure substance consists of a single type of particles. It always has the same colour, taste or texture at a given temperature and pressure. For example, pure water is always colourless, odourless and tasteless.

(c) Colloid: A colloid is a solution in which the size of solute particles is bigger than those of a true solution. These particles cannot be seen with naked eyes as they are stable, e.g., ink, blood, smoke, milk, fog and cloud.

(d) Suspension: Suspension is a heterogeneous mixture. The particles of a suspension are greater than 1000 nm and are visible to naked eyes, e.g., chalk powder in water, paints, etc.

Question 5.
Classify each of the following as a homogeneous or a heterogeneous mixture: soda water, wood, air, soil, vinegar, filtered tea.
Answer:

1. Homogeneous mixtures: Soda water, vinegar, filtered tea.
2. Heterogeneous mixtures: Wood, soil, air.

Question 6.
How would you confirm that a colourless liquid given to you is pure water?
Answer:
By finding the boiling point of the given colourless liquid. If the given colourless liquid boils at exactly 373K at 1 atmosphere pressure, then it is pure water. This is because pure substances have fixed melting and boiling points.

Question 7.
Which of the following materials fall in the category of a ‘pure substance’?
(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air
Answer:
Pure substances are: ice (a compound), iron (an element), hydrochloric acid (a compound), calcium oxide (a compound) and mercury (an element).

Question 8.
Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water
Answer:
Solutions are sea water, soda water and air.

Question 9.
Which of the following will show “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution
Answer:
Milk and starch solution are colloids and will show Tyndall effect.

Question 10.
Classify the following into elements, compounds and mixtures.
Sodium, Soil, Sugar solution, Silver, Calcium carbonate, Tin, Silicon, Coal, Air, Soap, Methane, Carbon dioxide, Blood
Answer:

• Elements: Sodium, Silver, Tin, Silicon
• Compounds: Calcium carbonate, Meth – ane, Carbon dioxide
• Mixtures: Sugar solution, Soil, Coal, Air, Blood, Soap

Question 11.
Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle
Answer:
Chemical changes are growth of a plant, rusting of iron, cooking of food, digestion of food, burning of a candle.

JAC Class 9 Science Solutions