Q1: Explain how sound is produced by your school bell.
Ans: The sound of the bell depends upon the vibration of the bell when it is rung. When the bell starts ringing, it drives the molecules around the air to vibrate. This produces the wave. So the compression is produced, however, the rarefaction makes the sound echo through the air.
Since, the vibrating molecules put pressure on one another, in turn, air particles are disturbed and start moving forward and backward.  

Q2: Why are sound waves called mechanical waves?
Ans: Mechanical Waves are waves that generate through a medium (solid, liquid, or gas) at a wave speed. A sound wave is an example of a mechanical wave. Sound waves cannot travel through a vacuum, It requires some medium to propagate, which could be air, water, or metal similar to mechanical waves. That’s why; a sound wave is called a mechanical wave.

Q3: Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Ans: A sound wave needs a medium to propagate. Be it a solid, gas, or liquid medium. The moon resides in a vacuum; since it has no air that defines it has no medium to generate the sound waves. So, it is not feasible to hear any sound on the moon.

Q4: Which wave property determines
(a) Loudness,
Ans: The loudness of the sound is a decibel unit (dB). Loudness is defined as the intensity of the wave or the molecule. The determination of the loudness depends on the magnitude of the wave. Precisely, the amplitude of the wave determines the loudness of sound.
(b) Pitch
Ans: Pitch is defined as the response of the sound by ear which goes hand in hand. Higher the pitch, the higher the frequency. So, the pitch of the sound is dependent on the frequency of the sound.

Q5: How are the wavelength and frequency of a sound wave related to its speed?
Ans: Longer the wavelength, the longer will be the frequency waves. In the context of speed, it is dependent upon the medium through which the sound wave is traveling. The more inflexible, inelastic the medium, the faster will the movement of sound.
The equation forms as VW = fλ, where VW is the speed of sound, f is its frequency, and λ is its wavelength.

Q6: Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Ans: The wave with the greatest frequency has the shortest wavelength.
We have to calculate the wavelength of the sound with the frequency of 220 Hz and the speed will be 440m/s.
Given:  v=440 m/s , f = 220Hz

The wavelength is $\mathrm{2m.<span\; style="box-sizing:\; border-box;\; font-family:\; lato,\; sans-serif;\; letter-spacing:\; inherit;\; text-align:\; inherit;\; color:\; rgb(0,\; 0,\; 0)\; !important;\; line-height:\; 1.8;\; font-size:\; 20px;\; word-break:\; break-word;"></span>}$

Q7: 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?
Ans: The time interval between successive compressions from the source

T = 1/v = 1/500 = 0.002 second.

Q8: Distinguish between loudness and intensity of sound.
Ans: Differences between loudness and intensity of sound are:

Q9: An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Ans:

Speed of sound = distance/time

Therefore, distance travelled by sound during echo = speed × time = 342 × 3 = 1026 m

 so the distance of reflecting surface = 1026/2 = 513 m

Q10: Why are the ceilings of concert halls curved?
Ans: The curved architecture of any structure helps the sound to reach every end. Concert halls are very big, so the sound might not reach every corner of the hall. This is achieved when the sound generates the reflection technique to reach every corner.

Q11: What is the range of frequencies associated with
(a) Infrasound?
Ans: Infrasound is used for the sound below 20 Hz. Sound at 20-200 Hz is called low-frequency sound. The range of frequency is less than 20 Hz.

(b) Ultrasound?
Ans: Ultrasound adds biologically significant sounds ranging from 15 kHz or so up to 200 kHz, which is too high in frequency. So the range for Ultrasound = greater than 20 KHz

Q12: A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in saltwater is1531 m/s, how far away is the cliff?
Ans: Distance travelled by a sonar pulse = speed of sound in saltwater × time = 1531×1.02 = 1561.62 m

Therefore, the distance of cliff from submarine =1561.62/2 = 780.81 m

Q13: What is sound and how is it produced?
Ans: Sound is the type of energy that is defined by vibrations between any state of medium (Gas, Solid, Liquid). Vibration is the movement of air particles. The oscillation of a molecule propagates the sound.

Q14: Describe with the help of a diagram, how compressions and rarefactions are produced in the air near a source of the sound.
Ans: Compression and rarefaction are produced because of the movement in the medium caused by sound waves.

Q15: Why is a sound wave called a longitudinal wave?

Ans: The movement of the particle is called vibration. A medium can be anything – a liquid (such as water), a solid (such as the seafloor), or a gas (such as air). A sound wave is called a compressional or longitudinal wave when it vibrates parallel to the direction in which the sound wave moves.

Q16: Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?
Ans: There are many times we have observed that the thunder is heard a few seconds later after the flash

This happens because the speed of light in the atmosphere for air is

3 × 108 m/s² which are very high, to that of sound which is only 330 m/s.

This is the reason; the sound of thunder reaches us later than the flash.

Q17: Differentiate between longitudinal and transverse waves?
Ans: The difference between longitudinal and transverse waves are as follows

Q18: Define the terms:
Ans: 

(a) Wavelength: It is the distance between two successive crests or troughs of a wave. The direction will be the same as the wave

(b) Frequency: It is defined as the number of waves that pass a fixed place in a given amount of time. The Hertz measurement, abbreviated Hz, is the number of waves that pass by per second.

Q19: What is an echo? Name two areas of its application

Ans: Echo is a sound repetition by sound wave reflection, having a lasting or far-reaching impact.

Application of echo: In the medical field, echo uses sound waves to create pictures of the heart’s chambers, valves, walls, and the blood vessels (aorta, arteries, and veins) attached to the heart for testing purposes. Also, it is used in SONAR and detecting flaws in metal objects.

Q20: Why are sound waves called mechanical waves?
Ans: Sound needs a medium to propagate, as it does not generate in a vacuum. So, all sound waves are examples of mechanical waves. Sound waves are called mechanical waves as it is a wave that is an oscillation of matter.

Q21: Define (a) Time Period (b) Amplitude of a wave
Ans: 

(a) Time period is defined as the time taken by a complete cycle of the wave to pass a particular area.

The formula for time is: T (period) = 1 / f (frequency).

λ = c / f = wave speed i.e.  c (m/s) / frequency f (Hz).

(b) Amplitude of a wave is the maximum amount of displacement of a particle on the medium from its static position. Amplitude measures how far a wave rises and dips.

Q22: What do you understand by the loud and soft sound?

Ans:  Sound is a type of vibrating pressure that is transmitted in waves.

• Loud sound: Higher the sound, the higher will be the amplitude referred to as loud sound.
• Soft sound: lower the sound, less will be the amplitude that defines the soft sound.

Q23: What do you understand by low-pitched and high-pitched sound?
Ans: Low pitched sound: It refers to the low sound that defines slower oscillation. A sound that is low-pitched is deep.
High pitched sound: The vibration is that of a pure tone with a frequency equal to 3000 Hz, which is considered a high-pitched sound as it completes a large number of vibrations in a given time.

Q24: Why do we see light first and hear the sound later during a thunderstorm?
Ans: There are many times we have observed that the thunder is heard a few seconds later after the flash. This happens because the speed of light in the atmosphere for air is  3× 108 ms⁻² which is very high, to that of sound which is only 330 ms⁻¹.  This is the reason; the sound of thunder reaches us later than the flash.

Q25:  Why are the ceilings of concert halls curved?
Ans: The curved architecture of any structure helps the sound to reach every end. Concert halls are very big, so the sound might not reach every corner of the hall.  This is achieved when the sound generates the reflection technique to reach every corner.

Q26: How does the sound produced by a vibrating object in a medium reach your ear?
Ans: Sound waves enter the outer ear and travel through a narrow passageway called the ear canal, which leads to the eardrum. When the sound waves fall on the eardrum, the eardrum starts vibrating back and forth rapidly.
The sound produced by a vibrating object reaches our ear through sound waves which travel in the medium as a series of compressions and rarefactions. The process is repeated further and as result sound waves propagate in the form of compressions and rarefactions to the listener’s ear.

Q27: What are the wavelength, frequency, time period, and amplitude of a sound wave?
Ans: The wavelength of a sound wave is defined as the distance between the identical parts of the wave also called crests and troughs.
The wavelength of the sound wave is calculated as:
Wavelength = velocity of sound / frequency

• Frequency is defined as the number of vibrations or oscillations per second i.e. it is the number of complete waves or cycles produced in one second. It refers to how rapidly or slowly the oscillations occur.
• The time period is the time taken to complete one vibration/oscillation/complete wave is called the time period. It is measured in seconds.
• Amplitude is the utmost rearrangement of the particles of the medium from their actual static position.

Q28: Cite an experiment to show that sound needs a material medium for its propagation.
Ans: The Bell-jar experiment shows that sound needs a medium for its propagation. An electric bell and an airtight glass bell jar are required. 

• The electric bell is suspended inside the airtight bell jar. The bell jar is connected to a vacuum pump if you press the switch you will be able to hear the bell. 
• Now start the vacuum pump. When the air in the jar is pumped out slowly, the sound becomes dimmer, although the same current is passing through the bell.
• After some time when less air is left inside the bell jar, you will hear a very feeble sound. Now if we evacuate the bell jar no sound is heard.

Q29: What happens when sound travels in the air?
Ans: The air is made up of many tiny particles. The movement propagates through a medium, and then alternate regions of pressure variations are created.
The region where particles come closer to each other (high density) and the pressure of air is high is called compression. The region where particles are far apart from each other (low density) and pressure of air are less are called rarefaction compression and rarefactions always occur together.

Q30: Sound requires a medium to travel? Justify experimentally.
Ans: Sound requires a medium for propagation and it can beproved by the following experiment:
(i) Take a bell jar and suspend an electric bell in it.
(ii) Thebell jar is connected to a vacuum pump. Till the air is in the bell jar, the sound of the electric bell is louder.
(iii) Now, withthe help of a vacuum pump, pump out the air gradually
(iv) Now as air is pumped out, the sound of the bell gets fainter and fainter.
(v) Now, when the bell jar is completely vacuumed no sound is heard.
(vi) This shows that air is required for the propagation of sound.

Q31: Discuss briefly the structure and working of the human ear?
Ans: Ear collects sound waves and channels them into the ear canal (external auditory meatus), where the sound is amplified. Sound waves cause the eardrum to vibrate.

• Structure of the Human Ear: The outer ear is called pinna followed by an auditory canal in which ends in a tympanic membrane. The tympanic membrane is then connected to three bones, hammer, anvil, and stirrup. After that, there is a cochlea connected to an auditory nerve.
• Working of the Human Ear: The auricle or the pinna collects the sound and the collected sound passes through and reaches the auditory nerve. After which it forces the eardrum (tympanic membrane) to vibrate. The vibrations are then amplified by 3 bones and the pressure variations reach the inner ear after which the cochlea converts them to electrical signals. The auditory nerve carries the electrical signals to the brain and the brain interprets them as sound.

Q32: What is SONAR? Write the working?
Ans: SONAR is an abbreviation of Sound Navigation and Ranging. It is the method used for echoing. Dinars are used to find the depth of the sea or to locate underwater things like shoals of fish, enemy submarines, etc.
It is used to navigate or detect and communicate with the objects present on underwater surfaces such as oceans by using sound propagation. Sonar works by sending short bursts of ultrasonic sound from a ship into the sea and then picking up the echo produced by the reflection of ultrasound from underwater objects like the bottom of the sea.

Q1: When the work is said to be done?
Ans: When a force acts on an object and moves it in the same direction that of force then work is said to be done.

Q2: What will be the expression for the work done when a force acts on an object in the direction of its motion.
Ans: Work done = Force × Displacement
If W is the work done, F is the force applied on object and d is the displacement, then the expression of work done will be
W = F × d

Q3: Explain 1 joule of work done.
Ans: When a force of 1 N (Newton) is applied on an object and that object displaces upto a distance of 1 m (meter) in the same direction of its displacement, then 1 joule (J) of work is done on the object.

Q4: How much work is done in ploughing a 15 m long field when a pair of bullocks applies a force of 140 N on the plough?
Ans: Since Work done (W) = Force (F) × Displacement (d)
Hence, Work done in ploughing (W) = 140 N × 15 m =  2100 J

Q5: The force acting on the object is 7 N, and the displacement of the object occurs in the direction of the force is 8 m. Suppose that force acts on the object through displacement, then how much work was done in this case?
Ans:  As we know, Work done (W) = Force (F) × Displacement (d)
Thus, Work done in the given case (W) = 7 N × 8 m =  56 J

Q6: Define kinetic energy of an object.
Ans: The kinetic energy of an object is a kind of mechanical energy that exists in the object due to its state of motion (movement).

Q7: Write down the kinetic energy expression of an object.
Ans: If m is the mass of an moving object and v is its velocity, then the expression of its kinetic energy (KE) will be
K.E = 1/2mv²

Q8: Define power.
Ans: The rate by which work is done refers to power. It is expressed by P.
Power = Work done/Time
P = W/t

Q9: What is 1 watt of power?
Ans: When an object is doing work at the rate of 1 J/s, then the power of that body or object is 1 watt (where watt is the unit of power).

Q10: An object is thrown at an angle to the ground, moves along a curve and falls back to the ground. The start and end points of the object path are on the same horizontal line. How much work is done by the gravity on that object?  
Ans: There must be a displacement to calculate the work, but since the vertical displacement in this case is zero (because the start and end points are on the same horizontal line), the work done by gravity is zero.

Q11: How does the state of energy get changed when a battery lights up a bulb?
Ans: The chemical energy of the battery is converted into heat and light energy of the bulb in the given case.

Q12: Calculate the work done by the force that changes the velocity of a moving body from 5 ms^-1 to 2 ms^-1. The body has a mass of 20 kg.
Ans: Since work done by force = Change in the kinetic energy of the moving body
Therefore, Work done by force = 

= 1/2 x 20 ( 5 ²– 2²) = 10 x (25-4) = 10 x 21 = 210 J

Q13: An object having 10 kg weight is moved from point A to point B on the table. If the distance between A and B is horizontal, what work does gravity do to the object?  Give the reason for the answer.
Ans: Since the work done by gravity on the object depends on the change in the vertical height of the object, the vertical height of the object will not change. Because the connection level of A and B is at the same height, the work done is zero.

Q14: The potential energy of an object decreases gradually in a free fall. How does this violate the law of conservation of energy?
Ans: This does not violate the law of conservation of energy, because the potential energy of an object in free fall gradually decreases with gradual changes until the kinetic energy of the object maintains the state of free fall, that is, the total energy of the object remains conserved.

Q15: What energy conversion occurs when riding a bicycle?  
Ans: Our muscle energy is converted into mechanical energy while riding a bicycle.

Q16:  Does energy transfer occur when you push a huge rock with all your strength without moving it? Where did the energy you applied go?  
Ans: As long as you push a big rock with all your strength and do not move it, energy transfer will not occur, because cell energy is only used for muscle contraction and relaxation, and also for releasing heat (sweating).

Q17:A household uses 250 units of energy in a month. How much energy is used  by that house in joules?
Ans: Energy consumption by a house = 250 kWh
Since, 1 kWh = 3.6 × 10⁶ J
hence, 250kWh= 250 × 3.6 × 10⁶ = 9 × 10⁸ J

Q18: The output power of the electric heater is 1500 watts. How much energy does it consume in 10 hours?  
Ans: Power of electric heater (p) = 1500W = 1.5kW
Energy = Power × Time = 1.5kW × 10 hours = 15 kWh

Q19: An object of mass m moves at a constant speed v. How much work does the subject need to do to make it stable?  
Ans: For an object to be stationary, the work done must be equal to the kinetic energy of the moving object.
The kinetic energy of any object is equal to
K.E=1/2mv², where m is the mass of the body and v is its velocity.

Q20: Sony said that even if different forces act on the object, the acceleration of the object can be zero. Do you agree with her, if yes, why?  
Ans: Yes, we agree with Soni, because the displacement of an object becomes zero when many balancing forces act on that object.

Q21: Calculate the energy (in kilowatt hours) consumed by four 500 W devices in 10 hours.
Ans: Since, Energy = Power × Time
Hence, Energy consumed by four 500 W devices in 10 hours
= 4 × 500 × 10
= 20000 Wh
= 20 kWh

Q22: Free-falling objects will eventually stop when they hit the ground. What will happen to their kinetic energy?
Ans: The object will eventually stop after it hits the ground in free fall, because its kinetic energy will be transferred to the ground when it hits the ground.

Q23: A large force acting on an object, and the displacement of that object is zero, what will be the work done?  
Ans: The work done on the body is defined as the force exerted on the body that causes a net displacement of the body.  
Work done = Force x Displacement  
If the force does not cause any displacement, the work done to the object is zero.

Q24: Write some differences between kinetic and potential energy.
Ans: Differences between kinetic and potential energy:

Q25: Describe the law of conservation of energy. 
Ans: The law of conservation of energy says that:

• Energy cannot be produced or destroyed. It can only be transformed from one form to another.  
• The energy of the universe is constant.

Q26: A person weighing 50 kg climbs the stairs with a height difference of 5 meters, within 4 seconds.  
(a) What kind of work is done by that person?  
(b) What is the average power of that person?
Ans: Mass of the man = 50 Kg
Distance moved by that man = 5 meter
Time taken to cover the given distance = 4s
(a) Work Done = Force  Acceleration
In this case, the increase in Potential energy = Work done =Mgh
=50×10×5
=2500 J
(b) Power = Work Done /Time Taken  =2500/ 4=625 Watts

Q27: Write differences between power and energy.
Ans: Differences between power and energy are given below:


Q28: Write down the expressions for
(a) Potential energy of an object
(b) Kinetic energy of an object
Ans: 
(a) The expression for Potential energy of an object = P.E = mgh
Where, m = Mass of Body
g = Acceleration due to gravity
h = Height
(b) The expression for Kinetic energy of an object = 1/2mv²
Where, m = Mass of body
v = Velocity of body

Q29: If a force of 12.5 N is applied to complete a work of 100 J, what is the distance covered by the force?
Ans:  W = Work = 100 J
F = Force = 12.5 N
And S is the distance moved or displacement
Since, Work done = Force  Displacement
W = FS
100 =12.5 × S
100/12.5  = S
8 m=S (Displacement)

Q30: A car weighing 1800 kg is moving at a speed of 30 m/s when braking. If the average braking force is 6000 N, it is determined that the vehicle has traveled to a standstill distance. What is the distance at which it becomes stable?
Ans: M = Mass of the car = 1800 Kg
V = Velocity of the car = 30 m/s
F = Force applied while braking = 6000 N
KE=1/2mv²
KE =121800×900
KE=810000 J
KE of car = Work done by the car = Force  Displacement
810000=6000× Displacement
810000/6000= Displacement
135 m= Displacement

Q31: What do you understand about average power?
Ans: The agent may not always be able to complete the same amount of work in a given time period. In other words, the power of this work will change over time. Therefore, in this case, we can take the average power of the work done by the body per unit time (that is, the total energy consumed divided by the total time).

Q32: Take a look at the steps below. Based on your understanding of the word “work”, prove whether the work will proceed.  

• Suma swims in the pond.  
• The donkey carries a heavy load.  
• The windmill draws water from the well.  
• Green plants perform photosynthesis.  
• The trains are pulled by engines. 
• Drying food grains in the sun.  
• Sailing boats are powered by wind.

Ans: The work is said to be done when a force acts on an object and moves in the direction of the force. According to this explanation, the following activities were taken in which work will be proceeded:

• Suma swims in the pond.  
• The donkey carries a heavy load.  
• The windmill draws water from the well.  
• The trains are pulled by engines. 
• Sailing boats are powered by wind.

Q33: The law of conservation of energy is explained by discussing the energy changes that occur when we move the pendulum laterally and swing it. Why does the pendulum eventually stop? What happens to the energy and does it violate energy conservation law?
Ans: Bob will eventually stop due to the friction created by the air and the rigid support that holds the thread in place. This does not violate the law of conservation of energy, because mechanical energy can be converted into another unusable form of energy for some useful work. This energy loss is called energy dissipation.

Q34: Get the expression of the potential energy of an object. Calculate PE for a body of 10 kg which is resting at a height of 10 m.
Ans: The potential energy of an object with mass = m kg, at height above the ground =h m
Gravitational force of attraction on that body = mgN
To lift that body to B height at h  m above the ground.
Force applied to lift this body with a constant velocity =mgN
Distance moved by the body after applying force = hm
Work done in lifting the body from a to B distance = Force × Distance
Energy cannot be destroyed, hence, this energy is stored as potential energy in the stone.
m = 10Kg
g = 10 m/s²
h = 10 m
PE = mgh
PE = mgh
= 10 × 10 × 10
= 1000Joules

Q1: What do you mean by free fall?
Ans: It is the object falling towards earth under the influence of attraction force of earth or gravity.

Q2: What do you mean by acceleration due to gravity?
Ans: During free fall any object that has mass experiences force towards centre of earth and hence an acceleration works as well. “acceleration experienced by an object in its freefall is called acceleration due to gravity.” It is denoted by g.

Q3: Why is it difficult to hold a schoolbag having a strap made of a thin and strong string?
Ans: It is difficult to hold a schoolbag having a strap made of a thin and strong string because a bag of that kind will make its weight fall over a small area of the shoulder and produce a greater pressure that makes holding the bag difficult and painful.

Q4: What do you mean by buoyancy?
Ans: It is the upward force experienced by an object when it is immersed into a fluid.

Q5: You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Ans: Mass will be slightly more than 42 kg.

Q6: You have a bag of cotton and anir on bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?
Ans: The bag of cotton is heavier since volume of cotton bag is greater than iron bar, so the up thrust is larger in case of cotton hence real mass of cotton bag is more and it is heavier.

Q7: How does the force of gravitation between two objects change when the distance between them is reduced to half?
Ans: The force of gravitation between two objects is inversely proportional to the square of the distance between them therefore the gravity will become four times if distance between them is reduced to half.

Q8: Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than alight object?
Ans: In free fall of objects the acceleration in velocity due to gravity is independent of mass of those objects hence a heavy object does not fall faster than alight object.

Q9: What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 x 1024 kg and radius of the earth is 6.4 x 106 m.)
Ans: 

Q10: 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?
Ans: The earth and the moon are attracted to each other by same gravitational force because for both of them formula to calculate force of attraction is the same 

d is also same for both.

Q11: If the moon attracts the earth, why does the earth not move towards the moon?
Ans: Earth does not move towards moon because mass of moon is very small as compared to that of earth.

Q12: 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.]
Ans: Since  W = m x g and given in the question that value of g is greater at the poles than at the equator, hence weight of same amount of gold will be lesser at equator than it was on the poles. Therefore, the friend will not agree with the weight of gold bought.

Q13: Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans: A greater surface area offers greater resistance and buoyancy same is true in the case of a sheet of paper that has larger surface area as compared to paper crumpled into a ball. So sheet of paper falls slower.

Q14: 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’s of a 10 kg object on the moon and on the earth?
Ans: value of gravity on earth = 9.8 m/s²
value of gravity on moon = l/6^th of earth = 9.8/6 = 1.63 ms2
weight of object on moon = m x 1 .63 = 10 x 1.63 = 16.3 N
weight of object on earth = m x. 9.S = 10 x 9.8 = 98 N

Q15: A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.
Ans: (i) v = u + gt
0 = 49 + (-9.8) x t
9.8t = 49
t = 49/9.8 = 5 s

= 49 x 5 + 1/2 x 9.8 x 25
= 245 – 122.5
= 122.5
(ii) total time taken to return = 5 + 5 = 10 s

Q16: Why does a block of plastic released under water come up to the surface of water?
Ans: Since of plastic has density very less as compared to water i.e. weight of plastic is less than the buoyant force experienced by it therefore a block of plastic released under water come up to the surface of water/floats.

Q17:  The volume of 50 g of a substance is 20 cm³ . If the density of water is 1g /cm³ , will the substance float or sink?
Ans: Density of that substance (d) = mass/volume = 50/20 = 2.5 g /cm³
since the density of substance (2.5) is greater than the density of water (1) therefore it will sink.

Q18: State the Universal law of Gravitation?
Ans. According to Universal law of Gravitation every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Let M₁, M₂ = Masses of two bodies
r = Distance of separation
f = force acting b/w them


Q19: If heavier bodies are attracted more strongly by the earth, why do they not fall faster to the ground?
Ans: Heavier bodies do not fall fast on the ground even though they are attracted by the earth strongly because of their larger mass, the acceleration produced in them by the force of earth will be less as
F = m
F/m = a
m-mass, F = force, a = Acceleration
so if Mass is more, Acceleration will be less

Q20: How does acceleration due to gravity change with the shape of earth?
Ans: Since earth is not a perfect sphere, it is flattened from the top and bulges at the centre and acceleration due to gravity (g) is inversely proportional to the radius of earth so, g is more at poles because of lesser radius and less at equator because of greater radius.

Q21: What do you understand by the gravitational force of earth and weight?
Ans: Gravitational force of earth is the force by which earth exerts on any object towards itself.
Weight is the force which the object exerts on the earth.

Q22: A man of mass 60 Kg is standing on the floor holding a stone weighing 40 N. What is the force with which the floor is pushing him up?
Ans: The gravitational pull on the man = Mg
= 60 x 10 = 600.V
The weight he is carrying = 40 N
The total downward force on the floor = 40 N + 600 N
= 640 N
The Gravitational force and upward force of the floor is an action – reaction pair.
The force with which the floor pushes the man = 640 N.

Q23: What is acceleration due to gravity and how is it different from acceleration?
Ans: Acceleration due to gravity is the acceleration produced in the object when it falls freely under the effect of gravitational force of earth only. Acceleration is produced when any external force applied on the body makes it to move.

Q24: What is the importance of the universal law of Gravitation?
Ans: The importance of universal law of gravitation:

• The force that binds us to the earth.
• The motion of the moon around the earth.
• The motion of the planet around the sun
• The tides due to the moon and the sun

Q25: Define Pressure? How is thrust different from Pressure?
Ans: The pressure due to a force is defined as the force acting or unit area.

A unit of Pressure is N / m²
Thrust is also the pressure but it is the force acting on a surface normal to its area.

Q26: What are fluids? What are the factors on which the upward pressure at a point on a  fluid depends?
Ans: Fluids are that which flow and it includes both liquids and gases.
Factors on which the upward pressure at a point of the fluid depends are:

• the depth of the point from the surface of the liquid.
• the density of the liquid
• the acceleration due to gravity.

Q27: State the universal law of gravitation.
Ans: According to Newton’s universal law of gravitation:
Every mass in this universe attracts every other mass with a force which is directly proportional to the product of two masses and inversely proportional to the square of the distance between them.

Q28: Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Ans: The formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth is given below:

F = magnitude of gravitational force
G = Universal gravitation constant
M = mass of earth
m = mass of object
d = distance of object from the centre of earth

Q29: What are the differences between the mass of an object and its weight?
Ans:

Q30: Why is the weight of an object on the moon 1/6th its weight on the earth?
Ans. since we know FF = mx g Mass of object remains the same whether on earth or moon but the value of acceleration on moon is 1/6th of the value of acceleration on earth. Because of this weight of an object on moon is 1/6th its weight on the earth.

Q1: An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling 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.
Ans: 

• Yes, it is possible for an object to be traveling with a non-zero velocity if it experiences a net zero external unbalanced force. This is based on Newton’s First law of motion, which states that an object will change its state of motion if and only if the net force acting on it is non-zero.  
• Thus, in order to change its motion or to bring about motion, some external unbalanced force is required. 
• In this case, the object experiences a net zero unbalanced force, which will cause it to move with some non-zero velocity provided that the object was previously moving with a constant velocity.

Q2: When a carpet is beaten with a stick, dust comes out of it. Explain.
Ans: When a carpet is beaten with a stick, dust is released due to the principle of inertia, which is part of Newton’s First Law of Motion. Here’s how it works:

• The carpet and dust particles are initially at rest.
• Beating the carpet causes it to move.
• Dust particles resist this change in motion because of their inertia.
• As the carpet moves forward, it exerts a backward force on the dust.
• This force causes the dust to be expelled from the carpet.

Thus, the dust comes out when the carpet is struck.

Q3: Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ans: It is important to tie any luggage kept on the roof of a bus with a rope due to the principles of Newton’s First Law of Motion, also known as the law of inertia. Here’s why:

• When the bus is in motion, the luggage also moves along with it.
• If the bus suddenly stops, the luggage will continue moving forward due to its inertia, which can cause it to fall off.
• Similarly, if the bus slows down or turns, the luggage will resist these changes in motion, increasing the risk of it falling off.

By securing the luggage with a rope, you can prevent accidents and ensure that it stays in place during the journey.

Q4: A stone of 1kg is thrown with a velocity of 20ms⁻¹ across the frozen surface of a lake and comes to rest after traveling a distance of 50m. What is the force of friction between the stone and the ice?
Ans: Given:
Mass of stone: m = 1kg
Initial velocity of stone: u = 20ms⁻¹
Final velocity of stone: v = 0ms⁻¹ (as it comes to rest)
Distance traveled on ice: s = 50m
To find: Force of friction between stone and ice.
First, we need to find the deceleration:
It is known that – v² = u² + 2as
Thus, 0² = (20)² + 2a(50)
⇒ 0  = 400 + 100a
⇒ −400 = 100a
⇒ a = −4ms⁻²
The negative sign implies deceleration.
Next, finding the frictional force:
F = ma
⇒ F = (1)(−4)
⇒ F = −4N
Thus, the force of friction between stone and ice is −4N.

Q5: 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 1.7ms⁻²?
Ans: Given:
Mass of vehicle: m = 1500kg
Negative acceleration: a = −1.7ms⁻²
To find: Force of friction between road and vehicle.
It is known that – F = ma
⇒ F = (1500)(−1.7)
⇒ F = −2550N
Thus the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7ms⁻² is −2550N.

Q6: An object of mass 100kg is accelerated uniformly from a velocity of 5ms−1 to 8ms−1 in 6s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Ans: Given:
Mass of object: m = 100kg
Initial velocity of object: u = 5ms⁻¹
Final velocity of object: v = 8ms⁻¹
Time duration of acceleration: t = 6s
To find: 

• Initial momentum
• Final momentum
• Force exerted on the object

It is known that – momentum = mass × velocity
Initial_momentum = mass × initial_velocity
⇒ Initial_momentum = 100 × 5
⇒ Initial_momentum = 500kgms⁻¹
Final_momentum = mass × final_velocity
⇒ Final_momentum = 100 × 8
⇒ Final_momentum = 800kgms⁻¹
Now, the force – F = ma

Thus,

• Initial momentum:500kgms⁻¹
• Final momentum: 800kgms⁻¹
• Force exerted on object: 50N

Q7: Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Ans: Akhtar, Kiran, and Rahulwere riding in a fast-moving motorcar when an insect hit the windshield and got stuck. They discussed the incident, leading to different viewpoints:

• Kiran argued that the insect experienced a greater change in momentum than the motorcar because its velocity changed significantly upon impact.
• Akhtar believed that the motorcar, moving at a higher velocity, exerted a larger force on the insect, which caused its death.
• Rahul suggested that both the motorcar and the insect experienced the same force and change in momentum.

Here are some comments on their suggestions:

• Kiran’s point is valid; the insect’s momentum change is indeed greater due to its sudden stop.
• Akhtar’s assertion about force is partially correct; while the motorcar does exert a larger force, it is not solely responsible for the insect’s death.
• Rahul’s claim that both experienced the same force is misleading; the forces they experience differ due to their masses and the nature of the collision.

In summary, while all three perspectives have merit, the nuances of momentum and force in collisions highlight the complexity of the situation.

Q8: State Newton’s second law of motion?
Ans: Newton’s Second Law of Motion states that when an unbalanced external force acts on an object, its velocity changes, resulting in acceleration. This law can also be expressed as the time rate of change of momentum of a body being equal, in both magnitude and direction, to the force applied on it.
Mathematical Representation:
F = ma, where ‘F’ denotes the force, ‘m’ the mass of the object, and ‘a’ the acceleration produced. This relationship indicates that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

Q9:  What is the momentum of a body of mass 200g moving with a velocity of 15ms⁻¹?
Ans: Given:
Mass of body: m = 200g = 0.2kg
Velocity of body: v = 15ms⁻¹
To find: Momentum of the body.
It is known that – momentum=mass×velocity
⇒ momentum = 0.2 × 15
⇒ momentum = 3kgms−1
Thus, the momentum of the body is 3kgms⁻¹.

Q10: Define force and what are the various types of forces?
Ans: Force is defined as the push or pull exerted on an object that produces a change in its state or shape. It can also cause modifications in speed and/or direction of motion.
The various types of force are:

• Mechanical force
• Gravitational force
• Frictional force
• Electrostatic force
• Electromagnetic force
• Nuclear force

Q11: A force of 25N acts on a mass of 500g resting on a frictionless surface. What is the acceleration produced?
Ans: Given:
Mass: m = 500g = 0.5kg
Force exerted: F = 25N
To find: Acceleration.
It is known that – F = ma
⇒ a = Fm
⇒ a = 25 / 0.5
⇒ a = 50ms⁻²
Thus, the acceleration produced is 50ms⁻².

Q12: State Newton’s first law of Motion?
Ans: Newton’s first law of motion is also called the law of Inertia. It states that an object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. Or, an object rest will continue to be at rest and an object in motion will continue to be in motion until and unless it is acted upon by an external force.

Q13: A body of mass 5kg starts and rolls down 32m of an inclined plane in 4s. Find the force acting on the body?
Ans: Given:
Mass of body: m = 5kg
Initial velocity of body: u = 0ms⁻¹ (as it is starting to roll)
Distance travelled on inclined plane: s = 32m
Time duration of rolling: t = 4s
To find: Force acting on the body.
First we need to find the acceleration:
It is known that 

Next, finding the force:
F = ma
⇒ F = (5 × 4)
⇒ F = 20N
Thus, the force acting on the body is 20N.

Q14: On a certain planet, a small stone tossed up at 15ms⁻¹ vertically upwards takes 7.5s to return to the ground. What is the acceleration due to gravity on the planet?
Ans: Given:
Initial velocity of stone:u = 15ms⁻¹
Final velocity of stone: v = 0ms⁻¹ (as it becomes zero at the highest point)
Total time duration of flight (tossed up and falling down to the ground): t = 7.5s
To find: Acceleration due to gravity of the planet.
It is known that – v = u + at
Thus, 0 = 15 + (at)

 this denotes the time for one-half of the entire flight.
Thus the total duration of the flight is twice this duration.

Now, the acceleration due to gravity is –

Thus, the acceleration due to gravity of the planet is −4ms⁻²

Q15: Why does the passenger sitting in a moving bus are pushed in the forward direction when the bus stops suddenly?
Ans: The passengers sitting in a moving bus are pushed forward when the bus stops suddenly due to inertia. This occurs because:

• The upper body of the passenger continues moving forward.
• The lower body, which is in contact with the seat, remains at rest.
• This difference causes the upper body to lurch forward in the direction the bus was travelling.

Q16: Why does the boat move backward when the sailor jumps in the forward direction?
Ans: The boat moves backward when the sailor jumps forward due to Newton’s third law of motion. This law states:

• For every action, there is an equal and opposite reaction.
• When the sailor jumps forward, he exerts a force on the boat.
• As a result, the boat pushes back with an equal force, causing it to move backward.

This interaction illustrates how forces work between two objects, leading to the boat’s backward movement.

Q17: An astronaut has 80kg mass on earth.
(i) What is his weight on earth?
Ans: Given:
Mass of astronaut: m = 80kg
To find his weight on earth.
It is known that,
Acceleration due to gravity on earth: g_e = 10ms⁻²
Acceleration due to gravity on mars: g_m = 3.7ms⁻²
Weight: w = m × g
Weight on earth: w_e = m × g_e
⇒ w_e = 80 × 10
w_e = 800N
(ii) What will be his mass and weight on mars with gm=3.7ms⁻²?
Ans: Given:
Mass of astronaut: m = 80kg
To find his mass and weight on mars.
It is known that,
Acceleration due to gravity on earth: g_e = 10ms⁻²
Acceleration due to gravity on mars: g_m = 3.7ms⁻²
Weight: w = m × g
Weight on mars: wm=m×gm
⇒ w_m = 80 × 3.7
w_m = 296N
The mass of astronauts remains the same on mars because it is a constant value.
Thus, mass on mars is m = 80kg.

Q18: Which of the following has more inertia:
(a) A rubber ball and a stone of the same size?
Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a rubber ball and a stone of the same size, it is clear that the stone will have greater inertia than the ball. It is because, despite being the same size, the stone weighs more than the rubber ball.
(b) A bicycle and a train?
Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a bicycle and a train, it is clear that the train will have greater inertia than the bicycle because the train weighs more than the bicycle.
(c) A five rupees coin and a one-rupee coin?
Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a five rupees coin and a one-rupee coin, the five rupees coin will have greater inertia than the one-rupee coin because five rupees coin weighs more than a one-rupee coin.

Q19: 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.

Ans: In the example, the number of times the velocity of football changes is four.

(i) The velocity of the football changes first when a player kicks the ball towards another player on his team. Here, the agent supplying the force is the foot of the football player who is kicking the ball.
(ii) The velocity of the football changes second when that another player kicks the ball towards the goal. Here, the agent supplying the force is the foot of the other player who is now kicking the ball towards the goal.
(iii) The velocity of the football changes for the third time when the goalkeeper of the opposite team stops the football by collecting it. Here, the agent supplying the force are the hands of the goalkeeper who collects the ball.
(iv) The velocity of the football changes for the fourth time when the goalkeeper kicks it towards a player of his team. Here, the agent supplying the force is the foot of the goalkeeper who is now kicking the ball towards his teammate.

Q20: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Ans: Some leaves may detach from a tree when its branch is vigorously shaken due to the following reasons:

• The branches move while the leaves tend to remain still because of their inertia.
• When the branch shakes, the force causes a rapid change in direction.
• This sudden force can exceed the attachment strength of the leaves, leading to their detachment.

In summary, the combination of inertia and the force of shaking results in some leaves falling off the tree.

Q21: Why do you fall in the forward direction when a moving bus breaks to a stop and fall back when it accelerates from rest?
Ans: The behaviour of passengers in a moving bus can be explained by the principle of inertia. When the bus suddenly stops, the following occurs:

• The passengers’ upper bodies continue moving forward because they are in a state of motion.
• The lower part of their bodies, which is in contact with the seat, remains at rest.
• This difference causes the upper body to be pushed forward, in the direction the bus was moving.

Conversely, when the bus accelerates from a stop:

• The lower part of the passengers’ bodies begins to move forward with the bus.
• The upper bodies, however, tend to stay at rest due to inertia.
• This results in the upper bodies being pushed backward, opposite to the direction of the bus’s acceleration.

Q22: If action is always equal to the reaction, explain how a horse can pull a cart.
Ans: The ability of a horse to pull a cart can be explained through the principles of action and reactionas outlined in Newton’s third law of motion. Here’s how it works:

• When the horse pulls the cart, it exerts a force on the cart.
• According to Newton’s third law, the cart exerts an equal and opposite force back on the horse.
• The horse’s muscles generate enough force to overcome the cart’s resistance and any friction with the ground.
• The ground also plays a crucial role; as the horse pushes against it, the ground pushes back, allowing the horse to move forward.
• This interaction enables the horse to pull the cart despite the equal and opposite forces at play.

In summary, the horse can pull the cart because:

• The horse applies a force to the cart.
• The reaction force from the cart does not prevent the horse from moving forward.
• The ground provides necessary traction for the horse to exert force effectively.

Q23: Explain why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Ans: It is difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity because of Newton’s third law of motion. In this case, the water being ejected out in the forward direction with great force (action) will create a backward force that results in the backward movement (reaction) of the hose pipe. As a result of this backward force and movement, it will be difficult for the fireman to hold the hose properly with stability.

Q24: From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35ms⁻¹. Calculate the initial recoil velocity of the rifle.
Ans: Given:
Mass of rifle: m₁= 4kg
Mass of bullet: m₂ = 50g = 0.05kg
Initial velocity of rifle: u₁= 0ms⁻¹ (it is stationary during firing)
Initial velocity of bullet: u₂ = 0ms⁻¹(it starts from rest, inside the barrel of the rifle)
Fired velocity of bullet: v₂ = 35ms⁻¹
To find: Recoil velocity of rifle:v₁
By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get –
m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂
Here, m₁u₁ + m₂u₂ is the total sum of momentum of rifle and bullet before firing and m₁v₁ + m₂v₂ is the total sum of momentum of rifle and bullet after firing.
Substituting the values in – m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂

⇒ v₁ = −4.375ms⁻¹ (The negative sign indicates the backward direction in which the rifle moves when it recoils)
Thus, the recoil velocity of the rifle is 4.375ms⁻¹.

Q25: An 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40000N and the track offers a friction force of 5000N, then calculate:
(a) The net accelerating force
Ans: Given:
Force exerted by engine on wagons: F = 40000N
Frictional exerted on wagons: f = 5000N
Mass of engine: me = 8000kg
Mass of each wagon: m_w=2000kg
Mass of all five wagons: m_W = 5 × m_w = 5 × 2000 = 10000kg
Mass of entire train: m_T=m_e + m_W = 8000 + 10000 = 18000kg
To find accelerating force.
Net accelerating force can be found by subtracting the frictional force from the force exerted by the engine on the wagons.
Thus, Net Accelerating Force=Force Of Engine − Frictional Force
⇒ Net Accelerating Force = F − f
⇒ Net Accelerating Force = 40000 − 5000
⇒ Net Accelerating Force = 35000N
(b) The acceleration of the train
Ans:  Given:
Force exerted by engine on wagons: F = 40000N
Frictional exerted on wagons: f = 5000N
Mass of engine: m_e = 8000kg
Mass of each wagon: m_w = 2000kg
Mass of all five wagons: m_W = 5 × m_w = 5 × 2000 = 10000kg
Mass of entire train: m_T = m_e + m_W=8000 + 10000=18000kg
To find the acceleration of the train.
It is known that – F = ma


Q26: wo objects, each of mass 1.5kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5ms−1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Ans: Given:
Mass of object 1: m₁= 1.5kg
Mass of object 2: m2 = 1.5kg
Initial velocity of object 1: u₁= 2.5ms⁻¹
Initial velocity of object 2: u₂ = −2.5ms⁻¹ (negative sign because it is moving in the opposite direction)
Mass of combined object after collision: m = m₁ + m₂ = 1.5 + 1.5 = 3kg
To find: Final velocity of the combined object after collision:v
By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get –
m_v = m₁u₁ + m₂u₂
Here, m₁u₁ + m₂u₂ is the total sum of the momentum of objects before the collision and mv is the total momentum of the combined objects after the collision.
Substituting the values in – m_v = m₁u₁ + m₂u₂
⇒ (3 × v) = (1.5 × 2.5) + (1.5 × −2.5)
⇒ (3 × v) = (3.75) + (−3.75s)
⇒ (3 × v) = 0
⇒ v = 0ms⁻¹
Thus, the velocity of the combined object after collision is 0ms⁻¹.

Q27: According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Ans: The student’s reasoning that the two opposite and equal forces cancel each other is not entirely correct. Here’s a clearer explanation of why the truck does not move:

• Third Law of Motion: According to this law, when we push an object, it pushes back with an equal and opposite force.
• Massive Truck: A truck is very heavy, which means it has a large mass. The force required to move it must overcome its inertia.
• Balanced Forces: When you push the truck, the force you apply is equal to the force the truck exerts back on you. However, this does not mean the truck will move. If the frictional force between the truck’s wheels and the ground is greater than the force you apply, the truck will remain stationary.
• Friction: The friction between the truck’s tyres and the road is a significant factor. It resists movement and must be overcome for the truck to start moving.
• Net Force: For an object to move, there must be a net unbalanced force acting on it. In this case, if your push does not exceed the frictional force, the net force is zero, and the truck does not move.

Q28: A hockey ball of mass 200g traveling at 10ms⁻¹ is struck by a hockey stick so as to return it along its original path with a velocity at 5ms⁻¹. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans: Given:
Mass of hockey ball: m = 200g = 0.2kg
Initial velocity of hockey ball: u = 10ms⁻¹
Final velocity of hockey ball: v = −5ms⁻¹ (because it moves back in its original direction)
To find: Change in momentum of hockey ball due to the force of hockey stick
ChangeOfMomentum = m_v − m_u
Here, m_u is the initial momentum of the hockey ball and m_v is the final momentum of the hockey ball.
Substituting the values in – Change Of Momentum = m_v − m_u
⇒ ChangeOfMomentum = (0.2 × −5) − (0.2 × 10)
⇒ ChangeOfMomentum = (−1) − (2)
⇒ ChangeOfMomentum = −3kgms⁻¹
Thus, the change in momentum of hockey ball due to the force of hockey stick is −3kgms⁻¹.

Q29: A bullet of mass 10g traveling 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.
Ans: Given:
Mass of bullet: m = 10g = 0.01kg
Initial velocity of bullet: u = 150ms⁻¹
Final velocity of bullet: v = 0ms⁻¹ (as it comes to rest after penetration)
Time duration of bullet travel: t = 0.03s
To find:

• Distance of penetration of bullet into the block
• Force exerted by the block on the bullet

(a) Distance of penetration:
It is known that – v = u + at
Thus, 0 = 150 + (a × 0.03)
⇒ (a × 0.03) = −150
⇒ a = −5000ms⁻²
Now,

(b) Next, finding the force:
F = ma
⇒ F = (0.01 × −5000)
⇒ F = −50N
Thus,

• Distance of penetration of bullet into the block is 2.25m
• Force exerted by the block on the bullet is −50N

Q30: Differentiate between mass and weight?
Ans: The difference between mass and weight is given below,


Q31: A scooter is moving with a velocity of 20ms−1when brakes are applied. The mass of the scooter and the rider is 180kg. The constant force applied by the brakes is 500N.
(a) How long should the brakes be applied to make the scooter comes to a halt?
Ans:  Given:
Mass of scooter and rider: m = 180kg
Initial velocity of scooter: u = 20ms−1
Final velocity of scooter: v = 0ms⁻¹(as it halts after applying the brake)
Force of the brake: F = −500N(as it opposes the motion)
To find the time duration over which brake should be applied to stop the scooter.
Now, the force – F = ma

Rearranging,

(b) How far does the scooter travel before it comes to rest?
Ans: Given:
Mass of scooter and rider: m = 180kg
Initial velocity of scooter: u = 20ms⁻¹
Final velocity of scooter: v = 0ms⁻¹(as it halts after applying the brake)
Force of the brake: F = −500N(as it opposes the motion)
To find distance travelled by scooter before coming to halt.

Acceleration is negative because it is retarding the motion of the scooter.

Q32: State Newton’s third law of motion and how does it explain the walking of man on the ground?
Ans: Newton’s third law of motion states that for every action, there is an equal and opposite reaction acting on different bodies. This implies the existence of the action-reaction force pair. That is, for every action force created an equal and opposite reaction force will be created.
The walking of a man on the ground can be explained with Newton’s third law of motion. During walking on the ground, the man creates an active force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the man forward enabling him to walk.

Q33: State Newton’s second law of motion and derive it mathematically?
Ans: Newton’s second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Mathematical derivation:
Say we have an object of mass m that is moving along a straight line with an initial velocity, u.
It is then uniformly accelerated to velocity, v in time, t  by the application of a constant force, F throughout the time, t.
Initial Momentum of object: p₁= m × u
Final Momentum of object: p₂= m × v
Now, the change of momentum is the Final momentum subtracted by the Initial momentum
Thus, Δp = p₂ − p₁= m_v − m_u = m(v − u)
⇒ Δp = m(v − u)
The rate of change of momentum is Δp / t

We know that the applied force is proportional to the rate of change of momentum of the object.

Using these, we get
⇒ F = ma
The SI unit of force is Newton (Kgms⁻²)
The second law of motion gives a method to measure the force acting on an object as a product of its mass and acceleration.

Q34: Why does a person while firing a bullet holds the gun tightly to his shoulders?
Ans: A person while firing a bullet holds the gun tightly to his shoulder because of the recoil of the gun when the bullet is fired. This is in accordance with Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction, that acts on different bodies. 

• When a bullet is fired, the forward motion of the bullet (action) creates a recoil or backward motion of the gun (reaction). The action force being much greater will create an equivalent recoil force in the backward direction. 
• If the person who holds the gun does not hold it properly to his shoulders that can result in injury. This is because the shoulder absorbs most of the force during recoil that enables the shooter to take a steady shot.
• Thus, if not held tightly to his shoulders, the shot will not be precise and this can also cause the gun to fly away from his hands.

Q35: A car is moving with a velocity of 16ms−1 when brakes are applied. The force applied by the brakes is 1000N. The mass of the car its passengers is 1200kg.
(a) How long should the brakes be applied to make the car come to a halt?
Ans: Given:
Mass of car and passengers: m = 1200kg
Initial velocity of car: u = 16ms⁻¹
Final velocity of car: v = 0ms⁻¹(as it halts after applying the brake)
Force of the brake: F = −1000N(as it opposes the motion)
To find time duration over which brake should be applied to stop the car.
Now, the force – F = ma

Rearranging,

(b) How far does the car travel before it comes to rest?
Ans: Given:
Mass of car and passengers: m = 1200kg
Initial velocity of car: u = 16ms⁻¹
Final velocity of car: v = 0ms⁻¹(as it halts after applying the brake)
Force of the brake: F = −1000N(as it opposes the motion)
To find distance travelled by car before coming to halt.
Acceleration – 

Acceleration is negative because it is retarding the motion of the scooter.

Q1: Distinguish between speed and velocity.
Ans: Speed of a body is the distance travelled by a body  per unit time while velocity is the displacement travelled by a body  per unit time.

Q2: Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Ans: If the distance travelled by a body is equal to the displacement, then the magnitude of the average velocity of an object will be equal to its average speed.

Q3: What does the odometer of an automobile measure?
Ans: The odometer of an automobile is used to measure the distance covered by an automobile.

Q4: What does the Graph of an object look like when it is in uniform motion?
Ans: Graphically it will be linear; it looks like a straight line when it is in uniform motion.

Q5: 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 × 10⁸ms⁻¹
Ans: The given data is that time is five minutes and speed is(3 × 10⁸ms⁻¹)
Distance = Speed × Time
⇒ 5min × (3 × 10⁸ms⁻¹)
⇒( 5 × 60)sec × (3 × 10⁸ms⁻¹)
⇒ 300sec × (3 × 10⁸ms⁻¹)
⇒ 900 × 10⁸ms⁻¹ = 9 × 10¹⁰m
∴ Distance = 9 × 10⁷km

Q6: When will you say a body is in
(i) Uniform acceleration?
Ans: When an object travels in a straight line and its velocity changes by equal amount in an equal interval of time, it is said to have uniform acceleration.
(ii) Non-uniform acceleration?
Ans:  Non-uniform acceleration is also called variable acceleration. When the velocity of an object changes by unequal amounts in equal intervals of time, it is said to have non-uniform acceleration.

Q7: A bus decreases its speed from 80kmh⁻¹ to 60kmh⁻¹ in 5s. Find the acceleration of the bus.
Ans: Initial speed of bus (u)  = 80kmh⁻¹

Acceleration (a) =  

∴ Acceleration (a) = −1.11m/s²

Q8: What is the nature of the distance time graphs for uniform and non-uniform motion of an object?
Ans: If an object has a uniform motion then the nature of distance time graph will be linear, that is it would in a straight line and if it has non-uniform motion then the nature of the distance-time graph will be a curved line.

Q9: What is the quantity which is measured by the area occupied below the velocity-time graph?
Ans: The area occupied below the velocity-time graph measures the displacement moved by any object.

Q10: A bus starting from rest moves with a uniform acceleration of 0.1ms⁻² for 2 minutes . Find
(a) The speed acquired,    
Ans: u = 0, a = 0.1ms⁻², t = 2min = 120sec
v = u + at = 0 + 0.1 × 120=12ms⁻¹
Speed acquired = v = 12ms⁻¹
(b) The distance travelled.
Ans: s = ut + 1/2at² = 0 × 120 + 1/2 x 0.1 × 120² = 720m

Q11: A trolley, while going down an inclined plane, has an acceleration of 2cm s⁻² . What will be its velocity 3s after the start?
Ans: Given : u = 0, a = 2cm/s², t = 3s
v = u + at = 0 + 2 × 3
= 6cm/s

Q12: A racing car has a uniform acceleration of4ms⁻² . What distance will it cover in 10s  after start?
Ans: Given: u =0, a = 4m/s², t = 10s
s = ut + 1/2at²
s = 0 × 10 + 1/2 × 4 × 10²
∴ s = 200m

Q13: Differentiate between distance and displacement?
Ans: The difference between distance and displacement is as below, 

Q14: Derive mathematically the first equation of motion V = u + at?
Ans: Acceleration is defined as the rate of change of velocity.
Let V = final velocity; V_o = initial velocity, T =  time, a = acceleration.
So by definition of acceleration

If V_o = u =  initial velocity, then  [V = u + at]

Q15: Calculate the acceleration of a body which starts from rest and travels 87.5m in 5sec?
Ans: Given Data: u = 0 (starts from rest) u =  initial velocity
a =  acceleration = ?
t = 5sec, t = time
S = 87.5m (S = distance)
From second equation of motion

∴ a = 7m/s² 

Q16: Define uniform velocity and uniform acceleration?
Ans: 

• Uniform velocity:- A body is said to move with uniform velocity if equal displacement takes place in equal intervals of time, however small these intervals may be. 
• Uniform acceleration:- A body is said to move with uniform acceleration if equal changes in velocity take place in equal intervals of time, however, small intervals may be.

Q17: The velocity-time graph of two bodies A and B traveling along the +x direction are given in the (a) Are the bodies moving with uniform acceleration?
Ans: Yes the bodies are moving with uniform acceleration.
(b) Which body is moving with greater acceleration A or B?
Ans: Body A is moving with greater acceleration.

Q18: Calculate the acceleration and distance of the body moving with 5m/s which comes to rest after traveling for 6sec?
Ans: Acceleration = a = ?
Final velocity = V = o (body comes to rest)
Distance = s     =?
Time  = t =  6 sec
From,    V = u + at

∴ s = 15m

Q19: A body is moving with a velocity of 12m/s and it comes to rest in 18m, what was the acceleration?
Ans: Initial velocity = u = 12m/s
Find velocity  = V = 0
S = distance=18m
A= acceleration =?
From 3^rd equation of motion;

Q20: A body starts from rest and moves with a uniform acceleration of 4m/s² until it travels a distance of 800m, find the find velocity?
Ans: Initial velocity =u=0
Final velocity = v = ?
Acceleration=a=4m/s²
Distance = s = 800m
v² − u² = 2as
v² − (0) = 2 × 4 × 800
v = 80m/s

Q21: Differentiate between scalars and vectors?
Ans: The difference between scalars and vectors is as below,

Q22: An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
Ans: Yes, if an object is moved through a distance it can have zero displacement because displacement of an object is the actual change in its position when it moves from one position to the other position. So if an object travels from point A to B and then returns back to point A again, the total displacement will be zero.

Q23: A train starting from a railway station and moving with uniform acceleration attains a speed 40kmh⁻¹ in 10min. Find its acceleration.
Ans: Since the train starts from rest (railway station) = u = zero
Final velocity of train = v = 40kmh⁻¹

time(t) = 10min = 10 × 60 = 600 seconds
Since

Q24:What can you say about the motion of an object whose distance time graph is a straight line parallel to the time axis?
Ans:    If the object’s distance time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest that is not moving.

Q25: What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis?
Ans:  Such a graph indicates that the object is travelling with uniform velocity.

Q26: A train is travelling at a speed of 90kmh⁻¹ . Brakes are applied so as to produce a uniform acceleration of−0.5ms⁻² . Find how far the train will go before it is brought to rest.
Ans: 

Given: a = −0.5ms⁻², v = 0 (train is brought to rest)


Q27: A stone is thrown in a vertically upward direction with a velocity of5ms⁻¹ . If the acceleration of the stone during its motion is10ms⁻¹ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Ans:     Given: u = 5ms⁻¹, a = −10ms⁻²
v = 0   (since at maximum height its velocity will be zero)
v = u + at = 5 + (−10) × t


Q28: Two cars A and B are moving along in a straight line.    Car A is moving at a speed of 80kmph  while car B is moving at a speed of 50kmph in the opposite direction, find 
(a) The relative velocity of car B with respect to A.
(b)  Relative velocity of car B with respect to A

Ans:
(a) The relative velocity of car B with respect to A.
Velocity of car A = 80kmph
 Velocity of Car B = − 50  kmph  (-ve sign indicates that Car B is moving in the opposite direction to Car A )
The relative velocity of car A with respect to B
velocity of  car A + (- velocity    of car B)
⇒ 80 + (−(−50))
⇒ 80  + 50
⇒ +130kmph
shows that for a person in car B, car A will appear to move in the opposite direction with a speed of the sum of their individual speed.
(b)  Relative velocity of car B with respect to A
Ans: 
 ⇒ Velocity of car B+ (- velocity of car A)
⇒ −50  + (−80)
⇒ −130kmph
It shows that car B will appear to move with 130   kmph in opposite direction to car A

Q29: A ball starts from rest and rolls down 16m down an inclined plane in 4s.
(a) What is the acceleration of the ball?
Ans:  Given: u= initial velocity = 0 (body starts from rest) S= distance = 16 m
T= time = 4s

(b) What is the velocity of the ball at the bottom of the incline?
Ans: From, v = u + at
v = 0 + 2 × 4
[v = 8m/s]

Q30: 

Fig 2.26 shows the displacement-time graph for the motion of two boys A and B along a straight road in the same direction. Answer the following:
(i) When did B start after A ?
(ii) How far away was A from B when B started?
(iii) Which of the two has greater velocity ?
(iv) When and where did B overtake A ?

.
Ans:(i) B started his motion 2 h later from the start of A.
(ii) When B started, A was at distance 10 km away from B.
(iii) B has greater velocity than A since the straight line on graph for B has greater slope than that for A.
(iv) B overtook A at the instant when both were at the same place. This position is at the point where the two straight lines meet each other. For this point, distance from the starting point is 20 km and time is 4 h. Thus B overtook A when A has travelled for 4 h (or B has travelled for 4 – 2 = 2 h) at distance 20 km from the starting point

Q31: A body is dropped from a height of320m. The acceleration due to the gravity is 10m/s²?
(a) How long does it take to reach the ground?
Ans:  Given Data: Height = h
Distance = s = 320m
Acceleration due to gravity = g = 10m/s²
Initial velocity  = u = 0

(b) What is the velocity with which it will strike the ground?
Ans: From v = u + at
v = 0 + 10 × 8
v = 80m/s

Q32: Derive third equation of motion v² − u² = 2as numerically?
Ans: We know,

When, v= final velocity
u= initial velocity
a = acceleration
t = time
s = distance
From equation (i) 
Put the value of t in equation (ii)


Q33: The velocity-time graph of the runner is given in the graph.
(a) What is the total distance covered by the runner in16s?
Ans:  We know that area under v-t graph gives displacement:
So, Area = distance = s = area of triangle + area of rectangle Area of triangle = 1/2 × base × height

∴Area of triangle = 30m
Area of rectangle= length × breadth
⇒ (16 − 6)×10
⇒ 10 × 10
⇒ 100m
Total area = 130m
Total distance = 130m
(b) What is the acceleration of the runner at t = 11s?
Ans: Since at t = 11sec , particles travel with uniform velocity so, there is no change in velocity hence acceleration = zero.

Q34: A boy throws a stone upward with a velocity of 60m/s .
(a) How long will it take to reach the maximum height(g=−10m/s²)?
Ans: u = 60 m/s; g = -10m/s²; v=0
 The time to reach maximum height is
v = u + at = u + gt
0 = 60 − 10t
t = 60 / 10
=6s
(b) What is the maximum height reached by the ball?
Ans: The maximum height is:

(c) How long will it take to reach the ground?
Ans: The time to reach the top is equal to the time taken to reach back to the ground. Thus, the time to reach the ground after reaching the top is 6s or the time to reach the ground after throwing is 6 + 6 = 12s.

Q35: Derive the third equation of motion  v² − u² = 2as  as graphically?
Ans: Let at time t = 0, the body moves with initial velocity u and time at ‘t’ has final velocity ‘v’ and in time ‘t’ covers a distance ‘s’

Put the value of ‘t’ in equation (i)

Third equation of motion.

Q1: Who discovered cells, and how?
Ans: Robert Hooke, an English scientist, discovered cells in 1665. While examining a tiny slice of cork under his self-designed microscope, he saw a honeycomb-like structure.

Q2:  Why is the cell called the structural and functional unit of life?
Ans: Because all living organisms are made up of cells, the cell is the basic construction unit of a living organism, and all of a living organism’s activities are the sum of activities conducted by its cells, the cell is referred to as the structural and functional unit of life.

Q3: Why is the plasma membrane called a selectively permeable membrane?
Ans: The plasma membrane is a very distinct structure. It is made up of lipids and proteins that selectively allow the entry of some molecules into the cell while preventing the exit of others, making it selectively permeable.

Q4: If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen?
Ans: Because all of a cell’s components are digested by its lysosomes, if the cell’s organisation is damaged by some physical or chemical action, the cell will not survive.

Q5: Why are lysosomes known as suicide bags?
Ans: Lysosomes are organelles inside cells that contain hydrolytic (digestive) enzymes. When a cell is injured, its lysosomes may burst, allowing enzymes to digest the cell itself. As a result, we can call lysosomes “suicide bags.”

Q6: What would happen if the plasma membrane ruptures or breaks down?
Ans: The rupture or breakdown of a cell’s plasma membrane signals that the cell has been injured, and in this case, the damaged cell’s lysosomes may burst, causing the digestive enzymes inside those lysosomes to eat their own cell. The cell will die as a result of this.

Q7: What would happen to the life of a cell if there was no Golgi apparatus?
Ans: The preservation, modification, and packaging of products in particles are all tasks of the Golgi apparatus. All types of storage, modification, packaging and dispatch of materials within and beyond the cell would be impossible if there was no Golgi apparatus for the cell.

Q8: Which organelle is known as the powerhouse of the cell? Why?
Ans: The cell’s Mitochondria, also known as the cell’s powerhouse, synthesises energy in the form of ATP during respiration, which is essential for many living processes.

Q9: Where do the lipids and proteins constituting the cell membrane get synthesised?
Ans: There are two types of endoplasmic reticulum:
(i) The smooth endoplasmic reticulum (SER) is in charge of producing the lipids that make up the cell membrane.
(ii) The ribosomes are housed in the rough endoplasmic reticulum (RER), which is responsible for the production of proteins that make up the cell membrane.

Q10: How do substances like CO₂ and water move in and out of the cell? Discuss.
Ans: The exchange of gases CO₂ and O₂  between cells takes place by a diffusion process.
Diffusion is the migration of a chemical from a high-concentration region to a low-concentration zone. CO₂ is produced inside the cell as a result of respiration and accumulates in the cell, resulting in a high concentration of CO₂ in the cell compared to the outside environment. However, because O₂ is used inside the cell during respiration, its concentration declines inside the cell while remaining relatively high in the environment. As a result, CO₂ diffuses out of the cell and O₂ diffuses in.

Q11: Fill in the gaps in the following table illustrating the differences between prokaryotic and eukaryotic cells.
Ans: The difference between prokaryotic cell and eukaryotic cell as follows:

Q12: Make a comparison and write down ways in which plant cells are different from animal cells.
Ans: The difference between plant cells and animal cells is as follows:

Q13: How is a prokaryotic cell different from a eukaryotic cell?
Ans: The difference between prokaryotic cell and eukaryotic cell as follows:

Q14: How does an Amoeba obtain its food?
Ans: Amoeba feeds on planktonic bacteria that float in water. It grows artificial feet, or pseudopodia, to encircle the meal and then catches it in a sac-like structure called the food vacuole, within which food digestion occurs.Process showing Amoeba obtaining its food
Q15: What is osmosis?
Ans: The passage of water (solvent) through a semipermeable membrane from a location of high-water concentration to a region of low water concentration is known as osmosis. It can only happen in a liquid medium; it cannot happen in solids or gases. Plant roots, for example, absorb water from the earth.

Q16: Carry out the following osmosis experiment:
Take four peeled potato halves and scoops 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 of sugar in the boiled potato cup D.
Keep these for two hours. Then observe the four potato cups and answer the following:
(i) Explain why water gathers in the hollowed portion of B and C.
Ans: When we add one teaspoon of sugar in cup B and one teaspoon salt in cup C, we create a hypertonic solution within, which allows water from outside to enter through osmosis and accumulate in the hollowed area of cups B and C.
(ii) Why is potato A necessary for this experiment?
Ans: To observe osmosis, you’ll need Potato A.
(iii) Explain why water does not gather in the hollowed-out portions of A and D.
Ans: Because there is no solution (liquid medium) in the hollowed-out portions of A and D, osmosis cannot occur, and water does not collect.

Q1: On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Ans: As per Thomson’s model of an atom, the number of electrons and the number of protons are equal in an atom. Electrons are positively charged and protons are negatively charged, hence the + and – charges are neutralized by each other that makes atoms neutral as a whole.

Q2: On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nucleus of an atom?
Ans: The subatomic particle present in the nucleus of an atom, according to Rutherford’s model, is the proton.

Q3: Draw a sketch of Bohr’s model of an atom with three shells.
Ans: Bohr’s model of an atom with three shells is as follows:

Q4: 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?
Ans: If the α− particle scattering experiment is carried out using a foil of a metal other than gold we will get a different observation. Overall, the results would depend on the properties of the metal used.

Q5: Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Ans: The atomic mass of an atom is the sum of masses of protons and neutrons present in its nucleus.
Given that the mass of the helium atom is 4 u and two protons present in its nucleus.
So the number of neutrons will be
Number of neutrons = atomic mass − number of protons
⇒ Number of neutrons = 4−2
∴ Number of neutrons = 2
Therefore, the helium atom has 2 neutrons.

Q6:  Write the distribution of electrons in carbon and sodium atoms.
Ans: Distribution of electrons in carbon and sodium atoms is as follows:

Carbon:

• Atomic number: 6
• Electron distribution: 2, 4

Sodium:

• Atomic number: 11
• Electron distribution: 2, 8, 1

Q7: If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Ans: K shell contains total 2 electrons and L shell contains maximum 8 electrons. If K and L shells of an atom are full, then the total number of electrons in the atom will be 10.

Q8: If 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?
Ans: The atomic number of an atom is equal to the number of protons or electrons present in its nucleus. So the atomic number of an atom with 8 electrons and 8 protons is 8.
(b) What is the charge on the atom?
Ans: A single electron contains one negative charge and one single proton contains one positive charge. There are equal numbers of electrons and protons in an atom so they neutralize each other. The atom will be neutral.

Q9:  Write the electronic configuration of following ions:
(a) Cl⁻
Ans: Electronic configuration of Cl⁻ ion is 2,8,8.
(b) Mg
Ans: Electronic configuration of Mg ion is 2,8,2.
(c) Al³⁺
Ans: Electronic configuration of Al³⁺ ion is 2,8.
(d) O
Ans: Electronic configuration of O is 2,6.

Q10: What are the limitations of J.J. Thomson’s model of the atom?
Ans: The J.J. Thomson’s atomic model failed to explain the organization of electrons in an atom.

Q11: Na⁺ has completely filled K and L shells. Explain.
Ans: Sodium (Na) has atomic number 11, so the electronic configuration of Na is 2,8,1.
It has a single electron in the outermost shell, when it gives away that electron it becomes Na⁺ and has electronic configuration 2,8. Also the K shell contains a total 2 electrons and the L shell contains a maximum of 8 electrons. So Na⁺ has completely filled K and L shells.

Q12: If z = 3, what would be the valency of the element? Also, name the element.
Ans: z = 3 represents that element has 3 electrons in its shells. The electronic configuration is 2,1. It means the outermost shell electron has 1 electron, so its valency is 1. The element is Lithium.

Q13: 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.
Ans: False
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
Ans: True
(c) The mass of an electron is about 12000 times that of proton.
Ans: True
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Ans: False

Q14: Explain the formation of Al³⁺ ion and why is it formed?
Ans: Aluminum has an atomic number of 13. The electronic configuration of Al is 2,8,3. It has 3 electrons in the outermost shell and to become stable it needs to complete its octet. In the outermost shell, the maximum number of electrons must be 8. So it is easy to lose 3 electrons and complete the octet. By giving away the 3 outermost electrons it becomes  Al³⁺ ion.

Q15: Why metals are electropositive and non-metals are electronegative in nature?
Ans: Metals are electropositive in nature because all metals give away electrons from their outermost shell in order to complete the octet and become stable. So metals become positively charged. 
Non-metals are electronegative in nature because all non-metals gain electrons in order to complete the octet and become stable. So non-metals become negatively charged.

Q16: Write the postulates of Bohr theory?
Ans: The postulates of Bohr’s theory are as follows:

• Electrons revolve around the nucleus in specific paths called orbits or shells.
• Each orbit has a fixed amount of energy.
• The energy increases from the inner shells to the outer shells, with the innermost shell having the lowest energy.
• If energy is supplied, an electron can move from a lower orbit to a higher orbit.

Q17: Compare the three major particles in atoms with respect to their mass and charge?
Ans: Comparison of three major particles proton, neutron and electron with respect to their mass and charge is as follows:

Inside an atom electron revolves around the nucleus in a circular path. Protons and neutrons are present inside the nucleus.

Q18: In a gold – foil experiment:
(a) Why did many α− particles pass through the gold foil undeflected?
Ans: Most of the space within the atom was empty so many α− particles passed through the gold foil undeflected.
(b) Why did few α− particles deflect through small angles.
Ans: In a gold foil at center there is a positive charge so few α− particles deflect through small angles.
(c) Why did few α− particles, after striking the gold foil, retrace their path.
Ans:  Very few α-particles retraced their path after striking the foil, indicating that the positively charged nucleus is very small.

Q19: Define valency by taking examples of silicon and oxygen.
Ans: The valency of electrons is determined by electrons present in the outermost shell of an atom. Electrons present in the outermost shell of an atom are known as the valence electrons. Electrons gain or lose electrons to complete its octet. 

Atomic number of silicon is 14 and electronic configuration is 2,8,4. This means it can either gain or lose 4 electrons. So the valency of silicon is 4. 
Atomic number of oxygen is 8 and the electronic configuration is 2,6. To complete its octet oxygen gains 2 electrons hence the valency of oxygen is 2.

Q20: The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes  and in the sample?
Ans: Average atomic mass of sample is given as

We get

The percentage of isotopes is 

Q21: Define isotopes.

Ans: Isotopes are atoms which have identical atomic numbers but different mass numbers. Examples of isotopes are 

Q22: Which of the following electronic configuration are wrong and why?
(a) 2,8,2
(b) 2,8,8,2
(c) 2,8,9,1.
Ans: From the given electronic configuration, 2,8,9,1 is wrong because in the third shell the maximum number of electrons is 8. The correct electronic configuration is 2,8,8,2.

Q23: What are ions? What are its two types?
Ans: When one or more electrons are detached from a neutral atom, a positively charged particle is formed and called an ion. Ions may be cations and anions.

Q24: Show diagrammatically the formation of O²⁻ ion?
Ans: Atomic number of oxygen is 8 and its electronic configuration is 2,6. In the outermost shell oxygen has 6 electrons. To complete its octet and become stable it needs 2 electrons. By gaining 2 electrons it becomes O²⁻ ion.
Diagrammatic representation of formation of O²⁻ ion is as follows:

Q25: Define isobars.
Ans: Isobars are atoms that have different atomic numbers but the same mass number. Examples of isobars are 

Q26: For the symbol H, D and T tabulate three subatomic particles found in each of them.
Ans: H represents the hydrogen atom, D represents the deuterium atom and T represents the tritium atom. Three subatomic particles present in each of them is represented as follows:

Q27: Write the electronic configuration of any one pair of isotopes and isobars.
Ans: Electronic configuration of pairs of isotopes of carbon is  . Isotopes have the same number of electrons and protons.
Electronic configuration of a pair of isobars of argon and calcium is 

Q28: Compare the properties of electrons, protons and neutrons.
Ans: Comparison of electrons, protons and neutrons is as follows:

Q29: What are the limitations of Rutherford’s model of the atom?
Ans: Rutherford’s model of the atom has several limitations:

• The model does not explain the stability of atoms. It suggests that electrons orbit the nucleus, which would cause them to emit energy.
• As electrons lose energy, their orbits would shrink, eventually leading them to collide with the nucleus. This implies that atoms should be unstable, contradicting the observed stability of matter.

Q30: If bromine atom is available in the form of, say, two isotopes  Calculate the average atomic mass of bromine atom.
Ans: The average atomic mass of bromine is

Average atomic mass of bromine atoms is 80 u.

Q1: Define the atomic mass unit.
Ans: One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12^th) the mass of one atom of carbon-12. Relative atomic masses of all elements are measured with respect to the carbon-12 standard.
According to IUPAC (International Union of Pure and Applied Chemistry), the atomic mass unit (written as u, the unified mass unit) is defined as the mass of one-twelfth of a carbon-12 atom.
1 amu = 1/12^th mass of C₆¹²

Q2: Write down the formulae of 

(a) Sodium oxide
Ans:  Sodium oxide – Na₂O

(b) Aluminium chloride
Ans:  Aluminium chloride – AlCl₃

(c) Sodium sulphide
Ans:  Sodium sulphide – Na₂S

(d) Magnesium hydroxide
Ans:  Magnesium hydroxide – Mg(OH)₂

Q3: Write down the names of compounds represented the following formulae:
(a) Al₂(SO₄)₃
Ans: Al₂(SO₄)₃ – Aluminium sulphate
(b) CaCl₂
Ans: CaCl₂ – Calcium chloride
(c) K₂SO₄
Ans:  K₂SO₄ – Potassium sulphate
(d) KNO₃
Ans:  KNO₃ – Potassium nitrate
(e) CaCO₃
Ans:  CaCO₃ – Calcium carbonate

Q4: What is meant by the term chemical formula?
Ans: The chemical formula of a compound is a symbolic representation showing the types of atoms present and the number of each type in a single molecule (or formula unit) of that compound. It uses atomic symbols and numbers (subscripts) to indicate composition.

• They provide information on the elements that constitute the molecules of a compound and the ratio in which the atoms of those elements combine to form the molecules.
• Example: A water molecule contains two hydrogen atoms and one oxygen atom. Its chemical formula is H₂O.

Q5: What are polyatomic ions? Give examples.
Ans: Polyatomic ions are groups of two or more atoms covalently bonded together that carry a net electrical charge and act as a single ion in chemical reactions. They behave as one charged unit.
Examples:

• Ammonium – NH₄⁺
• Hydroxide – OH^–
• Nitrate – NO₃^–
• Hydrogen carbonate – HCO₃^–

Q6: Write the chemical formulae of the following.
(a) Magnesium chloride
Ans: Magnesium chloride – MgCl₂
(b) Calcium oxide
Ans: Calcium oxide – CaO
(c) Copper nitrate
Ans: Copper nitrate – Cu(NO₃)₂ (copper commonly exists as Cu²⁺ in this salt)
(d) Aluminium chloride
Ans: Aluminium chloride – AlCl₃
(e) Calcium carbonate
Ans:  Calcium carbonate – CaCO₃

Q7: Give the names of the elements present in the following compounds.
(a) Quick lime
Ans: Quick lime – CaO
Elements present – Calcium, Oxygen
(b) Hydrogen bromide
Ans: Hydrogen bromide – HBr
Elements present – Hydrogen, Bromine
(c) Baking powder
Ans: Baking powder (sodium hydrogen carbonate+mild acid like tartaric acid) – NaHCO₃
Elements present – Sodium, Hydrogen, Carbon, Oxygen
(d) Potassium sulphate
Ans:  Potassium sulphate – K₂SO₄
Elements present – Potassium, Sulphur, Oxygen

Q8: What is the mass of –
Atomic mass of –
S = 32u,  Al = 27u, Na = 23u, N = 14u, O = 16u
(a) 1 mole of nitrogen atoms?
Ans: Given the atomic mass of nitrogen is 14 u, 1 mole of nitrogen atoms has a mass of 14 g.
(b) 4 moles of aluminium atoms (Atomic mass of aluminium is 27)?
Ans: Mass of 1 mole of aluminium = 27 g. Therefore mass of 4 moles = 27 × 4 = 108 g.

Q9: Convert into mole.
Atomic mass of – C = 12u, H = 1u, O = 16u
(a) 12 g of oxygen gas
Ans: Molar mass of O₂ = 16 × 2 = 32 g/mol
⇒ Number of moles = mass / molar mass
 = 12 / 32 = 0.375 mol
(b) 20 g of water
Ans: Molar mass of H₂O = (1 × 2) + 16 = 18 g/mol
⇒ Number of moles = mass / molar mass
⇒ Number of moles = 20 / 18 ≈ 1.11 mol
(c) 22 g of carbon dioxide
Ans: Molar mass of CO₂ = 12 + (16 × 2) = 44 g/mol
⇒ Number of moles = mass / molar mass
⇒ Number of moles = 22 / 44 = 0.50 mol

Q10: State the Postulates of Dalton Theory?
Ans: Dalton’s atomic theory proposed that matter is made of small particles called atoms. Its main postulates are:

• All matter is composed of very small particles called atoms.
• Atoms are indivisible in chemical processes and cannot be created or destroyed in a chemical reaction.
• Atoms of a given element are identical in mass and chemical properties.
• Atoms of different elements have different masses and chemical properties.
• Atoms combine in simple whole-number ratios to form compounds.
• A given compound always contains the same relative number and kinds of atoms (constant composition).

Q11: Find the percentage of water of crystallization in FeSO₄.7H₂O.
Ans: Atomic masses:
 Fe = 55.9 u,
 S = 32 u, 
H = 1 u,
 O = 16 u.
Molar mass of FeSO₄·7H₂O
= Fe + S + (O × 4) + 7 × (H₂O)
= 55.9 + 32 + (16 × 4) + 7 × [(1 × 2) + 16 × 1]
= 55.9 + 32 + 64 + 7 × 18
= 151.9 + 126 = 277.9 g/mol
So, 1 of FeSO₄ contains 126/277.6 g water of crystallization.
Mass of water of crystallization = 7 × 18 = 126 g per mole of the hydrated salt.
Percentage of water = (mass of water / molar mass of hydrated salt) × 100%

Thus, we get 126/277.6 x 100 = 0.4534 x 100 = 45.34%
The percentage of water of crystallization in FeSO₄·7H₂O is approximately 45.36%.

Q12: 2.42g  of copper gave 3.025g of a black oxide of copper, 6.49g of a black oxide, on reduction with hydrogen, gave 5.192g of copper. Show that these figures are in accordance with the law of constant proportion?
Ans: Given:
Case A –
Mass of copper = 2.42 g
Mass of copper oxide = 3.025 g
Case B –
Mass of black copper oxide = 6.49 g
Mass of copper after reduction = 5.192 g
To verify the law of constant proportion, calculate the percentage of copper in the oxide in both cases.
Percentage of copper in Case A = (mass of copper / mass of copper oxide) × 100% = (2.42 / 3.025) × 100 = 80.00%

Percentage of copper in Case B = (mass of copper / mass of copper oxide) × 100% = (5.192 / 6.49) × 100 = 80.00%

Both samples contain the same percentage of copper (80.00%). This shows that copper combines with oxygen in a constant proportion, confirming the law of constant proportions.

Q13: A compound was found to have the following percentage composition by mass Zn = 22.65%, S = 11.15% , H = 4.88% , O = 61.32% . The relative molecular mass is 287 g/mole . Find the molecular formula of the compound, assuming that all the hydrogen in the compound is present in water of crystallization.
Ans: Given percentages (per 100 g of compound): 
Zn = 22.65 g,
S = 11.15 g,
H = 4.88 g
O = 61.32 g.
Atomic masses: 
Zn = 65.4 u
S = 32 u
H = 1 u
 O = 16 u.
⇒  Number of atoms
= percentage of element present in a compound × mass of compound / atomic mass × 100
Using the formula above

Empirical atom counts ≈ Zn₁S₁H₁₄O₁₁.
Assuming all hydrogen is in water of crystallization,
H₁₄ corresponds to 7 molecules of H₂O. 
These 7 waters contribute 7 oxygen atoms,
 leaving O atoms in the anhydrous part = 11 – 7 = 4.
Thus the formula for the anhydrous part is ZnSO₄, with 7 H₂O as waters of crystallization.
Therefore the compound is ZnSO₄·7H₂O.

Q14: Which element will be more reactive and why – the element whose atomic number is 10 or the one whose atomic number is 11?
Ans: The element with atomic number 11 is more reactive than the element with atomic number 10.
Reason: The element with atomic number 11 has electronic configuration (2, 8, 1). It has one electron in its outermost shell and can lose that electron easily to attain a stable noble-gas configuration, so it is more reactive (typical of alkali metals). The element with atomic number 10 has configuration (2, 8) and a full outer shell; it is already stable and thus much less reactive (a noble gas).

Q15: What are the failures of Dalton’s Atomic theory?
Ans: Dalton’s atomic theory was an important step, but it had several limitations:

• It did not account for subatomic particles: Atoms are not indivisible; electrons, protons and neutrons were later discovered.
• It did not account for isotopes: Atoms of the same element can have different mass numbers
  Example: hydrogen ¹₁H, deuterium ²₁H , and tritium³₁H, have the same atomic number, but different mass numbers.
• It did not account for isobars: Different elements can have atoms with the same mass number 
  Example: Ar₄₀¹⁸ andCa₄₀²⁰, they have different atomic numbers, but the same mass number.
• It assumed elements always combine in simple whole-number ratios; some complex molecules have larger whole-number ratios , though they still use whole numbers.
  Example: sucrose C₁₂H₂₂O₁₁
• It could not explain allotropy: Different forms of the same element (diamond and graphite for carbon) have different properties despite containing only one type of atom.

Q16: Calculate the Molecular Mass of
Atomic mass of – S = 32u, H = 1u, C = 12u, N = 14u, O =16u
(a) Ammonium sulphate (NH₄)₂SO₄
Ans: Molar mass of (NH₄)₂SO₄
= 2 × [N + (H × 4)] + S + (O × 4)
= 2 × [14 + (1 × 4)] + 32 + (16 × 4)
= 2 × 18 + 32 + 64 = 36 + 96 = 132 g/mol
(b) Penicillin C₁₆H₁₈N₂SO₄
Ans: Molar mass = (12 × 16) + (1 × 18) + (14 × 2) + 32 + (16 × 4)
= 192 + 18 + 28 + 32 + 64 = 334 g/mol
(c) Paracetamol C₈H₉NO₂
Ans: Molar mass = (12 × 8) + (1 × 9) + 14 + (2×16)
= 96 + 9 + 14 + 32 = 151 g/mol

Q17: Write an experiment to show that cathode rays travel in a straight line?
Ans: An experiment to show that cathode rays travel in a straight line can be performed using a fluorescent coated discharge tube and a source of cathode rays, an opaque object, and a high voltage source.
Set-up and procedure:

• Start the discharge so that cathode rays are produced inside the tube; the fluorescent coating will glow where rays strike.
• Place an opaque object in the path of the rays between the cathode and the fluorescent screen.
• When cathode rays strike against the screen, they produce fluorescence. But due to the placement of the opaque object, we will observe a sharp shadow being formed on the screen in the shape of the object. 
• A sharp shadow of the object appears on the glowing screen behind it.
• A sharp shadow is produced only if the rays travel in straight lines; if the rays bent around the object no clear shadow would form.

This observation shows that cathode rays travel in straight lines.

Q18: What is radioactivity? What are the applications of radioisotopes?
Ans: Radioactivity is the spontaneous emission of radiation (alpha particles, beta particles or gamma rays)in the form of particles or high-energy photons from the nuclei of unstable atoms as they transform to more stable forms.
Applications of radioisotopes:

• Co-60 emits γ-radiation used in radiotherapy to treat cancer.
• I-131 is used in the diagnosis and treatment of thyroid disorders.
• P-32 is used in treating certain types of leukaemia and as a tracer in medical research.
• C-14 is used as a tracer in biochemical studies and for dating formerly living materials.

Q19: There are two elements C and B. C emits an α – particle and B emits a β – particle. How will the resultant elements change?

Ans: When an element emits an α particle, its atomic number decreases by 2 and its mass number decreases by 4 (an α particle is a He nucleus: 2 protons and 2 neutrons).

• So element C after α emission: atomic number → (Z – 2), mass number → (A – 4).
• When an element emits a β particle (electron), a neutron in the nucleus converts to a proton. Thus the atomic number increases by 1 while the mass number remains unchanged.
• So element B after β emission: atomic number → (Z + 1), mass number → A.

Q20: What are isotopes? Name the isotopes of hydrogen and draw the structure of their atoms.
Ans: Isotopes are atoms of the same element that have the same atomic number (same number of protons) but different mass numbers (different numbers of neutrons).
Example – Isotopes of hydrogen:

• Protium: ¹₁H (one proton, no neutron)
• Deuterium: ²₁H (one proton, one neutron)
• Tritium: ³₁H  (one proton, two neutrons)

Structure of Isotopes of Hydrogen:

Q21: 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 + Ethanoic acid → Sodium ethanoate + Carbondioxide + Water
Ans: The law of conservation of mass states mass is neither created nor destroyed in a chemical reaction. 
So mass of reactants = mass of products.
Given:
Mass of sodium carbonate = 5.3 g
Mass of ethanoic acid = 6.0 g
Total mass of reactants = 5.3 + 6.0 = 11.3 g
Mass of products: sodium ethanoate = 8.2 g, CO₂ = 2.2 g, H₂O = 0.9 g
Total mass of products = 8.2 + 2.2 + 0.9 = 11.3 g
Mass of reactants = Mass of products = 11.3 g. This confirms the law of conservation of mass.

Q22: 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.
Ans: Given:
Mass of sample = 0.24 g
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
Percentage of boron = (mass of boron / mass of sample) × 100%

Percentage of oxygen = (mass of oxygen / mass of sample) × 100%

The compound contains 40% boron and 60% oxygen by mass.

Q23: If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of one atom of carbon?
Ans:
1 mole of carbon = 6.023 × 10²³ atoms = 12 g
Mass of one carbon atom = 12 g / (6.023 × 10²³) ≈ 1.993 × 10^-23 g

Short Ans Type:

Ques 1. Suggest separation technique(s) one would need to employ to separate the following mixtures.
(a) Mercury and water
Ans: Use a separating funnel.

Explanation: Mercury and water are immiscible liquids with different densities (mercury is denser, ~13.6 g/cm³, compared to water, ~1 g/cm³). A separating funnel allows the denser mercury to settle at the bottom, and the water can be drained off from the top, effectively separating the two liquids.

(b) Potassium chloride and ammonium chloride
Ans: Sublimation.

• Ammonium chloride sublimes (turns directly from solid to gas) when heated, while potassium chloride does not sublime and remains as a solid. By heating the mixture, ammonium chloride will vaporize and can be collected separately upon cooling, leaving potassium chloride behind.

Fig: Sublimation(c) Common salt, water and sand
Ans: Sedimentation, decantation, filtration and evaporation
Sedimentation is the process where solid particles settle at the bottom of a liquid. This is followed by decantation, which involves carefully pouring off the liquid, leaving the solid behind. Next, filtration is used to separate solids from liquids using a filter paper. Finally, evaporation removes the liquid, leaving behind any dissolved solids.

• Sedimentation: Particles settle at the bottom.
• Decantation: Pouring off the liquid.
• Filtration: Using filter paper to separate solids.
• Evaporation: Removing liquid to leave solids.

(d) Kerosene oil, water and salt
Ans: Use a separating funnel followed by evaporation. Kerosene oil and water are immiscible liquids, so a separating funnel can separate them, with kerosene forming the upper layer (density ~0.8 g/cm³) and water (with dissolved salt) the lower layer (density ~1 g/cm³). After separating kerosene, the salt water solution is heated to evaporate the water, leaving the salt behind.

Ques  2. Which of the tubes in figure  (a) and (b) will be more effective as a condenser in the distillation apparatus?

Ans: Condenser (a) will be more effective because it provides more surface area for cooling of the vapours passing through it. 

Ques  3. Salt can be recovered from its solution by evaporation. Suggest some other technique for the same?
Ans: Salt can be recovered from its solution by crystallisation. This method is often preferred over evaporation because:

• It effectively removes soluble impurities.
• It produces pure salt crystals.

In contrast, evaporation may leave behind some impurities, making crystallisation a more efficient technique.

Ques  4. The ‘sea-water’ can be classified as a homogeneous as well as a heterogeneous mixture. Comment.
Ans: Sea-water can be classified as both a homogeneous and a heterogeneous mixture due to the following reasons:

• It is considered homogeneous because it contains dissolved salts, creating a uniform solution.
• It is also seen as heterogeneous because it includes various insoluble components, such as sand, microbes, and shells.

Ques  5. While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice.
Ans: Acetone is soluble in water, creating a homogeneous mixture. Therefore, separation using a separating funnel is not effective. The best method to recover acetone is through simple distillation due to the significant difference in boiling points:

• Boiling point of acetone: 56°C
• Boiling point of water: 100°C

In a distillation flask:

• Acetone will boil at 56°C and convert into vapour.
• The vapour can then be condensed and collected in a separate flask.

Ques  6. What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature.
Ans: When a saturated solution of potassium chloride is prepared at 60°C and then allowed to cool to room temperature, the following observations can be made:

• The solution remains saturated at the higher temperature.
• As the temperature decreases, the solubility of potassium chloride also decreases.
• Consequently, some of the potassium chloride will settle at the bottom of the container.
• This occurs because the solution can no longer hold all the dissolved solute at the lower temperature.

(b) an aqueous sugar solution is heated to dryness.
Ans: When an aqueous solution of sugar is heated to dryness:

• The water evaporates, leaving sugar behind in the container.
• If heated further, the sugar may become charred.

(c) a mixture of iron filings and sulphur powder is heated strongly.
Ans: When a mixture of iron filings and sulphur powder is heated strongly, a chemical reaction occurs, resulting in the formation of ferrous sulphide.

Ques  7. Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do?
Ans: The particles in a colloidal solution do not settle down when left undisturbed due to several reasons:

• Size and Weight: Colloidal particles are smaller and lighter compared to those in a suspension.
• Brownian Movement: These particles are in constant zig-zag motion, known as Brownian movement, which counteracts the force of gravity.
• Charge and Repulsion: Colloidal particles carry a charge that causes them to repel each other, preventing them from clumping together and settling.

In contrast, particles in a suspension are larger and heavier, resulting in less movement. This allows them to settle under the influence of gravity.

Ques  8. Smoke and fog both are aerosols. In what way are they different?
Ans:  Smoke and fog are both types of aerosols, but they differ in their composition:

• The dispersion medium for both is air.
• In smoke, solid carbon particles are dispersed in the air.
• In fog, liquid water droplets are dispersed in the air.

Ques  9. Classify the following as physical or chemical properties
(a) The composition of a sample of steel is: 98% iron, 1.5% carbon and 0.5% other elements.
Ans: The composition of steel (98% iron, 1.5% carbon, 0.5% other elements) reflects the chemical identity and proportion of elements in the alloy, which is a chemical property, as it involves the substance’s chemical makeup rather than a physical characteristic like colour or density.

(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
Ans: The reaction of zinc with hydrochloric acid produces hydrogen gas, indicating a chemical change. This demonstrates a key chemical property of zinc. 

Fig:  Reaction of zinc with hydrochloric acid

(c) Metallic sodium is soft enough to be cut with a knife.
Ans: Metallic sodium is soft enough to be cut with a knife. This characteristic demonstrates the softness of sodium, which is a physical property.

(d) Most metal oxides form alkalis on interacting with water.
Ans: Most metal oxides form alkalis on interacting with water. This characteristic demonstrates the reaction between metal oxides and water, which is a chemical property.

Ques  10. The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100g of water while ‘C’ dissolved 50g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?

Ans: Concentration is the relative percentage of solute compared to the total volume of the solution and it is calculated by dividing mass by volume.
In the case of A, since 50g of NaOH has been dissolved in 100 mL of water, the total volume of solution became about 150 mL, thus concentration of NaOH would be less than 50%.
In the case of B, since 50g of NaOH has been dissolved in 100g of water, therefore, total volume of the solution would become 150 mL, consequently concentration of NaOH would again less than 50%.
In the case of C, 50g of NaOH has been dissolved in water and then volume of the solution made to 100mL, thus concentration of NaOH would become 50%.Thus, C made the solution of NaOH having concentration equal to 50%. 

Ques  11. Name the process associated with the following
(a) Dry ice is kept at room temperature and at one atmospheric pressure.
Ans: Sublimation is the process that occurs when dry ice is left at room temperature and one atmospheric pressure. During this process:

• Dry ice, which is solid carbon dioxide, turns into gas.
• This transition happens without passing through a liquid state.
• Sublimation is common in substances that can easily change from solid to gas.

Fig: Subliming dry ice

(b) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
Ans: When a drop of ink is placed on the surface of water, it spreads out and eventually mixes with the water. This occurs due to the motion of particles, which is known as diffusion.

(c) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
Ans: When potassium permanganate crystals are placed in a beaker and water is added while stirring:

• The potassium permanganate particles mix with the water.
• Stirring increases the motion of the particles, which helps them to mix more quickly.
• This process is known as diffusion.

(d) Milk is churned to separate cream from it.
Ans: When milk is churned, the process separates the cream from the milk. This separation occurs due to the action of centrifugal force, which is the principle behind centrifugation.

(e) Settling of sand when a mixture of sand and water is left undisturbed for some time.
Ans: When mixture of sand and water is left undisturbed, the sand settle at the bottom of water, thus this is the process of sedimentation.

Fig: Mixture of sand and water

(f) Fine beam of light entering through a small hole in a dark room, illuminates the particles in its paths.
Ans: When a fine beam of light enters a dark room through a small hole, it illuminates particles in its path. This occurs due to:

• The collision of air and dirt particles.
• The scattering of light, which makes the dust particles appear to dance.
• This phenomenon is known as the Tyndall effect.

Ques  12. You are given two samples of water labelled as ‘A’ and ‘B’. Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment.
Ans: Since impurities in water raise its boiling point, thus water in sample B is impure. Hence it will not freeze at 0⁰C because of impurities since impAns: Since impurities in water raise its boiling point, thus water in sample B is impure. Hence it will not freeze at 0⁰C because of impurities since impurities decreases the freezing point below the 0⁰C, this is the cause that’s why sea water remain liquid below the 0⁰C.

Ques  13. An element is sonorous and highly ductile. Under which category would you classify this element? What other characteristics do you expect the element to possess?
Ans: Since the element is sonorous and ductile, it can be classified as a metal. Other expected characteristics of metals include:

• Good conductivity of heat and electricity
• Lustrous appearance
• Malleability (can be shaped into thin sheets)

Ques  14. Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures.
(a) A volatile and a non-volatile component.
Ans: The mixture of acetone and water. In this acetone is volatile and water is non-volatile. The mixture of water and acetone can be separated by the process of distillation.

(b) Two volatile components with appreciable difference in boiling points.
Ans: Mixture of acetone and ethanol. The boiling point of acetone is 56⁰C and that of ethyl alcohol is 78.4⁰C.
The mixture of acetone and ethanol can Ans: Mixture of acetone and ethanol. The boiling point of acetone is 56⁰C and that of ethyl alcohol is 78.4⁰C.
The mixture of acetone and ethanol can be separated using fractional distillation. Since they are two immiscible liquids, thus their mixture can be separated using separating funnel.Fig: Fractional distillation(c) One of the components changes directly from solid to gaseous state.
Ans: The mixture of salt and ammonium chloride. In this mixture ammonium chloride changes from solid to gaseous state directly.The mixture of salt and ammonium chloride can be separated by the process of sublimation.

(d) Two or more coloured constituents soluble in some solvent.
Ans: The ink is the mixture of dyes of many colours. The different dyes of ink can be separated using chromatography.

Ques  15. Fill in the blanks

(a) A colloid is a __________ mixture and its components can be separated by the technique known as _________.

Ans: heterogeneous, centrifugation

(b) Ice, water and water vapour look different and display different _________ properties but they are ___________ the same.
Ans: Physical, chemically

(c) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of________ and the lower layer will be that of ___________.
Ans: Upper layer: Water

Lower layer: Chloroform

(d) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called____________.
Ans: Fractional distillation

(e) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the _________ of light by milk and the phenomenon is called _________ . This indicates that milk is a ________ solution.
Ans: scattering, Tyndall Effect, colloidal

Ques  16. Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it be a pure substance or a mixture? Give reasons for the same.
Ans: Pure substance, since it contains a single component, i.e. sucrose.

Fig: Structure of sucrose

Ques  17. Give some examples of Tyndall effect observed in your surroundings?
Ans: Examples of Tyndall Effect:

• When a sunbeam shines through a window, dust particles in the air make the beam visible.
• Milk appears faint blue in a glass due to its colloidal nature, which scatters light.
• Sunlight filtering through clouds looks bright because of light scattering by water droplets.
• In a dense forest, mist with tiny water droplets creates a visible beam of light.

Ques  18. Can we separate alcohol dissolved in water by using a separating funnel? If yes, then describe the procedure. If not, explain.
Ans: The mixture of alcohol and water cannot be separated using a separating funnel, since these are not immiscible liquids. The mixture of alcohol and water can be separated by the process of distillation.

Fig: Distillation process

Ques  19. On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.
(a) Is this a physical or a chemical change?
Ans: The conversion of calcium carbonate into calcium oxide and carbon dioxide is a chemical change.

(b) Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.
Ans: Yes one acidic and one basic solution can be formed by the calcium oxide and carbon dioxide, which are product formed in the above process. Since metallic oxides are basic and non-metallic oxides are acidic in nature.Calcium oxide is a metallic oxide. Hence by dissolving it in water a basic solution is formed because of the formation of calcium hydroxide. The reaction involved Calcium oxide is a metallic oxide. Hence by dissolving it in water a basic solution is formed because of the formation of calcium hydroxide. The reaction involved in this can be written as follows:

CaO + H₂O → Ca(OH)₂

Carbon is a non metal hence carbon dioxide is acidic in nature. When it is dissolved in water an acidic solution is formed.

CO₂+H₂O→ H₂CO₃

Ques  20. Non metals are usually poor conductors of heat and electricity. They are non-lustrous, non-sonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
Ans: Graphite.

Fig: Graphite

(b) Name a non-metal which exists as a liquid at room temperature.
Ans: Bromine

(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
Ans: Graphite. Graphite is a good conductor of electricity. It is an allotropic form of carbon.

(d) Name a non-metal which is known to form the largest number of compounds.
Ans: Carbon is a non-metal. It is known to form the largest number of compounds.

(e) Name a non-metal other than carbon which shows allotropy.
Ans: Sulphur is a non-metal which shows allotropy. Disulphur and trisulphur are some of the allotropes of sulphur.

(f) Name a non-metal which is required for combustion.
Ans: Oxygen

Ques  21. Classify the substances given in Figure into elements and compounds

Ans: Elements: Cu, Zn, O₂, F₂, Hg, Diamond

Compound: CaCO3, NaCl, H₂O,

Ques  22. Which of the following are not compounds?

(a) Chlorine gas
(b) Potassium chloride
(c) Iron
(d) Iron sulphide
(e) Aluminium
(f) Iodine
(g) Carbon
(h) Carbon monoxide
(i) Sulphur powder
Ans: Chlorine gas, iron, aluminium, iodine, carbon, and sulphur powder are not compounds.

• Chlorine gas is a diatomic molecule, consisting of two chlorine atoms.
• Iron and aluminium are both elements, not chemically combined with others.
• Iodine is also an elemental substance.
• Carbon can exist as a pure element, such as in graphite or diamond.
• Sulphur powder is a form of elemental sulphur.