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Students can Download Chapter 10 Mechanical Properties of Fluids Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Plus One Physics Mechanical Properties of Fluids One Mark Questions and Answers

Question 1.
Water is flowing through a very narrow tube. The velocity of water below which the flow remains a streamline flow is known as
(a) relative velocity
(b) terminal velocity
(c) critical velocity
(d) particle velocity
Answer:
(c) critical velocity
Critical velocity is that velocity of liquid flow, upto which the flow of liquid is a streamlined and above which its flow becomes turbulent.

Question 2.
Bernoulli’s equation for steady, non-viscous, imcompressible flow expresses the
(a) conservation of angular momentum
(b) conservation fo density
(c) conservation of momentum
(d) conservation of energy
Answer:
(d) conservation of energy

Question 3.
When cooking oil is heated in a frying pan, the oil moves around in the pan more easily when it is hot. The main reason for this is that with rise in temperature, there is a decrease in
(a) surface tension
(b) viscosity
(c) angle of contact
(d) density
Answer:
(d) density

Question 4.
At what temperature density of air is maximum?
Answer:
4°C

Question 5.
A thin glass plate is lying on a wet marble floor, It is difficult to pull the glass plate because of
(i) surface tension
(ii) Viscosity
(iii) friction
(iv) atmosphere
(v) Gravity
Answer:
(i) Viscosity

Question 6.
Why do clouds float in the sky?
Answer:
Zero terminal velocity

Question 7.
A spinning cricket ball does not follow parabolic path. Why?
Answer:
Due to the magnus effect.

Question 8.
The deep water runs slow. Why?
Answer:
According to equation of continuity velocity is inversly proportional to velocity (AV = constant). Hence deep water runs slow.

Question 9.
Why dust generally settles down in closed room?
Answer:
The dust particles (tiny spheres) acquire terminal velocity as it fall through air. The terminal velocity is directly proportional to square of radius. Hence terminal velocity of dust particle is very small. So they settle down gradually.

Question 10.
Why more viscous oil is used in summer than in winter in scooters?
Answer:
The viscosity decreases with increase in temperature.

Question 11.
Why is sand drier than clay?
Answer:
Capillary action

Question 12.
Why cotton dress is preferred in summer?
Answer:
Cotton dresses have fine pores which act as capillaries for the sweat.

Question 13.
Why oil is poured to calm the sea?
Answer:
When oil is poured in water, the surface tension of water is reduced and water spreads over large area of sea.

Question 14.
How plants draw water from ground?
Answer:
The capillary action.

Question 15.
How do insects run on the surface of water?
Answer:
Because of surface tension, the surface of water behaves like stretched membrane hence it can support weight of small insects.

Question 16.
How ploughing a field helps to retain moisture?
Answer:
When field is ploughed, capillaries are broken and hence water can not rise up and retains moisture.

Question 17.
Hot soup tastes better than cold soup. Why?
Answer:
The surface tension of hot soap is less compared to cold soap. So hot soap spreads larger area.

Plus One Physics Mechanical Properties of Fluids Two Mark Questions and Answers

Question 1.
Remya found that a piece of metal weighs 210 g in air and 180 g when it is immersed in water. Determine the density of the metal piece.
Answer:
Relative density,
R.D = \(\frac{\text { Weight in air }}{\text { Loss of Weight in water }}\)
= \(\frac{210}{30}\) = 7.

Question 2.
Why is hot soup tastier than cold one?
Answer:
When temperature increases, the surface tension of soap decrease. Hence hot soap can enter into tiny pours of taste buds.

Question 3.
Why straws are used to drink soft drinks?
Answer:
When we suck the straw, pressure inside straw becomes lower than atmospheric pressure. This pressure difference cause the soft drink to rise through the straw.

Question 4.
Why new earthen pots keeps water more cool than old earthen pots?
Answer:
The capillaries of old earthen pots will get blocked with passage of time. For new earthen pots, water oozes out through capillaries, gets evaporated at the surface and makes it cool.

Plus One Physics Mechanical Properties of Fluids Three Mark Questions and Answers

Question 1.
Air is blown in between two pith balls suspended freely.

1. What will happen to the balls?
   • They repel each other
   • They attract each other
   • They start oscillating
   • They remain in their initial position They fall on the ground
2. Give your explanation

Answer:

1. They attract each other
2. When air is blown in between two pith balls, the pressure between the balls decreases. Due to this decrease in pressure between the balls, they attract each other.

Question 2.
A child dipped two identical capillary tubes, one in a beaker containing mercury. He observed that water and mercury have risen through the tubes to a certain heights.

1. Name this phenomenon.
2. What difference did he observe in the shape of the meniscus of the two liquids in the tubes?
3. If he plots a graph connecting the radius of the capillary tube and capillary height, what will be the shape of the graph?

Answer:
1. Capillary rise.

2. The shape of the water meniscus in the tube becomes concave upwards. But the shape of mercury measures in the tube become convex upward.

3. h α \(\frac{1}{r}\), this is in the form y α \(\frac{1}{x}\)
Hence when we draw graph between ‘h’ and ‘r’ we get a graph of hyperbola.

Question 3.
Bernoulli’s theorem is a consequence of energy conservation principle. Using this theorem explain the working of atomiser.
Answer:
Atomiser (application of Bernoulli’s theorem)

Atomizer is used for getting a fine spray of perfumer or insecticide. It consists of a cylinder with a piston. A small vessel containing liquid to be sprayed is attached to the cylinder. When the piston is moved forward air is blown out through a small opening of the cylinder.

As the velocity of flow of air increases, the pressure at the opening decreases. Due to the lower pressure at the opening, the liquid rises through the narrow tube and gets sprayed out along with air.

Question 4.
Surface tension is numerically equal to the surface energy.

1. Difine surface tension.
2. Derive an expression for the rise of liquid in a capillary tube.

Answer:
1. Liquids acquire a free surface when poured in a container. These surfaces possess some additional energy. This phenomenon is known as surface tension.

2. When a drop is split into tiny droplets, the surface area increases. So work has to be done for splitting the drop. Let R be radius of the drop and r the radius of the droplets: R = 1 × 10^-3m
surface area of the drop = 4πR²
= 4π × (1 × 10^-3)²
= 4π × 10^-6m²
Volume of the drop = Volume of 10⁶ droplets

∴ Surface area of million droplets =10⁶ × 4π²
= 10⁶ × 4π(1 × 10^-5)²
= 4π × 10^-4 m²
∴ Increase in surface area = 4π × 10^-4 – 4π × 10^-6
= 3.96π × 10^-4m²
∴ Energy expended = 3.967π × 10^-4 × S
= 3.96π × 10^-4 × 72 × 10^-3J
= 8.95 × 10^-5J.

Question 5.
Match the following

1. Pascal’s law	a. \( \sqrt{2 g h}\)
2. Bernoulli’s theorem	b. a1v1= a2v2
3. Surface tension	c. Hydraulic jack
4. Velocity of efflux	d. Reynolds number
5. Equation of continuity	e. Angle of contact
6.  Viscosity	f. Ventiurimeter

Answer:
1 – C, 2 – f, 3 – e, 4 – a, 5 – b, 6 – d.

Question 6.
Give reasons for the following cases.

1. It is easier to swim in sea water than in river walls.
2. The passangers are advised to remove ink from pen while going up in an aeroplane.

Answer:
1. The density of sea water is more than that of river water due to the presence of salt. Hence sea water offers more upthrust and only a very small portion of human body will be in sea water compared to river water.

2. In ink pen, ink is filled in atmospheric pressure. As we go higher pressure decreases and hence ink will have a tendancy to come out in order to equalise the pressure.

Plus One Physics Mechanical Properties of Fluids Four Mark Questions and Answers

Question 1.
A large tank containing water has a small hole near the bottom of the tank 1.5 m below the surface of water.

1. What is the velocity of the water flowing from the hole?
2. Explain the principle used in deriving the velocity of water flowing from the hole.
3. Where must a second hole to be drilled so that the velocity of water leaving this hole is half of water flowing through the first hole.

Answer:
1. Velocity of water flowing through the hole
u = \( \sqrt{2 g h}\)
= \(\sqrt{2 \times 10 \times 1.5}\) = 5.47m/s.

2. Bernoulli’s theorem
As we move along a streamline the sum of the pressure (p), the kinetic energy per unit volume \(\frac{\rho v^{2}}{2}\) and the potential energy per unit volume (ρgh) remains a constant.

(OR)

Mathematically Bernoulli’s theorem can be written as
P + \(\frac{1}{2}\)ρv² + ρgh = constant.

3.

h₂ = \(\frac{1.5}{4}\) m = 0.375 m, from the top side of tank.

Question 2.
Rain drops have an average size of 1 mm when it is formed at the upper atmosphere.

1. Why the velocity of the rain drop is uniform?
2. Derive an expression for the terminal velocity of the drop in terms of coefficient of viscosity of air.
3. If the size of the rain drop become half, then what happens to its terminal speed?

Answer:
1. Due to viscous force acting on the raindrop, it moves with uniform speed.

2. Viscous force, boyancy force and weight of the body
Expression for terminal velocity:
Consider a sphere of radius ‘a’ densitity σ falling through a liquid of density a and viscocity η. The viscous force acting on the sphere can be written as
F = 6πaηv
Where v is the velocity of sphere. This force is acting in upward direction. When the viscous force is equal to the weight of the body in the medium, the net force on the body is zero. It moves with a constant velocity called the terminal velocity.
The weight of a body in a medium,

When body has terminal velocity, we can write.

Question 3.

1. Fill in the blanks using the word from the list appended with each statement.
   • Viscosity of gases_____with temperature (increase/decrease)
   • For a fluid in steady flow, the increase in flow speed at a constriction follows from_____ (conservation of mass/Bernoulli’s theorem)
   • The working of a hydraulic lift is based on (Pascal’s Law/ principle of Conservation of Energy)
   • Small insects can walk over the surface of water. It is due to the_____(surface tension of water/viscosity of water)
2. A girl dips a thin capillary tube in water. Water rises through it.
   • Name the phenomenon.
   • How does this rise vary with the diameter of the tube?

Answer:
1. Fill in the blanks :

• increases
• Conservation of mass
• Pascals law
• Surface tension

2. A girl dips a thin capillary tube in water:

• Capillary rise
• h α \(\frac{1}{r}\) ie. when diameter of tube increases, the. capillary rise decreases.

Question 4.
The schematic diagram of a sprayer or atomiser is given below.

1. Name the principle of working of this device from the following:
   • Surface tension
   • Viscosity
   • Bernoulli’s principle
   • Archimedes’ principle
2. Write its mathematical expression.
3. Wings of an aeroplane are curved outwards while flattened inwards. Why?

Answer:
1. Bernollis principle

2. P + \(\frac{1}{2}\)ρv² + ρgh = constant.

3. When the aeroplane moves forward, the air blown in the form of stream lines over the wings of aeroplane is shown figure.

As the upper surface of wing is more curved than its lower surface, the speed of air above the wings is larger than the speed of the air below the wings.

Hence the pressure above the wings becomes less than the pressure below the wings. Due to this pressure difference the aeroplane will get upward force to overcome gravitational force.

Question 5.
During windstorms, roofs of certain houses are blown off without damaging other parts of the houses.

1. Name the theorem which explains this phenomenon.
2. State the theorem.
3. Explain this phenomenon on the basis of this theorem.

Answer:

1. Bernoulli’s theorem
2. For a small amount of liquid in stream line flow, between two points, the total energy is constant.
3. When windstorm blown off, the pressure on the top side of roof decreases. Hence a pressure difference is developed in between roof. Due to this pressure difference, roof of certain houses are blown off.

Question 6.
Two thin evacuated (one end closed) glass take A and B are carefully immersed in a beaker containing mercury such a way that there is no chance to get air in to the tubes. A is stand vertically and B is making an angle θ with the vertical.

1. Is any rise of mercury in the tubes?
2. Is any height difference of mercury levels in tube A and B? Justify your answer.
3. When the doctors are measuring body pressure, it is advisable to lie on a table. Why?

Answer:

1. Yes
2. No. Pressure is same at same level. To get same pressure, height of mercury becomes same.
3. When we lie on the table, the pressure of our body will be same at all points.

Question 7.
A small metal sphere is falling through a caster oil.

1. Name the forces acting on the metal sphere?
2. Which of these forces change? Why?
3. Name the velocity of the sphere when the unbalanced force on it is zero?
4. Write down the expression for this velocity in terms of coefficient of viscosity?

Answer:

1. forces acting on the metal sphere:
   • Weight of the body (mg)
   • Buoyant force or up thrust
   • Viscous force
2. Viscous force. Viscous force is the friction offered by the liquid. It is a self adjusting force.
3. Terminal velocity
4. Terminal velocity, V = \(\frac{2}{9} a^{2}\left(\frac{f-N}{\eta}\right) g\).

Plus One Physics Mechanical Properties of Fluids Five Mark Questions and Answers

Question 1.
A capillary tube when dipped into water, it is commonly observed that water will rise through the tube.

1. Which of the following is responsible for this?
   • Gravitational force
   • Viscous force
   • Nuclear force
   • Surface tension
   • Elastic force
2. Derive an expression for the capillary rise.
3. If the radius of the tube becomes doubled, then what happens to the height of water column in the tube?

Answer:
1. Surface tension.

2.

Consider a capillary tube of radius ‘a’ dipped in a liquid of density ρ and surface tension S. If the liquid has a concave meniscus it will rise in the capillary tube. Let h be the rise of the liquid in the tube. Let p₁ be the pressure on the concave side of the meniscus and p₀, that on the other side. The excess pressure on the concave side of the meniscus can be written as
p₁ – p₀ = \(\frac{2 \mathrm{S}}{\mathrm{R}}\)
Where R is the radius of the concave meniscus. The tangent to the meniscus at the point A makes an angle θ with the wall of the tube.
In the right angled triangle ACO

substituting the values of R in the equation (1)

Considering two points M and N in the same horizontal level of a liquid at rest,
pressure at N = pressure at M
But pressure at M = p_i, the pressure over the concave meniscus and pressure at N = p_o + hρg
∴ P_i = P_o + hρg
or P_i – P_o = hρg ……..(3)
From equations (2) and (3), we get

3. We know

The capillary rise decreases to half of the original value.

Question 2.

1. Find the odd one out and justify your answer Atomiser, venturi meter, aeroplane, hydraulic lift
2. Mention one use of venturi meter.
3. Explain the working of the odd one which you have selected in question (a)

Answer:
1. Hydraulic lift – It is based on pascals law.

2. Venturimeter can be used to find velocity of flow of fluid through a pipe.

3.

A hydraulic lift is used to lift heavy load. Consider a liquid enclosed in a vessel with two cylinders C₁ and C₂ attached as shown in the figure. The cylinders are provided with two pistons having areas A₁ and A₂ respectively.
If F₁ is the force exerted on the area A₁,
pressure P₁ = \(\frac{F_{1}}{A_{1}}\).
If F₂ is the force exerted on the area A₂,
pressure P₂ = \(\frac{F_{2}}{A_{2}}\).
According to pascal’s law P₁ = P₂.

Using this method we can lift heavy load by applying small force.

Question 3.
When a capillary tube of radius ‘r’ is dipped in water, the water rises through it up to height ‘h’.

1. Which of the following is responsible for the above phenomenon?
   • Viscous force
   • elastic force
   • surface tension
   • gravitational force
   • negative force
2. To what height will water rise in a glass tube with a bore of radius 0.1 mm (take the angle of contact of glass with 0°, surface tension S = 0.0728 N/m)
3. If the length of tube is less that the length of capillary rise, will it overflow. Justify your answer.

Answer:
1. Surface tension

2.

3. The water will never overflow. If the tube is of insufficient length, the radius of curvature of liquid meniscus goes on increasing, making it more and more flat till water is in equilibrium.

Question 4.
A steel ball of radius 1 mm is falling vertically through a tank of oil at 30°C.

1. After some time the ball attains a constant velocity called_____
2. What are the forces acting on the ball and give their directions?
3. Write down the expression for resultant force acting on the ball?)
4. If the density of oil is 2 × 10³kg/m³, density of steel is 8 × 10² Kg/m³ and ‘η’of oil 2NS/m², What will be the constant velocity attained by the ball?

Answer:
1. Terminal velocity.

2. Weight of body (down ward), bouyanant force (up ward), Viscous force (upward).

3. Resultant force = weight of body – buoyant force.

4. Terminal velocity,

Plus One Physics Mechanical Properties of Fluids NCERT Questions and Answers

Question 1.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer:
Force, F = Weight of girl
= mg = 50 × 9.8N = 490N
Radius, r = 0.5 × 10^-2m
Area A = πr² = \(\frac{22}{7}\)(0.5 × 10^-2)² m²
Pressure

Question 2.
Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmospheric pressure.
Answer:
p = hρg, h = \(\frac{p}{\rho g}=\frac{1.01 \times 10^{5}}{984 \times 9.8}\)m = 10.47m.

Question 3.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 of spirit in the other. What is the specific gravity of spirit?

Answer:
Since the mercury columns in the two arms are at the same level,
∴ pressure due to water column = pressure due to spirit column
∴ h_wρ_wg = h_sρ_sg
or h_wρ_w = h_sρ
But h_w = 10 cm,
ρ_w = 1 gcm^-3,
h_s = 12.5cm
∴ 10 × 1 = 12.5 × ρ_s
or ρ_s = \(\frac{10}{12.5}\)gcm^-3
= 0.8cm^-3
∴ Specific gravity of spirit = 0.8.

Question 4.
Figs, (a) and (b) refer to the steady flow of a non-viscous liquid. Which of the two figures is incorrect? Why?

Answer:
Fig (a) is incorrect. This is because at a constriction (ie., where the area of cross-section of the tube is smaller), the flow speed is larger due to mass conservation. Consequently, pressure there is smaller according to Bernoulli’s equation. We assume the fluid to be incompressible.

Question 5.
What is the prssure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^-1Nm^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Excess pressure = \(\frac{2 \sigma}{R}\)

Total pressure = 1.01 × 10⁵ + \(\frac{2 \sigma}{R}\)
= 1.01 × 10⁵ + 310
= 1.0131 × 10⁵Pa
Since data is correct upto three significant figures. We should write total pressure inside the drop as 1.1 × 10⁵Pa.

Question 6.
During blood transfusion, the needle is inserted in a vein where the guage pressure is 2000 Pa, at what height must the blood container be placed so that blood may just enter the vein? Given: density of whole blood = 1.06 × 10³kgm^-3
Answer:
Guage pressure,
p = hρg, h

= 0.19m.

Students can Download Chapter 1 Physical World Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 1 Physical World

Summary
What is Physics?

a. Science And Scientific Method
Science is exploring, experimenting and predicting from what we see around us. ie. It is basically an systematic attempt to understand natural phenomena.

b. Physics
Two approaches are used in Physics: unification and reduction. In unification diverse physical phenomena are explained in terms of a few concepts or laws. For example all electric and magnetic phenomena can be explained by laws of electromagnetism (Maxwell’s equations).

In reduction, we derive properties of complex (bigger) system from properties and interactions of constituent parts. For example, the temperature of system is related to average kinetic energy of molecule of system.

Scope And Excitement Of Physics
The different subdisciplines of physics belongs to two domains: microscope domain and macroscopic domain. The macroscopic domain includes phenomena at laboratory, terrestrial and astronomic scales.

The microscopic domain of physics deals with constitution and structure of matter and their interaction with elementary particles like electron, proton, photon etc.

Physics covers a wide range of magnitude of physical quantities like length, time, mass, energy, etc. Physics includes phenomena involving elementary particles like electron, proton etc. whose range is 10^-14m.

It also deals with astronomical phenomena at the scale of even the entire universe (10⁺²⁶m). The range of time extends from 10^-22 s to 10¹⁸s. The range of mass goes from 10^-30 kg (mass of electron) to 10⁵⁵kg (mass of entire universe).

Physics, Technology And Society
The relation between Physics, technology and society can be seen in many examples. The steam engine has an important role in the Industrial Revolution in England in eighteenth century. The discovery of basic laws of electricity and magnetism contributed wireless communication technology.

Fundamental Forces Of Nature
There occur four fundamental forces in nature. They are gravitational force, electromagnetic force, strong nuclear force and weak nuclear force.

1. Gravitational Force:
It is a universal force. Gravitational force is the attractive force existing between any two bodies by virtue of its mass.

2. Electromagnetic Force:
The electromagnetic force exist between charged bodies. The electrostatic force of attraction or repulsion exist between charges at rest. A moving charge has magnetic effect in addition to electric effect. The electric and magnetic effects are inseparable and hence force experienced by charge is called electromagnetic force.

3. Strong Nuclear Force:
The strong nuclear force binds the nucleons (protons and neutrons) inside the nucleus. It is the strongest of all fundamental forces. The range of nuclear force is 1o^-15m (fermi) and it is charge independent.

4. Weak Nuclear Force:
The range of weak nuclear force is 10^-16m. This force exists only in few nuclear, reactions like b-decay.

5. Towards Unification of Forces:
Isac Newton unified terrestrial and celestial domain by applying law of gravitation in two domains. Oersted and Faraday showed that electric and magnetic phenomena are inseparable. Maxwell unified electromagnetism and optics by showing light is an electromagnetic wave.

1.5 Nature Of Physical Laws
The physical quantity that remains unchanged in process is called conserved quantity. Some of the conservation laws in nature are laws of conservation of mass, energy, linear momentum, angular momentum, charge etc.

Conservation laws have a deep connection with symmetries of nature. The symmetry of nature w.r.t. translation in time is equivalent to conservation of energy. Similarly the symmetry of nature w.r.t. translation in space is equivalent to conservation of linear momentum.

Symmetries of space and time and other types of symmetry play an important role n modern theories of fundamental forces in nature.

Students can Download Chapter 1 The Discipline of Computing Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 1 The Discipline of Computing

Summary
Computing milestones and machine evolution:
People used pebbles and stones for counting earlier days. They draw lines to record information. eg: 1 line for one, 2 lines for two, 3 lines for three, etc. In this number system the value will not change if the lines are interchange. This type of number system is called non positional number system.

Counting and the evolution of the positional number system:
In positional number system, each and every number has a weight. Earlier sticks are used to count items such as animals or objects. Around 3000 BC the Egyptians use number systems with radix 10(base-the number of symbols or digits used in the number system) and they write from right to left.

Later Sumerian/Babylonian use number system with largest base 60 and were written from left to right. They use space for zero instead of a symbol, 0. In 2500 BC, the Chinese use simple and efficient number system with base 10 very close to number system used in nowadays.

In 500 BC, the Greek number system known as Ionian, it is a decimal number system and used no symbols for zero. The Roman numerals consists of 7 letters such as l, V, X, L, C, D, M. The Mayans used number system with base 20 because of the sum of the number of fingers and toes is 10 + 10 = 20.

It is called vigesimal positional number system. The numerals are made up of three symbols; zero (shell shape, with the plastron uppermost), one (a dot) and five (a bar or a horizontal line). To represent 1 they used one dot, two dots for 2, and so on

The Hindu – Arabic number system had a symbol(0)for zero originated in India 1500 years ago. Consider the table to compare the number system

Roman Numerals	Decimal / Hindu – Arabic number
I	1
V	5
X	10
L	50
C	100
D	500
M	1000

Evolution of the computing machine:
(a) Abacus:
In 3000 BC Mesopotamians introduced this and it means calculating board or frame. It is considered as the first computer for basic arithmetical calculations and consists of beads on movable rods divided into two parts. The Chinese improved the Abacus with seven beads on each wire. Different Abacus are given below.

(b) Napier’s bones:
A Mathematician John Napier introduced this in AD 1617.

(c) Pascaline:
A French mathematician Blaise Pascal developed this machine that can perform arithmetical operations.

(d) Leibniz’s calculator:
In 1673, a German mathematician and Philosopher Gottfried Wilhelm Von Leibniz introduced this calculating machine.

(e) Jacquard’s loom:
In 1801, Joseph Marie Jacquard invented a mechanical loom that simplifies the process of manufacturing textiles with complex pattern. A stored program in punched cards was used to control the machine with the help of human labour. This punched card concept was adopted by Charles Babbage to control his Analytical engine and later by Hollerith.

(f) Difference engine:
The intervention of human beings was eliminated by Charles Babbage in calculations by using Difference engine in 1822. It could perform arithmetic operations and print results automatically

(g) Analytical engine:
In 1833. Charles Babbage introduced this. Charles Babbage is considered as the “Father of computer It is considered as the predecessor of today’s computer. This engine was controlled by programs stored in punched cards. These programs were written by Babbage’s assistant, Augusta Ada King, who was considered as the first programmer in the World.

(h) Hollerith’s machine:
In 1887, Herman Hollerith an American made first electromechanical punched cards with instructions for input and output. The.card contained holes in a particular pattern with special meaning. The Us Census Bureau had large amount of data to tabulate, that will take nearly 10 years.

By this machine this work was completed in one year. In 1896, Hollerith started a company Tabulating Machine Corporation. Now it is called International Business Machines(IBM).

(i) Mark-1:
In 1944 Howard Aiken manufactured automatic electromechanical computer in collaboration with engineers at IBM that handled 23 decimal place numbers and can perform addition, subtraction, multiplication and subtraction.

Generations of computers:
There are five generations of computers from 16^th century to till date.

First generation computers (1940 – 1956):
Vacuum tubes were used in first generation computers. The input was based on punched cards and paper tapes and output was displayed on printouts. The Electronic Numerical Integrator and Calculator(ENIAC) belongs to first generation was the first general purpose programmable electronic computer built by J. Presper Eckert and John V. Mauchly.

It was 30-50 feet long, weight 30 tons, 18,000 vacuum tubes, 70,000 registers, 10,000 capacitors and required 1,50,000 watts of electricity. It requires Air Conditioner. They later developed the first commercially successful computer, the Universal Automatic Computer(UNIVAC) in 1952. Von Neumann architecture

The mathematician John Von Neumann designed a computer structure that structure is in use nowadays. Von Neumann structure consists of a central processing unit(CPU), Memory unit, Input and Output unit. The CPU consists of arithmetic logical unit(ALU) and control unit(CU).

The instructions are stored in the memory and follows the “Stored Program Concept”. Colossus is the secret code breaking computer developed by a British engineer Tommy Flowers in 1943 to decode German messages.

Second generation computers (1956 -1963):
Transistors, instead of Vacuum tubes, were used in 2^nd generation computers hence size became smaller, less expensive, less electricity consumption and heat emission and more powerful and faster.

A team contained John Bardeen, Walter Brattain and William Shockley developed this computer at Bell Laboratories. In this generation onwards the concept of programming language was developed and used magnetic core (primary) memory and magnetic disk(secondary) memory.

These computers used high level languages(high level language means English like statements are used)like FORTRAN (Formula translation) and COBOL(Common Business Oriented Language). The popular computers were IBM 1401 and 1620.

Third generation computers (1964 – 1971):
Integrated Circuits(IC’s) were used. IC’s or silicon chips were developed by Jack Kilby, an engineer in Texas Instruments. It reduced the size again and increased the speed and efficiency. The high level language BASIC(Beginners All purpose Symbolic Instruction Code) was developed during this period.

The popular computers were IBM 360 and 370. Due to its simplicity and cheapness more people were used. The number of transistors on IC’s doubles approximately every two years. This law is called Moore’s Law, it is named after Gordon E Moore. It is an observation and not a physical or natural law.

Fourth generation computers (1971 onwards):
Microprocessors are used hence computers are called microcomputers. Microprocessor is a single chip which contains Large Scale of IC’s(LSI) like transistors, capacitors, resistors,etc due to this a CPU can place on a single chip. Later LSI were replaced by Very Large Scale Integrated Circuits(VLSI). The popular computers are IBM PC and Apple II.

Fifth generation computers (future):
Fifth generation computers are based on Artificial Intelligence(AI). Al is the ability to act as human intelligence like speech recognition, face recognition, robotic vision and movement etc. The most common Al programming language are LISP and Prolog.

Evolution of computing:
Computing machines are used for processing or calculating data, storing and displaying information. In 1940’s computer were used only for single tasks like a calculator. But nowadays computer is capable of doing multiple tasks at a time.

The “Stored Program Concept” is the revolutionary innovation by John Von Neumann helped storing data and information in memory. A program is a collection of instructions for executing a specific job or task.

Augusta Ada Lowelace: She was the Countess of Lowelace and she was also a mathematician and writer. She is considered as the first lady computer programmer.

Programming languages:
The instructions to the computer are written in different languages. They are Low Level Language(Machine language), Assembly Language(Middle level language) and High Level Language(HLL).

In Machine Language 0’s and 1 ’s are used to write program. It is very difficult but this is the only language which is understood by the computer. In assembly language mnemonics (codes) are used to write programs

Electronic Delay Storage Automatic Calculator(EDSAC) built during 1949 was the first to use assembly language. In HLL English like statements are used to write programs. A-0 programming language developed by Dr. Grace Hopper, in 1952, for UNIVAC-I is the first HLL.

A team lead by John Backus developed FORTRAN @IBM for IBM 704 computer and ‘Lisp’ developed by Tim Hart and Mike Levin at Massachusetts Institute of Technology. The other HLLs are C, C++, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

Usually programmers prefer HLL to write programs because of its simplicity. But computer understands only machine language. So there is a translation needed. The program which perform this job are language processors.

Algorithm and computer programs:
The step-by-step procedure to solve a problem is known as algorithm. It comes from the name of a famous Arab mathematician Abu Jafer Mohammed Ibn Musaa Al-Khowarizmi, The last part of his name Al-Khowarizmi was corrected to algorithm.

Theory of computing:
It deals with how efficiently problems can be solved by algorithm and computation. The study of the effectiveness of computation is based upon a mathematical abstraction of computers is called a model of computation, the most commonly used model is Turing Machine named after the famous computer scientist Alan Turing.

1. Contribution of Alan Turing:
He was a British mathematician, logician, cryptographer and computer scientist. He introduced the concept of algorithm and computing with the help of his invention Turing Machine.

He asked the question Can machines think’ led the foundation for the studies related to the computing machinery and intelligence. Because of these contributions he is considered as the Father of Modern Computer Science as well as Artificial Intelligence.

2. Turing Machine:
In 1936 Alan Turing introduced a machine, called Turing Machine. A Turing machine is a hypothetical device that manipulates symbols on a strip of tape according to a table of rules. This tape acts like the memory in a computer. The tape contains cells which starts with blank and may contain 0 or 1.

So it is called a 3 Symbol Turing Machine. The machine can read and write, one cell at a time, using a tape head and move the tape left or right by one cell so that the machine can read and edit the symbol in the neighbouring cells. The action of a Turing machine is determined by

1. the current state of the machine
2. the symbol in the cell currently being scanned by the head and
3. a table of transition rules, which acts as the program.

3. Turing Test:
The Turing test is a test of a machine’s ability to exhibit intelligent behaviour equivalent to, or indistinguishable from, that of a human. The test involves a human judge engages in natural language conversations with a human and a machine designed to generate performance indistinguishable from that of a human being.

All participants are separated from one another. If the judge cannot reliably tell the machine from the human, the machine is said to have passed the test. The test does not check the ability to give the correct answer to questions; it checks how closely the answer resembles typical human answers. Turing predicted that by 2000 computer would pass the test.

Students can Download Chapter 13 Kinetic Theory Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 13 Kinetic Theory

Plus One Physics Kinetic Theory One Mark Questions and Answers

Question 1.
The value of \(\frac{P V}{T}\) for one mole of an ideal gas is nearly equal to
(a) 2 Jmol^-1K^-1
(b) 8.3 Jmol^-1K^-1
(c) 4.2 Jmol^-1K^-1
(d) 2 cal mol^-1K^-1
Answer:
(d) 2 cal mol^-1K^-1
The value of \(\frac{P V}{T}\) for one mole of an ideal gas = gas constant = 2 cal mol^-1K^-1.

Question 2.
Mean free path of a gas molecule is
(a) inversely proportional to number of molecules per unit volume
(b) inversely proportional to diameter of the molecule
(c) directly proportional to the square root of the absolute temperature
(d) directly proportional to the molecular mas
Answer:
(a) inversely proportional to number of molecules per unit volume.

Question 3.
If for a gas \(\frac{R}{C_{v}}\) = 0.67, this gas is made up of molecules which are
(a) monoatomic
(b) diatomic
(c) Polyatomic
(d) mixture of diatomic and polyatomic molecules
Answer:
(a) monoatomic
For a gas, we know \(\frac{R}{C_{v}}\) = γ – 1
or, 0.67 = γ – 1, or γ = 1.67
Hence the gas is monoatomic.

Question 4.
According to kinetic theory of gases, molecules of a gas behave like
(a) inelastic spheres
(b) perfectly elastic rigid spheres
(c) perfectly elastic non-rigid spheres
(d) inelastic non-rigid spheres
Answer:
(b) According to kinetic theory of gases, gas molecules behave as a perfectly elastic rigid spheres.

Question 5.
Which one of the following is not an assumption of kinetic theory of gases?
(a) The volume occupied by the molecules of the gas is negligible.
(b) The force of attraction between the molecules is negligible.
(c) the collision between the molecules are elastic.
(d) All molecules have same speed.
Answer:
(d) Molecules of an ideal gas moves randomly with different speeds.

Question 6.
What is the shape of graph between volume and temperature, if pressure is kept constant?
Answer:
PV = nRT. Hence graph will be straight line.

Question 7.
What is the shape of graph between pressure p and I/V for a perfect gas at constant temperature?
Answer:
Straight line

Question 8.
Identify the minimum possible temperature at which all molecular motion ceases.
Answer:
Absolute temperature (OK or – 273.15°C).

Question 9.
What is the formula for average translational kinetic energy of a gas molecule?
Answer:
3/2 K_BT

Plus One Physics Kinetic Theory Two Mark Questions and Answers

Question 1.
Mention the conditions under which the real gases obey ideal gas equation.
Answer:
Low pressure and high temperature.

Question 2.
Why the temperature rises when gas is suddenly compressed?
Answer:
The work done on gas during compression increases the kinetic energy of molecules and hence temperature of gas rises.

Question 3.
Why evaporation causes cooking?
Answer:
During evaporation, fast moving molecules escape from liquid. Hence average kinetic energy of molecules left behind is decreaesd. This will reduce temperature & causes cooling.

Question 4.
When automobile travels long distance air pressure in tyres increases slightly. Why?
Answer:
As automobile moves, work is being done against force of friction. This work is converted in to heat and it increases the temperature. As P a T, increase in temperature will increase pressure.

Question 5.
PV = µ RT is the ideal gas equation. Real gas obeys ideal behaviour at high temperature and at low pressure.

1. Give an example for ideal gas
2. Why real gases obey ideal gas equation at high temperature and at low pressure.

Answer:

1. Hydrogen
2. The interaction between molecules can be neglected at high T and at low temperature.

Question 6.
A vessel of volume V contains a gas of µ moles at a temperature T.

1. What is the ideal gas equation if gas is considered to be ideal one?
2. The variation of pressure P with number of moles per unit volume in a vessel is shown in the graph. Analyse the graph and choose the correct one and justify your answer.

(i) The temperature inside the vessel decreases.
(ii) The temperature inside the vessel increases.
Answer:
1. PV = µRT

2. P = \(\frac{1}{3}\)nmc^-2
In this case, when n increases, (mc^-2) decreases to maintain P as constant. Temperature is directly proportional to mc^-2. Hence we can say that temperature inside the vessel decreases.

Plus One Physics Kinetic Theory Three Mark Questions and Answers

Question 1.
1 mole of ideal gas is taken in vessel.

1. State the following statements as true or false.
   • In gas equation R is constant.
   • All real gas obeys gas equation at all temperature and pressures.
2. Draw the variation of R with pressure for the above ideal gas.
3. Draw the variation of R with volume for this ideal gas.

Answer:
1. Following statements as true or false:

• True
• False

2.

3.

Question 2.

1. Air pressure in a car tyre increase during driving. Why?
2. Air is filled in a vessel at 60°C. To what temperature should it be heated in order that 1/3^rd of air may escape out of the vessel? (Expansion of air may be neglected).

Answer:
1. During driving the temperature of air inside the tyre increases due to motion.

2. T₁ = 60 + 273 = 333K
V₁ = V; T₂ = ? V₂ = V+ V/3
V₂ = \(\frac{4}{3}\)V

Question 3.
Find the degrees of freedom of the following.

1. A body is confined to move in a straight line
2. A body moves in a plane
3. A body moves in a space

Answer:

1. 1
2. 2
3. 3

Plus One Physics Kinetic Theory Four Mark Questions and Answers

Question 1.
An enclosed vessel contains many number of molecules moving in random direction.

1. Explain the term pressure in terms molecular concept.
2. Derive an expression for the pressure exerted by the gas molecules by assuming postulates of kinetic theory of gases.

Answer:
1. Pressure P = \(\frac{2}{3} n \overline{K E}\)
Where ‘n’ is the number of gas molecules per unit volume. \(\overline{\mathrm{KE}}\) is the average kinetic energy of a gas molecules moving in random direction.

2.

Consider molecules of gas in a container. The molecules are moving in random directions with a velocity V. This is the velocity of a molecule in any direction.

The velocity V can be resolved along x, y and z directions as V_x, V_y, and V_z respectively. If we assume a molecule hits the area A of container with velocity V_x and rebounds back with -V_x.

The change in momentum imparted to the area A by the molecule = 2mV_x. The molecules covers a distance V_xt along the x-direction in a time t. All the molecules within the volume AV_xt will collide with the area in a time t.

If ‘n’ is the number of molecules per unit volume, the total number of molecules hitting the area A, N = AV_xt n.
But on an average, only half of those molecules will be hitting the area, and the remaining molecules will be moving away from the area. Hence the momentum imported to the area in a time t.

Q = 2mv_x × \(\frac{1}{2}\) AV_xt n.
= nmV_x2 At
The rate of change of momentum,
\(\frac{Q}{t}\) = nmV_x² A
But rate of change of momentum is called force, ie. force F = nmV_x2A
∴ pressure P =nmV_x 2 (P = \(\frac{F}{A}\))
Different molecules move with different velocities. Therefore, the average value V2_x is to be taken. If \(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) isthe average value then the pressure.
p = nm\(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) ………(1)
\(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) is known as the mean square velocity.
Since the gas is isotropic (having the same properties in all directions), we can write

Hence the eq (1) can be written as

But nm = ρ, the density of gas
∴ P = \(\frac{F}{A}\) ρ\(\overline{\mathbf{V}}^{2}\).

Question 2.
1. Fill in the blanks

2. What happens to the value of ratio of specific heat capacity, if we consider all rotational degrees of freedom of a 1-mole diatomic molecule?
Answer:
1.

2. Total degrees of freedom = 3 (trans) + 3 (Rot) = 6
∴ C_V = 3R, C_P = 4R
Ratio of specific heat γ = \(\frac{4}{3}\)
Ratio of specific heat capacity decreases.

Plus One Physics Kinetic Theory NCERT Questions and Answers

Question 1.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer:
PV= µRT or V = \(\frac{\mu \mathrm{RT}}{\mathrm{P}}\)

= 22.4 × 10^-3 m³ = 22.4 litre.

Question 2.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m³ at a temperature of 27°C and 1-atmosphere pressure.
Answer:
V = 25.0m³, T = (27 + 273), K = 300 K, k = 1.38 × 10^-23JK^-1
PV= nRT = n(Nk)T = (nN)kT = N_tkT
Here N_t represents the total number of air molecules in the given gas.

Question 3.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³s^-1. Identify the gas.
Answer:
According to Graham’s law of diffusion of gases, the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass. If R₁ and R₂ be the rates of diffusion of two gases having molecular masses M₁ and M₂ respectively, then

Question 4.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.
Answer:
Consider one mole of oxygen gas at STP. It occupies 22.4 litre of volume which will contain 6.023 × 10²³. (ie. Avogadro number) molecules. Considering spherical shape of molecule, volume of oxygen molecule

Volume of 6.023 × 10²³ molecules
= \(\frac{4}{3}\) × 3.142(1.5)³ × 10^-30 × 6.02 × 10²³ m³
= 85.1 × 10^-7m³
= 8.51 × 10^-6m³ = 8.51 × 10^-3
litre Molecular volume of one mole of oxygen (∵ 1m³ = 10³ litre)
∴ Molecular volume of one mole of oxygen = 8.51 × 10^-3 litre
Actual volume occupied by one mole of oxygen at STP = 22.4 litre
Fraction of molecular volume to actual volume

Students can Download Chapter 15 Waves Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 15 Waves

Plus One Physics Waves One Mark Questions and Answers

Question 1.
Velocity of sound in vacuum is
(a) 330ms^-1
(b) 165ms^-1
(c) zero
(d) 660ms^-1
Answer:
(c) zero
Sound requires a material medium for propation. Hence, velocity of sound in vacuum is zero.

Question 2.
The physical quantity that remains unchanged when a sound wave goes from one medum to another is
(a) amplitude
(b) speed
(c) wavelength
(d) frequency
Answer:
(d) frequency
When a sound wave goes from one medium to another the frequency of the wave remains unchanged.

Question 3.
What is the range of frequency of audible sound?
Answer:
20Hz to 20KHz.

Question 4.
Why does sound travel faster in iron than in air?
Answer:
Because solids are more elastic compared to air.

Question 5.
What kind of waves help the bats to find their way in dark?
Answer:
Ultrasonic wave

Question 6.
In which gas, hydrogen and oxygen will the sound have greater velocity?
Answer:
As velocity, v α \(\sqrt{1 / ρ}\), velocity of sound will be greater in hydrogen gas.

Question 7.
Why transverse waves can not setup in gas?
Answer:
The rigidity modulus of gas is zero.

Question 8.
What is the effect of pressure on the velocity of sound waves?
Answer:
No effect

Question 9.
Why bells are made up of metal and not wood?
Answer:
The wood causes high damping.

Question 10.
What is the velocity of sound in perfect rigid body?
Answer:
The velocity is infinite cause young’s modulus of perfect rigid body is infinite.

Plus One Physics Waves Two Mark Questions and Answers

Question 1.
If a tuning fork is held above a resonance column, then maximum sound can be heard at certain height of the air column.

1. Name the type of wave produced in the air col¬umn.
2. What do you mean by beats?

Answer:

1. Longitudinal
2. Periodic variation of intensity of sound is called beats.

Question 2.
Why bells of colleges and temple are of large size?
Answer:
Larger, the area of source of sound more is the energy transmitted into the medium. Hence intensity of sound is more and loud sound is heard.

Plus One Physics Waves Three Mark Questions and Answers

Question 1.
A sound wave of frequency 400 Hz is travelling in air at a speed of 320m/s.

1. The speed of sound wave in vaccum is______
   • 320m/s
   • more than 320m/s
   • less than320m/s
   • Zero
2. What is the wavelength of the above wave?
3. Calculate the diffference in phase between two points on the wave 0.2m apart in the direction of travel.

Answer:
1. Zero

2. v = fλ
λ = \(\frac{v}{f}=\frac{320}{400}\) = 0.8m

3. At 0.2 m apart, the phase differnce’is given by
∆Φ = \(\frac{0.2}{0.8} \times 2 \pi=\frac{\pi}{2} \mathrm{rad}\).

Question 2.
A string fixed one end is suddenly brought in to up and down motion.

1. What is the nature of the wane produced in the string and name the wave.
2. A brass wire 1 m long has a mass 6 × 10^-3 kg. If it is kept at a tension 60N, What is the speed of the wave on the wire.

Answer:
1. Transverse wave

2.

Plus One Physics Waves Four Mark Questions and Answers

Question 1.
A sonometer wire of length 30cm vibrates in the second overtone

1. Represent it pictorially
2. What is the distance between two points in the string which has a phase difference of n
3. A violin string resonates in its fundamental frequency of 196hz. Where along the string must you place your finger so that the fundamental frequency becomes 440Hz, If the length of violin string is 40cm.

Answer:
1.

2. 0.30 = \(\frac{3}{2}\)λ
λ = \(\frac{2 \times 0.3}{3}\) = 0.2
We know 2π = λ
ie. π radian = \(\frac{λ}{2}\) wave length
= \(\frac{0.2}{2}\) = 0.1m
∴ The distance between two points is 0.1 m (for a a phase difference of π radian).

3. For fundamental mode of vibration.

The finger must be placed 37.5 from one end.

Question 2.
A horizontal metal wire is fixed on a state of tension between two vertical supports when plucked it gives a fundamental frequency f_O.

1. Obtain a mathematical expression for f_O
2. A 5.5 m wire has a mass of 0.035 kg. If the tension of the string is 77N, the speed of wave on the string is
   • 110 ms^-1
   • 11\(\sqrt{10}\)ms^-1
   • 77 ms^-1
   • 11 ms^-1
   • 102 ms^-1
3. What change, if any, will be observed in the fundamental frequency if the wire is now immersed in water and plucked again?

Answer:
We know velocity on a string, V = \(\sqrt{T / m}\)
But V = λf

2. Mass per unit length,m = \(\frac{M}{\ell}=\frac{0.035}{5.5}\)
= 6.36 × 10^-3kg/m
T = 77N

3. Frequency does not change.

Question 3.
When a pebble is dropped to the surface of water, certain waves are formed on the water surface.

1. What type of wave is it?
2. Is it a progressive wave? Explain?
3. The equation for such a wave is y = 4sinπ (2t – 0.01x). where ‘y’ and ‘x’ are in cm. and ‘t’ in sec. find its
   • Amplitude
   • Wavelength
   • initial phase
   • Frequency

Answer:
1. Transverse wave.

2. It is a progressive wave. It moves from one point to another point.

3. y = 4sinπ(2t – 0.01x)
y = 4 sin (2πt-0.01πx)
= -4 sin (0.01 πx – 2πt)
Comparing with standard wave equation, y = A sin (kx – ωt), we get

• Amplitude A = 4 m
• Kx = 0.01 × πx
  \(\frac{2 \pi}{\lambda}\)x = 0.01 πx, λ = 200m
• Initial phase = 0
• ωt = 2 πt, 2 π f t = 2 πt, f = 1 Hz.

Question 4.
The speed of a wave along a stretched string depends only on the tension and the linear mass density of the string and does not dependent on the frequency of the wave.

1. Give the equation of speed of transverse wave along a stretched string.
2. Why the speed does not depend on the frequency of the wave.
3. A steel wire 0.72 long has a mass of 5 × 10^-3Kg. If the wire is under a tension 60N, what is the speed of transverse wave on the wire?

Answer:

1. v = \(\sqrt{\frac{T}{m}}\) Where T is the tension and m is the mass per unit length.
2. The frequency of the wave is determined by the source that generates the wave.
3. 

Question 5.
While discussing the propagation of sound through atmospheric air, one argued that the velocity of sound is 280 ms-1 and said that he calculated it using Newton’s formula. But another learner argued that velocity of sound is 330 ms1. He justified his argument by saying that he has applied Laplace corrected formula.

1. Write the formula used by the second learner.
2. Using the above relation, show that velocity depends on temperature and humidity while is independent of pressure.
3. “Sound can be heard over longer distance on rainy days.” Justify.

Answer:
1. a = \(\sqrt{\frac{2 \lambda P}{\rho}}\)

2.

If temperature remain same
PV = constant
∴ v is independent of p.

3. During rainy day as ρ decreases. Hence v increase and sound propagate longer distance.

Question 6.
When a stone is dropped in to the river, certain waves are formed on its surface.

1. What type of wave it is?
2. Is it a progressive wave? Explain.
3. If yes, derive a mathematical expression for the above wave.

Answer:
1. Transverse wave.

2. Yes, because each particles of the medium vi¬brates simple harmonically.

3. Consider a harmonic wave travelling along the +ve x-direction with a speed V. Let ‘0’ be the particle in the medium. Its displacement at any instant of time may be written as y = A sin ωt

Consider another particle ‘p’ at a distance x from ‘0’ to its right. The displacement of p at any instant.
y = A sin (ωt – α) ______(1)
α is the phase difference between 0 and P. Here
α = \(\frac{2 \pi}{\lambda}\)x equation (1) becomes

Plus One Physics Waves Five Mark Questions and Answers

Question 1.
A boy standing near a railway track found that the pitch of the siren of a train increases as it approaches him

1. State the phenomenon behind it?
2. List any two applications of the same Phenomenon
3. Obtain an expression for the apparent frequency of the siren as heard by the boy.

Answer:
1. Doppler effect.

2. Doppler effect can be used to find the speed of moving object. Dopplar effect in light is used to find speed of galaxies.

3. The apparent change in the frequency of sound wave due to the relative motion of source or listener or both is called Doppler effect. It was proposed by John Christian Doppler and it was experimentally tested by Buys Ballot.

Considers source is producing sound of frequency v. Let V be the velocity of sound in the medium and λ the wavelength of sound when the source and the listener are at rest.

The frequency of sound heard by the listener is ν = \(\frac{v}{\lambda}\). Let the source and listener be moving with velocities v_s and v_l in the direction of propogation of sound from source to listener. (The direction S to L is taken as positive).

The relative velocity of sound wave with respect to the source = V – V_s
Apparent wavelength of sound,
λ¹ = \(\frac{V-V_{s}}{v}\) _____(1)
Since the listener is moving with velocity v^, the relative velocity of sound with respect to the listener,
V¹ = V – V_l ______(2)
Apparent frequency of sound as heard by the listener is given by
ν = \(\frac{v^{1}}{\lambda^{1}}\) ______(3)
Sub (1) and (2) in eq.(3) we get

Question 2.

1. Waves are means of transferring energy from one point to another. Distinguish between longitudinal and transverse waves.
2. What is a plane progressive wave? Arrive at an expression for the displacement of a particle on the path of the wave, advancing in the positive x-direction.
3. The velocity of sound is greater in solids than in gases. Explain.

Answer:
1.

Longitudinal wave	Transverse wave
1. Can’t be polarized 2. Direction of propagation is parallel to the direction of vibration of particles.	1. Can be polarized 2. Direction of propagation is perpendicular to the direction of vibration of particles.

2. Out of syllabus

3. Solids are highly elastic as compared to liquids and gases.

Question 3.

1. Sound produced by an open pipe contains:
   • Fundamental component only
   • Odd harmonics only
   • All the harmonics
   • Even harmonics only
2. A pipe 30 cm long is open at both ends. Which harmonic mode of the pipe is resonantly exerted by a 1.1 kHz source?
3. Will resonance with the same source be observed if one end of the pipe is closed? (Take the speed
   of sound in air to be 330 ms^-1)

Answer:
1. All the harmonics.

2. We know frequency of oscillation in the Pipe, f = \(\frac{n V}{2 L}\)
Substituting the values v, L and f we get

n = 2 means that the oscillation is second harmonics.

3. The condition for second harmonics in closed pipe is

The frequency required for resonance in a closed pipe is 825 Hz. Hence we do not get resonance at 1.1 Khz.

Question 4.
A boy plucks at the centre of a stretched string of length 1 m and abserves a wave pattern.

1. Which type of wave is produced on the string?
2. What are the conditions for the formation of the above mentioned wave?
3. The distance between consecutive nodes is
   1. λ
   2. λ/2
   3. 2λ
   4. λ/4
4. A steel rod100 cm long is clamped at its middle. The fundamental frequency of longitudiral vibrations of the rod is given to be 2.5kHz. What is the speed of sound in steel?

Answer:
1. Standing wave (or) stationary wave

2. Same frequency, same amplitude, travelling in opposite direction.

3. λ/2

4. Length of the rod l = \(\frac{\lambda_{1}}{2}\)
λ₁ = 2l =2m
v = ν₁λ₁
v = 2500 × 2 = 5000 m/s.

Plus One Physics Waves NCERT Questions and Answers

Question 1.
A string of mass 2.50kg is under a tension of200N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Tension T = 200N
Length I = 20.0m; Mass M = 2.50kg

Question 2.
A stone droped from the top of a tower of height 300m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top. Given that the speed of sound in air is 340ms^-1? (g = 9.8ms^-2)
Answer:
Time after which the splash is heard at the top is equal to the sum of the time t₁ taken by the stone to fall down and the time t₂ taken by the sound to travel from bottom to top.
Using S = ut + \(\frac{1}{2}\) at², we 9et S = \(\frac{1}{2}\)gt₁²
(∵ u = 0 and a =g)

Question 3.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7kms^-1? The operating frequency of the scanner is 4.2MHz.
Answer:
λ = ? u = 1.7kms^-1 = 1700ms^-1
v = 4.2 × 10⁶ Hz, u = vλ = or λ = \(\frac{u}{v}\)
or λ = \(\frac{1700}{4.2 \times 10^{6}}\) m = 4.05 × 10^-4 m = 0.405mm.

Question 4.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45Hz. The mass of the wire is 3.5 × 10^-2kg and its linear mass density is 4.0 × 10^-2kgm^-1. What is

1. the speed of a transverse wave on the string and
2. the tension in the string?

Answer:
mass of wire M = 3.5 × 10^-2kg
Linear density µ = mass / length
= M/l = 4.0 × 10^-2kg
∴ Length of wire l = \(\frac{M}{\mu}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}\) m = 0.875m
In the fundamental mode,
λ =2l = 2 × 0.875m = 1.75m

1. Speed of transverse vyaves u = v λ
= 45 × 1.75ms^-1 = 78.75ms^-1

2. u= \(\sqrt{\frac{T}{\mu}}\) or T = µu² = 4 × 10^-2(78.75)²N
= 278.06N.

Question 5.
A steel rod 100cm long is clamped at its middle. The fundemental frequency of longitudinal vibrations of the rod are given to be 2.53kHz. What is the speed ‘ of sound in steel?
Answer:
l = 1m, v = 2.53 × 10³HZ, \(\frac{\lambda}{2}\) = I or λ = 2m
u = v λ = 2.53 × 10³ × 2ms^3-1 = 5.06 × 10³ ms^-1
= 5.06kms^-1

Students can Download Chapter 2 Data Representation and Boolean Algebra Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 2 Data Representation and Boolean Algebra

Number System:
It is a systematic way to represent numbers in different ways. Each number system has its own Base, that is a number and that number of symbols or digits used.

1. Most Significant Digit (MSD): The digit with most weight is called MSD. MSD is also called Left Most Digit(LMD)
2. Least Significant Digit (LSD): The digit with least weight is called LSD. LSD is also called Right Most Digit(RMD)
   • eg: 106 : Here MSD : 1 and LSD : 6
   • 345.78: Here MSD : 3 and LSD : 8
3. A Binary Digit is also called a bit.
4. The weight of each digit of a number can be represented by the power of its base.

Number conversions:
In general, to convert a decimal number into another number system(binary or octal or hexadecimal) do the following. Divide the number successively by the base of the number system do you want to convert and write down the remainders from bottom to top.

To convert a decimal fraction into another number system .multiply the number by the base of the number system do you want to convert then integer part and fractional part are separated again multiply the fractional part by the base and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Decimal to Binary:
Divide the number by the base 2 successively and write down the remainders from bottom to top.

Decimal fraction to binary:
multiply the number by the base 2 then integer part and fractional part are separated again multiply the fractional part by the base 2 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Decimal to Octal:
Divide the number by the base 8 successively and write down the remainders from bottom to top.

Decimal fraction to octal:
multiply the number by the base 8 then integer part and fractional part are separated again multiply the fractional part by the base 8 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.
eg: (55)₁₀ = ()₈

(0.140625)₁₀ = (0.11)₈

Decimal to Hexadecimal:
Divide the number by the base 16 successively and write down the remainders from bottom to top.

Decimal fraction to hexadecimal:
multiply the number by the base 16 then integer part and fractional part are separated again multiply the fractional part by the base 16 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Converting a number from any number system into decimal: For this multiply each digit by its corresponding weight and sum it up.

Binary to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 2.

Converting binary fraction to decimal

101.101 = 1 × 2² + 0 × 2¹ + 1 × 2⁰ + 1 × 2^-1 + 0 × 2^-2 + 1 × 2^-3
= 4 + 0 + 1 + 1/2 + 0 + 1/8
= 5 + 0.5 + 0.125
(101.101)₂ = (5.625)₁₀

Octal to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 8.
Eg: (1007)₈ =()₁₀?

1 × 8³ + 0 × 8² + 0 × 8¹ + 7 × 8⁰
= 512 + 0 + 0 + 7
=(519)₁₀
Converting octal fraction to decimal (600.005)₈ =()₁₀?

= 6 × 8² + 0 × 8¹ + 0 × 8⁰ + 0 × 8^-1 + 0 × 8^-2 + 5 × 8^-3
= 384 + 0 + 0 + 0 + 0 + 0.009765625
= (384.009765625)₁₀

Hexadecimal to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 16.
Eg: (100)₁₆ = ()₁₀?

= 1 × 16² + 0 × 16¹ + 0 × 16⁰
= 256 + 0 + 0
= (256)₁₀
Converting Hexadecimal fraction to decimal (60A.4)₈ =()₁₀?

= 6 x 16² + 0 x 16¹ + 10 x 16⁰ + 4 x 16^-1
= 1536 + 0 + 0 + .25
= (1536.25)₁₀

Octal to binary conversion:
Convert each octal digit into its 3 bit binary equivalent. Consider the following table

Hexadecimal to binary conversion:
Convert each Hexadecimal digit into its 4 bit binary equivalent. Consider the following table

1010 1011 1100 (ABC)₁₆=(101010111100)₂

Binary to octal conversion
Divide the binary number into groups of 3 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding octal equivalent.
eg: (10100110)₂= ()₈?
Insert one zero in the left side to form 3 bits group

(10100110)₂= (246)₈

Binary to Hexadecimal conversion:
Divide the binary number into groups of 4 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding Hexadecimal equivalent.
eg: (100100110)₂ = ()₁₆?
Insert 3 zeros in the left side to form 4 bits group

(100100110)₂ = (126)₁₆

Octal to Hexadecimal conversion:
First convert octal number into binary(see 1.6.7), then convert this binary into hexadecimal(also see 1.6.10)
eg: Convert (67)₈ = ( )₁₆
Step I: First convert this number into binary equivalent for this do the following:

Step II: Next convert this number into hexadecimal equivalent for this do the following.

So the answer is (67)₈ = ( 37)₁₆

Hexadecimal to octal conversion:
First convert Hexadecimal to binary(see 1.6.8), then covert this binary into octal(also see 1.6.9)
eg: Convert (A1)₁₆ = ( )₈?
Step I: First convert this number into binary equivalent. For this do the following

Step II. Next convert this number into octal equivalent. For this do the following.
So the answer is (A1)₁₆ = (241)₈

Data representation:
The data stored in the computer memory is in the form of binary.

Representation of integers
There are three ways to represent integers in computer. They are as follows:

1. Sign and Magnitude Representation (SMR)
2. 1’s Complement Representation
3. 2’s Complement Representation

1. SMR:
Normally a number has two parts sign and magnitude, eg: Consider a number+5. Here + is the sign and 5 is the magnitude. In SMR the most significant Bit (MSB) is used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is -ve. eg: If a computer has word size is 1 byte then

Here MSB is used for sign then the remaining 7 bits are used to represent magnitude. So we can , represent 2⁷ = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to +127 and 0 to -127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.

2. 1’s Complement Representation:
To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but +ve number has no 1 ’s complement,
eg:
(i) To find the 1’s complement of -21
+21 = 00010101
To get the 1’s complement change all 0 to 1 and.all 1 to 0.
-21 = 11101010
1’s complement of-21 is 11101010

(ii) Find the 1’s complement of +21. Positive numbers are represented by using SMR.
+21 = 00010101 (No need to take the 1’s complement)

3. 2’s Complement Representation:
To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg: To find the 2’s complement of -21
+21 = 00010101
First take the 1’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of -21 is 1110 1011

Representation of floating point numbers:
A real number consists of an integer part and fractional part and represent by using Exponent and Mantissa method. This method is also used to represent too big numbers as well as too small numbers.
Eg: .0000000000000000000000001 can be represented easily as 1 × 10^-25. Here T is the Mantissa and -25 is the exponent.

A computer with 32 bit word length is used 24 bits for mantissa and the remaining 8 bits used to store exponent.

Representation of characters:
1. ASCII(American Standard Code for Information Interchange):
It is 7 bits code used to represent alphanumeric and some special characters in computer memory. It is introduced by U.S. government. Each character in the keyboard has a unique number.

Eg: ASCII code of ‘a’ is 97, when you press ‘a’ in the keyboard , a signal equivalent to 1100001 (Binary equivalent of 97 is 1100001) is passed to the computer memory. 2⁷ = 128, hence we can represent only 128 characters by using ASCII. It is not enough to represent all the characters of a standard keyboard.

2. EBCDIC(Extended Binary Coded Decimal Interchange Code):
It is an 8 bit code introduced by IBM(lnternational Business Machine). 2⁸ = 256 characters can be represented by using this.

3. ISCII(lndian Standard Code for Information Interchange):
It uses 8 bits to represent data and introduced by standardization committee and adopted by Bureau of Indian Standards(BIS).

4. Unicode:
The limitations to store more characters is solved by the introduction of Unicode. It uses 16 bits so 2¹⁶ = 65536 characters (i.e, world’s all written language characters) can store by using this.

Binary arithmetic:
Binary addition:
The rules for adding two bits

eg: Find the sum of binary numbers 110011 and 100001.

Binary subtraction:
The rules for subtracting a binary digit from another digit.

Subtraction using 1’s complement:
The steps are given below:

Step 1: Add 0s to the left of the subtrahend, to make two numbers with same number of bits.
Step 2: Find 1’s complement of subtrahend.
Step 3: Add the complement with minuend.
Step 4: If there is a carry, ignore the carry, the result is positive then add the carry 1 to the result.
eg: Subtract 1101 from 111100 using 1’s complement method.
Step 1: Insert two Os to the left of 1101. Hence the subtrahend is 001101.
Step 2: 1’s complement of 001101 is 110010
Step 3: Add this to the minuend.

Step 4: Ignore the carry the result is positive and add add the carry 1 to 101110

Hence the result is 101111.

Subtraction using 2’s complement:
The steps are given below :
Step 1: Add 0s to the left of the subtrahend, to make two numbers with same number of bits.
Step 2: Find 2’s complement of subtrahend.
Step 3: Add the complement with minuend.
Step 4: If there is a carry, ignore the carry, the result is positive.
eg: Subtract 1101 from 111100 using 2’s complement method.
Step 1: Insert two 0s to the left of 1101. Hence the subtrahend is 001101.
Step 2: Find the 2’s complement of 001101.
1’s complement is 110010.
2’s complement is 110010 + 1 = 110011
Step 3: Add this to the minuend.

Step 4: Ignore the carry the result is positive. Hence the result is 101111.

Introduction to Boolean algebra:
The name Boolean Algebra is given to honour the British mathematician George Boole. Boolean algebra deals with two states true or false otherwise Yes or No and numerically either 0 or 1.

Binary valued quantities:
A logical decision which gives YES or No values is a binary decision, A statement which gives YES or NO values(TRUE or FALSE) is a logical statement or truth function. A variable which can assign TRUE or FALSE (1 or 0) values is a logical variable

Boolean operators and logic gates:
Logical Operators are AND, OR and NOT. A logical gate is a physical device (electronic circuit)that can perform logical operations on one or more logical inputs and produce a single logical output. A table represents the set f all possible values and the corresponding results in a statement is called truth table.
1. The OR operator and OR gate:
The OR operator gives a 1 either one of the operands is 1. If both operands are 0, it produces 0. The truth table of X OR Y is

The logical OR gate is given below.

The truth table and the gate for the Boolean expression Y = A + B + C

2. The AND operator and AND gate:
The AND operator gives a 1 if and only if both operands are 1. If either one of the operands is 0, it produces 0 The truth table of X AND Y is

The logical AND gate is given below.

The truth table and the gate for the Boolean expression Y = A . B . C

3. The NOT operator and NOT gate:
It produces the vice versa. NOT gate is also called inverter. It is a unary operator that means it has only one input and one output. The truth table of NOT X is

Basic postulates of Boolean algebra:
Boolean algebra consists of some fundamental laws. These laws are called postulates.
Postulate 1: Principles of 0 and 1
If A ≠ 0 , then A = 1 and A 1, then A = 0

Principle of Duality:
When changing the OR(+) to AND(.), AND (.) to OR(+), 0 to 1 and 1 to 0 in a Boolean expression we will get another Boolean relation which is the dual of the first, this is the principle of duality.

Basic theorems of Boolean algebra:
There are some standard and accepted rules in every theory, these rules are known as axioms of the theory.

Identity law:
If X is a Boolean variable, the law states that

1. 0 + X = X
2. 1 + X = 1 (these are additive identity law)
3. 0 . X = 0
4. 1 . X = X (these are multiplicative identity law)

Following are the truth tables

Idempotent law:
This law states that

1. X + X = X
2. X . X = X

Involution law:
This states that
\(\overline{\overline{\mathrm{X}}}=\mathrm{x}\)
The compliment of compliment of a number is the number itself.

Complimentary law:
This law states that

1. \(x+\bar{x}=1\)
2. \(x \cdot \bar{x}=0\)

The truth table is given below

Commutative law:
This law allows to change the position of variable in OR and AND

1. X + Y = Y + X
2. X . Y = Y . X

The truth table is given below

Associative law:
It allows grouping of variables differently

1. X + (Y + Z) = (X + Y) + Z
2. X . (Y . Z) = (X . Y) . Z

The truth table is given below

Distributive law:
This law allows expansion of multiplication over addition and also allows addition operation over multiplication.

1. X . (Y + Z) = X . Y + X . Z
2. X + Y . Z = (X + Y) . (X + Z)

The truth table is given below

Absorption law:
It is a kind of distributive law in which two variables are used and result will be one of them

1. X + (X . Y) = X
2. X . (X + Y) = X

The truth table is given below

De Morgan’s theorem:
Demorgan’s first theorem states that
\(\overline{\mathrm{X}+\mathrm{Y}}=\overline{\mathrm{X}} \cdot \overline{\mathrm{Y}}\)
ie. the compliment of sum of two variables equals product of their compliments.

The second theorem states that
\(\overline{\mathrm{X} . {\mathrm{Y}}}=\overline{\mathrm{X}}+\overline{\mathrm{Y}}\)
ie. The compliment of the product of two variables equals the sum of the compliment of that variables.

Circuit designing for simple Boolean expressions:
By using basic gates such as AND, OR and NOT gates we can create logic circuits.

Universal gates:
By using NAND and NOR gates only we can create other gate hence these gates are called Universal gate.

NAND gate:
The output of AND gate is inverted by NOT gate is the NAND gate

NOR gate:
The output of OR gate is inverted by NOT gate is the NOR gate.

Students can Download Chapter 1 Some Basic Concepts of Chemistry Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

INTRODUCTION
Chemistry is the branch of science which deals with the composition, properties and transformation of matter. These aspects can be best understood in terms of basic constituents of matter: atoms and molecules. That is why chemistry is called the sci-ence of atoms and molecules.

IMPORTANCE OF CHEMISTRY
Chemistry plays an important role in almost all walks of life. In recent years chemistry has tackled with a fair degree of success some of the pressing aspects of environmental degradation. Safer alternatives to environmentally hazardous refrigerants like CFCs, responsible for ozone depletion in the stratosphere, have been successfully synthesised.

NATURE OF MATTER
Matter is anything which has mass and occupies space. Matter can exist in three physical states: solid, liquid and gas.

At the macroscopic or bulk level, matter can be clas-sified as mixtures or pure substances.
A material containing only one substance is called a pure substance. Materials containing more than one substance are called mixtures. Pure substances are further classified into two types: elements and com¬pounds. Mixtures are also of two types: homoge¬neous mixtures and heterogeneous mixtures. A mix¬ture is said to be homogeneous if it has same com¬position throughout. Some examples of homoge¬neous mixtures are air, gasoline, kerosene, milk, alloys, etc. Heterogeneous mixtures are the mixtures which have different composition in different parts. Some examples of heterogeneous mixtures are iron and sulphur, muddy water, etc.

PROPERTIES OF MATTER AND THEIR MEASUREMENT
Every substances has characteristic properties. These are classified into two categories – physical properties and chemical properties.
Physical properties are those properties which can be measured or observed without changing the iden¬tity or the composition of the substance. Some ex¬amples of physical properties are colour, odour, melt¬ing point, boiling point, density, etc. chemical properties are characteristic reactions of different substances; these include acidity or basic¬ity, combustibility, etc.

THE INTERNATIONAL SYSTEM OF UNITS (SI UNITS)
A unit may be defined as the standard of reference chosen to measure any physical quantity. There are many different systems of units.
The improved metric system of units accepted inter-nationally is called International System of Units or SI units. The SI system has seven base units from which all other units are derived.

UNCERTAINTY IN MEASUREMENT
Scientific measurements involving some measuring devices have some degree of uncertainty. The magnitude of uncertainty depends on the accuracy of the measuring device and also on the skill of its operator.
The closeness of a set of values obtained from identical measurements of a quantity is known as precision of the measurement.
The term accuracy is defined as the closeness of a measurement or a set of measurements to its true value.

SIGNIFICANT FIGURES
The total number of digits in a measurement is called the number of significant figures. It includes the num¬ber of figures that are known with certainty plus the last uncertain digit, beginning with the first non-zero digit.
1) All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.

2) Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.
3) Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.

4) Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example,0.200 g has three significant figures. But, if otherwise, the zeros are not significant. For example, 100 has only one significant figure.

5) Exact numbers have an infinite number of significant figures. For example, in 2 balls or 20 eggs, there are infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e.,2 = 2.000000 or 20 = 20.000000

LAWS OF CHEMICAL COMBINATIONS
Law of Conservation of Mass
It states that matter can neither be created nor destroyed.
This law was put forth by Antoine Lavoisier in 1789.

Law of Definite Proportions
This law was given by, a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight. It is sometimes also referred to as Law of definite composition.

Law of Multiple Proportions
This law was proposed by Dalton in 1803. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

Gay Lussac’s Law of Gaseous Volumes
This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Avogadro Law
In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules Avogadro made a distinction between atoms and molecules which is quite understandable in the present times.

DALTON’S ATOMIC THEORY
The main postulates of the theory are:-

1. Matter is made up of extremely small, indivisible particles called atoms.
2. Atoms of the same element are identical in all respects i.e. size and mass.
3. Atoms of different elements are different, i.e., they possess different sizes, shapes, masses, and chemical properties.
4. Atoms of different elements may combine with each other in a simple whole number ratio to form compound atoms or molecules.
5. Atoms can neither be created nor destroyed, i.e., atoms are indestructible.

ATOMIC AND MOLECULAR MASSES
ATOMIC MASS
The mass of an atom is extremely small. These are very inconvenient for calculations. This difficulty was overcome by expressing atomic masses as relative masses, i.e., with respect to the mass of an atom of a standard substance. The scale in which the relative masses are expressed is called atomic mass unit scale or amu scale. One atomic mass unit (amu) is equal to one-twelfth (1/ 12) the mass of an atom of carbon-12. Recently the symbol ‘u’ (unified mass) is used in place of amu.

AVERAGE ATOMIC MASS
The atomic masses of many elements have fractional values because they exist as mixture of isotopes.
In the case of such elements, the atomic mass is taken as the average of the atomic masses of the various isotopes. For example, ordinary chlorine is a mixture of two isotopes with atomic masses 35 u and 37 u and they are present in the ration 3:1. Therefore,
Atomic mass of Chlorine = \(\frac{35 \times 3+37 \times 1}{3+1}\) = 35.5u

MOLECULAR MASS
Molecular mass is the sum of atomic masses of the elements present in a molecule.
For example, molecular mass of water
= 2 x atomic mass of hydrogen + 1 x atomic mass of oxygen
= 2 × (1.008u) + 1 × 16.00 u
= 18.02 u

FORMULA MASS
In ionic compounds, we use formula mass instead of molecular mass. Formula mass of an ionic compound is the sum of the atomic masses of all atoms in a formula unit of the compound.

MOLE CONCEPT AND MOLAR MASSES
‘Mole’ was introduced as the seventh base quantity for the amount of substance in SI system. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the ¹²C isotope. This number is known as ‘Avogadro constant’ (N_A = 6.022 × 10²³). The mass of one mole of a substance in grams is called its molar mass. The molar mass is numerically equal to atomic/ molecular/ formula mass in u.

PERCENTAGE COMPOSITION
One can check the purity of a sample by analysing its mass percentage
Mass % of an element in a compound =

EMPIRICAL FORMULA FOR MOLECULAR FORMULA
An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

Problem 1.2
A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulae?
Solution:
Step 1. Conversion of mass per cent to grams.
Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen is present, 24.27g carbon is present and 71.65 g chlorine is present.

Step 2. Convert into number moles of each element
Divide the masses obtained above by respective atomic masses of various elements.

Step 3. Divide the mole value obtained above by the smallest number
Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:CI.
In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4. Write empirical formula by mentioning the numbers after writing the symbols of respective elements.
CH₂Cl is, thus, the empirical formula of the above compound.

Step 5. Writing molecular formula
a) Determine empirical formula mass. Add the atomic masses of various atoms present in the empirical formula.
ForCH₂Cl, empirical formula mass is 12.01 + 2 1.008 + 35.453 = 49.48 g

b) Divide Molar mass by empirical formula mass

c) Multiply empirical formula by ‘n’ obtained above to get the molecular formula
Empirical formula = CH₂Cl, n = 2.
Molecular formula = 2 × [CH₂Cl]=C₂H₄Cl₂

STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
‘Stoichiometry’ deals with the calculation of masses (sometimes volumes also) of the reactants and prod¬ucts involved in a chemical reaction. The coefficients of reactants and products in a balanced chemical equation is called the stoichiometric coefficients.

LIMITING REAGENT
The amount of the product obtained in a chemical reaction is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is com¬pletely consumed in a reaction containing two or more reactants.

REACTIONS IN SOLUTIONS
1. Mass per cent :
It is obtained by the following relation:

2.Mole Fraction :
It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB respectively; then the mole fractions of A and B are given as

3. Molarity :
It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution.

4. Molality :
It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by ‘m’.

Note :
Molarity of a solution changes with temeprature. But molality of a solution does not change with temperature since mass remains unaffected with temperature.

Students can Download Chapter 14 Oscillations Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 14 Oscillations

Plus One Physics Oscillations One Mark Questions and Answers

Question 1.
Fill in the blanks :
A girl is swinging on a swing in a sitting position. When she stands up, the period of the swing will______.
Answer:
Decreases

Question 2.
A particle executes a simple harmonic motion with a frequency f. What is the frequency with which its kinetic energy oscillates?
Answer:
Frequency of oscillation of kinetic energy is 2f.

Question 3.

Answer:
a. 7/2R
b. ω = \(\sqrt{k / m}\)

Question 4.
Can a simple pendulum vibrate at centre of earth?
Answer:
No. Because ‘g’ at centre of earth is zero.

Question 5.
A glass window may be broken by a distant explosion. Why?
Answer:
The sound waves can cause forced vibrations in glass due to difference between frequency of sound wave and natural frequency of glass. This can break the glass window.

Plus One Physics Oscillations Two Mark Questions and Answers

Question 1.
A simple pendulum is transferred from earth to moon. Will it go faster or slower?
Answer:
The value of g at moon is low compared to earth. The decrease in g will increase time period of simple pendulum. Hence pendulum will vibrate slower.

Question 2.
Here five examples of accelerated motion are given in first column. Match each examples given in the second column.

Answer:

Question 3.
A girl is swinging a swing in sitting position. What shall be the effect of frequency of oscillation if

1. if she stands up
2. if another girl sits gently by her side

Answer:
1. If the system is considered as simple pendulum, length of pendulum is reduced as girl stands up.

So frequency of oscillation is increased.

2. The time period and hence frequency of simple pendulum is independent of mass. Hence there is no change in frequency.

Plus One Physics Oscillations Three Mark Questions and Answers

Question 1.
A student is advised to study the variation of period of oscillation with the length of a simple pendulum in the laboratory. According he recorded the period of oscillation for different lengths of the pendulum.

1. If he plots a graph between the length and period of oscillation, what will be the shape of the graph?
2. How would you determine the value of acceleration due to gravity using l – T² graph?

Answer:
1.

This is the equation parabola. Hence the shape of graph between period(T) and length (l) will be parabola.

2. Find slope of l – T² graph. The acceleration due to gravity can be found using formula g = 4P² × slope of (l – T²) graph.

Question 2.
A simple pendulum has a bob of mass m is suspended from the ceiling of a lift which is lying at the ground floor of a multistoried building.

1. Find the period of oscillation of pendulum when the lift is stationary.
2. What is the tension of the string of the pendulum when it is ascending with an acceleration ‘a’?
3. What is the period of oscillation of the pendulum while the lift is ascending?

Answer:

1. T = \(2 \pi \sqrt{l /g}\)
2. Tension, T= m (g + a)
3. T = 2π\(\sqrt{\frac{\ell}{g+a}}\)

Question 3.
A body tied a spherical pot with a string and suspended it on a clamp. He then filled it with water. Length of the string if 90 cm and diameter of the pot is 20 cm. The pot is slightly displaced to one side and leave it to oscillate. Considering the above example as a simple pendulum (g = 9.8 ms^-2)

1. What is the length of the Pendulum
2. Calculate the period of oscillation of the pendulum.

Answer:
1. Length of pendulum l = 90 + 10 = 100 cm

2.

Question 4.

1. Motion repeated at regular intervals of time is called periodic. Explain the simple harmonic motion with a figure.
2. A particle executes a simple harmonic motion with a period 2 seconds, starting from its equilibrium at time t = 0. Find the minimum time in which it is displaced by half the amplitude.

Answer:
1. A periodic motion in which acceleration is directly proportional to displacement but opposite in direction is called SHM.

The graphical variation of simple harmonic motion with time given above.

2. y = a sin wt

Sin π/6 = sin wt
wt = π/6
\(\frac{2 \pi}{T}\)t = π/6
But T = 2s. Hence we get

Question 5.
A spring of spring constant ‘k’ is used to suspend a mass ‘m’ at its free end while the other end of the spring is rigidly fixed.

1. If the mass is slightly depressed and released, then name the motion of the mass.
2. Write down the expression for the period of oscillation of the mass.
3. If this system is taken into outer space then what happens to its period? Why?

Answer:

1. Simple Harmonic Motion
2. T = \(2 \pi \sqrt{m / k}\)
3. Period of oscillation does not change.

Plus One Physics Oscillations Four Mark Questions and Answers

Question 1.
A simple harmonic motion is represented by x(t) = Acosωt.

1. What do you mean by simple harmonic motion.
2. An SHM has amplitude A and time period T, What is the time taken to travel from x = A to x = A/2

Answer:
1. Simple harmonic motion is the simplest form of oscillatory motion. The oscillatory motion is said to be simple harmonic motion if the displacement ‘x’ of the partide from the origin varies with time as

2. x = Acosωt

Question 2.

1. Define simple harmonic motion for a particle moving in a straight line.
2. Use your definition to explain how simple harmonic motion can be represented by the equation.
3. Show that the above equation is dimensionally correct.
4. A mechanical system is known to perform simple harmonic motion. What quantity must be measured in order to determine frequency for the system?

Answer:

1. A periodic motion in which acceleration is directly proportional to displacement and opposite in direction is called simple harmonic oscillation.
2. Mathematically, a simple harmonic oscillation can be expressed as
   y = a sin wt (or) y = a cos wt
3. y = a sin w t
   Sin wt has no dimension. Hence we need to consider dimension of ‘a’ only, ie, y = a, L = L
4. Its period is determined.

Question 3.
A particle executes simple harmonic motion according to the equation x = 5sin\(\left(\frac{2 \pi}{3} t\right)\)

1. find the period of the oscillation
2. What is the minimum time required for the particle to move between two points 2.5cm on either side of the mean position?

Answer:
1. x = 5sin\(\left(\frac{2 \pi}{3} t\right)\), when we compare this equation
with standard S.H.M, x = a sin wt.
We get wt = \(\frac{2 \pi}{3} t\)

2. y = a sin wt
2.5 = 5 sin w × t

Time taken to travel 2.5 from the mean position is 0.25 sec. Hence time taken to travel 2.5 cm on either side of the mean position is 0.5 sec.

Question 4.
A mass m is suspended at one end of a spring and the other end of the spring is firmly fixed on the ceiling. If the mass is slightly depressed and released it will execute oscillation.

1. Write down the expression for the frequency of oscillation of the mass.
2. If the spring is cut into two equal halves and one half of the spring is used to suspend the same mass then obtain an expression for the ratio of periods of oscillation in two cases.
3. If this system is completely immersed in water then what happens to the oscillation?

Answer:
1. f = \(\frac{1}{2 \pi} \sqrt{k / m}\)

2. f₁ = [atex]\frac{1}{2 \pi} \sqrt{k / m}[/latex] ____(1)

3. When spring is cut in to half,spring constant becomes 2k. Hence frequency,

4. If this system is immersed in water, the amplitude of oscillation decrease quickly. Hence system comes to rest quickly.

Question 5.
Starting from the origin, a body oscillates simple harmonically with an amplitude of 2m and a period of 2s.

1. What do you mean simple harmonic motion.
2. Draw the variation of displacement with time for the above motion.
3. After what time, will its kinetic energy be 75% of the total energy?

Answer:
1. The oscillatory motion is said to be simple harmonic motion if the displacement ‘x’ of the particle from the origin varies with time as

where
x(t) = displacement x as a function of time t
A = amplitude
ω = angular frequency
(ω t + Φ) = phase (time-dependent)
Φ = phase constant or initial phase

2.

3. y = a sin ω t
ν = aω cos ωt
Kinetic energy,

Question 6.
A body of mass 16 kg is oscillating on a spring of force constant 100 N/m.

1. What do you mean by spring constant.
2. Derive a general expression for period of oscillating spring.

Answer:
1. Spring constant is the force required for unit extension.

2.

Consider a body of mass m attached to a massless spring of spring constant K. The other end of spring is connected to a rigid support as shown in figure. The body is placed on a frictionless horizontal surface.

If the body be displaced towards right through a small distance Y, a restoring force will be developed.

Comparing this equation with standard differential equation \(\frac{d^{2} x}{d t^{2}}\) + ω²x = 0
We get ω² = k/m

Question 7.
A particle execute simple harmonic motion according to the equation, x = 5sin \(\left(\frac{2 \pi}{3}\right) t\)

1. What is the period of the oscillation?
2. Write an expression for velocity and acceleration of the above particle.

Answer:
1. When x = 5sin\(\left(\frac{2 \pi}{3}\right) t\) with standard equation.
x = asin ω t we get

2. Velocity V = \(\frac{d}{d t}\) 5sin\(\left(\frac{2 \pi}{3}\right) t\)

Question 8.

1. Arrive the differential equation of SHM.
2. What do you mean by seconds pendulum.

Answer:
1. The force acting simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion.

Comparing this equation with standard differential equation \(\frac{d^{2} x}{d t^{2}}\) + ω²x = 0,
We get ω² = k/m.

2. A pendulum having period 2 sec is called seconds pendulum.

Question 9.
SHM is a type of motion in which both speed and acceleration change continuously.

1. Which of the following condition is sufficient for SHM?
   • a = ky,
   • a = ky
   • a = ky²
2. Draw a graph of SHM between
   • displacement-time
   • speed – time
   • acceleration -time

Answer:
1. a = ky

2. Variation of displacement Y with time t

Variation of velocity Y with time t

Variation of acceleration with time

Question 10.

1. Is oscillation of a mass suspended by a spring is simple harmonic?
2. Write period of oscillation of the spring?
3. There are two springs, one delicate and another stout. For which spring, the frequency of oscillation will be more?
4. Two un equal springs of same material are loaded with same load, which one will have large value of time period?

Answer:
1. Yes

2. T = \(2 \pi \sqrt{\frac{m}{k}}\)

3. When a stout spring loaded with mass ‘m’, the extension (l) produced is large.
∴ T is large, because T = \(2 \pi \sqrt{\frac{m}{k}}\),
T is small, i.e, frequency is large. Stout spring oscillate with larger frequency.

4. When a longer spring is locked with weight mg, the extension T is more
∴ T is large, because T = \(2 \pi \sqrt{\frac{l}{g}}\)
So longer spring will have a large value of period.

Question 11.
A simple pendulum consists of a metallic bob suspended from a long straight thread whose one
end is fixed to a rigid support.

1. What is the time period of second’s pendulum?
2. Derive an expression for period of simple pendulum.

Answer:
1. 2 sec

2.

Consider a mass m suspended from one end of a string of length L fixed at the other end as shown in figure. Suppose P is the instantaneous position of the pendulum. At this instant its string makes an angle θ with the vertical.

The forces acting on the bob are (1) weight of bob F_g(mg) acting vertically downward. (2) Tension T in the string.

The gravitational force F_g can be divided into a radial component F_gCosθ and tangential component F_gSinθ. The tangential component F_gSinθ produces a restoring torque.
Restoring torque τ = – F_g sinθ . L
τ = – m_g sinθ . L _____(1)
-ve sign shown that the torque and angular displacement θ are oppositely directed. For rotational motion of bob,
τ = Iα ______(2)
Where I is moment of inertia about the point of suspension and α is angular acceleration. From eq (1) and eq (2).
Iα = – mgsinθ . L
If we assume that the displacement θ is small, sinθ ≈ θ
∴ Iα = -mgθ . L
Iα + mgθ L = 0

Comparirig eq (3) with standard differential equation


∴ period of simple pendulum,

for simple pendulum I = mL²
Substituting I = mL² we get

The above equation gives that T is indepent of mass.

Plus One Physics Oscillations Five Mark Questions and Answers

Question 1.

1. Which of the following condition is sufficient for the simple harmonic motion?
   • a = ky
   • a = ky²
   • a = -ky
   • a = -ky²
     Where ‘a’ – acceleration, y – displacement
2. Prove that simple harmonic motion is the projection of uniform circular motion on any diameter of the circle.
3. Represent graphically the variations of potential energy, kinetic energy and total energy as a function of position ‘x’ for a linear harmonic oscillator. Explain the graph.

Answer:
1. a = -ky

2.

Consider a particle moving along the circumference of a circle of radius ‘a’ and centre O, with uniform angular velocity w. AB and CD are two mutually perpendicular diameters along X and Y axis. At time t = 0.

let the particle be at P₀ so that ∠P₀OB = Φ. After time ‘t’ second, let the particle reach P so that ∠POP₀ = ω t. N is the foot of the perpendicular drawn from P on the diameter CD.

Similarly M is the foot of the perpendicular drawn from P to the diameter AB. When the particle moves along the circumference of the circle, the foot of the perpendicular executes to and fro motion along the diameter CD or AB with O as the mean position.

From the right angle triangle O MP, we get
Cos (ωt + Φ) = \(\frac{O M}{O P}\)
∴ OM = OPcos(ωt + Φ)
X= a cos (ωt + Φ) _______(1)
Similarly, we get
Sin (ωt + Φ) = \(\frac{y}{a}\) (or)
Y = a sin (ωt + Φ) _______(2)
Equation (1) and (2) are similar to equations of S.H.M. The equation(1) and (2) shows that the projection of uniform circular motion on any diameter is S.H.M.

3. KE = PE
\(\frac{1}{2}\)mω²(a² – x²) = \(\frac{1}{2}\) = mω²x²
Solving we get, x = \(\frac{a}{\sqrt{2}}\)
where a is the amplitude of oscillation.

Question 2.
The spring has a scale that reads from zero to 30 kg. The length of the scale is 30cm.

1. Calculate the force constant K
2. If the period of oscillation is 1 sec. Calculate mass of the body attached to the spring.
3. If the spring is cut into two halves, What is the force constant of each half?

Answer:
1. 30 cm = 30kg
1 cm = \(\frac{30}{30}\) = 1kg
∴ Spring constant K = 1 kg/cm = 10N/cm =1000N/m.

2.

3. If spring of spring constant K is cut in to half, spring constant of each half became, 2K

Plus One Physics Oscillations NCERT Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Answer:
(b) A freely suspended bar magnet displaced from its N-S direction and released and
(c) A hydrogen molecule rotating about its centre of mass

Question 2.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = 200x²
(c) a = 10x
(d) a = 100x²
Answer:
Only (c) ie. a = -10x represent SHM. This is because acceleration is proportional and opposite to displacement (x).

Question 3.
A spring having spring constant 1200Nm^-1 is mounted on a horizontal table as shown in figure. A mass of 3 kg is attached to the free end of the spring. The mass is the pulled sideways to a distance of 2.0cm and released.

Determine

1. the frequency of oscillations,
2. maximum acceleration of the mass and
3. the maximum speed of the mass.

Answer:
k = 1200Nm^-1, m = 3kg, a = 2.0cm = 2 × 10^-2m
1.

2. Maximum acceleration = ω² a = \(\frac{k}{m}\)a

3. Maximum speed = aω = a\(\sqrt{\frac{K}{m}}\)

Question 4.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0m. If the position moves with simple harmonic motion with an angular frequency of 200 rev/min, what is its maximum speed?
Answer:
a = \(\frac{1}{2}\)m, ω = 200 rev/min
U_max = aω = \(\frac{1}{2}\) × 200m/min = 100 m/min.

Question 5.
A circular disc of mass 10kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = -αθ, where J is the restoring couple and θ the angle of twist).
Answer:

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Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Plus One Chemistry Environmental Chemistry One Mark Questions and Answers

Question 1.
The greatest affinity for haemoglobin is shown by
a) NO
b) CO
c) O₂
d) CO₂
Answer:
b) CO

Question 2.
London smog is ___________ in nature.
Answer:
reducing

Question 3.
Addition of phosphate fertilizers into water leads to
a) Increased growth of decomposers
b) Reduced algal growth
c) Increased algal growth
d) Nutrient enrichment
Answer:
d) Nutrient enrichment

Question 4.
Which of the following is a viable particulate?
a) Algae
b) Smoke
c) Mist
d) Fumes
Answer:
a) Algae

Question 5.
Which of the following is not a greenhouse gas?
a) CO₂
b) CH₄
c) O₂
d) Water vapour
Answer:
c) O₂

Question 6.
Methyl isocyanate is prepared by the action of ___________ on methyl amine
Answer:
COCl₂

Question 7.
In a photo chemical smog the gas that causes eye irritation is ___________ .
a) CO₂
b) CH₄
c) PAN
d) Acrolein
e) both (c) and (d)
Answer:
PAN

Question 8.
Super sonic jet planes can contribute to zone depletion by introduction of the gas straight to the stratosphere
Answer:
NO

Question 9.
The difference between the amounts of dissolved oxygen in a sample of water saturated with oxygen and that after incubation for a period of 5 days is known as.
Answer:
BOD

Question 10.
Excess of nitrate ions in drinking water causes.
Answer:
Methenoglobinemia

Plus One Chemistry Environmental Chemistry Two Mark Questions and Answers

Question 1.
Write the equation for the combustion of ethane.
a) Complete combustion
b) Incomplete combustion
Answer:
a) 2C₂H₆(g) +7O₂ → 4CO₂(g) + 6H₂O(v)
b) 2C₂H₆(g)+5O₂ → 4CO(g) + 6H₂O(v)

Question 2.
Write the three common pollutants.
Answer:
1. Gases such as CO and SO₂
2. Compounds of metals like lead, mercury, zinc.
3. Radioactive substance

Question 3.
Three equations are given below. After studying it, write about Acid rain.
(1) CO₂(g) + H₂O(l) → 2H+ (aq) + CO₃^2-(aq)
(2) 2SO₂(g) + O₂(g) +2H₂O(l) → 2H₂SO₄(aq)
(3) 4NO₂(g) + O₂(g) + 2H₂O(l) → 2HNO₃ (aq)
Answer:
Oxides of Sulphur and Nitrogen, mist of HCI and Phosphoric acid, etc. present in polluted airdissolve in rain water making it more acidic. This is known as acidic rain. SO₂ and NO₂ present in polluted air are the major contributors of acid rain.
2SO₂(g) +O₂(g) + 2H₂O(l) → 2H₂SO₄(aq)
4NO₂(g) +O₂(g) + 2H₂O(l) → 2HNO₃(aq)

Question 4.
Marble of Taj Mahal reacts with acid rain. What is its result?
Answer:
The marble with which Taj Mahal is made is continuously eaten away by acid rain. It is due to the presence of chemical factories in Agra. The acids present in acid rain react with marble and making the marble dull and rough.
CaCO₃ + H₂SO₄ → CaSO₄ +H₂O + CO₂

Question 5.
Why usage of chlorofluorocarbons being discouraged? or Explain ozone layer depletion?
Answer:
The decomposition of CFC’s destroying ozone. CF₂Cl₂(g) + hv → Cl(h) + CF₂Cl(g)
The reactive Cl atom reacts with O₃ to form ClO radical. Cl(g) + O₃ → ClO(g) + O₂(g)
Thus each Cl atom produced, can destroy many O₃ molecules. This leads to ozone depletion.

Question 6.
What is meant by BOD?
Answer:
It is biochemical oxygen demand. It is the amount of dissolved oxygen required by micro-organisms to oxidise organic and inorganic matter present in polluted water.

Question 7.
1. How we can control air pollution?
2. How does the soil pollution occur?
Answer:
1. Air pollution can be controlled by the following ways.
• Exhaust to oxidize CO to CO₂.
• CO₂ level can be maintained by planting trees.
• Hydrogen gas is looked upon as pollution less future fuel.
• The large amount of nitrogen oxides emitted from power plant can be removed by scrubbing it with H₂SO₄.
2. It occurs due to
• Indiscriminate use of fertilizers, pesticides etc.
• Dumping of waste materials.
• Deforestation

Question 8.
What do you mean by greenhouse effect?
Answer:
Global warming is caused by the excess amount of CO₂ in the atmosphere. When CO₂ accumulate due to the decrease of trees it will increase the temperature of earth. This leads to melting of ice. So the sea level rises.

Question 9.
What do you mean by global warming?
Answer:
As more and more infrared radiations are trapped, the atmosphere becomes hotter and the global temperature rises up. This is known as global warming. There has been a marked increase in the levels of CO₂ in the atmosphere due to severe deforestation and burning of fossil fuels.

Question 10.
Give two examples in which green chemistry has been applied.
Answer:
1) Oxidative cracking process in the formation of ethylene is a significant step.
2) Fuel cells for cellular phones have been developed. This cell last for the full life time of the phone.

Question 11.
Oxygen play a key role in the troposphere while ozone in the stratosphere.
i) How is ozone formed in the atmosphere?
ii) What are the causes of depletion of ozone layer?
Answer:
i) By electric discharge of O₂ during lightning.
ii) UV radiations enter to the earth surface and it causes skin diseases.

Question 12.
No new industries are allowed in thickly populated cities by Government order. Why such an order is issued?
Answer:
The city will be destroyed due to acid rain caused by industrial pollution. The Govt, order was to protect the environment (orthe city).

Question 13.
Discuss the importance of dissolved oxygen in water.
Answer:
It helps living organisms in water. They inhale dissolved oxygen in water.

Question 14.
Define environmental chemistry.
Answer:
Environmental chemistry is defined as the branch of science which deals with the chemical processes occurring in the environment. It involves the study of origin, transport, reactions, effects and the fates of chemical species in the environment.

Question 15.
Explain tropospheric pollution.
Answer:
Tropospheric pollution occurs due to the presence of gaseous and the particulate pollutants.
• Gaseous air pollutants. These include mainly oxides of sulphur (SO₂, SO₃), oxides of nitrogen (NO, NO₂) and oxides of carbon (CO, CO₂) in addition , to hydrogen sulphide (H₂S), Hydrocarbons, ozone and other oxidants.
• Particulate pollutants. These include dust, mist, fumes, smoke, smog, etc.

Question 16.
Which gases are responsible for greenhouse effect? List some of them.
Answer:
The main gas responsible for greenhouse effect is CO₂. Other greenhouse gases are methane, nitrous oxide, water vapours, chlorofluorocarbons (CFC’s) and ozone.

Question 17.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.
Answer:
Excessive phytoplankton (organic pollutants such as leaves, grass, trash, etc) present in water is biodegradable. A large population of bacteria decomposes this organic matter in water. During this process they consume the oxygen dissolved in water. Water has already limited dissolved oxygen (≈10 ppm) which gets is further depleted. When the level of dissolved oxygen falls below 6 ppm, the fish cannot survive. Hence, they die and float dead in water.

Question 18.
How can domestic waste be used as mannure?
Answer:
Domestic waste comprises two types of materials, biodegradable such as leaves, rotten food, etc, and non-biodegradable such as plastics, glass, metal scrap, etc. The non-biodegradable waste is sent to industry for recycling. The biodegradable waste should be deposited in the land fills. With the passage of time, it is converted into compost manure.

Question 19.
For your agricultural field or garden, you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Answer:
The compost producing pit should be set up at a suitable place or in a tin to protect ourselves from bad odour and flies. It should be kept covered so that flies cannot make entry into it and the bad odour is minimized. The recyclable material like plastics, glass, newspapers, etc., should be sold to the vendor who further sells it to the dealer. The dealer further supplies it to the industry involved in recycling process.

Plus One Chemistry Environmental Chemistry Three Mark Questions and Answers

Question 1.
Fish grow in cold water as well as in warm water.
1. Do you agree? What is the reason?
2. pH of rain water is 5.6. Is it true? What is the reason?
Answer:
1. No. Fish grow in cold water. Warm water contains less dissolved O₂ than cold water,

2. Normally rain water contains dissolved CO₂ and hence it shows acidic pH of 5.6.
H₂O + CO₂ \(\rightleftarrows \) H₂CO₃ or 2H⁺ + CO₂
When pH of rain water became below 5.6, it becomes acid rain.

Question 2.
Write the mechanism of formation of photochemical smog.
Answer:
At high temp, the petrol and diesel engines, N₂ & O₂ combine to form NO which is emitted into atmosphere. NO is then oxidised in airto form NO₂ which absorbs sunlight and form NO and free O atom.

The O atoms being reactive and combine with O₂ to form O₃.
O₂ (g) + O(g) → O₃(g)
The O₃ react with NO formed by the photochemical decomposition of NO₂.
NO(g) + O₃(g) → NO₂(g) + O₂(g)
NO₂ and O₃ are good oxidising agents and they react with unburnt hydrocarbons in the polluted airto form substances such as acrolein and formaldehyde. These are the main substances of photochemical smog.

Question 3.
1. What are the major pollutants of water?
2. What is meant by eutrophication?
Answer:
1. Micro-organism present in domestic sewages, organic wastes, plant nutrients, toxic metals, sediments, pesticides and radioactive substances,

2. Addition of P to water as PO₄^3- ion encourages the formation of algae which reduces the dissolved oxygen content of water. This is called eutrophication.

Question 4.
1. Acid rain causes extensive damage to vegetation and aquatic life. )
i) What do you mean by acid rain?
ii) Name the chemicals responsible for acid rain.
2. List gases which are responsible for greenhouse effect.
Answer:
1. i) When the pH of the rain water drops below 5.6, it is called acid rain.
ii) Oxides of nitrogen and sulphur mist of hydrochloric acid and phosphoric acid etc.

2. Such as carbon dioxide, methane, ozone, chlorofluorocarbon compounds (CFC).

Question 5.
Statues and monuments in India such as Tajmahal are affected by acid rain. How?
Answer:
The statues of monuments in India are affected by acid rain. For example, the air in the vicinity of Taj Mahal contains very high levels of oxides of sulphur and nitrogen. This results in acid rain which reacts with marble of Taj Mahal causing pitting.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
CaCO₃ + 2HNO₃ → Ca(NO₃)2 +H₂O + CO₃.
As a result, the monument is being slowly eaten away and the marble is getting decolourised and lustreless. Thus, acid rain is considered as a threat to Taj Mahal.

Question 6.
1. Writethree major consequences of air pollution.
2. Write any two suitable methods to control air pollution.
Answer:
1. We cannot get pure oxygen for inspire.
Air pollution causes global warming.
It leads to acid rain.
It leads to diseases.

2. Plant trees.
Reduce the use of motor vehicles.
Do not burn plastics.

Question 7.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
CO combines with haemoglobin to forms a complex entity, carboxyhaemoglobin which is about 300 times more stable than oxy-haemoglobin. In blood, when the concentration of carboxyhaemoglobin reaches 3 -4%, the oxygen-carrying capacity of the blood is significantly reduced. In other words, the body becomes oxygen-starved. This results into headache, nervousness, cardiovascular disorder, weak eyesight etc.

Plus One Chemistry Environmental Chemistry Four Mark Questions and Answers

Question 1.
Classify the following pollutants:
(a) Carbon monoxide (CO)
(b) Detergents
(c) Plastic
(d) DDT
(e) Sewage
(f) Cigarette smoke
Answer:

Type of Pollution
Air	Water	Soil
CO Cigarette smoke	DDT Detergent Sewage	DDT Plastic

Question 2.
As a result of stratospheric pollution, the ozone layer is destructed.
a) What is ozone layer?
b) How is it useful to us?
c) Write a harmful effect of ozone layer depletion.
Answer:
a) The layer of ozone seen in stratosphere is called ozone layer.
b) Ozone layer plays a significant role in protecting earth from UV rays.
c) Due to ozone layer depletion agricultural crops were found to give reduced yields. Small aquatic organisms which are very sensitive are destroyed due to the increase in the level of UV radiation.

Question 3.
a) What is smog?
b) What are the adverse effects of photochemical smog?
c) Write any two methods to control photochemical smog.
Answer:
a) Smog is a mixture of smoke and fog. This is the most common example of air pollution that occurs in many cities throughout the world. There are two types of smog:
1) Classical smog
2) Photochemical smog

b) Adverse effects of photochemical smog:
• Eye irritants
• Nose irritation
• Head ache
• Chest pain
• Dryness of throat
• Cough
• Difficulty in breathing

c) 1) Use catalytic converters in automobiles.
2) Plant certain plants (e.g. Pinus) which can metabolise nitrogen oxide.

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Kerala Plus One Computer Science Notes Chapter 7 Control Statements

Summary
These are classified into two decision making and iteration statements

Decision making statements:
if statement:
Syntax: if (condition)
{
Statement block;
}
First the condition is evaluated if it is true the statement block will be executed otherwise nothing will be happened.

if…else statement:
Syntax: if (condition)
{
Statement block1;
}
else
{
Statement block2;
}

Nested if:
An if statement contains another if statement completely then it is called nested if.
if (condition 1)
{
if (condition 2)
{
Statement block;
}
}
The statement block will be executed only if both the conditions evaluated are true.

The else if ladder:
The syntax will be given below
if (expression1)
{
statement block1;
}
else if (expression 2)
{
statement block 2;
}
else if (expression 3)
{
statement block 3;
}
……..
else
{
statement block n;
}
Here firstly, expression 1 will be evaluated if it is true only the statement blockl will be executed otherwise expression 2 will be evaluated if it is true only the statement block2 will be executed and so on. If all the expression evaluated is false then only statement block n will be executed

switch statement:
It is a multiple branch statement. Its syntax is given below.
switch(expression)
{
case value: statements;break;
case value: statements;break;
case value: statements;break;
case value: statements;break;
case value: statements;break;
………
default: statements;
}
First expression evaluated and selects the statements with matched case value. If all values are not matched the default statement will be executed.

Conditional operator:
It is a ternary operator hence it needs three operands. The operator is “?:”.
Syntax:
expression ? value if true : value if false. First evaluates the expression if it is true the second part will be executed otherwise the third part will be executed.

Iteration statements:
If we have to execute a block of statements more than once then iteration statements are used.

while statement:
It is an entry controlled loop. An entry controlled loop first checks the condition and execute(or enters in to) the body of loop only if it is true. The syntax is given below
Loop variable initialised
while(expression)
{
Body of the loop;
Update loop variable;
}
Here the loop variable must be initialised before the while loop. Then the expression is evaluated if it is true then only the body of the loop will be executed and the loop variable must be updated inside the body. The body of the loop will be executed until the expression becomes false.

for statement:
The syntax of for loop is
for(initialization; checking ; update loop variable)
{
Body of loop;
}
First part, initialization is executed once, then checking is carried out if it is true the body of the for loop is executed. Then loop variable is updated and again checking is carried out this process continues until the checking becomes false. It is an entry controlled loop.

do-while statement:
It is an exit controlled loop. Exit control loop first execute the body of the loop once even if the condition is false then check the condition.
do
{
Statements
} while(expression);
Here the body executes at least once even if the condition is false. After executing the body it checks the expression if it false it quits the body otherwise the process will be continue.