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Students can Download Chapter 14 Oscillations 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 14 Oscillations

Introduction
In this chapter, we study oscillatory motion. The description of an oscillatory motion requires some fundamental concept like period, frequency, displacement, amplitude and phase.

Periodic And Oscillatory Motions
Periodic Motion:
A motion that repeats itself at regular intervals of time is called periodic motion.
Example:

• The orbital motion of planets in the solar system.
• The piston in a steam engine going back and forth

Oscillations or Vibrations:
A body executes to and fro motion at regular intervals of time is called oscillatory (or) vibratory motion.

Note:
(1) When the frequency is small, we call it oscillation. When the frequency is high, we call it vibration.
Period and frequency:
Period (T):
Time taken to complete one oscillation is called period

Frequency (n):
The number of oscillations per second is called frequency.
frequency v = \(\frac{1}{T}\)Hz

Displacement:
Displacement of oscillation means the change of any physical property with time under consideration.

Explanation:
For example, consider the oscillation of block attached to a spring. In this case the displacement of block with time is referred to as displacement.

In the case if oscillation of simple pendulum, the angle from the vertical as a function of time may be regarded as a displacement variable. The voltage across a capacitor, changing with time in an a.c. circuit is also a displacement variable

Note:
The displacement variable may take both positive and negative values.

Mathematical expression for displacement:
The displacement of a periodic function can be written as

Simple Harmonic Motion (S.H.M.)
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 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

Graphical Variation of S.H.M.

The above graph shows the graphical representation of x(t) = A coswt with time.
(Initial phase Φ = 0)

Amplitude of S.H.M.
The maximum displacement of S.H.M. from mean position is called the amplitude of S.H.M.
Note:
1.

Two simple harmonic motions having, same w and Φ but different amplitudes A and B as shown in the above figure.

2.

Two simple harmonic motions having the same A and w but different phase angle Φ as shown in the above figure.

Simple Harmonic Motion And Uniform Circular Motion

Question 1.
Show that the projection of uniform circular motion on any diameter of the circle is S.H.M.

Answer:
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 Y second, let the particle reach P so that ∠POP₀ = ωt. 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) = \(\frac{\mathrm{OM}}{\mathrm{OP}}\)
∴ OM = OP cos (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.
At t = 0, if the particle is at B, then Φ = 0. Then equations (1) and (2) reduce to
x = a cos ωt ………………….. (3)
y = a sin ωt …………………….(4)

Velocity And Acceleration In Simple Harmonic Motion B
Velocity Of S.H.M.
The y displacement of S.H.M. is given by
y = a sin ω t
∴ velocity in y direction

Case – 1
At the mean position y=0, therefore velocity is maximum. The maximum velocity is given by

Case – 2
At the extreme position, y = a
∴ V_minimum = \(\sqrt{a^{2}-a^{2}}\) = 0
So the velocity of a S.H.M. varies between o and wa

Acceleration of S.H.M
We know y = a sin ω t
Velocity v = \(\frac{d y}{d t}\) = a ω cos ω t
Acceleration a = \(\frac{d^{2} y}{d t^{2}}\) = -aω² sin ωt
a = -aω² sin ωt

This equation shows that acceleration of a SHM is directly proportional to the displacement and opposite to the displacement.

Variation of displacement Y with time t:

Variation of velocity (v) with time:

Variation of acceleration with time:

Force Law For Simple Harmonic Motion
According to Newton’s second law of motion F = ma
But a = -ω²y(t)
ie. force acting on the S.H.M. in Y direction
F = -mω²y(t)
(or) F = -ky(t)
Where k= mω²
From the above equation (1), we can take an alternative definition of simple harmonic motion.

Statement:
Simple harmonic motion is the motion executed by a particle subject to a force, which is proportional to the displacement of the particle and is directed towards the mean position.

Energy In Simple Harmonic Motion
A simple harmonically moving particle possesses both potential energy and kinetic energy. Potential energy is due to its displacement against restoring force. Kinetic energy is due to its motion.

Total energy of the S.H.M. is the sum of the kinetic energy and potential energy. Total energy remains a constant throughout its motion.

Expression for Kinetic energy:
Let m be the mass of the particle executing SHM. Let V be the velocity at any instant,

Expression for potential energy:
Potential energy is work required to take the particle against the restoring force.

Work done to displace the particle through a small distance dy, dw = force × displacement
= mω²y × dy [force = ω²y ]. Therefore total work done to take the particle from o to y.


This work done is stored in the particle as its potential energy.

At extreme position y = a

At equilibrium position y = 0
∴ PE = 0
Total energy of a S.H.M.
Total energy = PE + kE

∴ Total energy = maximum KE = maximum PE

Graphical variation of PE, KE and TE of S.H.M.

Some Systems Executing Simple Harmonic Motion
Oscillations due to a spring Hooks Law:
The force acting simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion.
F α – x (or) F= kx
where k is called spring constant

Period of oscillation of a spring:

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 ‘x’, a restoring force will be developed.

The Simple Pendulum:

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_gCos θ and tangential component F_gSin θ. The radial component is cancelled by the – tension T. But the tangential component F_gSin θ produces a restoring torque.

Restoring torque τ = – F_g sin θ . L.
τ = -mg sin θ.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 a moment of inertia about the point of suspension and α is angular acceleration. From eq (1) and eq (2).
Iα = -mg sin θ.L
If we assume that the displacement θ is small, sin θ ≈ θ.
∴ Iα = -mg θ.L
Iα + mg θ.L = 0

Damped Simple Harmonic Motion
Periodic oscillations of decreasing amplitude due to the presence of resistive forces of the medium are called damped oscillations.

Question 2.
Derive a differential equation for a damped simple harmonic oscillation.

Answer:
Consider a block of mass ‘m’ connected to one end of a massless spring of spring constant K. The other end of spring is connected to rigid support. The block is connected to a vane through a rod (The vane and rod are massless). The vane is submerged in a liquid.

Let the equilibrium position of block be ‘O’. If this block is moved along downward direction through a distance x, a damping force will be developed on vane due to liquid. This damping force is proportional to velocity of vane ie; damping force F_d α – v (or) F_d = -bv

where b is called damping constant. The value of b depends on the characteristics of the liquid and the vane.

The restoring force on the block due to spring. F_s = -kx. where x is the displacement of the mass from its equilibrium position.
∴ Total force on the block, F = -bv + -kx
ma = -bv + -kx
ma + bv + kx = 0
\(m \frac{d^{2} x}{d t^{2}}+b \frac{d x}{d t}+k x=0\)
This is the differential equation of S.H.M.

The motion of damped harmonic oscillator:
The solution of the above differential equation of damped harmonic oscillator is

Case – 1
b = 0 (There is no damping force). In this case, we get

The above result shows that, if there is no damping force (b = 0). The oscillator behaves like a undamped oscillator.

Case – 2
If b is small, the amplitude of the oscillator decreases continuously with time. The motion is approximately a periodic. The Variation of displacement x (t) with time T is shown below.

Case – 3
If damping constant b is large, the amplitude of the oscillator decreases to zero very quickly. The motion is not a periodic motion. The variation of displacement x (t) with time T is shown below.

The Energy variation of damped oscillator:
The energy of an undamped oscillator, E = \(\frac{1}{2}\)kA².
where A is the amplitude of an undamped oscillator. But for the damped oscillator, amplitude = Ae^-bt/2m.

The above equation shows that the energy of a damped oscillator decreases exponentially with time, which is shown below

Note:
Small damping means that the dimensionless ratio
\((b / \sqrt{k m})\) is much less than 1.

Forced oscillations and resonance
Free oscillation:
When a body oscillates in the absence of external forces (eg. friction etc.) the oscillations are said to be free oscillations.

Forced oscillation:
When an external periodic force is applied to a damped harmonic oscillator, the oscillator will vibrate with a constant amplitude and frequency of vibration will be that of the applied periodic force. This type of oscillation is called forced oscillation.

The differential equation of forced oscillator:
Let F(t) = F₀ cosω_dt is an external force applied to a damped oscillator.

The total force acting on the damped oscillator,
F = – bv – kx + F₀ cosω_dt
where -bv is the damping force and -kx is the linear restoring force.
F + bv + kx = F₀ cosω_dt
\(m \frac{d^{2} x}{d t^{2}}+b \frac{d x}{d t}+k x=F_{0} \cos \omega_{d} t\)
This is the differential equation of an oscillator of mass m on which a periodic force of (angular) frequency ω_d is applied. The oscillator initially oscillates with its natural frequency w. When we apply the external periodic force, the oscillation with the natural frequency die out, and the body oscillates with the (angular) frequency of the external periodic force.

The motion of forced oscillator:
The solution (displacement) of the above differential equation of forced oscillator can be written as
x(t) = Acos(ω_dt + Φ)

where m is the mass of the particle v₀ and x₀ are the velocity and the displacement of the particle at time t = 0, (which is the moment when we apply the periodic force)

Case -1
When b = 0 (damping force = 0) and ω = ω_d,
we get A = \(\frac{F_{0}}{0}\)
A = ∞
This is an ideal case. This case never arises in a real situation as the damping is never perfectly zero.

Case – 2
(Small damping, driving frequency far from natural frequency).
In this ω_db << m(ω² – ω_d²). Hence we can neglect ω_db. Hence amplitude of oscillation can be written as.

Case – 3
(Small damping, driving frequency close to natural frequency).
In this case m(ω² – ω_a²) << ω_db. Hence we can neglect m (ω² – ω_a²). Hence amplitude of oscillation can be written as

This equation shows that the maximum amplitude fora given driving frequency is governed by the driving frequency and damping constant.

Resonant Oscillation
When the frequency of the external periodic force is varied, it is found that the amplitude of the forced vibration increases and reaches a maximum value and then decreases.

The amplitude will be maximum when the frequency of the applied periodic force is equal to the natural frequency of the vibration. Such oscillations are called resonant oscillations and the phenomenon is called resonance.

Graphical variation amplitude with driving frequency:

Examples of resonance:
All mechanical structures have one or more natural frequencies, and if a structure is subjected to a strong external periodic driving force that matches one of these frequencies, the resulting oscillations of the structure may rupture it.

The Tacoma Narrows Bridge at Puget Sound, Washington, USA was opened on July 1, 1940. Four months later winds produced a pulsating resultant force in resonance with the natural frequency of the structure.

This caused a steady increase in the amplitude of oscillations until the bridge collapsed. It is for the same reason the marching soldiers break steps while crossing a bridge. Aircraft designers make sure that none of the natural frequencies at which a wing can oscillate match the frequency of the engines in flight! Earthquakes cause vast devastation.

In an earthquake, short and tall structures remain unaffected while the medium height structures fall down. This happens because the natural frequencies of the short structures happen to be higher and those of taller structures lower than the frequency of the seismic waves.

Students can Download Chapter 6 Depreciation, Provisions and Reserves Notes, Plus One Accountancy Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Notes Chapter 6 Depreciation, Provisions and Reserves

Summary:
Meaning of depreciation:
Depreciation is decline in the value of a tangible fixed asset. In accounting depreciation is the process of allocating depreciable cost over useful life of a fixed asset.

Depreciation and similar terms:
Depreciation term is used in the context of tangible fixed assets. Depletion (in the context of extractive industries), and amortisation (in the context of intangible assets) are other related terms.

Factors Affecting Depreciation:

• Wear and Tear due to use and/or passage of time
• Expiration of Legal Rights
• Obsolescence

Importance of depreciation:

• Depreciation must be charged to ascertain true and fair profit or loss.
• Depreciation is a non-cash operating expense.

Methods of charging depreciation:
Depreciation amount can be calculated using:

• Straight line method, or
• Written down value method

Methods of recording depreciation:
In the books of account there are two types of arrangements for recording depreciation on fixed assets.

• Charging depreciation to asset account or
• Creating provision for depreciation/ accumulated depreciation account.

Charging depreciation to asset account:
According to this arrangement depreciation is deducted from the depreciable cost of the asset (credited to the asset account) and charged (or debited) to profit and loss account. Journal entries under this recording method are as follows:

2. Following two entries are recorded at the end of every year.
(a) For deducting depreciation amount from the cost of the asset.

(b) For charging depreciation to profit and loss account

3. Balance Sheet Treatment
When this method is used, the fixed asset appears at its net book value (ie. cost less depreciation charged till date) on the asset side of the balance sheet and not at its original cost (also known as historical cost).

Creating Provision for depreciation account/ Accumulated depreciation Account:
This method is designed to accumulate the depreciation provided on an asset in a separate account generally called “Provision for Depreciation” or “Accumulated Depreciation” account.

The following journal entries are recorded under this method:

2. The following two journal entries are recorded at the end of each year
(a) For crediting depreciation amount to provision for depreciation account.

(b) For charging depreciation to profit and loss account

3. Balance Sheet treatment
In the balance sheet the fixed asset continues to appear at its original cost on the asset side. The depreciation charged till that date appears in the provision for depreciation account, which is shown either on the “liabilities side” of the balance sheet or by way of deduction from the original cost of the asset concerned on the asset side of the balance sheet.

Disposal of Asset:
Disposal of asset can take place either

• At the end of its useful life or
• During the useful life (due to absolescence or any other abnormal factors). In this case the following journal entries are recorded.

1. For the sale of asset as scrap

2. For transfer of balance in asset account

In case, however “the provision for depreciation account” has been in use for recording the depreciation, then before passing the above entries transfer the balance of the provision for depreciation account to the asset account by recording the following journal entry.

Factors affecting the amount of depreciation:

• Depreciation amount is determined by Original cost
• Salvage value, and
• Useful life of the asset

Provisions and Reserves:
A provision is a charge against profit. It is created for a known current liability the amount of which is uncertain. Reserve on the other hand, is an appropriation of profit. It is created to strengthen the financial position of the business.

Types of Reserves:
Reserves may be:

• General reserve and specific reserve;
• Revenue reserve and capital reserve.

Secret Reserve:
When the total depreciation charged is higher than the total depreciable cost, the Secret reserve is created. The secret reserve is not explicitly shown in the balance sheet.

Students can Download Chapter 13 Kinetic Theory 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 13 Kinetic Theory

Summary
Introduction
The kinetic theory was developed in the nineteenth century by Maxwell, Boltzmann and other. It gives a molecular interpretation of pressure and temperature of a gas. It also explains gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It help us to find molecular sizes and masses.

Molecular Nature Of Matter
The scientific atomic theory is credited to John Dalton. Atomic theory is not the end of quest, but the beginning. Atoms consist of a nucleons and electrons. The nucleous itself is made up of protons and neutrons. The protons and neutrons are again made up of quarks. Even quarks may not be the end of the story.

There may be string-like elementary entities. In this chapter we shall limit ourselves to understanding the behavior of gases.

Behavior Of Gases
Gas Laws
1. Boyles law
The law states that at a given temperature, the volume of a given mass of gas varies inversely as its pressure.
It can be written as
\(P a \frac{1}{V}\) (at constant T)
PV = constant
PV = μ RT
μ → No. of moles, R → universal gas constant, T → temperature
Boyles law is not obeyed by gasses at all temperatures and pressure. Usually, Boyles, law is obeyed by gases at high temperature and low pressure (Graph is given below).

A real gas which obey this law is called ideal or perfect gas.

Variation of ‘R’ \(\left(=\frac{\mathrm{PV}}{\mathrm{T}}\right)\) with Pressure (for different temperatures)

Variation of R \(\left(=\frac{\mathrm{PV}}{\mathrm{T}}\right)\) with Pressure for temperature T₁ > T₂ > T₃ is shown in the above graph. The above graph shows that, all real gases approach ideal gas behavior at low pressure and high temperature.

At low pressure or high temperature, the molecules are for apart and molecular interactions are negligible. Without interactions the gas behaves like an ideal gas.

Note:
R = 8.324 J mol^-1 k^-1.
Variation of V with P for different temperature:

The above graph shows experimental PV curves for steam at three temperature The dotted line are the theoretical curves. (According to Boyles law). The theoretical value and experimental value comes in agreement at high temperatures and low pressures.

2. Charles law
Charles law states that the volume of a given mass of gas is proportional to its temperature when its pressure is kept constant.
V a T (P is constant)
i.e.,

The graph between V and T:

The above graph shows experimental T-V curves (solid lines) for Co₂ at three pressures compared with Charles law (dotted lines). T is in unit of 300 k and V in units of 0.13 litres.

Question 1.
Why theoretical value does not agree with experimental value?
Answer:
According to Charles law the graph between T and V is straight line. It means that when temperature decreases, the volume of gas decreases and finally becomes zero.

Practically volume will not be zero. Because the molecules require some finite space to exist. This implies that we cannot reduce its temperature to zero value. A zero kelvin temperature is only an idealized concept.

Dalton’s law of Partial Pressures:
It states that, the total pressure of a mixture of ideal gases is the sum of partial pressures.

Proof:
Consider a mixture of non-interacting ideal gases. μ₁, moles of gas 1, μ₂ moles of gas 2 etc. in a vessel of volume V at temperature T and pressure P. Using Boyles law, we can write
PV = (μ₁ + μ₂ +…………..)RT
P = \(\frac{\mu_{1} \mathrm{RT}}{\mathrm{V}}+\frac{\mu_{2} R T}{V}+\ldots \ldots \ldots\)
P = P₁ + P₂ +……………………

Kinetic Theory Of An Ideal Gas
The kinetic theory of gases has been developed by Clausius, Maxwell, Boltzmann and others. The theory is based on the following postulates.

• The gas is a collection of large number of molecules. The molecules are perfectly elastic hard spheres.
• The size of a molecule is negligible compared with the distance between the molecules.
• The molecules are always in random motion
• During their motion, the molecules collide with each other and with the walls of the containing vessel.
• The collisions are elastic and hence the total K.E energy and the total momentum of the colliding molecules before and after collisions are the same.
• The kinetic energy of a molecule is proportional to the absolute temperature of the gas.
• There is no force of attraction or repulsion between molecules.

The pressure of an ideal gas:

Consider molecules of gas in a container. The molecules are moving in random directions with 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 the container with velocity V_x and rebounds back with -V_x. (The velocities V_x and V_y do not change because this collision is perfectly an elastic one).

Therefore, the change in momentum imparted to the area A by the molecule
= mv_x^– – mV_x
= 2mV_x

To find the total number of collisions taking place in a time t, consider the motion of the molecules towards the wall. 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_xtn.
= nmV_x²At
The rate of change of momentum,
\(\frac{Q}{t}\) = nmV_x²A
But rate of change of momentum is called force, ie. force F = nmV_x²A
∴ Pressure P = nmV_x² (P = \(\frac{F}{A}\))
Different molecules move with different velocities. Therefore, the average value V_x² is to be taken. If \(\overline{\mathbf{v}}_{\mathbf{x}}^{2}\) is the average value then the pressure.

\(p=n m \bar{v}_{x}^{2}\) ……………….. (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

Kinetic Interpretation of gas laws:

Question 2.
Derive the ideal gas equation from P = \(\frac{1}{3} \mathrm{nm} \overline{\mathrm{v}}^{2}\)
Answer:
The average kinetic energy of the molecule is
KE = \(\frac{1}{2} m \bar{v}^{2}\) …………………….(3)
The eq (2) can be modified as

The average Kinetic energy of a molecule remains constant when the temperature is constant. That is when the temperature varies, \(\overline{\mathrm{KE}}\) also varies accordingly. The kinetic energy of a molecule is related to its absolute temperature by an equation
\(\overline{\mathrm{KE}}\) = \(\frac{3}{2}\)K_BT
Substitute equation (5) in equation (6); we get

N = μ N₀
P V = μ N₀K_BT
P V = μ R T…………………… (8)
(N_BK_B = R)
This is the ideal gas equation

Deduction of Boyles law:
If the temperature is constant for gas, the eq (8) can be written as
PV = Constant
This is called Boyles law

Deduction of Charles law:
If pressure of a gas is constant, the eq (8) can be written as
V a T

This is called Charles law.

Deduction of Avogadro’s Hypothesis:
If P₁, T₁, and V constant, N₀ will be constant, ie. equal Volumes of all gases, under the same conditions of pressure and temperature will contain the same number of molecules. This is known as Avogadro’s hypothesis.

Law Of Equipartion Of Energy
Degrees of freedom:
Degrees of freedom is number of independent ways by which a molecule can possess kinetic energy of translation, rotation and vibration.

Law of equipartition energy:
The total kinetic energy of a molecule is equally divided among the different degrees of freedom.

K.E. per degree of freedom:
The average energy per degree of freedom

Where K_B is Boltzman constant.

Degrees of freedom and energy of monoatomic gas:
A monoatomic atom has 3 degrees of freedom, ie; it can move in x, y and z-direction. The average energy of single monoatomic gas in the x-direction,

Note:
1. If a molecule is restricted to move in plane. It has 2 degrees of freedom.
2. If a molecule is restricted to move in a line, it has only 1 degrees of freedom.

Degrees of freedom and energy of single diatomic molecule (rigid rotator)

Consider a diatomic molecule as a rigid rotator (Rigid rotator means that the molecule does not vibrate). A rigid diatomic molecule has 3 translation degrees of freedom and 2 rotational degrees of freedom.
(Rotational degrees of freedom is shown in the above figure).
∴ Total average energy of diatomic rigid rotator,

Note:
A diatomic molecule has 3 rotational degrees of freedom. But we consider only 2 degrees of freedom. We neglect rotation along the line joining the atoms. Because it has very small moment of inertia.

Degrees of freedom and energy of single diatomic molecule of nonrigid rotator:
Molecules like ‘co’ even at moderate temperatures have a mode of vibration. The vibration energy of a diatomic molecule.

Where K is the force constant of the oscillator and y the vibrational coordinate. The vibration energy mode contain two terms (1) Kinetic energy (2) Potential energy. Hence a single mode of vibration of molecule is considered as 2 degrees of freedom (potential energy and kinetic energy)
∴ The total vibrational energy of a single-mode
= 2 × \(\frac{1}{2}\)K_BT
= K_BT
∴ The total energy of diatomic nonrigid rotator

Degrees of freedom and energy of polyatomic molecule, (non- rigid rotator):
If a polyatomic molecule has ‘f’ modes of vibration, total number of degrees freedom = 3 vibration + 3 rotator+f vibration.
∴ Total average energy of single polyatomic molecule,

Specific Heat Capacity
Monoatomic Gases (Molar specific heat capacity):
The energy of a single monoatomic gas = 3 × \(\frac{1}{2}\) K_BT
The energy of one mole monoatomic gas = 3 × \(\frac{1}{2}\) K_BT × N_A
[one-mole atom contain Avogadro number (N_A) of atoms]

A Diatomic Gas (Molar specific heat capacity):
A rigid diatomic molecule has 5 degrees of freedom : 3 translational and 2 rotational.
∴ energy of single diatomic (rigid) molecule = 5 × \(\frac{1}{2}\) K_BT
for one mole of diatomic molecule, energy U = 5 × \(\frac{1}{2}\) K_BT × N_A

Nonrigid diatomic molecule having a vibrational mode (Molar specific heat capacity):
If the diatomic molecule is not rigid but has a vibration mode. The energy of one mole,

Polyatomic Gas (Molar specific heat capacity):
In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (t) of vibrational modes.

The energy of one mole polyatomic gas,

Note: The experimental value of C_P and C_V of polyatomic gases are greater than the predicted values. The theoretical value and experimental value will be equal when we include vibrational modes of motion in the calculation.

Specific heat capacity of solids:
Consider a solid of N atoms. Each atom is vibrating about its mean position. Each vibration mode has two degrees of freedom (corresponding to potential energy and kinetic energy). Hence an oscillation in one dimension has average energy of

2 × \(\frac{1}{2}\) K_BT = K_BT
∴ Total energy in three dimension
= 3 × K_BT
= 3K_BT.

For one mole of solid, total energy,
U = 3K_BT × N_A
[N_A = Avagadro number]
= 3N_AK_BT
U = 3RT …………………. (1) [∴ R = K_BN_A].
We know specific heat capacity,

Note:
In solids, we do not consider the translational and rotational degrees of freedom. We consider only vibrational degrees of freedom.

Specific heat capacity of water:
We treat water like a solid. A water molecule has 3 atoms. Each atom in the molecule is vibrating about its mean position. A single vibration mode has 2 degrees of freedom (1) potential energy (2) kinetic energy.

ie; The energy of one atom in one dimensional vibration mode =
2 × \(\frac{1}{2}\) K_BT = K_BT

The energy of one atom having 3 dimensional vibration mode = 3 × K_BT
The energy of one H₂0 molecule having 3 dimension vibration mode
U = 3 × 3K_BT × N_A
U = 9RT [∵ R = K_B N_A]
∴ Specific heat capacity,
\(\mathrm{c}=\frac{\mathrm{dU}}{\mathrm{dT}}=\frac{\mathrm{d}}{\mathrm{dT}}(\mathrm{RT})\)
C = 9R.

Mean Free Path
Molecules in a gas have large speeds. Yet a gas leaking from a cylinder in a kitchen takes considerable time to diffuse to the other corners of the room. Why?

The molecules in a gas have a finite size. So they collide with other molecules during their motion. As a result, they cannot move straight like path. Their paths are continuously deflected.

Mean free path:
The mean free path is the average distance covered by a molecule between two successive collisions.

Expression for mean free path:

Suppose the molecules of a gas are spheres with diameterd. Let 〈v〉 be the average velocity of the molecule.
The volume covered by a molecule during its motion, in a time Δt = πd² 〈v〉 Δt.
If ‘n’ is the number of molecules per unit volume, the total number of molecules in the above volume
= πd² 〈v〉 Δt n.
The number collisions in a time Dt,
= πd² 〈v〉 Δt n.
Number of collisions in one second,
= n π d² 〈v〉
∴ The time between two successive collisions,
\(\tau=\frac{1}{n \pi d^{2}\langle v\rangle}\).
The average distance between two successive collisions,

In this derivation, we imagined the other molecules to be rest. But actually all other molecules are moving. Hence we must take relative velocity 〈v_r〉 instead of 〈v〉. A more exact treatment gives
\(\ell=\frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^{2}} \dots(1)\)
The mean free path given by the above equation depends inversely on the number density and the size of the molecule.

Students can Download Chapter 5 Digestion and Absorption Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 5 Digestion and Absorption

What is digestion?
This process of conversion of complex food substances to simple absorbable forms is called digestion.

DIGESTIVE SYSTEM:
It includes

• Alimentary canal
• Associated glands.

Alimentary canal:

The human digestive system
1. The alimentary canal begins with the mouth, and it leadsto the buccal cavity or oral cavity.

2. The oral cavity has a number of teeth and a musculartongue. Each tooth is embedded in a socket of jaw bone. This type of attachment is called thecodont.

3. Human being forms two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

Dental formula of adult human
An adult human has 32 permanent teeth which are of four different types (Heterodont dentition).

• incisors (I)
• canine (C)
• premolars (PM)
• molars (M).

Arrangement of teeth in each half of the upper and lower jaw in the order I, C, PM, M is represented by a dental formula which in human is
\(\frac{2123}{2123}\)

• The hardest part of teeth is made up of enamel, helps in the mastication oWood.
• The tongue is attached to the floor of the oral cavity by the frenulum.
• The upper surface of the tongue has small projections called papillae, some of which bear taste buds.
• The oral cavity leads pharynx which serves as a common passage for food and air.
• The oesophagus and the trachea (wind pipe) open into the pharynx.
• A cartilaginous flap called epiglottis prevents the entry of food into the glottis – opening of the wind pipe – during swallowing.
• The oesophagus is a thin, long tube which passing through the neck, thorax and diaphragm and leads to a ‘J’ shaped bag like structure called stomach.
• A muscular sphincter (gastro-oesophageal) regulates the opening of oesophagus into the stomach.

The stomach, located in the upper left portion of The abdominal cavity, has three major parts:

1. A cardiac portion into which the oesophagus opens
2. A fundic region and
3. A pyloric portion which opens into the first part of small intestine

Small intestine is distinguishable into three regions:

1. A ‘U’ shaped duodenum
2. A long coiled middle portion jejunum and
3. A highly coiled ileum

The opening of the stomach into the duodenum is guarded by the pyloric A sphincter. Ileum opens into the large intestine.

Large intestine:
It consists of

• caecum
• colon
• rectum.

Caecum:
It is a small blind sac consists of some symbiotic micro-organisms. A narrow finger-like tubular projection, the vermiform appendix which is a vestigial organ, arises from the caecum. The caecum opens into the colon.

colon:
It is divided into three parts-an ascending, a transverse and a descending part. The descending part opens into the Rectum.

Rectum: It opens out through the anus.
Wall layers of alimentary canal:
It consists of four layers namely

1. Serosa
2. muscularis
3. sub-mucosa &
4. mucosa.

1. Serosa is the outermost layer and is made up of a thin mesothelium with some connective tissues.
2. Muscularis is formed by smooth muscles
3. The submucosal layer is formed of loose connective tissues containing nerves, blood and lymph vessels.
4. The innermost layer lining of the alimentary canal is the mucosa.
   • In duodenum, glands are also present in sub-mucosa.
   • Mucosa forms irregular folds (rugae) in the stomach and small finger-like foldings called villi in the small intestine
   • The cells lining the villi produce numerous projections called microvilli giving a brush border appearance.
   • These modifications increase the surface area.
   • Villi are supplied with a network of capillaries and a large lymph vessel called the lacteal.
   • Mucosal epithelium has goblet cells which secrete mucus that help in lubrication.
   • Mucosa also forms glands in the stomach (gastric glands) and crypts in between the bases of villi in the intestine (crypts of Lieberkuhn).

Digestive Glands:
It includes

1. salivary glands
2. liver
3. pancreas.

1. Salivary glands:
Saliva is mainly produced by three pairs of salivary glands

• parotids (cheek)
• sub-maxillary/sub-mandibular (lower jaw)
• sublinguals (belowthe tongue).

The duct systems of liver, gall bladder and pancreas These glands situated just outside the buccal cavity secrete salivary juice into the buccal cavity.

2. Liver:

• It is the largest gland of the body weighing about 1.2 to 1.5 kg in an adult human.
• It is situated in the abdominal cavity, just below the diaphragm and has two lobes.
• The hepatic lobules are the structural andfunctional units of liver containing hepatic cells arranged in the form of cords.
• Each lobule is covered by a thin connective tissue sheath called the Glisson’s capsule.
• The bile secreted by the hepatic cells passes through the hepatic ducts and is stored in gall bladder.
• The duct of gall bladder (cystic duct) along with the hepatic duct from the liver forms the common bile. duct
• The bile duct and the pancreatic duct open together into the duodenum which is guarded by sphincter of Oddi.

3. Pancreas:
It is a heterocrine (both exocrine and endocrine) elongated organ situated between the limbs of the ‘U’ shaped duodenum.

Secretions of exocrine and endocrine:

• The exocrine portion secretes an alkaline pancreatic juice containing enzymes
• The endocrine portion secretes hormones, insulin and glucagons.

DIGESTION OF FOOD:

1. The process of digestion is accomplished by mechanical and chemical processes.
2. The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. *Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus.

Contents of saliva:
It contains electrolytes (Na⁺, K+⁺, Cl^–, HCO^–“) and enzymes, salivary amylase and lysozyme.
The chemical process of digestion is initiated in the oral cavity by the carbohydrate splitting enzyme, the salivary amylase( pH 6.8).

Digestion of starch:
About 30 percent of starch is hydrolysed here by salivary amylase (optimum pH 6.8) into a disaccharide -maltose.

Enzyme for preventing infections:
Lysozyme present in saliva acts as an antibacterial agent that prevents infections.

• The bolus is then passed into the pharynx and then into the oesophagus by swallowing or deglutition.
• The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis.
• The gastro-oesophageal sphincter controls the passage of food into the stomach.

Mucosa of stomach and gastric glands:
Gastric glands have three major types of cells namely

(i) mucus neck cells which secrete mucus (ii) peptic or chief cells which secrete the proenzyme pepsinogen and (iii) parietal or oxyntic cells which secrete HCI and intrinsic factor (factor essential for absorption of vitamin B12).

Digestion in stomach:
1. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme.

2. The proenzyme pepsinogen, in the presence hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach.

3. Pepsin converts proteins into proteoses and peptones (peptides).

4. The mucus and bicarbonates play an important role in lubrication and protection of the mucosal epithelium from concentrated hydrochloric acid- pH (pH 1.8).

Special type of proteolytic enzyme in infants:
Rennin is a proteolytic enzyme found in gastric juice of infants which helps in the digestion of milk proteins. The bile, pancreatic juice and the intestinal juice are the secretions released into the, small intestine.

Pancreatic juice:
It contains inactive enzymes

1. Trypsinogen, 2. chymotrypsinogen, 3. procarboxypeptidases, 4. amylases, lipases and 5. nucleases.

Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into aqtive trypsin.

Contents of Bile and functions:
The bile released into the duodenum contains bile pigments (bilirubin and bili-verdin), bile salts, cholesterol and phospholipids but no enzymes Bile helps in emulsification of fats, i.e., breaking down of the fats into very smalfmicelles.

Bile also activates lipases. The intestinal mucosal epithelium has goblet cells which secrete mucus. The secretions of the brush border cells of the mucosa alongwith the mucus constitute the intestinal juice or succu sentericus.

Intestinal juice/succus entericus:
It contains enzymes like

1. disaccharidases (eg: maltase)
2. dipeptidases
3. lipases,
4. nucleosidases etc.

How does the intestine protect itself from digestion?
1. The mucus alongwith the bicarbonates from the pancreas protects the intestinal mucosa from acid as well as provide an alkaline medium (pH 7.8) for enzymatic activities.

2. Sub-mucosal glands (Brunner’s glands) also help in this.

The breakdown of biomacromolecules occurs in the duodenum region of the small intestine. The simple substances thus formed are absorbed in the jejunum and ileum regions of the small intestine. The undigested and unabsorbed substances are passed on to the large intestine.

Functions of large intestine:

1. Absorption of some water, minerals and certain drugs
2. Secretion of mucus which helps in adhering the waste (undigested) particles together and lubricating it for an easy passage.

The undigested, unabsorbed substances called faeces enters into the caecum of the large intestine through ileo-caecal valve, which prevents the back flow of the faecal matter. It is temporarily stored in the rectum till defaecation.

• The sight, smell and the presence of food in the oral cavity can stimulate the secretion of saliva.
• Gastric and intestinal secretions are also stimulated by neural signals.
• The muscular activities of different parts of the alimentary canal controlled by neural mechanisms, both local and through CNS.

ABSORPTION OF DIGESTED PRODUCTS:

1. The end products of digestion passthrough the intestinal mucosa into the blood or lymph.
2. It is carried out by passive, active or facilitated transport mechanisms.
3. Small amounts of monosacharides like glucose, amino acids and some of electrolytes like chloride ions are generally absorbed by simple diffusion.
4. Fructose and some amino acids are absorbed with the help of the carrier ions like Na⁺. This mechanism is called the facilitated transport.
5. Transport of water depends upon the osmotic gradient.
6. Various nutrients like amino acids, monosacharides like glucose,electrolytes like Na⁺ are absorbed into the blood by active transport and hence requires energy.

How does fat absorption occur?
Fatty acids and glycerol being insoluble, cannot be absorbed into the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa. Then, the fat globules are coated with small protein called as chylomicrons which are transported into the lymph vessels (lacteals) in the villi.

These lymph vessels ultimately release the absorbed substances into the blood stream. The maximum absorption occurs in the small intestine. The absorbed substances finally reach the tissues which utilise them for their activities. This process is called assimilation.

The digestive wastes, solidified into faeces in the rectum initiate a neural reflex causing an urge for its removal. The egestion of faeces to the outside through the anal opening (defaecation) is a voluntary process and is carried out by a mass peristaltic movement.

DISORDERS OF DIGESTIVE SYSTEM:
Jaundice:
The liver is affected, skin and eyes turn yellow due to the deposit of bile pigments.

Vomiting:
It is the ejection of stomach contents through the mouth. This reflex action is controlled by the vomit centre in the medulla. „

Diarrhoea:
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea. It reduces the absorption of food.

Constipation:
In constipation, the faeces are retained within the rectum as the bowel movements occur irregularly. ‘

Indigestion:
In this condition, the food is not properly digested leading to a feeling of fullness. The causes are inadequate enzyme secretion, anxiety, food poisoning, overeating, and spicy food.

NCERT SUPPLEMENTARY SYLLABUS
Calorific value of carbohydrate, protein and fat:
Carbohydrates, proteins and fats are chief sources of energy in humans. These are oxidized and released energy stored in ATP, it is used for many activities of the cell.

calorific value kcal = 4.184kJ):
It is defined as the amount of heat produced in calories (cal) or in joules (J) from complete combustion of 1 gram food in a bomb calorimeter (a closed metal chamber filled with 02). One kilocalorie is the amount of heat energy needed to raise the temperature of one kilogram of water through 100C(1.80F).

The calorific values of carbohydrates, proteins and fats are 4.1 kcal /g, 5.65 kcal /g and 9.45 kcal/g, respectively. The actual amounts of energy liberated in the body by these nutrients referred to as the physiologic value of the food, and are 4.0 kcal /g, 4.0 kcal Ig and 9.0 kcal /g respectively.

DEFICIENCY DISEASES:
The low amount of nutrients (Vitamin A, iron and iodine) in the diet cause deficiency disorders The important among them is protein energy malnutrition (PEM). lt is major health and nutritional problems in India.

Young children (0-6 years) require more protein for each kilogram of body weight than adults. So they are more vulnerable to malnutrition. Malnutrition leads to permanent impairment of physical and mental growth and childhood mortality and morbidity. The details of the disorders are given below.

The child suffering from PEM can recover if adequate quantities of protein and carbohydrate rich food are given.

Students can Download Chapter 7 Body Fluids and Circulation Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 7 Body Fluids and Circulation

Blood
Blood is a special connective tissue contains

• Fluid matrix
• Plasma
• Formed elements.

Plasma:
It constitute nearly 55 per cent of the blood. 90 – 92 percent of plasma is water and proteins (Fibrinogen, globulins and albumins) contribute 6 – 8 percent of it. Fibrinogens are needed for clotting or coagulation of blood.

Defence mechanism and osmotic balance:
Globulins primarly are involved in defense mechanisms of the body and the albumins help in osmotic balance. Plasma also contains small amounts of minerals like Na⁺, Ca⁺⁺, Mg⁺⁺, HCO₃^–, CI^–, etc. Factors for coagulation or clotting of blood are also present in the plasma in an inactive form.

What is serum?
Plasma without the clotting factors is called serum.

Formed Elements

1. Erythrocytes
2. Leucocytes
3. Platelets

They constitute nearly 45 per cent of the blood.
1. Erythrocytes or red blood cells (RBC):
They are the most abundant and on an average, 5 millions to 5.5 millions of RBCs mm-3 of blood. RBCs are formed in the red bone marrow in the adults. RBCs are devoid of nucleus in most of the mammals and are biconcave in shape. They have a red coloured, iron containing complex protein called haemoglobin.

A healthy individual has 12 – 16 gms of haemoglobin in every 100 ml of blood. These molecules play a significant role in transport of respiratory gases. RBCs have an average life span of 120 days after which they are destroyed in the spleen (graveyard of RBCs).

2. Leucocytes:
They are also known as white blood cells, colourless, nucleated and are relatively lesser in number which averages 6000 – 8000 mm-3 of blood. Leucocytes are generally short lived. The two main categories of WBCs granulocytes and agranulocytes.

Neutrophils, eosinophils and basophils are different types of granulocytes, while lymphocytes and monocytes are the agranulocytes.

• Neutrophils are the most abundant cells (60 – 65 percent) of the total WBCs and basophils are the least (0.5 – 1 percent) among them.
• Neutrophils and monocytes (6 – 8 per cent) are phagocytic cells which destroy foreign organisms entering the body.
• Basophils secrete histamine, serotonin, heparin, etc., and are involved in inflammatory reactions.
• Eosinophils (2 – 3 per cent) resist infections and are also associated with allergic reactions.

Lymphocytes (20-25 percent) are of two major types – ‘B’ and ‘T’ forms. Both B and T lymphocytes are responsible for immune responses of the body.

3. Platelets:
They are also called thrombocytes, are cell fragments produced from megakaryocytes (special cells in the bone marrow). Blood normally contains 1,500,00 – 3,500,00 platelets mm-3. Platelets are involved in the coagulation or clotting of blood. A reduction in their number can lead to clotting disorders which will lead to excessive loss of blood from the body.

Blood Groups
The ABO and Rh- are widely used all over the world.

ABO grouping
It is based on the presence or absence of two surface antigens on the RBCs namely A and B. The plasma of different individuals contain two natural antibodies (proteins produced in response to antigens). There are four groups of blood, A, B, AB and O.

Who are called as universal donor and receipient?
The group ‘O’ blood can be donated to persons with any other blood groupand hence ‘O’group individuals are called ‘universal donors’. Persons with ‘AB’ group can accept blood from persons with AB as well as the other grbups of blood, such persons are called ‘universal recipients’.

Rh grouping
Rh antigen is observed on the surface of RBCs of majority (nearly 80 percent) of humans. Such individuals are called Rh positive (Rh+ve) and without antigen are called Rh negative (Rh-ve). Rh group should also be matched before transfusions.

What is Rh incompatibility?
It is the mismatching of blood between the Rh-ve blood of a pregnant mother with Rh+ve blood of the foetus. Rh antigens of the foetus do not get exposed to the Rh-ve blood of the mother in the first pregnancy as the two bloods are well separated by the placenta.

But during the delivery of the first child, there is a possibility of mixing of the maternal blood to small amounts of the Rh+ve blood from the foetus. In such cases, the mother starts preparing antibodies against Rh in her blood. In subsequent pregnancies, the

Rh antibodies from the mother (Rh-ve) can leak into the blood of the foetus (Rh+ve) and destroy the foetal RBCs. This causes severe anaemia and jaundice to the baby. This condition is called Erythroblastosis foetalis.

Solving of Rh incompatibility:
This can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Coagulation of Blood
Blood coagulates or clots in response to an injury or trauma to prevent excessive loss of blood from the body. The network of threads called fibrins in which dead and damaged formed elements of blood are trapped.

Fibrins are formed by the conversion of inactive fibrinogens in the plasma by the enzyme thrombin. Thrombins are formed from another inactive substance present in the plasma called prothrombin.

An enzyme complex, thrombokinase, is required forthe above reaction. An injury stimulates the platelets in the blood to release certain factors which activate the mechanism of coagulation .Calcium ions play a very important role in clotting.

Lymph (tissue fluid)
As the blood passes through the capillaries in tissues, fluid released out is called the interstitial fluid or tissue fluid. It has the same mineral distribution as that in plasma. An elaborate network of vessels called the lymphatic system collects this fluid and drains it back to the major veins.

The fluid present in the lymphatic system is called the lymph. Lymph is a colourless fluid which are responsible forthe immune responses of the body. Lymph is also an important carrier for nutrients, hormones, etc. Fats are absorbed through lymph in the lacteals present in the intestinal villi.

Circulatory Pathways
The circulatory patterns are of two types – open or closed.

Open circulatory system:
It is present in arthropods and molluscs in which blood pumped by the heart passes through large vessels into open spaces or body cavities called sinuses.

Closed circulatory system:
It is present in Annelids and chordates in which the blood is pumped by the heart circulated through a closed network of blood vessels.
All vertebrates possess a muscular chambered heart.
1. Fishes have a 2-chambered heart with an atrium and a ventricle.

2. Amphibians and the reptiles (except crocodiles) have a 3-chambered heart with two atria and a single ventricle.

3. Crocodiles, birds and mammals possess a 4-chambered heart with two atria and two ventricles. In fishes the heart pumps out deoxygenated blood which is oxygenated by the gills and supplied to the body parts from where deoxygenated blood is returned to the heart (single circulation).

In amphibians and reptiles, the left atrium receives oxygenated blood from the gills/lungs/skin and the right atrium gets the deoxygenated blood from other body parts. However, they get mixed up in the single ventricle which pumps out mixed blood (incomplete double circulation).

In birds and mammals, oxygenated and deoxygenated blood received by the left and right atria respectively passes on to the ventricles of the same sides. The two separate circulatory pathways are present in these organisms, hence the animals have double circulation.

Human Circulatory System
Heart:
It is situated in the thoracic cavity, in between the two lungs. It is protected by a double walled membranous bag,pericardium, enclosing the pericardial fluid. It has four chambers, two small upper chambers called atria and two larger lower chambers called ventricles.

A thin, muscular wall called the inter atrial septum separates the right and the left atria, whereas a thick- walled, the inter-ventricular septum, separates the left and the right ventricles. The atrium and the ventricle of the same side are separated by a thick fibrous tissue called the atrioventricular septum.

Tricuspid and bicuspid valve:
The opening between the right atrium and the right ventricle is guarded by a valve formed of three muscularflaps or cusps, the tricuspid valve. Bicuspid or mitral valve guards the opening between the left atrium and the left ventricle.

Semilunar valve:
The openings of the right and the left ventricles into the pulmonary artery and the aorta respectively are provided with the semilunar valves.

Function of semilunar valve:
It allows the flow of blood only in one direction, i.e., from the atria to the ventricles and from the ventricles to the pulmonary artery or aorta. These valves prevent any backward flow.

SAN & AVN:
A specialised cardiac musculature called the nodal tissue is also distributed in the right upper corner of the right atrium called the sino-atrial node (SAN). Another mass of this tissue is seen in the lower left comer of the right atrium close to the atrio-ventricular septum called the atrio-ventricular node (AVN).

A bundle of nodal fibres, atrioventricular bundle (AV bundle) continues from the AVN which passes through the atrio-ventricular septa to emerge on the top of the interventrical sepyum and divides into a right and left bundle. These branches give rise to minute fibres throughout the ventricular musculature, they are called purkinje fibres.

These fibres alongwith right and left bundles are known as bundle of HIS. The nodal musculature has the ability to generate action potentials without any external stimuli.

What is the pacemaker of heart?
The SAN can generate the maximum number of action potentials, i.e., 70 – 75 min^-1, and is responsible for initiating and maintaining the rhythmic contractile activity of the heart. Hence it is called the pacemaker. Our heart normally beats 70 – 75 times in a minute (average 72 beats min^-1).

Cardiac Cycle
As the tricuspid and bicuspid valves are open, blood from the pulmonary veins and vena cava flows into the left and the right ventricle respectively. The semilunar valves are closed at this stage. The action potential is conducted to the ventricular side by the AVN and AV bundle from where the bundle of HIS transmits it through the entire ventricular musculature.

This causes the ventricular muscles to contract, (ventricular systole), the atria undergoes relaxation (diastole). As the ventricular pressure increases the semilunar valves open, allowing the blood in the ventricles to flow through these vessels into the circulatory pathways.

The ventricles relax (ventricular diastole) and the ventricular pressure falls causing the closure of semilunar valves which prevents the backflow of blood into the ventricles.

This sequential event in the heart which is cyclically repeated is called the cardiac cycle and it consists of systole and diastole of both the atria and ventricles. The heart beats 72 times per minute,i.e., that many cardiac cycles are performed per minute.

The duration of a cardiac cycle is 0.8 seconds. During a cardiac cycle, each ventricle pumps out approximately 70 mL of blood which is called the stroke volume. The stroke volume multiplied by the heart rate (no. of beats per min.) gives the cardiac output.

Cardiac output is the volume of blood pumped out by each ventricle per minute and averages 5000 mL or 5 litres in a healthy individual.

Sound produced in heart:
For example, the cardiac output of an athlete will be much higherthanthat of an ordinary man. During each cardiac cycle sounds are produced, the first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas the second heart sound (dub) is associated with the closure of the semilunar valves. These sounds are of clinical diagnostic significance.

Electrocardiograph (ECG)
ECG is a graphical representation of the electrical activity of the heart during a cardiac cycle. Each peak in the ECG is identified with a letter from P to T that corresponds to a specific electrical activity of the heart.
1. The P-wave represents the electrical excitation Diagrammatic presentation of a standard ECG (or depolarisation) of the atria, which leads to the
contraction of both the atria.

2. The QRS complex represents the depolarisation of the ventricles, which initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole.

3. The T-wave represents the return of the ventricles from excited to normal state (repolarisation). The end of the T-wave marks the end of systole.

By counting the number of QRS complexes that occur in a given time period, one can determine the heart beat rate of an individual.

Since the ECGs obtained from different individuals have the same shape any deviation from this shape indicates abnormality or disease.

Doube Circulation
It involves two types of circulation,
1. Pulmonary circulation:
The deoxygenated blood pumped into the pulmonary artery is passed on to the lungs from where the oxygenated blood is carried by the pulmonary veins into the left atrium. This pathway constitutes the pulmonary circulation.

2. Systemic circulation:
The oxygenated blood entering the aorta is carried by a network of arteries, arterioles and capillaries to the tissues from where the deoxygenated blood is collected by a system of venules, veins and vena cava and emptied into the right atrium. This is the systemic circulation.

The systemic circulation provides nutrients, O₂ and other essential substances to the tissues and takes CO₂ and other harmful substances away for elimination. A vascular connection between the digestive tract and liver called hepatic portal system.

The hepatic portal vein carries blood from intestine to the liver before it is delivered to the systemic circulation. In Coronary system of blood vessels is present in our body exclusively for the circulation of blood to and from the cardiac musculature.

Regulation Of Cardiac Activity
Normal activities of the heart are auto regulated by specialised muscles (nodal tissue), hence the heart is called myogenic. Medulla oblangata control the cardiac function through autonomic nervous system (ANS).

Neural signals through the sympathetic nerves (part of ANS) increase the rate of heart beat. On the other hand, parasympathetic neural signals decrease the rate of heart beat. Adrenal medullary hormones can also increase the cardiac output.

Disorders Of Circulatory System
High Blood Pressure (Hypertension):
Hypertension is the term for blood pressure that is higherthan normal (120/80). In this measurement 120 mm Hg (millimetres of mercury pressure) is the systolic, or pumping, pressure and 80 mm Hg is the diastolic, or resting, pressure. High blood pressure leads to heart diseases and also affects vital organs like brain and kidney.

Coronary Artery Disease (CAD):
Coronary Artery Disease, often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. It is caused by deposits of calcium, fat, cholesterol and fibrous tissues, which makes the lumen of arteries narrower.

Angina:
It is also called ‘angina pectoris’. A symptom of acute chest pain appears when no enough oxygen is reaching the heart muscle. It occurs due to conditions that affect the blood flow.

Heart Failure:
It is the state of heart when it is not pumping blood effectively enough to meet the needs of the body. Heart failure is not the same as cardiac arrest (when the heart stops beating) or a heart attack (when the heart muscle is suddenly damaged by an inadequate blood supply).

Students can Download Chapter 15 Waves 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 15 Waves

Summary
The wave is the propagation of disturbance that carries energy from one point to another point, without translatory motion of particles in the medium. There are three types of wave.
1. Mechanical Wave: Requires medium for propagation.
Eg: Sound wave, Matter wave, Seismic wave, etc.

2. Electromagnetic Wave: No medium for propagation.
Eg: Light, X-rays, UV ray, etc.

3. Matter Wave: Wave associated with moving particles (microscopic particle).
Eg: Wave of moving electron, proton, etc.

Generation of longitudinal waves by tuning for k
When prongs of tuning fork moves outward, it compresses the surrounding air and a region of increased pressure is formed. This region is called condensation. When the prongs move inward a region of low pressure called rarefraction is formed. Thus condensations and rarefractions are alternately produced.

Expression for progressive wave (Displacement relation)
A plane progressive wave propagating along positive direction of ‘x’ is given by

The wave propagating along negative ‘x’ direction is given by y (x, t) = a sin (kx + ωt + Φ).
y(x, t) gives the transverse displacement of element at position x at time t.

Crust and Trough:
Crust is the point of maximum positive displacement of wave. Trough is the point of maximum negative displacement of wave.

Transverse and Longitudinal Wave
Based on the direction of propogation and vibration wave can be of two types.

Transverse wave	Longitudinal wave
1. The direction of vibrations of particles of medium is perpendicular to direction of propogation of wave. 2. They travel in the form of crust and troughs 3. Can be polarised Eg: Vibrations in stretched string, light etc.	1.  The direction of vibration of particles of medium is in the direction of propogation of wave. 2. They travel in the form of condensations and rare fractions. 3. Cannot be polarised Vibrations of tuning fork, sound wave, etc.

Parameters of wave
Amplitude:
The magnitude of maximum displacement of element of the wave from initial position is called amplitude (a).
Phase and initial phase:
The value (kx + ωt + Φ) is called phase and f is the initial phase. The phase gives the state of motion of wave at position ‘x’ and at time V. Initial phase gives initial state of wave.
Wave length (λ):
The linear distance travelled by the wave in one complete oscillation(or vibration). Or it can be defined as distance between two conT secutive crusts or troughs. It is the distance travelled during time period T.
Wave number/Angular wave number/propagation constant:
Wave number ‘k’ is defined as

Its unit is radian/m
Time period (T):
Time for one complete oscillation/vibration is called time period.
Frequency (ν):
The number of oscillations/vibrations in one second is called frequency. Its unit is S^-1 or Hz (Hetz).
ν = \(\frac{1}{T}\)
Angular velocity or angular frequency (ω):
Angular displacement per unit time is called angular velocity or angular frequency.

The speed of travelling wave [Relation connecting V, ν and λ]

The wave propagating along x direction is represented by y = A sin (kx – ωt + Φ). As wave moves, each point on the wave from (like A) retains its displacement. This is possible only if (kx – ωt) is constant. As the wave moves both x and t are changing to keep (kx – ωt) as constant (x increase with t).

The velocity depends on wavelength and frequency. The wavelength and frequency of wave depends on the properties of medium, i.e. velocity of wave in a medium is determined by

• linear mass density
• Elastic properties.

Speed of wave in stretched string (Transverse wave)
The velocity of wave in stretched string depends on

• linear mass density (µ)
• The tension (T)

Speed of sound wave (Longitudinal wave)
The speed of sound in medium depends on

• density of medium (ρ)
• Modulus of elasticity

Case: 1( In solid )
If solid has Young’s modulus ‘Y’

Case: 2( In liquid )
If liquid has Bulk modulus ‘B’

Case: 3 In gas
The speed of sound waves in gas was determined by Newton. According to Newton, condensations and rare fractions are isothermal processes. Hence modulus of elasticity is equal to pressure.

This is called Newton’s formula.
Correction in Newton’s formula.

Question 1.
Find velocity of sound in air using Newton’s formula (P = 1.013 × 10⁵ ρ =1.239 kg m^-3)
Answer:

Note: The velocity of sound at STP is found to be 332 ms^-1
Laplace’s Formula:
Laplace corrected Newton’s formula taking condensation and rare fraction as adiabatic process. The
modulus of elasticity is now ‘γP‘ where γ = \(\frac{C_{p}}{C_{v}}\). C_p is specific heat capacity at constant pressure and C_v is specific heat capacity at constant volume.

The principle of superposition of waves
It states that when two or more waves pass through a media the net displacement of particle at any time is the algebraic sum of displacements due to each wave.

(or)

The overlapping waves algebraically add to produce a resultant wave.

Reflection of waves
Reflection from rigid boundary:
When a travelling wave is reflected by a rigid boundary, phase reversal (phase difference of π or 180°) will take place.

Reflection from open boundary:
When a travelling wave is reflected by an open boundary, no phase change will happen. The incident and reflected wave superimpose to give maximum displacement at boundary.

Standing waves

When two waves of same amplitude and frequency travelling in opposite direction superimpose the resulting wave pattern does not move to either sides. This pattern is called standing wave.
The wave travelling in positive direction of x axis y₁(x, t) = a sin(kx – ωt)
The wave travelling in negative direction of x axis y₂(x, t) = a sin(kx + ωt)
According to superposition Principle, the combined wave is
y(x, t) = y₁(x, t) + y₂(x, t)
y(x, t) = a sin(kx – ωt) + a sin(kx + ωt)
But sin A + sin B = 2 sin \(\frac{(A+B)}{2} \cos \frac{(A-B)}{2}\)
Hence we get,combined wave as

This wave has an amplitude of ‘2asinkx’, and it is not a moving wave.
Nodes & Antinodes:
The position of maximum amplitude in a standing wave is termed as anti node and position of minimum amplitude (zero) is termed as node.
Node: The amplitude of standing wave is ‘2 a sin kx’. It is zero when kx = 0, π, 2π…. etc.
ie kx = nπ.

Antinode:
The amplitude has maximum value 2a when (2a sin kx = 2a)
sinkx = 1;
ie; kx = π/2, 3π/2, 5π/2 …… etc

But k = \(\frac{2 \pi}{\lambda}\)
Hence we get

n = 0, 1, 2, 3 etc.

1. Standing Waves In Stretched String & Modes Of Vibration Of String:
A string of length L is fixed at two ends. The position of one end is chosen as x = 0, then the position of other end will be x = L. At x = 0, there will be node. To occur node at x = L, it must satisfy

The frequency of vibrations of stretched string of length L is

n = 1,2, 3…etc.
This set of frequencies at which the string can vibrate are called natural frequencies or modes of vibration or harmonics. The above equation shows that the modes of vibration (natural frequencies) of string are integral multiple of lowest frequency
n = \(\frac{V}{2 L}\) (for n = 1)
Fundamental mode(or) First harmonics:
For n = 1
ν₁ = \(\frac{V}{2 L}\)
This is the lowest frequency with which string vibrates. This is called fundamental mode or first harmonic of vibration.
Relation between I and L for first harmonics:

The string vibrate in a single segment as shown in figure.

Second harmonic:

For n = 2

Relation between I and L for second harmonics

Third harmonic:

For n = 3, there is 3^rd harmonic. Thus collection of all possible mode is called harmonic series and n is called harmonic number.

2. Vibrational modes of an air column:
(a) In closed tube:
In closed tube one end is closed and other end is open. Air column in a glass tube partially filled with water is an example of closed system. The air column in tube can be set into vibrations with the help of air excited by tuning fork.

The longitudinal waves thus generated is reflected at the closed end and a node is formed there [reflected and incident wave are out of phase and at the closed end, they superimpose to give minimum displacement]. At the open end, the displacement is maximum and antinode is formed. If L is the length of air column, anti node occurs at x = L.
We know the condition for antinode x = (n + 1/2) λ/2
Therefore L = (n + 1/2) λ/2
for n = 0, 1, 2, …….etc.
The wavelength, λ = \(\frac{2 \mathrm{L}}{(\mathrm{n}+1 / 2)}\) _____(1)
for n = 0, 1, 2, …. etc.
The frequency ν = (n + 1/2) \(\frac{V}{2 L}\) ____(2)
for n = 0, 1, 2, …….etc.
From this equation, it is clear that the air column can vibrate with different modes of frequencies (normal modes or harmonics)
Fundamental Mode(or) First harmonic:
We get fundamental mode when n=0 Substitute this in eq(2), we get
ν₁ = \(\frac{V}{4 L}\)
This is the fundamental frequency. The higher frequencies are odd harmonics of fundamental frequency ie;

(b) Open tube:
In open tube, both ends are open. At both ends, antinodes are formed.
The condition to get antinode x = n λ/2

The frequency (fundamental frequency):
We will get fundamental frequency in open tube,when n = 1. Substitute this in eq(4)
ν₁ = \(\frac{V}{2 L}\)
In open pipe all harmonies are generated whereas in closed pipe only odd harmonies are generated.
For n = 2, ν₂ = 2\(\frac{V}{2 L}\). This is second harmonic.
For n = 3, ν₃ = 3\(\frac{V}{2 L}\). This is 3rd harmonic.
ν₂ = 2 ν₁ & ν₃ = 3 .ν₁. Thus both odd and even harmonics are generated.

Beats:
When two sound waves of nearly same frequency and amplitude travelling in same direction super imposed and periodic variation of sound intensity (wavering of sound or waxing and waning of sound) is produced. This is called beats.
Explanation:
If two tuning forks of slightly different frequencies are sounded together, a regular rise and fall of sound can be heard. The sound travels in the form of condensation and rarefactions.

When two condensations due to two notes reach our ear at the same time, they superimpose to get maximum intensity (waxing of sound). If two rarefactions are reached simultaneously, they superimpose to get minimum intensity (waning of sound).

Analytical treatment of beats
Beats frequency
Suppose two sound waves of u₁ and u₂ propagating in the same direction through a medium. For simplicity let the listener be situated at x = 0 and the amplitudes of waves to be equal ie. a₁ = a₂ = a.
The displacements y₁ and y₂ due to each wave are given by
y₁ = a sin2pu₁t and y₁ = a sin2pu₂t
According to super position principle, the resultant displacement at the same time t is
y = y₁ + y₂
= a sin2pu₁t + a sin2pu₂t
y = a [sin2pu₁t + sin2pu₂t]

It is clear that, the amplitude of resultant wave (A) changes with time. It shows maxim and minima.
The resultant amplitude will be maximum, if,

Hence the amplitude of the resultant wave will be maximum at times

Time interval between successive maxima = \(\frac{1}{v_{1}-v_{2}}\)
resultant amplitude will be minimum if

Time internval between two consecutive minima = \(\frac{1}{v_{1}-v_{2}}\)
Frequency of minima = ν₁ – ν₂
Graphical representation of beats

Doppler Effect
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 n. Let V be the velocity of sound in the medium and I 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 f, the rela¬tive velocity of sound with respect to the listener,
V¹ = V – V_l _____(2)

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.

Students can Download Chapter 6 Breathing and Exchange of Gases Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 6 Breathing and Exchange of Gases

What is respiration?
The process of exchange of O₂ from the atmosphere with CO₂ produced by the cells is called breathing, commonly known as respiration.

RESPIRATORY ORGANS:

1. Lower invertebrates like sponges, coelenterates, flatworms, etc. exchange O2 with CO2 by simple diffusion over their entire body surface. 2. Earthworms use their moist cuticle and insects have a network of tubes to transport atmospheric air within the body. 3. Gills are used by most of the aquatic arthropods and molluscs whereas lungs are used by the terrestrial forms for the exchange of gases. 4. Fishes use gills whereas reptiles, birds and mammals respire through lungs. 5. Frogs can respire through their moist skin also. 6. Mammals have a well developed respiratory system.

Human Respiratory System:
The nostrils leads to a nasal chamber through the nasal passage. The nasal chamber opens into nasopharynx the common passage for food and air. Nasopharynx opens through glottis of the larynx region into the trachea.

Sound box in respiratory system:
Larynx is a cartilaginous box which helps in sound production called as the sound box. Glottis is covered by a thin elastic cartilaginous flap called epiglottis to prevent the entry of food into the larynx. Trachea is a straight tube which divides into a right and left primary bronchi.

Each bronchi undergoes repeated divisions to form the secondary and tertiary bronchi and bronchioles ending up in very thin terminal bronchioles.
Each terminal bronchiole gives rise to vascularised bag-like structures called alveoli. The branching network of bronchi, bronchioles and alveoli comprise the lungs.

The two lungs which are covered by a double layered pleura, with pleural fluid between them. It reduces friction on the lung surface. The conducting part transports the atmospheric air to the alveoli. Exchange part is the site of diffusion of O₂ and CO₂ between blood and atmospheric air.

Where is lung fitted in human body?
The lungs are situated in the thoracic chamber which is formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs and on the lower side by the dome-shaped diaphragm. The anatomical setup of lungs in thorax is the arrangement essential for breathing, directly alterthe pulmonary volume.

Respiration involves the following steps:

• Breathing or pulmonary ventilation by which atmospheric air is drawn in and CO₂ rich alveolar air is released out.
• Diffusion of gases (O₂ and CO₂) across alveolar membrane.
• Transport of gases by the blood.
• Diffusion of O₂ and.CO₂ between blood and tissues.
• Utilisation of O₂ by the cells for catabolic reactions and resultant release of CO₂

MECHANISM OF BREATHING:
Breathing involves two stages:

1. Inspiration during which atmospheric air is drawn in and
2. Expiration by which the alveolar air is released out.

How does increase or decrease of pulmonary volume occur?
Inspiration is initiated by the contraction of diaphragm which increases the volume of thoracic chamber .The contraction of external inter-costal muscles lifts up the ribs and the sternum causing an increase in the volume of the thoracic chamber. It increase the pulmonary volume.

An increase in pulmonary volume decreases the intra-pulmonary pressure to less than the atmospheric pressure which forces the airfrom outside to move into the lungs,i.e., inspiration Relaxation of the diaphragm and the inter-costal muscles returns the diaphragm and sternum to their normal positions and reduce the thoracic volume.

It reduce the pulmonary volume. This leads to an increase in intra-pulmonary pressure to slightly above the atmospheric pressure causing the expulsion of airfrom the lungs, i.e., expiration.

Instrument used for measuring breathing movements:
A healthy human breathes 12 – 16 times/minute. The volume of air involved in breathing movements can be estimated by using a spirometer.

Respiratory Volumes and Capacities:

Tidal Volume (TV): Volume of air inspired or expired during a normal respiration. It is approx. 500 mL, i.e. a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute.

Inspiratory Reserve Volume (IRV): Additional volume of air, a person can inspire by a forcible inspiration. This averages 2500 mL to 3000 mL.

Expiratory Reserve Volume (ERV): Additional volume of air, a person can expire by a forcible expiration. This averages 1000 mL to 1100 mL.

Residual Volume (RV): Volume of air remaining in the lungs even after a forcible expiration. This average 1100 mL to 1200 mL.

Expiration Capacity (EC): Total volume of air a person can expire after a normal inspiration. This includes tidal volume and expiration reserve volume (TV + ERV).

Inspiratory Capacity (IC): Total volume of air a person can inspire after a normal expiration. This includes tidal volume and inspiratory reserve volume (TV+IRV).

Functional Residual Capacity (FRC): Volume of air that will remain in the lungs after a normal expiration. This includes ERV + RV.

Vital Capacity (VC): The maximum volume of air a person can breathe in after a forced expiration. This includes ERV, TV and IRV or the maximum volume of air a person can breathe out after a forced inspiration.

Total Lung Capacity: Total volume of air accommodated in the lungs at the end of a forced inspiration. This includes RV, ERV, TV and IRV or vital capacity + residual volume.

EXCHANGE OF GASES:
Alveoli are the sites of exchange of gases. O₂ and CO₂ are exchanged in these sites by simple diffusion.
Rate of diffusion:
Solubility of the gases and the thickness of the membranes are the important factors that affect the rate of diffusion.

Partial pressure of gases:
Pressure contributed by an individual gas in a mixture of gases is called partial pressure and is represented as pO₂ for oxygen and pCO₂ for carbon dioxide. As the solubility of CO₂ is 20 – 25 times higher than that of O₂, the amount of CO₂ that can diffuse through the diffusion membrane.

As the solubility of CO₂ is 20 – 25 times higher than that of O₂ the amount of CO₂ that can diffuse through the diffusion membrane.

The diffusion membrane is made up of three major layers such as

1. The thin squamous epithelium of alveoli, 2. The endothelium of alveolar capillaries and 3. The basement substance in between them.

Its total thickness is much less than a millimetre.

TRANSPORT OF GASES:
Blood is the medium of transport for 02 and C02.

1. About 97 per cent of O₂ is transported by RBCs in the blood.
2. The remaining 3 per cent of O₂ is carried in a dissolved state through the plasma.
3. Nearly 20 – 25 per cent of CO₂ is transported by RBCs whereas 70 per cent of it is carried as bicarbonate.
4. About 7 per cent of CO₂ is carried in a dissolved state through plasma.

Transport of Oxygen:
O₂ bind with haemoglobin to form oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four molecules of O₂. Binding of oxygen with haemoglobin is related to partial pressure of O₂.

A sigmoid curve is obtained when percentage saturation of haemoglobin with O2 is plotted against the pO2. This curve is called the Oxygen dissociation curve.

Oxygen dissociation curve is useful in studying the effect of factors like pCO₂, H⁺ concentration, etc., on binding of O₂ with haemoglobin.

In the alveoli, high pO2, low pCO2, lesser H+ concentration and lower temperature are all favourable for the formation of oxyhaemoglobin.

In the tissues, low pO2, high pCO2, high H+ concentration and higher temperature exist are favourable for dissociation of oxygen from the oxyhaemoglobin.

Therefore O₂ gets bound to haemoglobin in the lung surface and gets dissociated at the tissues.

Transport of Carbon dioxide:
CO₂ is carried by haemoglobin as carbamino-haemoglobin (about 20 – 25 per cent). This binding is related to the partial pressure of CO₂. When pCO₂ is high and pO₂ is low as in the tissues, more binding of carbon dioxide occurs whereas, when the PCO₂ is low and pCO₂ is high as in the alveoli, dissociation of CO₂ from carbamino-haemoglobin takes place. RBCs contain a very high concentration of the enzyme, carbonic anhydrase which facilitates the reaction in both directions.

At the tissue site where partial pressure of CO₂ is high due to catabolism, CO₂ diffuses into blood (RBCs and plasma) and forms HCO₂ and H⁺. At the alveolar site where pCO₂ is low, the reaction proceeds in the opposite direction leading to the formation 0f CO₂ and H₂O. Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO₂ to the alveoli.

Students can Download Chapter 16 Probability Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 16 Probability

Plus One Maths Probability Three Mark Questions and Answers

Question 1.
Two dice are thrown. The events A, B and C are as follows:

1. A: getting an even number on the first die.
2. B: getting an odd number on the first die.
3. C: getting sum of the numbers on the dice ≤ 5.

Describe the events.
Answer:

1. A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
2. B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
3. C = {(1, 1), (1, 2), (2, 1), (2, 2), (1, 3), (3, 1), (2, 3), (3, 2), (1, 4), (4, 1)}

Question 2.
A bag contains 6 red and 12 green balls. Two balls are drawn. What is the probability that one is red and other is green?
Answer:
Here total number of balls = 6 + 12 = 18 Two balls from 18 can be drawn in
¹⁸C₂ = \(\frac{18 \times 17}{1 \times 2}\) = 153
One red ball out of 6 red can be drawn in ⁶C₁ = 6 ways. One green balls from 12 green may be done in ¹²C₁ = 12 ways.
Therefore, number of favorable cases
= 6 × 12 = 72
The probability that one is red and other is green \(=\frac{72}{153}=\frac{8}{17}\).

Question 3.
In class XI of a school 40% of the students study Mathematics and 30% study Biology, 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Answer:
Let M- Mathematics and B – Biology be the events.
P(M) = \(\frac{40}{100}\) = \(\frac{2}{5}\); P(B) = \(\frac{3}{10}\);
P(M ∩ B) = \(\frac{1}{10}\)
P(M ∪ B) = P(M) + P(B) – P(M ∩ B)

Plus One Maths Probability Four Mark Questions and Answers

Question 1.
Two die are rolled, A is the event that the sum of the numbers shown on the two die is 7 and B is the event that at least one of the die shows up 2. Are the two events A and B

1. Mutually exclusive
2. Exhaustive.

Answer:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5),(2,6), (3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A = {(6, 1), (1, 6), (4, 3), (3, 4), (2, 5), (5, 2)}
B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
1. Now; A ∩ B = {(2, 5), (5, 2)} ≠ Φ
Therefore not mutually exclusive.

2. A ∪ B = {(6, 1), (1, 6), (4, 3), (3, 4), (2, 5), (5, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (1, 2), (3, 2), (4, 2), (6, 2)} ≠ S
Therefore not mutually exhaustive.

Question 2.
A letters of the word ASSASSINATION are randomly chosen. Find the probability that letter is

1. a vowels. (2)
2. a consonant (2)

Answer:
There 13 letter in the word, with 6 vowels and 7 consonants.
One letter is selected out of 13 in ¹³C₁ = 13 ways.
1. One vowel is selected out of 6 in ⁶C₁ = 6 ways.
Thus the probability of a vowel = \(\frac{6}{13}\).

2. One consonant is selected out of 7 in ⁷C₁ = 7 ways.
Thus the probability of a consonant = \(\frac{7}{13}\).

Question 3.
If E and F are events such that P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\) and P(E and F) = \(\frac{1}{8}\). Find

1. P(E or F)
2. P(not E and not F)

Answer:
P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\); P(E ∩ F) = \(\frac{1}{8}\)
1. P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

2. P(not E and not F) = P(E’ ∩ F’) = P(E ∪ F)’
= 1 – P(E ∪ F) = 1 – \(\frac{5}{8}\) = \(\frac{3}{8}\)

Plus One Maths Probability Practice Problems Questions and Answers

Question 1.
A die is thrown. Describe the following events: (1 score each)

1. A: a number less than 7.
2. B: a number greater than 7.
3. C: a multiple of 3:
4. D: a number less than 4.
5. E: An even number greater than 4.
6. F: a number not less than 3.

Also find
A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E F’, E ∩ F’.
Answer:

1. A = {1, 2, 3, 4, 5, 6}
2. B = Φ
3. C = {3, 6}
4. D = {1, 2, 3}
5. E = {6}
6. F = {3, 4, 5, 6}

Now; A ∪ B = {1, 2, 3, 4, 5, 6}
A ∩ B = Φ; B ∪ C = {3, 6}, E ∩ F = {6}
F’ = {1, 2}, E ∩ F’ = Φ.

Question 2.
Describe the sample space for the following events: (2 score each)

1. A coin is tossed and a die is thrown.
2. A coin is tossed and then a die is rolled.
3. 2 boys and 2 girls are in a Room X and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.
4. One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its upper most face is noted.
5. An experiment consists of tossing a coin and then throwing it second time if a head occur. If a tail occurs on the first toss, then a die is rolled once.
6. A coin is tossed. If the outcome is a head, a die is thrown. If the die shows up shows up an head, a die is thrown. If the die shows up an even number, the die is thrown again.

Answer:
1. A coin is tossed then S₁ = {H, T}
A die is rolled then S₂ = {1, 2, 3, 4, 5, 6}
Hence sample space
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

2. A coin is tossed then S₁ = {H, T}
A die is rolled then S₂ = {1, 2, 3, 4, 5, 6}
Hence sample space S = {H1, H2, H3,H4, H5,H6, T}

3. Let B₁, B₂ denote the boys and G₁, G₂ girls in room X, B₁ denote the boys and G₃, G₄, G₅ girls in room Y. Hence sample space
S = {XB₁, XB₂, XG₁, XG₂, YB₁, YG₃, XG₄, XG₅}

4. Three die are R,W, and B, then S₁ = {R, W, B)
A die is rolled then S₂ = {1, 2, 3, 4, 5, 6}
Hence sample space S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

5. A coin is tossed then S₁ = {H, T}
A coin tossed again then S₂ = {H, T}
A die is rolled then S₃ = {1, 2, 3, 4, 5, 6}
Hence sample space
S = {HH, HT, T1, T2, T3, T4, T5, T6}

6. A coin is tossed then S₁ = {H, T}
When the coin shows T, then there is no action. When the coin shows H, a die is thrown
S₂ = {1, 2, 3, 4, 5, 6}
When the die shows {1, 3, 5}, then there is no action.
When the die shows {2, 4, 6}, then a die is thrown again.
Hence;
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46,H61, H62, H63, H64, H65, H66}.

Question 3.
A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?
Answer:
When die is rolled the sample space is S = {1, 2, 3, 4, 5, 6}.
E: die shows 4 = {4}
F: die shows even number = {2, 4, 6}
Now E ∩ F = {4} ≠ Φ
Thus E and F are not mutually exclusive.

Question 4.
Two die are thrown simultaneously. Find the probability of getting 4 as the product.
Answer:
n(S) = 36
A = {(1, 4), (4, 1), (2, 2)}
P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{36}=\frac{1}{12}\).

Question 5.
A coin is tossed, twice, what is the probability that atleast one tail occurs?
Answer:
Here, S = {HH, HT, TH, TT}; n(S) = 4
A = {TH, HT, TT}; n(A) = 3
P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{4}\).

Question 6.
A die is rolled, find the probability of following events: (2 score each)

1. A prime number will appear.
2. A number greater than or equal to 3 will appear.
3. A number less than or equal to one will appear.
4. A number more than 6 will appear.
5. A number less than 6 will appear.

Answer:
Here sample space is S = {1, 2, 3, 4, 5, 6}
1. A: A prime number will appear A = {2, 3, 5};
P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

2. B: A number greater than or equal to 3 will appear. B = {3, 4, 5, 6};
P(B) = \(\frac{n(B)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

3. C: A number less than or equal to one will appear. C = {1};
P(C) = \(\frac{n(C)}{n(S)}=\frac{1}{6}\)

4. D: A number more than 6 will appear.
D = Φ; P(D) = \(\frac{n(D)}{n(S)}=\frac{0}{6}=0\)

5. E: A number less than 6 will appear. E = {1, 2, 3, 4, 5, 6};
P(E) = \(\frac{n(E)}{n(S)}=\frac{6}{6}=1\).

Question 7.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that (2 score each)

1. The card drawn is black.
2. The card drawn is a king.
3. The card drawn is black and a king.
4. The card drawn is either black ora king.

Answer:
1. A pack of 52 cards has 26 black cards. So one black card may be drawn in 26 ways. Therefore; Probability of card drawn is black
= \(\frac{26}{52}=\frac{1}{2}\)

2. A pack of 52 cards has 4 kings. So one king card may be drawn in 4 ways. Therefore; Probability of drawing a king
= \(\frac{4}{52}=\frac{1}{13}\)

3. A pack of 52 cards has 2 black king cards. So one black king card may be drawn in 2 ways. Therefore; Probability of drawing a king
= \(\frac{2}{52}=\frac{1}{26}\)

4. A pack of 52 cards has 26 black cards which include 2 black king and 2 red king cards. So number of cards which are black or king cards
= 26 + 2 = 28.
Therefore; Probability of drawing either a black or king card = = \(\frac{28}{52}=\frac{7}{13}\).

Question 8.
GivenP(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\). Find P(A or B), if A and B are mutually exclusive events.
Answer:
Here; P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\)
Since A and B mutually exclusive events
P(A ∪ B) = P(A) + P(B)

Students can Download Chapter 4 Principles of Programming and Problem Solving Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 4 Principles of Programming and Problem Solving

Plus One Principles of Programming and Problem Solving One Mark Questions and Answers

Question 1.
The process of writing program is called _______.
Answer:
programming or coding

Question 2.
One who writes program is called _________
Answer:
Programmer

Question 3.
The step by step procedure to solve a problem is known as ________
Answer:
Algorithm

Question 4.
Diagrammatic representation of an algorithm is known as __________
Answer:
Flow Chart

Question 5.
Program errors are known as _________
Answer:
bugs

Question 6.
Process of detecting and correcting errors is called ________
Answer:
debugging

Question 7.
Mr. Ramu represents an algorithm by using some symbols. This representation is called _______.
Answer:
Flow Chart

Question 8.
Your computer teacher asked you that which symbol is used to indicate beginning or ending flow chart? What is your answer?
(a) Parallelogram
(b) Rectangle
(c) Oval
(d) Rhombus
Answer:
(c) Oval

Question 9.
You are suffering from stomach ache. The doctor prescribes you for a scanning. This process is related in a phase in programming. Which phase is this?
Answer:
Problem identification

Question 10.
Your friend asked you a doubt that to draw a flow chart parallelogram is used for what purpose?
Answer:
To input/output

Question 11.
Mr. Anil wants to perform a multiplication which symbol is used to represent in a flow chart.
Answer:
It is a processing so rectangle is used

Question 12.
Mr. George wants to check a number is greater than zero and to perform an operation while drawing a flow chart which symbol is used for this?
Answer:
Rhombus

Question 13.
ANSI means ________
Answer:
American National Standards Institute

Question 14.
To indicate the flow of an operation which symbol is used to draw a flow chart.
Answer:
Flow lines with arrow heads

Question 15.
Mr. Johnson is drawing a flow chart but it is not fit in a single page. Which symbol will help him to complete the flow chart?
Answer:
Connectors

Question 16.
Mr. Ravi developed a s/w student information system, He wants to protect the s/w from unauthorized copying. There is an act what is it?
Answer:
Copy right act

Question 17.
Odd man out.
(а) Parallelogram
(b) Oval
(c) Rectangle
(d) Star
Answer:
(d) Star, others are flowchart symbols

Question 18.
Odd man out
(a) Oval
(b) Rhombus
(c) Connector
(d) Triangle
Answer:
(d) Triangle, Others are flowchart symbols

Question 19.
Raju wrote a program and he wants to check the errors and correct if any? What process he has to do for this?
Answer:
debugging

Question 20.
Odd one out.
(a) Problem identification
(b) Translation
(c) debugging
(d) copyright
Answer:
(d) Copy right, others are phases in programming

Question 21.
Odd man out.
(a) syntax error
(b) logical error
(c) Runtime error
(d) printer error
Answer:
(d) Printer error, Others are different types of errors

Question 22.
A computerized system is not complete after the execution and testing phase? What is the next phase to complete the system?
Answer:
Documentation.

Question 23.
Mr. Sathian takes a movie DVD from a CD library and he copies this into another DVD without permission. This process is called _______________
Answer:
Piracy

Question 24.
Mr. Santhosh purchased a movie DVD and he takes several copies without permission. He is a _________
(a) Programmer
(b) Administrator
(c) Pirate
(d) Organizer
Answer:
(c) Pirate

Question 25.
The symbol used for copy right is a _______
(a) @
(b) Copy
(c) &
(d) ©
Answer:
(d) ©

Question 26.
Following are the advantages of flowcharts one among them is wrong. Find it.
(a) Better communication
(b) Effective analysis
(c) proper program documentation
(d) Modification easy
Answer:
(d) Modification easy. It is a disadvantage.

Question 27.
Which flow chart symbol has one entry flow and two exit flows?
Answer:
Diamond

Question 28.
Which flow chart symbol is always used in pair?
Answer:
connector

Question 29.
Program written in HLL is known as ____________
Answer:
Source code

Question 30.
Some of the components in the phases of programming are given below. Write them in order of their occurrence. (1)

1. Translation
2. Documentation
3. Problem identification
4. Coding of a program

Answer:
The chronological order is as follows

1. Problem Identification
2. Coding of a program
3. Translation
4. Documentation

Question 31.
_________ is the stage where programming errors are discovered and corrected.
Answer:
Debugging or compiling

Question 32.
Ramesh has written a C++ program. During compilation and execution there were no errors. But he got a wrong output. Name the type of error he faced.
Answer:
Logical Error

Question 33.
Pick out the software which rearranges the scattered files in the hard disk and improves the performance of the system.
(a) Backup software
(b) File compression software
(c) Disk defragmenter
(d) Antivirus software
Answer:
(c) Disk defragmenter

Question 34.
Some phases in programming are given below.

1. Source coding
2. Execution
3. Translation
4. Problem study

These phases should follow a proper order. Choose the correct order from the following:
(a) 4 → 2 → 3 → 1
(b) 1 → 3 → 2 → 4
(c) 1 → 3 → 4 → 2
(d) 4 → 1 → 3 → 2
Answer:
(d) 4 → 1 → 3 → 2

Question 35.
Which one of the following errors is identified at the time of compilation?
(a) Syntax error
(b) Logical error
(c) Run-time error
(d) All of these
Answer:
(a) Syntax error

Question 36.
Pick the odd one out and give a reason for your finding

Answer:
c. This has one entry flow and more than one exit flow.

OR

b. Used for both input and output.

Plus One Principles of Programming and Problem Solving Two Mark Questions and Answers

Question 1.
A debate on ‘Whether Free Software is to be promoted’ is planned in your class. You are asked to present points in support of Free Software. What would be your arguments, (at least three)?
Answer:
Freedom to use Comparatively cheap Freedom to modify and redistribute

Question 2.
Mr. Roy purchased a DVD of a movie and he found that on the cover there is a sentence copyright reserved and a mark ©. What is it? Briefly explain?
Answer:
It is under the act of copyright and the trademark is © copyright is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator.

Question 3.
Can a person who knows only Malayalam talk to a person who knows only Sanskrit normally consider the corresponding situation in a computer program and justify your answer?
Answer:
Normally it is very difficult to communicate. But it is possible with the help of a translator. Translation is the process of converting programs written in High Level Language into Low Level Language (machine Language). The compiler or interpreter is used for this purpose. It is a program.

Question 4.
Define the term, debugging. Write the names of two phases that are included in debugging. (2)

OR

Define the different types of errors that are encountered during the compilation and running of a program.
Answer:
Debugging:
The program errors are called ‘bugs’ and the process of detecting and correcting errors is called debugging. Compilation and running are the two phases.

OR

In general there are two types of errors syntax errors and logical errors. When the rules or syntax of the language are not followed then syntax errors occurred and it is displayed after compilation.

When the logic of a program is wrong then logical errors occurred and it is not displayed after compilation but it is displayed in the execution and testing phase.

Question 5.
Write an algorithm to input the scores obtained in three unit tests and find the average score.

OR

Explain the flowchart and predict the output.
Answer:

• Step 1: Start
• Step 2: Read S1, S2, S3
• Step 3: avg = S1 + S2 + S3/3
• Step 4: Print avg
• Step 5: Stop

OR

This flowchart is used to print the numbers as 1,2, 3, ………, 10.

Question 6.
Differentiate between top down design and bottom up design in problem sloving.
Answer:
Bottom up design:
Here also larger programs are divided into smaller ones and the smaller ones are again subdivided until the lowest level of detail has been reached. We start solving from the lowest module onwards. This approach is called Bottom up design.

Question 7.
Answer any one question from 5 (a) and 5 (b).
1. Draw a flowchart for the following algorithm.

• Step 1: Start
• Step 2: Input N
• Step 3: S = 0, K = 1
• Step 4: S = S + K
• Step 5: K = K + 1
• Step 6: If K < = N Then Go to Step 4
• Step 7: Print S
• Step 8: Stop

OR

2. Name the two stages in programming where debugging process is involved. What kinds of errors are removed in each of these stages?
Answer:
1.

2. The two stages are compile time and run time. In the debugging process can remove syntax error, logical error and runtime error.

Question 8.
Answer any one question from 7(1) and 7(2)
1. Observe the following portion of a flowchart. Fill in the blank symbols with proper instructions to get 321 as the output.

2. The following flowchart can be used to print the numbers from 1 to 100. Identify another problem that can be solved using this flowchart and write the required instructions in the symbols.

Answer:
1.

2. The following flowchart can be used to store another problem such as used to print odd numbers lessthan 200.

Question 9.
Write an algorithm to print the numbers upto 100 in reverse order, That is the output should be as 100, 99, 98, 97, …………., 1

OR

Draw a flow chart to check whether the given number is positive, negative or zero.
Answer:

• Step 1: Start
• Step 2: Set i ← 100
• Step 3: if i <= 0 then go to step 6
• Step 4: Print i
• Step 5: Set i ← i — 1 go to step 3
• Step 6: Stop

OR

Plus One Principles of Programming and Problem Solving Three Mark Questions and Answers

Question 1.
When you try to execute a program, there are chances of errors at various stages, Mention the types of errors and explain.
Answer:
1. Syntax error,
eg: 5 = x

2. Logic error:
If the programmer maks any logical mistakes, it is known as logical error.
eg: To find the sum of values A and B and store it in a variable C you have to write C = A + B. Instead of this if you write C = A × B, it is called logic error.

3. Runtime error:
An error occured at run time due to inappropriate data.
eg: To calculate A/B a person gives zero to B. There is an error called division by zero error during run time.

Question 2.
Following is a flow chart to find and display the largest among three numbers. Some steps are missing in the flowchart. Redraw the flow chart by adding necessary steps and specify its purpose. How can this flow chart be modified without using a fourth variable?
Answer:

Question 3.
A flow chart is given below.

1. What will be the output of the above flow chart?
2. How can you modify the above flow chart to display the even numbers upto 20, starting from 2.

Answer:
1. 1, 2, 3, 4, 5

2.

Question 4.
Write an algorithm to check whether the given number is even or odd.
Answer:

• Step 1: Start
• Step 2: Read a number to N
• Step 3: Divide the number by 2 and store the remainder in R.
• Step 4: If R = O Then go to Step 6
• Step 5: Print “N is odd” go to step 7
• Step 6: Print “N is even”
• Step 7: Stop

Question 5.
Write an algorithm to find the largest of 2 numbers?
Answer:

• Step 1: Start
• Step 2: Input the values of A, B Compare A and B.
• Step 3: If A > B then go to step 5
• Step 4: Print “B is largest” go to Step 6
• Step 5: Print “A is largest”
• Step 6: Stop

Question 6.
Write an algorithm to find the sum of n natural numbers and average?
Answer:

• step 1: Start
• Step 2: Set i ←1, S 0
• Step 3: Read a number and set to n
• Step 4: Computer i and n if i > n then go to step 7.
• Step 5: Set S ← S + i
• Step 6: i ← i + 1 go to step 4
• Step 7: avg ← S/n
• Step 8: Print “Sum = S and average = avg”
• Step 9: Stop

Question 7.
Write an algorithm to find the largest of 3 numbers.
Answer:

• Step 1: Start
• Step 2: Read 3 numbers and store in A, B, C
• Step 3: Compare A and B. lf A > Bthengotostep 6
• Step 4: Compare B and C if C > B then go to step 8
• Step 5: print “B is largest” go to step 9
• Step 6: Compare A and C if C > A then go to step 8
• Step 7: Print”A is largest” go to step 9
• Step 8: Print “C is largest”
• Step 9: Stop

Question 8.
Write an algorithm to calculate the simple interest (I =P × N × R/100)
Answer:

• Step 1: Start
• Step 2: Read 3 values for P, N, R
• Step 3: Calculate I ← P × N × R/100
• Step 4: Print “The simple interest = l”
• Step 5: Stop

Question 9.
Write an algorithm to calculate the compound interest (C.l = P × (1 + r/100)n – P)
Answer:

• Step 1: Start
• Step 2: Read 3 number for p, n, r
• Step 3: Calculate C.I = p × (1 + r/100)n – p
• Step 4: Print “The compound Interest = C.l”
• Step 5: Stop

Question 10.
Write an algorithm to find the cube of first n natural numbers (eg: 1, 8, 27, …., n3)
Answer:

• Step 1: Star
• Step 2: Set i ← 1
• Step 3: Read a number and store in n
• Step 4: Compare i and n if i > n then go to step 7
• Step 5: Print i × i × i
• Step 6: i ← i + 1 go to step 4
• Step 7: Stop

Question 11.
Write an algorithm to read a number and find its factorial (n ! = n × (n – 1) × (n – 2) ×………..3 × 2 × 1)
Answer:

• Step 1: Start
• Step 2: Fact ← 1
• Step 3: Read a number and store in n
• Step 4: If n = 0 then go to step 7
• Step 5: Fact ← Fact × n
• Step 6: n ← n – 1 go to step 4
• Step 7: Print “Factorial is fact”
• Step 8: Stop

Question 12.
Draw a flow chart to find the sum of n natural numbers and average.
Answer:

Question 13.
Draw a flow chart to find the largest of 3 numbers.
Answer:

Question 14.
Draw a flow chart to find the largest of 2 numbers.
Answer:

Question 15.
Draw a flow chart to check whether the given number is even or odd.
Answer:

Question 16.
Draw a flow chart to calculate simple interest.
Answer:

Question 17.
Draw a flow chart to calculate compound interest.
Answer:

Question 18.
Draw a flow chart to find the cube of n natural numbers.
Answer:

Question 19.
Draw a flow chart to read a number and find its factorial.
Answer:

Question 20.
Mr. Vimal wants to represent a problem by using a flowchart, which symbols are used for this. Explain.
Answer:
Flow chart symbols are explained below
1. Terminal (Oval)

It is used to indicate the beginning and ending of a problem

2. Input/Output (parallelogram)

It is used to take input or print output.

3. Processing (Rectangle)

It is used to represent processing. That means to represent arithmetic operation such as addition, subtraction, multiplication.

4. Decision (Rhombus)

It is used to represent decision making. One exit path will be executed at a time.

5. Flowlines (Arrows)

It is used to represent the flow of operation

6. Connector

These symbols will help us to complete the flow chart, which is not fit in a single page. A connector symbol is represented by a circle and a letter or digit is placed within the circle to indicate the link.

Question 21.
Jeena uses an algorithm to represent a problem while Neena uses flowchart which is better? Justify your answer?
Answer:
Flowchart is better. The advantages of flow chart is given below.
1. Better communication:
A flow chart is a pictorial representation while an algorithm is a step by step procedure to solve a program. A programmer can easily explain the program logic using a flow chart.

2. Effective analysis:
The program can be analyzed effectively through the flow chart.

3. Effective synthesis:
If a problem is big it can be divided into small modules and the solution for each module is represented in flowchart separately and can be joined together final system design.

4. Proper program documentation:
A flow chart will help to create a document that will help the company in the absence of a programmer.

5. Efficient coding:
With the help of a flowchart it is easy to write program by using a computer language.

Question 22.
A flow chart is a better method to represent a program. But it has some limitation what are they?
Answer:
The limitations are given below

• To draw a flowchart, it is time consuming and laborious work.
• If any change or modification in the logic we may have to redraw a new flow chart.
• No standards to determine how much detail can include in a flow chart.

Question 23.

1. Problem identification	a. Flowchart
2. Steps to obtain. the solution	b. Syntax Error
3. Coding	c. Runtime Error
4. Translation	d. COBOL
5. Debugging	e. X-ray
6. Execution & Testing	f. Compiler

Answer:
1 – e
2 – a
3 – d
4 – f
5 – b
6 – c

Question 24.
Alvis executes an error free program but he got an error. Explain different types of error in detail.
Answer:
There are two types of errors in a program before execution and testing phase.They are syntax error and logical error. When the programmer violates the rules or syntax of the programming language then the syntax error occurred.
eg: It involves incorrect punctuation.

Keywords are used for other purposes, violates the structure etc,… It detects the compiler and displays an error message that include the line number and give a clue of the nature of the error, When the programmer makes any mistakes in the logic, that types of errors are called logical error. It does not detect by the compiler but we will get a wrong output.

The program must be tested to check whether it is error free or not. The program must be tested by giving input test data and check whether it is right or wring with the known results. The third type of errors are Runtime errors.

This may be due to the in appropriate data while execution. For example consider B/C. If the end user gives a value zero for c, the execution will be interrupted because division by zero is not possible. These situation must be anticipated and must be handled.

Question 25.
The following are the phases in programming. The order is wrong rearrange them in correct order.

1. Debugging
2. Coding
3. Derive the steps to obtain the solution
4. Documentation
5. Translation
6. Problem identification
7. Execution and testing

Answer:
The correct order is given below.

1. Problem identification
2. Derive the steps to obtain the solution
3. Coding
4. Translation
5. Debugging
6. Execution and testing
7. Documentation

Question 26.
Draw a flow chart to input ten different numbers and find their average.
Answer:

Question 27.
Draw the flow chart to find the sum of first N natural numbers.
Answer:

Question 28.
Make a flow chart using the given labelled symbols, for finding the sum of all even numbers upto ‘N’

OR

Write an algorithm to accept an integer number and print the factors of it.

Answer:
Draw flowchart in any of the following order
e, c, d, f, i, h, a, b, g
e, d, c, f, i, h, a, b, g
e, c, d, a, f, i, h, b, g
e, d, c, a, f, i, h, b, g

• Step 1: Start
• Step 2: Input n
• Step 3: i = I
• Step 4: if i <= n/2 then repeat step 5 & 6
• step 5: if n % i == 0 print i
• Step 6: i = i + I
• Step 7: Stop

Question 29.
List the two approaches followed in problem solving or programming. How do they differ?
Answer:
Approaches in problem solving:
(a) Top down design:
Larger programs are divided into smaller ones and solve each tasks by performing simpler activities. This concept is known as top down design in problem solving

(b) Bottom up design:
Here also larger programs are divided into smaller ones and the smaller ones are again subdivided until the lowest level of detail has been reached. We start solving from the lowest module onwards. This approach is called Bottom up design.

Phases in Programming
1. Problem identification:
This is the first phase in programming. The problem must be identified then only it can be solved, for this we may have to answer some questions.

During this phase we have to identify the data, its type, quantity and formula to be used as well as what activities are involved to get the desired out put is also identified for example if you are suffering from stomach ache and consult a Doctor.

To diagnose the disease the Doctor may ask you some question regarding the diet, duration of pain, previous occurrences etc, and examine some parts of your body by using stethoscope X-ray, scanning etc.

2. Deriving the steps to obtain the solution. There are two methods, Algorithm and flowchart, are used for this.
(a) Algorithm:
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.

(b) Flowchart:
The pictorial or graphical representation of an algorithm is called flowchart.

Question 30.
Write an algorithm to find the sum of the squres of the digits of a number. (For example, if 235 is the input, the output should be 2² + 3² + 5² = 38)
Answer:

• Step 1: Start
• Step 2: Set S = O
• Step 3: Read N
• Step 4: if N > O repeat step 5, 6 and 7
• Step 5: Find the remainder. That is rem = N % 10
• Step 6 : S = S + rem × rem
• Step 7 : N = N/10
• Step 8 : Print S
• Step 9 : Stop

Question 31.
Consider the following algorithm and answer the following questions:

• Step 1: Start
• Step 2 : N = 2, S = 0
• Step 3: Repeat Step 4, Step 5 while N <= 10
• Step 4: S = S + N
• Step 5: N = N + 2
• Step 6: Print S
• Step 7: Stop

1. Predict the output of the above algorithm.
2. Draw a flowchart for the above algorithm

Answer:
1. The output is 30

2.

Question 32.
“It is better to give proper documentation within the program”. Give a reason.
Answer:
It is the last phase in programming. A computerized system must be documented properly and it is an ongoing process that starts in the first phase and continues till its implementation. It is helpful for the modification of the program later.

Plus One Principles of Programming and Problem Solving Five Mark Questions and Answers

Question 1.
Mr. Arun wants to develop a program to computerize the functions of supermarket. Explain different phases he has to undergo is detail.

OR

Briefly explain different phases in programming.
Answer:
The different phases in programming is given below:
1. Problem identification:
This is the first phase in programming. The problem must be identified then only it can be solved, for this we may have to answer some questions.
During this phase we have to identify the data, its type, quantity and formula to be used as well as what activities are involved to get the desired output is also identified for example if you are suffering from stomach ache and consult a Doctor.

To diagnose the disease the doctor may ask you some question regarding the diet, duration of pain, previous occurrences, etc and examine some parts of your body by using stethoscope X-ray, scanning, etc.

2. Deriving the steps to obtain the solution.
There are two methods, Algorithm and flowchart, are used for this.
(a) Algorithm:
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 Ibu Musaa Al-Khowarizmi. The last part of his name Al-Khowafizmi was corrected to algorithm.

(b) Flowchart:
The pictorial or graphical representation of an algorithm is called flowchart.

3. Coding:
The dummy codes (algorithm)or flow chart is converted into program by using a computer language such s Cobol, Pascal, C++, VB, Java, etc.

4. Translation:
The computer only knows machine language. It does not know HLL, but the human beings HLL is very easy to write programs. Therefore a translation is needed to convert a program written in HLL into machine code (object code).

During this step, the syntax errors of the program will be displayed. These errors are to be corrected and this process will be continued till we get “No errors” message. Then it is ready for execution.

5. Debugging:
The program errors are called ‘bugs’ and the process of detecting and correcting errors is called debugging. In general there are two types of errors syntax errors and logical errors. When the rules or syntax of the language are not followed then syntax errors occurred and it is displayed after compilation.

When the logic of a program is wrong then logical errors occurred and it is not displayed after compilation but it is displayed in the execution and testing phase.

6. Execution and Testing:
In this phase the program will be executed and give test data for testing the purpose of this is to determine whether the result produced by the program is correct or not. There is a chance of another type of error, Run time error, this may be due to inappropriate data.

7. Documentation:
It is the last phase in programming. A computerized system must be documented properly and it is an ongoing process that starts in the first phase and continues till its implementation. It is helpful for the modification of the program later.

Question 2.
Briefly explain the characteristic of an algorithm.
Answer:
The following are the characteristics of an algorithm

1. It must starts with ‘start’ statement and ends with ‘stop’ statement.
2. it should contains instructions to accept input and these are processed by the subsequent instructions.
3. Each and every instruction should be precise and must be clear. The instructions must be possible to carry out, for example consider the examples the instruction “Go to market” is precise and possible to carried out but the instruction “Go to hell” is also precise and can not be carried out.
4. Each instruction must be carried out in finite time by a person with paper and pencil.
5. The number of repetition of instructions must be finite.
6. The desired results must be obtained after the algorithm ends.

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

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Plus One Data Representation and Boolean Algebra One Mark Questions and Answers

Question 1.
___________ is a collection of unorganized fact.
Answer:
Data

Question 2.
Data can be organized into useful ____________
Answer:
Information

Question 3.
___________ is used to help people to make decision.
Answer:
Information

Question 4.
Processing is a series of actions or operations that convert inputs into __________
Answer:
Output

Question 5.
The act of applying information in a particular context or situation is called ____________
Answer:
Knowledge

Question 6.
What do you mean by data processing?
Answer:
Data processing is defined as a series of actions or operations that converts data into useful information.

Question 7.
Odd man out and justify your answer.
(a) Adeline
(b) 12
(3) 17
(d) Adeline aged 17 years is in class 12.
Answer:
(d) Adeline aged 17 years is in class 12. This is information. The others are data.

Question 8.
Raw facts and figures are known as _______
Answer:
data

Question 9.
Processed data is known as _______
Answer:
Information

Question 10.
Which of the following helps us to take decisions?
(a) data
(b) information
(c) Knowledge
(d) intelligence
Answer:
(b) information

Question 11.
Manipulation of data to get information is known as ___________
Answer:
Data processing

Question 12.
Arrange the following in proper order
Process, Output, Storage, Distribution, Data Capture, Input.
Answer:

1. Data Capture
2. Input
3. Storage
4. Process
5. Output
6. Distribution

Question 13.
Pick the odd one out and give reason:
(a) Calculation
(b) Storage
(c) Comparison
(d) Categorization
Answer:
(b) Storage. It is one of the data processing stage the others are various operations in the stage Process

Question 14.
Information may act as data. State true or False.
Answer:
False

Question 15.
Complete the Series.

1. (101)₂, (111)₂, (1001)₂, ……….
2. (1011)₂, (1110)₂, (10001)₂, ………

Answer:

1. 1011, 1101
2. 10101, 10111

Question 16.
What are the two basic types of data which are stored and processed by computers?
Answer:
Characters and number

Question 17.
The number of numerals or symbols used in a number system is its _______________
Answer:
Base

Question 18.
The base of decimal number system is ________
Answer:
10

Question 19.
MSD is ________
Answer:
Most Significant Digit

Question 20.
LSD is _________
Answer:
Least Significant Digit

Question 21.
Consider the number 627. Its MSD is _________
Answer:
6

Question 22.
Consider the number 23.87. Its LSD is __________
Answer:
7

Question 23.
The base of Binary number system is ___________
Answer:
2

Question 24.
What are the symbols used in Binary number system?
Answer:
0 and 1

Question 25.
Complete the following series,
(101)₂, (111)₂, (1001)₂
Answer:
1011, 1101

Question 26.
State True or False. In Binary, the unit bit changes either from 0 to 1 or 1 to 0 with each count.
Answer:
True

Question 27.
The base of octal number system is ________
Answer:
8

Question 28.
Consider the octal number given below and fill in the blanks.
0, 1, 2, 3, 4, 5, 6, 7, __
Answer:
10

Question 29.
The base of Hexadecimal number system is ________
Answer:
16

Question 30.
State True or False.
In Positional number system, each position has a weightage.
Answer:
True

Question 31.
In addition to digits what are the letters used in Hexadecimal number system.
Answer:
A(10), B(11), C(12), D(13), E(14), F(15)

Question 32.
Convert (1110.01011)₂ to decimal.
Answer:
1110.01011 = 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 + 0 × 2 – 1 + 1 × 2 – 2 + 0 × 2 – 3 + 1 × 2 – 4 + 1 × 2 – 5
= 8 + 4 + 2 + 0 + 0 + 0.25 + 0 + 0.0625 + 0.03125
= (14.34375)10

Question 33.
1 KB is bytes.
(a) 2⁵
(b) 2¹⁰
(c) 2¹⁵
(d) 2²⁰
Answer:
(b) 2¹⁰

Question 34.
The base of hexadecimal number system is ________.
Answer:
16

Question 35.
A computer has no _________
(a) Memory
(b) l/o device .
(c) CPU
(d) IQ
Answer:
(d) IQ

Question 36.
Pick the odd man out.
(AND, OR, NAND, NOT)
Answer:
NOT

Question 37.
Select the complement of X + YZ.
(a) \(\bar{x}+\bar{y}+\bar{z}\)
(b) \(\bar{x} .\bar{y}+\bar{z}\)
(c) \(\bar{x} \cdot(\bar{y}+\bar{z})\)
(d) \(\bar{x}+\bar{y} \cdot \bar{z}\)
Answer:
(c) \(\bar{x} \cdot(\bar{y}+\bar{z})\)

Question 38.
Select the expression for absorption law.
(a) a + a = a
(b) 1 + a = 1
(c) o . a = 0
(d) a + a . b = a
Answer:
(a) a + a . b = a

Question 39.
What is the characteristic of logical expression?
Answer:
Logical expressions yield either true or false values

Question 40.
Name the table used to define the results of Boolean operations.
Answer:
Truth Table

Question 41.
According to ________ law, \(\bar{x}+\bar{y}=\overline{x y}\) and \(\overline{x y}=\bar{x}+\bar{y}\)
Answer:
De Morgan’s law

Question 42.
A NOR gate is ON only when all its inputs are
(a) ON
(b) Positive
(c) High
(d) OFF
Answer:
(d) OFF

Question 43.
The only function of a NOT gate is __________
Answer:
Invert an output signal

Question 44.
NOT gate is also known as _________
Answer:
Inverter

Question 45.
What is the relation between the following statements.
x + 0 = x and x . 1 = x
Answer:
One is the dual of the other expression.

Question 46.
The algebra used to solve problems in digital systems is called __________
Answer:
Boolean Algebra

Question 47.
Pick the one which is not a Basic Gate.
(AND, OR, XOR, NOT)
Answer:
XOR

Question 48.
Select the universal gates from the list. (NAND, NOR, NOT, XOR)
Answer:
NAND, NOR

Question 49.
Which is the final stage in data processing?
Answer:
Distribution of information is the final stage in data processing

Question 50.
Fill up the missing digit.
(41)₈ = ( )₁₆
Answer:

• Step 1: Divide the number into one each and write down the 3 bits equivalent.
• Step 2: Then divide the number into group of 4 bits starting from the right then write its equivalent hexa decimal.

Answer:
So the answer is 21.

Question 51.
Real numbers can be represented in memory by using __________
Answer:
Exponent and Mantissa

Question 52.
Consider the number 0.53421 x 10^-8 Write down the mantissa and exponent.
Answer:
Mantissa: 0.53421
Exponent: -8

Question 53.
Characters can be represented in memory by using _________
Answer:
ASCII Code

Question 54.
ASCII Code of A’ is __________
Answer:
(100 0001)₂ = 65

Question 55.
ASCII Code of’a’ is __________
Answer:
(110 0001)₂ = 97

Question 56.
Define the term ‘bit’?
Answer:
A bit stands for Binary digit. That means either 0 or 1.

Question 57.
Find MSD in the decimal number 7854.25
Answer:
Because it has the most weight

Question 58.
ASCII stands for __________.
Answer:
American Standard Code for Information Interchange

Question 59.
List any two image file formats.
Answer:
BMP, GIF

Question 60.
Name the operator which performs logical multiplication.
Answer:
AND

Question 61.
Name a gate which is ON when all its inputs are OFF .
Answer:
NAND or NOR

Question 62.
Specify the laws applied in the following cases.

1. a (b + c) = ab + ac
2. (a + b) + c = a + (b + c)

Answer:

1. Distribution law
2. Associative law

Question 63.
Pick the correct Boolean expression from the following.
(a) \(A +\bar{A}=163.\)
(b) \(\text { A. } \bar{A}=1\)
(c) \(A \cdot \overline{A B}=A + B\)
(d) A + AB = A
Answer:
(a) & (d)

Question 64.
1’s complement of the binary number 110111 is _________
Answer:
Insert 2 zeroes in the left hand side to make the binary number in the 8 bit form 00110111
To find the 1’s complement, change all zeroes to one and all ones to zero. Hence the answer is 11001000

Plus One Data Representation and Boolean Algebra Two Mark Questions and Answers

Question 1.
Why do we store information?
Answer:
Normally large volume of data has to be given to the computer for processing so the data entry may be taken more days, hence we have to store the data. After processing these stored data, we will get information as a result that must be stored in the computer for future references.

Question 2.
What is source document.
Answer:
Acquiring the required data from all the sources for the data processing and by using this data design a document, that contains all relevant data in proper order and format. This document is called source document.

Question 3.
Briefly explain data, information and processing with real life example.
Answer:
Consider the process of making coffee. Here data is the ingredients – water, sugar, milk and coffee powder
Information is the final product i.e. Coffee Processing is the series of steps to convert the ingredients into final product, Coffee. That is mix the water,sugar and milk and boil it. Finally pour the coffee powder.

Question 4.
ASCII is used to represent characters in memory. Is it sufficient to represent all characters used in the written languages of the world? Propose a solution. Justify.
Answer:
No It is not sufficient to represent all characters used in the written languages of the world because it is a 7 bit code so it can represent 2⁷ = 128 possible codes. To represent all the characters Unicode is used because it uses 4 bytes, so it can represent 232 possible codes.

Question 5.
The numbers in column A have an equivalent number in another number system of column B.
Find the exact matvh

A	B
(12)8	(1110)2
F16	25
(19)16	10
(11)8	(13)16
	(17)8
	9

Answer:

A	B
12	10
F	(17)8
(19)16	25
(11)8	9

Question 6.

1. Name various number systems commonly used in computers.
2. Include each of the following numbers into all possible number systems
   123 569 1101

Answer:

1. The number system are binary, octal, decimal and hexa decimal
2. All possible number systems are
   • 123 Octal, decimal and hexa decimal
   • 569 Decimal, hexa decimal
   • 1101 Binary, Octal, Decimal, Hexa decimal

Question 7.
Fill up the missing digit. (Score 2)
If (220)_a = (90)_b then (451)_a = (  )₁₀
Answer:
It contains 2 & 9, so a and b 2, b 8. The values of a can be 8 or 10. The values of b can be 10 or 16. L.H.S > R.H.S. a < b and a b also.
The possible values of a and b are given below.

Question 8.
Convert (106)₁₀ = (  )₂?
Answer:

Question 9.
Convert (106)₁₀ = (  )₈?
Answer:

Question 10.
(106)₁₀ = (  )₁₆?
Answer:

Question 11.
Convert (55.625)₁₀ = (  )₂?
Answer:
First convert 55, for this do the following

Write down the remainders from bottom to top.
(55)₁₀ = (110111 )₂
Next convert 0.625, for this do the following.

Write down the remainder from top to bottom. So the answer is
(55.625)₁₀ = (110111.101)₂

Question 12.
Convert (55.140625)₁₀ = (  )₈?
Answer:
First convert 55, for this do the following.

Write down the remainders from bottom to top.
(55)₁₀ = (67)₈
Next convert 0.140625, for this do the following.

Write down the remainders from top to bottom. So the answer is
(55.140625)₁₀ = (67.11 )₈

Question 13.
Convert (55.515625)₁₀ = (  )₁₆
Answer:
First convert 55, for this do the following.

Write down the remainders from bottom to top.
ie. (55)₁₀ = (37)₁₆
Next convert .515625

So the answer is
(55.515625)₁₀ = (37.84)₁₆

Question 14.
Convert (101.101)₂ = ( )₁₀?
Answer:
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

Question 15.
Convert (71.24)₈ = (  )₁₀?
Answer:
71.24 = 7 × 8¹ + 1 × 8⁰ + 2 × 8^-1 + 4 × 8⁼²
= 56 + 1 + 2/8 + 4/82
= 57 + 0.25 + 0.0625 (71.24)8
(71.24)₈ = (57.3125)₁₀

Question 16.
Convert (AB.88)₁₆ = (  )₁₀?
Answer:

= 160 + 11 + 0.5 + 0.03125
(AB.88)₁₆ = (171.53125)₁₀

Question 17.
Convert (1011)₂ = (  )₈?
Answer:
Step I: First divide the number into groups of 3 bits starting from the right side and insert necessary zeroes in the left side.
0 0 1 | 0 1 1

Step II: Next write down the octal equivalent.

So the answer is (1011)₂ = (13)₈.

Question 18.
Convert (110100)₂ = (  )₁₆
Answer:

• Step I: First divide the number into groups of 4 bits starting from the right side and insert necessary zeroes in the left side.
• Step II: Next write down the hexadecimal equivalent.

So the answer is (110100)₂ = (34)₁₆.

Question 19.
(72)₈ = ( )₂?
Answer:
Write down the 3 bits equivalent of each digit.

So the answer is (72)₈ = (111010)₂.

Question 20.
Convert (AO)₁₆ = (  )₂ ?
Answer:
Write down the 4 bits equivalent of each digit

So the answer is (AO)₁₆ = (1010 0000)₂.

Question 21.
Convert (67)₈ = (  )₁₆?
Answer:
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

Question 22.
Convert (A1)₁₆ = (  )₈?
Answer:
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.

Question 23.
What is the use of the ASCII Code?
Answer:
ASCII means American Standard Code for Information Interchange. It is a 7 bit code. Each and every character on the keyboard is represented in memory by using ASCII Code.
eg: A’s ASCII Code is 65 (1000001), a’s ASCII Code is 97 (1100001)

Question 24.
Pick invalid numbers from the following.

1. (10101)₈
2. (123)₄
3. (768)₈
4. (ABC)₁₆

Answer:

1. (10101)₈ – Valid
2. (123)₄ – Valid
3. (768)₈ – Invalid. Octal number system does not contain the symbol 8
4. (ABC)₁₆ – Valid

Question 25.
Convert the decimal number 31 to binary.
Answer:

(31)₁₀ = (11111)₂

Question 26.
Find decimal equivalent of (10001 )₂
Answer:

= 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 16 + 0 + 0 + 0 + 1
= (17)₁₀

Question 27.
If (X)₈ =(101011 )₂ then find X.
Answer:
Divide the binary number into groups of 3 bits and write down the corresponding octal equivalent.

X = 53

Question 28.
Fill the blanks:
(a) (………..)₂ = (AB)₁₆
(b) (——D—–)₁₆ = (1010 1000)₂
(c) 0.25₁₀ = (—–)₂

Answer:
Write down the 4bit equivalent of each digit
(a) (………..)₂ = (AB)₁₆

= (10101011)₂
(b) (——D—–)₁₆ = (1010 1000)₂
(A D 8)₁₆ =(1010 1101 1000)₂

(c) 0.25₁₀ = (—–)₂

0.25₁₀ = (0.01)₂

Question 29.
Which is the MSB of representation of -80 in SMR?
Answer:
It is 1 because In SMR if the number is negative then the MSB is 1.

Question 30.
Write 28.756 in Mantissa exponent form.
Answer:
28.756 = .28756 × 100
= .28756 × 102
= .28756 E + 2

Question 31.
Represent -60 in 1’s complement form.
Answer:
+60 = 00111100
Change all 1 to 0 and all 0 to 1 to get the 1’s complement.
-60 is in 1’s complement is 11000011

Question 32.
Define Unicode.
Answer:
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.

Question 33.
Substract 1111 from 10101 by using 2’s complement method.
Answer:
To subtract a number from another number find the 2’s complement of the subtrahend and add it with the minuend. Here the subtrahend is 1111 and minuend is 5 bits. So insert a zero. So subtrahend is 01111. First take the 1’s complement of subtrahend and add 1 to it

Here is a carry. So ignore the carry and the result is +ve.
So the answer is 110

Question 34.
You purchased a soap worth Rs. (10010)₂ and you gave Rs. (10100)₂ and how much rupees will you get back in binary.
Answer:
Substract (10010)₂ from (10100)₂

You will get rupees (10)₂

Question 35.
Draw the logic circuit diagram for the following Boolean expression.
Answer:

Question 36.
Simplify the expression using basic postulates and laws of Boolean algebra.

1. \(\bar{x}+x \cdot \bar{y}\)
2. \(x(y+y . z)+y(\bar{x}+x z)\)

Answer:

1. \(\bar{x}+\bar{y}\)
2. y

Question 37.
Show \(A(\bar{B}+C)\) using NOR gates only.
Answer:

Question 38.
The following statement Demorgan’s theorem of Boolean algebra. Identify and state ‘Break the line, change the sign’.
Answer:
Demorgan’s theorems,
Demorgan’s first theorem,
\(\overline{x + y}\) = \(\bar{x} . \bar{y}\)
Demorgan’s second theorem,
\(\overline{x – y}\) = \(\bar{x} + \bar{y}\)

Question 39.
Prove algebraically that
\(x \cdot y + x \cdot \bar{y} \cdot z\) = x . y + x . z
Answer:
\(x \cdot y + x \cdot \bar{y} \cdot z\) = \(x(y+\bar{y} . z)\)
= x . (y + z) = x . y + x . z
Hence proved.

Question 40.
State which of the following statements are logical statements.
(a) AND is a logical operator
(b) ADD 3 to y
(c) Go to class
(d) Sun rises in the west.
(e) Why are you so late?
Answer:
(a) and (d) are logical statements because these statements have a truth value which may be true or false.

Question 41.
Express the integer number -39 in sign and magnitude represnetation.
Answer:
First find the binary equivalent of 39 for this do the following

In sign and magnitude representation -39 in 8 bit form is (10100111)₂.

Question 42.

1. Which logic gate does the Boolean expression \(\overline{\mathrm{AB}}\) represent?
2. Some NAND gates are given. How can we construct AND gate, OR gate and NOT gate using them?

Answer:
1. NAND

2. AND gate

Question 43.
Perform the following number conversions.

1. (110111011.11011)₂ = (….)₈
2. (128.25)₁₀= (…..)₈

Answer:
1. 110111011.11011
Step 1: Insert a zero in the right side of the above number and divide the number into groups of 3 bits as follows
110 111 011 . 110 110

Step 2: Write down the corresponding 3 bit binary equivalent of each group
6 7 3 .6 6
Hence the result is (673.66)₈

2. It consists of 2 steps.
Step 1: First convert 128 into octal number for this do the following

Write down the remainders from bottom to top.
(128)₁₀ = (200)₈

Step 2: Then convert .25 into octal number for this do the following

(0.25)₁₀ = (0.2)₈.
Combine the above two will be the result.
(128.25)₁₀= (200.2)₈

Question 44.
Represent -38 in 2’s complement form.
Answer:
+38 = 00100110
First take the 1 ’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of -38 is (11011010)₈.

Plus One Data Representation and Boolean Algebra Three Mark Questions and Answers

Question 1.
Differentiate manual data processing and electronic data processing?
Answer:
In manual data processing human beings are the processors. Our eyes and ears are input devices. We get data either from a printed paper, that can be read using our eyes or heard with ears. Our brain is the processor and it can process the data, and reach in a conclusion known as result. Our mouth and hands are output devices.

In electronic data processing the data is processing with the help of a computer. In a super market, key board and hand held scanners are used to input data, the CPU process the data, monitor and printers (Bill) are output devices.

Question 2.
Complete the series.

1. 3248, 3278 ,3328, …., ….
2. 5678, 5768, 605s, ……, …..

Answer:
1. (324)₈ = 3 × 82 + 2 × 81 + 4 × 80
= 3 × 64 + 2 × 8 + 4 × 1
= 192 + 16 + 4 = (212)10

(327)₈ = 3 × 82 + 2 × 81 + 7 × 80
= 192 + 16 + 7 = (215)₁₀

(332)₈= 3 × 82 + 3 × 81 +2 × 80
= 192 + 24 + 2 = (218)₁₀
So the missing terms are (221)₁₀, (224)₁₀ we have to convert this into octal.

So the missing terms are (335)₈, (340)₈

2. (567)₈ = 5 x 82 + 6 x 81 + 7 x 80
= 5 x 64 + 6 x 8 + 7 x 1
= 320 + 48 + 7 = (375)₁₀

(576)₈ = 5 x 82 + 7 x 81 + 6 x 80
= 320 + 56 + 6 = (382)₁₀

(605)₈ = 6 x 82 + 0 x 81 + 5 x 80
= 6 + 64 + 0 + 5
= 384 + 0 + 5 = (389)₁₀
So the missing terms are (396)₁₀, (403)₁₀ we have to convert this into octal.

So the missing terms are (614)₈, (623)₈

Question 3.
Fill up the missing digits.

1. (4……)₈ = (……110)₂
2. (…….7……)₈ = (100…….110)₂

Consider the following

Answer:
1. 4…….100 and 110……….6
So (46)₈ = (100 110)₂

2. 100…….4
7……111
110………6
So (476)₈ = (100 111 110)₂

Question 4.
Fill up the missing numbers.

1. (A…….)₁₆ = (……..1001)₂
2. (…….B…….)₁₆ = (1000………1111)₂

Consider the following:


Answer:
1. A……..1010
1001………9
So (A9)₁₆ = (1010 1001)₂

2. B…….1011
1000………8
1111………F
So (8BF)₁₆ = (1000 1011 1111)₂

Question 5.
Complete the Series.

1. 6ADD, 6ADF, 6AE1, ……., …….
2. 14A9, 14AF, 14B5, …….., ……

Answer:
1. Consider the sequence
6ADD, 6ADF, 6AE1, ………….
Here the ‘numbers’ are
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11,——–
The difference between 6ADD & 6ADF is 2

Similarly 6ADF & 6AE1 is 2

So Add 2 to 6AE1 we will ge 6AE3 Then add 2 to 6AE3 we will get 6AE5 Therefore the missing terms 6AE3, 6AE5

2. Consider the sequence.
14A9, 14AF, 14B5, ———
The difference between 14A9 and 14AF is 6.
The normal sequence is

The difference between 14AF and 14B5 is also 6.

The normal sequence is

Similarly the next 6 terms in the sequence are given below.

Similarly the next 6 terms are

So the missing terms are 14BB and 14C1

Question 6.
Find the octal numbers corresponding to the following numbers using shorthand method.

1. (ADD)₁₆
2. (DEAD)₁₆

Answer:
1. (ADD)₁₆
Step 1: Write down the 4 bit binary equivalent of each digit

Step 2: Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

(ADD)(ADD)₁₆ = (5335)(ADD)₈

2. (DEAD)₁₆
Step 1: Write down the 4 bit binary equivalent of each digit.

Step 2: Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

(DEAD)₁₆ = (157255)₈

Question 7.
If (126)x = (56)y, then find x and y.
Answer:
L.H.S contains 2 & 6, so x ≠ 2
R.H.S contains 5 & 6, so y ≠ 2
L.H.S > R.H.S
So x < y and x ≠ y also The possible values of x and y are given below.

Case I:
Let x = 8 then y = 10

It is grater than (56)₁₀
so when x = 8 then y ≠ 10

case II:
let x = 8 and y = 16

Question 8.
If (102)x = (42)y then (154)x = (  )y.
Answer:
L.H.S contains 2, so x ≠ 2
R.H.S contains 5 & 4, so y ≠ 2
L.H.S > R.H.S
So x < y and x ≠ y also
The possible values of x and y are given below.

case I:
let x = 8 and y = 10

So when x = 8 then y ≠ 10

case II:
let x = 8 and y = 16

So x = 8 and y = 16
then we have fo find the hexadecimal equivalent of (154)₈ For this first convert this into binary thus again convert it into hexadecimal. First write down the 3 bit equivalent of 154.

Then divide this number into groups of 4 bits starting from the right and write down the hexa decimal equivalent.

so the result is (154)₈ = (6C)₁₆

Question 9.
Fill up the missing digit.
If (121)_a = (441)_b then (121)_b = (  )₁₀
Answer:
L.H.S. contains 2, so a ≠ 2
R.H.S. contains 4, so b ≠ 2
L.H.S. < R.H.S. So a > b and a b also.
Hence the values of a can be 10 or 16.
The values of b can be 8 or 10.
The possible values of a and b are given below.

Case I:
Let a = 16 and b = 10
(121 )₁₆ = (289)_10, so b ≠ 10

Case II:
Let a = 16 and b = 8
(121)₁₆ = (289)₁₀
(441)₈ = 4 × 8² + 4 × 8¹ + 1 × 8⁰
= 256 +32 + 1
= (289)₁₀.
So a = 16 and b = 8.
Then (121)₈ = 1 × 82 + 2 × 81 + 1 × 80
= 64 + 16 + 1 = (81)₁₀

Question 10.
Fill up the missing digit. (Score 3)
If (128)_a = (450)_b then (16)_a = (  )₁₀
Answer:
L.H.S. contains 2 & 8, so a 2 and a ≠ 8.
R.H.S. contains 4 and 5, so b ≠ 2.
L.H.S. < R.H.S. so a > b and a ≠ b also.
The possible values of a and b are given below.

Case I:
a = 16 and b = 8
(128)₁₆ = (296)₁₀
(450)₈ = (296)₁₀ So a = 16 and b = 8.
Then (16)₁₆ = 1 × 16 + 6 × 16⁰ = (22)₁₀

Question 11.
Fill up the missing digit.
(3A.6D)₁₆ = (  )₈
Answer:
Step I: Write down the 4 bits equivalent of each digit.

Step II: Divide this number into groups of 3 bit starting from the right side of the left side of the decimal point and starting from the left side of the right side of the decimal point.
So 00/111/010.011/011/010

Step III: Write the octal equivalent of each group. SO we will get. (72.332)₈.
(3A.6D)₁₆ = (72.332)₈

Question 12.
What are the various ways to represent integers in computer?
Answer:
There are three ways to represent integers in computer. They are as follows:

1. Sign 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: 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

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

Question 13.
Write short notes about Unicode (3)
Answer:
It is like ASCII Code. By using ASCII, we can represent limited number of characters. But using Unicode we can represent all of the characters used in the written languages of the world.
eg: Malayalam, Hindi, Sanskrit .

Question 14.
Match the following.

1. (106)10	a. (171.53125)10
2. (71.24)8	b. (6a)16
3. (AB.88)16	c. (20)8
4. (10)16	d. (10000000)2
5. (128)10	e. (10)16
6. (16)10	f. (57.3125)10

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

Question 15.
Find the largest number in the list.

1. (1001)₂
2. (A)₁₆
3. (10)₈
4. (11)₁₀

Answer:
Convert all numbers into decimal
1. (1001)₂ = 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 0 + 0 + 1
= (9)₁₀

2.) (A)₁₆ = (10)₁₀

3. (10)₈ = 1 × 8¹+0 × 8⁰
= (8)₁₀
So the largest number is 4 – (11)₁₀

Question 16.
Subtract 10101 from 1111 by using 2’s complement method.
Answer:
To subtract a number from another number find the 2’s complement of the subtrahend and add it with the minuend. Here subtrahend is 10101 and minuend is 1111 First take the 1’s complement of subtrahend and add 1 to it.
1’s complement of 10101 is 01010 add 1 to it

Here is no carry. So the result, is -ve and take the 2’s complement of 11010 and put a -ve symbol. So 1’s complement of 11010 is 00101 add 1 to this

So the result is -00110

Question 17.
Mr. Geo purchased (10)₂ kg sugar @Rs. (110 10)₂ and (1010)₂ kg Rice @Rs. (10100)₂. So how much rupees he has to pay in decimal.
Answer:
Convert each into decimal number system multiply and sum it up.
(10)₂ = (2)₁₀

(11010)₂ = 1 × 2⁴ + 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2¹
= 16 +8 + 0 +2 + 0
= (26)₁₀

(1010)₂ = 1 × 2³ + 0 × 2² + 1 × 2¹ +0 × 2⁰
= 8 + 0 + 2 + 0
= (10)₂

(10100)₂ = 1 × 2⁴ + 0 × 2³ + 1 × 2² +0 × 2¹ + 0 × 2⁰
= 16 + 0 + 4 + 0 + 0
= (20)₂
therfore 2 × 26 + 10 × 20
= 52 + 200
= 252
So Mr. Geo has to pay Rs. 252/-

Question 18.
Mr. Vimal purchased a pencil @ Rs. (101)2, a pen @ Rs. (1010)2 and a rubber @ Rs. (10)2. So how much rupees he has to pay in decimal.
Answer:
Add 101 + 1010 + 10

then convert (10001)₂ into decimal
(10001)₂ = 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 16 + 0 + 0 + 0 + 1
= (17)₁₀
So Mr. Vimal has to pay Rs. 17/-

Question 19.
Mr. Antony purchased 3 books worth Rs. a total of (1100100)₂. Atlast he returned a book worth Rs. (11001)₂. So how much amount he has to pay for the remaining two books in decimal number sys¬tem.
Answer:

then convert (1001011 )₂ into decimal
(1001011)₂ = 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 1 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰
=64 + 0 + 0 + 8 + 0 + 2 + 1
= (75)₁₀
So he has to pay Rs. 75/-

Question 20.
Mr. Leones brought two products from a super market a total of Rs. (11010010)₂ and he got a dicount of Rs. (1111)₂ So how much he has to pay for this products in decimal number system.
Answer:
Substract (1111)₂ from (11010010)₂

then convert (11000011)₂ into decimal
(11000011)₂ = 1 × 2⁷ + 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰
=128 + 64 + 0 + 0 + 0 + 0 + 2 + 1
= (195)₁₀

Question 21.
A textile showroom sells shirts with a discount of Rs. (110010)₂ on all barads. Mr. Raju wants to buy a shirt worth Rs. (11111000)₂. So after discount how much amount he has to pay in decimal.
Answer:
Substract (110010)₂ from (11 111 000)₂

then convert (11000110)₂ into decimal
(11000110)2 = 1 × 2⁷ + 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 0 × 2⁰
= 128 + 64 + 0 + 0 + 0 + 4 + 2 + 0
= (198)₁₀

Question 22.
Mr. Lijo purchased a product worth Rs. (1110011)₂ and he has to pay VAT @ Rs. (1100)₂. Then calculate the total amount he has to pay in decimal.
Answer:
Add (1110011)₂ and (1100)₂

then convert (1111111)₂ into decimal
(1111111)₂ = 1 × 2⁶ + 1 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰
= 64 + 32 + 16 + 8 + 4 + 2 + 1
= (127)₁₀

Question 23.
By using truth table, prove the following laws of Boolean Algebra.

1. Idempotent law
2. Involution law

Answer:
1. A + A = A
A = A = A

2. (A¹)¹ = A

Question 24.
Consider the logical gate diagram.

1. Find the logical expression for the circuit given.
2. Find the compliment of the logical expression.
3. Draw the circuit diagram representing the compliment.

Answer:
1. \((x+\bar{y}) \cdot z\)

2. \((\bar{x} . y)+\bar{z}\)
3.

Question 25.
Draw the logic circuit diagram for the following Boolean expression.
\(A \cdot(\bar{B} + \bar{C})+\bar{A} \bar{B} \bar{C}\)
Answer:

Question 26.
Consider a bulb with three switches x, y and z. Write the Boolean expression representing the following states.

1. All the switches x, y and z are ON
2. x is ON and y is OFF or Z is OFF
3. Exactly one switch is ON.

Answer:

1. xy . z
2. \(x \bar{y}+\bar{z}\)
3. \(x . \bar{y} . \bar{z}+\bar{x} . y . \bar{z}+\bar{x} . \bar{y} . \bar{z}\)

Question 27.
Match the following.

A	B
i. Idem potent law	a. x + (y + z)=(x + y)+z
ii. Involution law	b. x + xy = x
iii. Complementarity law	c. x + y = y + x
iv. Commutative law	d. xx- 0
v. Absorption law	e. x = x
vi. Associative law	f. x + x = x

Answer:
i – f, ii – e, iii – d, iv – c, v – b, vi – a

Question 28.
Explain the principle of duality.
Answer:
It states that, starting with a Boolean relation, another Boolean relation can be derived by

1. Changing each OR sign (+) to a AND sign (.)
2. Changing each AND sign (.) to an OR sign (+)
3. Replacing each 0 by 1 and each 1 by 0.

The relation derived using the duality principle is called the dual of the original expression,
eg: x + 0 = x is the dual of x . 1 = x

Question 29.
Draw the circuit diagram for \(F=A \bar{B} C+\bar{C} B\) using NAND gate only.
Answer:
\(F=A \bar{B} C+\bar{C} B\)
= (A NAND (NOT B) NAND C) NAND ((NOT C) NAND B)

Question 30.
Draw a logic diagram for the function f = YZ + XZ using NAND gates only.
Answer:
f = YZ + XZ
= (Y NAND Z) NAND (X NAND Z)

Question 31.
How do you make various basic logic gates using NAND gates.
Answer:
1. AND operation using NAND gate,
A.B = (A NAND B) NAND (A NAND B)

2. OR operation using NAND gate,
A + B = (A NAND A) NAND (B NAND B)

3. NOT operation using NAND gate,
NOT A = (A NAND A)

Question 32.
Which of the following Boolean expressions are correct? Write the correct forms of the incorrect ones.

1. A + A¹ = 1
2. A + 0 = A
3. A . 1 = A
4. A . A¹ = 1
5. A + A . B = A
6. A . (A + B) = A
7. A + 1=1
8. \((\overline{\mathrm{A} . \mathrm{B}})=\overline{\mathrm{A}} . \overline{\mathrm{B}}\)
9. A + A¹B = A + B
10. A + A = A
11. A + B . C = (A+B) . (B+C)

Answer:

1. Correct
2. Correct
3. Correct
4. Wrong, A . A¹ = 0
5. Correct
6. Correct
7. Correct
8. Wrong \(\overline{\mathrm{A} . \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
9. Correct
10. Correct
11. Wrong, A + B . C = (A + B) . (A + C)

Question 33.
Prove algebraically that (x + y)’ . (x’ + y’) = x’ . y’
Answer:
LHS = (x + y)’ . (x’ + y’)
= (x’ . y’) . (x’ . y’)
= x’ . y’ . x’ + x’ . y’ . y’
= x’ . y’ + x’ . y’
= x ‘. y’ = RHS
Hence proved.

Question 34.
Give the complement of the following Boolean Expression.

1. (A + B) . (C + D)
2. (P + Q) + (Q + R) . (R + P)
3. (B + D’) . (A + C’)

Answer:
1. ((A+B) . (C+D))1 = (A+B)’ + (C+D)’
= A’ . B’ + C’ . D’

2. ((P+Q) + (Q+R) . (R.P))’ = (P+Q) ‘. ((Q+R) . (R+P))’
= P’ . Q’ . (Q+R)’ + (R+P)’
= P’ . Q’ . (Q’ . R’ + R’ . P’),

3. ((B+D’).(A+C’))’ = (B+D’)’0 + (A+C’)’
= B’ . D” + A’ . C”
= B’ . D + A’ . C

Question 35.
State and prove the idempotent law using truth table. Idempotent law
Answer;
Idempotent law states that

1. A + A = Aand
2. A . A = A Proof

1. A + A = A
Truth table is as follows:

ie. A + A = A as it is true for both values of A. Hence proved.

2. A . A = A
Truth table is as follows:

ie. A . A = A itself. It is true for both values of A. Hence proved.

Question 36.
State the Absorption laws of Boolean algebra with the help of truth tables.
Answer:
Absorption law states that
A + A . B = A and A . (A + B) = A
Proof:
The Truth table of the expression A + A . B=A is as follows.

Here both columns A and A + A . B are identical. Hence proved.
For A . (A + B) = A, the truth table is as follows:

Both columns A & A . (A + B) are identical. Hence proved

Question 37.
State Demorgen’s laws. Prove anyone with truth table method.
Answer:
Demorgan’s first theorem states that (A + B)’ = A’ . B’
ie. the complement of sum of two variables equals product of their complements,

The second theorem states that (A . B)’ = A’ + B’
ie. The complement of the product of two variables equals the sum of the complement of that variables.
Proof:
Truth table of first one is as follows:

From the truth table the columns of both (A + B)’ and A’ . B’ are identical. Hence proved.

Question 38.
Fill in the blanks:

1. (0.625)₁₀ = (……….)₂
2. (380)₁₀ = (……..)₁₆
3. (437)₈ = (………)₂

Answer:

1. (0.101)₂
2. (17C)₁₆
3. (100 011 111)₂

Question 39.
What do you mean by universal gates? Which gates are called Universal gates? Draw their symbols.

OR

Construct a logical circuit for the Boolean expression \(\bar{a} \cdot b+a \cdot \bar{b}\). Also write the truth table.
Answer:
Universal gates:
By using NAND and NOR gates only we can create other gate hence these gates are called Universal gate.
NAND gate:

NOR gate:

Truth table:

Logical circute:

Question 40.
Computers uses a fixed number of bits to respresent data which could be a number, a character, image, sound, video etc. Explain the various methods used to represent characters in memory.
Answer:
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 the 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(International Business Machine). 2⁸ = 256 characters can be represented by using this.

3. ISCII(Indian 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.

Question 41.
Draw the logic circuit for the function
\(f(a, b, c)=a . b . c+\bar{a} . b+a . \bar{b}+a . b . \bar{c}\)

OR

Prove algebrically.
\(x . y+x . \bar{y} . z=x . y .+x . z\)
Answer:

OR

\(x \cdot y+x \cdot \bar{y} \cdot z=x \cdot(y+\bar{y} \cdot z)\)
= x . (y + z) = x . y + x . z
Hence proved.

Question 42.
Following are the numbers in various number systems. Two of the numbers are same. Identify them:

1. (310)₈
2. (1010010)₂
3. (C8)₁₆
4. (201)₁₀

OR

Consider the following Boolean expression:
(B’ + A)’ = B . A’
Identify the law behind the above expression and prove it using algebriac method.
Answer:
1. (310)₈ = 3 * 8² + 1 * 8¹ + 0 * 8⁰
= 192 + 8 + 0
= (200)₁₀

2. (1010010)₂ = 1 × 2⁶+ 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰
= 64 + 0 + 16 + 0 + 0 + 2 + 0
= (82)₁₀

3. (C8)₁₆ = C × 16 + 8 × 16⁰
= 12 × 16 + 8 × 1
= 192 + 8
= (200)₁₀
Here (a) (310)₈ and (C8)₁₆ are same

OR

This is De Morgan’s law (B’ + A’) = (B’)’ . A’
= B . A’
Hence it is proved

Question 43.
Find the decimal equivalent of hexadecimal number (2D)₁₆. Represent this decimal number in 2’s complement form using 8 bit word length.
Answer:
Convert (2D)₁₆ to binary number for this write down the 4 bit binary equivalent of each number

(2D)₁₆ = (00101101 )₂
First find the 1’s complement of (00101101 )₂ and add 1 to it

Hence 2’s complement is (11010011)₂

Question 44.
Answer any one question from 15(a) and 15(b).
1. Draw the logic circuit for the Boolean expression:
\((A+\overline{B C})+\overline{A B}\)

2. Using algebraic method prove that
\(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y \cdot Z+Y=1\)
Answer:
1.

OR

2. L.H.S. = \(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y Z+Y\)
= \(\bar{y} \cdot(\bar{z}+z)+y \cdot(z+1)\)
= \(\bar{y}. 1+\bar{y} \cdot 1=y \cdot y=1\)

Question 45.
With the help of a neat circuit diagram, prove that NAND gate is a universal gate.
Answer:
1. AND operation using NAND gate,
A . B = (A NAND B) NAND (A NAND B)

2. OR operation using NAND gate,
A + B = (A NAND A) NAND (B NAND B)

3. NOT operation using NAND gate,
NOT A = (A NAND A)

Question 46.
Boolean expression:
\((A+\overline{B C})+\overline{A B}\)

OR

Using algebraic method, prove that
\(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y \cdot Z+Y=1\)
Answer:

OR

= Y . Z + Y . Z + Y . Z + Y
= Y . (Z + Z) + Y . (Z + 1)
= Y . 1 + Y. 1
= Y + Y
= 1
Hence the result.

Plus One Data Representation and Boolean Algebra Five Mark Questions and Answers

Question 1.
Explain the components of Data processing.
Answer:
Data processing consists of the techniques of sorting, relating, interpreting and computing items of data in orderto convert meaningful information. The components of data processing are given below.

1. Capturing data: In this step acquire or collect data from the user to input into the computer.
2. Input: It is the next step. In this step appropriate data is extracted and feed into the computer.
3. Storage: The data entered into the computer must be stored before starting the processing.
4. Processing/Manipulating data: It is a laborious work. It consists of various steps like computations, classification, comparison, summarization, etc. that converts input into output.
5. Output of information: In this stage we will get the results as information after processing the data.
6. Distribution of information: In this phase the information(result) will be given to the concerned persons/computers.

Question 2.
Define computer. What are the characteristics?
Answer:
A computer is an electronic device used to perform operations at very high speed and accuracy.
Following are the characteristics of the computer.

1. Speed: It can perform operations at a high speed.
2. Accuracy: It produces result at a high degree of accuracy.
3. Diligence: Unlike human beings, a computer is free from monotony, tiredness, lack of concentration etc. We know that it is an electronic ma chine. Hence it can work four hours without making any errors.
4. Versatility: It is capable of performing many tasks. It is useful in many fields.
5. Power of Remembering: A computer consists of huge amount of memory. So it can store and recall any amount of information. Unlike human beings, it can store huge amount of data and can be retrieved when needed.

Disadvantages of computer:

1. No. IQ: It has no intelligent quotient. Hence they are slaves and human beings are the masters. It can’t take its own decisions.
2. No feelings: Since they are machines they have no feelings and instincts. They can perform tasks based upon the instructions given by the humans (programmers)