Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 2 Quantum Mechanical Model of Atom Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

11th Chemistry Guide Quantum Mechanical Model of Atom Text Book Back Questions and Answers

Textual Questions:

I. Choose the best Answer:

Question 1.
Electronic configuration of species M²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ and its atomic weight is 56. The number of neutrons in the nucleus of species M is
(a) 26
(b) 22
(c) 30
(d) 24
Answer:
(c) 30

Question 2.
The energy of light of wavelength 45nm is
(a) 6.65 × 10¹⁵ J
(b) 6.67 × 10¹¹ J
(c) 4.42 × 10^-18 J
(d) 4.42 × 10^-5 V
Answer:
(c) 4.42 × 10^-18 J

Question 3.
The energies E₁ and E₂ of two radiation are 25 eV and 50 eV respectively. The radiation between their wavelengths ie., λ₁ and λ₂ will be
(a) \(\frac{\lambda_{1}}{\lambda_{2}}\) = 1
(b) λ₁ = 2 λ₂
(c) λ₁ = \(\sqrt{225 \times 50}\) λ₂
(d) 2λ₁ = λ₂
Answer:
(b) λ₁ = 2 λ₂

Question 4.
Splitting of spectral lines in an electric field is called
(a) Zeeman effect
(b) shielding effect
(c) Compton effect
(d) stark effect
Answer:
(d) stark effect

Question 5.
Based on equation E = -2.178 × 10^-18 J(z²/n²), certain conclusions are written. Which of them is not correct?
(a) Equation can be used to calculate the change in energy when the electron changes orbit
(b) For n – 1 , the electron has a more negative energy then it does for n = 6 which means that the eiectron is more loosely bound in the smallest allowed orbit
(c) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance ffome nucleus.
(d) Larger the value of n, the larger is the orbit radius.
Answer:
(b) For n – 1 , the electron has a more negative energy then it does for n = 6 which means that the eiectron is more loosely bound in the smallest allowed orbit.

Question 6.
According to the Bohr Theory , which of the following transitions in the hydrogen atom will give rise to least energetic photon?
(a) n = 6 to n = 1
(b) n = 5 to n = 4
(c) n = 5 to n = 3
(d) n = 6 to n = 5
Answer:
(d) n = 6 to n = 5

Question 7.
Assertion:
The spectrum of He⁺ is expected to be similar to that of hydrogen
Reason: He⁺ is also one electron system.
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reasons is not the correct explanation of assertion.
(c) If assertion is true but reason is false
(d) If both assertion and reason are false
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.

Question 8.
Which of the following pairs of d-orbitals will have electron density along the axes?
(a) dz², dxz
(b) dxz, dyz
(c) dx², dx² – y²
(d) dxy, dx² – y²
Answer:
(c) dx², dx² – y²

Question 9.
Two electron occupying the same orbital are distinguished by
(a) Azimuthal quantum number
(b) Spin quantum number
(c) Magnetic quantum number
(d) Orbital quantum number
Answer:
(b) Spin quantum number

Question 10.
The electronic configuration of Eu (Atomic no, 63), Gd (Atomic no . 64), and Tb (Atomic no. 65) are
(a) [Xe] 4f⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(b) [Xe] 4f⁷ 6s², [Xe] 4f¹ 5d¹ 6s² and [Xe] 4f⁹ 6s²
(c) [Xe] 4f⁷ 6s², [Xe] 4f⁸ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(d) [Xe] 4f⁶ 5d¹ 6s², [Xe] 4f¹ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Answer:
(b) [Xe] 4f⁷ 6s², [Xe] 4f¹ 5d¹ 6s² and [Xe] 4f⁹ 6s²

Question 11.
The maximum number of electrons in a sub shell is given by the expression
(a) 2n²
(b) 2l + 1
(c) 4l + 2
(d) none of these
Answer:
(c) 4l + 2

Question 12.
For d-electrons, the orbit angular momentum is
(a) \(\frac{\sqrt{2} h}{2 \pi}\)

(b) \(\frac{\sqrt{2 h}}{2 \pi}\)

(c) \(\frac{\sqrt{2 \times 4} \mathrm{~h}}{2 \pi}\)

(d) \(\frac{\sqrt{6} \mathrm{~h}}{2 \pi}\)
Answer:
(d) \(\frac{\sqrt{6} \mathrm{~h}}{2 \pi}\)

Question 13.
What is the maximum number electrons that car be associated with following set of quantum numbers? n = 3, l = 1 and m = -1
(a) 4
(b) 6
(c) 2
(d) 10
Answer:
(c) 2

Question 14.
Assertion:
The number of radials and angular nodes for 3p orbital are I, 1 respectively.
Reason:
The number of radials and angular nodes depends only one the quantum number.
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion
(c) Assertion is true but the reason is false
(d) Both assertion and reason are false
Answer:
(c) Assertion is true but the reason is false

Question 15.
The total number of orbitals associated with the principal quantum number n = 3 is
(a) 9
(b) 8
(c) 5
(d) 7
Answer:
(a) 9

Question 16.
If n = 6, the sequence for filling electrons will be,
(a) ns → (n – 2)f → (n – 1)d → np
(b) ns → (n – 1 )d → (n – 2)f → np
(c) ns → {n – 2)f → np → (n – 1 )d
(d) none of these are correct
Answer:
(a) ns → (n – 2)f → (n – 1)d → np

Question 17.
Consider the following sets of quantum numbers:

Which of the following sets of quantum numbers is not possible?
(a) (i), (ii) and (iv)
(b) (ii), (iv) and (v)
(c) (i) and (iii)
(d) (ii), (iii) and (iv)
Answer:
(b) (ii), (iv) and (v)

Question 18.
How many electrons in an atom with atomic number 105 can have (n + l) = 8?
(a) 30
(b) 17
(c) 15
(d) unpredictable
Answer:
(b) 17

Question 19.
Electron density in the yz plane of 3d_{x² – y²} orbital is
(a) zero
(b) 0.50
(c) 0.75
(d) 0.90
Answer:
(a) zero

Question 20.
If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is
(a) \(\frac{1}{m} \sqrt{\frac{h}{\pi}}\)

(b) \(\sqrt{\frac{\mathrm{h}}{\pi}}\)

(c) \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

(d) \(\frac{\mathrm{h}}{4 \pi}\)
Answer:
(c) \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

Question 21.
A macroscopic particle of mass 100 g and moving at a velocity of 100 cm s^-1 will have a de Broglie wavelength of
(a) 6.6 × 10^-29 cm
(b) 6.6 × 10^-30 cm
(c) 6.6 × 10^-31 cm
(d) 6.6 × 10^-32 cm
Answer:
(c) 6.6 × 10^-31 cm

Question 22.
The ratio of de Brogue wavelengths of a deuterium atom to that of an α – particle, when the velocity of the former is five times greater than that of later, is
(a) 4
(b) 0.2
(c) 2.5
(d) 0.4
Answer:
(d) 0.4

Question 23.
The energy of an electron in the 3rd orbit of a hydrogen atom is -E. The energy of an electron in the first orbit will be
(a)-3E
(b) -E/3
(c) -E/9
(d) -9E
Answer:
(d) -9E

Question 24.
Time independent Schnodinger wave equation is
(a) Hψ = Eψ
(b) ∆²ψ + 8π²m(E + V)ψ
(c) \(\frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}+\frac{2 m}{h^{2}}(\mathrm{E}-\mathrm{V}) \psi=0\)
(d) all of these
Answer:
(a) Hψ = Eψ

Question 25.
Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?
(a) ∆x . ∆p ≥ \(\frac{h}{4}\)
(b) ∆x . ∆v ≥ \(\frac{h}{4 \pi m}\)
(c) ∆E . ∆t ≥ \(\frac{h}{4 \pi}\)
(d) ∆E . ∆x ≥ \(\frac{h}{4 \pi}\)
Answer:
(d) ∆E . ∆x ≥ \(\frac{h}{4 \pi}\)

II. Write brief answers to the following questions:

Question 26.
Which quantum number reveals information about the shape, energy, orientation, and size of orbitals?
Answer:
Magnetic quantum numbers reveal information about the shape, energy, orientation, and size of orbitals.

Question 27.
How many orbitals are possible for n = 4?
Answer:
When n = 0, l = 0, 1,2 and 3. Hence, there are four subshells namely, s, p, d and f
l = 0, m₁ = 0; one 4s orbital, l = 1, m = -1, 0, +1; three 4p orbitals,
l = 2, m₁ = – 2, -1, 0, +1, +2; five 4d orbitals and
l = 3, m₁ = – 3, -2, -1, 0, 1, 2, 3; seven 4f orbitals. Hence, the number of possible orbitals when n = 4 are sixteen.

Question 28.
How many radial nodes for 25, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
Answer:
The number of radial nodes is equal to (n – l – 1) and angular nodes is l.

Orbital	N	l	Radial node (n – l – 1)	Angular node, l
2s	2	0	1	0
4p	4	1	2	1
5d	5	2	2	2
4f	4	3	0	3

The number of radial nodes for 2s, 4p, 5d, and 4f orbitals are respectively 1,2,2 and 0 and the number of angular nodes for 2s, 4p, 5d, and 4f orbitals respectively are 0, 1, 2, and 3.

Question 29.
The stabilization of a half-filled d – orbital is more pronounced than that of the p-orbital. Why?
Answer:
The exactly half-filled orbitals have greater stability. The reason for their stability are –

1. symmetry
2. exchange energy.

(1) Symmetry: The half-filled orbitals are more symmetrical than partially filled orbitals and this symmetry leads to greater stability.

(2) Exchange energy: The electrons with the same spin in the different orbitals of the same sub-shell can exchange their position. Each such exchange releases energy and this is known as exchange energy. Greater the number of exchanges, the greater the exchange energy, and hence greater the stability. In d-orbital, 10 exchanges are possible but in p-orbital 6 exchanges are possible. So, d – orbital with 5 unpaired electrons (10 exchanges)n i.e. half-filled is more stable than p – orbital with 3 unpaired electrons (6 exchanges).

Question 30.
Consider the following electronic arrangements for the d⁵ configuration.

(i) Which of these represents the ground state?
(ii) Which configuration has the maximum exchange energy?
Answer:
(i) The ground state electronic configuration is

(ii) The configuration has the maximum exchange energy is

Question 31.
State and explain Paull’s exclusion principle.
Answer:
Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”.
Illustration: H(Z = 1) 1s¹.
One electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 and s = + \(\frac {1}{2}\). For helium Z = 2. He: 1s². In this one electron has the quantum number same as that of hydrogen, n = 1,l = 0, m = 0 and s = +½ For other electron, fourth quantum number is different, i.e. n = 1, l = 0, m = 0 and s = – ½.

Question 32.
Define orbital. What are ‘n’ and ‘l’ values for 3px and 4d _{x² – y²} electron?
Answer:
Orbital is a three-dimensional space in which the probability of finding the electron is maximum. The values of ‘n’ and ‘l’ for 3px orbital are n = 3 and l = 1, 4d _{x² – y²} orbital are n = 4 and l = 2.

Question 33.
Explain briefly the time-independent Schrodinger wave equation.
Answer:
Erwin Schrodinger expressed the wave nature of electrons in terms of a differential equation. This equation determines the change of wave function in space depending on the field of force in which the electron moves. The time-independent Schrodinger equation can be expressed as
Hψ = Eψ, where H is called Hamiltonian operator, ψ is the wave function and is a function of position coordinates of the particle and is denoted as ψ(x, y, z), E is the energy of the system.

The above Schrodinger wave equation does not contain time as a variable and is referred to as time-independent Schrodinger wave equation. This equation can be solved only for certain values of E, the total energy, i.e., the energy of the system is quantized. The permitted total energy values are called eigenvalues and corresponding wave functions represent the atomic orbitals.

Question 34.
Calculate the uncertainty in position of an electron, if ∆v = 0.1 % and υ = 2.2 × 10⁶ ms^-1.
Answer:
Heisenberg’s Uncertainty Principle is ∆x. ∆v > h/4πm.
Given:
∆v = 0.1%,
υ = 22 × 10⁶ ms^-1.
h = 6.626 × 10^-34 kgm²s^-1 .
m = 9.1 X 10 21 kg.
∆v = \(\frac{0.1 \times 2.2 \times 10^{6} \mathrm{~ms}^{-1}}{100}\)

= 2.2 × 10³ ms^-1
Uncertainty in position,
∆x ≥ \(\frac{h}{4 \pi m}\)
∆x ≥ \(\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{9.1 \times 10^{-31} \mathrm{~kg} . \times 2.2 \times 10^{3} \mathrm{~ms}^{-1}}\)

∆x ≥ 2.64 × 10^-8 m.

Question 35.
Determine the values of all the four quantum numbers of the 8th electron in the O – atom and 15th electron in the Cl atom and the last electron in Chromium.
Answer:
(1) O (Z = 8) 1s² 2s² 2p_x² 2p_y¹ 2p_z¹
Four quantum numbers for 2p_x¹ electron in oxygen atom:
n = principal quantum number = 2
l = azimuthal quantum number =1
m = magnetic quantum number =+1
s = spin quantum number = +\(\frac {1}{2}\)

(2) Cl (Z = 17) 1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹
Four quantum numbers for 15^th electron in chlorine atom:
n = 3, l = 1, m = 0, s = + ½

(3) Cr (Z = 24) 1s² 2s² 2p² 3s² 3p² 3d² 4s¹
n = 3, l = 2, m = +2, s = + ½

Question 36.
The quantum mechanical treatment of the hydrogen atom gives the energy value:
E_n = –\(\frac{-13.6}{n^{2}}\) eV/atom
(i) Use this expression to find ∆E between n = 3 and n = 4.
(ii) Calculate the wavelength corresponding to the above transition.
Answer:
Energy of the electron in the nth orbit is
E_n = –\(\frac{-13.6}{n^{2}}\)eV/atom.

When n = 3,
E₃ = \(\frac{-13.6 \mathrm{eV} / \text { atom }=-1.51 \mathrm{eV} / \text { atom }}{9}\)

When n = 4,
E₄ = \(\frac{-13.6 \mathrm{eV} / \text { atom }=-0.85 \mathrm{eV} / \text { atom }}{16}\)

∆E = (E₄ – E₃) = (-0.85) – (-1.51) = 0.66 eV/atom
Wavelength corresponding to this transition,

Question 37.
How fast must a 54g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400 Å?
Answer:
de Broglie wavelength,
λ = \(\frac{h}{m v}\)
Given:
de Broglie wavelength, λ = 5400 Å and mass, m = 54 g.
Velocity of the tennis ball
v = \(\frac{h}{m \lambda}\)

v = \(\frac{6.626 \times 10^{-34} J s}{54 \times 10^{-3} k g \times 5400 \times 10^{-10} m}\)

v = 2.27 × 10^-26 ms^-1

Question 38.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals,
(i) n = 1, l = 2
(ii) n = 5, l = 3
(iii) n = 7, l = 0
Answer:

Question 39.
Give the electronic configuration of Mn²⁺ and Cr³⁺.
Answer:
1. Mn (Z = 25)
Mn → Mn²⁺ + 2e^–
Mn²⁺ electronic configuration is 1s 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

2. Cr (Z = 24)
Cr → Cr³⁺ + 3e^–
Cr³⁺ electronic configuration is Is² 2s² 2p⁶ 3s²3p⁶ 3d³

Question 40.
Describe the Aufbau principle.
Answer:
Aufbau Principle states that “In the ground state of the atoms, the orbitals are filled in the order of their increasing energies”. That is the electrons first occupy the lowest energy orbital available to them. Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. The order of filling of various orbitals as per the Aufbau principle is given in the figure. which is in accordance with the (n + l) rule.

Question 41.
An atom of an element contains 35 electrons and 45 neutrons. Deduce
(i) the number of protons
(ii) the electronic configuration for the element
(iii) All the four quantum numbers for the last electron.
Answer:
(i) Atomic Number of the element, z = No. of protons or No. of electrons. = 35
Mass number of the element, A = No. of protons + No of neutrons = 35 + 45 = 80
Number of Protons = 80 – 45 = 35.

(ii) Electronic configuration of the element (z = 35)
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵

(iii) Quantum number for the last electron (4p_z),
n = 4, l = 1, m = +1, or -1 s = +1/2

Question 42.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the nucleus.
Answer:
According to the de Broglie concept, the electron that revolves around the nucleus exhibits both particle and wave character. In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave, Otherwise, the electron wave is out of phase.
Circumference of the orbit = nλ
2πr = nλ
2πr = \(\frac{n h}{m v}\)
Rearranging,
mvr = \(\frac{n h}{m v}\)
Angular momentum = \(\frac{n h}{2 \pi}\)
The above equation was already predicted by Bohr.
Hence, de Broglie and Bohr’s concepts are in agreement with each other.

Question 43.
Calculate the energy required for the process.
He⁺_(g) → He²⁺_(g) + e^–
The ionization energy for the H atom in its ground state is — 13.6 eV/atom.
Answer:
He⁺_(g) → He²⁺_(g) + e and
E_n = – 13.6z²/n²
E_l = – \(\frac{13.6(2)^{2}}{(1)^{2}}\) = -56.4 eV

E_∞ = \(\frac{-13.6(2)^{2}}{(\infty)^{2}}\) = 0

Required energy for the given process is,
E_∞ – E_l = 0 – (-56.4) = 56.4 eV.

Question 44.
An ion with mass number 37 possesses unit negative charge. It the ion contains 11.1% more neutrons than electrons. Find the symbol of the ion.
Answer:
Let the number of electrons in an ion = x
number of neutrons = n = x + \(\frac{11.1}{100}\) eV = 1.111 x
(As the number of neutrons is 11.1% more than the number of electrons)
In the neutral of the atom, a number of electrons.
e^– = x – 1 (as the ion carries -1 charge)
Similarly number of protons = P = x – 1
Number of protons + number of neutrons = mass number = 37
(x – 1) + 1.111 x = 37 .
2.111 x = 37 +1
2.111 x = 38
x = \(\frac{38}{2.111}\) = 18.009 = 18
∴ Number of protons = atomic number – 1 = 18-1 = 17
∴ The symbol of the ion = \(_{17}^{37} \mathrm{Cl}\).

Question 45.
The Li²⁺ ion is a hydrogen-like ¡on that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4th orbit.
Answer:

Question 46.
Protons can be accelerated in particle accelerators. Calculate the wavelength (in Å) of such accelerated proton moving at 2.85 × 10⁸ms^-1 (mass of proton is 1.673 × 10^-27 kg).
Answer:
Given:
velocity, v = 2.85 × 10⁸ms^-1.
mass, m = 1.673 × 10^-27 kg
λ = \(\frac{h}{m v}\)

λ = \(\frac{6.626 \times 10^{-34} \mathrm{kgms}^{-1}}{1.673 \times 10-27 \mathrm{~kg} \times 2-85 \times 10^{8} \mathrm{~ms}^{-1}}\)

λ = 1.389 × 10^-8 Å

Question 47.
What is the de Broglie wavelength (in cm) of a 160g cricket ball travelling at 140 Km hr^-1.
Answer:
Given:
velocity, v = 140 km/hr. = \(\frac{140 \times 10^{3}}{60 \times 60 \mathrm{~ms}^{-1}}\)

mass, m = 160g = 160 × 10^-3 kg
λ = \(\frac{h}{m v}\)
λ =\(\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} \mathrm{~s}^{-\mathrm{le}}}{160 \times 10^{-3} \mathrm{~kg} \times 3.88 \mathrm{~ms}^{-1}}\)

λ = 1.605 × 10^-34 m.

Question 48.
Suppose that the uncertainty in determining the position of an electron in an orbit is 0.6 Å. What is the uncertainty in its momentum?
Answer:
Heisenberg Uncertainity Principle is ∆x . ∆p ≥ \(\frac{h}{4 \pi}\)
Given:
∆x = 0.6 Å = 0.6 × 10^-10 m
h = 6.626 × 10^-34 kgm²s^-1
Uncertainity in momentum,
∆p ≥ \(\frac{h}{4 \pi \Delta x}\)

∆p ≥ \(\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{4 \times 3.14 \times 0.6 \times 10^{-10} \mathrm{~m}}\)

∆p ≥ 8.8 × 10^-25 kgms^-1

Question 49.
Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity (∆V) is equal to 1/4π of its velocity(V)
Answer:

Question 50.
What is the de Brogue wave length of an electron, which is accelerated from the rest, through a potential difference of 100V?
Answer:
Potential difference = 100 V = 100 × 106 × 10^-19 J
λ = \(\frac{h}{\sqrt{2} \text { mev }}\)

λ = \(\frac{6.626 \times 10^{-34} \mathrm{Kgm}^{2} \mathrm{~s}}{\sqrt{2} \times 9.1 \times 10^{-31} \mathrm{~kg} \times 100 \times 1.6 \times 10^{-19} \mathrm{~J}}\)

λ = 1.22 × 10^-10

Question 51.
Identify the missing quantum numbers and the sub energy level.

Answer:

11th Chemistry Guide Quantum Mechanical Model of Atom Additional Questions and Answers

I. Choose the best Answer:

Question 1.
The angular momentum of the electron in the nth orbit is
(a) \(\frac{\pi h}{2 n}\)

(b) \(\frac{2 n h}{\pi}\)

(c) \(\frac{n h}{2 \pi}\)

(d) \(\frac{2 \pi}{n h}\)
Answer:
(c) \(\frac{n h}{2 \pi}\)

Question 2.
The frequency of radiation emitted when an electron jumps from higher energy state (E₂) to a lower energy state (E₁) is given by
(a) v = \(\frac{\left(E_{2}+E_{1}\right)}{h}\)

(b) v = \(\frac{\left(E_{1}+E_{2}\right)}{h}\)

(c) v = \(\frac{\left(E_{1}-E_{2}\right)}{h}\)

(d) v = \(\frac{\left(E_{2}-E_{1}\right)}{h}\)
Answer:
(d) v = \(\frac{\left(E_{2}-E_{1}\right)}{h}\)

Question 3.
Splitting of spectral lines in the presence of a magnetic field is called
(a) Zeeman effect
(b) Stark effect
(e) shielding effect
(d) Compton effect
Answer:
(a) Zeeman effect

Question 4.
Which one of the following has zero rest mass?
(a) electron
(b) proton
(c) neutron
(d) photon
Answer:
(d) photon

Question 5.
For a microscopic particle such as an electron, the mass is of the order of
(a) 10^-29 kg
(b) 10^-31 kg
(c) 10³¹ kg
(d) 10^-30 kg
Answer:
(b) 10^-31 kg

Question 6.
Which one of the following has insignificant de Broglie wavelength?
(a) electron
(b) proton
(e) neutron
(d) iron ball
Answer:
(d) iron ball

Question 7.
Which of the following statements are true about de Broglie wavelength?
(i) The particle travels at a speed much higher than the speed of light.
(ii) The particle can have high linear momentum.
(iii) The mass of the particle is of the order of 10^-30 kg.
(iv) The particle travels at speed much less than the speed of light.
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iv)
Answer:
(b) (iii) and (iv)

Question 8.
The correct mathematical expression/s for Heisenberg Uncertainty principle is
(i) ∆x . ∆p ≥ \(\frac{h}{4 \pi}\)

(ii) ∆x . ∆p ≥ \(\frac{h}{4 \pi m}\)

(iii) ∆x . ∆v ≥ \(\frac{h}{4 \pi m}\)

(iv) ∆x . ∆v ≥ \(\frac{h}{4 \pi}\)
(a) (i) and (ii)
(b) (ii) and(iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iii)

Question 9.
The wave nature of electron was experimentally confirmed by
(a) Louis de Broglie
(b) Davisson and German
(c) Schrodinger
(d) Niels Bohr
Answer:
(b) Davisson and German

Question 10.
The term ‘ ψ ’ in the Schrodinger equation is
(a) Eigenvalue
(b) Hamiltonian operator
(c) Wave function
(d) all of these
Answer:
(c) Wave function

Question 11.
The permitted total energy values in the Schrodinger equation are called
(a) eigenvalues
(b) eigen functions
(c) wave functions
(d) Hamiltonian operator
Answer:
(a) eigenvalues

Question 12.
_______ is a three-dimensional space in which the probability of finding the electron is maximum.
(a) orbit
(b) Orbital
(c) wave function
(d) eigenvalue
Answer:
(b) Orbital

Question 13.
Which of the following has always positive value?
(a) ψ
(b) ψ²
(c) both ψ and ψ²
(d) ψ³
Answer:
(b) ψ²

Question 14.
The maximum number of electrons that can be accommodated in M shell is
(a) 8
(b) 32
(c) 16
(d) 18
Answer:
(d) 18

Question 15.
The maximum number of electrons that can be accommodated in a given subshell is
(a) (2l + 1)
(b) 4l + 2
(c) l + 2
(d) 2(l + 1)
Answer:
(b) 4l + 2

Question 16.
The region where the probability density function reduces to zero is called
(a) wave function
(b) orbital
(c) nodal surface
(d) probability density region
Answer:
(c) nodal surface

Question 17.
Number of subshells and electrons associated with n = 4 respectively are
(a) 4, 16
(b) 32, 64
(c) 16, 32
(d) 8, 16
Answer:
(c) 16, 32

Question 18.
The radial wave function in the ψ(r, θ, φ) = R(r), f(θ), g(φ) is
(a) f(θ)
(b) g(φ)
(c) R(r)
(d) f(θ) and g(φ)
Answer:
(c) R(r)

Question 19.
The plot of _______ shows the maximum probability that occurs at a distance of 0.52 Å from the nucleus.
(a) 4π²r² vs ψ²
(b) 4π² r² ψ² vs r²
(c) 4π² ψ² vs r
(d) 4π² ψ² vs r²
Answer:
(d) 4π² ψ² vs r²

Question 20.
The number of radial nodes for a 3s orbital is
(a) 3
(b) 2
(c) 1
(d) 0
Answer:
(b) 2

Question 21.
The number radial nodes for a ‘nd’ orbital is
(a) (n – 1)
(b) (n – l)
(c) (n – l + 1)
(d) (n – l – 1)
Answer:
(d) (n – l – 1)

Question 22.
The order of the effective nuclear charge felt by an electron in an orbital within a shell is
(a) s < p < d < f
(b) s > p > d > f
(c) s < p ≈ d < f
(d) s ≈ p > d ≈ f
Answer:
(b) s > p > d > f

Question 23.
Which of the following sequences shows the correct increasing order of energy’?
(a) 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p
(b) 3s, 3p, 4s, 4p, 3d, 5s, 4d, 5p
(c) 3s, 3p, 4s, 4p, 3d, 5s, 5p, 4d
(d) 3s, 3p, 4s, 4p, 3d, 4d, 5s, 5p
Answer:
(a) 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p

Question 24.
The orbital with following quantum numbers (i) n = 4,1 = 3 and (ii) n = 3 and 1 = 2 are
(a) 3d, 4f
(b) 3d, 4d
(c) 3f, 4f
(d) 3p, 4f
Answer:
(c) 3f, 4f

Question 25.
The energy of one photon of a beam of light with wavelength 3.31 × 10^-6 m is
(a) 6 × 10^-20 J
(b) 2 × 10^-20 J
(c) 6 × 10^-21 J
(d) 2 × 10^-21 J
Answer:
(a) 6 × 10^-20 J

Question 26.
The ratio of energies of two radiations with wavelengths of 300 nm and 900 nm is
(a) 1 : 3
(b) 2 : 1
(c) 3 : 1
(d) 1 : 2
Answer:
(c) 3 : 1

Question 27.
The momentum of a particle which has a wavelength of 6.62 Å is (in kg ms^-1)
(a) 10^-20
(b) 10^-22
(c) 10^-34
(d) 10^-24
Answer:
(d) 10^-24

Question 28.
If the energy difference between the ground state of an atom and its excited state is 3.31 × 10^-19 J, the wavelength of the photon required to produce this transition is
(a) 3 × 10^-9 m
(b) 6 × 10^-7 m
(c) 3 × 10^-8 m
(d) 6 × 10^-6 m
Answer:
(b) 6 × 10^-7 m

Question 29.
The spectrum of He⁺ is similar to
(a) Li⁺
(b) H⁺
(c) He
(d) Li²⁺
Answer:
(c) He

Question 30.
An atom has two electrons in the K shell, eight electrons in the L shell, and eight electrons in the M shell. The number of ‘s’ electrons present in that element is
(a) 18
(b) 10
(c) 6
(d) 4
Answer:
(c) 6

Question 31.
The atomic number of an element ‘A’ is 25. How many electrons are present in the third shell of the element in its A³⁺ state?
(a) 10
(b) 12
(c) 15
(d) 4
Answer:
(b) 12

Question 32.
Which of the following sets of quantum numbers are not possible?

(a) (i), (iii) and (iv)
(b) (i), (iv) and (v)
(c) (iii), (iv) and (v)
(d) (i), (ii) and (iii)
Answer:
(c) (iii), (iv) and (v)

Question 33.
ψ² is always
(a) negative
(b) positive
(c) either positive or negative
(d) none of these
Answer:
(b) positive

Question 34.
Which of the following sets of quantum numbers represents the highest energy of an atom?
(a) n = 4, l = 1, m = 0, s = +1/2
(b) n = 3, l = 0, m = 0, s = +1/2
(c) n = 3, l = 0, m = 1, s = +1/2
(d) n = 4, l = 0, m = 0, s = +1/2
Answer:
(a) n = 4, l = 1, m = 0, s = +1/2

Question 35.
The number of possible exchanges for [Ar] 3d⁴ 4s² configuration is
(a) 10
(b) 6
(c) 4
(d) 1
Answer:
(b) 6

II. Very Short Question and Answers (2 Marks):

Question 1.
Write a note about J.J. Thomson’s atomic model.
Answer:

• J.J. Thomson’s cathode ray experiment revealed that atoms consist of negatively charged particles called electrons.
• He proposed that an atom is a positively charged sphere in which the electrons are embedded like the seeds in the watermelon.

Question 2.
What is the Zeeman effect?
Answer:
The splitting of spectral lines in the presence of a magnetic field is called the Zeeman effect.

Question 3.
What is Stark effect?
Answer:
The splitting of spectral lines in the presence of electric field is called Stark effect.

Question 4.
Write the limitation of Bohr’s postulates?
Answer:
Bohr’s postulates are applicable to one electron speciesvsuch as H, He⁺ and Li²⁺ etc.,

Question 5.
Write the formulae for the radius and energy of electron in the nth orbit of electron.
Answer:
Radius of electron in the nth orbit,

Question 6.
Write de Broglie equation and explain the terms in it.
Answer:
de Beoglie equation is λ = \(\frac{h}{m v}\)
where λ = de Broglie wavelength of matter waves, m = mass of the particle and v = velocity.

Question 7.
What are quantum numbers?
Answer:

• The electron in an atom can be characterized by a set of four quantum numbers, namely principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m), and spin quantum number (s).
• When the Schrodinger equation is solved for a wave function T, the solution contains the first three quantum numbers n, l, and m.
• The fourth quantum number arises due to the spinning of the electron about its own axis.

Question 8.
State Heisenberg Uncertainty principle.
Answer:
It is impossible to accurately determine both the position as well as the momentum of a microscopic particle simultaneously.

Question 9.
What is classical mechanics?
Answer:
The motion of objects that we come across in our daily life can be well described based on Newton’s law of motion is called classical mechanics.

Question 10.
What is the limitation of classical mechanics?
Answer:
Classical mechanics does not consider the dual nature of the matter which is significant for microscopic particles. As a consequence, it fails to explain the motion of microscopic particles.

Question 11.
What is quantum mechanics?
Answer:
Based on Heisenberg’s principle and the dual I nature of the microscopic particles, a new mechanics I developed to study the motion of microscopic particles is called quantum mechanics.

Question 12.
What is Schrodinger’s equation?
Answer:
Erwin Schrodinger expressed the wave nature of electron in terms of the differential equation is called Schrodinger equation. This equation determines the change of wave equation in space depending on the field of force in which the electron moves.

Question 13.
Write the Schrodinger equation in terms of the operator?
Answer:
The time-independent Schrodinger equation can be expressed as Hψ = Eψ
where H is called Hamiltonian operator, ψ is the wave function and E is the energy of the system.

Question 14.
Write the Schrodinger equation in terms of a differential equation?
Answer:
The time independent Schrodinger equation in terms of a differential equation is
\(\frac{\partial^{2} \Psi}{\partial x^{2}}+\frac{\partial^{2} \Psi}{\partial y^{2}}+\frac{\partial^{2} \Psi}{\partial z^{2}}+\frac{8 \pi^{2} m}{h^{2}}(\mathrm{E}-\mathrm{V}) \Psi=0\)

Question 15.
What is a Hamiltonian operator?
Answer:
The total energy operator is called Hamiltonian operator and it can be expressed as
Ĥ = [latex]\frac{-h^{2}}{8 \pi^{2} m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)+V[/latex]

Question 16.
Define an orbital.
Answer:
Orbital is a three-dimensional space in which the probability of finding the electron is maximum.

Question 17.
Calculate the total number of angular nodes and radial nodes present in 4p and 4d orbitals.
Answer:
1. For 4p orbital:
Number of angular nodes = l
For 4p orbital 7 = l
Number of angular nodes = l
Number of radial nodes = n – l – 1 = 4 -1 -1 = 2
Total number of nodes = n -1 = 4 – 1 = 3
1 angular node and 2 radial nodes.

2. For 4d orbital:
Number of angular nodes = l
For 4d orbital l = 2
Number of angular nodes = 2
Number of radial nodes = n – l – 1 = 4 – 2 – 1 = 1
Total number of nodes = n – l = 4 – l = 3
1 radial nodes and 2 angular node.

Question 18.
The energies of the same orbital decrease with an increase in the atomic number. Justify this statement.
Answer:
The energy of the 2s orbital of a hydrogen atom is greater than that of 2s orbital of lithium and that of lithium is greater than that of sodium and so on because of H (Z =1), Li (Z = 3), and Na (Z = 11). When the atomic number increases, the energies of the same orbital decrease. E_2s(H) > E_2s(Li) > E_2s(Na) > E_2s(K) ………….

Question 19.
What is the principal quantum number?
Answer:
The principal quantum number represents the energy level in which an electron revolving around the nucleus and is denoted by the symbol ‘n’.

Question 20.
What is a nodal surface or a radial node?
Answer:
The region where this probability density function reduces to zero is called nodal surface or a radial node.

Question 21.
What is the significance of the solution to the Schrodinger equation?
Answer:
The solution to the Schrodinger equation gives the permitted energy values called eigenvalues and the wave functions corresponding to the eigenvalues are called atomic orbitals.

Question 22.
What are radial and angular wave functions?
Answer:
The solution of the Schrodinger wave equation for one electron system can be expressed in spherical polar coordinates as ψ(r, θ, φ) = R(r). f(θ). g(φ) Where R(r) is called a radial wave function, f(θ) and g(φ) are called angular wave functions.

Question 23.
What is the ground state?
Answer:
The electron present in the lowest energy state is called the ground state.

Question 24.
State (n + l) rule.
Answer:
The lower the value of (n + l) for an orbital, the lower is its energy. If two orbitals have the same value of (n + l), the orbital with lower value of ‘n’ will have the lower energy.

III. Short Question and Answers (3 Marks):

Question 1.
What are the conclusions of Rutherford’s α – rays scattering experiment?
Answer:

• Rutherford bombarded a thin gold foil with a stream of fast-moving α – particles.
• It was observed that most of the a-particles passed through the foil.
• Some of them were deflected through a small angle.
• Very few α- particles were reflected back by 180°.
• Based on these observations, he proposed that in an atom, there is a tiny positively charged nucleus and the electrons are moving around the nucleus with high speed.

Question 2.
Write the assumptions of Bohr’s atom model.
Answer:
Bohr’s atom model is based on the following assumptions:
(i) The energies of electrons are quantized.
(ii) The electron is revolving around the nucleus in a certain fixed circular path called stationary orbit.
(iii) Electron can revolve only in those orbitsJn which the angular momentum {mvr) of the electron must be equal to an integral multiple of \(\frac{h}{2 \pi}\)
i,e., mvr = \(\frac{n h}{2 \pi}\), where n = 1, 2, 3, … etc.,
(iv) As long as an electron revolves in the fixed orbit, it does lose its energy. However, when an electron jumps from higher energy state (E₂) to a lower energy state (E¹), the excess energy is emitted as radiation.
The frequency of the emitted radiation is (E₂ – E₁) = hυ.

Frequency, υ = \(\frac{\left(E_{2}-E_{1}\right)}{h}\)
Conversely, when suitable energy is supplied to an electron, it will jump from lower energy orbit to a higher energy orbit.

Question 3.
Explain Davisson and Germer’s experiment.
Answer:

• The wave nature of electrons was experimentally confirmed by Davisson and Germer.
• They allowed the accelerated beam of electrons to fall on a nickel crystal and recorded the diffraction pattern.
• The resultant diffraction pattern is similar to the X-ray diffraction pattern.
• The finding of wave nature of electron leads to the development of various experimental techniques such as electron microscope, low energy electron diffraction, etc.

Question 4.
Show that de Broglie and Bohr’s concepts are in agreement with each other.
Answer:
According to the de Broglie concept, the electron that revolves around the nucleus exhibits both particle and wave character. In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave, Otherwise, the electron wave is out of phase.
Circumference of the orbit = nλ
2πr = nλ
2πr = \(\frac{n h}{m v}\)
Rearranging,
mvr = \(\frac{n h}{2 \pi}\)
Angular momentum = \(\frac{n h}{2 \pi}\)
The above equation was already predicted by Bohr. Hence, de Broglie and Bohr’s concepts are in agreement with each other.

Question 5.
Write a note about the principal quantum number.
Answer:

• The principal quantum number represents the energy level in which an electron revolves around the nucleus and is denoted by the symbol ‘n’.
• The ‘n’ can have the values 1, 2, 3,… n = 1 represents K shell; n=2 represents L shell and n = 3, 4, 5 represent the M, N, O shells, respectively.
• The maximum number of electrons that can be accommodated in a given shell is 2n².
• ‘n’ gives the energy of the electron,

E_n = \(\frac{(-1312.8) Z^{2}}{n^{2}}\) KJ mol^-1 and the distance of the electron from the nucleus is given by r_n = \(\frac{(-0.529) n^{2}}{Z}\) A.

Question 6.
Write the significance of principle quantum number.
Answer:
Principle quantum number represents the energy level in which electron revolves around the nucleus and is denoted by the symbol ‘n’.
(i) The ‘n’ can have the values 1, 2, 3, … n = 1 represent K shell: n = 2 represents L shell and n = 3, 4, 5 represent the M, N, O shells respectively.
(ii) The maximum number of electrons that can be accommodated in a given shell, is 2n².
(iii) ‘n’ gives the energy of the electron,
E_n = \(\frac{(-1312.8) Z^{2}}{n^{2} k J m o l^{-1}}\)

and the distance of the electron from the nucleus is given by
r_n = \(\frac{(0.529) n^{2}}{Z}\) Å.

Question 7.
Write notes on Azimuthal Quantum number.
Answer:
(i) Azimuthal quantum number is represented by the letter ‘l’ and can take integral values from zero to (n – l), where n is the principle quantum number.
(ii) Each l value represents a subshell. l = 0, 1,2, 3 and 4 represents the s, p, d, f and g orbitals respectively.
(iii) The maximum number of electrons that can be accommodated in a given subshell is 2 (2l + 1).
(iv) It is used to calculate the orbital angular momentum using the expression
Angular momentum = \(\frac{\sqrt{l(l+1) h}}{2 \pi}\)

Question 8.
Write the significance of magnetic quantum numbers.
Answer:
(i) Spin quantum number is denoted by the letter ‘ml’. It takes integral values ranging from -l to +l through 0. i.e., if l = 1: m = -1, 0 and +1.
(ii) Different values of m for a given l value, represent different orientations of orbitals in space.
(iii) The Zeeman effect provides the experimental justification for this quantum number.
(iv) The magnitude of the angular momentum is determined by the quantum number ‘l’ while its direction is given by magnetic quantum number.

Question 9.
Write notes on spin quantum number.
Answer:
(i) The spin quantum number represents the spin of the electron and is denoted by the letter ‘m_s‘.
(ii) The electron in an atom revolves not only around the nucleus but also spins. It is usual to write this as an electron spins about its own axis either in a clockwise direction or in an anti-clockwise direction. The visualization is not true. However, spin is to be understood as representing a property that revealed itself in magnetic fields.
(iii) Corresponding to the clockwise and anti¬clockwise spinning of the electron, a maximum of two values are possible for this quantum number.
(iv) The values of ‘m_s‘ is equal to -1/2 and +1/2.

Question 10.
Write the significance of ψ and ψ².
Answer:
The wave function ψ has no physical meaning and the square of the wave function |ψ|² is related to the probability of finding the electrons within a given volume of space. |ψ|² varies with the distance from nucleus (radial distribution of the probability) and the direction from the nucleus (angular distribution of the probability).

Question 11.
Show that the probability of finding the electron is independent of the direction from the nucleus.
Answer:
The variation of the probability of locating the electron on a sphere with nucleus at its centre depends on the azimuthal quamtum number of the orbital in which the electron is present. For 1s orbital, l = 0, and m = 0.
f(θ) = \(\frac{1}{\sqrt{2}}\) and g(φ) = \(\frac{1}{\sqrt{2 \pi}}\)

Therefore, the angular distribution function is equal to \(\frac{1}{2 \sqrt{\pi}}\) i.e., it is independeent of the angle θ and φ. Hence, the probability of finding the electron is independent of the direction from the nucleus.

Question 12.
Sketch the shapes of 1s, 2s and 3s orbitals.
Answer:

Question 13.
Sketch and explain the shapes of p-orbitals.
Answer:
The shape of the p-orbitais are shown in figure, For p orbitaIs, l = 1 and the corresponding m values are -1, 0 and +1. The three different ‘m’ values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as p_x, p_y and p_z and the angular distribution for these orbitals shows that the lobes are along the x, y and z axis respectively.

Question 14.
What are ground and excited states?
Answer:
The electron in the hydrogen atom occupies the ly orbital that has the lowest energy. This state is called ground state. When this electron gains some energy, it moves to the higher energy orbitals such as 2s, 2p etc., These states are called excited states.

Question 15.
Explain the significance of effective nuclear charge.
Answer:
In a multi-electron atom, in addition to the electrostatic attractive force between the electron and nucleus, there exists a repulsive force among the electrons. These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron.
The net charge experienced by the electron is called effective nuclear charge. The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number l. The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > l.

Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d <f.

Question 16.
State and explain Hund’s rule.
Answer:
Hund’s rule of maximum multiplicity states that- electron pairing in the degenerate orbitals does not take place until all the available orbitals contains one electron each. Consider the carbon atom which has six electrons. According to Aufbau principle, the electronic configuration is 1s², 2s², 2p².
It can be represented as below,

In this case, in order to minimize the electron- electron repulsion, the sixth electron enters the unoccupied 2py orbital as per Hund’s rule, i.e., it does not get paired with the fifth electron already present in the 2px orbital.

IV. Long Question and Answers:

Question 1.
Derive de Broglie equation.
Answer:
Albert Einstein proposed that light has dual nature. i.e., light photons behave both like a particle and as a wave. Louis de Broglie extended this concept and proposed that all forms of matter showed dual character. To quantify this relation, he derived an equation for the wave length of a matter wave. He combined the following two equations of energy of which one represents wave character (hυ) and the other represents the particle nature (mc²).
Planck’s quantum hypothesis is
E = hυ ………..(1)
Einstein’s mass-energy relationship,
E = mc²
From (1) and (2), hυ = mc²
\(\frac{h c}{\lambda}\) = mc²
λ = \(\frac{h}{m c}\)
Equation (3) represents the wavelength of photons whose momentum is given by mc. For a particle of matter with mass ‘m’ and moving with a velocity ‘v’, the equation (3) can be written as
λ = \(\frac{h}{m v}\) …………..(4)

The equation (4) is called de Broglie equation for matter waves and this is valid only when the particle travels at speeds much less than the speed of light. This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.

For a particle with high linear momentum the wavelength will be so small and cannot be observed. For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.

Question 2.
Write the main features of the quantum mechanical model of atom.
Answer:
(i) The energy of electrons in atoms is quantized.
(ii) The existence of quantized electronic energy levels is a direct result of the wave-like properties of electrons.
(iii) According to Heisenberg uncertainty principle, the exact position and momentum of an electron cannot be determined with absolute accuracy. As a consequence, quantum mechanics introduced the concept of orbital. Orbital is a three dimensional space in which the probability of finding the electron is maximum.
(iv) The solution of Schrodinger wave equation for the allowed energies of an atom gives the wave function, ψ, which represents an atomic orbital. The wave nature of electron present in an orbital can be well defined by the wave function ψ.
(v) The wave function, ψ, itself has no physical meaning. However, the probability of finding the electron in a small volume dxdydz around a point (x, y, z) is proportional to |ψ(x, y, z)|² dxdydz. |ψ(x, y, z)|² is known as probability density and is always positive.

Question 3.
Describe the radial distribution function of 1s and 2s orbitals of hydrogen atom.
Answer:
Consider a single electron of hydrogen atom in the ground state for which the quantum numbers are n = 1 and l = 0. i.e., it occupies l.y orbital. The plot R(r)² versus r for is orbital given in figure.

The graph shows that as the distance between the electron and the nucleus decreases, the probability of finding the electron increases. At r = 0, the quantity R(r)² is maximum. The maximum value for |ψ|² is at the nucleus. However, probability of finding the electron in a given spherical shell around the nucleus is important. Let us consider the volume (dV) bounded by two spheres of radii r and r + dr.

The above plot shows that the maximum probability occurs at distance of 0.52 Å. from the nucleus. This is equal to the Bohr radius. It indicates that the maximum probability of finding the electron around the nucleus is at this distance. However, there is a probability to find the electron at other distances also. The radial distribution function of 2s, orbital of the hydrogen atom represented as follows.

Question 4.
Sketch and explain the shapes of d-orbitals.
Answer:
For ‘d’ orbital l = 2 and the corresponding ‘m’ values are -2, -1, 0, +1, +2. The shape of the ‘d’ orbital looks like a ‘clover leaf’. The five m values give rise to five d orbitals namely d_xy, d_yz, d_zx, d_{x² – y²}, and d_z². The 3d orbitals contain two nodal planes as shown in figure.

Question 5.
Sketch and explain the shapes of f-orbitals.
Answer:
For ‘f orbitals, l = 3 and the m values are -3, -2, -1, 0, +1, +2, +3 corresponding to seven f orbitals which are shown in figure. There are three nodal planes in the f orbitals.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 13 Introduction to Object-Oriented Programming Techniques Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 13 Introduction to Object-Oriented Programming Techniques

11th Computer Science Guide Introduction to Object-Oriented Programming Techniques Text Book Questions and Answers

Book Evaluation

Part I

Choose The Correct Answer

Question 1.
The term is used to describe a programming approach based on classes and objects is
a) OOP
b) POP
c) ADT
d) SOP
Answer:
a) OOP

Question 2.
The paradigm which aims more at procedures.
a) Object Oriented Programming
b) Procedural programming
c) Modular programming
d) Structural programming
Answer:
b) Procedural programming

Question 3.
Which of the following is a user defined data type?
a) class
b) float
c) int
d) object
Answer:
a) class

Question 4.
The identifiable entity with some characteristics and behaviour is.
a) class
b) object
c) structure
d) member
Answer:
b) object

Question 5.
The mechanism by which the data and functions are bound together into a single unit is known as
a) Inheritance
b) Encapsulation
c) Polymorphism
d) Abstraction
Answer:
b) Encapsulation

Question 6.
Insulation of the data from direct access by the program is called as
a) Data hiding
b) Encapsulation
c) Polymorphism
d) Abstraction
Answer:
a) Data hiding

Question 7.
Which of the following concept encapsulate all the essential properties of the object that are to be created?
a) class
b) Encapsulation
c) Polymorphism
d) Abstraction
Answer:
d) Abstraction

Question 8.
Which of the following is the most important advantage of inheritance?
a) data hiding
b) code reusability
c) code modification
d) accessibility
Answer:
b) code reusability

Question 9.
“Write once and use it multiple time” can be achieved by
a) redundancy
b) reusability
c) modification
d) composition
Answer:
b) reusability

Question 10.
Which of the following supports the transitive nature of data?
a) Inheritance
b) Encapsulation
c) Polymorphism
d) Abstraction
Answer:
a) Inheritance

Part – II

Very Short Answer

Question 1.
How is modular programming different from the procedural programming paradigm?
Answer:
Modular programming:

• Emphasis on the algorithm rather than data.
• Programs are divided into individual modules.
• Each modules are independent of each other and have their own local data.
• Modules can work with their own data as well as with the data passed to it.

Procedural programming:

• Programs are organized in the form of subroutines or subprograms.
• All data items are global.
• Suitable for a small-sized software applications.
• Difficult to maintain and enhance the program code as any change in data type needs to be propagated to all subroutines that use the same data type.

Question 2.
Differentiate classes and objects.
Answer:
Class: A Class is a construct in C++ which is used to bind data and its associated function together into a single unit using the encapsulation concept. Class is a user-defined data type.

Object: An identifiable entity with some characteristics and behaviour is called an object.

Question 3.
What is polymorphism?
Answer:
Polymorphism is the ability of a message or function to be displayed in more than one form.

Question 4.
How are encapsulation and abstraction are interrelated?
Answer:
The mechanism by which the data and functions are bound together into a single unit is known as Encapsulation. It implements abstraction.
Abstraction refers to showing only the essential features without revealing background details.

Question 5.
Write the disadvantages of OOP.
Answer:

1. Size: Object-Oriented Programs are much larger than other programs.
2. Effort: Object-Oriented Programs require a lot of work to create.
3. Speed: Object-Oriented Programs are slower than other programs, because of their size.

Part – III

Short Answers

Question 1.
What is a paradigm? Mention the different types of paradigm.
Answer:
The paradigm means organizing principle of a program. It is an approach to programming.
There are different approaches available for problem-solving using computers. They are:

• Procedural programming,
• Modular Programming and
• Object-Oriented Programming.

Question 2.
Write a note on the features of procedural programming.
Answer:
Important features of procedural programming

1. Programs are organized in the form of subroutines or subprograms
2. All data items are global
3. Suitable for small-sized software application
4. Difficult to maintain and enhance the program code as any change in data type needs to be propagated to all subroutines that use the same data type. This is time-consuming.
5. Example: FORTRAN and COBOL.

Question 3.
List some of the features of modular programming
Answer:
Important features of modular programming:

• Emphasis on the algorithm rather than data.
• Programs are divided into individual modules.
• Each modules are independent of each other and have their own local data.
• Modules can work with its own data as well as with the data passed to it.
• Example: Pascal and C.

Question 4.
What do you mean by modularization and software reuse?
Answer:

1. Modularization: where the program can be decomposed into modules.
2. Software reuse: where a program can be composed of existing and new modules.

Question 5.
Define information hiding.
Answer:
The data is not accessible to the outside world, and only those functions which are wrapped in the class can access it.These functions provide the interface between the object’s data and the program.
This encapsulation of data from direct access by the program is called data hiding or information hiding.

Part-IV

Explain in Detail

Question 1.
Write the differences between object-oriented programming and procedural programming.
Answer:
Object-Oriented Programming:

• Emphasizes data rather than algorithms.
• Data abstraction is introduced in addition to procedural abstraction.
• Data and its associated operations are grouped into a single unit.
• Programs are designed around the data being operated.
• Example: C++, Java, VB.Net, Python

Procedural Programming:

• Programs are organized in the form of subroutines or subprograms.
• All data items are global.
• Suitable for a small-sized software application.
• Difficult to maintain and enhance the program code as any change in data type needs to be propagated to all subroutines that use the same data type.
• Example: FORTRAN and COBOL

Question 2.
What are the advantages of OOPs?
Answer:
Advantages of OOP Re-usability:
“Write once and use it multiple times” you can achieve this by using class.

Redundancy:
Inheritance is a good feature for data redundancy. If you need the same functionality in multiple classes you can write a common class for the same functionality and inherit that class to sub-class.

Easy Maintenance:
It is easy to maintain and modify existing code as new objects can be created with small differences to existing ones.

Security:
Using data hiding and abstraction only necessary data will be provided thus maintains the security of data.

Question 3.
Write a note on the basic concepts that support OOPs?
Answer:
Object-Oriented Programming has been developed to overcome the drawbacks of procedural and modular programming. It is widely accepted that object-oriented programming is the most important and powerful way of creating software.

The Object-Oriented Programming approach mainly encourages:

1. Modularization: where the program can be decomposed into modules.
2. Software reuse: where a program can be composed of existing and new modules.

Main Features of Object-Oriented Programming:

1. Data Abstraction
2. Encapsulation
3. Modularity
4. Inheritance
5. Polymorphism

Encapsulation:
The mechanism by which the data and functions are bound together into a single unit is known as Encapsulation. It implements abstraction. Encapsulation is about binding the data variables and functions together in class. It can also be called data binding. Encapsulation is the most striking feature of a class.

The data is not accessible to the outside world, and only those functions which are wrapped in the class can access it. These functions provide the interface between the object’s data and the program. This encapsulation of data from direct access by the program is called data hiding or information hiding.

Data Abstraction:
Abstraction refers to showing only the essential features without revealing background details. Classes use the concept of abstraction to define a list of abstract attributes and function which operate on these attributes. They encapsulate all the essential properties of the object that are to be created. The attributes are called data members because they hold information. The functions that operate on these data are called methods or member functions.

Modularity:
Modularity is designing a system that is divided into a set of functional units (named modules) that can be composed into a larger application.

Inheritance:
Inheritance is the technique of building new classes (derived class) from an existing class (base class). The most important advantage of inheritance is code reusability.

Polymorphism:
Polymorphism is the ability of a message or function to be displayed in more than one form.

11th Computer Science Guide Introduction to Object-Oriented Programming Techniques Additional Questions and Answers

Choose The Correct Answer (1 Mark)

Question 1.
In procedural programming all data items are ……………….
(a) Cobol
(b) global
(c) fortran
(d) class
Answer:
(b) global

Question 2.
The object-oriented paradigm allows us to organize software as a collection of objects that consist of …………………..
a) Data
b) Behaviour
c) Both data and behaviour
d) None of these
Answer:
c) Both data and behaviour

Question 3.
………………. is an example of object-oriented programming.
(a) Python
(b) Java
(c) VB.Net
(d) All the above
Answer:
(d) All the above

Question 4.
Paradigm means ………………..
a) Organizing principle of a program
b) An approach to programming
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 5.
………………. is about binding the data variables and functions together in class.
(a) Data abstraction
(b) Modularization
(c) Redundancy
(d) Encapsulation
Answer:
(d) Encapsulation

Question 6.
In ………………. approach programs are organized in the form of subroutines or subprograms.
a) Modular Programming
b) Procedural Programming
c) Object-Oriented Programming
d) All the above
Answer:
b) Procedural Programming

Question 7.
…………………. approach of the program is time-consuming.
a) Modular Programming
b) Procedural Programming
c) Object-Oriented Programming
d) All the above
Answer:
b) Procedural Programming

Question 8.
……………… language is based on procedural programming.
a) FORTRAN
b) COBOL
c) Both A and B
d) C++
Answer:
c) Both A and B

Question 9.
…………….. paradigm consists of multiple modules; each module has a set of functions of related types.
a) Modular Programming
b) Procedural Programming
c) Object-Oriented Programming
d) All the above
Answer:
a) Modular Programming

Question 10.
In ……………… approach data is hidden under the modules
a) Modular Programming
b) Procedural Programming
c) Object-Oriented Programming
d) All the above
Answer:
a) Modular Programming

Question 11.
______ language is based on modular programming.
a) Pascal
b) C
c) Both A and B
d) C++
Answer:
c) Both A and B

Question 12.
………………. paradigm emphasizes the data rather than the algorithm.
a) Modular Programming
b) Procedural Programming
c) Object-Oriented Programming
d) All the above
Answer:
c) Object-Oriented Programming

Question 13.
………………… implements programs using classes and objects.
a) Modular Programming
b) Procedural Programming
c) Object-Oriented Programming
d) All the above
Answer:
c) Object-Oriented Programming

Question 14.
A …………….. is a construct in C++ which is used to bind data and its associated function together into a single unit.
a) Class
b) Structure
c) Array
d) None of these
Answer:
a) Class

Question 15.
A Class is a construct in C++ which uses the ……………….. concept.
a) Polymorphism
b) Encapsulation
c) Abstraction
d) Inheritance
Answer:
b) Encapsulation

Question 16.
Class is a ……………… data type.
a) User-defined
b) Derived
c) Primitive
d) None of these
Answer:
a) User-defined

Question 17.
…………………. can be defined as a template or blueprint representing a group of objects that share common properties and relationship.
a) Class
b) Structure
c) Array
d) None of these
Answer:
a) Class

Question 18.
……………….. are the basic unit of OOP.
a) Attributes
b) Objects
c) Members
d) None of these
Answer:
b) Objects

Question 19.
The class variables are called……………………
a) Instances
b) Objects
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 20.
An identifiable entity with some characteristics and behaviour is called ………………
a) Instances
b) Objects
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 21.
In ……………… method, programs are designed around the data being operated.
a) Modular Programming
b) Procedural Programming
c) Object-Oriented Programming
d) All the above
Answer:
c) Object-Oriented Programming

Question 22.
………………… language is based on object-oriented programming.
a) C++, Java
b) VB.Net
c) Python
d) All the above
Answer:
d) All the above

Question 23.
……………… is widely accepted that object-oriented programming is the most important and powerful way of creating software.
a) Modular Programming
b) Procedural Programming
c) Object-Oriented Programming
d) All the above
Answer:
c) Object-Oriented Programming

Question 24.
The Object-Oriented Programming approach mainly encourage ______
a) Modularisation
b) Software re-use
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 25.
…………………….. means the program can be decomposed into modules.
a) Modularisation
b) Software re-use
c) Both A and B
d) None of these
Answer:
a) Modularisation

Question 26.
………………… means, a program can be composed of existing and new modules.
a) Modularisation
b) Software re-use
c) Both A and B
d) None of these
Answer:
b) Software re-use

Question 27.
The main feature of Object-Oriented Programming is ………………….
a) Data Abstraction and Encapsulation
b) Modularity
c) Inheritance and Polymorphism
d) All the above
Answer:
d) All the above

Question 28.
……………………. implements abstraction.
a) Polymorphism
b) Encapsulation
c) Abstraction
d) Inheritance
Answer:
b) Encapsulation

Question 29.
……………….. can be called data binding.
a) Polymorphism
b) Encapsulation
c) Abstraction
d) Inheritance
Answer:
b) Encapsulation

Question 30.
……………………… is the most striking feature of a class,
a) Polymorphism
b) Encapsulation
c) Abstraction
d) Inheritance
Answer:
b) Encapsulation

Question 31.
The encapsulation of data from direct access by the program is called …………………….
a) Data hiding
b) Information hiding
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 32.
…………………… refers to showing only the essential features without revealing background details,
a) Polymorphism
b) Encapsulation
c) Abstraction
d) Inheritance
Answer:
c) Abstraction

Question 33.
The attributes are called ……………………..
a) Data members
b) Methods
c) Member functions
d) Either B or C
Answer:
a) Data members

Question 34.
…………………….. hold information.
a) Data members
b) Methods
c) Member functions
d) Either B or C
Answer:
a) Data members

Question 35.
The functions that operate on data numbers are called …………………..
a) Data members
b) Methods
c) Member functions
d) Either B or C
Answer:
d) Either B or C

Question 36.
………………. is designing a system that is divided into a set of functional units.
a) Polymorphism
b) Modularity
c) Abstraction
d) Inheritance
Answer:
b) Modularity

Question 37.
……………………. is the technique of building new classes from an existing class,
a) Polymorphism
b) Modularity
c) Abstraction
d) Inheritance
Answer:
d) Inheritance

Question 38.
In inheritance, the existing class is called as …………………….. class.
a) Base
b) Derived
c) Abstract
d) None of these
Answer:
a) Base

Question 39.
In inheritance, the newly created class is called as ………………… class.
a) Base
b) Derived
c) Abstract
d) None of these
Answer:
b) Derived

Question 40.
……………… is the ability of a message or function to be displayed in more than one form.
a) Polymorphism
b) Modularity
c) Abstraction
d) Inheritance
Answer:
a) Polymorphism

Question 41.
……………. means, write once and use it multiple times.
a) Re-usability
b) Redundancy
c) Security
d) None of these
Answer:
a) Re-usability

Question 42.
If we need the same functionality in multiple class you will write a common class for the same functionality and inherit that class to subclass is called ……………………
a) Re-usability
b) Redundancy
c) Security
d) None of these
Answer:
b) Redundancy

Question 43.
………………. is a good feature for data redundancy.
a) Polymorphism
b) Modularity
c) Abstraction
d) Inheritance
Answer:
d) Inheritance

Question 44.
…………………… programs are much larger than other programs.
a) Modular
b) Procedural
c) Object-Oriented
d) All the above
Answer:
c) Object-Oriented

Question 45.
………………….. program requires a lot of work to create.
a) Modular
b) Procedural
c) Object-Oriented
d) All the above
Answer:
c) Object-Oriented

Question 46.
……………… programs .are slower than other programs, because of their size.
a) Modular
b) Procedural
c) Object-Oriented
d) All the above
Answer:
c) Object-Oriented

Very Short Answers 2 Marks

Question 1.
Write a note on an object-oriented program.
Answer:
Object-Oriented Programming (OOP) is the term used to describe a programming approach based on classes and objects. The object-oriented paradigm allows us to organize software as a collection of objects that consist of both data and behaviour.

Question 2.
What is modularity?
Answer:
Modularity is designing a system that is divided into a set of functional units (named modules) that can be composed into a larger application.

Question 3.
Define Data binding.
Answer:
Encapsulation is about binding the data variables and functions together in class. It can also be called data binding.

Short Answers (3 Marks)

Question 1.
Write about objects.
Answer:
Objects: Represents data and its associated function together into a single unit. Objects are the basic unit of OOP. Basically, an object is created from a class. They are instances of class also called class variables. An identifiable entity with some characteristics and behaviour is called an object.

Question 2.
What are Encapsulation and data binding?
Answer:
The mechanism by which the data and functions are bound together into a single unit is known as Encapsulation. Encapsulation is the most striking feature of a class. The data is not accessible to the outside world, and only those functions which are wrapped in the class can access it. These functions provide the interface between the object’s data and the program. This encapsulation of data from direct access by the program is called data hiding or information hiding.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 12 Arrays and Structures Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 12 Arrays and Structures

11th Computer Science Guide Arrays and Structures Text Book Questions and Answers

Book Evaluation

Part I

Choose The Correct Answer

Question 1.
which of the following is the collection of variables of the same type that are referenced by a common name ?
a) int
b) float
c) Array
d) class
Answer:
c) Array

Question 2.
Array subscripts is always starts with which number ?
a) -1
b) 0
c) 2
d) 3
Answer:
b) 0

Question 3.
int age[ ]={6,90,20,18,2}; How many elements are there in this array?
a) 2
b) 5
c) 6
d) 4
Answer:
b) 5

Question 4.
cin>>n[3]; To which element does this statement accepts the value?
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Question 5.
By default, the string end with which character?
a) \0
b) \t
c) \n
d) \b
Answer:
a) \0

Part – II

Very Short Answers

Question 1.
What is Traversal in an Array?
Answer:
Accessing each element of an array at least once to perform any operation is known as “Traversal”. Displaying all the elements in an array is an example of “traversal”.

Question 2.
What is Strings?
Answer:
A string is defined as a sequence of characters where each character may be a letter,’ number or a symbol.
Every string is terminated by a null (\0) character which must be. appended at the end of the string.

Question 3.
What is the syntax to declare two – dimensional array? ‘
Answer:
The declaration of a 2-D array is
datatype array_name[row – size] [col – size];
In the above declaration, data-type refers to any valid C++ data – type, array _ name refers to the name of the 2 – D array, row – size refers to the number of rows and col-size refers to the number of columns in the 2 – D array.

Part – III

Short Answers

Question 1.
Define an Array. What are the types?
Answer:
An array is a collection of variables of the same type that are referenced by a common name.
There are different types of arrays used in C++. They are:

• One-dimensional arrays
• Two-dimensional arrays
• Multi-dimensional arrays

Question 2.
With note an Array of strings.
Answer:
An array of strings is a two – dimensional character array. The size of the first Index (rows) denotes the number of strings and the size of the second index (columns) denotes the maximum length of each string. Usually, an array of strings are declared in such a way to accommodate the null character at the end of each string. For example, the 2-D array has the declaration:
char name [7][10];
In the above declaration,
No. of rows = 7;
No. of columns = 10;
We can store 7 strings each with a maximum length of 10 characters.

Question 3.
Write a C++ program to accept and print your name?
Answer:
PROGRAM
#include<iostream>
using namespace std;
int main( )
{
char str[100];
cout<< “Enter your name :”;
cin.get(str,100);
cout<< “You name is : ” << str <<endl;
return 0;
}
Output

Part – IV

Explain In Detail

Question 1.
write a c++ program to find the difference between two matrix.
Answer:
PROGRAM
#include<iostream>
#include<conio.h>
using namespace std;
int main( )
{
int row, col, m1[10][10], m2[10][10], sum[10][10];
cout<<“Enter the number of rows : “;
cin>>row;
cout<<“Enter the number of columns :”;
cin>>col;
cout<< “Enter the elements of first matrix: “<<endl;
for (int i = 0;i<row;i++ )
for (int j = 0;j <col;j++ )
cin>>m1[i][j];
cout< < “Enter the elements of second matrix: “<<endl;
for (int i = 0;i<row;i++ )
for (int j = 0;j<col;j++ )
< cin>>m2[i][j];
cout<<“Output: “<<endl;
for (int i = 0;i<row;i++ )
{
for (int j = 0;j<col;j++ )
{
sum[i][j]=m1[i][j] – m2[i][j]); cout<<sum[i][j]<<“”;
}
cout<<endl<<endl;
}
getch( );
return 0;
}
Output

Question 2.
How will you pass two dimensional array to a function explain with example.
Answer:
Passing 2″D array to a function:
C++ program to display values from two dimensional array.
#include<iostream>
using namespace std;
void display (int n[3][2]);
int main( )
{
int num[3][2] = { {3, 4}, {9, 5>, {7, 1} } ;
display(num);
return 0;
}
void display(int n[3][2])
{
cout << “\n Displaying Values” << endl; for (int i=0; i<3; i++)
{
for (int j=0; j<2; j++)
{
cout << n[i][j] << “”;
}
cout << endl << endl;
}
}
Output
Displaying Values
3 4
9 5
7 1
In the above program, the two-dimensional array num is passed to the function display
( ) to produce the results.

11th Computer Science Guide Arrays and Structures Additional Questions and Answers

Choose The Correct Answer (1 Mark)

Question 1.
The size of the array is referred to as its ………………..
(a) dimension
(b) direction
(c) location
(d) space
Answer:
(a) dimension

Question 2.
…………… is an easy way of storing multiple values of the same type referenced by a common name.
a) Variables
b) Literals
c) Array
d) None of these
Answer:
c) Array

Question 3.
Displaying all the elements in an array is an example of ………………..
(a) memory allocation
(b) call by reference
(c) traversal
(d) none of these
Answer:
(c) traversal

Question 4.
A (n) ………………….. is a collection of variables of the same type that are referenced by a common name.
a) Structures
b) Array
c) Unions
d) None of these
Answer:
b) Array

Question 5.
During ……………….. the array of elements cannot be initialized more than its size.
(a) declaration
(b) initialization
(c) assigning
(d) execution
Answer:
(b) initialization

Question 6.
The …………….. of the array is referred to as its dimension.
a) Elements
b) Size
c) Format
d) None of these
Answer:
b) Size

Question 7.
Pass an array to a function in C++, the function needs the array name as ………………..
(a) a function
(b) an argument
(c) global object
(d) string
Answer:
(b) an argument

Question 8.
…………………. is a type of arrays used in C++.
a) One-dimensional array
b) Two-dimensional array
c) Multidimensional array
d) All the above
Answer:
d) All the above

Question 9.
A structure without a name tag is called ………………..
(a) homogenous structure
(b) anonymous structure
(c) array of structure
(d) dynamic memory
Answer:
(b) anonymous structure

Question 10.
A one-dimensional array represents values that are stored in a single …………….
a) Row
b) Column
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 11.
Array size should be specified with …………………
a) square brackets[ ]
b) Parenthesis ( )
c) Curly braces{ }
d) Angle brackets < >
Answer:
a) square brackets[ ]

Question 12.
In an array declaration, ……………… defines how many elements the array will hold.
a) array_name
b) array_size
c) data_type
d) None of these
Answer:
b) array_size

Question 13.
In an array declaration, …………………. declares the basic type of the array, which is the type of each element in the array.
a) array_name
b) array_size
c) data type
d) None of these
Answer:
c) data type

Question 14.
In an array declaration, ……………….. specifies the name with which the array will be referenced
a) array_name
b) array_size
c) data type
d) None of these
Answer:
a) array_name

Question 15.
The array subscript always starts with ………………
a) 0
b) 1
c) -1
d) None of these
Answer:
a) 0

Question 16.
The subscript always is a(n) ………………value.
a) Integer
b) Unsigned integer
c) Signed integer
d) float
Answer:
b) Unsigned integer

Question 17.
In Turbo C++, int data type requires …………….. bytes of memory.
a) 4
b) 8
c) 2
d) 1
Answer:
c) 2

Question 18.
In Dev C++, int data type requires ………………. bytes of memory,
a) 4
b) 8
c) 2
d) 1
Answer:
a) 4

Question 19.
The memory space allocated for an array can be calculated using the -formula.
a) Number of bytes allocated for the type of array + Number of elements
b) Number of bytes allocated for the type of array x Number of elements
c) Number of bytes allocated for the type of array – Number of elements
d) None of these
Answer:
b) Number of bytes allocated for the type of array x Number of elements

Question 20.
An array can be initialized at the time of its ………………
a) Declaration
b) Process
c) Either A or B
d) None of these
Answer:
a) Declaration

Question 21.
Unless an array is initialized, all the array elements contain ……………..values.
a) Null
b) Default
c) Garbage
d) None of these
Answer:
c) Garbage

Question 22.
While declaring and Initializing values In an array, the values should be given within the ……………..
a) square brackets[ ]
b) Parenthesis ()
c) Curly braces{ }
d) Angle brackets < >
Answer:
c) Curly braces{ }

Question 23.
The …………….. of an array may be optional when the array is initialized during declaration.
a) Data type
b) Size
c) Name
d) None of these
Answer:
b) Size

Question 24.
The subscript in the bracket can be a(n) ……………….
a) Variable
b) Constant
c) Expression that evaluates to an integer
d) Either A or B or C
Answer:
d) Either A or B or C

Question 25.
Accessing each element of an array at least once to perform any operation is known as ………………..
a) Traversal
b) Process
c) Reference
d) None of these
Answer:
a) Traversal

Question 26.
…………………. is a process of finding a particular value present in a given set of numbers.
a) Filtering
b) Searching
c) Seeking
d) None of these
Answer:
b) Searching

Question 27.
The ………………. compares each element of the list with the value that has to be searched until all the elements in the array have been traversed and compared.
a) Linear search
b) Sequential search
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 28.
A ……………. is defined as a sequence of characters where each character may be a letter, number, or a symbol.
a) String
b) Character
c) Literal
d) Identifier
Answer:
a) String

Question 29.
In C++, there is no basic data type to represent a …………….
a) String
b) Character
c) Literal
d) float
Answer:
a) String

Question 30.
How much memory required for the following array?
char country[6];
a) 12 bytes
b) 6 bytes
c) 60 Bytes
d) 24 bytes
Answer:
b) 6 bytes

Question 31.
…………….. is a way of initializing the character array:
a) char country[6]={T, ‘N’ ‘D’ ‘I’, ‘A’, ‘\0’};
b) char country[ ]=”INDIA”;
c) char country[ ]={T, ‘N’, ‘D’, T, ‘A’ ‘\0’};
d) Either A or B or C
Answer:
d) Either A or B or C

Question 32.
Which is a correct statement from the following?
a) At the end of the string, a null character is automatically added by the compiler.
b) If the size of the array is not explicitly mentioned, the compiler automatically calculates the size of the array based on the number of elements in the list and allocates space accordingly.
c) In the initialization of the string, if all the characters are not initialized, then the rest of the characters will be filled with NULL.
d) All the above
Answer:
d) All the above

Question 33.
In C++,………………. is used to read a line of text including blank spaces.
a) cin.get( )
b) cin
c) put( )
d) None of these
Answer:
a) cin.get( )

Question 34.
………………. function can read the characters till it encounters a newline character or a delimiter specified by the user.
a) getline( )
b) put
c) puts( )
d) None of these
Answer:
a) getline( )

Question 35.
…………………. arrays are a collection of similar elements where the elements are stored in a certain number of rows and columns.
a) One dimensional
b) Two dimensional
c) Multidimensional
d) All the above
Answer:
b) Two dimensional

Question 36.
In two dimensional arrays, …………………. size is compulsory.
a) Column
b) Row
c) Both A and B
d) None of these
Answer:
a) Column

Question 37.
In two dimensional arrays, …………….. size is optional.
a) Column
b) Row
c) Both A and B
d) None of these
Answer:
b) Row

Question 38.
int A[3][4 j; How many elements are there in the array “A”?
a) 3
b) 4
c) 7
d) 12
Answer:
d) 12

Question 39.
The two-dimensional array uses ………………. index values to access a particular element in it.
a) two
b) one
c) three
d) many
Answer:
a) two

Question 40.
In a two dimensional array, the first index specifies the …………….. value.
a) Column
b) Row
c) Both A and B
d) None of these
Answer:
b) Row

Question 41.
In a two dimensional array, the second index specifies the …………… value.
a) Column
b) Row
c) Both A and B
d) None of these
Answer:
a) Column

Question 42.
The two-dimensional array can be viewed as a …………………
a) List
b) Matrix
c) Linear block
d) None of these
Answer:
b) Matrix

Question 43.
There are …………… types of 2-D array memory representations.
a) two
b) one
c) three
d) many
Answer:
a) two

Question 44.
……………… is a type of 2-D array memory representation.
a) Row-Major order
b) Column-Major order
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 45.
In row-major order, all the elements are stored ……………… in contiguous memory locations.
a) Column by Column
b) Row by Row
c) Both A and B
d) None of these
Answer:
b) Row by Row

Question 46.
In column-major order, all the elements are stored ………………. in contiguous memory locations.
a) Column by Column
b) Row by Row
c) Both A and B
d) None of these
Answer:
a) Column by Column

Question 47.
An array of strings is a ………………. dimensional character array,
a) Two
b) One
c) Multi
d) None of these
Answer:
a) Two

Question 48.
In a two-dimensional character array, the size of the first index (rows) denotes the ……………….
a) Maximum length of each string
b) Number of strings
c) Maximum characters
d) None of these
Answer:
b) Number of strings

Question 49.
In a two-dimensional character array, the size of the second index (rows) denotes the ……………….
a) Maximum length of each string
b) Number of strings
c) Maximum characters
d) None of these
Answer:
a) Maximum length of each string

Question 50.
char Name[6][10]; How many strings can be stored in the above array?
a) 10
b) 60
c) 6
d) 16
Answer:
c) 6

Question 51.
To pass an array to a function in C++, the function needs the ………….. as an argument,
a) Array name
b) Array type
c) Dimension
d) None of these
Answer:
a) Array name

Very Short Answers (2 Marks)

Question 1.
What is the formula to calculate the memory space allocated for an array?
Answer:
A number of bytes allocated for the type of array x Number of elements.

Question 2.
What are the types of arrays?
Answer:
There are different types of arrays used in C++. They are:

1. One-dimensional arrays
2. Two-dimensional arrays
3. Multi-dimensional arrays

Question 3.
Write about returning structures from functions.
Answer:
A structure can be passed to a function through its object. Therefore, passing a structure to a function or passing a structure object to a function is the same because the structure object represents the structure. Like a normal variable, a structure variable(structure object) can be passed by value or by references/addresses. Similar to built-in data types, structures also can be returned from a function.

Question 4.
Write the syntax and example for one-dimensional declaration and initialization.
Answer:
Syntax:
[size] = {value-1,value-2,………….,value-n};
Example:

Question 5.
What is a global object?
Answer:
Objects declared along with structure definition are called global objects.

Question 6.
Write a note on strings.
Answer:
strings:
A string is defined as a sequence of characters where each character may be a letter, number or symbol. Each element occupies one byte of memory. Every string is terminated by a null C\0′ ASCII code 0) character which must be appended at the end of the string.

Question 7.
Differentiate array and structure.
Answer:
Array:

• An array is a collection of variables of the same data type.
• Array data are accessed using index.
• Array allocates static memory
• Array element access takes lesser time.

Structure:

• A structure is a collection of variables of different data type.
• Structure elements are accessed using operator.
• Structures allocate dynamic memory.
• Structure elements take more time.

Short Answers (3 Marks)

Question 1.
Write about the initialization of 2 – D array.
Answer:
The array can be initialized in more than one way at the time of 2-D array declaration.
For example
int matrix[4][3] = {
{10,20,30},// Initializes row 0
{40,50,60},// Initializes row 1
{70,80,90},// Initializes row 2
{100,110,120}// Initializes row 3
};
int matrix[4][3] = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120};
Array’s row size is optional but column size is compulsory.

Question 2.
What is subscript? Give its rules,
Answer:
subscript:
Each element has a unique index number starting from 0 which is known as a subscript.
Rules for subscript:

• The subscript always starts with 0.
• It should be an unsigned integer value.
• Each element of an array is referred to by its name with a subscript index within the square bracket.

For example:
num[3] refers to the 4th element in the array.

Question 3.
How memory locations are allocated to a one-dimensional array?
Answer:
Memory representation of a one-dimensional array:
The amount of storage required to hold an array is directly related to type and size.
The following figure shows the memory allocation of an array with five elements.

The above figure clearly shows that the array num is an integer array with 5 elements. Here, there is a total of 5 elements in the array, where for each element, 4 bytes in Dev C++ (or)2 bytes in Turbo C++ will be allocated. Totally 20 bytes will be allocated for this array (In Turbo C++ 10 Bytes).

Question 4.
Tabulate the memory requirement of data type in Turbo C++ and Dev C++.
Answer:

Data type	Turbo C++	Dev C++
char	1	1
int	2	4
float	4	4
long	4	4
double	8	8
long double	10	10

Question 5.
Explain character Array (String) creation with syntax and example.
Answer:
To create any kind of array, the size (length) of the array must be known in advance, so that the memory locations can be allocated according to the size of the array. Once an array is created, its length is fixed and cannot be changed during run time, This is shown in the following figure.

Syntax of array declaration is:
char array_name[size];
In the above declaration, the size of the array must be an unsigned integer value.

For example,
char country[6];
Here, the array reserves 6 bytes of memory, for storing’ a seqi characters. The length of the string cannot be more than 5 characters and one location is reserved for the null character at the end.

Question 6.
What is an array of strings?
Answer:
An array of strings is a two – dimensional character array. The size of the first index (rows) denotes the number of strings and the size of the second index (columns) denotes the maximum length of each string. Usually, an array of strings are declared in such a way to accommodate the null character at the end of each string. For example, the 2 – D array has the declaration: char Name[6][10];
In the above declaration, the 2 – D array has two indices which refer to the row size and column size, that is 6 refers to the number of rows, and 10 refers to the number of columns.

Question 7.
Write note on getline( ) function.
Answer:
In C++, getline( ) is also used to read a line . of text from the input stream. It can read the characters tilt it encounters a newline character or a delimiter specifier’ by the user. This function is available in the header.

Question 8.
How memory represented for a 2-D array?
Answer:
There are two types of 2-D array memory representations. They are:

• Row Major order
• Column Major order

For example:
int A[4][3]={ { 8,6,5}, { 2,1,9}, {3,6,4}, {4,3,2} };
Row Major order:
In row-major order, all the elements are stored row by row in continuous memory locations, that is, ail the elements in first row, then in the second row and soon. The memory representation of row-major order is as shown below.

Question 9.
How 2-D character array initialized and stored in memory? Explain.
Answer:
2D char array initialization:

For example:
char Name[6][10] = {“Mr. Bean’; “Mr.Bush”
“Nicole”, “Kidman” “Arnold”, “Jodie”};
In the above’ example, the 2-D array is initialized with 6 strings, where the string is a maximum of 9 characters long since the last character is null. The memory arrangement of a 2-D array is shown below and ail the strings are stored in continuous locations.

Question 10.
What is called nested structure? Give example.
Answer:
The structure declared within another structure is called a nested structure. Nested structures act as members of another structure and the members of the child structure can be accessed as parent structure name. Child structure name. Member name, struct dob
{
int date;
char month[3];
int year;
} ;
Values can be assigned to this structure as follows.
dob = {25, “DEC”, 2017}

Question 11.
Write a program to pass an array as an argument to a function.
Answer:
In C++, arrays can be passed to a function as an argument.
To pass an array to a function in C++, the function needs the array name as an argument.
C++ program-to display marks of 5 students (one-dimensional array)
#indude<iostream>
using namespace std;
void display (int m[5]);
int main( )
{
int marks[5]={88, 76, 90, 61, 69};
display(marks);
return 0;
}
void’ display (int m[5])
{
cout << “\n. Display Marks: ” << end!; for (int i=0; i<5; i++)
{
cout <-< “Student” << i+1 << ” << m[i]<<endl;
}
}
Output
Display Marks:
Student 1: 88
Student 2: 76
Student 3: 90
Student 4: 61
Student 5: 69

Question 12.
Write a C++ program to pass a 2-D array to a function.
Answer:
C++ program to display values from two dimensional array:
#include<iostream>
using namespace std;
void display (int n[3][2]);
int main( )
{
int num[3][2] = { {3, 4}, (9, 5}, {7,1} };
display(num);
return 0; ‘
}
void display(int n[3][2])
{
cout << “\n Displaying Values” << endl;
for (int i=0; i<3; i++)
{
for (int j=0; j<2; j++)
{
cout << n[i][j] << ”
}
cout << endl << endl;
}
}
Output
Displaying Values
3 4
9 5
7 1

Question 13.
Write a user-defined function to return the structure after accepting value through the keyboard. The structure definition is as follows: struct Item{int item no; float price;};
Answer:
item accept (item i)
{

cout << “\n Enter the Item No:”; cin >> i.no;
cout << “\n Enter the Item Price:”; cin >> i.price;
return i;

}

Explain in Detail (5 Marks)

Question 1.
How will you search an element in a one-dimensional array? Explain linear search.
Answer:
Searching in a one-dimensional array:
Searching is a process of finding a particular value present in a given set of numbers. The linear search or sequential search compares each element of the list with the value that has to be searched until all the elements in the array have been traversed and compared.

Program for Linear Search
#include<iostream>
using namespace std;
int Search(int arr[ ], int size, int value)
{
for (int i=0; <size; i++)
{
if (arr[i] == value)
return i; // return index value
}
return -1;
}
int main( )
{
int num[10], val, id;
for (int i=0; i< 10; i++) .
{
cout<<“\n Enter value” << i+1 <<“=”;
cin>>num[i];
}
cout<< “\n Enter a value to be searched: “;
cin>>val;
id=Search(num,10,val);
if(id==-1) .
cout<< “\n Given value is not found in the array..”;
else
cout<< “\n The value is found at the position” << id+1;
return 0;
}
The above program reads an array and prompts for the values to be searched. It calls the Search() function which receives array, size and value to be searched as parameters. If the value is found, then it returns the array index to the called statement; otherwise, it returns -1.

Question 2.
Explain two-dimensional array with syntax and example.
Answer:
Two-dimensional array:
Two-dimensional (2D) arrays are collections of similar elements where the elements are stored in a certain number of rows and columns. An example m x n matrix where m denotes the number of rows and n denotes the number of columns is shown in the following figure, int arr[3][3];
2D array conceptual memory representation

The array arr can be conceptually viewed in matrix form with 3 rows and columns. point to be noted here is since the subscript starts with 0 arr [0][0] represents the first element.

The syntax for declaration of a 2-D array is:
data-type array_name[row-size][col-size];
In the above declaration, data-type refers to any valid C++ data-type, array_name refers to the name of the 2-D array, row-size refers to the number of rows and col-size refers to the number of columns in the 2-D array.

For example:
int A[3][4];
In the above example, A is a 2-D array, 3 denotes the number of rows and 4 denotes the number of columns. This array can hold a maximum of 12 elements.

Case Study

Question (1)
Write a program to accept the marks of 10 students and find the average, maximum and minimum marks.
Answer:
PROGRAM
#include<iostream>
using namespace std;
int main( )
{
int marks[10], sum=0,max,min;
float avg;
cout<< “\n Enter Mark 1” << “=”;
cin>> marks[0];
max=marks[0];
min=marks[0];
for(int i=1; i< 10; i++)
{
cout<< “\n Enter Mark” << i+1 << “=”;
cin>> marks[i];
sum=sum+marks[i];
if (marks[i]>max)
max = marks[i];
if (marks[i]<min)
min = marks[i];
}
avg=sum/10.0;
cout<< “\n The Total Marks: ” << sum;
cout<< “\n The Average Mark: ” <<avg;
cout<< “\n The Maximum Mark: ” <<max;
cout<< “\n The Minimum Mark: ” <<min;
}
Output

Question (2)
Write a program to accept rainfall recorded in four metropolitan cities of India and find the city that has the highest and lowest rainfall.
Answer:
PROGRAM
#include<iostream>
#include<string.h>
using namespace std;
int main( )
{
int rate[10],min,n;
char shop[5][20];
char minshop[20];
cout<<“\nHow many Shops
cin>>n;
for(int i=0;i<n;i++)
{
cout<<“\nEnter Name of the Shop “<<i+1<<” :”;
cin> >shop[i];
cout<<“\nEnter price of Product-X at “<<shop[i]<<” :”;
cin>>rate[i];
}
min=rate[0];
strcpy(minshop,shop[0]);
for(int i=l;i<n;i++)
{
if (rate[i]<min)
{
min = rate[i];
strcpy(minshop,shop[i]);
}
}
cout< < “\n The Lowest cost of the Proudct=X at the” “<<minshop <<“is”<<min;
}
Output

Question (3)
Survey your neighboring shops and find the price of any particular product of interest and suggest where to buy the product at the lowest cost.
Answer:
PROGRAM
#include<iostream>
#include<string.h>
using namespace std;
int main( )
{
int rate[10],min,n;
char shop[5][20];
char minshop[20];
cout<<“\nHow many Shops “;
cin>>n;
for(int i=0;i<n;i++)
{
cout<<“\nEnter Name of the Shop “<<i+l<<” :”;
cin>>shop[i];
cout<<“\nEnter price of Product-X at “<<shop[i]<<“:”;
cin> >rate[i];
}
min=rate[0];
strcpy(nninshop,shop[0]);
for(int i=l;i<n;i++)
{
if (rate[i]<min)
{
min = rate[i];
strcpy(minshop,shop[i]);
}
}
cout< < “\n The Lowest cost of the Proudct=X at the ” <<minshop <<“is”<<min;
}
Output

STRUCTURES

Book Evaluation

Part -I

Choose The Correct Answer

Question 1.
The data elements in the structure are also known as ……………..
a) objects
b) members
c) data
d) records
Answer:
b) members

Question 2.
Structure definition is terminated by ……………..
a) :
b) }
c) ;
d) ::
Answer:
c) ;

Question 3.
What will happen when the structure is declared?
a) it will not allocate any memory
b) it will allocate the memory
c) it will be declared and initialized
d) it will be only declared
Answer:
b) it will allocate the memory

Question 4.
What is the out of this program?
#include<iostream>
#include<string.h>
using namespace std;
int main( )
{
struct student
{
int n;
char name[10];
};
students;
s.n = 123;
183
strcpy(s.name, “Balu”);
cout<<s.n;
co
ut<< s.name <<endl;
return 0;
}
a) 123Balu
b) BaluBalu
c) Balul23
d) 123 Balu
Answer:
a) 123Balu

Question 5.
A structure declaration is given below.
struct Time
{
int hours;
int minutes;
int seconds;
}t;
Using the above declaration which of the following refers to seconds.
a) Time.seconds
b) Time::seconds
c) seconds
d) t. seconds
Answer:
d) t. seconds

Question 6.
What will be the output of this program?
#include <iostream>
using namespace std;
struct ShoeType
{
string name; double price;
>; , ‘ ‘
int main()
184 {
ShoeType shoe1, shoe2;
shoe1.name = “Adidas”;
shoe1.price = 9.99;
cout<< shoe1.name<< ” #. “<< shoe1.
price<<endl;
shoe2 = shoe1;
shoe2.price = shoe2.price / 9;
cout<< shoe2.name<< ” # “<< shoe2.price;
return 0;

a) Adidas # 9.99 Adidas #1.11	b) Adidas # 9.99 Adidas # 9.11
c) Adidas # 9.99 Adidas # 11.11	d) Adidas #9.11 Adidas # 11.11

Answer:
a) Adidas # 9.99 Adidas #1.11

Question 7.
Which of the following is a properly defined structure?
a) struct {int num;}
b) struct sum {int num;}
c) struct sum int sum;
d) struct sum {int num;};
Answer:
d) struct sum {int num;};

Question 8.
A structure declaration is given below.
struct employee
{
int empno;
char ename[10];
} e[5];
Using the above declaration which of the following statement is correct.
a) cout<<e[0].empno<<e[0].ename;
b) cout<<e[0].empno<<ename;
c) cout< <e[0]->empno< <e[0]->ename;
d) cout<<e.empno<<e.ename;
Answer:
a) cout<<e[0].empno<<e[0].ename;

Question 9.
Which of the following cannot be a structure member?
a) Another structure
b) Function
c) Array
d) variable of double datatype
Answer:
b) Function

Question 10.
When accessing a structure member, the identifier to the left of the dot operator is the name of
a) structure variable
b) structure tag
c) structure member
d) structure-function
Answer:
a) structure variable

Part – II

Very Short Answers

Question 1.
Define structure. What is its use?
Answer:
Definition:
The structure is user-defined which has the combination of data items with different data types. This allows to a group of variables of mixed data types together into a single unit.
Use: The structure provides a facility to store different data types as a part of the same logical element in one memory chunk adjacent to each other.

Question 2.
To store 100 integer numbers which of the following is good to use?
Answer:
Array or Structure State the reason.
In any situation when more than one variable is required to represent objects of uniform data types, an array can be used. If the elements are of different data types, then the array cannot support them.
The structure provides a facility to store different data types as a part of the same logical element in one memory chunk adjacent to each other. So, to store 100 integer numbers Array is good.

Question 3.
What is the error in the following structure definition.
Answer:
struct employee {inteno; charename[20]; char dept;}
Employee e1,e2;

Error:
In the above code objects for employee strcture is created at the end of structure definition. So,structure name is not needed. It may written as:
struct employee
{
int eno;
char ename[20];
char dept;
}e1,e2;

Question 4.
Write a structure definition for the structure student containing exam no, name and an array for storing five subject marks.
Answer:
Structure definition:
struct student
{
int examno;
char sname[30];
int mark [5];
};

Question 5.
Why for passing a structure to a function call by reference is advisable to us?
Answer:
In call by reference, any change made to the contents of structure variable inside the function are reflected back to the calling function. Otherwise changes in the structure content within the function will not implement in the structure content.

Question 6.
What is the size of the following highlighted variable in terms of byte if it is compiled in dev C++
Answer:
struct A{ float f[3]; char ch[5];long double d;>;
struct B{ A a; int arr[2][3];}b[3]
struct A {float f[3]; char ch[5]; long double d;>;
Memory requirements (Using Dev C++):

Variable	Memory requirement
float f(3);	12 Bytes
char ch[5];   ,	5 Bytes
double d;	8 Bytes

Using Turbo C++ also the same memory is needed. (Only int data type requirement is differ. Here no int members. So, no change in the requirement.)
struct B { A a; int arr[2][3];}b[3];
Memory requirements (Using Dev C++):

Variable	Memory requirement
A a; (Which is a structure variable)	25 Bytes
int arr[2][3];	24 Bytes
b[3] which is a structure variable of structure B	49 Bytes for each element of b. So, Total requirement of b[3] is 147 Bytes.

Memory requirements (Using Turbo C++):

Variable	Memory requirement
A a; (Which is a structure variable)	25 Bytes
int arr[2][3];	12 Bytes
b[3] which is a structure variable of structure B	37 Bytes for each element of b. So, Total requirement of b[3] is 111 Bytes.

Question 7.
Is the following snippet is fully correct. If not identify the error.
Answer:
struct sum1{ int n1,n2;}s1;
struct sum2{int n1,n2}s2;
cin>>s1.n1>>s1.n2;
s2=s1;
No. It is not fully correct. The following errors are there.
Errors:

Statement	Error
struct sum1 {int n1,n2;}sl;	No Error
struct sum2 {int n1,n2}s2;	Semicolon missing at the end of declaration statement;
cin>>s1.n1>>s1. n2;	No Error
s2-s1;	Objects of different structures can not be directly copied.

Question 8.
Differentiate array and structure.
Answer:
In any situation when more than one variable is required to represent objects of uniform data¬types, array can be used. If the elements are of different data types, then array cannot support. If more than one variable is used, they can be stored in memory but not in adjacent locations. It increases the time consumption while searching.

The structure provides a facility to store different data types as a part of the same logical element in one memory chunk adjacent to each other.

Question 9.
What are the different ways to initialize the structure members?
Answer:
Values can be assigned to structure elements similar to assigning values to variables.
Example:
Consider the following structure: struct Student
{
long int rollno;
int age;
float weight;
}
balu ;
We can initialize as given below:
balu.rollno= 702016;
balu. age=18; ‘
balu.weight= 48.5;
Also, values can be assigned directly as similar to assigning values to Arrays.
balu={702016, 18, 48.5};

Question 10.
What is wrong with the following C++ declarations?
Answer:
A. struct point ( double x, y)
B. struct point { double x, double y >;
C. struct point { double x; double y >
D. struct point { double x; double y; };
E. struct point { double x; double y; >

Errors:
A. struct point (double x, y)
Structure definition must be terminated with;. It is missing. Members must be enclosed in { }. So it may be written as
struct point { double x, y };

B. struct point {double x, double y>;
Structure member declaration must be terminated with; It is missing. So, it may corrected as
struct point {double x; double y;>; or struct point {double x, y;>;

C. struct point {double x; double y>
Structure member declaration and definition must be terminated with ; It is missing. So, it may corrected as
struct point {double x; double y;>; or
struct point { double x, y; >;

D. struct point {double x; double y;};,
No error .
E. struct point {double x; double y;>
– Structure definition must be terminated with ; it is missing. So, it may be corrected as
struct point {double x; double y;>;

Part – III

Short Answers

Question 1.
How will you pass a structure to a function ?
Answer:
A structure variable can be passed to a function in a similar way of passing any argument that is of built-in data type,

1. If the structure itself is an argument, then it is called “call by value”.
2. If the reference of the structure is passed as an argument then it is called, “call by reference”.

Call by value
When a structure is passed as argument to a function using call by value method,any change made to the contents of the structure variable inside the function to which it is passed do not affect the structure variable used as an argument.

Call by reference
In this method of passing the structures to functions, the address of a structure variable / object is passed to the function using address of(&) operator. So any change made to the contents of structure variable inside the function are reflected back to the calling function.

Question 2.
The following code sums up the total of all students name starting with ‘S’ and display it. Fill in the blanks with required statements.
Answer:
struct student {int exam no, lang, eng, phy, che, mat, esc, total; char name[15];>;
int main( )
{
student s[20];
for(int i=0;i<20;i++)
{
…………………. //accept student details
}
for(int i=0;i<20;i++)
{
……………. //check for name starts with’ letter “S”
…………….. // display the detail of the checked name
}
return 0;
}
MODIFIED PROGRAM
using namespace std;
#include<iostream>
struct student
{
int exam_.no,lang,eng,phy,che,mat,esc,total; char name[15];
};
int main( )
{
student s[20];
for(int i=0;i<20;i++)
{
//accept student details
cout<<“\nEnter student name”;
cin>>s[i].name;
cout<<“\nEnter student exam number”;
cin>>s[i].exam_no;
cout<<“\nEnter Languge Mark”;
cin>>s[i].lang;
cout<<“\nEnter Engjish Mark”;
cin>>s[i].eng;
cout<<“\nEnter Physics Mark”;
cin>>s[i].phy;
cout<<“\nEnter Chemistry Mark”;
cin>>s[i].che;
cout<<“\nEnter Maths Mark”;
cin>>s[i].mat;
cout<<“\nEnter Computer Science Mark”;
cin>>s[i].csc;
s[i].total = s[i].lang + s[i].eng + s[i]. phy+s[i].che+s[i].mat+s[i]. csc;
}
cout<<“\nDetails of student name starts with letter S “<<endl;
for(int i=0;i<20;i++)
{
//check for name starts with letter “S” if(s[i].name[0] == ‘S’)
{
// display the detail of the checked name

cout<<“\nName : “<<s[i].
name<<“Total Mark”<<s[i].
total <<endl;
}
}
return 0;
}
Output

Question 3.
What is called nested structure. Give example
Answer:
The structure declared- within another structure is called a nested structure.
Example:
struct Student
{
int age;
float height, weight;
struct dob
{
int date;
char month[4];
int year;
};
}mahesh;
The nested structures act as members of another structure and the members of the child structure can be accessed as parent structure name. Child structure name. Member name.
Example:
mahesh. dob. date
Here

• mahesh is a parent structure name.
• dob is a child structure name
• date is a member of child structure

Question 4.
Rewrite the following program after removing the syntactical error(s), if any.
Answer:
Underline each correction.
struct movie
{
charm_name[10];
charm-Lang[10];
float ticket cost =50;};
Movie;
void main( )
{
gets(m_name);
cin>>m-lang;
return 0; ,
}
MODIFIED PROGRAM
PROGRAM (USING DEV C++)
#include<iostream>
#include<conio.h>
#include<stdio.h>
struct movie
{
char m_name[10];
char m_lang[10];
float ticket_cost;
} Movie;
int main( )
{
std::cout<<“\nEnter Move! Name ;
gets(Movie.m_name);
std::cout<<:”\nEnter Movie Language ,:
std: :cin>>Movie.m -lang;
std::cout<<“\nEnter Ticket Cost: “;
std::cin>>Movie.ticket_cost;
std::cout<<“\nMovie name is : “<
std::cout< <“\nMovie Language is : “<<Movie.m-lang;
std::cout<<“\nTicket cost is : “<<Movie.ticket_cost;
}
Output

Question 5.
What ¡s the difference among the following two programs?
a) #include<iostream.h>
struct point
{
double x;
double y;
};
int main()
{
struct point test;
testx = .25; test.y = .75;
cout< <test.x< ctest.y;
return 0;
}
b) #include
struct .
{
double x;
double y;
} Point;
int main(void)
{
Point test={.25,.75>;
return 0;
}
Answer:
The method of structure initialisation is different.

Question 6.
How to access members of a structure? Give f example.
Answer:
Once the two objects of student structure type are declared, their members can be accessed directly. The syntax for that is using a dot (.) between the object name and the member name.
Syntax is :
Object name. Member

For example:
struct Student
{
longrollno;
int age;
float weight;
}balu, frank;
The elements of the structure Student can be accessed as follows: balu.rollno balu.age balu. weight frank, rollno frank.age frank.weight
balu.rollno
balu.age
balu.weight
frank.rollno
frank.age
frank.weight

Question 7.
Write the syntax and an example for structure.
Answer:
Structure is declared using the keyword ‘struct’. The syntax of creating a structure is given below.
struct structure_name
{
type member_name1;
type member_name2;
} reference_name;
An optional field reference_name can be used to declare objects of the structure type directly.

Example:
struct Student
{
longrollno;
int age;
float weight;
}balu, frank;

Question 8.
For the following structure, definition writes the user-defined function to accept data through the keyboard.
struct date
{
int dd,mm,yy
};
struct item
{
int item id;
char name[10];
float price;
date date_manif;
}
Answer:
USER DEFINED FUNCTION
item accept( )
{
item obj;
cout<<“\n Enter Item Id “;
cin> > obj. item id;
cout<<“\nEnter Name “;
cin>obj.name;
cout<<“\nEnter Price “;
cin>>obj.price;
cout<<“\nEnter Date: date, month and year
cin>>obj.date_manif.dd >> obj.date_manif. mm >> obj.date_manif.yy;
return(obj);
}

Question 9.
What is called an anonymous structure? Give an example.
Answer:
A structure without a name/tag is called anonymous structure, struct
{
long rollno;
int age;
float weight;
} student;
The student can be referred to as a reference name to the above structure and the elements can be accessed like
student, roll no
student.age and
student.weight.

Question 10.
Write a user-defined function to return the structure after accepting value through the keyboard. The structure definition is as follows:
struct Item{int item no;float price;};
Answer:
USER DEFINED FUNCTION
Item accept( )
{
Item obj; ,
cout<<‘AnEnter Item Number and Price “; cin>>obj.item_no>>obj, price.
return(obj);
}

Part-IV

Explain in Detail

Question 1.
Explain the array of structures with examples.
Answer:
A class may contain many students. So, the definition of structure for one student can also be extended to all the students. If the class has 20 students, then 20 individual structures are required. For this purpose, an array of structures can be used.
An array of structures is declared in the same way as declaring an array with built-in data types like int or char.
The following program reads the details of 20 students and prints the same.
#include<iostream>
using namespace std;
struct Student
{
int age;
float height, weight;
char name[30];
};
void main()
{
Student std[20];
int i;
cout<< ” Enter the details for 20 students”<<endl;
for(i=0;i<20;i++)
{
cout<< ” Enter the details of student”<<i+1<<endl;
cout<< ” Enter the age:”<<endl; cin>>std[i].age;
cout<< “Enter the height:”<<endl; cin>>std[i].height;
cout<< “Enter the weight:”<<endl; cin>>std[i].weight;
}
cout<< “The values entered for Age, height and weight are”<<endl;
for(i=0;i<20;i++)
cout<< “Student “<<i+l<< “\t”< <std[i]. age<< “\t”<<std[i].height<< “\t”<<std[i]. weight;
}
Output
Enter the details of 20 students
Enter the details for student1
Enter the age:
18
Enter the height:
160.5
Enter the weight:
46.5
Enter the details for student2
Enter the age:
18
Enter the height:
164.5
Enter the weight:
61.5
………………………………….
…………………………
The values entered for Age, height and weight are
Student 1 18 160.5 46.5
Student 2 18 164.5 61.5
………………………

Question 2.
Explain call by value with respect to structure.
Answer:
A structure variable can be passed to a function in a similar way of passing any argument that is of built-in data type. If the structure itself is an argument, then it is called “call by value”.

Call by value
When a structure is passed as argument to a function using call by value method,any change made to the contents of the structure variable inside the function to which it is passed do not affect the structure variable used as an argument.
Example:
#include<iostream>
using namespace std;
struct Employee
{
char name[50]; ,
int age;
float salary;
};
void printData(Employee); // Function declaration
int main( )
{
Employee p;
cout<< “Enter Full name: “;
cin>>p.name;
cout<< “Enter age: “;
cin>>p.age;
cout<< “Enter salary: “;
cin>>p.salary;
// Function call with structure variable as an argument
printData(p);
return 0;
}
void printData(Employee q)
{
cout<< “\nDisplaying Information.” <<endl;
cout<< “Name: ” << q.name <<endl;
cout<<“Age: ” <<q.age<<endl;
cout<< “Salary:” <<q.salary;
}
Output
Enter Full name: Kumar
Enter age: 55
Enter salary: 34233.4
Displaying Information.
Name: Kumar
Age: 55
Salary: 34233.4
In the above example, a structure named Employee is declared and used. The values that are entered into the structure are name, age and salary of a Employee are displayed using a function named printData( ).

Question 3.
How call by reference is used to pass structure to a function .Give an Example
Answer:
Call by reference
In this method of passing the structures to functions ,the address of a structure variable / object is passed to the function using address of(&) operator. So any change made to the contents of structure variable inside the function are reflected back to the calling function
Example:
#include<iostream>
using namespace std;
struct Employee
{
char name[50];
int age;
float salary;
};
void readData(Employee &);
void printData(Employee);
int main()
{
Employee p;
readData(p);
printData(p);
return 0;
}
void readData(Employee &p)
{
cout<< “Enter Full name: “;
cin.get(p.name, 50);
cout<< “Enter age:
cin>>p.age; ,
cout<< “Enter salary:
cin>>p. salary;
}
void printData(Employee p)
{
cout<< “\nDisplaying Information.” <<endl;
cout<< “Name: ” << p.name <<endl;
cout<<“Age:” <<p.age<<endl;
cout<< “Salary:” <<p.salary;
}
Output
Enter Full name: Kumar
Enter age: 55
Enter salary: 34233.4
Displaying Information.
Name: Kumar
Age: 55
Salary: 34233.4
Structures are usually passed by reference method because it saves the memory space and executes faster.

Question 4.
Write a C++ program to add two distances using the following structure definition
Answer:
struct Distance
{
int feet;
float inch;
}d1, d2, sum;
PROGRAM
#include<iostream>
using namespace std;
struct Distance
{
int feet;
float inch;
} d1, d2, sum;
int main( )
{
cout<<“\nEnter distance 1: feet and inch : “;
cin>>d1.feet>>d1.inch;
cout<<“\nEnter distance 2: feet and inch :
cin>>d2.feet>>d2.inch;
sum.feet = d1.feet + d2.feet;
sum.inch =(d1.inch + d2.inch) – (int) (d1.inch + d2,inch) / 12 * 12;
sum.feet = sum.feet + (d1.inch + d2.inch)/12;
cout<<“\nDitance 1: Feet = “<<d1.feet<<”
Inch = “<<d1.inch;
cout<<“\nDitance 2: Feet = “<<d2.feet<<”
Inch = “<<d2.inch;
cout<<“\nSum of Distance 1 and 2 : Feet = “<<sum.feet<<” Inch = “<<sum.inch;
}
Output

Question 5.
Write a C++ Program to Add two Complex Numbers by Passing Structure to a Function for the following structure definition
Answer:
struct complex
{
float real;
float imag;
};
The prototype of the function is complex add Complex Numbers(complex, complex);
PROGRAM
#include<iostream>
using namespace std;
struct complex
{
float real;
float imag;
};
complex add Complex Numbers (complex obj1,complex obj2)
{
complex obj3;
obj3.real = obj1.reai+obj2,real;
obj3.imag = obj1.imag + obj2.imag;
return(obj3);
}
int main( )
{
complex c1,c2,c3;
cout<<“\nEnter First complex number real and imaginary :”;
cin>>c1,real>>c1.imag;
cout<<“\nEnter Second complex number real and imaginary :”;
cin>>c2.real>>c2.imag;
c3 = addComplexNumbers(c1,c2);
cout<<“\nFirst Complex Number is : “< }
Output

Question 6.
WrIte a C++ Program to declare a structure book containing name and author as character array of 20 elements each and price as Integer. Declare an array of book. Accept the name, author, price detail for each book. Define a user defined function to display the book details and calculate the total price. Return total price to the calling function.
Answer:
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
#include<istdio.h>
struct book
{
char bname[20],author[20];
int price;
};
int display(book x[5])
{
int total;
for(int i=0;i<5;i++)
{
total = total + x[i].price;
cout<<“\nBook name : “<<x[i].
bname;
cout<<“\nAuthor name : “<<x[i].
author;
cout<<“\nPrice of the book: “<<x[i].
price;
}
return(total);C++C
}
int main( )
{
book. b[5];
for(int i=0;i<5;i++)
{
cout<<“\nEnter Book name “;
cin>>b[i].bname;
cout<<“\nEnter Author name “;
cin>>b[i].author;
cout<<“\nEnter Book Price “;
cin > > b[i]. price;
}
cout<<“\nDetails of Books”<<endl;
cout<<“\nTotal cost of the book is : Rs.”<<display(b);
}
Output

Question 7.
Write a C++ program to declare and accept an array of professors. Display the details of the department=”COMP.SCI” and the name of the professors start with ‘A’. The structure “college” should contain the following members.
prof-id as integer ,
name and Department as character array
Answer:
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
#include<string.h>
struct college .
{
char pname[20],department[20]; int prof-id;
};
void display(college x[5])
{
for(int i=0;i<5;i++)
{
if((strcmp(x[i]. department,”COMP. SCI”)==0) && x[i].pname[0]==’A’)
cout<<x[i].pname«endl;
}
return;
}
int main( )
{
college c[5];
for(int i=0;i <5;i++)
{
cout<<“\nEnter Professor Name “;
cin>>c[i].pname; ,
cout<<“\nEnter Department Name “;
cin> >c[i] .department;
cout< <“\nEnter Professor Id
cin>>c[i].prof-id;
}
cout<<“\nDetails of Comp. Sci. . Dept, Professors whose name starts. with A “<<endl; . display(c);
}
Output

Question 8.
Write the output of the following C++ program
Answer:
#incluçle<iostream>
#include<stdio>
#include<string>
#include<conio>
using namespace std;
struct books
{
char name[20), author[20];
} a[50];
int main ( )
{
clrscr( );
cout<< “Details of Book No ” << 1 << “\n”;
cout<< “———————\n”;
cout<< ‘Book Name :”<<strcpy(a[0].
name,”Programming “)<<endl;
cout<< “Book Author :”<<strcpy(a[0].
author,”Dromy”)<<endl;
cout<< “\nDetails of Book No ” << 2 <<”\n”;
cout<< ” ——————- \n”;
cout<<- “Book Name :”<<strcpy(a[l].
name,”C++programming” )<<endl;
cout<< “Book Author :”<<strcpy(a[l].
author,”BjarneStroustrup “)< <endl;
cout<<“\n\n”;
cout<< “======================
============================
\n”; ‘
cout<< ” S.No\t| Book Name\t|author\n”;
cout<< “=======================
=============================
for (int i = 0; i < 2; i++)
{
cout<< “\n ” << i + 1 << “\t|” << a[i].name << “\t| ” << a[i],author;
}
cout<< “\n=========-==========
===========================
==”;
return 0;
}
Output

Question 9.
Write the output of the following C++ program
Answer:
#include<iostream>
#include<string>
using namespace std;
struct student
{
introll_no;
char name[10];
long phone_number;
};
int main() .
{
student p1 = {1,”Brown”,123443};
student p2, p3;
p2.roll_no = 2;
strcpy(p2.name,”Sam”);
p2.phone_number = 1234567822;
p3.roll_no = 3;
strcpy(p3.name,”Addy”);
p3.phone_number = 1234567844;
cout<< “First Student” <<endl;
cout<< “roll no : ” << p1.roll_no <<endl;
cout<< “name : ” << p1.name <<endl;
cout<< “phone no ; ” << p1.phone_number <<endl;
cout<< “Second Student” <<endl;
cout<< “roll no : ” << p2.roll_no <<endl;
cout<< “name : ” << p2.name <<endl;
cout<< “phone no : ” << p2.phone_number <<endl;
cout<< “Third Student” <<endl;
cout<< “roll no : ” << p3.roll_no <<endl;
cout<< “name : ” << p3.name <<endl;
cout<< “phone no : ” << p3.phone_number <<endl;
return0;
}
Output

Question 10.
Debug the error in the following program.
Answer:



MODIFIED PROGRAM
#include<iostream.h>
struct PersonRec
{
char lastName[10];
char firstName[10];
int age;
}
Person Rec PeopleArrayType[ 10];
void LoadArray(PeopleRec peop[10]);
void main( )
{
PersonRec people [10];
for (i = 0; i < 10; i++)
{
cout< < people[i].firstName< < ” <<peopie[i].lastName <<setw(10) <<people[i].age;
}
}
LoadArray(PersonRec peop[10])
{
for (int i = 0; i < 10; i++)
{
cout<< “Enter first name: “;
cin<<peop[i].firstName;
cout<< “Enter last name: “;
cin>>peop[i].lastName;
cout<< “Enter age: “;
cin>> peop[i].age;←
}

11th Computer Science Guide Arrays and Structures Additional Questions and Answers

Choose The Correct Answer 1 mark

Question 1.
…………….. is a user-defined which has the combination of data items with different data types.
a) Structure
b) Class
c) Union
d) Array
Answer:
a) Structure

Question 2.
…………………. allows to group of variables of mixed data types together into a single unit.
a) Structure
b) Class
c) Union
d) Array
Answer:
a) Structure

Question 3.
If the elements are of different data types,then ……………… cannot support.
a) Structure
b) Class
c) Union
d) Array
Answer:
d) Array

Question 4.
Structure is declared using the keyword …………………
a) Structure
b) struct
c) Stru
d) STRUCT
Answer:
b) struct

Question 5.
Identify the structure name from the following; struct Student balu;
a) balu
b) struct
c) Student
d) None of these
Answer:
c) Student

Question 6.
Identify the structure variable (object) from the following: struct Student balu;
a) balu
b) struct
c) Student
d) None of these
Answer:
a) balu

Question 7.
How much of memory is required for the object of the following structure in Dev C++.
struct Student
{
long rollno;
int age;
float weight;
}balu, frank;
a) 12 Bytes
b) 8 Bytes
c) 10 Bytes
d) None of these
Answer:
a) 12 Bytes

Question 8.
How much of memory is required for the object of the following structure in Turbo C++.
struct Student
{
long rollno;
int age;
float weight;
}balu, frank;
a) 12 Bytes
b) 8 Bytes
c) 10 Bytes
d) None of these
Answer:
c) 10 Bytes

Question 9.
If the members are pointer types then ……………….. is used to access the structure members.
a) →
b) ←
c) .(dot)
d) None of these
Answer:
a) →

Question 10.
A structure without a …………….. is called an anonymous structure.
a) Name
b) Tag
c) Name/Tag
d) None of these
Answer:
c) Name/Tag

Question 11.
…………………… operator is used to accessing structure member.
a) ::
b) @
c) .(dot)
d) $
Answer:
c) .(dot)

Question 12.
The structure declared within another structure is called a …………….. structure.
a) Nested
b) Group
c) Block
d) None of these
Answer:
a) Nested

Question 13.
……………. structures act as members of another structure.
a) Nested
b) Group
c) Block
d) None of these
Answer:
a) Nested

Question 14.
The members of the child structure can be accessed as …………………….
a) Child structure name. Parent structure name. Member name
a) Parent structure name. Child structure name. Member name
a) Parent structure name. Member name. Child structure name
a) Member name. Parent structure name. Child structure name
Answer:
a) Child structure name. Parent structure name. Member name

Question 15.
A structure variable can be passed to a function in a similar way of passing any argument that is of ……………………. data type.
a) Derived
b) User-defined
c) Built-in
d) None of these
Answer:
c) Built-in

Question 16.
If the structure itself is an argument to a function, then it is called ………………..
a) Call by value
b) Call by Reference
c) Call by Expression
d) None of these
Answer:
a) Call by value

Question 17.
If the reference of the structure is passed as an argument then it is called ……………….
a) Call by value
b) Call by Reference
c) Call by Expression
d) None of these
Answer:
b) Call by Reference

Question 18.
When a structure is passed as an argument to a function using ……………… method, any change made to the contents of the structure variable inside the function to which it is passed do not affect the structure variable used as an argument.
a) Call by value
b) Call by Reference
c) Call by Expression
d) None of these
Answer:
a) Call by value

Question 19.
In ……………….method of passing the structures to functions, any change made to the contents of structure variable inside the function are reflected back to the calling function.
a) Call by value
b) Call by Reference
c) Call by Expression
d) None of these
Answer:
b) Call by Reference

Question 20.
Structures are usually passed by reference method because ……………..
a) It saves the memory space
b) Executes faster
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 21.
Identify the correct statement from the following.
a) A structured object can also be assigned to another structure object only if both the objects are of the same structure type.
b) The structure elements can be initialized either by using separate assignment statements or at the time of declaration by surrounding its values with braces.
c) Array of structure variable can also be created.
d) All the above
Answer:
d) All the above

Very Short Answers (2 Marks)

Question 1.
What is the drawback of an array?
Answer:
In any situation when more than one variable is required to represent objects of uniform data-types, array can be used. If the elements are of different data types, then array cannot support.
If more than one variable is used, they can be stored in memory but not in adjacent locations. It increases the time consumption while searching.

Question 2.
What is the advantage of structure?
Answer:
The structure provides a facility to store different data types as a part of the same logical element in one memory chunk adjacent to each other.

Question 3.
What are global objects?
Answer:
Objects declared along with structure definition are called global objects

Question 4.
Write note on the anonymous structure.
Answer:
A structure without a name/tag is called an anonymous structure.

Example:
struct
{
long rollno;
int age;
float weight;
} student;
The student can be referred to as a reference name to the above structure and the elements can be accessed like a student, roll no, student.age and student.weight

Question 5.
Define Nested structure.
Answer:
The structure declared within another structure is called a nested structure.

Question 6.
Give an example of nested structure.
Answer:
struct Student
{
int age;
float height, weight;
struct dob
{
int date;
char month[4];
int year;
};
}mahesh;

Question 7.
How to access nested member structure?
Answer:
The nested structures act as members of another structure and the members of the child structure can be accessed as
parent structure name. Child structure name. Member name
Example:
mahesh. dob. date

Short Answers 3 Marks

Question 1.
Write the syntax of defining a structure? Give an example.
Answer:
Structure is declared using the keyword ‘struct’. The syntax of creating a structure is given below.
struct structure_name
{
type member-name1;
type member_name2;
} reference_name;
An optional field reference_name can be used to declare objects of the structure type directly.
Example:
struct Student
{
long rollno;
int age;
float weight;
};

Question 2.
How memory is allocated for a structure?
Answer:
Consider the following structure :
struct Student
{
long rollno;
int age;
float weight;
};
In the above declaration of the struct, three variables rollno, age and weight are used. These variables (data element) within the structure are called members (or fields). In order to use the Student structure, a variable of type Student is declared and the memory allocation is shown in the following figure.

Question 3.
How will you refer structure elements?
Answer:
Referencing Structure Elements
Once the two objects of student structure type are declared, their members can be accessed directly.
The syntax for that is using a dot (.) between the object name and the member name. For example, the elements of the structure Student can be accessed as follows:
balu.rollno
frank, rollno
balu.age
frank.age
balu.weight
frank.weight

Explain In Detail 5 Marks

Question 1.
Explain structure assignments in detail.
Answer:
Structure Assignments
Structures can be assigned directly instead of assigning the values of elements individually.
Example:
If Mahesh and Praveen are same age and same height and weight then the values of Mahesh can be copied to Praveen struct Student
{
int age;
float height, weight;
}mahesh;
The age of Mahesh is 17 and the height and weights are 164.5 and 52.5 respectively.The following statement will perform the assignment.
mahesh = {17, 164.5, 52.5};
praveen = mahesh;
will assign the same age, height and weight to Praveen.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 2 Number Systems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 2 Number Systems

11th Computer Science Guide Number Systems Text Book Questions and Answers

Part – I

I. Choose The Correct Answer:

Question 1.
Which refers to the number of bits processed by a computer’s CPU? .
a) Byte
b) Nibble
c) Word length
d) Bit
Answer:
c) Word length

Question 2.
How many bytes does 1 KiloByte contain?
a) 1000
b) 8
c) 4
d) 1024
Answer:
d) 1024

Question 3.
Expansion for ASCII
a) American School Code for Information Interchange
b) American Standard Code for Information Interchange
c) All Standard Code for Information Interchange
d) American Society Code for Information Interchange
Answer:
b) American Standard Code for Information Interchange

Question 4.
2^50 is referred as
a) Kilo
b) Tera
c) Peta
d) Zetta
Answer:
c) Peta

Question 5.
How many characters can be handled in Binary Coded Decimal System?.
a) 64
b) 255
c) 256
d) 128
Answer:
a) 64

Question 6.
For 11012 whai is the Hexadecimal equivalent?
a) F
b) E
c) D
d) B
Answer:
c) D

Question 7.
What is the 1’s complement of 00100110?
a) 00100110
b) 11011001
c) 11010001
d) 00101001
Answer:
b) 11011001

Question 8.
Which amongst this is not an Octal number?
a) 645
b) 234
c) 876
d) 123
Answer:
c) 876

Part II

Very Short Answers.

Question 1.
What is data?
Answer:
The term data comes from the word datum which means a raw fact. The data is a fact about people, places, or some objects.
Example: Rajesh, 16, XI.

Question 2.
Write the l’s complement procedure.
Answer:
The steps to be followed to find l’s complement of a number:
Step 1: Convert given Decimal number into Binary
Step 2: Check if the binary number contains 8 bits, if less add O at the left most bit, to make it as 8 bits.
Step 3: Invert all bits (i.e. Change 1 as 0 and 0 as 1)

Example: Find 1’s complement for (-24)₁₀

Question 3.
Convert (46)₁₀ into a Binary number.
Answer:

Question 4.
We cannot find l’s complement for (28)₁₀ State reason.
Answer:
Since it is a positive number. 1’s complement will come only for negative numbers.

Question 5.
List the encoding systems for characters in memory.
Answer:
There are several encoding systems used for computers. They are

• BCD – Binary Coded Decimal
• EBCDIC – Extended Binary Coded Decimal Interchange Code
• ASCII – American Standard Code for Information Interchange
• Unicode
• ISCII – Indian Standard Code for Information Interchange

Part III

III. Very Short Answers

Question 1.
What is radix of a number system? Give example.
Answer:
Each number system Is uniquely Identified by Its base value or radix. Radix or base Is the count of number of digits In each number system. Radix or base is the general Idea behind positional numbering system. Ex.

Number system	Base / Radix
Binary	2
Octal	8
Decimal	10
Hexadecimal	16

Question 2.
Write a note on the binary number system.
Answer:
There are only two digits in the Binary system, namely, 0 and 1. The numbers in the binary system are represented to the base 2 and the positional multipliers are the powers of 2.

The leftmost bit in the binary number is called the Most Significant Bit (MSB) and it has the largest positional weight. The rightmost bit is the Least Significant Bit (LSB) and has the smallest positional weight.

Question 3.
Convert (150)₁₀ into Binary, then convert that Binary number to Octal.
Answer:
Decimal to Binary conversion 150

Binary to octal conversion
LSB to MSB divide the number into three-digit binary and write the equivalent octal digit

Question 4.
Write a short note on ISCII.
Answer:
ISCII – Indian Standard Code for Information Interchange (ISCII) is the system of handling the character of Indian local languages. This is an 8 – bit coding system. Therefore it can handle 256 (2⁸) characters. It is recognized by the Bureau of Indian Standards (BIS). It is integrated with Unicode.

This system is formulated by the Department of Electronics in India in the year 1986-88 and recognized by the Bureau of Indian Standards (BIS). Now, this coding system is integrated with Unicode.

Question 5.
Add : a) -22₁₀ + 15₁₀ b) 20₁₀ + 25₁₀.
Answer:
a) -22₁₀ + 15₁₀

Answer in 2’s complement form . 11111001 is 2’s complement of 7 which is the answer.

Part IV

IV. Detail Answers.

Question 1.
a) Write the procedure to convert fractional Decimal to Binary.
Answer:
Conversion of fractional Decimal to Binary
The method of repeated multiplication by 2 has to be used to convert such kinds of decimal fractions.

The steps involved in the method of repeated multiplication by 2:

Step 1. : Multiply the decimal fraction by 2 and note the integer part. The integer part is either 0 or 1.
Step 2: Discard the integer part of the previous product. Multiply the fractional part of the previous product by 2. Repeat Step 1 until the same fraction repeats or terminates (0).
Step 3: The resulting integer part forms a sequence of 0’s and 1’s that becomes the binary equivalent of a decimal fraction.
Step 4: The final answer is to be written from the first integer part obtained till the last integer part obtained.
Integer part (last integer part obtained)

Write the integer parts from top to bottom to obtain the equivalent fractional binary number.
Hence
(0.2)₁₀ = (0.00110011.,.)₂ = (0.00110011)₂.

b) Convert (98.46)₁₀ to Binary.
Convert (98.46)₁₀ to Binary
Procedure: Conversion of an integral part:

(98)₁₀ = (1100010)₂ Conversion of fractional part:

Question 2.
Find l’s Complement and 2’s Complement for the following Decimal number.
a) -98
b) -135
Answer:
a) Conversion of (98)₁₀ into binary

Conversion of (135)₁₀ into binary

Question 3.
a) Add 1101010₂ + 101101)₂
Answer:
a) Add 1101010₂ + 101101₂

b) Subtract 1101011₂ – 111010₂.
Subtract 1101011₂ – 111010₂

11th Computer Science Guide Number Systems Additional Questions and Answers

Part I

Choose The Correct Answer.

Question 1.
The simplest method to represent a negative binary number is called ………………..
(a) signed magnitude
(b) sign bit or parity bit
(c) binary
(d) decimal
Answer:
(a) signed magnitude

Question 2.
Computer understand ________________language.
a) High level
b) Assembly
c) Machine
d) All the above
Answer:
c) Machine

Question 3.
Expansion for BCD ………………..
(a) Binary coded decimal
(b) binary complement decimal
(c) binary computer decimal
(d) binary convert decimal
Answer:
(a) Binary coded decimal

Question 4.
__________is the basic unit of data in computer.
a) BIT
b) BYTE
c) NIBBLE
d) WORD
Answer:
a) BIT

Question 5.
The ……………….. operator is defined in boolean algebra by the use of the dot (.) operator.
(a) AND
(b) OR
(c) NOT
(d) NAND
Answer:
(a) AND

Question 6.
Binary digit means __________
a) 0
b) 1
c) either 0 or 1
d) None of these
Answer:
c) either 0 or 1

Question 7.
The convert (65)₁₀ into its equivalent octal number ………………..
(a) (101)₈
(b) (101)₁₀
(c) (101)₁₂
(d) (101)₄
Answer:
(a) (101)₈

Question 8.
A collection of 8 bits is called __________
a) BIT
b) BYTE
C) NIBBLE
d) WORD
Answer:
b) BYTE

Question 9.
……………….. is the general idea behind the positional numbering system.
(a) Radix
(b) Computer memory
(c) Binary number
(d) Decimal number
Answer:
(a) Radix

Question 10.
__________refers to the number of bits processed by a computer’s CPU.
a) Word length
b) Nibble
c) Word size
d) None of these
Answer:
a) Word length

Question 11.
Bit means ………………..
(a) nibble
(b) byte
(c) word length
(d) binary digit
Answer:
(d) binary digit

Question 12.
__________is a valid word length of a computer.
a) 64
b) 32
c) 16
d) All the above
Answer:
d) All the above

Question 13.
The computer can understand ……………….. languages.
(a) computer
(b) machine
(c) post
(d) pre
Answer:
(b) machine

Question 14.
1 KiloByte equals to __________bytes.
a) 1024
b) 256
c) 1000
d) 128
Answer:
a) 1024

Question 15.
How many bytes does 1 zettabyte contain?
(a) 2⁹⁰
(b) 2⁸⁰
(c) 2⁷⁰
(d) 2⁶⁰
Answer:
(c) 2⁷⁰

Question 16.
1024 MegaBytes equals to _________
a) 1 GigaByte
b) 1 TeraByte
c) 1 YottaByte
d) None of these
Answer:
a) 1 GigaByte

Question 18.
1-kilo byte represents ……………….. bytes.
(a) 512
(b) 256
(c) 1024
(d) 64
Answer:
(c) 1024

Question 18.
1Kb equals to _________bytes.
a) 2¹⁰
b) 2²⁰
c) 2³⁰
d) 2⁴⁰
Answer:
a) 2¹⁰

Question 19.
How many megabytes does 1 GB contain?
(a) 2²⁰
(b) 2¹⁰
(c) 2³⁰
(d) 2⁴⁰
Answer:
(b) 2¹⁰

Question 20.
1 GB equals to ________ bytes.
a) 2¹⁰
b) 2²⁰
c) 2³⁰
d) 2⁴⁰
Answer:
c) 2³⁰

Question 21.
What is the 1’ s complement of 11001?
(a) 11100110
(b) 01010101
(c) 11110000
(d) 100100111
Answer:
(a) 11100110

Question 22.
1 PetaByte(PB) equals to _________bytes.
a) 2⁵⁰
b) 2⁶⁰
c) 2⁷⁰
d) 2⁸⁰
Answer:
a) 2⁵⁰

Question 23.
The hexadecimal equivalent of 15 is ………………..
(a) A
(b) B
(c) E
(d) F
Answer:
(d) F

Question 24.
1 ZettaByte (1ZB) equals to _______ bytes.
a) 2⁵⁰
b) 2⁶⁰
c) 2⁷⁰
d) 2⁸⁰
Answer:
c) 2⁷⁰

Question 25.
The radix of a hexadecimal number is ………………..
(a) 2
(b) 8
(c) 16
(d) 10
Answer:
(c) 16

Question 26.
Computer memory is normally represented in terms of ________ bytes.
a) Kilo
b) Mega
c) Kilo or Mega
d) None of these
Answer:
c) Kilo or Mega

Question 27.
The most commonly used number system is ………………..
(a) binary
(b) decimal
(c) octal
(d) hexadecimal
Answer:
(b) decimal

Question 28.
The most commonly used coding scheme to represent character set and the number is ________
a) BCD
b) ASCII
c) EBCDIC
d) All the above
Answer:
b) ASCII

Question 29.
What does MSB mean?
(a) Major sign bit
(b) Most sign bit
(c) Minor sign bit
(d) Most significant bit
Answer:
(d) Most significant bit

Question 30.
The ASCII value for blank space is _________
a) 43
b) 42
c) 32
d) 62
Answer:
c) 32

Question 31.
The binary equivalent of hexadecimal number B is ………………..
(a) 1011
(b) 1100
(c) 1001
(d) 1010
Answer:
(a) 1011

Question 32.
The most commonly used numbering system in real life is the _________number system.
a) Hexadecimal
b) Octal
c) Binary
d) Decimal
Answer:
d) Decimal

Question 33.
What is the range of ASCII values for lower case alphabets?
(a) 65 to 90
(b) 65 to 122
(c) 97 to 122
(d) 98 to 122
Answer:
(c) 97 to 122

Question 34.
_________is the count of number of digits in each number system.
a) base
b) radix
c) base or radix
d) symbols
Answer:
c) base or radix

Question 35.
What is the ASCII value for blank space?
(a) 8
(b) 2
(c) 18
(d) 32
Answer:
(d) 32

Question 36.
Identify the true statement from the following.
a) In the positional number system, each decimal digit is weighted relative to its position in the number.
b) A numbering system is a way of representing numbers.
c) The speed of a computer depends on the number of bits it can process at once.
d) All the above
Answer:
d) All the above

Question 37.
Which one of the following bits has the smallest positional weight?
(a) MSB
(b) LSB
(c) UPS
(d) USB
Answer:
(b) LSB

Question 38.
The rightmost bit in the binary number is called as the __________
a) MSB
b) LSB
c) FSB
d) None of these
Answer:
b) LSB

Question 39.
Name the person who proposed the basic principles of Boolean Algebra?
(a) Wiliam Boole
(b) George Boole
(c) James Boole
(d) Boolean George
Answer:
(b) George Boole

Question 40.
_______ numbers are used as a shorthand form of a binary sequence.
a) Hexadecimal
b) Octal
c) Decimal
d) None of these
Answer:
a) Hexadecimal

Question 41.
What is the other name for a logical statement?
(a) Truth values
(b) Truth functions
(c) Truth table
(d) Truth variables
Answer:
(b) Truth functions

Question 42.
In hexadecimal number system letter ‘E’ represents _______
a) 12
b) 13
c) 14
d) 15
Answer:
c) 14

Question 43.
The NOT operator is represented by the symbol.
(a) over bar
(b) single apostrophe
(c) a and b
(d) plus
Answer:
(c) a and b

Question 44.
_______is a method to convert decimal number to binary number.
a) Repeated division by 2
b) Sum of powers of 2
c) Repeated addition by 2
d) Either A or B
Answer:
d) Either A or B

Question 45.
The output for the AND operator is ………………..
(a) A + B
(b) –
(c) A.B
(d) AB + C
Answer:
(c) A.B

Question 46.
Computer can handle _______ numbers.
a) signed
b) unsigned
c) signed and unsigned
d) None of these
Answer:
c) signed and unsigned

Question 47.
Which gate takes only one input?
(a) OR
(b) AND
(c) NOT
(d) XOR
Answer:
(c) NOT

Question 48.
In the signed magnitude method, the leftmost bit is called _______bit.
a) sign
b) parity
c) sign or parity
d) None of these
Answer:
c) sign or parity

Question 49.
Which is not a derived date?
(a) AND
(b) NAND
(c) NOR
(d) XOR
Answer:
(a) AND

Question 50.
The numbers are represented in computers in _______method.
a) Signed magnitude representation
b) 1’s complement
c) 2’s complement
d) All the above
Answer:
d) All the above

Question 51.
The statement “C equal the complement of A or B” means
(a) C = A + B
(b) C = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
(c) C = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)
(d) C = \(\overline{\mathrm{A}\mathrm{B}}\)
Answer:
(a) C = A + B

Question 52.
If the number has_______sign, it will be considered as negative in signed magnitude representation.
a) +
b) no
c) –
d) A or B
Answer:
c) –

Question 54.
What is the output of the XOR gate?
(a) C = A% B
(b) C = A \(\otimes\) A
(c) C = A \(\odot\) B
(d) C = A \(\oplus\) B
Answer:
(d) C = A \(\oplus\) B

Question 54.
2’s complement of (0001i000)2 is_______
a) 11100111
b) 00011001
c) 11101000
d) None of these
Answer:
c) 11101000

Question 55.
Find A + \(\overline{\mathrm{A}}\) .B = ………………..
(a) A + B
(b) A.B
(c) \(\overline{\mathrm{A}}\).B
(d) A.\(\overline{\mathrm{B}}\)
Answer:
(d) A.\(\overline{\mathrm{B}}\)

Question 56.
When two binary numbers are added _______will be the output.
a) sum
b) carry
c) sum and carry
d) None of these
Answer:
c) sum and carry

Question 57.
When subtracting 1 from 0, borrow 1 from the next _______
a) LSB
b) MSB
c) either A or B
d) None of these
Answer:
b) MSB

Question 58.
Find the wrong pair from the following:
(a) Null element : A + 1 = 1
(b) Involution : \(\overset { = }{ A }\) = A
(c) Demorgan’s : \(\overline{\mathrm{A+B}}\) =\(\overline{\mathrm{A}}\) . \(\overline{\mathrm{A}}\)
(d) Commutative : A + B = B . A
Answer:
(d) Commutative : A + B = B . A

Question 59.
_______ is the character encoding system.
a) BCD and ISCII
b) EBCDIC
c) ASCII and Unicode
d) All the above
Answer:
d) All the above

Question 60.
With 2 inputs in the truth table, how many sets of values will be obtained.
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(a) 4

Question 61.
EBCDIC stands for _______
a) Extensive Binary Coded Decimal Interchange Code
b) Extended Binary Coded Decimal Interchange Code
c) Extended Binary Coded Digit Interchange Code
d) Extended Bit Coded Decimal Interchange Code.
Answer:
b) Extended Binary Coded Decimal Interchange Code

Question 62.
ASCII stands for _______
a) Arithmetic Standard Code for Information Interchange
b) American Structured Code for Information Interchange
c) American Standard Code for Information Interchange
d) American Standard Code for Instant Interchange
Answer:
c) American Standard Code for Information Interchange

Question 63.
ISCII stands for_______
a) International Standard Code for Information Interchange
b) Indian Structured Code for Information Interchange
c) India’s Standard Code for Information Interchange
d) Indian Standard Code for Information Interchange
Answer:
d) Indian Standard Code for Information Interchange

Question 64.
BCD is _______bit code.
a) 6
b) 7
c) 8
d) None of these
Answer:
a) 6

Question 65.
EBCDIC is_______ bit code
a) 6
b) 7
c) 8
d) None of these
Answer:
c) 8

Question 66.
ASCII is________ bit code
a) 6
b) 7
c) 8
d) None of these
Answer:
b) 7

Question 67.
Unicode is _______ bit code
a) 16
b) 7
c) 8
d) None of these
Answer:
a) 16

Question 68.
ISCII is_______ bit code
a) 16
b) 7
c) 8
d) None of these
Answer:
c) 8

Question 69.
_______coding system is formulated by IBM.
a) BCD
b) EBCDIC
c) ISCII
d) None of these
Answer:
b) EBCDIC

Question 70.
IBM stands for_______
a) Indian Business Machine
b) International Basic Machine
c) International Business Method
d) International Business Machine
Answer:
d) International Business Machine

Question 71.
_______is the system of handling the characters of Indian local languages.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
d) ISCII

Question 72.
ISCII system is formulated fay the _______ in India.
a) Department of Electronics
b) Department of Electricity
c) Department of E-commerce
d) Department of Economics
Answer:
a) Department of Electronics

Question 73.
SCO system can handle___________characterscharacters.
a) 64
b) 128
c) 256
d) 65536
Answer:
a) 64

Question 74.
EBCDIC system can handle _______ characters.
a) 64
b) 128
c) 256
d) None of these
Answer:
c) 256

Question 75.
ASCII system can handle _______characters.
a) 64
b) 128
c) 256
d) None of these
Answer:
c) 256

Question 76.
Unicode system can handle _______ characters.
a) 64
b) 128
c) 256
d) 65536
Answer:
d) 65536

Question 77.
ISCII system can handle _______ characters.
a) 64
b) 128
c) 256
d) 65535
Answer:
c) 256

Question 78.
__________ language characters are not represented by ASCII.
a) Tamil
b) Malayalam
c) Telugu and Kannada
d) All the above
Answer:
d) All the above

Question 79.
Tamil, Malayalam, Telugu, and Kannada language characters are represented by _______ code.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 80.
_______scheme is denoted by hexadecimal numbers
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 81.
ISCII code was formulated in the year_______
a) 1986 – 88
b) 1984 – 86
c) 1988
d) 1987
Answer:
a) 1986 – 88

Question 82.
_______coding system is integrated with Unicode.
a) ASCII
b) EBCDIC
c) BCD
d) ISCII
Answer:
d) ISCII

Question 83.
_______was generated to handle all the coding system of Universal languages.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 84.
The popular coding scheme after ASCII is_______
a) EBCDIC
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 85.
BCD system is_______bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁴
Answer:
c) 2⁶

Question 86.
EBCDIC system is _______bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁷
Answer:
a) 2⁸

Question 87.
ASCII system is a bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁷
Answer:
d) 2⁷

Question 88.
Unicode system is _________bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁷
Answer:
b) 2¹⁶

Question 89.
ISCII svstem is _________bit encoding system.
a) 2⁸
b) 2¹⁶
c) 2⁶
d) 2⁷
Answer:
a) 2⁸

Question 90.
The input code in ASCII can be converted into _________system.
a) EBCDIC
b) Unicode
c) BCD
d) ISCII
Answer:
a) EBCDIC

Question 91.
What is ASCII value for ‘A’ in a decimal number,
a) 97
b) 65
c) 98
d) 32
Answer:
b) 65

Question 92.
What is the ASCII value for ‘A’ in a binary number?
a) 01100001
b) 01000001
c) 01100010
d) 00100000
Answer:
b) 01000001

Question 93.
What is the ASCII value for ‘A’ in an octal number?
a) 141
b) 101
c) 142
d) 40
Answer:
b) 101

Question 94.
What is the ASCII value for ‘A’ in hexadecimal numbers?
a) 61
b) 41
c) 62
d) 20
Answer:
b) 41

Question 95.
Find the false statement in the following.
a) Computers can handle positive and negative numbers.
b) MSB is called a sign bit
c) LSB is called a parity bit
d) All the above
Answer:
c) LSB is called a parity bit

Question 96.
Match the following.
a) 78 – (1) Binary number
b) linn – (2) Octal number
c) CAFE – (3) Decimal number
d) 71 – (4) Hexadecimal number

a) 3, 1, 4, 2
b) 4, 3, 2, 1
c) 1, 3, 2, 4
d) 3, 1, 2, 4
Answer:
a) 78 – (1) Binary number

Question 97.
In signed magnitude representation,_________ in the sign bit represents negative number.
a) 0
b) 1
c) No symbol
d) None of these
Answer:
b) 1

Question 98.
In signed magnitude representation, __________in the sign bit represents positive number.
a) 0
b) 1
c) No symbol
d) None of these
Answer:
a) 0

Question 99.
The term data comes from the word __________
a) datum
b) date
c) fact
d) None of these
Answer:
a) datum

Part II

Very Short Answers.

Question 1.
What is nibble?
Answer:
Nibble is a collection of 4 bits. A nibble is half a byte.

Question 2.
Define information.
Answer:
Information is a processed fact and obtained from the computer as output. It conveys meaning.

Question 3.
What is radix?
Answer:
The base value of a number is also known as the radix.

Question 4.
Define Bit and Byte.
Answer:
Bit: A bit is the short form of a Binary digit which can be ‘0’ or ‘1’. It is the basic unit of data in computers.
Byte: A collection of 8 bits is called Byte. It is the basic unit of measuring the memory size in the computer.

Question 5.
Expand: BCD, EBCDIC, ASCII
Answer:
BCD – Binary Coded Decimal; EBCDIC – Extended Binary Coded Decimal Interchange Code; ASCII – American Standard Code for Information Interchange.

Question 6.
What are the different types of coding schemes to represent the character sets?
Answer:
The different coding schemes are

• BCD – Binary Coded Decimal
• EBCDIC – Extended Binary Coded Decimal Interchange Code
• ASCII – American Standard Code for Information Interchange
• Unicode
• ISCII – Indian Standard Code for Information Interchange.

Question 7.
What are the methods of converting a number from decimal to binary?
Answer:

1. Repeated division by two.
2. Sum of powers of 2.

Question 8.
What does base or radix mean?
Answer:
Radix or base is the count of a number of digits in each number system. Radix or base is the general idea behind the positional numbering system.

Question 9.
What are the various ways for Binary representation of signed numbers?
Answer:

1. Signed magnitude representation
2. 1’s complement
3. 2’s complement

Question 10.
Write a note on the decimal number system.
Answer:
It consists of 0,1,2,3,4,5,6,7,8,9(10 digits). It is the oldest and most popular number system used in our day-to-day life. In the positional number system, each decimal digit is weighted relative to its position in the number.
Its base or radix is 10.

Question 11.
Write about the octal number system.
Answer:
Octal number system uses digits 0,1,2,3,4,5,6 and 7 (8 digits). Each octal digit has its own positional value or weight as a power of 8. Its base or radix is 8.

Question 12.
How will you convert decimal to hexadecimal?
Answer:
To convert Decimal to Hexadecimal, “Repeated division by 16” method can be used) In this method, we have to divide the given number by 16.
Example: Convert (31)10 into its equivalent hexadecimal number.

Question 13.
Give the procedure to Octal to Binary.
Answer:
Procedure: For each octal digit in the given number write its 3 digits binary equivalent using positional notation.
Example: Convert (6213)₈ to equivalent Binary number.

Question 14.
How will you convert Hexadecimal to Binary?
Answer:
Procedure: Write 4 bits Binary equivalent for each Hexadecimal digit for the given number using the positional notation method.
Example:
Convert (8BC)₁₆ into an equivalent Binary number.

Question 15.
Write short note on Binary Coded Decimal (BCD).
Answer:
This is 2⁶ bit encoding system. This can handle 2⁶ = 64 characters only. This encoding system is not
in the practice right now.

Question 16.
Write note on EBCDIC encoding system.
Answer:
Extended Binary Coded Decimal Interchange Code (EBCDIC) is similar to ASCII Code with 8 bit representation. This coding system is formulated by International Business Machine (IBM). The coding system can handle 256 characters. The input code in ASCII can be converted to EBCDIC system and vice – versa.

Question 17.
Write a note on the ISCII encoding system.
Answer:
ISCII is the system of handling the character of Indian local languages. This is an 8-bit coding system. Therefore it can handle 256 (28) characters. This system is formulated by the Department of Electronics in India in the year 1986-88 and recognized by the Bureau of Indian Standards (BIS). Now, this coding system is integrated with Unicode.

Part III

III. Very Short Answers

Question 1.
Write about the binary number system.
Answer:
There are only two digits in the Binary system, namely, 0 and 1. The numbers in the binary system are represented to the base 2 and the positional multipliers are the powers of 2.
The leftmost bit in the binary number is called the Most Significant Bit (MSB) and it has the largest positional weight. The rightmost bit is the Least Significant Bit (LSB) and has the smallest positional weight.

Question 2.
What is the octal number system?
Answer:
The octal number system uses digits 0, 1, 2, 3, 4, 5, 6, and 7 (8 digits): Each octal digit has its own positional value or weight as a power of 8.
Example: The Octal sequence (547)₈ has the decimal equivalent:

Question 3.
Give the procedure to convert decimal to octal.
Answer:
To convert Decimal to Octal, “Repeated Division by 8” method can be used) In this method, we have to divide the given number by 8.
Example:
Convert (65)₁₀ into its equivalent Octal number.

Question 4.
Give the procedure to convert Octal to Decimal
Answer:
To convert Octal to Decimal, we can use positional notation method)

• Write down the Octal digits and list the powers of 8 from right to left (Positional Notation).
• For each positional notation of the digit write the equivalent weight.
• Multiply each digit with its corresponding weight.
• Add all the values.

Example:
Convert (1265)₈ to equivalent Decimal number.

(1265)₈ = 512 x 1 + 64 x 2 + 8 x 6 + 1 x 5
= 512 + 128 + 48 + 5
(1265)₈ = (693)₁₀

Question 5.
How will you convert Hexadecimal to Decimal?
Answer:
To convert Hexadecimal to Decimal we can use the positional notation method.

• Write down the Hexadecimal digits and list the powers of 16 from right to left (Positional Notation)
• For each positional notation written for the digit, now write the equivalent weight.
• Multiply each digit with its corresponding weight
• Add all the values to get one final value.

Example:
Convert (25F)₁₆ into its equivalent Decimal number.

(25F)₁₆ = 2 x 256 + 5 x 16 + 15 x 1
= 512 + 80 + 15 (25F)₁₆
= (607)₁₀

Question 6.
Write about binary representation for signed numbers.
Answer:
Computers can handle both positive (unsigned) and negative (signed) numbers. The simplest method to represent negative binary numbers is called Signed Magnitude. In signed magnitude method, the leftmost bit is the Most Significant Bit (MSB), which is called the sign bit or parity bit.
The numbers are represented in computers in different ways:

• Signed Magnitude representation
• 1’s Complement
• 2’s Complement

Question 7.
Explain ASCII code in detail.
Answer:
This is the most popular encoding system recognized by the United States. Most of the computers use this system. Remember this encoding system can handle English characters only. This can handle 27 bit which means 128 characters.

In this system, each character has an individual number. The new edition ASCII -8, has 28 bits and can handle 256 characters are represented from 0 to 255 unique numbers.

The ASCII code equivalent to the uppercase letter ‘A’ is 65. The binary representation of the ASCII (7 bit) value is 1000001. Also 01000001 in ASCII-8 bit.

Question 8.
Explain Unicode in detail.
Answer:
This coding system is used in most modern computers. The popular coding scheme after ASCII is Unicode. ASCII can represent only 256 characters. Therefore English and European Languages alone can be handled by ASCII. Particularly there was a situation when the languages like Tamil, Malayalam, Kannada, and Telugu could not be represented by ASCII.

Hence, Unicode was generated to handle all the coding system of Universal languages. This is a 16-bit code and can handle 65536 characters. The unicode scheme is denoted by hexadecimal numbers.

Part IV

IV. Detail Answers.

Question 1.
Explain decimal to binary conversion using Repeated Division by 2 methods.
Answer:
To convert Decimal to Binary “Repeated Division by 2” method can be used. Any Decimal number divided by 2 will leave a remainder of 0 or 1. Repeated division by 2 will leave a sequence of 0s and Is that become the binary equivalent of the decimal number.

Suppose it is required to convert the decimal number N into binary form, dividing N by 2 in the decimal system, we will obtain a quotient N1 and a remainder Rl, where R1 can have a value of either 0 or 1. The process is repeated until the quotient becomes 0 or 1. When the quotient is ‘0’ or ‘1’, it is the final remainder value. Write the final answer starting from the final remainder value obtained to the first remainder value obtained.

Example:
Convert (65)₁₀ into its equivalent binary number

Question 2.
Explain decimal to binary conversion using Sum of powers of 2 methods.
Answer:
A decimal number can be converted into a binary number by adding up the powers of 2 and then adding bits as needed to obtain the total value of the number.
a) Find the largest power of 2 that is smaller than or equal to 65.
6510 > 6410

b) Set the 64’s bit to 1 and subtract 64 from the original number
65 – 64 = 1

c) 32 is greater than the remaining total.
Therefore, set the 32’s bit to 0.

d) 16 is greater than the remaining total.
Therefore, set the 16’s bit to 0
.
e) 8 is greater than the remaining total.
Therefore, set the 8’s bit to 0.

f) 4 is greater than the remaining total.
Therefore, set the 4’s bit to 0.

g) 2 is greater than the remaining total.
Therefore, set the 2’s bit to 0.

h) As the remaining value is equivalent to l’s bit, set it to 1.
1 – 1 = 0
Conversion is complete 65₁₀ = (1000001)₂

Example:
The conversion steps can be given as follows:
Given Number: 65
Equivalent or value less than the power of 2 is: 64
(1) 65 – 64 = 1
(2) 1 – 1= 0

65₁₀ = (1000001)₂.

Question 3.
Explain the procedure to convert fractional decimal to Binary.
Answer:
The method of repeated multiplication by 2
has to be used to convert such kind of decimal fractions.
The steps involved in the method of repeated multiplication by 2:

Step 1: Multiply the decimal fraction by 2 and note the integer part. The integer part is either 0 or 1.
Step 2: Discard the integer part of the previous product. Multiply the fractional part of the previous product by 2. Repeat Step 1 until the same fraction repeats or terminates (0).
Step 3: The resulting integer part forms a sequence of Os and Is that becomes the binary equivalent of a decimal fraction.
Step 4: The final answer is to be written from the first integer part obtained till the last integer part obtained.

Write the integer parts from top to bottom to obtain the equivalent fractional binary number. Hence (0.2)₁₀= (0.00110011…)₂ = 0.00110011)₂

Question 4.
How will you convert Binary to Decimal?
Answer:
To convert Binary to Decimal we can use the positional notation method.
Step 1: Write down the Binary digits and list the powers of 2 from right to left (Positional Notation)
Step 2: For each positional notation written for the digit, now write the equivalent weight.
Step 3; Multiply each digit with its corresponding weight
Step 4: Add all the values.

Example:
Convert (111011)2 into its equivalent decimal number.

32 + 16 + 8 + 0 + 2 + 1 = (59)₁₀
(111011)₂ = (59)₁₀

Question 5.
How will you convert Binary to Octal?
Answer:
Step 1.: Group the given binary number into 3 bits from right to left.
Step 2: You can add preceding O to make a group of 3 bits if the leftmost group has less than 3 bits.
Step 3: Convert equivalent octal value using “2’s power positional weight method”

Example
Convert (11010110)₂ into an octal equivalent number

Step 1: Group the given number into 3 bits from right to left.
011 010 110
The left-most groups have less than 3 bits, so 0 is added to its left to make a group of 3 bits.

Step-2: Find the Octal equivalent of each group.

Question 6.
Give the procedure to convert Binary to Hexadecimal.
Answer:
Step 1: Group the given number into 4 bits from right to left.
Step 2: You can add preceding 0’s to make a group of 4 bits if the leftmost group has less than 4 bits.
Step 3: Convert equivalent Hexadecimal value using “2’s power positional weight method”.

Example
Convert (1111010110)₂ into Hexadecimal number
Step 1: Group the given number into 4 bits from right to left. 1
0011 1101 0110
0’s are added to the leftmost group to make it a group of 4 bits.

Question 7.
Give the procedure to convert fractional Binary to Decimal equivalent.
Answer:
The steps to convert fractional Binary number to its decimal equivalent:
Step 1 : Convert an integral part of Binary to Decimal equivalent using positional notation method.
Step 2 : To convert the fractional part of binary to its decimal equivalent.
Step 2,1 : Write down the Binary digits in the fractional part.
Step 2,2 : For all the digits write powers of 2 from left to right starting
from 2^-1, 2^-2, 2^-3 2^-n, now write the equivalent weight.
Step 2.3 : Multiply each digit with its corresponding weight.
Step 2.4 : Add all the values which you obtained in Step 2.3.

Step 3 : To get final answer write the integral
part (after conversion), followed by a decimal point(.) and the answer arrived at Step 2.4

Example:
Convert the given Binary number (11.011)₂ into its decimal equivalent Integer part (11)₂ = 3

Question 8.
Explain the method of representing signed binary numbers in the Signed Magnitude representation.
Answer:
The value of the whole numbers can be determined by the sign used before it. If the number has a ‘+’ sign or no sign it will be considered as positive. If the number has signed it will be considered negative.

Example:
+ 43 or 43 is a positive number
– 43 is a negative number

In signed binary representation, the leftmost bit is considered as a sign bit. If this bit is 0, it is a positive number and if it 1, it is a negative number. Therefore a signed binary number has 8 bits, only 7 bits used for storing values (magnitude), and 1 bit is used for signs.

Question 9.
Explain the method of representing signed binary numbers in l’s complement representation.
Answer:
This is an easier approach to represent signed numbers. This is for negative numbers only i.e. the number whose MSB is 1.

The steps to be followed to find l’s complement of a number:

Step 1: Convert given Decimal number into Binary
Step 2: Check if the binary number contains 8 bits, if less add 0 at the left most bit, to make it as 8 bits.
Step 3: Invert all bits (i.e. Change 1 as 0 and 0 as 1)

Example: Find 1’s complement for (-24)₁₀

Question 10.
Explain the method of representing signed binary numbers in 2’s complement representation.
Answer:
The 2’s-complement method for the negative number is as follows:
a) Invert all the bits in the binary sequence (i.e., change every 0 to 1 and every 1 to 0 ie.,l’s complement)
b) Add 1 to the result to the Least Significant Bit (LSB).
Example: 2’s Complement represent of (-24)₁₀

Question 11.
Explain binary addition with a suitable example.
Answer:
The following table is useful when adding two binary numbers.

In 1 + 1 = 10, is considered as sum 0 and the 1 as carry bit. This carry bit is added with the previous position of the bit pattern.
Example: Add: 1011₂ + 1001₂

Example: Perform Binary addition for the following:
23₁₀ + 12₁₀
Step 1: Convert 23 and 12 into binary form

Step 2: Binary addition of 23 and 12:

Question 11.
Explain binary subtraction with a suitable example.
Answer:
The table for Binary Subtraction is as follows:

When subtracting 1 from 0, borrow 1 from the next Most Significant Bit, when borrowing from the next Most Significant Bit, if it is 1, replace it with 0. If the next Most Significant Bit is 0, you must borrow from a more significant bit that contains 1 and replace it with 0 and 0s upto that point become Is.
Example : Subtract 1001010₂ — 10100₂.

Example: Perform Binary addition for the
following:
(-21)₁₀ + (5)₁₀
Step 1: Change -21 and 5 into binary form

Workshop

Question 1.
Identify the number system for the following numbers.
Answer:

Question 2.
State whether the following numbers are valid or not. If invalid, give a reason.
Answer:

Question 3.
Convert the following Decimal numbers to their equivalent Binary, Octal, Hexadecimal.
i) 1920
ii) 255
iii) 126
Answer:
i) 1920

ii) 255

iii) 126

Question 4.
Convert the given Binary number into its equivalent Decimal, Octal, and Hexadecimal numbers.
i)101110101
ii) 1011010
iii) 101011111
Answer:
i) 101110101
Binary to Decimal (Multiply by positional value and then add)

ii) 1011010
Binary to decimal

iii) 101011111
Binary to decimal

Question 5.
Convert the following Octal numbers into Binary numbers.
a) 472
b) 145
c) 347
d) 6247
e) 645
Answer:
Procedure: Write three digits binary number for every octal digit that will give the equivalent binary number.

Ans.
(472)₈ = (100111010)₂

Answer:
(145)₈ = (001100101)₂

Answer:
(347)₈ = (011100111)₂

Answer:
(6247)₈ = (110010100111)₂

Answer:
(645)₈ = (110100101)₂.

Question 6.
Convert the following Hexadecimal numbers to Binary numbers
a) A6
b) BE
c) 9BC8
d) BC9
Answer:
Procedure: Write four digits binary number for every Hexadecimal digit that will give the equivalent binary number.

Answer:
(A6)₁₆ = (10100110)₂.

Answer:
(BE)₁₆ = (1011 1110)₂

Answer:
(9BC8)₁₆ = (1001101111001000)₂

Answer:
(BC9)₁₆ = (101111001001)₂

Question 7.
Write the l’s complement number and 2’s complement number for the following decimal numbers:
Perform the following binary computations:
a) -22
b) -13
c) -65
d) -46
Answer:
a) -22

Question 8.
a) 10₁₀ + 15₁₀
b) – 12₁₀ + 5₁₀
c) 14₁₀ – 12₁₀
d) (-2)₁₀ – (-6)₁₀
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 8 Iteration and Recursion Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 8 Iteration and Recursion

11th Computer Science Guide Iteration and Recursion Text Book Questions and Answers

Part I

Choose The Correct Answer :

Question 1.
A loop invariant need not be true.
a) at the start of the loop
b) at the start of each iteration
c) at the end of each iteration
d) at the start of the algorithm
Answer:
d) at the start of the algorithm

Question 2.
We wish to cover a chessboard with dominoes, I I 1 the number of black squares and the number of white squares covered by dominoes, respectively, placing a domino can be modeled by
a) b ; = b + 2
b) w := w + 2
c) b, w : = b + 1, w + 1
d) b : = w
Answer:
d) b : = w

Question 3.
If m x a + n x b is an invariant for the assignment a, b : -a + 8, b + 7, the values of m and n are
a) m = 8, n = 7
b) m – 1, n = -8
c) m = 7, n = 8
d) m – 8, n = -7
Answer:
b) m – 1, n = -8

Question 4.
Which of the following is not an invariant of the assignment? m, n ; = m + 2, n + 3.
a) m mod 2
b) n mod 3
c) 3 x m – 2 x n
d) 2 x m – 3 x n
Answer:
d) 2 x m – 3 x n

Question 5.
If Fibonacci number is defined recursively as

to evaluate F(4), how many times F( ) is applied?
a) 3
b) 4
c) 8
d) 9
Answer:
a) 3

Question 6.
Using this recursive definition

how many multiplications are needed to calculate a¹⁰?
a) 11
b) 10
c) 9
d) 8
Answer:
c) 9

Part – II

Short Answers

Question 1.
What is an invariant?
Answer:
An expression involving variables, which remain unchanged by an assignment to one of these variables is called an invariant of the assignment.

Question 2.
Define a loop invariant.
Answer:
Each time the loop body is executed, the variables which remains unchanged by the execution of the loop body is called the loop invariant.

Question 3.
Does testing the loop condition affect the loop invariant? Why?
Answer:
No, the loop condition does not affect the loop invariant. Because the loop invariant is true at four points.

1. At the start of the loop.
2. At the start of each iteration.
3. At the end of each iteration.
4. At the end of the loop.

Question 4.
What is the relationship between loop invariant, loop condition, and the input-output recursively?
Answer:

• Establish the loop invariant at the start of the loop.
• The loop body should so update the variables as to progress toward the end and maintain the loop invariant, at the same time.
• When the loop ends, the termination condition and the loop invariant should establish the input-output relation.

Question 5.
What is recursive problem-solving?
Answer:
Recursion is a method of solving problems that involves breaking a problem down into smaller and smaller subproblems until the user gets into a small problem that can be solved trivially. Usually, recursion involves a function calling itself. While it may not seem like much on the surface, recursion allows us to write elegant solutions to problems that may otherwise be very difficult to program.

Question 6.
Define factorial of a natural number recursively.
Answer:
factoriai(n)
— inputs: n
— outputs: f
if n = 0
f = 1
else
f = n x factorial(n -1) → Recursive process
Example:
To calculate 5 factorial factorial(5)
= 5 x factorial(4)
= 5 x 4 x factorial(3)
= 5 x 4 x 3 x factorial(2)
= 5 x 4 x 3 x 2 x factorial (1) =5 x 4 x 3 x 2 x 1 = 120

Part – III

Explain In Brief

Question 1.
There are 7 tumblers on a table, all standing upside down. You are allowed to turn any 2 tumblers simultaneously in one move. Is it possible to reach a situation when all the tumblers are right side up? (Hint: The parity of the number of upside-down tumblers is invariant).
Solution:
Let u – No. of tumblers right side up
v – No. of tumblers upside down
Initial stage : u = 0, v = 7 (All tumblers upside down)
Final stage output: u = 7, v = 0 (All tumblers right side up)

Possible Iterations:
(i) Turning both up side down tumblers to right side up
u = u + 2, v = v – 2 [u is even]

(ii) Turning both right side up tumblers to upside down.
u = u – 2, v = v + 2 [u is even]

(iii) Turning one right side up tumblers to upside down and other tumblers from upside down to right side up.
u = u + 1 – 1 = u, v = v + 1 – 1 = v [u is even]

Initially u = 0 and continues to be even in all three cases. Therefore u is always even. Invariant: u is even (i. e. No. of right side up tumblers are always even)
But in the final stage (Goal), u = 7 and v = 0 i. e. u is odd.
Therefore it is not possible to reach a situation where all the tumblers are right side up.

Question 2.
A knockout tournament is a series of games. Two players compete in each game; the loser is knocked out (i.e. does not play anymore), the winner carries on. The winner of the tournament is the player that is left after all other players have been knocked out. Suppose there are 1234 players in a tournament. How many games are played before the tournament winner is decided?
Solution :
Suppose there are 2 players A and B competing in a knockout match, so the number of possible matches is – A Vs B
The answer is Only One.
Suppose there are 3 players A, B and C competing in a knockout match, so the Number of possible matches is – 2.
A Vs B , and A/B( Whoever wins with plays with C) vs C.
Suppose there are 4 players A, B, C & D competing in a knockout match, so the Number of possible matches is – 3.
A Vs B , C Vs D A/B Vs C/D
Thus the General Observations is –
When there are 2 players – 1 Match
When there are 3 players – 2 Matches
When there are 4 players – 3 Matches
When there are n players – ( n – 1) Matches
Hence when there are 1234 players there will be 1233 matches.
The number of games in a single-elimination tournament is always 1 less than the number of players/teams.

Question 3.
King Vikramaditya has two magic swords. With one, he can cut off 19 heads of a dragon, but after that, the dragon grows 13 heads. With the other sword, he can cut off 7 heads, but 22 new heads grow. If all heads are cut off, the dragon dies. If the dragon has originally 1000 heads, can it ever die? (Hint: The number of heads mod 3 is invariant.)
Answer:
No. of heads of dragon = 1000
sword 1: cuts 19 heads but 13 heads grow back.
sword 2: cuts 7 heads but 22 heads grow back.
Let n be the number of heads of the dragon at the initial state.

Case 1: King uses Sword 1
Sword 1 cuts off 19 heads but 13 heads grow back.
n : = n – 19 + 13 = n – 6 No. of heads are reduced by 6.

Case 2: King uses Sword 2
Sword 2 cuts 7 heads but 22 heads grow back.
n : = n – 7 + 22 = n + 15
No. of heads are increased by 15.

Note:
In the above two cases, either 6 heads are removed or 15 heads added. Both 6 and 15 are multiples of 3.
Therefore repeating case 1 and case 2 recursively will either reduce or increase dragon heads in multiples of 3.
That is invariant is n mod 3.
If n mod 3 = 0 then there is a possibility that the dragon dies.
But 1000 is not a multiple of 3 1000 mod 3 = 1 ≠ 0
It is not possible to kill the dragon. The dragon never dies.

Part IV

Explain In Detail

Question 1.
Assume an 8 x 8 chessboard with the usual coloring. “Recoloring” operation changes the color of all squares of a row or a column. You can recolor repeatedly. The goal is to attain just one black square. Show that you cannot achieve the goal. (Hint: If a row or column has b black squares, it changes by (|8 – b) – b |).
Solution:
Let us start with a normal coloured chessboard, with a number of black squares B=32 and the number of white squares W=32.
So W – B = 0, which is divisible by 4 and W + B = 64.
W-B = 0 mod 4

Whenever we change the colours of a row or column, we change the colour of 8 squares. Let this row (or column) have w white squares + b black squares w + b = 8 squares.

If this operation B increases (or decreases) by 2n, then W decreases (or increases) by 2n so that W + B = 64, but B – W will change by 4n and it will remain divisible by 4.
W – B = 0 mod 4

After every operation, “B – W mod 4” can have no other values.
But the required state has 63 white squares and 1 black square, so it requires
W-B = 63-1 = 62 = 2 mod 4 which is impossible.

Question 2.
Power can also be defined recursively as

Construct a recursive algorithm using this definition. How many multiplications are needed to calculate a¹⁰?
Recursive Algorithm
power(a,n)
— inputs: n is an integer, n ≥ 0
— outputs: aⁿ
if n = 0 → Base case
1
else
if(n%2 = 0)
a x power(a, a^n-1) → Recursion step for odd number else
a x power(a^n/2 x a^n/2) → → Recursion step for even number

To calculate a¹⁰
power(a, 10)
= a x power(a⁵ x a⁵)
= a x power(a, a⁴) x power(a, a⁴)
= a x a x power(a² x a²) x power(a, a⁴)
= a x a x a x power(a, a⁰) x power(a, a⁰ ) x
power(a, a⁴)
= a x a x a x a x a x power(a² x a²)
= a x a x a x a x a x a x a x power(a, a⁰) x a x power(a, a⁰)
=a x a x a x a x a x a x a x a x a x a
There are nine multiplications are needed.

Question 3.
A single-square-covered board is a board of 2ⁿ x 2ⁿ squares in which one square is covered with a single square tile. Show that it is possible to cover this board with triominoes without overlap.
Solution :
A triamine is an L-shaped tile formed with three adjacent squares.
Corner-covered board and triamine

Cover the corner-covered board with the L-shaped triominoes without overlap, Triominoes can be rotated as needed.
The size of the problem is n (board of size 2ⁿ x 2ⁿ). We can solve the problem by recursion. The base case is n = 1. It is a 2 x 2 corner-covered board. We can cover it with one triamine and solve the problem.

In the recursion step, divide the corner-covered board of size 2n x 2n into 4 sub-boards, each of size 2^n-1 x 2^n-1, by drawing horizontal and vertical lines through the center of the board. Place a triamine at the center of the entire board so as to not cover the corner-covered sub-board, as shown in the left-most board of the given figure below. Now, we have four corner-covered boards, each of size 2^n-1 x 2^n-1.
Recursive process of covering a corner-covered board of size 2 x 2³

We have 4 sub-problems whose size is strictly smaller than the size of the given problem. We can solve each of the sub-problems recursively. tile corner_covered board of size n
if n = 1 — base case
cover the 3 squares with one triominoe
else — recursion step
divide board into 4 sub_boards of size n – 1
place a triominoe at centre of the board
leaving out the corner_covered sub-board
tile each sub-board of size n-1

The resulting recursive process for covering a 2³ x 2³ corner-covered board is illustrated in the above Figure.

11th Computer Science Guide Iteration and Recursion Additional Questions and Answers

Part I

Choose The Correct Answer:

Question 1.
………………… is an algorithm design technique, closely related to induction.
(a) Iteration
(b) Invariant
(c) Loop invariant
(d) Recursion
Answer:
(d) Recursion

Question 2.
Each time the loop body is executed, the variables are _________
a) updated
b) unchanged
c) neither A nor B
d) destroyed
Answer:
a) updated

Question 3.
In a loop, if L is an invariant of the loop body B, then L is known as a …………………
(a) recursion
(b) variant
(c) loop invariant
(d) algorithm
Answer:
(c) loop invariant

Question 4.
_________ is more powerful than the iteration.
a) Recursion
b) Specification
c) Composition
d) None of these
Answer:
a) Recursion

Question 5.
The unchanged variables of the loop body are…………………
(a) loop invariant
(b) loop variant
(c) condition
(d) loop variable
Answer:
(a) loop invariant

Question 6.
_________ is a recursive solver case.
a) Base case
b) Recursion steps
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 7.
If L is a loop variant, then it should be true at ………………… important points in the algorithm.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 8.
In_________, the size of moot to a sub-problem must be strictly smaller than the size of the given input.
a) Recursion
b) Specification
c) Composition
d) None of these
Answer:
a) Recursion

Question 9.
In an expression, if the variables have the same value before and after an assignment, then it is of an assignment.
(a) variant
(b) Invariant
(c) iteration
(d) variable
Answer:
(b) Invariant

Question 10.
An expression involving variables, which remains unchanged by an assignment to one of these variables, is called _________ of the assignment.
a) an invariant
b) variant
c) neither A nor B
d) None of these
Answer:
a) an invariant

Question 11.
When the solver calls a sub solver, then it is called …………………
(a) Iterative call
(b) solver call
(c) recursive call
(d) conditional call
Answer:
(c) recursive call

Question 12.
_________ Is an algorithm design technique
to execute the same action repeatedly.
a) iteration
b) recursion
c) Both iteration and recursion
d) None of these
Answer:
c) Both iteration and recursion

Question 13.
Which of the following is updated when each time the loop body is executed?
(a) data type
(b) subprogram
(c) function
(d) variable
Answer:
(d) variable

Question 14.
Recursion is an algorithm design technique, closely related to _________
a) induction
b) introduction
c) intuition
d) None of these
Answer:
a) induction

Part II

Short Answers

Question 1.
When the loop variant will be true?
Answer:
The loop invariant is true before the loop body and after the loop body, each time.

Question 2.
How problems are solved using recursion?
Answer:
Using recursion, we solve a problem with a given input, by solving the same problem with a part of the input and constructing a solution to the original problem from the solution to the partial input.

Question 3.
If L is a loop variant, then where it is true in the algorithm?
Answer:
It is true in the following four points of an algorithm.

1. at the start of the loop.
2. at the start of each iteration.
3. at the end of each iteration.
4. at the end of the loop.

Part – III

Explain In Brief

Question 1.
Write a note on Recursion.
Answer:
Recursion:
Recursion is another algorithm design technique, closely related to iteration, but more powerful. Using recursion, we solve a problem with a given input, by solving the same problem with a part of the input and constructing a solution to the original problem from the solution to the partial input.

Question 2.
Write the recursive algorithm for the length of a sequence.
Answer:
The recursive algorithm for the length of a sequence
can be written as
length (s)
— inputs: s
— outputs : length of s
if s has one customer — base case
1
else
1 + length(tail(s)) — recursion step.

Part IV

Explain In Detail

Question 1.
Explain loop Invariant in detail.
Answer:
In a loop, if L is an invariant of the loop body B, then L is known as a loop invariant,
while C
— L
B
— L
The loop invariant is true before the loop body and after the loop body, each time. Since L is true at the start of the first iteration, L is true

at the start of the loop also (just before the loop). Since L Is true at the end of the last iteration, L is true when the loop ends also (just after the loop). Thus, if L is a loop variant, then it is true at four important points in the algorithm, as annotated in the algorithm and shown in Figure 3.1.

i) at the start of the loop (just before the loop)
ii) at the start of each iteration (before loop body)
iii) at the end of each iteration (after loop body)
iv) at the end of the loop (just after the loop)

i) — L, start of loop
while
C

ii) — L, start of iteration
B

iii) –L, end of Iteration

iv) — L, end of loop

The points where the loop invariant is true

Question 2.
Design an iterative algorithm to compute aⁿ.
Answer:
Let us name the algorithm power(a, n).
For example,
power(10, 4) = 10000
power (5,3) = 125 .
power (2,5) = 32
Algorithm power(a, n) computes aⁿ by multiplying a cumulatively n times.

The specification and the loop invariant are shown as comments.

power (a, n)
— inputs: n is a positive integer
— outputs: p = an
p, i := 1,0
while i ≠ n
— loop invariant: p = aⁱ
p, i: = p x a, i + 1
The step by step execution of power (2, 5) is shown in the following Table

Question 3.
Explain the outline of the recursive problem-solving technique.
Answer:
The outline of the recursive problem-solving technique is shown below.

solver (input)
if the input is small enough
construct solution
else
find sub_problems of reduced input
solutions to sub_problems = solver for each sub_problem
construct a solution to the problem from
solutions to the sub_problems

Whenever we solve a problem using recursion, we have to ensure these two cases: In the recursion step, the size of the input to the recursive call is strictly smaller than the size of the given input, and there is at least one base case.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 5 Taxonomy and Systematic Botany Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 5 Taxonomy and Systematic Botany

11th Bio Botany Guide Taxonomy and Systematic BotanyText Book Back Questions and Answers

Part – A

Choose the right answer:

Question 1.
Specimen derived from non-original collection serves as the nominclatural type, when original specimen is missing it is known as
a) Holotype
b) Neotype
c) Isotype
d) Paratype
Answer:
b) Neotype

Question 2.
Phylogenilic classification is the most favoured classification because it reflects,
a) Comparative anatomy
b) Number of flowers produced
c) Comparative cytology
d) Evolutionary relationships
Answer:
d) Evolutionary relationships

Question 3.
The taxonomy which involves the similarities and dissimilarities among the immune system of different taxa is termed as
a) Chemotaxonomy
b) Molecular systematics
c) Sero taxonomy
d) Numerical taxonomy
Answer:
c) Sero taxonomy

Question 4.
Which of the following is not a flowering plant with root nodules containing filamentous nitrogen-fixing micro-organisms?
a) Crotalaria juncea
b) Cycas revoluta
c) Cice rarietinum
d) Casuarina equisetifolia
Answer:
b) Cycas revoluta

Question 5.
Flowers are weakly zygomorphic in
a) Cerapegia
b) Thevelia
c) Datura
d) Solanum
Answer:
c) Datura

Two marks

Question 1.
What is the role of national gardens in conserving biodiversity
Answer:
Botanical Gardens play the following important roles.

1. Gardens with an aesthetic value attract a large number of visitors. For example, the Great Banyan Tree (Ficus benghalensis) in the Indian Botanical Garden at Kolkata
2. Gardens have a wide range of species and supply taxonomic material for botanical research.
3. Garden is used for self-instruction or demonstration purposes.

Question 2.
Where will you place the plants which contain two cotyledons with cup-shaped thalamus
Answer:

1. Two cotyledonous plants are kept under Dicotyledonou
2. Cup-shaped thalamus is a unique feature of the series Thalamiflorae of subclass

Polypetalae:

• Class
• Dicot
• Subclass
• Polypetalae
• Series
• Thalamiflorae

Question 3.
How do molecular markers work to unlock the evolutionary history of organisms?
Answer:
Molecular taxonomy using molecular markers like RAPD’s etc helps in establishing the relationship between the members of different taxonomic groups at the DNA level. Thus it helps to unlock the evolutionary history of organisms.

Question 4.
Give the F.D of Clutoria ternalia
Answer:

Question 5.
How will you distinguish Solanaceae members from Liliaceae members.
Answer:

JSolanaceae (Dicot) Morphology	Liliaceae (Monocot) Morphology
Bulbous stem / rhizome / corm absent but stem tuber present eg. Solarium tuberosum	Bulbous Stem – Lilium Rhizome -Polygonatum Corm – Colchicum Occur
Radical leaves Cariscup } Absent from roots	Radical leaves present eg. Lilium
Leaves alternate & exstipulate	Slipules absent exstipulate fleshy and hollow

Flowers – Pentamerous	Flowers – Trimerous
Calyx_ – Persistent -Solanum melongena	Calyx , Corolla – Absent
Corolla of various shapes present	Perranth is only present
Stamens – 5 – Epipetalous	Stamens – 6 – In a whorl of three each apostamenous
Carpels: Ovary superior bicarpellary, bilocular Carpels oblipuely placed bicarpellary later become tetra carpellary due to the formation of false septa	Ovary superior tricarpellary trilocular
Fruits – Berry / capsule	Fruits – Loculierdal capsula
Anatomy: Bi-collateral Vascular Bundles occur Cambium present Secondary growth present	Anatomy: Conjoint collateral Vascular bundles Cambium absent No secondary growth

Part – B

11th Bio Botany Guide Taxonomy and Systematic Botany Additional Important Questions and Answers

Choose the correct answer:

Question 1.
Who is called the father of Botany?
(a) Linnaeus
(b) Theophrastus
(c) Darwin
(d) Thales
Answer:
(b) Theophrastus

Question 2.
The latest ICBN was held at
a) Cambridge – England
b) Leningrad – Russia
c) Shenzhen – China
d) Rio-de genero – America
Answer:
cl Shenzhen China

Question 3.
Who proposed the concept of “idos” or species?
(a) Theophrastus
(b) Stebbins
(c) Darwin
(d) Plato
Answer:
(d) Plato

Question 4.
The floral of Tamil Nadu Carnatic by K.M. Mathew (1983) and Flora of Madras Presidency by J.S Gamble and Fischer belong to
a) Monograph
b) Catalogue
c) Flora
d) Regional flora
Answer:
d) Regional flora

Question 5.
Who first proposed the early elementary rule of naming plants?
(a) A.P.de Candolle
(b) Linnaeus
(c) Alphonse de Candolle
(d) Simpson
Answer:
(b) Linnaeus

Question 6.
Botanical Garden of New South Wales is located at
a) Brazil
b) New York
c) Sydney
d) Rio de Janeiro
Answer:
c) Sydney

Question 7.
18^th International Botanical congress was held at …………….
(a) Sydney
(b) Leningrad
(c) Melbourne
(d) London
Answer:
(c) Melbourne

Question 8.
Plants with incomplete flowers either a petalous or with undifferentiated calyx and corolla are placed under the sub class
a) Heterornerae
b) Monochlarnydeae
c) Bicarpellatae
d) Monocotyledonae
Answer:
b) Monochlamvdeae

Question 9.
Adolph Engler and Karl – A – Prantl are
a) Americal taxonomists
b) French taxonomists
c) Chinese taxonomists
d) German taxonomists
Answer:
d) German taxonomists

Question 10.
…………… is a descriptive phrase of a plant.
(a) Vernacular name
(b) Binomial
(c) Polynomial
(d) Botanical name
Answer:
(c) Polynomial

Question 11.
The fruit is geocarpic in
a) Vigna radiate
b) Cablab purpuris
c) Araehis hypogea
d) Cicer arietinum
Answer:
c) Arachis hypogea

Question 12.
The term biosystematics was introduced by
a) Chrispeels and Gartner
b) Camp and Gilly
c) Engler and Prantl
d) Bentham and Hooker
Answer:
b) Camp and Gilly

Question 13.
…………… are the tools for identifying unfamiliar plants.
(a) Flora
(b) Keys
(c) Monograph
(d) Catalogues
Answer:
(b) Keys

Question 14.
Most of the seeds are reniform in
a) Fabaceae
b) Solanaceae
c) Asteraceae
d) Liliaceae
Answer:
a) Fabaceae

Question 15.
Plants with incomplete flowers either petalous or with undifferentiated calyx and corolla are placed under
a) Monocotyledonae
b) Polypetalae
c) Monochlamydeae
d) Gamopetalae
Answer:
c) Monochlamydeae

Question 16.
First modern botanical garden was established by …………… .
(a) Theophrastus
(b) Linnaeus
(c) Luca Ghini
(d) Stebbins
Answer:
(c) Luca Ghini

Question 17.
Phenology is the study
a) Pollen grains structure
b) Development of gametes
c) Study of climate and weather on plants
d) Study of functional aspects of plants
Answer:
c) Study of climate and weather on plants

Question 18
Tephrosia purpurea, Indigofera tinctoria are plants used as
a) Biofuel or bioenergy
b) Bio manure
c) Biomedicine
d) Biopesticides
Answer:
b) Bio manure.

Question 19.
Carpels are obliquely placed in the family
a) Fabaceae
b) Solanaceae
C) Liliaceae
d) Malvaceae
Answer:
b) Solanaceae

Question 20.
Which series includes epigynous flowers with an inferior ovary?
(a) Heteromerae
(b) Disaflorea
(c) Inferae
(d) Thalanifloreae
Answer:
(c) Inferae

II. Fill up the blanks in the given Tabulation

Question 1.
Name of the Taxonomist BOOK No. of plants

Name of the Taxonomist	BOOK	No. of plants
i. Theophrastus (372 – 287 BC)	De Historia Pfcntarum	………………….
ii. ……………….	Materia Medica	600 plants
iii. Carolus Linnaeus	Species Plantarum	………….
iv. ………………	3 volumes of Genera Plantarum	97205 species (202 families)

Answer:
i. 500 Plants
ii. Discoredes (62 – 127 AD)
iii. 7300 Species
iv. Bentham & Hooker

Question 2.

Taxon	Definition	Example
i. Family	Comprises a no of genera which share some similarities	…………………..
ii. …………………	Group of families shows fewer similarities among themselves	Malvales
iii. Class	Group of order which share few similarities	…………………..
iv. ………………….	Number of classes	Magnoliophyta

Answer:
(i) Asteraceae
(ii) order
(iii) Asteropsida
(iv) Sub – Division

Question 3.

Name of the IBC	Place	Year
i. 5th International Botanical Congress	………………………..	1930
ii. 12th International Botanical Congress	Leningrad – Russia	…………………..
iii. 18th International Botanical Congress	……………………..	2011
iv. …………………	Shenzhen – China	2017

Answer:
b) Cambridge – England 1975 Melbourne Australia 19th I BC

III. Find out the correct statement.

Question 1.
Find out the Correct Statements the given below.
a) Scientific Names are treated as Latin regardless of their derivation
b) Cryptogams include non-flowering plants
c) Linnaeus system of classification is known as the Natural system of classification
d) According to APG IV Monocots contain 10 orders and 37 families
(I) a & b
(II) b & c
(III) c & d
(IV) a & d
Answer:
(I) a & b

Question 2.
Find out the correct statements from the given below
a) Botanical name of chilly – Capsicum esculentum
b) Ashwagandha is also known as Amukkura
c) An alkaloid colchicine is got from Colchicum luteum
d) Glycine max is the botanical name of the Soya bean
(I) a & b
(II) b & d
(III) a & c
(IV) c & d
Answer:
(II) b & d

IV. Assertion and reason

Question 1.
Assertion: Classification is essential in biology because there is a vast diversity of organisms to sort out and compare
Reason: Unless they are organized into manageable Categories, It will be difficult for the identification
(A) Assertion and Reason correct. The reason is explaining Assertion
(B) Assertion and Reason are correct but Reason not explaining Assertion
(C) Assertion is true, but Reason is wrong
(D) Assertion is true, but Reason is not explaining Assertion
Answer:
a – Assertion and Reason correct. Reason is explaining Assertion

Question 2.
Assertion: Cronquist system of classification could not persist for a long time.
Reason: The system is not very useful for identification and cannot be adopted in herbaria.
Answer:
d) Assertion ‘A’ is true Reason ‘R’ is not explaining Assertion

Question 3.
Assertion : Monograph is a complete global account of a faxon of any rank family genus or species at a given time
Reason : It has books of libraries rich in botanical litles.
Answer:
d) Assertion correct but Reason not explaining Assertion

Question 4.
Assertion A: Chemotaxonomy is the scientific approach to the classification of plants on the basis of their biochemical constituents
Reason R: Proteins, amino acids, nucleic acids, and peptides, etc are the most studied chemicals in chemotaxonomy
Answer:
a) Assertion and Reason ‘R’ correct Reason is explaining Assertion

V. Match the following and find the correct

Question 1
Name of Herbarium No. of specimens
i) Presidency college Herbarium – Chennai – A. 4,08,776
ii) Central National Herbarium – Calcutta – B. 30,500
iii) Madras Herbarium – Coimbatore – C. 15,000
iv) Jawaharlal Nehru Tropical Botanical Garden – D. 2,00,000
and Reserach Institute Trivandrum

Answer:
a) C D A B

Question 2.
Botanical Name Common Name
(I) Glycirrhiza glabra A. Thanneer Muttan
(II) Withania somnifera B. Athimaduram
(III) Asperagus racemosus C. Senkandal
(IV) Gloriosa superba D. Amukkara

Answer:
a) B D A C

VI. Choose the wrong answer

Question 1.
(i) Karyology – Study of Chromosomes
(ii) Palynology – Study of Pollen
(iii) Serology – Study of Antibiotics
(iv) Paleology – Study of Fossils
Answer:
(iii) – Serology study of antibiotics

Question 2.
Type Example
(i) Tree – Solanum violceum
(ii) Prickles on the Body of the plant – Solanum xantho carpum
(iii) Vines – Lycium sinensis
(iv) Herb – Solanum nigrum
Answer:
(II) prickles on the body of the plant – body Solanum xantho carpum

Question 3.
Botanical Garden Major Attraction
(i) Nation Botanical Garden Lucknow – Germplasm collection & exsitu conversation
(ii) JNTBGRI Trivandrum (Kerala) – Bambusetum
(iii) National orchidarium Yercaud – Fernery
(iv) Indian Botanical Garden Kolkata – The great Banyan tree
Answer:
(iii) National orchidarium Yercard – Fernery

Question 4.
Choose the right answer :
(i) Plants having Dome-shaped thalamus – Gamopetalae
(ii) Plants having Cup-shaped thalamus – Calyciflorae
(iii) Plants having epigynous flowers – Thalamiflorae
(iv) Plants with united petals & sepals – Inferae
Answer:
(ii) Plants having cup-shaped thalamus Calyciflorae

VII. Match correctly & give the correct answer

Question 1.
(i) Systema Naturae -A. 1862-63
(ii) Philosophia botanica -B.1753
(iii) Species plantarum -C. 1735
(iv) General plantarum -D.1737

Answer:
c) C – D – B – A

Question 2.
(i) Library of British Museum – A. Revision
(ii) Malvaceae of India by T.K. Paul – B. Catalogue
(iii) Family of Lentibulariaceae
by M.K. Janarthanam & Henry – C. Botanical Garden Lucknow
(iv) 500 species of rose hybrids – D. Monograph

Answer:
a) B A D C

Question 3.
Botanical Name – Common Name
(i) Butea Frondosa – A. Rose Wood
(ii) Sesbania grandiflora – B. Garden pea
(iii) Dalbergia latefolia – C. Flame of the forest
(iv) Pisum sativum – D. Agathe

Answer:
b) C D A B

Question 4.
Common name Botanical name
(i) Rose wood – A. Pterocarpus dalbergioides
(ii) Red Sandal wood – B. Dalbergia latifolia
(iii) Padauk – C. Pterocarpus marsupium
(iv) Vengai – D. Pterocarpus santalinus

Answer:
c) B D A C

VIII. Find out the true and false statements from the following and on that basis find the correct answer:

Question 1.
(i) The evolution & classification of flowering plants – Arthur cronquist
(ii) Origin of species – Engler & prantl
(iii) Philosophia botanica – Linnaeus
(iv) Theorie elementaire de-botanique – A.P. de. Candolle

Answer:
a) True False True True

Question 2.
Find out the True and False statements from the following and on that basis find the correct answer:
(i) Documents of all plant species in a given geographical area is known as – Monograph
(ii) These are often descriptive & poetic references to plants – Vernacular name
(iii) A complete global account of a taxon of any rank – Flora
(iv) Tools of Identification implemented by Computer – Polyclave key

Answer:
b) False True False True

Two marks

Question 1.
Define Taxonomy.
Answer:
Taxonomy is “the science dealing with the study of classification including the bases, principles, rules and procedures”.

Question 2.
What are the characteristics of a species
Answer:

• Population of organism closely resemble each other
• Descend from common ancestor
• They sexually interbreed freely producing fertile offspring
• They have morphological resemblance in asexually reproducing organism
• In fossil organisms they are identified by their morphological & anatomical resemblance

Question 3.
Which is the lowest taxon in classification? Define.
Answer:
Species is the lowest taxon in classification. It is defined as the group of individuals which are closely resembling each other and interbreed among themselves producing fertile offspring.

Question 4.
Define Binomial Nomenclature.
Answer:

• Introduced by Gaspard Barhin
• Implemented by Carolus Linnaeus
• Scientific name of a plant consists of 2 words.
• First one is Genus name
• Second one is Species name.
• Genus Species

Eg: Mangifera – Genus – indica – Species

Question 5.
What are vernacular names? Give an example.
Answer:
Vernacular names are known as common names. Example: Albizia Amara L. is called as Usilai in South Tamil Nadu and Thurinji in North Tamil Nadu.

Question 6.
Biosystematics Define & Give its objectives.
Answer:
Definition:

Question 7.
What is the importance of serotaxonomy.
Answer:

Question 8.
When a neotype specimen is selected?
Answer:
Neotype Specimen is derived from non – original collection selected as the type when the original specimen is missing or destroyed.

Question 9.
Define karyotaxonomy.
Answer:

• Increased knowledge of chromosomes Have been used for extensive biosystematic studies & resolving many taxonomic problems.
• Cytological especially chromosomal characters such as number, size, morphology and behaviour during meiosis are of taxonomic value.

Question 10.
Differentiate Regional Flora from continental flora.
Answer:
Regional Flora from continental flora.

1. Regional Flora: Flora covering a large geographical area or a botanical region Ex: flora of Madras Presidency.
2. Continental Flora: Flora covering the entire continent. Ex: flora of Europaea.

Question 11.
Write down the Aims of Chemotaxonomy.
Answer:

• To develop taxonomic characters to improve, the existing system of plant classification
• To improve the present-day knowledge of phylogeny o plants.

Question 12.
Define Biosystematics
Answer:

• Introduced by Camp and Gilly in 1943
• It is an experimental, ecological cyto taxonomy through which life forms studied and their relationships defined

Question 13.
How Cronquist classified the angiosperms?
Answer:
Cronquist classified the angiosperms into two main classes Magnoliopsida and Liliopsida.

Question 14.
Why do we say that the development of fruit in Arachis hypogea is geocarpic?
Answer:

• In Arachis hypogea after fartilization, the stipe of ovary become meristematic and grows down into the soil.
• The ovary gets buried into the soil and so we call the fmit as groundnut.

Question 15.
Distinguish between Stipule, Stipel and Pulvinus in the leaf of Fabaceae.
Answer:

Stipule	Stipel	Pulvinus
The compound leaf has a stipule a green scale like structure from which the leaf originate	It is a small scaly structure at the base of the leaflet of compound leaf	The R achis and petcole of the leaf, and leaf lets have swollen, this condition is known as Pulvinus

Question 16.
Draw the structure of Papilionaceous Carolla of Fabacoae
Answer:

Question 17.
Differentiate between Phylloclade & Cladodo of Liliaceae.
Answer:

Phylloclade	Cladodo
(aerial sterm or branch modification) Eg. Ruscus Branch is modified . leaves reduced to scales	(aerial sterm modification) Eg Asparagus Aerial sterm is modified, leaves reduced to scales

Question 18.
Point out the aims of chemotaxonomy.
Answer:
The aims of chemotaxonomy:

1. To develop taxonomic characters which may improve the existing system of plant classification.
2. To improve present-day knowledge of the phylogeny of plants.

Question 19.
What is meant by Scapigerous Inflorescence?
Answer:
Here the inflorescence axis (peduncle) arising from the ground bearing a cluster of flowers at its apex, with pedicels of equal length.

Question 20.
Label the given diagram

Answer:
A- Pedicil
B – Perianth
C – Epipetalous stamens
D- Ovary

Question 21.

Answer:
A- Persistent calyx
B – Spiny outgrowth
C – Valves
D – Seed

Question 22.
Differentiate between Magnoliopsia and Liliopsida

Magnoliopsida	Liliopsida
6 subclasses	5 subclasses
64 orders	19 orders
318 families	65 families
165,000 species	50000 species

Demerits:

• Highly Phylogenetic could not persist.
• Not useful for identification & cannot be adopted.

Three marks

Question 1.
Compare the Gynoecium of Pisum sativum and Datura metal.
Answer:
Gynoecium of Pisum sativum:

1. Mono Carpellary
2. Unilocular
3. Ovules on marginal placentation
4. Feathery stigma

Gynoecium of Datura metal:

1. Bicarpellary
2. Tetralocular
3. Ovules on axile placentation
4. Bilobed stigma

Question 2.
Distinguish between Monophyletic, Paraphyletic & polyphyletic group.
Answer:

Question 3.
Why do we think that cladistics is of much needed and important today?
Answer:

• Commonly used & accepted for phylogenetic classifications.
• Produces a hypothesis about the relationship of organisms to predict the morphological characteristics of an organism.
• Help to elucidate the mechanism of evolution.

Question 4.
Give the systematic position of Pea family.
Answer:

APG Classification		Bentham & Hooker’s Classifications
Kingdom	Plantae	Kingdom	Plantae
Clade	Angiosperm	Class	Dicotyledonae
Clade	Eudicots	Sub class	Polypetatae
Clade	Rosids	Series	Calyaflorae
Order	Fabales	Order	Rosales
Family	Fabaceae	Family	Fabaceae

Question 5.
Differentiate between Taxonomy & Systematics.
Answer:
Taxonomy:

1. The discipline of classifying organisms into taxa
2. Governs the practices of naming, describing, identifying and specimen preservation.
3. Classification + Nomenclature = Taxonomy

Systematics:

1. Broad field of biology that studies the diversification of species
2. Governs the evolutionary history and phylogenetic relationship in addition to taxonomy
3. Taxonomy + Phylogeny = Systematics

Question 6.
Define Herbarium.
Answer:

• Collection of collected, pressed and dried plant specimens preserved, then mounted on a sheet of paper is referred to as Herbarium.
• It also refers to the Institution where many such Herbaria are preserved.
• Eg. Royal Botanical garden Kew London.

Question 7.
Linnaeus classification is also called sexual system of classification. Why?
Answer:
Linnaeus classification is mostly based on sexual characters like number, union, length and distribution of stamens and also on carpel characters. Hence it is called sexual system of classification.

Question 8.
Various types of habits in Fabaceae.
Answer:

Herb	Indigofera, crotalaria
Prostrate herbs	Indigofera enneaphylla
Erect herb	Crotalaria verrucosa
Shrubs	Cajanus cajan
Small trees	Sesbania
Climbers Large trees	Clitoria sp Pongamia, Dalbergia
Woody climber	Mucuna
Hydrophyte	Aeschyno mene aspera

Question 9.
Explain Androecium of Family Fabaceae
Answer:
1. Diadelphous – Stamens (9) + 1-9 Stamens united 1 free
2. Diadelphous – (5) + (5) – Stamens in 2 bundles of 5 each.
3. Monoadelphous but dimorphic – Out of 10 stamens 5 are with longer flaments longer anther 5 are with shorter Flaments and short anthers known as Dimorphic

Question 10.
Tabulate various types of Inflorescence of Solanaceae
Answer:

Solitary flower	Datura stramonium
Terminal cymose	Solanum
Extra axillary Scorpioid cymo or Rhipidium	Solanurn nigrurn
Helicoid cyme	Solanurn tuberosum
Umbellate cyme	Withania somnifera

Question 11.
Tabulate various types of petals of Solanaceae.
Answer:

5 petals – sympetalous

Rotate & tubular – Solanum

Bell-shaped – Atropa

I Infundibuliform  – Petunia

Bilipped & Zygomorphic – Schizanthus

Infundibuliform & Convolute – Datura

Question 12.
Tabulate ornamental plants from any 3 families you have studied.
Answer:

Family Fabaceae	Family  Solanaceae	Family Liliaceae
Butea frondosa (Flame of the forest) Clitoria tematea (sangu – flower)	Cestrium diumum – day Tulipa suaveolens — Tulip Jasmine)	Petunia hybrida Agapanthus african  us garden petunia (African only

Five Marks

Question 1.
What is meant by Taxonomical Aids. Explain any one of it
Answer:

• Tools aiding Taxonomical study are known as Taxonomical Aids
• There are many types of these Aids keys, Flora, Revisions Monograph, Catalogues,
• Herbarium and Botanical garden

Types:

• Local Flora – Covers limited area sate, Country, City mountain, etc Eg. Flora of Thiruvannamalai District
• Regional Flora – Include large geographical area Eg. Flora of Tamilnadu Carnatic by K.M. Mathew
• Continental Flora – Cover entire continent Eg. Flora of Europaea by D.A. Web
• Electronic Flora – Digital format of flora published online Eg. E Flora of China

Question 2.
Type Concept Explain also its types
Answer:
ICN’s second principle is that a specimen must be associated with a scientific name known as nomenclatural type (specimen, or its illustration)
Eg. Herbarium sheet: –

There are 7 types

1.  Holotype:
   The original Protologue of the author is a definite source of identity
   Citation & submission of it is one of the criteria for valid publication of a botanical name
2. Isotype: Duplicate of Holotypes
   • Same person on the same date with same field number
   • Reliable duplicates of holotype – to be distributed to various herbaria of various region
3. Lectotype: Specimen selected from original material may serve as Lectotype when holotype is missing or destroyed
4. Syntype: When the author cites more than one specimen in his Protologue without designating Holotype.
5. Neotype: Specimen from the non-original collection when original
6. Paratype: holotype missing or destroyed: specimen other than Holo, Iso, or Syntype
7. Epitope: Specimen or illustration serves as an interpretive type when all the above types are ambiguous.

Question 3.
Draw the outline classification of APG IV.
Answer:

Question 4.
Give an account of APG classification
Answer:

• Most recent classification of flowering plants
• Done in last decade of 20th century
  
• All these provide data with respect to DNA seqences of 2 chloroplast genes (extrachromosomal) (atp B and r bcL) and one nuclear gene (nuclear ribosomal 18 s DNA).

Question 5.
Tabulate Bentham & Hooker’s system of classification
Answer:

Question 6.
Define biosystematics & list out the aim of biosystematics.
Answer:
1. Biosystematics: Biosystematics is an “Experimental, ecological and cytotaxonomy” through which life forms are studied and their relationships are defined.
2. Aims of Biosystematics: The aims of biosystematics are as follows:

• To delimit the naturally occurring biotic community of plant species.
• To establish the evolution of a group of taxa by understanding the evolutionary and phylogenetic trends.
• To involve any type of data gathering based on modem concepts and not only on morphology and anatomy.
• To recognize the various groups as separate biosystematics categories such as ecotypes, ecospecies, cenospecies, and comparium.

Question 7.
Tabulate various International Herbarium
Answer:
International Herbarium

Herbarium	Year Established	Acronym	Number of specimens
1.Museum National d’Historie Naturelle, Paris, France	1635	P ,PC	10,000,000
2. New York Botanical Garden, Bronx, New York, U.S.A	1891	NY	72,00,000
3. Komarov Botanical Institute, St.Petersburg (Leningrad), Russia	1823	LE	71,60,000
4. Royal Botanic Gardens, Kew, England, U.K	1841	K	70,00,000

Question 8.
Write down the uses of Herbarium
Answer:

• Provides resource material for systematic research & studies
• A place for the orderly arrangement of voucher specimens
• Voucher specimen serves as a reference for comparing doubtful Newly collected fresh specimen
• Voucher specimens play a role in
  • Studies like floristic diversity
  • Environmental assessment
  • Ecological mechanisms &
  • Survey of unexplored area
• provides an opportunity for documenting biodiversity and studies related to the field of ecology & conservation biology.

Question 9.
Explain Chemotaxonomy
Answer:

• Study of various chemicals available in plants help to solve the certain taxonomical problem
• Chemotaxonomy scientific approach of classification of plants on the basis of their biochemical constituents
• Proteins – (more controlled by genes less subjected to natural selection)
• So used at all hierarchical level of classification starling from variety to division.
• Other chemicals studied are – Amino acids, nucleic acids – peptides

Question 10.
Explain Engler & Prantl’s Phylogenetic system of classification
Answer:
The two German scientists published their work in a monumental work.
Die Naturechen P flangen families in 23 volumes.

Question 11.
Write down the significance of Molecular taxonomy
Answer:

• Conserved molecular sequences helped to identify
  
• DNA data help in – investigation of evolutionary patterns
• DNA taxonomy – play vital role in, understanding
• phytogeography – help in genome mapping & bio deversity Conservation
• DNA based Molecular markers – used for designing DNA based molecular probes

Question 12.
What is DNA barcoding?
Answer:

• A scanner like the UPC of supermarket things, – DNA barcoding is a taxonomical device to distinguish one species from another.
• A very short genetic sequence from a standard part of a genome is used as a DNA tag or barcode to identify a plant
• Paul Hebert proposed it and so-known as the Father of barcoding.

Question 13.
Significance of DNA barcoding.
Answer:

• Helps in the identification and classification organism
• Aids in the mapping the extend of bio-diversity
• Eventhough it require a large data base of sequences for comparison & prior knowledge of the barcoding region, it is helpful tool to determine the authenticity of botanical material in whole, cut or powdered form.

Question 14.
Differences between Classical and Modern taxonomy.
Answer:

Question 15.
Give an account of Botanical description of clitoria ternatea (Sangupushpam).
Answer:
Habit – Twining climber
Root – Branched tap root system
Stem – Aerial weak stem, twiner
Leaf – Imparipmnately compound, Petcolate, Alternate, stipulate,
Leaf lets stipellate, stipels are pulvinate reticulate venation
Inflorescence – Solitary and Axillary
Flower -Zygomorphic Bracteate, Bracteolate (large) Bisexual complete
Dichlamydeous – Pedicellate, Pentarnerous & Hypogynous
Calyx – 5 sepals synsepalous valvate aestivation odd sepal anterior in position
Corolla – 5 petals apopetalous
1 standard petal -(Vexillum)
2 wing petals -(aiea)
2 keel petal(carina) – united at the base in descendingly imbricate aestivation

Androecium – 10 stamens diadeiphous
(9) + 1 (i.e.) nine united one free in 2 + bundleš
Anther – Dithecous , basifixed introse and devisce by Longitudinal Division
Gynoecium – Made up of ovary style & stigma
Ovary – Superior with a prominent stipe monocarpellary, unilocular with many ovules on marginal placentation

Style – simple incurved
Stigma – feathery
Fruit – Legume
Seed – non endospermous reniform

Question 16.
Describe any 3 Molecular method ¡n genetic analysis & phylogenetics.
Answer:

Question 17.
Give an account of the botanical description of Datura metal
Answer:
Habit – Large erect, stout, herb
Root – Branched tap root system
Stem – Hollow, herbaceous strong odour
Leaf – Simple, alternate, petiolate, entire or deeply lobed glabrous, exstipulate unicostate reticulate venation
Inflorescence – Solitary & Axillary cyme
Flower – Actinomorphic, (Regular) Bracteate, Bracteolate, Bisexual Complete Dichlamydeous Pentamerous, sessile & hypogynous
Calyx – 5 sepals synsepalous
Valvate aestivation persistant
Corolla – 5 petals synpetalous plicate 10 lobed Twisted aestivation funneishaped
Androecium – 5 stamens – epipetalous altemi petalous .
Anther – dithecous, basifixed, introse longitudinal dehiseence
Gynoecium – Superior – bicarpellary bilocular,
Ovary – syncarpous basically bilocular later become tetralocular due to the formation of false septa Carpels obliquely placed ovules on swollen axile placentation
Style – simple long flu form
Stigma – bibbed
Fruit – Spinesent capsule opening by four apical valves persistent calyx
Seed – Endospermous

Question 18.
Give an account of botanical description of Allium cepa.
Answer:
Habit – Perrennial herb with bulb
Root – Fibrous adventitious root system
Stem – Underground bulb
Leaf – Radical leaves cylindrical fleshy with sheathing leaf bases & parallel venation
Inflorescence – scafrigerous, pedicels of equal length arising from apex of peduncle
Flower – Small white
Actinomorphic,
Bracteate, eBracteolate Bisexual Complete
Monochiamydeous.
Trimerous – hypogynous
Flowers – Protandrous
Perianth – 6 Tetals in 2 whorls of 3 each syntepalous Valvate acstivatíon
Androecium – 6 stamens in a whorls of 3 each epipelatous
apostamenous
Anther – Dithecous basifixed, introse and longitudinal dehiscence
Gynoecium
Ovary – Superior, tricarpellary trilocular 2 ovules in each locule on axile
placentation
Style – simple, slender
Sligma – simple
Fruit – loculicidal capsule.

Question 19.
Give an account of the Economic importance of Fabaceae in the form of a Tabulation Economic Importance of the family Fabaceae
Economic
Answer:

Question 20.
Economic Importance of Solanaceae
Answer:

Question 21.
Tabulate economic Importance of the family Liliaceae Economic Importance of the family Liliaceae
Answer: