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 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 Computer Science Guide Pdf Chapter 7 Composition and Decomposition Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 7 Composition and Decomposition

11th Computer Science Guide Composition and Decomposition Text Book Questions and Answers

Part I

Choose the correct answer.

Question 1.
Suppose u, v = 10, 5 before the assignment. What are the values of u and v after the . sequence of assignments?
1, u := v
2. v := u
a) u, v = 5, 5
b) u, v = 5, 10
c) u, v = 10, 5
d) u, v = 10, 10
Answer:
a) u, v = 5, 5

Question 2.
Which of the following properties is true after the assignment (at line 3)?
1. — i + j = 0
2. i, j : = i + 1, j – 1
3. –?
a) i + j >0
b) i + j < 0
c) i + j = 0
d) i = j
Answer:
b) i + j < 0

Question 3.
If Cl is false and C2 is true, the compound statement,
1. if C₁
2. S₁
3. else
4. if C₂
5. S₂
6. else
7. S₃
executes
a) S₁
b) S₂
c) S₃
d) none
Answer:
b) S₂

Question 4.
If C is false just before the loop, the control flows through. –
1. S₁
2. while C
3. S₂
4. S₃
a) S₁ ; S₃
b) S₁ ; S₂ ; S₃
c) S₁ ; S₂ ; S₂ ; S₃
d) S₁; S₂ ; S₂ ; S₂ ; S₃
Answer:
a) S₁ ; S₃

Question 5.
If C is true, S₁ is executed in both the flowcharts, but S₂ is executed in

a) (1) only
b) (2) only
c) both (1) and (2)
d) neither (1) nor (2)
Answer:
a) (1) only

Question 6.
How many times the loop is iterated? i := 0
while i ≠ 5
i : = i + 1
a) 4
b) 5
c) 6
d) 0
Answer:
b) 5

Part II

Short Answer Questions

Question 1.
Distinguish between a condition and a statement.
Answer:

Condition	Statement
It is a phrase that describes a test of the state.	It is a phrase that commands the computer to do an action.

Question 2.
Draw a flowchart for a conditional statement.
Answer:

Question 3.
Both conditional statements and iterative statements have a condition and a statement. How do they differ?
Answer:
Conditional Statement:

• Statements are executed only once when the condition is true.
• If condition statement.

Iterative Statement:

• An iterative statement repeatedly evaluates a condition and executes a statement until it becomes false.
• While condition statement.

Question 4.
What is the difference between an algorithm and a program?
Answer:

ALGORITHM	PROGRAM
It is a step-by-step procedure to solve a problem.	It is a set of instructions to solve a problem by the computer.
No need to follow the grammar of a language	Follow strictly the grammar of a programming language.

Question 5.
Why is a function an abstraction?
Answer:
Once a function is defined, it can be used over and over and over again. Reusability of a single function several times is known as an abstraction.

Question 6.
How do we define a statement?
Answer:
In a refinement, starting at a high level, each statement is repeatedly expanded into more detailed statements in the subsequent levels.

Part III

Explain in brief

Question 1.
For the given two flowcharts write the pseudo-code.

Answer:
if condition is True
Execute Statement S₁
else
Execute Statement S₂
endif
PSEUDO CODE FOR SECOND FLOWCHART
if the condition is True
Execute Statement S₁
Execute Statement S₂
else
Execute Statement S₂
endif.

Question 2.
If C is false in line 2, trace the control flow in this algorithm.
1. S₁
2. — C is false
3. if C
4. S₂
5. else
6. S₃
7. S₄
Answer:
The control flow for the given algorithm is as follows:
S₁
S₃
S₄
The condition is false so it executes S₃. In this case S₂ skipped.

Question 3.
What is case analysis?
Answer:
The case Analysis statement generalizes the problem into multiple cases. Case Analysis splits the problem into an exhaustive set of disjoint cases.

Question 4.
Draw a flowchart for -3 case analysis using alternative statements.
Answer:

Where C₁, C₂ and C₃ are conditions
S₁, S₂, S₃ and S₄ are statements.

Question 5.
Define a function to double a number in two different ways: (1) n + n,(2)2 x n.
Answer:
1. Double (n)
– – inputs: n is a real number or an integer, n > 0
– – Outputs: y is a real number or an integer such that y = n + n

2. Double (n)
– – inputs: n is a real number or an integer, n > 0
– – Outputs: y is a real number or an integer such that y = 2 x n

Part IV

Explain in detail

Question 1.
Exchange the contents: Given two glasses marked A and B. Glass A is full of apple drink and glass B is full of grape drink. Write the specification for exchanging the contents of glasses A and B, and write a sequence of assignments to satisfy the specification.
Answer:
The sequence of assignments to satisfy the specification:
exchange(A, B)
–inputs : A, B are real and > 0
–outputs: A, B are real and > 0
State representation
T = A
A: = B
B: = T

Question 2.
Circulate the contents: Write the specification and construct an algorithm to circulate the contents of the variables A, B and C as shown below: The arrows indicate that B gets the value of A, C gets the value of B and A gets the value of C.

Specification :
Circulated,B,C)
– – inputs: A,B,C all are real numbers
– – outputs: A,B,C, all are real numbers
T: = C
C: = B
B: = A
A: = T

Algorithm :
i) circulated,B,C)
ii) – – A,B,C
iii) T = C
iv) C = B
v) B = A
vi) A: = T

Question 3.
Decanting problem. You are given three bottles of capacities 5, 8, and 3 litres. The 8L bottle is filled with oil, while the other two are empty. Divide the oil in an 8L bottle into two equal quantities. Represent the state of the process by appropriate variables. What are the initial and final states of the process? Model the decanting of oil from one bottle to another by assignment. Write a sequence of assignments to achieve the final state.
Answer:
To divide the oil equally in the oil bottle, the following 7 steps are needed:

• Pour 5L of oil into the second bottle. (3,5,0)
• Pour 3L from the second bottle to the third. (3,2,3)
• Pour 3L from the third to the first. (6,2,0)
• Pour all 2L from the second to the third bottle (6,0,2)
• Pour 5L from the first to the second bottle. (1.5.2)
• Pour 1 L from the second to the third bottle. (1.4.3)
• Finally, pour now all 3L from the third bottle to the first (4,4,0)

Assignment statement as follows:

• E,F,T = 8,0,0
• E,F,T = 3,5,0
• E,F,T = 3,2,3
• E,F,T = 6,2,0
• E,F,T = 6,0,2
• E,F,T = 1,5,2
• E,F,T = 1,4,3
• E,F,T = 4,4,0

Question 4.
Trace the step-by-step execution of the algorithm for factorial(4).
factorial(n)
— inputs i rs is an integer, n > 0
— outputs : f ‘=s n!
f, i := 1,1
while i ≤ n
f,i = 11= f x i, i+1
Answer:
Tracing steps:
factorial(5)
i) f = 1, i = 1
ii) f = 1 x 1, i = 2
iii) f = 1 x 2, i = 3
iv) f = 2 x 3, i = 4
v) f = 6 x 4, i = 5
vi) f = 24 x 5, i = 6
vii) condition 6 < 5 becomes false then display output f value as 24.

11th Computer Science Guide Composition and Decomposition Additional Questions and Answers

Part I

Choose the correct answer:

Question 1.
Which one of the following is odd?
(a) Python
(b) C++
(c) C
(d) Ctrl + S
Answer:
(d) Ctrl + S

Question 2.
_________ is a notation for representing algorithms.
a) programming language
b) pseudo code
c) flow chart
d) All the above
Answer:
d) All the above

Question 3.
There are important control flow statements.
(a) four
(b) three
(c) two
(d) five
Answer:
(b) three

Question 4.
_________ is a notation similar to programming languages.
a) programming language
b) pseudo code
c) flow chart
d) none of these
Answer:
b) pseudo code

Question 5.
A ………………. is contained in a rectangular box with a single outgoing arrow, which points to the box to be executed next.
(a) statement
(b) composition
(c) notation
(d) condition
Answer:
(a) statement

Question 6.
_________ is a diagrammatic notation for representing algorithms.
a) programming language
b) pseudo code
c) flow chart
d) none of these
Answer:
c) flow chart

Question 7.
The algorithm can be specified as ……………….
(a) monochromatize (a, b, c)
(b) a = b = 0
(c) C = A + B + C
(d) none
Answer:
(a) monochromatize (a, b, c)

Question 8.
An algorithm expressed in a_________ is called a program.
a) programming language
b) pseudo code
c) flow chart
d) none of these
Answer:
a) programming language

Question 9.
Which one of the following is the elementary problem-solving technique?
(a) Specification
(b) Abstraction
(c) Composition
(d) decomposition
Answer:
(d) decomposition

Question 10.
_________ is formal.
a) programming language
b) pseudo code
c) flow chart
d) none of these
Answer:
a) programming language

Question 11.
How many different notations are there for representing algorithms?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 12.
_________ do not allow the informal style of natural languages such as English or Tamil.
a) programs
b) pseudo code
c) flow charts
d) none of these
Answer:
a) programs

Question 13.
Which one of the following algorithmic notations is used for communication among people?
(a) Flow chart
(b) Pseudo code
(c) PL
(d) Interpreter
Answer:
(b) Pseudo code

Question 14.
_________ notation is not formal nor exact.
a) programs
b) pseudo code
c) flow charts
d) none of these
Answer:
b) pseudo code

Question 15.
The algorithmic notation similar to a Programming language is ……………….
(a) Flow chart
(b) Pseudo code
(c) C ++
(d) C
Answer:
(b) Pseudo code

Question 16.
In _________, there is no need to follow the rules of the grammar of a programming language.
a) programs
b) pseudo code
c) flow charts
d) none of these
Answer:
b) pseudo code

Question 17.
Which one is used for converting programs into computer-executable instructions?
(a) Converter
(b) Apps
(c) Translator
(d) exe files
Answer:
(c) Translator

Question 18.
_________ show the control flow of algorithms using diagrams in a visual manner.
a) programs
b) pseudo code
c) flow charts
d) none of these
Answer:
c) flow charts

Question 19.
The notation which is not formal nor exact is ……………….
(a) Flow chart
(b) Pseudocode
(c) Compiler
(d) Translator
Answer:
(b) Pseudocode

Question 20.
In flowcharts, _________ boxes represent conditions.
a) arrows
b) diamond shaped
c) rectangular shaped
d) none of these
Answer:
b) diamond shaped

Question 21.
Find the pair which is wrongly matched.
(a) Rectangular boxes – Statements
(b) Diamond boxes – Output
(c) Arrow – Control flow
(d) Parallelogram – Input
Answer:
(b) Diamond boxes – Output

Question 22.
A condition is contained in a diamond-shaped box with _________ outgoing arrows.
a) two
b) three
c) one
d) many
Answer:
a) two

Question 23.
The inputs and outputs are drawn using ……………….. boxes.
(a) rectangular
(b) diamond
(c) Parallelogram
(d) Oval
Answer:
(c) Parallelogram

Question 24.
In flow chart, _________ marked Start and the End are used to indicate the start and the end of an execution:
a) special box (oval)
b) diamond shaped
c) rectangular shaped
d) parallelogram
Answer:
a) special box (oval)

Question 25.
The flow of control is represented in the flowchart by ………………..
(a) arrow
(b) dot
(c) box
(d) plus
Answer:
(a) arrow

Question 26.
A_________ is a phrase that commands the computer to do an action.
a) statement
b) sentence
c) walkthrough
d) none of these
Answer:
a) statement

Question 27.
A condition is contained in a diamond-shaped box with ……………….. outgoing arrows.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 28.
Statements composed of other statements are known as_________ statements.
a) compound
b) bock
c) nested
c) none of these
Answer:
a) compound

Question 29.
_________ statement are compound statements.
a) input
b) output
c) comment
d) control flow
Answer:
d) control flow

Question 30.
How many outgoing arrows are needed for rectangular boxes in the flow chart?
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 31.
___________ is a control flow statements.
a) Sequential
b) Alternative
c) Iterative
d) All the above
Answer:
d) All the above

Question 32.
Which one of the following is not a control flow statement?
(a) Sequential
(b) Assignment
(c) Iterative
(d) Alternative
Answer:
(b) Assignment

Question 33.
Case analysis splits the problem into an exhaustive set of __________ cases.
a) disjoint
b) similar
c) recursive
d) none of these
Answer:
a) disjoint

Question 34.
Which one of the following statements are executed one after the other as written in the algorithm?
(a) Sequential
(b) Iterative
(c) Conditional
(d) Decisive
Answer:
(a) Sequential

Question 35.
An__________ executes the same action repeatedly, subject to a condition.
a) iterative process
b) condition
c) sequential
d) None of these
Answer:
a) iterative process

Question 36.
Case analysis statement generalizes the statement into ……………….. cases.
(a) 2
(b) 3
(c) 5
(d) multiple
Answer:
(d) multiple

Question 37.
Testing the loop condition and executing the , loop body once is called.
a) iteration
b) condition
c) sequential
d) none of these
Answer:
a) iteration

Question 38.
Which one of the following processes executes the same action repeatedly?
(a) Conditional
(b) Alternative
(c) Iterative
(d) None of these
Answer:
(c) Iterative

Question 39.
__________ breaking and combining the solutions of the smaller problems to solve the original problem.
a) Composition
b) Decomposition
c) Eliminating
d) None of these
Answer:
b) Decomposition

Question 40.
Testing the loop condition and executing the loop body once is called ………………..
(a) alternative
(b) conditional
(c) Iteration
(d) Decomposition
Answer:
(c) Iteration

Question 41.
A(n)__________ is like a sub-algorithm.
a) function
b) array
c) structure
d) None of these
Answer:
a) function

Question 42.
A __________ can be used as a black box in solving other problems.
a) function
b) array
c) structure
d) None of these
Answer:
a) function

Question 43.
Identify the correct statement from the following.
a) There is no need for the users to know how the function is implemented in order to use it.
b) An algorithm used to implement a function may maintain its own variables.
c) users of the function need only to know what the function does, and not how it is done by the function.
d) All the above
Answer:
d) All the above

Question 44.
Conditional statement is executed only if the condition is true. Otherwise, __________ .
a) terminates the algorithm
b) repeat the procedure
c) skip the loop
d) nothing is done
Answer:
d) nothing is done

Part II

Short Answers

Question 1.
What is programming language?
Answer:
A programming language is a notation for expressing algorithms to be executed by computers.

Question 2.
What is a Pseudocode?
Answer:
Pseudocode is a mix of programming languages like constructs and plain English. Algorithms expressed in Pseudocode are not intended to be executed by computers but for human readers to understand.

Question 3.
What is flowchart?
Answer:
A flowchart is a diagrammatic notation for representing algorithms. They give a visual intuition of the flow of control, when the algorithm is executed.

Question 4.
What is a control flow statement? Classify it.
Answer:
Control flow statements are compound statements. They are used to alter the control flow of the process depending on the state of the process. They are classified as:

1. Sequential
2. Alternative
3. Iterative

Question 5.
What are the constraints of a programming language?
Answer:

• Programming language is formal.
• Programs must obey the grammar of the programming language exactly. Even punctuation symbols must be exact.
• They do not allow the informal style of natural languages such as English or Tamil.

Question 6.
When a condition statement will be executed?
Answer:
It will be executed only when the condition statement is true.

Question 7.
What is the advantage of a flowchart?
Answer:
They show the control flow of algorithms using diagrams in a visual manner.

Question 8.
What is a statement?
Answer:
A statement is a phrase that commands the computer to do an action.

Question 9.
What do you mean by compound statement?
Answer:
Statements may be composed of other statements, leading to a hierarchical structure of algorithms. Statements composed of other statements are known as compound statements.

Question 10.
What are the control flow statements?
Answer:
There are three important control flow statements:

• Sequential
• Alternative
• Iterative

Question 11.
What happen when a control flow statement is executed?
Answer:
When a control flow statement is executed, the state of the process is tested, and depending on the result, a statement is selected for execution.

Question 12.
Write the specification for finding minimum of two numbers.
Answer:
The specification of algorithm minimum is minimum (a, b)
– – inputs: a , b
– – outputs: result = a ↓ b.

Part III

Explain in brief

Question 1.
How many notations are there for representing algorithms? Explain.
Answer:
There are mainly three different notations for representing algorithms.

• A programming language is a notation for expressing algorithms to be executed by computers.
• Pseudocode is a notation similar to programming languages. Algorithms expressed in pseudo-code are not intended to be executed by computers, but for communication among people.
• A flowchart is a diagrammatic notation for representing algorithms. They give a visual intuition of the flow of control, when the algorithm is executed.

Question 2.
What are the disadvantages of a flowchart?
Answer:
Flowcharts also have the following disadvantages:

• Flowcharts are less compact than the representation of algorithms in a programming language or pseudo code.
• They obscure the basic hierarchical structure of the algorithms.
• Alternative statements and loops are disciplined control flow structures.

Question 3.
Write a note on refinement.
Answer:
After decomposing a problem into smaller subproblems, the next step is either to refine the subproblem or to abstract the subproblem.
1. Each subproblem can be expanded into more detailed steps. Each step can be further expanded to still finer steps, and so on. This is known as refinement.

2. We can also abstract the subproblem. We specify each subproblem by its input property and the input-output relation. While solving the main problem, we only need to know the specification of the subproblems. We do not need to know how the subproblems are solved.

Question 4.
Write an algorithm that compares numbers and produces the result as

We can split the state into an exhaustive set of 3 disjoint cases: a < b, a = b, and a> b. Then we can define compare() using a case analysis. The algorithm is as given below.

1. compare(a, b)
2. case a < b
3. result := -1
4. case a = b
5. result := 0
6. else – – a > b
7. result: = 1

Question 5.
Construct specification and write an iterative algorithm to compute the quotient and remainder after dividing an integer A by another integer B.
Answer:
Specification: divide (A, B)
— inputs: A is an integer and B ≠ 0
— outputs : q and r such that A = q x B
+ r and 0 ≤ r < B

Algorithm:
divide (A, B)
— inputs: A is an integer and B * 0
— outputs : q and r such that
A = q X B + r and
— 0 < r < B
q := 0, A
while r ≥ B
q, r := q + 1, r – B

Question 6.
Explain Decomposition.
Answer:
It involves breaking down a problem into smaller and more manageable problems, and combining the solutions of the smaller problems to solve the original problem. Often, problems have structure.
We can exploit the structure of the problem and break it into smaller problems. Then, the smaller problems can be further broken until they become sufficiently small to be solved by other simpler means.
Their solutions are then combined together to construct a solution to the original problem.

Question 7.
Construct an iterative algorithm to compute the quotient and remainder after dividing an integer A by another integer B.
Answer:
divide (A, B)
– – inputs: A is an integer and B ≠ 0
– – Outputs: q and r such that A = q x B + r and 0 < r < B
q : = 0, A
While r ≥ B
q, r : = q + 1, r – B

Question 8.
Draw a flow chart for Eat breakfast.
Answer:
Flowchart for Eat breakfast

Part IV

Explain in detail

Question 1.
What is flowchart? Explain the various boxes (symbols) used in flowchart.
Answer:
A flowchart is a diagrammatic notation for representing algorithms. They show the control flow of algorithms using diagrams in a visual manner.

In flowcharts, rectangular boxes represent simple statements, diamond-shaped boxes
represent conditions, and arrows describe how the control flows during the execution of the algorithm. A flowchart is a collection of boxes containing statements and conditions which are connected by arrows showing the order in which the boxes are to be executed,

i) A statement is contained in a rectangular box with a single outgoing arrow, which points to the box to be executed next.

ii) A condition is contained in a diamond-shaped box with two outgoing arrows, labeled true and false. The true arrow points to the box to be executed next if the condition is true, and the false arrow points to the box to be executed next if the condition is false.

iii) Parallelogram boxes represent inputs given and outputs produced.

iv) Special boxes marked Start and the End is used to indicate the start and the end of execution.

Question 2.
Draw a flowchart to compute the quotient and remainder after dividing an integer A by another integer B.
Answer:
Flowchart for integer division

Question 3.
Explain sequential statement with suitable example.
Answer:
A sequential statement is composed of a sequence of statements. The statements in the sequence are executed one after another, in the same order as they are written in the algorithm, and the control flow is said to be sequential. Let S₁ and S₂ be statements. A sequential statement composed of S₁ and S₂ is written as
S₁
S₂
In order to execute the sequential statement, first, do S₁ and then do S₂.

The sequential statement given above can be represented in a flowchart as shown in in the following Figure. The arrow from S₁ to S₂ indicates that S₁ is executed, and after that, S₂ is executed.

Let the input property be P, and the input-output relation be Q, for a problem. If statement S solves the problem, it is written as
i) –P
ii) S
iii) — Q

Question 4.
Explain alternative statement with an example. A condition is a phrase that describes a test of the state. If C is a condition and both SI and S2 are statements, then

if C
S₁
else
S₂

is a statement, called an alternative statement, that describes the following action:
i) Test whether C is true Or false.
ii) If C is true, then do SI; otherwise do S2.

In pseudo-code, the two alternatives S1 and S2 are indicated by indenting them from the keywords if and else, respectively. Alternative control flow is depicted in the following flowchart.

Condition C has two outgoing arrows, labeled true and false. The true arrow points to the SI box. The false arrow points to the S2 box. Arrows out of S1 and S2 point to the same box, the box after the alternative statement.

Question 5.
Explain conditional statement with suitable example.
Answer:
Sometimes we need to execute a statement only if a condition is true and do nothing if the condition is false. This is equivalent to the alternative statement in which the else-clause is empty. This variant of alternative statement is called a conditional statement. If C is a condition and S is a statement, then
if C
S
is a statement, called a conditional statement, that describes the following action:

• Test whether C is true or false.
• If C is true then do S; otherwise do nothing.

The conditional control flow is depicted in the following flowchart.

Question 6.
Write a note on case analysis.
Answer:
An alternative statement analyses the problem into two cases. The case analysis statement generalizes it to multiple cases. Case analysis splits the problem into an exhaustive set of disjoint cases. For each case, the problem is solved independently. If C1, C2, and C3 are conditions, and S1, S2, S3, and S4 are statements, a 4 – case analysis statement has the form.

1. case C1
2. S1
3. case C2
4. S2
5. case C3
6. S3
7. else
8. S4

The conditions C1, C2, and C3 are evaluated in turn. For the first condition that evaluates to true, the corresponding statement is executed, and the case analysis statement ends. If none of the conditions evaluates to true, then the default case S4 is executed.

1. The cases are exhaustive: at least one of the cases is true. If all conditions are false, the default case is true.
2. The cases are disjoint: only one of the cases is true. Though it is possible for more than one condition to be true, the case analysis always executes only one case, the first one that is true.
3. f the three conditions are disjoint, then the four cases are (1) C1, (2) C2, (3) C3, (4) (not C1), and (not C2) and (not C3).

Question 7.
Explain iterative statement with suitable example.
Answer:
An iterative process executes the same action repeatedly, subject to a condition C.
If C is a condition and S is a statement, then
while C
S
is a statement, called an iterative statement, that describes the following action:

• Test whether C is true or false.
• If C is true, then do S and go back to step 1; otherwise do nothing.

The iterative statement is commonly known as a loop. These two steps, testing C and executing S, are repeated until C becomes false. When C becomes false, the loop ends, and the control flows to the statement next to the iterative statement.

The condition C and the statement S are called the loop condition and the loop body, respectively. Testing the loop condition and executing the loop body once is called an iteration, not C is known as the termination condition.
Iterative control flow is depicted in the following flowchart

Condition C has two outgoing arrows, true and false. The true arrow points to S box. If C is true, S box is executed and control flows back to C box. The false arrow points to the box after the iterative statement (dotted box). If C is false, the loop ends and the control flows to the next box after the loop.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 13 Photosynthesis Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis

11th Bio Botany Guide Photosynthesis Text Book Back Questions and Answers

Part-I

Questions 1.
Assertion (A): Increase in Proton gradient inside lumen responsible for ATP synthesis.
Reason (R): Oxygen evolving complex of PS I located on the thylakoid membrane facing Stroma, releases H⁺ ions.
(a) Both Assertion and Reason are True.
(b) Assertion is True and Reason is False.
(c) Reason is True and Assertion is False.
(d) Both Assertion and Reason are False.
Answer:
(a) Both Assertion and Reason are True.

Question 2.
Which chlorophyll molecule does not have a phytol tail?
a) Chl – a
b) Chl – b
c) Chl – c
d) Chl – d
Answer:
(c) Chl – c

Question 3.
The correct sequence of flow of electrons in the light reaction is:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.
(b) PS I, plastoquinone, cytochrome, PS II ferredoxin.
(c) PS II, ferredoxin, plastoquinone, cytochrome, PS I.
(d) PS I, plastoquinone, cytochrome, PS II, ferredoxin.
Answer:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.

Question 4.
For every CO₂ molecule entering the C₃ cycle, the number of ATP & NADPH required
a) 2ATP + 2NADPH
b) 2ATP + 3NADPH
c) 3ATP + 2NADPH
d) 3ATP + 3NADPH
Answer:
(c) 3ATP + 2NADPH

Question 5.
Identify true statement regarding light reaction of photosynthesis?
(a) Splitting of water molecule is associate with PS I.
(b) PS I and PS II involved in the formation of NDPH + H⁺.
(c) The reaction center of PS I is Chlorophyll a with absorption peak at 680 nm.
(d) The reaction center of PS II is Chlorophyll a with absorption peak at 700 nm.
Answer:
(b) PS I and PS II involved in the formation of NDPH + H⁺.

Question 6.
Two groups (A&B) of bean plants of similar size and same leaf area were placed in identical conditions. Group A was exposed to light of wavelength 400 – 450 nm & Group B to light of wavelength of 500 – 550nm. Compare the photosynthetic rate of the 2 groups giving reasons.
Answer:
‘A’ group of plants exposed to light of 400 – 450nm. Chlorophyll a shows maximum absorption peak at 450nm (blue region). Hence rate of photosynthesis was high.
‘B’ group of plants exposed to light of 500 – 550nm. This wavelength refers to green region of the spectrum. Chlorophyll does not absorb light in the green region but reflects green. So plants appear green rate of photosynthesis was negligible in these plants.

Question 7.
A tree is believed to be releasing oxygen during nighttime. Do you believe the truthfulness of this statement? Justify your answer by giving reasons?
Answer:
Yes, a tree is believed to be releasing O₂ during nighttime because at night CAM plants fix CO₂ with the help of phospho Enol Pyruvic acid and produce oxala acetic acid, which is converted into malic acid-like C₄ cycle.

Question 8.
Grasses have an adaptive mechanism to compensate for photorespiratory losses – Name and describe the mechanism.
Answer:
The photorespiratory losses are checked by certain grasses by having physiological adaptation. The process of photosynthesis occurs in mesophyll cells and bundle sheath cells.

Mesophyll cells:

• Initially, CO₂ is taken up by Phosphoenolpyruvate (PEPA) (3C) and changed to oxaloacetate (4C) in the presence of PEP carboxylase.
• Oxaloacetate is reduced to Malate/Aspartate. The product formed reaches the bundle sheath.

Bundle Sheath:

• The oxidation of Malate and Aspartate occurs with the release of carbon dioxide and the formation of Pyruvate (3C)
• Due to increased CO₂ concentration RUBISCO functions as a carboxylase and not as Oxygenase.
• The photosynthetic losses are prevented.
• RUBP operates now under the Calvin cycle and pyruvate transported back to Mesophyll cells is changed into Phosphoenolpyruvate to keep the cycle going.

Question 9.
In Botany class, teacher explains, Synthesis of one glucose requires 30 ATPs in C₄ plants and only 18 ATPs in C3 plants. The same teacher explains C₄ plants are more advantageous than C₃ plants. Can you identify the reason for this contradiction?
Answer:
C₄ Plants are more advantageous than C₃ plants because of the following reasons:

C4 Plants	C3 Plants
CO2 fixation occurs in mesophyll cells only	CO2 fixation occurs in mesophyll and bundle sheath cells
RUBP is the only CO2 acceptor	PEPA Phosphoenol pyruvate in mesophyll is the acceptor in the first phase
Fixation of CO2 occurs if the atmospheric concentration of C02 is 50 ppm only	It can fix carbon dioxide even if the atmospheric concentration of CO2 is below 10 ppm
Optimum temperature is 20° to 25°C	Optimum temperature is 30° to 45°C and is thus effective in tropical regions.
RUBP carboxylase enzyme also functions as oxygenase if the 0, concentration is higher than carbon dioxide	PEP carboxylase enzyme functions even at low carbon – dioxide concentrations.
Higher rate of photorespiration and hence rate of photosynthesis is reduced.	Minimal rate of photorespiration is seen is C4 plants.

Question 10.
When there is plenty of light and higher concentration of O₂, what kind of pathway does the plant undergo? Analyse the reasons.
Answer:
The rate of photosynthesis decreases when there is an increase in oxygen concentration. This Inhibitory effect of oxygen was first discovered by Warburg (1920) using green algae, Chlorella.

Part-II.

11th Bio Botany Guide Photosynthesis Additional Important Questions and Answers

I. Choose the Correct Answers

Question 1.
Photosynthesis is the major:
(a) endothermic reaction
(b) exothermic reaction
(c) endergonic reaction
(d) exergonic reaction
Answer:
(c) endergonic reaction

Question 2.
The physiological unit of photosynthesis is
a) 150-250 Chlorophyll molecules
b) 200-300 chlorophyll molecules
c) 440-660 chlorophyll molecules
d) 450- 650 chlorophyll molecules
Answer:
b) 200-300 chlorophyll molecules

Question 3.
How many million tonnes of dry matter produced annually by photosynthesis?
(a) 1700 million tonnes
(b) 1900 million tonnes
(c) 1400 million tonnes
(d) 2000 million tonnes
Answer:
(a) 1700 million tonnes

II. Match Correctly & Choose The Right Answer

Question 4.
I) Black Mann – A) the importance of Chlorophyll
II) Warburg – B) Law of limiting factor
III) Dustrochet – C) C4 cycle
IV) Hatch & Slack – Chlorella

Answer:
c) B-D-A-C

Question 5.
Thylakoid disc diameter is:
(a) 0.35 to 0.75 microns
(b) 0.25 to 0.8 microns
(c) 0.45 to 0.8 microns
(d) 0.50 to 0.9 microns
Answer:
(b) 0.25 to 0.8 microns

Question 6.
The pigment responsible for the yellowing of leaves during autumn season is
a) Violaxanthin
b) Fucoxanthin
c) Phycobillin
d) Lycopene
Answer:
d) Lycopene

Question 7.
The no of quanta of light required for the release of one oxygen molecule
a) 18 quanta
b) 8 quanta
c) 81 quanta
d) 19 quanta
Answer:
b) 8 quanta

Question 8.
Each pyrrole ring comprises of:
(a) six carbons and one nitrogen atom
(b) three carbons and one nitrogen atom
(c) four carbons and one nitrogen atom
(d) four carbons and two nitrogen atom
Answer:
(c) four carbons and one nitrogen atom

Question 9.
RUBISCO – Constitute …………. of chloroplast protein
a) 17%
b) 20%
c) 18%
d) 16%
Answer:
d) 16%

Question 10.
Pheophytin resembles chlorophyll ‘a’ except that it lacks:
(a) Fe atom
(b) Mn atom
(c) Mg atom
(d) Cu atom
Answer:
(c) Mg atom

Question 11.
According to Emerson the fall in quantum yield about 680 nm is called
a) Phoisynthtic drop
b) Emerson drop
c) Airburg effect
d) Red drop
Answer:
d) Red drop

Question 12.
Which one of the photosynthetic pigments is called shield pigment:
(a) carotenes
(b) chlorophyll ‘b’
(c) pheophytin
(d) carotenoids
Answer:
(d) carotenoids

Question 13.
Which of the following equation correctly sums up photosynthesis

Answer:

Question 14.
Photosynthetic rate of red light (650 nm) is equal to:
(a) 42.5
(b) 10.0
(c) 43.5
(d) 40.8
Answer:
(c) 43.5

Question 15.
Which photosystem is found to be located on the outer surface of the thylakoid?
a) PS I
b) P.S II
c) P. 890
d) Both (a) and (b)
Answer:
a) P.S I

Question 16.
Phosphorylation taking place during respiration is called as:
(a) Photophorylation
(b) Oxidative phosphorylation
(c) Reductive phosphorylation
(d) None of the above
Answer:
(b) Oxidative phosphorylation

Question 17.
The term Quantosome was coined by
a) Emerson
b) Liebig
c) Calvin & Melvin
d) Park & Biggins
Answer:
d) Park & Biggins

Question 18.
In bioenergetics of light reaction, to release one electron from pigment system it requires:
(a) two quanta of light
(b) four quanta of light
(c) one quantum of light
(d) eight quanta of light
Answer:
(a) two quanta of light

Question 19.
Photosynthesis produces
a) 1700 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year
b) 7100 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year
c) 1600 million tonnes of dry matter/week by fixing 56 x 12¹⁰ kg of carbon every week
d) 6100 million tonnes of dry matter/month by fixing 100 x 10¹² kg of carbon every month
Answer:
a) 1700 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year

Question 20.
In C₄ plants, how many ATPs and NADPH + H⁺ are utilised for the release of one oxygen molecule:
(a) 3 ATPs and 2 NADPH + H⁺
(b) 4 ATPs and 3 NADPH + H⁺
(c) 2 ATPs and 2 NADPH + H⁺
(d) 5 ATPs and 2 NADPH + H⁺
Answer:
(d) 5 ATPs and 2 NADPH + H⁺

Question 21.
Products of light reaction in photosynthesis are
a) ATP & NADPH₂
b) ADP & glucose
c) Ferredoxin and cytochrome b6
d) Cytochrome
Answer:
a) ATP & NADPH₂

Question 22.
In the sugarcane plant, the dicarboxylic acid pathway was first discovered by:
(a) Hatch and Slack
(b) Kortschak, Hart and Burr
(c) Calvin and Benson
(d) Mitchell and Root
Answer:
(b) Kortschak, Hart and Burr

Question 23.
Photosynthetic pigments in chloroplasts lie embedded in
a) Chloroplast envelope
b) Plastogloblue
c) matrix
d) thylakoids
Answer:
d) thylakoids

Question 24.
Indicate the correct answer:
(a) C₄ plants are adapted to only rainy conditions
(b) C₄ plants are partially adapted to drought condition
(c) C₄ plants are exclusively adapted to desert condition
(d) C₄ plants are adapted to aquatic condition
Answer:
(b) C₄ plants are partially adapted to drought condition

Question 25.
Carotenoids and Xanthophylls are also known as
a) Respiratory pigments
b) Accessory pigments
c) Photosynthetic pigments.
d) Photolytic pigments
Answer:
b) Accessory pigments

Question 26.
Which metal ion is a
a) Iron
b) cobalt
c) Magnesium
d) Zinc
Answer:
c) Magnesium

Question 27.
The important external factors affecting photosynthesis are:
(a) light, chlorophyll, temperature
(b) light, stomatal opening, oxygen
(c) light, protoplasmic factor, oxygen
(d) light, CO₂ and oxygen
Answer:
(d) light, CO₂ and oxygen

Question 28.
The process of photophosphorylation was discovered by
a) Priestly
b) Calvin
c) Amon
d) Warburg
Answer:
c) Arnon

Question 29.
Which of the following is a C4 plant
a) Potato
b) Sugarcane
c) Pea
d) Papaya
Answer:
b) Sugarcane

Question 30.
Splitting of water molecule (photolysis) produces:
(a) hydrogen and oxygen
(b) electrons, protons and oxygen
(c) electrons and oxygen
(d) hydrogen, carbon dioxide and oxygen
Answer:
(b) electrons, protons and oxygen

Question 31.
Dimorphism in chloroplasts is seen in
a) C4 plants
b) C₂ plants
c) CAM – plants
d) C3 plants
Answer:
a) C4 plants

Question 32.
Energy required for ATP synthesis in PSII comes from
a) Proton gradient
b) Electron gradient
c) Reduction of glucose
d) Oxidation of glucose
Answer:
a) Proton gradient

Question 33.
The by-product of Photosynthesis is
a) O₂
b) CO₂
c) Carbohydrate
d) H₂O
Answer:
a) O₂

Question 34.
Which of the following process is called reverse of Glycolysis?
a) CO₂ reduction
b) RUBP carboxylation
c) RUBP regeneration
d) ATP synthesis
Answer:
a) CO₂ reduction

Question 35.
The Dark reaction of Photosynthesis occurs in
a) Matrix
b) Grana
c) Stroma
d) Cytoplasm
Answer:
c) Stroma

Question 36.
A granal chloroplasts are characteristics of
a) Mesophyll of pea leaves
b) Bundle sheath of Mango leaves
c) Mesophyll of maize leaves
d) Bundle sheath of sugar cane leaves
Answer:
d) Bundle sheath of Sugar cane leaves

Question 37.
Which of the following plant is a better photosynthesis?
a) Mango
b) Sugarcane
c) Wheat
d) Rice
Answer:
b) Sugarcane

Question 38.
The enzyme that is not found in a C3 plant is
a) RUBP carboxylase
b) PEP carboxylase
c) NADP reductase
d) ATP synthase
Answer:
b) PEP carboxylase

Question 39.
Which of the following factors affect the rate of photosynthesis?
I. Light
II. Protoplasmic factor
III. Hormones Codes
IV. Haemoglobin
a) only III
b) I and II
c) only IV
d) I, II, and III
Answer:
d) I, II and III

Question 40.
Match the following columns

Column I	Column II
I. The 5C sugar that	A. RUBIS Co
II. 3 C sugar that gives Calvin cycle  its nickname	B. Glyceraldehyde 3 phosphate
III. Activated form of  3 PGA	C. 3 phosphoglyceric acid
IV. Huge enzyme complex that brings CO2 and 5C sugar together.	D. RUBP

Question 41.
One complete light reaction involves light energy.
a) 30 quanta
b) 48 quanta
c) 40 quanta
d) 25 quanta
Answer:
b) 48 quanta

Question 42.
During photosynthesis which of the following event does not take place?
a) oxidation of CO₂
b) Reduction of CO₂
c) oxidation of H₂O
d) Light absorption
Answer:
a) Oxidation of CO₂

Question 43.
The site of light trapping in the chloroplast is
a) Thylakoid membrane
b) Stroma
c) Plasma fluid
d) Stromal lamellae
Answer:
a) Thylakoid membrane

Question 44.
Kranz anatomy is traced in the Leaves of
a) Wheat
b) Potato
c) Mustard
d) Sugarcane
Answer:
d) Sugarcane

Question 45.
The intermediate got from Kreb’s cycle that is used for chlorophyll synthesis is
a) Citric acid
b) Isocitric acid
c) Succinic acid
d) Fumaric acid
Answer:
c) Succinic acid

Question 46.
The existence of light and dark reaction of photosynthesis was proved by
a) Blackman
b) Emerson
c) Warburg
d) Arnon
Answer:
a) Blackman

Question 47.
Choose the wrong match.
a) Hatch & Slack – Dicarboxylic acid pathway
b) Decker – PCO cycle
c) Ruben, Kamen – CAM cycle
d) Calvin Benson – PCR cycle
Answer:
c) Ruben, Kamen – CAM cycle

Question 48.
I) Primary CO, acceptor – PEPA
II) 4C compound produced 1 st – OAA
III) 1st carboxylation occurs in – Bundle sheath cells
IV) 2nd carboxylation occurs in – Mesophy 11 cells
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer:
c) III & IV

Question 49.
Say True or False with respect to C₂ cycle
I) RUBISCO has the most abundant protein on earth.
II) Photorespiration does not yield any free energy in the form of ATP
III) The end product is a 2 – c compound. So the cycle is known as C₂ cycle
IV) Under certain conditions 50% of the photosynthetic potential is lost because of photorespiration.

Answer:
b) True – True – False – True

Question 50.
Say True or False and choose the right option from the given choice.
I) PAR is between – 400 – 700 mil.
II) Heliophytes (Beans) require higher light intensity than sccophytes (oxalis)
III) Red light induces lowest rate of photosynthesis.
IV) Green light induces highest rate of photosynthesis.

Answer:
c) True – True – False – False

Question 51.
Choose the wrongly matched pair
a) Stephen hales – Father of plant physiology
b) Lavoisier – Purifying gas oxygen is produced in sun light
c) Vonmayer – Green plants convert solar energy into chemical energy
d) Emerson &Amoid – C4 cycle
Answer:
d) Emerson & Arnold – C4 cycle

Question 52.
Choose the rightly matched pair
a) Chlorophyll a – Accessory pigments and trap solar energy
b) Chlorophyll b – Differs from Chlorophyll a in having CH3 instead of CHO – at 3rd C atom
c) Chlorophyll c – Differs from Chlorophyll a by lacking a phytol tail
d) Chlorophyll d – It has CHO at 3rd at the 3rd carbon atom at 11 – pyrrole ring
Answer:
c) Chlorophyll c – Differs from chlorophyll a by lacking a phytol tail.

Question 53.
Choose the right matched pairs from the given options.
I) Green non sulphur bacteria – Clostridium & Lynbya
II) Green sulphur bacteria – Chlorobacterium & Chlorobium
III) Purple sulphur bacteria – Thiospirillum & Chromatium
IV) Purple non sulphur bacteria – Rhodopseudomonas & Rhodospirillum
a) I, II, & III
b) II, III & IV
C) I, II & IV
d) I, III & IV
Answer:
b) II, III & IV

II. Assertion (A) & Reason (R)

a) Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion
b) Both Assertion (A) and Reason (R) are true – Reason is not the correct explanation of Assertion.
c) Assertion (A) is false but Reason (R) is true.
d) Both Assertion (A) and Reason (R) are false.

Question 1.
Assertion (A): Chlorophyll appears green.
Reason (R): It absorbs light mainly in the region of green part of light spectrum.
Answer:
c) Assertion (A) is False but Reason (R) is true.

Question 2.
Assertion (A): Red of spectrum contains high energy.
Reason (R): Green light of Visible spectrum contain low energy than red light.
Answer:
c) Assertion (A) is False but Reason (R) is true.

Question 3.
Assetion (A): Non cyclic photo phosphorylation occurs in the stroma of chloroplasts
Reason (R): There is a continuous flow of electrons in this process.
Answer:
d) Both Assertion (A) and Reason (R) are false.

Question 4.
Assetion (A): Carotenes and Xanthophylls are soluble in either
Reason (R): These are accessory pigments of photosynthesis
Answer:
b) Both Assertion (A) and Reason (R) are true, Reason is not the correct explanation of Assertion

Question 5.
Assetion (A): Carotenoids are accessory pigments
Reason (R): Absorbed light energy is transferred to reaction centre by carotenoids.
Answer:
a) Assertion (A) and Reason (R) True and Reason is the correct explanation of Assertion.

III. 2 Marks Questions

Question 1.
What is the function of the plant in the universe?
Answer:
Plants are the major machinery which produces organic compounds like carbohydrates,lipids, proteins, nucleic acids and other biomolecules.

Question 2.
What is PAR?
Answer:
It refers to Photosynthetically Active Radiation, which is between 400 – 700 nm photosynthetic rate is maximum in blue and red light – Green light induces lowest rate of photosynthesis.

Question 3.
What is the site of photosynthesis?
Answer:
Chloroplasts are the main site of photosynthesis and both the energy-yielding process (Light reaction) and fixation of carbon dioxide (Dark reaction) that takes place in the chloroplast.

Question 4.
Name the photosynthetic pigments of Algae?
Answer:

• Chlorophyll b – Green Algae
• Chlorophyll c – Dianoflagellates, Diatoms & Brown Algae
• Chlorophyll d – Red Algae
• Chlorophyll e – Xantho phycean Algae.

Question 5.
Endosymbiotic hypothesis says that chloroplasts evolved from bacteria. Substantiate the statement.
Answer:
Presence of 70S ribosome and DNA gives them status of semi-autonomy and proves endosymbiotic hypothesis which says chloroplast evolved from bacteria.

Question 6.
Write down the significance of photorespiration.
Answer:

• Glycine and Serine synthesized during this process are precursors of many biomolecules like Chlorophyll, Proteins, Nucleotides.
• It consumes excess NADH + H⁺ generated.
• Glycolate protects cells from Photooxidation.

Question 7.
What are the conclusions of Hill’s Reaction?
Answer:

• During Photosynthesis oxygen is evolved from water.
• Electrons for the reduction of CO₂ are obtained from water.
• A reduced substance produced, later helps to reduce CO₂
• 2H₂O + 2A→ 2 AH₂ + O₂

Question 8.
What are Xanthophylls?
Answer:
Yellow (C₄₀H₅₆O₂) pigments are like carotenes but contain oxygen. Lutein is responsible for yellow colour change of leaves during autumn season. Examples: Lutein, Violaxanthin and Fueoxanthin.

Question 9.
Notes on Phycobillins.
Answer:

• They are proteinaceous pigments.
• They are soluble in water.
• Lack ‘Mg’ and phytol tail.
• There are 2 forms 1. Phycocyanin 2. Phycoerythrin.
• Phycocyanin occur in Cyanobacteria.
• Phyco erythrin occur in Rhodophycean Algae.

Question 10.
Define absorption spectrum.
Answer:
Pigments absorb different wavelengths of light. A curve obtained by plotting the amount of absorption of different wavelengths of light by a pigment is called its absorption spectrum.

Question 11.
Why do we call carotenoids shield pigments?
Answer:

• Carotenoids are yellow to orange pigments mostly tetraterpens and absorb light strongly in the blue to violet region of the visible spectrum.
• These pigments protect chlorophyll from photosynthetic oxidative damage.

Question 12.
What is known as substrate-level phosphorylation?
Answer:
Phosphorylation taking place during respiration is called oxidative phosphorylation and ATP produced by the breakdown of substrate is known as substrate-level phosphorylation.

Question 13.
What are Quantosomes?
Answer:

• They are physiological photosynthetic units, located on the inner membrane of thylakoid lamellae of size 180A X 160 A length & breadth.
• It was named by Park &Pickins( 1964).
• One quantosome contains about 230 chlorophyll molecules.
• It constitutes a photosynthetic unit responsible for the production of one O₂ molecule or reduction of one CO₂ molecule.

Question 14.
What is Bioluminescence?
Answer:
It is the special aspect of few living organism, in which there are some biochemical substances production is responsible for the emission of light by a living organism.

Question 15.
What is the significance of photorespiration?
Answer:
Significance of photorespiration:

1. Glycine and Serine synthesized during this process are precursors of many biomolecules like chlorophyll, proteins, nucleotides.
2. It consumes excess NADH + H⁺ generated.
3. Glycolate protects cells from Photooxidation.

Question 16.
Explain water oxidizing clock or S state mechanism.
Answer:
The splitting of water molecule, mechanism was studied by KoK et, al (1970).
It consists of a series of 5 states so, s₁, s₂, s₃, s₄.
Each sate acquires positive charge by a photon (hv) and after the state s₄ – if acquires 4 positive charges 4 electron and evolution of oxygen.

Two molecules of water go back to the so.
At the end of photolysis 4H⁺, 4e^– and O₂ are evolved from water.
4H₂O → 4H⁺+ + 40H^–
40H^{–  →} 2H₂O+O₂+4e^–
2H₂O → 4H⁺+ O₂ + 4e^–.

Question 17.
Distinguish between photophosphorylation and oxidative phosphorylation.
Answer:

Photophosphorylation	Phosphorylation
1. It is the process of synthesis of ATP from ADP by the addition of phosphate takes place with the help of photosynthesis light generated electron, which is known as photophosphorylation. It is of two types Cyclic and Non-cyclic photophosphorylation.	1. The process of production of ATP via terminal oxidation of reduced coenzymes during respiration is known as oxidative phosphorylation.

Question 18.
What are the air pollutants, that affect the rate of photosynthesis?
Answer:
Pollutants like SO₂, NO₂, O₃ (Ozone) and Smog affect the rate of photosynthesis.

Question 19.
What are the conditions for the occurrence of Non-cyclic photophosphorylation?
Answer:

• Non-Cyclic Photo Phosphorylation occurs, when.
• There is the availability of NADP⁺ for reduction.
• Two molecules of water go back to the so.
• When there is the splitting of water molecules.
• When both PSI and PS II are activated.

Question 20.
Name any three photosynthetic bacteria.
Answer:
Three photosynthetic bacteria:

1. Chlorobacterium
2. Thiospirillum
3. Rodhospirillum

Question 21.
What are the 2 phases of light reaction?
Answer:
The light reaction has 2 phases

1. Photooxidation phase
2.  Photochemical phase.

I) Photooxidation phase (POP):

•  Absorption of light energy.
• Transfer of energy from accessory pigments to the reaction centre.
• Activation of chlorophyll ‘a’ -molecule.

II) Photochemical phase (PCO):

• Photolysis of water and evolution of oxygen.
• Electron transport and synthesis of assimilatory power.

Question 22.
Greenlight induces lowest rate of photosynthesis justifies.
Answer:

• Yes green light induces the lowest rate of photosynthesis because it is not coming under photosynthetically
• Active reduction – (400 – 700 nm) known as PAR.
• PAR – (Photosynthetic rate is maximum in blue and red light not in green Light.

Question 23.
What will be the quanta requirement for the complete light reaction which releases 6 oxygen molecules?
Answer:

• Complet light reaction releases 6 oxygen molecules.
• If one molecule of oxygen evolution requires 8 quanta means for 6 oxygen molecules (6 x 8 = 48) quanta of light reaquired for a complete light reaction.

Question 24.
Give the balance sheet of Calvin or C3 cycle.
Answer:
One molecule of CO₂ is fixed in one turn of the Calvin or C³ cycle.
So 6 turns of cycle will be required to fix 6 molecules of C0₂ – (i.e) to form one molecule of Glucose C6H₁₂O6

In	Out
6CO2	1 Glucose
18 ATP	18 ADP
12 NADPH	12 NADP

IV. 3 Mark Questions

Question 1.
Mention any three significance of photosynthesis.
Answer:
Three significance of photosynthesis:

1. Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly.
2. It is the only natural process that liberates oxygen in the atmosphere and balances the oxygen level.
3. Photosynthesis balances the oxygen and carbon cycle in nature.

Question 2.
What are the properties of light.
Answer:

• Light is a transverse electromagnetic wave.
• It consists of ocillating electric and magnetic fields that are perpendicular to each other and perpenticular to the direction of propagation of the light.
• Light moves at a speed of 3 x 108 ms^-1
• Wave length is the distance between successive crests of the wave.
• Light as a particle is called photon. Each photon contains an amount of energy known as quantum.
• The energy of a photon depends on the frequency of the light.

Question 3.
Distinguish between Absorption spectrum & Action Spectrum.
Answer:

Absorption spectrum	Action spectrum
A curve obtained by plotting the amount of absorption of different wavelengths of light by a pigment is called the Absorption spectrum.	The curve showing the rate of photosynthesis at different wavelengths of light is called the action spectrum.

Question 4.
Distinguish between fluorescence and phosphorescence.
Answer:

Fluorescence	Phosphorescence
1. Immediate emission of absorbed radiations in the form of radiation energy (light) in the red region.	1. This is the delayed emission of absorbed radiations in the form of light in the red region.
2. The electrons move from S1 →SO	2. The electrons pathway is from S2 →S1 →T1→ SO

Question 5.
What is meant by the ground state?
Answer:
The action of photon plays a vital role in the excitation of pigment molecules to release an electron. When the molecules absorb a photon, it is in an excited state. When the light source turned off, the high-energy electrons return to their normal low-energy orbitals as the excited molecule goes back to its original stable condition known as the ground state.

Question 6.
Write down the significance of Photosynthesis.
Answer:

• Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly All other organism depend on them for energy.
• It liberates oxygen in the atmosphere and balances.
• Fuels such as coal, petroleum, and other fossil fuels are preserved forms got only from photosynthetic plants.
• It also provides fodder, fibre, firewood, timber useful medicinal products and these sources come by the act of photosynthesis.

Question 7.
What is DCMU?
Answer:

• It is a chemical herbicide having an inhibiting effect on photosynthesis.
• It is Dichloro phenyl D₁ Methyl Urea.
• It can inhibit electron flow during light reactions of photosynthesis.
• It is a herbicide that blocks the plastoquinone binding site of P.S II and inhibits electron flow from plastoquinone to cytochrome.

Question 8.
State black man’s law of limiting factor.
Answer:
It is a modified law proposed by Liebig’s law of minimum.

Definition:
According to Blackman at any given point of time, the lowest factor among essentials will limit the rate of Photosynthesis.

Example:
When in a condition, if light intensity is low also C0₂ concentration low, in this situation among the two factors which ever is the lowest is considered as the limiting factor here among the essentials CO₂ concentration is the limiting factor.

Question 9.
What is meant by dicarboxylic acid pathway?
Answer:
C₄ pathway is completed in two phases, first phase takes place in the stroma of mesophyll cells, where the CO₂ acceptor mblecule is 3 – Carbon compound, phosphoenolpyruvate (PEP) to form 4 – carbon Oxalo acetic acid (OAA). The first product is a 4 – carbon and so it is named as C₄ cycle. Oxalo acetic acid is a dicarboxylic acid and hence this cycle is also known as a dicarboxylic acid pathway.

Question 10.
State some interesting facts about C4 cycle.
Answer:

• C4 cycle is an alternative path way for CO2 fixation.
• It occur in nearly 1000 plant species 300 dicots but mostly 700 monocots (tropical and sub tropical grasses)
• It represent about 5% earths biomass and 1% of its known plants
• 30% terrestrial carbon fixation on earth is due to C4 So, if C4 plants on earth is increased, then by carbon sequestration by thus strategy severe climate change would be avoided in the near future.

Question 11.
What is Kranz Anatomy or what is meant by Dimorphism of Chloropiasts in C4 plants.
Answer:

C3 plants	C4 plants
C3 plants kranz Anatomy not seen	C4 plants show kranz Anatomy
C3 plants only one type of chloroplasts seen both in bundle sheath and mesophyll cells.	C4 plants Bundle sheath surrounding the vascular bundles have larger chloroplast and have thylakoids are free, not arranged in granum.
Thylakoids are arranged in granum as coins.	Mesophyll cells have smaller chloroplasts thylakoid arranged in granum

Question 12.
what is the significance of the CAM cycle?
Answer:
The significance of the CAM cycle:

1. It is advantageous for succulent plants to obtain CO₂ from malic acid when stomata are closed.
2. During daytime, stomata are closed and CO₂ is not taken but continues their photosynthesis.
3. Stomata are closed during the daytime and help the plants to avoid transpiration and water loss.

Question 13.
Compare the C3 and C4 on the basis of ATP production.
Answer:

C3 plants	C4 plants
The evolution of one oxygen molecule (4 electrons required) requires 8 quanta of light
C3 plants utilise 2 ATPs and 2 NAD PH+H+ to evolve one oxygen molecule	C4 plants utilise 5 ATPs and 2NADPH + H+ to evolve one oxygen molecule
To evolve 6 molecules of oxygen or 1 molecule of Glucose 8 ATPs and 12 NADPH + H+ are utilised.	To evolve 6 molecules of oxygen 30 ATPs and 12 NADPH + H+ are utilised.

Question 14.
What will be the quanta requirement for the complete light reaction which releases 6 oxygen molecules?
Answer:
The complete light reaction releases 6 oxygen molecules if one molecule of oxygen evolution, requires 81 quanta means for 6 oxygen molecules 6 x 8 = 48 quanta of light required for the complete light reaction.

Question 15.
Draw the Graphical representation of any 3 factors affecting photosynthesis.
Answer:

1. Light Intensity
2. CO₂ Concentration
3. Temperature

V. 5 Mark Questions

Question 1.
Distinguish between Photosystem – I and photosystem – II
Answer:
Photosystem – I:

1. The reaction centre is P700.
2. PS I is involved in both cyclic and non – cyclic.
3. Not involved in the photolysis of water and evolution of oxygen.
4. It receives electrons from PS II during non – cyclic photophosphorylation.
5. Located in unstacked region granum facing chloroplast stroma.
6. Chlorophyll and Carotenoid ratio is 20 to 30 : 1.

Photosystem – II:

1. Reaction centre is P680.
2. PS II participates in Non – cyclic pathway.
3. Photolysis of water and evolution of oxygen take place.
4. It receives electrons by photolysis of water.
5. Located in stacked region of thylakoid membrane facing lumen of thylakoid.
6. Chlorophyll and Carotenoid ratio is 3 to 7 : 1.

Question 2.
Structure of Chloroplast.
Answer:

Question 3.
Notes on photosystem and Reaction centre.
Answer:

• Thylakoid membrane contains photosystem I (PSI) and photosystem II (PSII)
• PS I is unstacked region of granum tàcing stroma ofchÍoroplast.
• PS II is found in stacked region of thylakoid membrane facing lumen of thylakoid.
• Each photosystem consists of central core complex (CC) and light harvesting complex (LHC) or Antenna molecules.
• The core complex consists of respective reaction centre associated with proteins, electron donors and acceptors.
• PSI – CCI consists of reaction centre P 700 and LHC – I
• PS II- CC II consists of reaction centre P680 and LHC – II
• Light harvesting complex consists of several chiorophylls, carotenoids and xanthophyll molecules.
• The main function of LHC is to harvesting light energy and transfer it to their respective reaction centre.

Question 4.
Difference between photosystem I and Photosystem II.
Answer:

Photosystem I	Photosystem II
1. The reaction centre is P700	1. Reaction centre is P 680.
2. PSI is involved in Photolysis of water and evolution.	2. PS II participates in Non – Cyclic pathway
3. Not involved in photolysis of water and evolution of oxygen.	3. Photolysis of water and evolution of oxygen take place.
4. It receives electrons from PSII during non – cyclic photophosphorylation.	4. It receives electrons by photolysis of water.
5. Located in unstacked region granum racing chloroplast stroma.	5. Located in stacked region of thylakoid membrane facing lumen of thylakoid.
6. Chlorophyll and carotenoid ratio is 20 to 30:1	6. Chlorophyll and carotenoid ratio is 3 to 7:1

Question 5.
Differences between Cyclic Photophosphorylation and Non – cyclic photophosphorylation.
Answer:

Cyclic Photophosphorylation	Non – Cyclic Photophosphorylation
1. PSI only involved	1. PS I and PS II involved.
2. Reaction centre is P 700	2. Reaction centre is P 680.
3. Electrons released are cycled back	3. Electron released are not cycled back.
4. Photolysis of water does not take place	4. Photolysis of water takes place
5. Only ATP Synthesized	5. ATP and NADPH + H+ are synthesized.
6. Phosphorylation takes place at two places	6. Phosphorylation takes place at only one place
7. It does not require an external electron donor.	7. Requires external electron donor like H2O or H2S
8. It is not sensitive to dichloro dimethyl urea (DCMU)	8. It is sensitive to DCMU and inhibits electron flow

Question 6.
Compare and contrast the photosynthetic processes in C₃ and C₄ plants.
Answer:
Contrast the photosynthetic processes in C₃ and C₄ plants:
C₃ Plants:

• CO₂ fixation takes place in mesophyll cells only.
• CO₂ acceptor is RUBP only.
• First product is 3C – PGA.
• Kranz anatomy is not present.
• Granum is present in mesophyll cells.
• Normal Chloroplast.
• Optimum temperature 20° to 25° C.
• Fixation of CO₂ at 50 ppm.
• Less efficient due to higher photorespiration.
• RUBP carboxylase enzyme used for fixation.
• 18 ATPs used to synthesize one glucose.
• Efficient at low CO₂.
• eg: Paddy, Wheat, Potato and so on.

C₄ Plants:

• CO₂ fixation takes place mesophyll and bundle sheath.
• PEP in mesophyll and RUBP in bundle sheath cells.
• First product is 4C – OAA.
• Kranz anatomy is present.
• Granum present in mesophyll cells and absent in bundle sheath.
• Dimorphic chloroplast.
• Optimum temperature 30° to 45° C.
• Fixation of CO₂ even less than 10 ppm.
• More efficient due to less photorespiration.
• PEP carboxylase and RUBP carboxylase used.
• 30 ATPs to produce one glucose.
• Efficient at higher CO₂.
• eg: Sugar cane, Maize, Sorghum, Amaranthus and so on.

Question 7.
Explain Non cyclic photophosphorylation.
Answer:
When PS II (P680) gets activated, electrons from a high energy state pass through a series of electron carriers like pheophytin, plastoquinone cytochrome complex, plastocyanin, and finally accepted by PS I (P700).

During this flow ATP is generated:

• PS. I (P 700) is activated by light electrons moved to high energy state and accepted by electron acceptor (FRS) Ferredoxin Reducing Substance, during downhill passes through Ferredoxin. During this process NADPH is reduced by H+ formed during photolysis.
• Electrons released from PS II are not cycled back but used in the reduction of NADPH+ into NADPH+H⁺.
• During electron transport, it generates ATP and this type of Phophorylation is called.

Non Cyclic Photophosphorylation:
The electron flow looks like the letter ‘Z’ so known as Z scheme. It has 3 stages.

1. Electron transport from water to P 680: Electrons lost by the PS II are replaced by electrons from splitting of the water molecule, producing electron, protons, and oxygen.
2. Electron transport from P680 to P 700: The flow, through various electron carrier molecules, like pheophytin, plastoquinone (PQ) cytochrome b6 – F complex, plastocyanin (PC) finally reaches P 700 (P.S.I)
3. Electron transport from P 700 to NADP⁺: PSI (P700) is excited now and the electrons pass through ferredoxin, NADP is reduced to NADPH + H⁺

Question 8.
Explain Calvin cycle or C₃ cycle?
Answer:

• It follows a light reaction.
• Utilises ATP and NADPH + H⁺ produced during the light reaction, and reduce carbon dioxide carbohydrate.
• These reactions do not require light so named as Dark reactions.
• The first formed product is a 3 carbon compound (Phospho Glyceric Acid) and so-known as C₃ cycle.
• It was found by Melvin, Calvin, and Benson – so known as the Calvin cycle.
• Occur in the stroma of the chloroplast.
• It is temperature-dependent, and so it is also called a thermochemical reaction.

Phase I carboxylation (Carbon fixation):

• The 5 C compound Ribulose 1 – 5 Bisphosphate (RUBP) with the help of (RUBISCO) enzyme accepts one molecule of carbon dioxide → 6 carbon compound (unstable)
• The 6c compound is broken into → 2 molecules of 3 c compound.

Phase II – Glycolytic Reversai/Reduction :

Phase III – Regeneration:

• The regeneration of RUBP involves several intermediate compounds of 6c, 5c, 4c, and 7c compounds.
• Fixation of one CO₂ require 3 ATPs + 2NADPH⁺ + H⁺
• Fixation of six CO₂ require I8 ATPs+ 12NADPH⁺ + H⁺
• 6 c compound is the net gain to form hexose sugar.

Question 9.
Draw the flow Chart of C₄ pathway?
Answer:

Question 10.
Explain CAM Cycle?
Answer:

• It is one of the carbon path ways in succulent plants growing in semi arid or xerophytic condition.
• The stomata are closed during day (scoto active) and open during night.
• This reverse rhythm help to conserve water loss through transpiration and will stop the fixation of CO₂ during day.
• At night time CAM plants fix CO, with help of (PEP) phospho Enol Pyruvic acid and produce (OAA) Oxalo Acetic Acid.
• Subsequently OAA is converted into a Malic acid-like C4 cycle and get accumulated in the vacuole, increasing the acidity.
• During daytime stomata, are closed and Malic acid is decarboxylated into pyruvic acid resulting in the decrease of acidity.
• CO₂ thus formed enters into the Calvin cycle and produces carbohydrates.

Question 11.
Give the flow chart of Photo respiration or C₂ cycle.
Answer:

Question 12.
Differentiate Photorespiration and Dark respiration.
Answer:

Photorespiration	Dark respiration
1. It takes place in photosynthetic green cells	It takes place in all living cells
2. It takes place only in the presence of light	It involves only Mitochondria
3. It involves Chloroplast, Peroxisome, and Mitochondria	It involves only Mitochondria
4. It does not involve Glycolysis, Kreb’s cycle, and ETS	It involves Glycolysis, Kreb’s cycle and ETS
5. Substrate is Glycolic acid	Substrate is Carbohydrates protein or fats
6. It is not essential for survival	Essential for survival
7. No phosphorylation and yield of ATP Phosphorylation	produces ATP energy
8. NADH2 is oxidised to NAD+	NAD+ is reduced to NADH2
9. Hydrogen peroxide is produced	Hydrogen peroxide is not produced
10. End products are CO2 and PGA	End products are CO2 and water

Question 13.
Give any 5 External factors affecting photosynthesis.
Answer:
1. Carbon dioxide:
330ppm or 0.3% of CO₂ is available in the atomsphere If there is an increase in CO₂ concentration the rate of Photosynthesis increases -If it increases beyond 500 PPm rate of photosynthesis will be inhibited.

2. Oxygen:
When there is increase in oxygen concentration there is unhibition of photosynthesis Warburg – studied this in chlorellain 1920. This effect is known as Warburg effect.

3. Temperature:

• Optimum temperature for photosynthesis vary from plant to plant
• Normally it is 25°C to 3 5°C
• In Opuntia, it is 55°C
• In Lichens it is 20°C
• In Algae growing in hot spring it is 75°C
• At high and low temperature the stomata will close also the enzymes get inactivated.

4. Water:

• Pholysis of water provide electrons and protons for the reduction of NADP – directly.
• Affect stomatal movement and hydration of protoplasm – indirectly.
• During water stress, supply of NADPH + H⁺ affected

5. Minerals:

Deficiency	Effect
Mg, Fe and N	Synthesis of chlorophyll
P	phosphorylation reactions
Mn, Cl-	photolysis of water
CU	Formation of plastocyanin

Question 14.
Explain the test tube funnel experiment.
Answer:
AIM: To proove that oxygen is evolved during Photosynthesis.
Procedure:
Take some hydrilla plant and place them at the bottom of a beaker containing water – Add, little NaHCO3 in to the water. Cover plant with an inverted funnel Invert a test tube over the funnel keep this set up in sun light.

Observation: Air bubbles are released from Hydrilla plant and collected in the test tube by downward displace ment of water. Take the test tube carefully by closing with a finger and then introduce a burning match stick, it bum brightly.
Inference: Hydrilla plant perform photosynthesis and oxygen is liberated during photosynthesis.

Question 15.
Explain the experiment to determine rate of photosynthesis by Witmott’s bubbler.
Answer:
Procedure:

• Wilmott’s bubbler consists of a wide mouth bottle fitted with a single holed cork, a glass tube with lower and having wider opening to insert hydrilla plant.
• The upper end is fitted to a narrow bottle with water.
• Fill the bottle with water and insert hydrilla living into wider part of the tube.
• Hydrilla plant should be cut inside the water to avoid entry of
  air bubbles.
• Fix the tube with jar which acts as water reservoir.
• Keep the apparatus in sunlight count the bubbles when they are in same size.

Question 16.
Differentiate photosynthesis in plants and Bacterial photosynthesis.
Answer:

Photosynthesis	Bacterial photosynthesis
1. Cyclic and Non – Cyclic phosphorylation takes place	Only cyclic phosphorylation takes place
2. Photosystem I and II involved	Photosystem I only involved
3.  Electron donor is water	Electron donor is H9S
4.  Oxygen is evolved	Oxygen is not evolved
5. Reaction centres are P700 and P680	Reaction centre is P890
6.  Reducing agent is NADPH + H+	Reducing agent is NADH + H+
7. PAR is 400 to 700 nm	PAR is above 700nm
8.  Chlorophyll, Carotenoid and Xanthophyll	Bacterio chlorophyll and Bacterio viridin
9. Photosynthetic apparatus – chloroplast	It is chromosomes and Chromatophores

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 6 Specification and Abstraction Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 6 Specification and Abstraction

11th Computer Science Guide Specification and Abstraction Text Book Questions and Answers

Part I

Choose the best answer

Question 1.
Which of the following activities is algorithmic in nature?
a) Assemble a bicycle
b) Describe a bicycle
c) Label the parts of a bicycle
d) Explain how a bicycle works
Answer:
a) Assemble a bicycle

Question 2.
Which of the following activities is not algorithmic in nature?
a) Multiply two numbers
b) Draw a kolam
c) Walk in the park
d) Braid the hair
Answer:
d) Braid the hair

Question 3.
Omitting details inessential to the task and representing only the essential features of the task is known as
a) specification
b) abstraction
c) composition
d) decomposition
Answer:
b) abstraction

Question 4.
Stating the input property and the as :-output relation a problem is known
a) specification
b) statement
c) algorithm
d) definition
Answer:
a) specification

Question 5.
Ensuring the input-output relation is
a) the responsibility of the algorithm and the right of the user
b) the responsibility of the user and the right of the algorithm
c) the responsibility of the algorithm but not the right of the user
d) the responsibility of both the user and the algorithm
Answer:
d) the responsibility of both the user and the algorithm

Question 6.
If i = 5 before the assignment i := i-1 after the assignment? the value of i is
a) 5
b) 4
c) 3
d) 2
Answer:
b) 4

Question 7.
If 0 < i before the assignment S := i-1 after the assignment, we cars conclude that
a) 0 < i
b) 0 < i c) i = 0 d) 0 >i
Answer:
b) 0 < i

Part II

Short Answers

Question 1.
Define an algorithm
Answer:
An algorithm is a step-by-step sequence of statements intended to solve a problem. An algorithm starts execution with the input data, executes the statements, and finishes execution with the output data.

Question 2.
Distinguish between an algorithm and a process.
Answer:
An algorithm is a sequence of instructions to accomplish a task or solve a problem.
An instruction describes an action. When the instructions are executed, a process evolves which accomplishes the intended task or solves the given problem.
We can compare an algorithm to a recipe, and the resulting process to cooking.

Question 3.
Initially, farmer, goat, grass, wolf = L, L, L, L, and the farmer crosses the river with a goat. Model the action with an assignment statement.
Answer:
The sequence of assignments for goat, grass, and wolf problem.

1. farmer, goat, grass, wolf = L, L, L, L

2. farmer, goat = R, R

3. farmer, goat, grass, wolf = R, R, L, L

4. farmer = L

5. farmer, goat, grass, wolf = L, R, L, L

6. farmer, grass = R, R

7. farmer, goat, grass, wolf = R, R, R, L

8. farmer, goat = L, L

9. farmer, goat, grass, wolf = L, L, R, L

10. farmer, wolf = R, R

11. farmer, goat, grass, wolf = R, L, R, R

12. farmer = L

13. farmer, goat, grass, wolf = L, L, R, R

14. farmer, goat =R, R

15. farmer, goat, grass, wolf = R, R, R, R

Question 4.
Specify a function to find the minimum of two numbers.
Answer:

1. – – minimum (a,b)
2. – – inputs: a, b are real numbers.
3. – – output: result: = minimum (a,b)

Question 5.
If √2 = 1.414, and the square_root() function returns -1.414, does it violate the following specification?
— square_root (x)
— inputs: x is a real number, x ≥ 0
— outputs: y is d real number such that y² = x
Answer:
— square_root (x)
— inputs: x is a real number, x > 0
— outputs: y is d real number such that y² = x
Yes. It violates the specification. For a positive input (x > 0), the output square-root value should also be positive.

Part III

Explain in brief.

Question 1.
When do you say that a problem is algorithmic in nature?
Answer:
A Problem is algorithmic in nature when its solution involves the construction of an algorithm. Also when the

1. Input data and output data of the problem are specified.
2. The relation between the input data and the output data is specified.

Question 2.
What is the format of the specification of an algorithm?
Answer:
An algorithm is specified by the properties of the given input and the relation between the input and the desired output. In simple words, the specification of an algorithm is the desired input-output relation.

Let P be the required property of the inputs and Q the property of the desired outputs. Then the algorithm S is specified as;

• algorithm_name (inputs)
• – inputs: P
• – outputs: Q

This specification means that if the algorithm starts with inputs satisfying P, then it will finish with the outputs satisfying Q.

Question 3.
What is abstraction?
Answer:
Abstraction:
It is the facility to define objects. It also involves the removal of unnecessary attributes and defining only essential attributes. For example, when we represent the state of a process we select only the variables essential to the problem and ignore inessential details.

Question 4.
How is the state represented in algorithms?
Answer:
Computational processes in the real-world have stated. As a process evolves, the state changes.
The state of a process can be represented by a set of variables in an algorithm. The state at any point of execution is simply the values of the variables at that point. As the values of the variables are changed, the state changes.

Example:
State: A traffic signal may be in one of the three states: green, amber, or red. The state is changed to allow a smooth flow of traffic. The state may be represented by a single variable signal which can have one of the three values: green, amber, or red.

Question 5.
What is the form and meaning of the assignment statement?
Answer:
Variables are named boxes to store values. The assignment statement is used to store a value in a variable. It is written with the variable on the left side of the assignment operator and a value on the right side.
variable := value

When this assignment is executed, the value on the right side is stored in the variable on the left side. The assignment
m := 2 stores value 2 in variable m.

If the variable already has a value stored in it, the assignment changes its value to the value on the right side. The old value of the variable is lost.
The right side of an assignment can be an expression.
variable := expression
In this case, the expression is evaluated and the value of the expression is stored in the variable.

Question 6.
What is the difference between assignment operator and equality operator?
Answer:

Assignment operator	Equality operator
It assign right-hand side of the assignment operator value(constant / variable/ expression) to the left-hand side variable.	It compares both operands and returns True if it is equal otherwise False. It is a relational operator.

Part IV

Explain in detail

Question 1.
Write the specification of an algorithm hypotenuse whose inputs are the lengths of the two shorter sides of a right-angled triangle, and the output is the length of the third side.
Answer:
hypotenuse(sl, s2).
— inputs: si and s2 both are real numbers
— outputs: I is a real number such that
l² = s1²+ S2².

Question 2.
Suppose you want to solve the quadratic equation ax² + bx + c = 0 by an algorithm.
quadratic_solve (a, b, c)
— inputs : ?
— outputs: ?
You intend to use the formula and you are prepared to handle only real number roots. Write a suitable specification.
\(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Solution;
quadratic_solve (a, b, c).
– inputs : a,b,c all are real numbers, a ≠ 0.
– outputs: x is a real number.
\(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
such that b² – 4ac > 0.

Question 3.
Exchange the contents: Given two glasses marked A and B Glass A is full of apple drink and glass B is full of grape drink. For exchanging the contents of glasses A and B, represent the state by suitable variables, and write the specification of the algorithm.
Answer:
The specification of the algorithm:
exchange (A, B)
–inputs: A,B are real and >0
–outputs: A,B are real and >0
State representation
TEMP:=A
A:=B
B := TEMP

11th Computer Science Guide Specification and Abstraction Additional Questions and Answers

Part I

Choose the best answer

Question 1.
Which one of the following is an example of a process?
(a) Braid the hair
(b) Adding three numbers
(c) Cooking a dish
(d) Walk in the Road
Answer:
(c) Cooking a dish

Question 2.
An instruction describes an __________
a) action
b) input
c) output
d) flow
Answer:
a) action

Question 3.
How many basic building blocks construct an algorithm?
(a) 3
(b) 4
(c) 5
(d) 8
Answer:
(b) 4

Question 4.
Instructions of a computer are also known as _________
a) statements
b) structures
c) procedures
d) None of these
Answer:
a) statements

Question 5.
……………….. how many control flow statement is there to alter the control flow depending on the state?
(a) 5
(b) 6
(c) 3
(d) 8
Answer:
(c) 3

Question 6.
__________ is a step in solving a mathematical problem suggested by G Polya.
a) Understand the problem and Devise a plan
b) Carry out the plan
c) Review your work
d) All the above
Answer:
d) All the above

Question 7.
……………….. statement is used to store a value in a variable.
(a) Assignment
(b) Sequential control flow
(c) Alternative control flow
(d) Iterative
Answer:
(a) Assignment

Question 8.
__________takes input data, process the data, and produce output data.
a) Function
b) Algorithm
c) Procedure
d) None of these
Answer:
b) Algorithm

Question 9.
Each part of the algorithm is known as ………………..
(a) input
(b) function
(c) variable
(d) program
Answer:
(b) function

Question 10.
When we do operations on data, we need to store the results in________
a) Function
b) Algorithm
c) Constant
d) Variable
Answer:
d) Variable

Question 11.
If i: = 3 before the assignment, i: = i + 1 after the assignment ………………..
(a) 3
(b) 4
(c) 5
(d) 0
Answer:
(b) 4

Question 12.
In algorithm, the statement to be executed next may depend on the __________
a) algorithm step
b) state of the process
c) user direction
d) None of these
Answer:
b) state of the process

Question 13.
If i: = 10 before the assignment, then i: = i % 2 after the assignment
(a) 10
(b) 5
(c) 0
(d) 1
Answer:
(c) 0

Question 14.
There are control flow statements to alter the control flow depending on the state.
a) two
b) five
c) four
d) three
Answer:
d) three

Question 15.
Initially the values of P and C are 4 and 5 respectively
– – P, C : = 4, 5
P : = C
C : = P. Then find P and C
(a) 4 and 4
(b) 5 and 4
(c) 5 and 5
(d) 4 and 5
Answer:
(c) 5 and 5

Question 16.
__________ control flow, a condition of the state is tested, and if the condition is true, one statement is executed; if the condition is false, n alternative statement is executed.
a) alternative
b) iterative
c) random
d) sequential
Answer:
a) alternative

Question 17.
How many Algorithmic designing techniques are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 18.
A ___________ takes an input and produces an output, satisfying a desired input-output relation.
a) function
b) union
c) structures
d) None of these
Answer:
a) function

Question 19.
which one of the following is the equality operator?
(a) =
(b) = =
(c) + +
(d) – –
Answer:
(b) = =

Question 20.
___________ is the basic principle and technique for designing algorithm.
a) Specification
b) Abstraction
c) Composition and Decomposition
d) All the above
Answer:
d) All the above

Question 21.
Which one of the following statements are not executed on the computers?
(a) Comment line
(b) Header file
(c) cin
(d) cout
Answer:
(a) Comment line

Question 22.
Ignoring or hiding unnecessary details and modeling an entity only by its essential properties is known as __________
a) data hiding
b) abstraction
c) data analysis
d) None of these
Answer:
b) abstraction

Question 23.
The values of the variables when the algorithm finishes is ………………..
(a) final stage
(b) final state
(c) last stage
(d) last state
Answer:
(b) final state

Question 24.
An algorithm is composed of__________
statement.
a) assignment
b) control flow
c) both A and B
d) None of these
Answer:
c) both A and B

Question 25.
Which one of the following is not a building block of the algorithm?
(a) data
(b) state
(c) variables
(d) functions
Answer:
(b) state

Question 26.
To solve a problem, we must state the problem__________
a) clearly and precisely
b) with ambiguity
c) with data hiding
d) None of these
Answer:
a) clearly and precisely

Question 27.
Reasoning:
I. We can store a value in a variable using the assignment operator.
II. We can change the value in a variable using the assignment operator.
III. Assignment operator is = =
(a) I and III are true
(b) I and II are true
(c) II and III are true
(d) I, II, III is true
Answer:
(b) I and II are true

Question 28.
Specification of an algorithm is the desired __________relation.
a) input-process
b) process-output
c) input-output
d) None of these
Answer:
c) input-output

Question 29.
In which one of the control flow statements, if the condition is false, then the alternative statement will be executed ………………..
(a) Sequential
(b) iterative
(c) selection
(d) alternative
Answer:
(d) alternative

Question 30.
The values of the variables when the algorithm starts are known as the __________state.
a) initial
b) final
c) intermediate
d) None of these
Answer:
a) initial

Question 31.
If the statement is executed one after the other, then it is a control flow.
(a) Sequential
(b) iterative
(c) selection
(d) alternative
Answer:
(a) Sequential

Question 32.
We can write the specification in a standard __________part format.
a) three
b) four
c) two
d) many
Answer:
a) three

Question 33.
Which one of the following is not a technique for designing algorithms?
(a) specifications
(b) abstraction
(c) encapsulation
(d) composition
Answer:
(c) encapsulation

Question 34.
__________defines the rights and responsibilities of the designer and the user.
a) Specification
b) Abstraction
c) Composition and Decomposition
d) All the above
Answer:
a) Specification

Question 35.
How many parts are there in the specification is ………………..
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 36.
In algorithms, the state of a computation is abstracted by a set of __________
a) expressions
b) variables
c) functions
d) None of these
Answer:
b) variables

Question 37.
Identify the statement which is not true?
(a) An instruction describes an object
(b) the specification is one of the algorithm design techniques
(c) An algorithm is a step by step sequence of instructions
Answer:
(a) An instruction describes an object

Question 38.
_________ is a basic and important abstraction.
a) process
b) state
c) addition
d) None of these
Answer:
d) None of these

Question 39.
Sequential, Alternative, and Iterative comes under the classification of ……………….. :
(a) Building blocks of the algorithm
(b) control flow statements
(c) Algorithm design techniques
(d) Abstraction
Answer:
(b) control flow statements

Part II

Short Answer Questions

Question 1.
Define State.
Answer:
The state of a process can be represented by a set of variables in an algorithm. The State at any point of execution is simply the values of the variables at that point.

Question 2.
What is the statement?
Answer:
Instructions of a computer are also known as statements.

Question 3.
Define variable.
Answer:
The data stored in a variable is also known as the value of the variable. We can store a value in a variable or change the value of a variable, using an assignment statement.

Question 4.
What are the building blocks of algorithms?
Answer:
The building blocks of algorithms are;

• Data.
• Variables.
• Control flow.
• Functions.

Question 5.
Define control flow.
Answer:
The order in which the statements are executed may differ from the order in which they are written in the algorithm. This order of execution of statements is known as the control flow.

Question 6.
What is a function?
Answer:
The parts of an algorithm are known as functions. A function is like a sub-algorithm. It takes an input, and produces an output, satisfying the desired input, output relation.

Question 7.
How will you specify the algorithm?
Answer:
The algorithm S is specified as;

• algorithm_name (inputs)
• inputs: P
• outputs: Q

This specification means that if the algorithm starts with inputs satisfying P, then it will finish with the outputs satisfying Q.

Question 8.
Define abstraction.
Answer:
Abstraction is the process of ignoring or hiding irrelevant details and modeling a problem only by its essential features.

Question 9.
What purpose abstraction is used?
Answer:
Abstraction is the most effective mental tool used for managing complexity. If we do not abstract a problem adequately, we may deal with unnecessary details and complicate the solution.

Part III

Explain in brief

Question 1.
Define Alternative control flow statement.
Answer:
In alternative control flow, a condition of the state is tested, and if the condition is true, one statement is executed; if the condition is false, an alternative statement is executed.

Question 2.
Justify goat, grass, and wolf problem is algorithmic.
Answer:
goat, grass, and wolf problem:
A farmer wishes to take a goat, a grass bundle, and a wolf across a river. However, his boat can take only one of them at a time. So several trips are necessary across the river.

Moreover, the goat should not be left alone with the grass (otherwise, the goat would eat the grass), and the wolf should not be left alone with the goat (otherwise, the wolf would eat the goat).

How can the farmer achieve the task? Initially, we assume that all the four are at the same side of the river, and finally, all the four must be on the opposite side. The farmer must be in the boat when crossing the river.
A solution consists of a sequence of instructions indicating who or what should cross. Therefore, this is an algorithmic problem.

Question 3.
Write the Iterative control flow statement.
Answer:
In iterative control flow, a condition of the state is tested, and if the condition is true, a statement is executed. The two steps of testing the condition and executing the statement are repeated until the condition becomes false.

Question 4.
Explain variable.
Answer:
Variables are named boxes for storing data. When we do operations on data, we need to store the results in variables. The data stored in a variable is also known as the value of the variable.
We can store a value in a variable or change the value of a variable, using an assignment statement.

Question 5.
Write the following
(i) initial state
(ii) final state
Answer:
The values of the variables when the algorithm starts are known as the initial state, and the values of the variables when the algorithm finishes are known as the final state.

Question 6.
When functions are needed?
Answer:
Algorithms can become very complex. The variables of an algorithm and dependencies among the variables may be too many. Then, it is difficult to build algorithms correctly.
In such situations, we break an algorithm into parts, construct each part separately, and then integrate the parts to the complete algorithm. The parts of an algorithm are known as functions.

Question 7.
Give an example for a function?
Answer:
Suppose we want to calculate the surface area of a cylinder of radius r and height h.
A = 2 πr² + 2πrh
We can identify two functions, one for calculating the area of a circle and the other for the circumference of the circle.
If we abstract the two functions as circle_area(r) and circle_circumference(r), then,
cylinder_area (r, h) can be solved as:
cylinder_area (r, h) = 2 X circle_area (r) + circle_circumference (r) X h.

Question 8.
What are the basic principles and techniques for designing algorithms?
Answer:
The basic principles and techniques for designing algorithms are:

• Specification.
• Abstraction.
• Composition.
• Decomposition.

Question 9.
Write the specification format and explain.
Answer:
Specification format:
We can write the specification in a standard three-part format:

• The name of the algorithm and the inputs.
• Input: the property of the inputs.
• Output: the desired input-output relation.

The first part is the name of the algorithm and the inputs. The second part is the property of the inputs. It is written as a comment which starts with – inputs: The third part is the desired input-output relation.
It is written as a comment which starts with — outputs: The input and output can be written using English and mathematical notation.

Question 10.
Write the specification of an algorithm for computing the square root of a number.
Answer:

• Let us name the algorithm square_root.
  it takes the number as the input. Let us name the input n. n should not be negative.
• It produces the square root of n as the output. Let us name the output y. Then n should be the square of y.

Now the specification of the algorithm is
a) square_root (n).
b) inputs: n is a real number, n > 0.
c) outputs: y is a real number such that y 2= n.

Question 11.
Give an example for abstraction.
Answer:

• A road map is designed for drivers. They do not usually worry about hills so most hills are ignored on a road map.
• A walker’s map is not interested in whether a road is a one-way street, so such details are ignored.

Part IV

Explain in detail

Question 1.
Explain the three control flow statement.
Answer:
There are three important control flow statements to alter the control flow depending on the state.

1. In sequential control flow, a sequence of statements is executed one after another in the same order as they are written.
2. In alternative control flow, a condition of the state is tested, and if the condition is true, one statement is executed; if the condition is false, an alternative statement is executed.
3. In iterative control flow, a condition of the state is tested, and if the condition is true, a statement is executed. The two steps of testing the condition and executing the statement are repeated until the condition becomes false.

Question 2.
What are the uses of the Operating System?
Answer:
The basic principles and techniques for designing algorithms are;

• Specification.
• Abstraction.
• Composition.
• Decomposition.

Specification: The first step in problem-solving is to state the problem precisely. A problem is specified in terms of the input given and the output desired. The specification must also state the properties of the given input, and the relation between the input and the output.

Abstraction: A problem can involve a lot of details. Several of these details are unnecessary for solving the problem. Only a few details are essential. Ignoring or hiding unnecessary details and modeling an entity only by its essential properties is known as abstraction. For example, when we represent the state of a process, we select only the variables essential to the problem and ignore inessential details.

Composition: An algorithm is composed of assignment and control flow statements. A control flow statement tests a condition of the state and, depending on the value of the condition, decides the next statement to be executed.

Decomposition: We divide the main algorithm into functions. We construct each function independently of the main algorithm and other functions. Finally, we construct the main algorithm using the functions.

Question 3.
Explain the specification format.
Answer:
Specification format: We can write the specification in a standard three-part format:

1. The name of the algorithm and the inputs
2. Input: the property of the inputs
3. Output: the desired input-output relation

The first part is the name of the algorithm and the inputs. The second part is the property of the inputs. It is written as a comment which starts with – inputs: The third part is the desired input-output relation. It is written as a comment which starts with outputs: The input and output can be written using English and mathematical notation.

Question 4.
Explain the state with a suitable example.
Answer:
State:
The state is a basic and important abstraction. Computational processes have stated. A computational process starts with an initial state. As actions are performed, its state changes. It ends with a final state.

The state of a process is abstracted by a set of variables in the algorithm. The state at any point of execution is simply the values of the variables at that point.

Example:
Chocolate Bars: A rectangular chocolate bar is divided into squares by horizontal and vertical grooves. We wish-to break the bar into individual squares.
To start with, we have the whole of the bar as a single piece. A cut is made by choosing a piece and breaking it along one of its grooves. Thus a cut divides a piece into two pieces. How many cuts are needed to break the bar into its individual squares? In this example, we will abstract the essential variables of the problem.

Essential variables:
The number of pieces and the number of cuts are the essential variables of the problem. We will represent them by two variables, p and c, respectively. Thus, the state of the process is abstracted by two variables p and c.

Irrelevant details:
The problem could be cutting a chocolate bar into individual pieces or cutting a sheet of postage stamps into individual stamps. It is irrelevant. The problem is simply cutting a grid of squares into individual squares.

The sequence of cuts that have been made and the shapes and sizes of the resulting pieces are irrelevant too. From p and c, we cannot reconstruct the sizes of the individual pieces. But, that is irrelevant to solving the problem.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 4 Reproductive Morphology Text Book Back Questions and Answers, Notes

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology

11th Bio Botany Guide Reproductive Morphology Text Book Back Questions and Answers

Part – I

Choose the Right Answer: 

Question 1.
Vexillary aestivation is characteristic of the family
a. Fabaceae
b. Asteraceae
c. Solanaceae
d. Brassieaceae
Answer:
a. Fabaceae

Question 2.
Gynoecium with united carpels is termed as
a. Apocarpous
b. Multicarpellary
c. Syncarpous
d. None of the above
Answer:
c. Syncarpous

Question 3.
Aggregate fruit develops from
a. Multicarpellary apocarpous ovary
b. Multicarpellary syncarpous ovary
c. Multicarpellary ovary
d. Whole Inflorescence
Answer:
a. multicarpellary apocarpous ovary

Question 4.
In an inflorescence where flowers are borne laterally in an aeropetal succession, the position of the youngest floral bud shall be
a. Proximal
b. distal
c. Intercalary
d. Anywhere
Answer:
a. Proximal

Question 5.
A true fruit is the one where
a. only ovary of the flower develops into fruit.
b. ovary and caly x of the flower develops into fruit.
c. ovary, caly x, and thalamus of the flower develops into fruit.
d. All floral whorls of the flower develops is to fruit.
Answer:
a.only ovary of the flower develops into the fruit

Question 6.
Find out the floral formula for a besexual flower with bract, regular, pentamerous, distinct caly x and corolla , superior ovary without bracteole.
Answer:

Question 7.
Giving the technical terms for the following.
a. A sterile stamen …………………..
b. Stamens are united in one bunch …………….
c. Stamens attached to the petals ……………….
Answer:
a. staminode
b. monodelphous
c. Epipetalous (petalostemonous)

Question 8.
Explain different types of placentation with example
Answer:
Marginal:

It is with the plaentae along the marging of a unicarpellate ovary.
Example-Fabaceae.

Axile:

The placentae arises from the column in a compound ovary with septa.
Example-Hibiscus, tomato lemon

Superficial:

Ovules arise from the surfae of the septa.
Example: Nymphaeceae

Parietal:

It is the placentae on the ovary walls or upon intruding partitions of a unilocular, compound Ovary.
Example: Mustard, Argemone, cucumber.

Free-central:

It is with the placentae along the column in a compound ovary without septa.
Example: Caryophyllaceae, Dianthus, Primrose

Basal:

It is the placenta at the base of the ovary.
Example: Sunflower (asrteraceae) Marigold.

Question 9.
Differences between aggregate fruit with multiple fruit.
Answer:

Question 10.
Explain different type of fleshy fruit with suitable example
Answer:

Part II

11th Bio Botany Guide Reproductive Morphology Additional Important Questions and Answers

Choose the Right Answer:

Question 1.
Placentation in tomato and lemon is …………….
(a) parietal
(b) marginal
(c) free – central
(d) axile
Answer:
(d) axile

Question 2.
This is not a racemose Inflorescence
a. Spite
b. Catkin
c. Spadix
d. Cauliflower
Answer:
d.Cauliflower

Question 3.
Geocarpic fruits are seen in …………… .
(a) carrot
(b) groundnut
(c) radish
(d) turnip
Answer:
(b) groundnut

Question 4.
Pendulose spikes occur in
a. Piper nigrum
b. Dry za sativa
c. Tridax sp
d. Zeamays
Answer:
a. Piper nigrum

Question 5.
When the calyx is coloured and showy, it is called …………… .
(a) petaloid
(b) sepaloid
(c) bract
(d) spathe
Answer:
(a) petaloid

Question 6.
Parietal placentation occurs in
a. Hibiscus
b. Nymphaeaceae
c. Cucumber
d. Fabaceae
Answer:
c. Cucumber

Question 7.
Trace the correct F.D of Jxora coccinea

Answer:

Question 8.
In Theobroma cocoa, the inflorescence arise from …………… .
(a) terminal shoot
(b) axillary part
(c) trunk of plant
(d) leaf node
Answer:
(c) trunk of plant

Question 9.
An example for a Pseudo fruit
a. Apple
b. Tomato
c. Pumpkin
d. Mango
Answer:
a. Apple

Question 10.
The fruit type intermediate between dehiscent and indehiscent is known as
a. Regma
b. Samara
c. Schizocarpic
d. Nut
Answer:
c. Schizocarpic

Question 11.
Thyrsus is a type of …………… inflorescence.
(a) raceme
(b) cyme
(c) mixed
(d) special
Answer:
(c) mixed

Question 12.
Br, Ebrl, O7 p3+3 A(3) Go – is
a. This F.D of male flower of musa
b. The F.D of crotalaria juncea
c. The F.D. of male flower of phyllanthus amaras
d. The F.D of male flower of cocos nucifera
Answer:
c.The F.D of male flower of phyllnthus amaras.

Question 13.
Calyz is distinctly leaf-like, large often orange or white coloured as in mussenda it is known as
a. Campanulate sepals
b. Tubular sepals
c. Petaloid sepals
d. Sepaloid petals
Answer:
c. Petaloid sepals

Question 14.
If unisexual and bisexual flowers are seen in same plant then the plant is said to be …………… .
(a) polyphyllous
(b) polygamous
(c) hermaphroditic
(d) dioecious
Answer:
(b) polygamous

III. Match the following

Question 1.
(I) Spathe – A) Hibiscus sp
(II) Spikelet – B) Musa sp
(III) Epicaly – C) Paddy
(IV) Pislillate flower – D) Cocas Nucifera

Answer:
b. D-C-A-B

Question 2.
(I) Epipetalous – A) G(2)
(II) Monoadeiphous – B) P(5)
(III) Inferior Ovary – C) A(a)
(IV) Gamophyllous – D) C(5)A5

Answer:
a. D-C-A-B

Question 3.
(I) Catkin – A) Cauliflower
(Il) Corymb – B) Mangifera indica
(III) Panicle – C) Coriandrumsatiuum
(IV) Umbel – C) Coriandrumsatiuum

Answer:
b. D-C-B-A

Question 4.
(I) A single female flower surrounded by a group of male – A) Coenanthum flowers-enclosed in an involucre
(II) Circular disclike fleshy open receptacle bearing – B) Hypanthodium pistillate at the centre & staminate flowers at periphery}
(III) Receptacle hollow male flowers towards ostiole female – C) Polychasialcyme and neutral in the middle
(IV) Central axis ends in a flower lateral axis branches repeatedly – D) cyathium

Answer:
b. D-A-B-C

Question 5.
(I) Persistent calyx – A) Calyx falls after the opening of a flower
(II) Deciduous calyx – B) Continue to grow with fruit and encloses it completely or partially
(III) Caduceus calyx – C) Calyx continues to be along with fruit forms a cup
(VI) Accresent calyx – D) Calyx falls during the early development stage of the flower

Answer:
C. C-A-D-B

IV. Choose the wrong pair

Question 1.
a. Follicle – Calotropis
b. Silicula – Capsella
c. Loculicidal capsule – Lady’s finger
d. Legume – Castor castor
Answer:
d. Legume

Question 2.
a. Raceme – Crotalaria
b. Cyme – Cyathium
c. Special type – Hypanthodium
d. Mixed type – Thyssus
Answer:
b. Cyme – Cyathium

Question 3.
a. Monoecious – Coconut
b. Dioecious – Musa
c. Polygamous – Mangifera
d. Bisexual – Brinjal
Answer:
b. Dioecius- Musa

V. Fill up the tabulation with the right answer.

Question 1.

Condition	Explanation	Example
1. Apostemonous	__________________	Cassia
2. Polyadelphons	Filaments connate into many bundles	__________________
3. __________________	Stamens adnate to petals	Datura
Syngenesious	__________________	Asteraceae

Answer:
1. Stamens distinct do not fuse with other parts
2. Citrus
3. Epipetalous
4. Anthers connate, filaments free

Question 2.

Attachment of Anther	Description	Example
1. Basiifixed	………………….	Datura
2. ………………….	The apex of filament is attached to the dorsal side of the anther	Hibiscus
3. Versatile	The filament is attached to the anther at the midpoint	……………..
4. …………..	The filament is continued from the base to the apex	Ranunculus

Answer:
1. Base of another is attached to the tip of the filament
2. Dorsifixed
3. Grasses
4. Adnate

VI. Choose the wrong pair

Question 1.
1. Ripened ovary – Seed
2. Ovary wall – Testa & tegmen
3. Ferlised ovule – Seed
4. Integuments of – Pericarp ovule
Answer:
3. Ferlized ovule – Seed

Question 2.
1. Cremocarp – groundnut
2. Carcerulus – coriander
3. Lomentum – abutilon
4. Regma – castor
Answer:
4. Regma – castor

Question 3.
1. Caryopsis – paddy
2. Cypsela – coriander
3. Cremocarp – groundut
4. Lomentum – sunflower
Answer:
1. Caryopsis – paddy

VII. Identify the diagram & Label it correctly

Question 1.
Cymose inflorescen or simple dichasium

Answer:
Cymose inflorescen or simple dichasium

A-Bract
B-Old flower
C-Young flower

Question 2.
This is Papilionaceous Corolla

Answer:
This is Papilionaceous Corolla

A-Standard petal or corolla
B-Wing petals or alae
C-Keel petals or camia

Question 3.
The given diagram is spadix inflorescence

Answer:
The given diagram is spadix inflorescene

A-Central axis
B- Female flower
C- Male flower

Question 4.
The given diagram is verticillaster

Answer:
The given diagram is verticillaster

A-Central axis
B-Monocharial scorpioid lateral branches
C-Blder flowers

Question 5.
The diagram represents the tetradynamous condition of stamen

Answer:
The diagram represents the tetradynamous condition of stamen

A-4 long stamens
B-2 long stamens

VIII. Find out the true or false

Question 1.
(i) Asymmetric flowers cannot be divided into equal halves in any plane
(ii) The calyx of tridax is modified into a tubular structure
(iii) Heterostemonous stamens, have different lengths in the same flower
(iv) Hypanthium is a fleshy elevated stamina disk which is nectariferous in nature.

Answer:
a. True – False – True – True

Question 2.
(i) Aeswation is the arrangement of sepals and petals in the flower when it open
(ii) Lodicule is the reduced scale-like perianth in the members of Poaceae
(iii) The walls of the ovary and septa form a cavity called locule
(iv) The branch that bears the flower is called the parental axis.

Answer:
c. False – True – True – False

IX. In the following diagram what are the parts.

Question 1.

Answer:
c. Peduncle – Involucre – Female flower – Male flower

X. Write down the edible parts of the following.

Question 1.

1. Apple …….?
2. Coconut …….?
3. Jack fruit ……?
4. Mango ……?
5. Tomato ……?
6. Orange …….?
7. Pomegranate ……..?

Answer:

1. Thalamus
2. Oily endosperm
3. Perianth
4. Fleshy juicy mesocarp
5. Epi, meso, and endocarp i.e (pericarp)
6. Juicy hairs
7. Testa of seed

XI. Read the following Assertion and Reason & Find the correct answer.

Question 1.
Assertion (A): Fruits are the products of pollination and fertilization
Reason (R): All floral whorls of a flower develop into a fruit
(a) A and R are correct. R is explaining A
(b) A and R are correct but R is not explaining Assertion
(c) A is true but R is wrong
(d) A is true but R is not explaining Assertion
Answer:
(c) A is true but R is wrong

Question 2.
Assertion (A): Homochlamydeous condition is prevalent in monocot
Reason (R) : Undifferentiated calyx and corolla is known as perianth
(a) A and R are correct and R is explaining A
(b) A and R are correct but R is not explaining Assertion
(c) A is true but R is wrong
(d) A is true but R is not explaining Assertion
Answer:
(a) A and R are correct and R is explaining A

Question 3.
Assertion (A): Almost all flowers are hermaphrodite
Reason (R): Male and female sex organs do not occur in the same flower
(a) A and R are correct R is explaining A
(b) A and R are correct but R is not explaining assertion
(c) A is true but R is wrong
(d) A is true but R is not explaining assertion
Answer:
(c) A is true but R is wrong

XII. Fill up the blanks by giving technical terms for the following.

Question 1.
a) The study of fruits …….
Answer:
Pomology

b) lkebana is an act of ……….
Answer:
Flower arrangement

c) The botanical name of Saffron flower ……………
Answer:
Crocus Sativum

d) The flower grows once in 12 years …………
Answer:
Kurinji (Strobilanthus kunthranus)

e) World’s largest fruit is …………………
Answer:
Lodoicea maldivica

f) King Herod’s palace, near dead sea, scientist have got a seed viable for …………. years
Answer:
20,000 years

g) The longest and largest inflorescence of any flowering plant is ………………
Answer:
Corypha umbraculifera (cudai palm)

h) The largest single flower is known sofae is ………………
Answer:
Rafflesia arnoldi

Give very short answers – 2 Marks

Question 1.
How will you define inflorescence?
Answer:
An inflorescence is a group of flowers arising from a branched or unbranched axis with a definite pattern.

Question 2.
Distinguish between Bract and Bracteole.
Answer:

Bract	Bracteole
Bract is that scale-like or structure leaf-like from which arises a flower The presence of bract can be denoted as Bracteate its absence known as a bracteole	It is the scale-like or leaf-like structures seen on the pedicel of the flower just above the Bract. The presence of bracteole in a flower is known as Bractolate, if it is absent it is known as Bracteolate

Question 3.
Distinguish between the Posterior and Anterior sides of a flower.
Answer:

Posterior side	Anterior side
The side of the flower facing the mother axis is called the Posterior side. It is also known the part towards the plan	The side of the flower facing away from the mother axis is called the anterior side. It is the part away from the plant.

Question 4.
Distinguish between Superior and Inferior Ovary.
Answer:

Superior Ovary	Inferior Ovary
It is the attachment of ovary relative to other floral parts – if the ovary with sepals, petals and stamens attached at the base of the ovary, e.g Hibiscus, Mangifera	If in the ovary the sepals petals and stamens attached at the base of the ovary it is called Inferior. e.g Ixora or Musa.

Question 5.
What is a sessile flower?
Answer:
A flower without a pedicel or stalk is said to be a sessile flower.

Question 6.
What is the use of Pappus?
Answer:
Pappus is the hair-like structures – (modification of calyx)
Pappus occur in Asteraceaemembers – They help in the dispersal of fruits.

Question 7.
Write the units of (a) Perianth and (b) Calyx.
Answer:
The units of (a) Perianth and (b) Calyx:

1. Perianth – tepals and
2. Calyx – sepals

Question 8.
Distinguish between Apocarpous & Syncarpous.
Answer:

Apocarpous	Syncarpous
A pistil containing two or more distinct carpels is known as apocarpous condition e.g Annona	A pistil containing two or more carpels which write or cannot-it is known as a syncarpous condition, e.g citus, tomato

Question 9.
Define a Carpel & Locule.
Answer:

Carpel	Locule
Components of gynoecium usually made of one or more carples they may be distinct or cannot Usually no. of carpel equals the no.of locule	The walls of the ovary and (crosswall of ovary from a cavity called lcoule Usually no.of locules equals the no of carples exception Bicarpellary unilocular condition in Asteraceae

Question 10.
Draw the structure of the Anthophore.
Answer:
Intermodal extension between Calyx and Coralla
A – Androecium
B – Gynoecium
C – Corolla
D – Anthophore
E – Calyx

Question 11.
Classify racemose Inflorescence.
Answer:

Question 12.
Draw the structure of Hypanthodium and label the parts.
Answer:

A – Ostiole
B – Male flowers
D – Neutral flower gall flower
C – Female flower
E – Receptacle

Question 13.
Compare Achene & Caryposis.
Answer:

Achene	Caryopsis
Indehiscent one-seeded fruit Develop from monocarpellary ovary Pericarp hard leathery-remain free from seed coat E.g. Clematis	Indehiscent one-seeded fruit Develop from monocarpellary ovary Pericarp fused with seed coat E.g. Paddy

Question 14.
Compare Legume and Follicle
Answer:

Question 15.
Differentiate between Dicotyledonous seed and Monocotyledonous seed.
Answer:

Dicot	Monocot
Two cotyledons occur Usually cotyledons store food and become thick and fleshy nourishes seedling during early development E.g.Pea	Only one cotyledon occur The endosperm persistent and nourishes the seedling during early development E.g. Castor

Question 16.
Differentiate between Albuminous and Non-albuminous seed.
Answer:

Albuminous seed	Non Albuminous seed
The cotyledons are then membranous and mature seeds have endosperm persistent and nourishes the seedling during its early development. Eg. Castor, Sunflower	Food is stored in cotyledons and mature seeds are without endosperm. Eg. Pea, Groundnut

Give Short Answers – 3 Marks

Question 1.
Differentiate between Racemose and Cymose Inflorescence.
Answer:

Characters	Racemose	Cymose
1. Main axis	Unlimited growth	Limited growth
2. Arrangement of flowers	Acropetal succession	Basipetal succession
3. Opening	Centripetal	Centrifugal
4. Oldest flower	At the base	At the top

Question 2.
What is meant by Salver shaped or Hypocrateriform corolla. Give Eg.
Answer:
Petals of a flower fused to form a long narrow tube with spreading limbs are called salver-shaped or hypocrateriform corolla.
E.g. Ixora, Catharanthus sp.

Question 3.
Differentiate between Corymbose and Umbellate inflorescence by drawing diagrams.
Answer:

Question 4.
Why do we call Thyrsus as Raceme of Cymes?
Answer:
a. The main axis is indefinite growth like raceme
b. But it bears pedicellate cymes on either side laterally. E.g. Ocimum sanctum.

Question 5.
Differentiate between Cyathium and Coenanthium.
Answer:

Cyathium	Coenanthium
A single female flower surrounded by many male flowers enclosed by a common involucre Flowers are naked-Aclamydeous Extrafloral nectary is present in involucre E.g. Euphorbia	Circular disc-like fleshy open receptacle bearing many pistillate or female flowers at the center surrounded by many male or staminate flowers at the periphery. Eg. Dorsenia

Question 6.
Differentiate between Homogamous head & Heterogamous head inflorescence.
Answer:

Homogamous Head	Heterogamous Head
Only one kind of florets 2 Types Has only tongue florets- E.g. Launaea Has only tube florets – E.g. Vernorua	It has 2 types of florets -Tongue of ray, Tube of Disc Tongue florets seen towards the periphery, and Tube florets located at the centre of the inflorescence E.g. Helianths & Tridax

Question 7.
Distinguish between the two types of Monochasial Cyme.
Answer:

Question 8.
Distinguish between Anthophore and Androphore, Gynophore and Gynandrophore
Answer:

Anthophore	Androphore	Gynophore	Gynandrophore
The intermodal elongation between calyx and corolla E.g. Silene conoidea	The internal elongation between coralla and Aroecium E.g. Grewia	The internal elongation between Androecium and cynoecium E.g. Capparis	The unified internal elongation between corolla and Androecium as well as between Androecium and gynoecium E.g gynandropsis

Question 9.
Distinguish between Connation & Adnation
Answer:

Question 10.
Differentiate between Didynamous and Tetradynamous condition of the stamen.
Answer:

Question 11.
Explain the various types of Schizocarpic fruit.
Answer:
This fruit type of intermediate between dehiscent and indehiscent fruit. The fruit instead of dehiscing rather splits into a number of segments, each containing one or more seeds. They are of the following types:

1. Cremocarp: Fruit develops from bicarpellary, syncarpous, inferior ovary and splitting into two one-seeded segments known as mericarps. e.g., Coriander and Carrot.
2. Carcerulus: Fruit develops from bicarpellary, syncarpous, superior ovary and splitting into four one-seeded segments known as nutlets, e.g., Leucas, Ocimum and Abutilon.
3. Lomentum: The fruit is derived from monocarpellary, unilocular ovary. A leguminous fruit, constricted between the seeds to form a number of one seeded compartments that separate at maturity, e.g., Desmodium, Arachis and Mimosa.
4. Regma: They develop from tricarpellary, syncarpous, superior, trilocular ovary and splits into one-seeded cocci which remain attached to carpophore, e.g., Ricinus and Geranium.

Question 12.
Identify the plant & Write down the floral formula of the given floral diagram.
Answer:

Question 13.
Draw the floral diagram of Ixora Coccinea flower and write down floral formula.
Answer:

Question 14.
Classify the anthers based on their mode of attachment.
Answer:
The anthers based on their mode of attachment:

1. Basifixed: (Innate) Base of anther is attached to the tip of filament, e.g., Brassica, Datura
2. Dorsifixed: Apex of filament is attached to the dorsal side of the anther, e.g. Citrus, Hibiscus
3. Versatile: Filament is attached to the anther at midpoint, e.g., Grasses
4. Adnate: Filament is continued from the base to the apex of anther, e.g. Verbena, Ranunculus, Nelumbo.

Essay Questions – 5 Marks

Question 1.
Distinguish between Monoecious – Dioecious &Polygamous.
Answer:

Monoecious	Dioecious	Polygamous
a.One house i.e male and female flowers present in the same flower . E.g. Coconut	Two house i.e male and and female flowers present on separate plants E.g. Papaya	Here male flowers(staminate) female flowers (pistillate) & bisexual flowers occur in a single plant E.g. Mangifera

Question 2.
List out the significance of fruits.
Answer:
The significance of fruits:

1. Edible part of the fruit is a source of food, energy for animals.
2. They are source of many chemicals like sugar, pectin, organic acids, vitamins and minerals.
3. The fruit protects the seeds from unfavourable climatic conditions and animals.
4. Both fleshy and dry fruits help in the dispersal of seeds to distant places.
5. In certain cases, fruit may provide nutrition to the developing seedling.
6. Fruits provide source of medicine to humans.

Question 3.
What are the various parts of a typical flower.
Answer:

Question 4.
Detìne Aestivation Give an account of various types of aestivation?
Answer:

• Aestivation: Arrangement of sepals and petals in the flowers bud.
•  Types : There are 5 types
•  Valvate : Margins of sepals and petals do not overlap but touch each other.
  Eg. Calyx — malvaceac inem bers .
• Twisted or convolute or contorted: One margin of each petal or sepal overlapping on
  the other petal – Eg. Corolla of Malvaceae (china i-ose)
• Imbricate : Sepals\ Petalsepals petaals – overlap irregularly one member of the whorl-
  exterior another interior other three one margin exterior other interior

3 types

1.  Ascending-imbricate Eg. Cassia,
2. . Descendingly-imbricate (vexillary aestivation) Eg. Clihoria,
3. . Quincuncial- Eg. Guava
   

Question 5.
Define placentation and explain the various types of placentation with diagrams.
Answer:

Marginal:

It is with the plaentae along the marging of a unicarpellate ovary.
Example-Fabaceae.
Axile:

The placentae arises from the column in a compound ovary with septa.
Example-Hibiscus, tomato lemon

Superficial:

Ovules arise from the surfae of the septa.
Example: Nymphaeceae

Parietal:

It is the placentae on the ovary walls or upon intruding partitions of a unilocular, compound Ovary.
Example: Mustard, Argemone, cucumber.

Free-central:

It is with the placentae along the column in a compound ovary without septa.
Example: Caryophyllaceae, Dianthus, Primrose

Basal:

It is the placenta at the base of the ovary.
Example: Sunflower (Asteraceae) Marigold.

Question 6.
Draw the floral diagram and flower of Cocos Nucifera and try to describe the flower with the floral diagram and floral formula.
Answer:

Male flower // Femle flower
Male flower of Cocos nucifera.

Male flower – Unisexual male Actinomorphic Bracteate, Bracteolate Incomplete
Perianth – 6 tepals – outer 3 and inner 3 two whorlsimbricate – apophyllous
Androecium – 6 stamens – outer 3 and inner 3 – free anther dithecous
Gynoecium – Absent – pistillode present

Female Flower – Unisexual female Actinomorphic Bracteole, ebracteolate, incomplete
Perianth – 6 tepals outer 3 -inner 3 outer valvate, inner imbricate apophyllous
Androecium – Absent staminode tricarpellary
Gynoecium – Ovary superior – tricarpellary trilocular syncarpous- ovules-Axile Placentation.

Question 7.
Describe the ovary types on the basis of its positive relative to other parts.
Answer:
The ovary can be divided into 3 types on this basis
Superior Ovary : (Flower Hypogynous)
It is the ovary with sepals, petals and stamens attached at the base of the ovary.

In feriror ovary : (Flower Epigynous)
It is the ovary with sepals, petals and stamens attached at the apex of the ovary.

Half inferior ovary : (Flower Perigynous) It is the ovary with sepals petals and stamens or hypanthium attached near the middle of the ovary

Question 8.
Give an account of Dry Dehiscent fruits.
Answer:

Question 9.
Give an account of dry indehiscent fruits.
Answer:

	Type of fruit	Nature of ovary	Special aspects	Example
1.	Achene	Monocarpellary superior ovary Apocarpous	Apocarpous Fruit wall pericarp is free from seed coat.	Clematis strawberry
2.	Cypsela	Bicarpellary inferior ovary syncarpous	Reduced scales Hairy or feathery – calyx lobes-Pappus	Tridax helianthus
3.	Caryopsis	Monocarpellary superior ovary	Fruit wall inseparably fused with seed	Oryza triticum
4.	Nut	Multicarpellary syncarpous superior ovary	Hard woody bony pericarp	Anacardium
5.	Samara	Monocarpellary superior ovary	Pericarp (ovary wall) Develop into then wing-like structure – help in fruit dispersal	Pterocarpus
6.	Utricle	Bicarpellary unilocular syncarpous superior ovary	Pericarp loosely encloses the seed.	Chenopodium.

Question 10.
Explain only the racemose type with the elongated main axis.
Answer:

Question 11.
Draw a chart depicting various types of Fruits.
Answer:

Question 12.
What is the significance of Seeds?
Answer:

Question 13.
Draw the tabulation showing various fruits & their edible part.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 10 Oscillations Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

11th Physics Guide Oscillations Book Back Questions and Answers

I. Multiple choice questions:

Question 1.
In a simple harmonic oscillation, the acceptation against displacement for one complete oscillation will be: (Model NSEP 2000 – 01)
(a) an ellipse
(b) a circle
(c) a parabola
(d) a straight line
Answer:
(d) a straight line

Hint:
The sketch between cause (magnitude of acceleration) and effect (magnitude of displacement) is a straight line.

Question 2.
A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3 s. The time period is:
(a) 15 s
(b) 6 s
(c) 12 s
(d) 9 s
Answer:
(c) 12 s

Hint:

The time period is the time taken by a particle to return to B.

Question 3.
The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on the surface of planet X such that the acceleration of the planet X is n times greater than the Earth is:
(a) 0.9 n
(b) \(\frac { 0.9 }{ n }\)m
(c) 0.9 n²m
(d) \(\frac{0.9}{n^{2}}\)
Answer:
(a) 0.9 n

Hint:

Question 4.
A simple pendulum is suspended from the roof of a school bus which moves in a horizontal direction with an acceleration a, then the time period is:
(a) T ∝ \(\frac{1}{g^{2}+a^{2}}\)
(b) T ∝ \(\frac{1}{\sqrt{g^{2}+a^{2}}}\)
(c) T ∝ \(\sqrt{g^{2}+a^{2}}\)
(d) T ∝ (g² + a²)
Answer:
(b) T ∝ \(\frac{1}{\sqrt{g^{2}+a^{2}}}\)

Hint: T = 2π\(\sqrt{\frac{l}{g}}\)
When a bus is moving
g’ = \(\sqrt{g^{2}+a^{2}}\)
∴ T ∝ \(\frac{1}{\sqrt{g^{2}+a^{2}}}\)

Question 5.
Two bodies A and B whose masses are in the rati0 1:2 are suspended from two separate massless springs of force constants k_A and k_B respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio 1:2, the ratio of the amplitude A to that of B is:
(a) \(\sqrt{\frac{k_{\mathrm{B}}}{2 k_{\mathrm{A}}}}\)
(b) \(\sqrt{\frac{k_{\mathrm{B}}}{8 k_{\mathrm{A}}}}\)
(c) \(\sqrt{\frac{2k_{\mathrm{B}}}{ k_{\mathrm{A}}}}\)
(d) \(\sqrt{\frac{8k_{\mathrm{B}}}{ k_{\mathrm{A}}}}\)
Answer:
(b) \(\sqrt{\frac{k_{\mathrm{B}}}{8 k_{\mathrm{A}}}}\)

Hint: v_A : v_B
Amplitude of A : Amplitude of B = \(\sqrt{\mathrm{K}_{\mathrm{B}}}: \sqrt{8 \mathrm{~K}_{\mathrm{A}}}\)

Question 6.
A spring is connected to a mass m Suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is:
(a) T’ = \(\sqrt{2}\)T
(b) T’ = \(\frac{\mathrm{T}}{\sqrt{2}}\)
(c) T’ = \(\sqrt{2T}\)
(d) T’ = \(\sqrt{\frac{\mathrm{T}}{2}}\)
Answer:
(b) T’ = \(\frac{\mathrm{T}}{\sqrt{2}}\)

Hint:
T = 2π\(\sqrt{\frac{\mathrm{m}}{k}}\)
When the spring is cut into two equal halves, then the force constant of each part is 2k.
When the mass is suspended from one of the halves.
Time period T’ = 2π\(\sqrt{\frac{\mathrm{m}}{2k}}\)
= \(\sqrt{\frac{\mathrm{T}}{2}}\)

Question 7.
The time period for small vertical oscillations of block of mass m when the masses of the pulleys are negligible and spring constant k₁ and k₂ is:

(a) T = 4π\(\sqrt{m\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\right)}\)
(b) T = 2π\(\sqrt{m\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\right)}\)
(c) T = 4π\(\sqrt{m\left(k_{1}+k_{2}\right)}\)
(d) T = 2π\(\sqrt{m\left(k_{1}+k_{2}\right)}\)
Answer:
(a) T = 4π\(\sqrt{m\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\right)}\)

Hint:
T = 2π\(\frac { m }{ k }\)
The given arrangement is similar to the combination of springs in series.

Question 8.
A simple pendulum has a time period T₁. When its point of suspension is moved vertically upwards according as y = kt², where y is vertical distance covered and k = 1 ms^-2, its time period becomes T₂. Then, \(\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}\) is (g = 10 ms^-2
(a) \(\frac { 5 }{ 6 }\)
(b) \(\frac { 11 }{ 10 }\)
(c) \(\frac { 6 }{ 5 }\)
(d) \(\frac { 5 }{ 4 }\)
Answer:
(c) \(\frac { 6 }{ 5 }\)

Hint:

Question 9.
An ideal spring of spring constant k, is suspended from the ceiling of a room and a block of mass m is fastened to its lower end. If the block is released when the spring is un-stretched, then the maximum extension in the spring is:
(a) 4\(\frac { mg }{ k }\)
(b) \(\frac { mg }{ k }\)
(c) 2\(\frac { mg }{ k }\)
(d) \(\frac { mg }{ 2k }\)
Answer:
(c) 2\(\frac { mg }{ k }\)

Hint:

Question 10.
A pendulum is hung in a very high building oscillates to and fro motion freely like a simple harmonic oscillator. If the acceleration of the bob is 16 ms^-2 at a distance of 4 m from the mean position, then the time period is: (NEET 2018 model)
(a) 2s
(b) 1s
(c) 2 πs
(d) πs
Answer:
(d) πs

Hint:
a = 16 m/s²; y = 4
a = – \(\frac { g }{ l }\)x = – ω²x
16 = \(\left|1-\omega^{2} \times 4\right|\)
∴ ω² = \(\frac { 16 }{ 4 }\) = 4
ω = 2
Time period T = \(\frac { 2π }{ ω }\) = πs

Question 11.
A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will:
(a) first increase and then decrease
(b) first decrease and then increase
(c) increase continuously
(d) decrease continuously
Answer:
(a) first increase and then decrease

Hint:
As the water flows out of the sphere, the time period first increases and then decreases. Initially, when the sphere is completely filled with water its centre of gravity (C.G) lies at its centre. As water flows out, the C.G begins to shift below the centre of the sphere.

The effective length of the pendulum increases and hence time period increases. When the sphere becomes more than half empty, its C.G begins to rise up. The effective length of the pendulum decreases and T decreases.

Question 12.
The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are: (AIPMT 2012)
(a) kg m s^-1
(b) kg m s^-2
(c) kg s^-1
(d) kg s
Answer:
(c) kg s^-1

Hint:
F_d ∝ v
F_d = – bv
F_d = kv
∴ k = \(\frac{\mathrm{F}_{d}}{\nu}\)
Units of k = \(\frac{\mathrm{kgms}^{-2}}{\mathrm{~m} / \mathrm{s}}\)
= kg^-1
Units of proportionality constant = kgs^-1

Question 13.
When a damped harmonic oscillator completes 100 oscillations, its amplitude is reduced to\(\frac { 1 }{ 3 }\) of its initial value. What will be its amplitude when it completes 200 oscillations?
(a) \(\frac { 1 }{ 5 }\)
(b) \(\frac { 2 }{ 3 }\)
(c) \(\frac { 1 }{ 6 }\)
(d) \(\frac { 1 }{ 9 }\)
Answer:
(b) \(\frac { 2 }{ 3 }\)

Hint:

Question 14.
Which of the following differential equations represents a damped harmonic oscillator?
(a) \(\frac{d^{2} y}{d t^{2}}\) + y = 0
(b) \(\frac{d^{2} y}{d t^{2}}\) + γ \(\frac { dy }{ dt }\) + y = 0
(c) \(\frac{d^{2} y}{d t^{2}}\) + k²y = 0
(d) \(\frac { dy }{ dt }\) + y = 0
Answer:
(b) \(\frac{d^{2} y}{d t^{2}}\) + γ \(\frac { dy }{ dt }\) + y = 0

Hint:
For a damped oscillator F ∝ v
Total restoring force F = – ky – bv
b – damping constant; y – displacement
If F = – ky – by
then m\(\frac{d^{2} y}{d t^{2}}\) = – ky – b\(\frac { dy }{ dt }\)
m\(\frac{d^{2} y}{d t^{2}}\) + ky + b\(\frac { dy }{ dt }\) = 0
÷ m we get
\(\frac{d^{2} y}{d t^{2}}\) + \(\frac { b }{ m }\)\(\frac { dy }{ dt }\) + \(\frac { k }{ m }\)y = 0
If k = m and \(\frac { b }{ m }\) = r
then the equation becomes
\(\frac{d^{2} y}{d t^{2}}\) + r\(\frac { dy }{ dt }\) + y = 0

Question 15.
If the inertial mass and gravitational mass of the simple pendulum of length l are not equal, then the time period of the simple pendulum is:
(a) T = 2π\(\sqrt{\frac{m_{i} l}{m_{g} g}}\)
(b) T = 2π\(\sqrt{\frac{m_{g} l}{m_{i} g}}\)
(c) T = 2π\(\frac{m_{g}}{m_{i}} \sqrt{\frac{l}{g}}\)
(d) T = 2π\(\frac{m_{i}}{m_{g}} \sqrt{\frac{l}{g}}\)
Answer:
(a) T = 2π\(\sqrt{\frac{m_{i} l}{m_{g} g}}\)

Hint:

II. Short Answers Questions:

Question 1.
What is meant by periodic and non-periodic motion? Give any two examples, for each motion.
Answer:
Periodic motion: Any motion which repeats itself in a fixed time interval is known as periodic motion. Examples: Hands in a pendulum clock, the swing of a cradle.
Non-Periodic motion: Any motion which does not repeat itself after a regular interval of time is known as non-periodic motion. Example: Occurrence of Earthquake, the eruption of a volcano.

Question 2.
What is meant by the force constant of a spring?
Answer:
Force constant is defined as force per unit length.

Question 3.
Define the time period of simple harmonic motion.
Answer:
The time period is defined as the time taken by a particle to complete one oscillation. It is usually denoted by T.

Question 4.
Define frequency of simple harmonic motion.
Answer:
The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is s^-1 or hertz (Hz).
Mathematically, frequency is related to time period by f = \(\frac{1}{\mathrm{T}}\)

Question 5.
What is an epoch?
Answer:
Initial phases of a particle is an epoch. At time t = 0 s (initial time), the phase φ = (φ₀ is called epoch (initial phase) where φ₀ is called the angle of epoch.

Question 6.
Write short notes on two springs connected in series.
Answer:
Let us consider two springs whose spring constant are k₁ and k₂ and which are connected to a mass m as shown in Figure.

Let F be the applied force towards right as shown in Figure. The net displacement of the mass point is
x = x₁ + x₂
From Hooke’s law, the net force,
F = – k_s(x₁ + x₂)
The effective spring constant can be calculated as

When n springs connected in series, the effective spring constant in series is

If all spring constants are identical
i.e., k₁ = k₂ = … = k_n = k then
\(\frac{1}{k_{s}}\) = \(\frac { n }{ k }\) ⇒ k_s = \(\frac { k }{ n }\)
This means that the effective spring constant reduces by the factor n. So, for springs in series connection, the effective spring constant is lesser than the individual spring constants.

Question 7.
Write short notes on two springs connected in parallel.
Answer:
Let us consider only two springs of spring constants k₁ and k₂ attached to a mass m as shown in Figure.

Net force for the displacement of mass m is F = – k_px … (1)
Where k_p is called effective spring constant. Let the first spring be elongated by a displacement x due to force F₁ and the second spring be elongated by the same displacement x due to force F₂, then the net force
F = – k₁x – k₂x … (2)
Equating equations (2) and (1), we get
k_p = k₁ + k₂ … (3)
Generalizing, for n springs connected in parallel,
k_p = \(\sum_{i=1}^{n} k_{i}\) … (4)
If all spring constants are identical i.e., k₁ = k₂ = … = k_n = k then
k_p = nk … (5)
It is implied that the effective spring constant increases by a factor n. So, for the springs in parallel connection, the effective spring constant is greater than the individual spring constant.

Question 8.
Write down the time period of simple pendulum.
Answer:
T = 2π\(\sqrt{\frac{l}{g}}\) in second.
where l – length of the pendulum.
g – acceleration due to gravity.

Question 9.
State the laws of simple pendulum.
Answer:
Laws of simple pendulum: The time period of a simple pendulum.

Depends on the following laws:
(i) Law of length: For a given value of acceleration due to gravity, the time period of a simple pendulum is directly proportional to the square root of length of the pendulum.
T ∝ \(\sqrt{l}\) … (1)

(ii) Law of acceleration: For a fixed length, the time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity.
T ∝ \(\frac{1}{\sqrt{g}}\) … (2)

Question 10.
Write down the equation of time period for linear harmonic oscillator.
Answer:
Time period T = \(\frac { 1 }{ f }\) = 2π \(\sqrt{\frac{m}{k}}\)s
where m is the mass, k is the spring constant.

Question 11.
What is meant by free oscillation?
Answer:
The oscillations in which the amplitude decreases gradually with the passage of time are called damped Oscillations.
Example:

1. The oscillations of a pendulum or pendulum oscillating inside an oil-filled container.
2. Electromagnetic oscillations in a tank circuit.
3. Oscillations in a dead beat and ballistic galvanometers.

Question 12.
Explain damped oscillation. Give an example.
Answer:
If an oscillator oscillates in a resistive medium, then its amplitude goes on decreasing. The motion of the oscillator is called damped oscillation.
Example:

• The oscillations of a pendulum (including air friction) or pendulum oscillating inside an oil-filled container.
• Electromagnetic oscillations in a tank circuit.

Question 13.
Define forced oscillation. Give an example.
Answer:
The body executing vibration initially vibrates with its natural frequency and due to the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.
Example: Soundboards of stringed instruments.

Question 14.
What is meant by maintained oscillation? Give an example.
Answer:
To avoid damping in an oscillation, energy is supplied from an external source, the amplitude of the oscillation can be made constant. Such vibrations are known as maintained vibrations.

Example:
The vibration of a tuning fork getting energy from a battery or from an external power supply.

Question 15.
Explain resonance. Give an example.
Answer:
The frequency of external periodic force (or driving force) matches with the natural frequency of the vibrating body (driven). As a result, the oscillating body begins to vibrate such that its amplitude increases at each step and ultimately it has a large amplitude. Such a phenomenon is known as resonance and the corresponding vibrations are known as resonance vibrations. Example: The breaking of glass due to sound

III. Long Answers Questions:

Question 1.
What is meant by simple harmonic oscillation? Give examples and explain why every simple harmonic motion is a periodic motion whereas the converse need not be true.
Answer:
Simple harmonic motion is a special type of oscillatory motion in which the acceleration or force on the particle is directly proportional to its displacement from a fixed point and is always directed towards that fixed point.

Example: Oscillation of a pendulum. SHM is a special type of periodic motion, where restoring force is proportional to the displacement and acts in the direction opposite to displacement.

Question 2.
Describe Simple Harmonic Motion as a projection of uniform circular motion.
Answer:
(i) Let us consider a particle of mass m moving with uniform speed v along the circumference of a circle whose radius is r in an anti-clockwise direction (as shown in Figure).

(ii) It is assumed that the origin of the coordinate System coincides with the center 0 of the circle.

(iii) If co is the angular velocity of the particle and 0 the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter a simple harmonic motion is obtained.

(iv) This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion.

(v) Conversely, any vibratory motion or revolution can be mapped to uniform circular motion. The position of a particle moving is projected on to its vertical diameter or on to a line parallel to vertical diameter.

(vi) Similarly, we can do it for horizontal axis or a line parallel to the horizontal axis.

Example: Let us consider a spring-mass system (or oscillation of pendulum) as shown in Figure. When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.

Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter) traces straight line motion that is simple harmonic in nature. The circle is known as the reference circle of the simple harmonic motion.

Question 3.
What is meant by angular harmonic oscillation? Compute the time period of angular harmonic oscillation.
Answer:

If a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation. The point at which the resultant torque acting on the body is taken to be zero. It is called mean position. If the body is displaced from the mean position, then the resultant torque acts such that it is proportional to the angular displacement. This torque has a tendency to bring the body towards the mean position.

Time period : Let \(\overline{θ}\) be the angular displacement of the body and the resultant torque \(\vec { τ }\) acting on the body is,
\(\vec { τ }\) ∝ \(\vec {θ}\) … (1)
\(\vec { τ }\) = – k\(\vec {θ}\) … (2)
K is the restoring torsion constant, that is torque per unit angular displacement. If I is the moment of inertia of the body and \(\vec { τ }\) ∝ \(\vec {α}\) is the angular acceleration then
\(\vec { τ }\) = I \(\vec{α}\) = k \(\vec {θ}\)
But \(\vec{α}\) = \(\frac{d^{2} \vec{\theta}}{d t^{2}}\) and therefore
\(\frac{d^{2} \vec{\theta}}{d t^{2}}\) = – \(\frac { k }{ I }\)\(\vec {θ}\) … (3)
This differential equation resembles simple harmonic differential equation.
By comparing equation (3) with simple harmonic motion given we get,
a = \(\frac{d^{2} y}{d t^{2}}\) = – ω²y, we have
ω = \(\sqrt{\frac{k}{I}}\) rads^-1 … (4)
The frequency of the angular harmonic motion is given equation ω = 2πf is
f = \(\frac { 1 }{ 2π }\) \(\sqrt{\frac{k}{I}}\)Hz … (5)
The time period is given equation is
T = \(\frac { 1 }{ f }\); T = 2π\(\sqrt{\frac{I}{k}}\) second
In angular simple harmonic motion, the displacement of the particle is measured in terms of angular displacement \(\vec {θ}\).

Question 4.
Write down the difference between simple harmonic motion angular simple harmonic motion.
Answer:
Comparison of simple harmonic motion and angular simple harmonic motion.

Question 5.
Discuss the simple pendulum in detail.
Answer:
Construction:
A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a massless and inextensible string. The other end is fixed on a stand as shown in Figure (a). At equilibrium, the pendulum does not oscillate and is suspended vertically downward.

Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released, the bob of the pendulum executes to and fro motion.

Calculation of time period: Let l be the length of the pendulum which is taken as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position, as shown in Figure (d),

• The gravitational force acting on the body (F = mg) acts vertically downwards.
• The tension in the string T acts along the string to the point of suspension.

Resolving the gravitational force into its components:

• Normal component: It is along the string but in opposition to the direction of tension, F_as = mg cosθ.
• Tangential component: It is perpendicular to the string i.e., along the tangential direction of arc of the swing, F_ps = mg sinθ.

Hence, the normal component of the force is, along the string,
T – W_as = m\(\frac{v^{2}}{l}\)
Here v is speed of bob
T – mg cosθ = m\(\frac{v^{2}}{l}\) … (1)
From the Figure, it is observed that the tangential component W_ps of the gravitational force always points towards the equilibrium position. This direction always points opposite to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is the restoring force. Applying Newton’s second law along tangential direction, we have

Where, s is the position of bob that is measured along the arc. Expressing arc length in terms of angular displacement i.e.,
s = lθ … (3)
Then its acceleration,
\(\frac{d^{2} \theta}{d t^{2}}\) = \(\frac{d^{2} \theta}{d t^{2}}\) … (4)
Substituting equation (4) in equation (2), we get,
l\(\frac{d^{2} \theta}{d t^{2}}\) = – g sin θ
\(\frac{d^{2} \theta}{d t^{2}}\) = – \(\frac { g }{ l }\) sin θ … (5)
Because of the presence of sinθ in the above differential equation, it is a nonlinear differential equation. It is assumed that “the small oscillation approximation”, sin θ ≈ θ, the above differential equation becomes linear differential equation.
\(\frac{d^{2} \theta}{d t^{2}}\) = – \(\frac { g }{ l }\) θ
It is known as oscillatory differential equation. Hence, the angular frequency of this oscillator (natural frequency of this system) is

Question 6.
Explain the horizontal oscillations of a spring.
Answer:

Let us consider a system containing a block of mass m fastened to a massless spring with stiffness constant or force constant or spring constant k placed on a smooth horizontal surface (frictionless surface) as shown in Figure. Let x₀ be the equilibrium position or mean position of mass m when it is left undisturbed.

When the mass is displaced through a small displacement x towards right from its equilibrium position and then released, it will oscillate back and forth about its mean position x₀. Let F be the restoring force (due to stretching of the spring) that is proportional to the amount of displacement of block: For one-dimensional motion, we get
F ∝ x
F = – kx
where negative sign implies that the restoring force will always act opposite to the direction of the displacement. This equation is called Hooke’s law. It is noticed that, the restoring force is linear with the displacement (i.e., the exponent of force and displacement are unity). This is not always true. If we apply a very large stretching force, then the amplitude of oscillations becomes very large.
m\(\frac{d^{2} x}{d t^{2}}\) = – kx
\(\frac{d^{2} x}{d t^{2}}\) = – \(\frac { k }{ m }\)x … (1)
Comparing the equation (1) with simple harmonic motion equation a = \(\frac{d^{2} y}{d t^{2}}\) = – ω²y, we get
ω² = \(\frac { k }{ m }\)
which means the angular frequency or natural frequency of the oscillator is
ω = \(\sqrt{\frac{k}{m}}\)rad s^-1 … (2)
The frequency of the oscillation is
f = \(\frac { ω }{ 2π }\) = \(\frac { 1 }{ 2π }\)\(\frac { k }{ m }\) Hertz … (3)
and the time period of the oscillation is
T = \(\frac { 1 }{ f }\) = 2π[/latex]\(\frac { m }{ k }\) seconds … (4)

Question 7.
Describe the vertical oscillations of a spring.
Answer:
Consider a massless spring with stiffness constant or force constant k attached to a ceiling as shown in Figure. Let the length of the spring before loading mass m be L. If the block of mass m is attached to the other end of the spring, then the spring elongates by a length l. Let F₁ be the restoring force due to the stretching of spring. Due to mass m, the gravitational force acts vertically downward. A free-body diagram is drawn for this system as shown in Figure. When the system is under equilibrium,

F₁ + mg = 0 … (1)
But the spring elongates by small displacement l,
∴ F₁ ∝ l ⇒ F₁ = – kl … (2)
Substituting equation (2) in equation (1), we get
– kl + mg = 0
mg = kl
(or) \(\frac { m }{ k }\) = \(\frac { 1 }{ g }\) … (3)
Suppose a very small external force is applied on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y + l) is
F₂ (y + 1)
F₂ = – k(y + 1) = – ky – kl … (4)
Since, the mass moves up and down with acceleration , by drawing the free body diagram for this case, we get
-ky – kl + mg = m\(\frac{d^{2} y}{d t^{2}}\) … (5)
The net force acting on the mass due to this stretching is
F = F₂ + mg
F = – ky – kl + mg … (6)
The gravitational force opposes the restoring force. Substituting equation (3) in equation (6), we get
F = – ky – kl + kl = ky
Time period:
Applying Newton’s law, we get
m\(\frac{d^{2} y}{d t^{2}}\) = – ky
\(\frac{d^{2} y}{d t^{2}}\) = – \(\frac { k }{ m }\)y … (7)
The above equation is in the form of simple harmonic differential equation. Hence the time period is
T = 2π \(\sqrt{\frac{m}{k}}\)second … (8)
The time period can be rewritten using equation (3) as
T = 2π \(\sqrt{\frac{m}{k}}\) = 2π \(\sqrt{\frac{l}{g}}\)second.

Question 8.
Write short notes on the oscillations of the liquid column in U-tube.
Answer:
Let us consider a U-shaped glass tube which consists of two open arms with uniform cross¬sectional area A. Let us pour a non-viscous uniform incompressible liquid of density p in the U-shaped tube to a height h as shown in the Figure. If the liquid and tube are not disturbed then the liquid surface will be in equilibrium position O.

It means the pressure as measured at any point on the liquid is the same and also at the surface on the arm (edge of the tube on either side), that balances the atmospheric pressure. Hence, the level of liquid in each arm will be the same. By blowing air one can provide sufficient force in one arm, and the liquid gets disturbed from equilibrium position O, It is meant, that pressure at blown arm is higher than the other arm.

A difference in pressure is created that will cause the liquid to oscillate for a very short duration of time about the mean or equilibrium position. Finally, it comes to rest.
Time period of the oscillation is
T= 2π \(\sqrt{\frac{l}{2g}}\)second

Question 9.
Discuss in detail the energy in simple harmonic motion.
Answer:
(i) Expression for Potential Energy:For the simple harmonic motion, the force and the displacement are related by Hooke’s law
\(\vec { F }\) = – k\(\vec { r }\)
F = – kx … (1)
the work done by the conservative force field is independent of path.
Calculation potential energy:
F = \(\frac { dU }{ dx }\) … (2)
Comparing (1) and (2), we get
\(\frac { dU }{ dx }\) = – kx
dU = – kx dx
This work done by the force F during a small displacement dx stores as potential energy

From equation,
ω = \(\sqrt{\frac{k}{m}}\) rad s^-1
By substituting the value of force constant k = mω² in equation (3), we get
U(x) = \(\frac { 1 }{ 2 }\)mω²x² … (4)
where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from the equation,
y = A sin ωt
we get,
x = A sin ωt
U(t) = \(\frac { 1 }{ 2 }\)mω²x² … (5)
This variation of U is shown below.

(ii) Expression for Kinetic Energy:
Kinetic energy
KE = \(\frac { 1 }{ 2 }\)mv_x² = \(\frac { 1 }{ 2 }\)m(\(\frac { dx }{ dt }\))² … (6)
Since the particle executes simple harmonic motion, from equation y = A sin ωt
x = A sin ωt
∴ velocity is

(iii) Expression for Total Energy: Total energy is the sum of kinetic energy and potential energy.
E = KE + U … (11)
E = \(\frac { 1 }{ 2 }\)mω²(A² – x²) + \(\frac { 1 }{ 2 }\)mω²x² … (12)
Hence, cancelling x² term,
E = \(\frac { 1 }{ 2 }\)mω²x² = r constant … (13)
Alternatively, from equation (5) and equation (10), we get the total energy as
E = \(\frac { 1 }{ 2 }\)mω²x² sin²ωt + \(\frac { 1 }{ 2 }\)mω²A²cos²ωt
= \(\frac { 1 }{ 2 }\)mω²A² (sin²ωt + cos²ωt)
From trigonometry identity,
(sin² ωt + cos² ωt) = 1
E = \(\frac { 1 }{ 2 }\)mω²A² = constant
which gives the law of conservation of total energy.

Question 10.
Explain in detail the four different types of oscillations.
Answer:
(i) Free oscillations: When the oscillator oscillates with a frequency that is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration.

Example:

• Oscillation of a simple pendulum,
• Vibration in a stretched string.

(ii) Damped oscillation: If an oscillator oscillates in a resistive medium, then its amplitude goes on decreasing. The energy of the oscillator is used to do work against the resistive medium. The motion of the oscillator is said to be a damped oscillation.

Example:

• The oscillations of a pendulum (including air friction) or pendulum oscillating inside an oil filled container.
• Electromagnetic oscillations in a tank circuit.

(iii) Forced oscillations: In this type of vibration, the body executing vibration initially vibrates with its natural frequency. Because of the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.

Example:
Soundboards of stringed instruments.

(iv) Maintained oscillations: The amplitude of the oscillation can be made constant. By supplying energy from an external source. Such oscillations are known as maintained oscillations.

Example:
The vibration of a tuning fork getting energy from a battery or from external power supply.

IV. Numerical Problems:

Question 1.
Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time Period is T = 2π\(\sqrt{\frac{\mathrm{R}}{g}}\)
Answer:

(Here, negative has no meaning. It can be neglected)
Comparing equation (1) & (2) we get,

Time period, T = 2π\(\sqrt{\frac{\mathrm{R}}{g}}\)
Hence proved.

Question 2.
Calculate the time period of the oscillation of a particle of mass m moving in the potential defined as
Answer:

In the problem consider oscillation of particle into 2 cases via x < 0 (consider it as SHM where the time period is considered as t₁) and another one as x > 0 (consider it as a motion under gravity, where time period is t₂)

Note:
We want to find the total time period, which will be T = t₁ + t₂.
According to the conservation of energy

In case: 2 (Motion under gravity)
t₂ = \(\frac { 2v }{ g }\) substituting equation (2) here we get,
t₂ = \(\frac { 2 }{ g }\)\(\sqrt{\frac{2 E}{m}}\)
⇒ 2\(\sqrt{\frac{2 \mathrm{E}}{m g^{2}}}\) … (4)
Adding equation (3) & (4)
Time period of oscillation,
T = t₁ + t₂
= π\(\sqrt{\frac{m}{k}}\) + 2\(\sqrt{\frac{2 \mathrm{E}}{m g^{2}}}\)

Question 3.
Consider a simple pendulum of length l = 0.9 m which is properly placed on a trolley rolling down on a inclined plane which is at θ = 45° with the horizontal. Assuming that the inclined plane is frictionless, calculate the time period of oscillation of the simple pendulum.
Answer:
The effective value of acceleration due to gravity (g) will be equal to the component of g normal to the inclined plane which is
g’ = g cos α
T = 2π\(\sqrt{\frac{l}{g^{\prime}}}\) = 2π\(\sqrt{\frac{l}{g \cos \theta}}\)
Length of the pendulum l = 0.9m
Angle of inclination θ = 45°

Question 4.
A piece of wood of mass m is floating erect in a liquid whose density is ρ. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is T = 2π\(\sqrt{\frac{m}{\mathrm{Ag} \rho}}\)
Answer:
When a wood is pressed and released,

Question 5.
Consider two simple harmonic motion along x and y-axis having same frequencies but different amplitudes as x = A sin (ωt + φ) (along x axis) and y = B sin ωt (along y axis). Then show that \(\frac{x^{2}}{\mathrm{~A}^{2}}+\frac{y^{2}}{\mathrm{~B}^{2}}-\frac{2 x y}{\mathrm{AB}}\) cosφ = sin²φ and also discuss the special cases when
(i) φ = 0
(ii) φ = π
(iii) φ = \(\frac { π }{ 2 }\)
(iv) φ = \(\frac { π }{ 2 }\) and A = B
(v) φ = \(\frac { π }{ 4 }\)
Note: when a particle is subjected to two simple harmonic motions at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.
Given:
x = A sin(ωt-φ) … (1)
y = B sin ωt … (2)
In equation (1) use,
sin (A – B) – sin A cos B + cos A sin B
(1) ⇒ x – A sin ωt. cos (φ) + A cos ωt. sinφ
x – A sin cat. cos φ = A cos cot sin φ
squaring on both sides we get,
(x – A sin cot. cos φ)² = A² cos²cot sin²φ … (3)
In equation (3) sin at can be re-written as, \(\frac { y }{ B }\) [from equation (2)]. Also, use
cos²cot = 1 – sin²ωt in equation (3)
∴ (3) becomes

Hence proved.

Special cases:
(i) φ = 0 in equation (5) we get,

The above equation resembles equation of a straight line passing through origin with positive slope.

(ii) φ = π in equation (5)

The above equation is an equation of a straight line passing through origin with a negative slope.

(iii) φ = \(\frac { π }{ 2 }\) in equation (5)
The above equation of an ellipse whose centre is origin.

(iv) φ = \(\frac { π }{ B }\) and A = B in equation (5)
\(\frac{x^{2}}{\mathrm{~A}^{2}}+\frac{y^{2}}{\mathrm{~A}^{2}}\) = 1
⇒ x² + y² = A²
The above equation of a circle whose centre is origin.

(v) φ = \(\frac { π }{ 4 }\), cos \(\frac { π }{ 4 }\) = \(\frac{1}{\sqrt{2}}\) sin\(\frac { π }{ 4 }\) = \(\frac{1}{\sqrt{2}}\) equation (5) we get,
\(\frac{x^{2}}{\mathrm{~A}^{2}}+\frac{y^{2}}{\mathrm{~A}^{2}}\) – \(\frac{(\sqrt{2}) x y}{A B}\) = \(\frac { 1 }{ 2 }\)
The above equation is an equation of tilted ellipse.

Question 6.
Show that for a particle executing simple harmonic motion,
(a) the average value of kinetic energy is equal to the average value of potential energy,
(b) average potential energy = average kinetic energy = \(\frac { 1 }{ 2 }\) (total energy)
Hint: average kinetic energy = < kinetic energy > = \(\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}(\text { Potential energy }) d t\) and average potential energy = < potential energy > = \(\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}(\text { Potential energy }) d t\)
Answer:
(a) Suppose a particle of mass m executes SHM of time period T. The displacement of the particle at any instant t is given by
y = A sin ωr … (1)
Velocity v = \(\frac { dy }{ dt }\) = \(\frac { d }{ dt }\)(sin ωt) = Aω cos ωt = ωA cosωt

Average potential energy over a period of oscillation is,

(b) Total energy
T.E = \(\frac { 1 }{ 2 }\) ω²y² + \(\frac { 1 }{ 2 }\)mω²(A² – y²)
But y = A sinωt
T.E = \(\frac { 1 }{ 2 }\)mω²A² ωt + \(\frac { 1 }{ 2 }\)mω²A² cos²ωt
= \(\frac { 1 }{ 2 }\)mω²A²(sin²ωt + cos²ωt)
From trignometry identity
sin² ωt + cos² ωt = 1

Question 7.
Compute the time period for the following system if the block of mass m is slightly displaced vertically down from its equilibrium position and then released. Assume that the pulley is light and smooth, strings and springs are light.
Answer:

Case (a):
When mass is slightly displaced vertically down: Now pulley is fixed rigidly here. When the mass is displaced by y and the spring will also be stretched by y.
Hence F = ky
Time period T = \(\sqrt{\frac{m}{k}}\)

case (b):
When the system is released: WTien mass is displaced by y, pulley is also displaced by 4y,
∴ F = 4 Icy
∴ T = 2π\(\sqrt{\frac{m}{4k}}\)