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TN State Board 11th Chemistry Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
40 ml of methane is completely burnt using 80ml of oxygen at room temperature. The volume of gas left after cooling to room temperature. The volume of gas left after cooling to room temperature is …………………….
(a) 40 ml of CO₂ gas
(b) 40 ml of CO₂ gas and 80 ml of H₂O gas
(c) 60 ml of CO₂ gas and 60 ml H₂O gas
(d) 120 ml of CO₂ gas
Solution:
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l)

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.
Answer:
(a) 40 ml of CO₂ gas

Question 2.
What are the values of n, l, m and s for 3p_x electron?
(a) 3, 2, 1,0
(b) 3, 1,-1, +\(\frac{1}{2}\)
(c) 3, 2, +1, –\(\frac{1}{2}\)
(d) 3, 0, 0, +\(\frac{1}{2}\)
Solution:
3p_x electron; n = 3 (main shell)
for p_x orbital, l = 1, m = -1, s = \(\frac{1}{2}\)
Answer:
(b) 3, 1,-1, +\(\frac{1}{2}\)

Question 3.
Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.
Answer:
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.

Question 4.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 5.
Lithium shows diagonal relationship with ………………………
(a) Sodium
(b) Magnesium
(c) Calcium
(d) Ahuninium
Answer:
(b) Magnesium

Question 6.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\) \((\frac { \partial V }{ \partial T } )\)
(a) T
(b) 1/T
(c) P
(d) None of these
Solution:

Answer:
(b) 1/T

Question 7.
Heat of combustion is always …………………….
(a) Positive
(b) Negative
(c) Zero
(d) Either positive or negative
Answer:
(b) Negative

Question 8.
If in a mixture where Q = K, then what happens?
(a) The reaction shift towards products
(b) The reaction shift towards reactants
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate
(d) Nothing happens
Answer:
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate

Question 9.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N₂
(b) He
(c) CO₂
(d) H₂
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.
Answer:
(c) CO₂

Question 10.
In the molecule O_A = C = O_B, the formal charge on O_A, C and O_B are respectively.
(a) – 1, 0, + 1
(b) + 1, 0, -1
(c) – 2, 0, + 2
(d) 0, 0, 0
Solution:

Formal charge of O_A/O_B = N_V – (N_e + \(\frac { N_{ b } }{ 2 } \)) = 6 – (4 + \(\frac{4}{2}\)) = 6 – 6 = 0
Formal charge of C = 4 – (0 + \(\frac{8}{2}\)) = 4 – 4 = 0
Answer:
(d) 0, 0, 0

Question 11.
In the hydrocarbon CH₃ – CH₃ – CH = CH – CH₂ – C = CH the state of hybridisation of carbon 1, 2, 3, 4 and 7 are in the following sequence.
(a) sp, sp, sp³, sp², sp³
(b) sp², sp, sp³, sp², sp³
(c) sp, sp, sp², sp, sp³
(d) None of these
Answer:
(a) sp, sp, sp³, sp², sp³

Question 12.
Enzyme present in apple is ………………………..
(a) Polyphenol oxidase
(b) Polyphenol reductase
(c) Polyphenol
(d) Polyphenol hydrolase
Answer:
(a) Polyphenol oxidase

Question 13.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is ………………………….
(a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 14.
In Finkelstein reaction, the mechanism followed is ……………………….
(a) S_N1
(b) E₁
(c) E₂
(d) S_N2
Answer:
(d) S_N2

Question 15.
Which sequence for greenhouse gases is based on GWP?
(a) CFC > N₂O > CO₂ > CH₄
(b) CFC > CO₂ > N₂O > CH₄
(c) CFC > N₂O > CH₄ > CO₂
(d) CFC > CH₄ > N₂O > CO₂
Answer:
(c) CFC > N₂O > CH₄ > CO₂

PART – II

Answer any six questions in which question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
Define Avogadro Number?
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 × 10²³.

Question 17.
Explain the meaning of the symbol 4P. Write all the four quantum numbers for these electrons?
Answer:
4f²: It means that the element has 2 electrons in outermost 4f shell.
Quantum number values are,
n = principal quantum number = 4
l = azimuthal quantum number = 3
m = magnetic quantum number = – 3, -2
s = spin quantum number = +\(\frac{1}{2}\), –\(\frac{1}{2}\)

Question 18.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”?
Answer:
No, It is not correct. Thq accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 19.
What are the uses of calcium hydroxide?
Calcium hydroxide is used
Answer:

1. In the preparation of mortar, a building material.
2. In white wash due to its disinfectant nature.
3. In glass making and tanning industry.
4. For the preparation of bleaching powder and for the purification of sugar.

Question 20.
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1. Calculate the melting point of sodium chloride?
Answer:
Given:
∆H_f = 30.4 kJ = 30400 J mol^-1
∆S_f (NaCl) = 28.4 kJ^-1 mol^-1
T_f = ?
∆S_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}}\); T_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{S}_{\mathrm{f}}}\)
T_f = \(\frac{30400 \mathrm{J} \mathrm{mol}^{-1}}{28.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}}\) = 1070.4 K

Question 21.
How is a gas-solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecules in the gaseous state and those dissolved in the liquid.
Example: In carbonate beverages the following equilibrium exists.
CO₂(g) ⇄ CO₂ (solution).

Question 22.
What type of hybridisations are possible in the following geometeries?

1. Octahedral
2. Tetrahedral
3. Square planar

Answer:

1. Octahedral geometry is possible by sp³d² (or) d²sp³ hybridisation.
2. Tetrahedral geometry is possible by sp³ hybridisation.
3. Square planar geometry is possible by dsp² hybridisation.

Question 23.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

1. A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
2. The tube is heated till reaction ceases and become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
3. The contents of the dist is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 24.
Complete the reactions and identify the products?

Answer:

PART – III

Answer any six questions in which question No. 32 is compulsory. [6 × 3 = 18]

Question 25.
The balanced equation for a reaction is given below 2x + 3y → 41 + m
When 8 moles of x react with 15 moles of y, then

1. Which is the limiting reagent?
2. Calculate the amount of products formed.
3. Calculate the amount of excess reactant left at the end of the reaction.

Answer:
2x + 3y → 41 + m
1. 2x reacts with 3y to give products.
8x reacts with 15y means, y is the excess because 8 moles of x should react with 4 × 3y = 12y moles of y to give products.
In this reaction 15y moles are used.
Therefore, 3 moles of y is excess and x is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 × 41 i.e. 161 and 4m as product.
8x+ 12y → 161 + 4m

3. At the end of the reaction, the excess reactant left is 3 moles of y.

Question 26.
Which would you expect to have a higher melting point, magnesium oxide or magnesium fluoride? Explain your reasoning.
Answer:

1. Magnesium oxide has very strong ionic bonds as compared to magnesium fluoride.
2. Mg²⁺ and O^2- have charges of +2 and -2, respectively.
3. Oxygen ion is smaller than fluoride ion.
4. The smaller the ionic radii, the smaller the bond length in MgO and the bond is stronger than MgF₂.
5. Due to more strong bond nature in MgO, it has high melting point than MgF₂.

Question 27.
A sample of solid KClO₃ (potassium chlorate) was heated in a test tube to obtain O₂ according to the reaction
2KClO_3(s) → 2KCl + 3O₂
The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:

Question 28.
List the characteristics of entropy?
Characteristics of entropy:
Answer:

• Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
• In general, the entropy of gaseous system is greater than liquids and greater than solids. The symbol of entropy is S.
• Entropy is defined as “for a reversible change taking place at a constant temperature (T), the change in entropy (∆S) of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).
  \(\Delta S_{\mathrm{sys}}=\frac{q_{\mathrm{rev}}}{T}\)
• If heat is absorbed, then ∆S is positive and there will be increase in entropy. If heat is evolved, ∆S is negative and there is a decrease in entropy,
• The change in entropy of a process represented by ∆S and is given by the equation,
  ∆S_sys = S_f – S_i
• If S_f > S_i, ∆S is positive, the reaction is spontaneous and reversible.
  If S_f < S_i, ∆S is negative, the reaction is non-spontaneous and irreversible.
• Unit of entropy: SI unit of entropy is J K^-1.

Question 29.
Derive the values of K_C and K_P for the synthesis of HI.
Answer:
H₂(g) + I₂(g) ⇄ 2HI(g)
Let us consider the formation of HI in which V moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’.
Let ‘x’ moles of each of H₂ and I₂ react together to form 2x moles of HI.

Applying mass of action

Caluculation of K_P: K_P = K_C.RT^∆ng
Here ∆n_g = n_p – n_r = 2 -2 = 0
Hence, K_P = K_C
K_P = \(\frac{4 x^{2}}{(a-x)(b-x)}\)

Question 30.
Describe the classification of organic compounds based on their structure?
Answer:
Classification of organic compounds based on the structure

Question 31.
For the following bond cleavages use curved-arrows to show the electron flow and classify each as homolytic or heterolytic fission. Identify reactive intermediate produced as free radical, carbocation and carbanion?

Answer:

Question 32.
An alkyl halide with molecular formula C₆H₁₂Br on dehydrohalogenation gave two isomeric alkenes X and,Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH₂CHO and (CH₃)₂ CHCHO. Find the alkyl halide?
Answer:

1. C₆ H₁₃ Br is 3 – Bromo – 4 – methylpentane.
2. 3 – Bromo – 4 – methylpentane on dehydration give two isomers X and Y as follows:

Therefore C₆H₁₃ Br is 3 – Bromo – 4 – methylpentane

Question 33.
What do you mean by ozone hole? What are its consequences?
Answer:
Depletion of ozone layer creates some sort of holes in the blanket of ozone which surrounds us in the atmosphere and this is known as ozone hole.

1. With the depletion of the ozone layer, UV radiations filters into the troposphere which leads to ageing of skin, cataract, sunburn, skin cancer etc.
2. By killing many of the phytoplanktons, it can damage the fish productivity.
3. Evaporation rate increases through the surface and stomata of leaves which can decrease the moisture content of the soil.

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) Balance the following equations by ion electron method?
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
(II) Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?

[OR]

(b)
(I) How many unpaired electrons are present in the ground state of Fe³⁺ (z = 26), Mn²⁺ (z = 25) and argon (z = 18)?
(II) Explain about the significance of de Broglie equation?
Answer:
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
Oxidation half reaction: (loss of electron)

Reduction of halfa reaction: (gain of electron)

Add H₂O to balance oxygen atoms

Add Hcl to balance hydrogen atoms
KMnO₄ + 5e^– + 8 HCl → MnCl₂ + 4H₂O …………………. (4)

To equalize the number of electrons equation (1) × 5 and equation (2) × 2

(II) Molecular mass of H₃BO₃ = (1 × 3) + (11 × 1) + (16 × 3) = 62
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass x mole
= 62 × 0.543
= 33.66 g

[OR]

(b) (I) Fe → Fe³⁺ + 3e^–
Fe (Z = 26)
Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25) Electronic configuration is
1s² 2s² 2p⁵ 3s² 3p⁶ 4s² 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

(II) Significance of de Broglie equation:

1. λ = \(\frac{h}{mv}\) This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
2. For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
3. 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.
4. For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 35 (a).
(I) Mention any two anomalous properties of second period elements?
(II) Arrange Na⁺, Mg²⁺ and Al³⁺ in the increasing order of ionic radii. Give reason?

[OR]

(b)
(I) How do you expect the metallic hydrides to be useful for hydrogen storage?
(II) Write a note about ortho water and para water?
Answer:
(a) (I) Anomalous properties of second period elements:

1. In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
2. In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

(II) Na⁺, Mg²⁺ and Al³⁺ are isoelectronic cations.

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is, \(\mathbf{r}_{\mathrm{Na}^{+}}\), > \(\mathbf{r}_{\mathrm{Mg}^{2+}}\), > \(\mathbf{r}_{\mathrm{Al}^{3+}}\).

[OR]

(b)
(I) In metallic hydrides, hydrogen is adsorbed as H-atoms. Due to the adsorption of H atoms the metal lattice expands and become unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.

(II)
1. Water exists in space in the interstellar clouds, in proto-planetary disks, in the comets and icy satellites on the solar system, and on the Earth.
2. In particular, the ortho-to-para ratio (OPR) of water in space has recently received attention. Like hydrogen, water can also be classified into ortho – H₂O and para – H₂O, in which the directions are antiparallel.

3. At the temperature conditions of the Earth (300 K), the OPR of H₂O is 3.
4. At low temperatures below (< 50 K) the amount of para – H₂O increases. It is known that the OPR of water in interstellar clouds and comets has more para – H₂O (OPR = 2.5) than on Earth.

Question 36 (a).
Derive the values of critical constants from the Van der Waals constants?

[OR]

(b) Derive the values of K_p and K_C for dissociation of PCl₅?
Answer:
(a) Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT for 1 mole.
From this equation, the values of critical constant P_C, V_C and T_C are derived in terms of a and b the Vander Waals constants.

The above equation (4) is an cubic equation of V, which can have three roots. At the critical point, all the three values of V are equal to the critical volume Vc. i.e. V = V_C„
i.e; V = V_C
V – V_C = 0 ………………… (5)
(V – V_C)³ ………………….. (6)
(V³ – 3V_CV²) + 3V_C²V – V_C³ …………………. (7)
As the equation (4) is identical with equation (7), comparing the ‘V’ terms in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

Substituting the values of V_C and P_C in equation (9)

Critical constants a and b can be calculated rising Vander Waals constants as follows:

(b) Consider that V moles of PCl₅is taken in container of volume‘V’
Let x moles of PCl₅ be dissociated into x moles of PCl₃ and x moles of Cl₂.

Applying law of mass action

K_p caluculation: K_p = K_C. \(\mathrm{RT}^{\mathrm{An}}\); ∆n_g = 2 – 1 = 1
We know that PV = nRT
RT = \(\frac{PV}{n}\)
Where ‘n’ is the total number of moles at equilibrium
n = a – x + x + x = a + x

Question 37 (a).
(I) Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement?
(II) Explain how non-ideal solutions shows positive deviation from Raoult’s law?

[OR]

(b)
(I) How will you distinguish between electrophiles and nucleophiles?
(II) Complete the following reactions and identify the products?
(a)
(b)
Answer:
(a) (I) When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.
(II)

1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water
2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alco¬hol and water-water interaction).
3. This results in the increased evaporation of both components from the aqueous solution of ethanol.
4. Consequently, the vapour pressure of the solution is greater than the vapour pressure
   predicted by Raoult’s law.
5. Here, the mixing process is endothermic i.e., DHmixing > 0 and there will be a slight increase in volume (DVmixing > 0)
6. 

(b)
(I)
Electrophiles:

1. They are electron deficient.
2. They are cations.
3. They are lewis acids.
4. Accept an electron pair.
5. Attack on electron rich sites.

Nucleophiles:

1. They are electron rich.
2. They are anions.
3. They are lewis bases.
4. Donate an electron pair.
5. Attack on electron deficient sites.

(II)
(a)
(b)

Question 38 (a).
(I) Is it possible to prepare methane by Kolbe’s electrolytic method?
(II) Explain how 2-butyne reacts with
(a) Lindlar’s catalyst
(b) Sodium in liquid ammonia.

[OR]

(b) (I) Discuss the aromatic nucleophilic substitution reactions of chlorobenzene?
(II) CCl₄ > CHCl₃ > CH₂C₁₂ > CH₃Cl is the decreasing order of boiling point in haloalkanes. Give reason?
Answer:
(a) (I) Kolbe’s electrolytic method is suitable for the preparation of symmetrical alkanes, that is alkanes containing even number of carbon atoms. Methane has only one carbon, hence it cannot be prepared by Kolbe’s electrolytic method.
(II) (a) 2-butyne reacts with Lindlar’s catalyst: 2-butyne can be reduced to cis – 2 butene using CaCO₃ supported in Pd -metal partially deactivated with sulphur.
This reaction is stereo specific giving only the cis – 2 – butene.

(b) 2-butyne reacts with sodium in liquid ammonia:
2-butyne can also be reduced to trans – 2 – butene using sodium in liquid ammonia. This reaction is stereospecific giving only the trans – 2 – butene.

[OR]

(b) (I) Aromatic nucleophilic substitution reactions:
Dow’s process:

1.

2.

3.

(II) The boiling point of chloro, bromo and iodoalkanes increases with increase in the number of halogen atoms. So the correct decreasing order of boiling point of haloalkanes is:
CCl₄ > CHCl₃ > CH₂Cl₂ > CH₃Cl.

Students can download Maths Chapter 8 Statistics and Probability Unit Exercise 8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Unit Exercise 8

Question 1.
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f₁ and f₂.

Answer:
Arithmetic Mean (\(\bar{x}\)) = 62.8
Sum of all the frequencies (Σf_i) = 50
Let the missing frequencies be f₁ and f₂


Substitute of the value of f₂ in (1)
f₁ + 12 = 20
⇒ f₁ = 20 – 12 = 8
The Missing frequency is 8 and 12.

Question 2.
The diameter of circles (in mm) drawn in the design are given below.

Calculate the standard deviation.
Answer:
Assumed mean = 42. 5

Question 3.
The frequency distribution is given below

In table k is a positive integer, has a variance of 160. Determine the value of k.
Answer:
Assumed mean = 3k



The value of k = 7

Question 4.
The standard deviation of some temperature data in degree Celsius (°C) is 5. If the data were converted into degree Fahrenheit (°F) then what is the variance?
Solution:
F° = (C° × 1.8) + 32
\(\begin{array}{l}{\sigma_{c}=5^{\circ} \mathrm{C}} \\ {\sigma_{\mathrm{F}}=\left(1.8 \times 5^{\circ} \mathrm{C}\right) \cdot 9^{\circ} \mathrm{F}}\end{array}\)
Adding value to data doesn’t affect standard deviation.
New variance = \(\sigma_{\mathrm{F}}^{2}=81^{\circ} \mathrm{F}\)

Question 5.
If for a distribution, Σ (x – 5) = 3, Σ (x – 5)² = 43, and total number of observations is 18, find the mean and standard deviation.
Answer:


(i) Arithmetic mean (\(\bar{x}\)) = 5.17
(ii) Standard deviation (σ) = 1.53

Question 6.
Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Answer:
Coefficients of variation of prices in city A.
Arrange in ascending order we get, 16, 19, 20, 22, 23
Assumed mean = 20

Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100\)
= \(\frac{2.45}{20} \times 100\)
= 12. 25%
Coefficient of variation = 12.25%
Coefficients of variation of prices in city B.
Arrange in ascending order we get, 10, 12, 15, 18, 20
Assumed mean = 15


Prices in city A is more stable (since 12.25 < 24.6 %)

Question 7.
If the range and coefficient of range of the data are 20 and 0.2 respectively, then find the largest and smallest values of the data.
Answer:
Range of the data (R) = 20
L – S = 20 ……(1)
Coefficient of range = 0.2
Coefficient of range = \(\frac{L-S}{L+S}\)
0.2 = \(\frac{20}{L+S}\)

substituting the value of L = 60 in (2)
60 + S = 100
S = 100 – 60 = 40
Largest value = 60
Smallest value = 40

Question 8.
If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.
Answer:
Sample space = {(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6), (3, 1), (3, 2), (3, 3),(3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1),(6, 2), (6, 3), (6, 4),(6, 5), (6, 6)}
n(S) = 36
(i) Let A be the event of getting product of face value 6.
A = {(1, 6), (2, 3), (3, 2) (6, 1)}
n(A) = 4
P (A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{4}{36}\)
(ii) Let B be the event of getting difference of face value is 5.
B = {(6, 1)}
n(B) = 1

The probability is \(\frac{1}{9}\)

Question 9.
In a two children family, find the probability that there is at least one girl in a family.
Answer:
Sample space (S) = {(Boy, Boy) (Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(S) = 4
Let A be the event of getting atleast one Girl
A = {(Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}\)
Probability of atleast one girl in a family is \(\frac{3}{4}\)

Question 10.
A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.
Answer:
Let the number of black balls be “x”
Sample space (S) = x + 5
n(S) = x + 5
Let A be the event of drawing a black ball
n (A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{x}{x+5}\)
Let B be the event of getting a white ball
n(B) = 5
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{5}{x+5}\)
By the given condition,
\(\frac{x}{x+5}=2 \times\left(\frac{5}{x+5}\right)\)
⇒ \(\frac{x}{x+5}=\left(\frac{10}{x+5}\right)\)
⇒ 10x + 50 = x² + 5x
⇒ x² + 5x – 10x – 50 = 0
⇒ x² – 5x – 50 = 0
⇒ (x – 10)(x + 5) = 0
⇒ x = 10 or x = -5
Number of balls will not be negative.
Number of black balls = 10

Question 11.
The probability that a student will pass the final examination in both English and Tamil is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Tamil examination?
Answer:
Let A be the event of getting student pass in English
Let B be the event of getting student pass in Tamil


Probability of passing the tamil examination is \(\frac{13}{20}\)

Question 12.
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5.
Answer:
Total number of cards = 52
3 cards are removed
Remaining number of cards = 52 – 3 = 49
n(S) = 49
(i) Let A be the event of getting a diamond card.
n(A) = 13
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{13}{49}\)
(ii) Let B be the event of getting a queen card.
n(B) = (4 – 1) = 3
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{3}{49}\)
(iii) Let C be the event of getting a spade card.
n(C) = (13 – 3) = 10
P(C) = \(\frac{n(C)}{n(S)}=\frac{10}{49}\)
(iv) Let D be the event of getting a 5 of heart card.
n(D) = 1
P(D) = \(\frac{n(\mathrm{D})}{n(\mathrm{S})}=\frac{1}{49}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.5

Multiple Choice Questions.

Question 1.
Which of the following is not a measure of dispersion?
(1) Range
(2) Standard deviation
(3) Arithmetic mean
(4) Variance
Answer:
(3) Arithmetic mean
Hint:
Measures of dispersion are,
(i) Range
(ii) Mean deviation
(iii) Quartile deviation
(iv) Standard deviation
(v) Variance
(vi) coefficient of variation

Question 2.
The range of the data 8, 8, 8, 8, 8.. . 8 is
(1) 0
(2) 1
(3) 8
(4) 3
Solution:
(1) 0
Hint
R = L – S = 8 – 8 = 0

Question 3.
The sum of all deviations of the data from its mean is _______
(1) always positive
(2) always negative
(3) zero
(4) non-zero integer
Answer:
(3) zero

Question 4.
The mean of 100 observations is 40 and their standard deviation is 3. The sum of squares of all deviations is
(1) 40000
(2) 160900
(3) 160000
(4) 30000
Solution:
(2) 160900
Hint:
σ = 3

Question 5.
Variance of first 20 natural numbers is ______
(1) 32.25
(2) 44.25
(3) 33.25
(4) 30
Answer:
(3) 33.25
Hint:
Variance of 20 natural numbers is

Question 6.
The standard deviation of a data Is 3. If each value is multiplied by 5 then the new variance is
(1) 3
(2) 15
(3) 5
(4) 225
Solution:
σ = 3. 1f each is multiplied by 5. The new standard variation is also multiplied by 3.
∴ The new S.D = 5 × 3 = 15
Variance = 15² = 225

Question 7.
If the standard deviation of x, y, z is p then the standard deviation of 3x + 5, 3y + 5, 3z + 5 is ________
(1) 3p + 5
(2) 3p
(3) p + 5
(4) 9p + 15
Answer:
(2) 3p
Hint:
(i) Each value is added by any constant there is no change in standard deviation.
(ii) Each value is multiplied by 3 standard deviations also multiplied by 3.
The standard deviation is 3p.

Question 8.
If the mean and coefficient of variation of a data are 4 and 87.5% then the standard
deviation is
(1) 3.5
(2) 3
(3) 4.5
(4) 2.5
Solution:
(1) 3.5
Hint:
\(\bar{x}\) = 4, coefficient of variation is = 87. 5%

Question 9.
Which of the following is incorrect?
(1) P(A) > 1
(2) 0 ≤ P(A) ≤ 1
(3) P(Φ) = 0
(4) P(A) + P(\(\bar{A}\)) = 1
Answer:
(1) P(A) > 1
Hint:
Probability is always less than one or equal to one.

Question 10.
The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is _________
(1) \(\frac{q}{p+q+r}\)
(2) \(\frac{p}{p+q+r}\)
(3) \(\frac{p+q}{p+q+r}\)
(4) \(\frac{p+r}{p+q+r}\)
Answer:
(1) \(\frac{q}{p+q+r}\)
Hint:
Sample spaces = p + q + r
Let A be the event of getting red
n(A) = p
P(A) = \(\frac{q}{p+q+r}\)

Question 11.
A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is _______
(1) \(\frac{3}{10}\)
(2) \(\frac{7}{10}\)
(3) \(\frac{3}{9}\)
(4) \(\frac{7}{9}\)
Answer:
(2) \(\frac{7}{10}\)
Hint:
Here n(S)= 10 (given digit at imit place. It has two digit)
n(A) = 7
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{7}{10}\)

Question 12.
The probability of getting a job for a person is \(\frac{x}{3}\). If the probability of not getting the job is \(\frac{2}{3}\) then the value of x is.
(1) 2
(2) 1
(3) 3
(4) 1.5
Solution:
(2) 1
Hint:
Probability of getting a job = \(\frac{x}{3}\)
Probability of not getting a job = 1 – \(\frac{x}{3}\)

Question 13.
Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \(\frac{1}{9}\), then the number of tickets bought by Kamalam is _______
(1) 5
(2) 10
(3) 15
(4) 20
Answer:
(3) 15
Hint:
n(S) = 135
P(A) = \(\frac{1}{9}\)

Question 14.
If a letter is chosen at random from the English alphabets {a, b, …, z}, then the probability that the letter chosen precedes x.
(1) \(\frac{12}{13}\)
(2) \(\frac{1}{13}\)
(3) \(\frac{23}{26}\)
(4) \(\frac{3}{26}\)
Solution:
(3) \(\frac{23}{26}\)
Hint:
n(S) = 26
Let A denote the letter chosen precedes x
A= {a, b, c, d, …, x}
n(A) = 23
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{23}{26}\)

Question 15.
A purse contains 10 notes of ₹ 2000, 15 notes of ₹ 500, and 25 notes of ₹ 200. One note is drawn at random. What is the probability that the note is either a ₹ 500 note or ₹ 200 note?
(1) \(\frac{1}{5}\)
(2) \(\frac{3}{10}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{4}{5}\)
Answer:
(4) \(\frac{4}{5}\)
Hint:
Sample space (S) = 10 + 15 + 25 = 50
n(S) = 50
Let A be the event of getting ₹ 500 note
n (A) = 15
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{15}{50}\)
Let B be the event of getting ₹ 200 note
n (B) = 25
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{25}{50}\)
Probability of the note is either a ₹ 500 note or ₹ 200 note
P(A) + P(B) = \(\frac{15}{50}+\frac{25}{50}\) = \(\frac{40}{50}\) = \(\frac{4}{5}\)

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is ……………………
(a) 3
(b) 0.75
(c) 0.075
(d) 0.3
Solution:

The amount of CO₂ released, x = 3.3 g
No. of moles of CO₂ released = 3.3/4.4 = 0.075 mol
Answer:
(c) 0.075

Question 2.
Two electrons occupying the same orbital are distinguished by …………………
(a) Azimuthal quantum number
(b) Spin quantum number
(c) Magnetic quantum number
(d) Orbital quantum number
Solution:
Spin quantum number For the first electron ms = + \(\frac{1}{2}\)
For the second electron ms = – \(\frac{1}{2}\)
Answer:
(b) Spin quantum number

Question 3.
Statement – 1: Ionization enthalpy of N is greater than that of O.
Statement – II: N has exactly half filled electronic configuration which is more stable than electronic configuration of O.
(a) Statement – I is wrong but statement – II is correct
(b) Statement – I is correct but statement – II is wrong.
(c) Statement – I and II are correct and statement – II is the correct explanation of statement – I.
(d) Statement – I and II are correct but statement – II is not the correct explanation of statement – I.
Answer:
(c) Statement – I and II are correct and statement – II is the correct explanation of statement – I.

Question 4.
Water gas is …………………….
(a) H₂O_(g)
(b) CO + H₂O
(c) CO + H₂
(d) CO + N₂
Answer:
(c) CO + H₂

Question 5.
Among the following the least thermally stable is ……………………
(a) K₂CO₃
(b) Na₂CO₃
(c) BaCO₃
(d) Li₂CO₃
Li₂CO₃ is least stable.
Answer:
(d) Li₂CO₃

Question 6.
Match the following.

Answer:

Question 7.
C(diamond) → C(graphite), ∆H = -ve, this indicates that …………………
(a) Graphite is more stable than diamond
(b) Graphite has more energy than diamond
(c) Both are equally stable
(d) Stability cannot be predicted
Answer:
(a) Graphite is more stable than diamond

Question 8.
In the equilibrium, 2A(g) ⇄ 2B(g) + C₂(g)
the equilibrium concentrations of A, B and C₂ at 400 K are 1 × 10^-4M, 2.0 × 10^-3M, 1.5 × 10⁴M respectively. The value of K_C for the equilibrium at 400 K is ……………………..
(a) 0.06
(b) 0.09
(c) 0.62
(d) 3 × 10^-2
Solution:
[A] = 1 × 10^-4M; [B] = 2 × 10^-3M; [C] = 1.5 × 10^-4M
2A(g) ⇄ 2B(g) + C₂(g)

= 6.0 × 10^-2 = 0.06
Answer:
(a) 0.06

Question 9.
Which of the following is a non-aqueous solution?
(a) Salt solution
(b) Sugar solution
(c) Br₂ in CCl₄
(d) Ethanol dissolved in water
Answer:
(c) Br₂ in CCl₄

Question 10.
Which of the following molecule does not exist due to its zero bond order?
(a) \(\mathrm{H}_{2}^{-}\)
(b) \(\mathrm{He}_{2}^{+}\)
(c) He₂
(d) \(\mathrm{H}_{2}^{+}\)
Answer:
(c) He₂

Question 11.
Which of the following is optically active?
(a) 3 – Chloropentane
(b) 2 – Chloropropane
(c) Meso – tartaric acid
(d) Glucose
Answer:
(d) Glucose

Question 12.
Which of the following represent a set of nucleophiles?
(a) BF₃, H₂O, NH^2-
(b) AlCl₃, BF₃, NH₃
(c) CN^–, RCH₂^–, ROH
(d) H⁺, RNH₃⁺, CCl₂
Answer:
(c) CN^–, RCH₂^–, ROH

Question 13.
Propyne on passing through red hot iron tube gives ……………………
(a)
(b)
(c)
(d) one of these
Answer:
(a)

Question 14.
Consider the following statements:

(I) E₂ reaction is a bimolecular elimination reaction of second order
(II) E₂ reaction takes place in two steps.
(III) E₂ reaction generally Jakes place in primary alkyl halides.

Which of the above statements is/are not correct?
(a) (I) only
(b) (II) only
(c) (III) only
(d) (I) & (III)
Answer:
(II) E₂ reaction takes place in two steps.

Question 15.
Photo chemical smog formed in congested metropolitan cities mainly consists of ………………….
(a) Ozone, SO₂ and hydrocarbons
(b) Ozone, PAN and NO₂
(c) PAN, smoke and SO₂
(d) Hydrocarbons, SO₂ and CO₂
Answer:
(b) Ozone, PAN and NO₂

PART – II

Answer any six questions in which question No. 23 is compulsory. [6 × 2 = 12]

Question 16.
Why interstitial hydrides have a lower density than the parent metal?
Answer:

1. d block elements form metallic or interstitial hydrides, on heating with dihydrogen under pressure.
2. Hydrogen atom being small in size occupy some in the metallic lattice producing distortion without any change in its type.
3. The densities of these hydrides are lower than those of metals from which they are formed since the crystal lattice expands due to the inclusion of dihydrogen.

Question 17.
Prove that calcium oxide is a basic oxide?
Answer:
Calcium oxide is a basic oxide. It combines with acidic oxides at high temperature.

Question 18.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means?
Answer:

1. The mathematical relationship between the volume of a gas and the number of moles is V ∝ n
2. \(\frac { V_{ 1 } }{ n_{ 1 } } \) = \(\frac { V_{ 2 } }{ n_{ 2 } } \) Constant, where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂
   and n₂ are the values of volume and number of moles of same gas at a different set of conditions.
3. If the volume of the gas increase then the number of moles of the gas also increases.
4. At a certain temperature and pressure, the volume of a gas is directly proportional to the number of the moles of the gas.

Question 19.
Why pressure has no effect on the synthesis of HI?
Answer:
When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium.
H₂(g) + I₂(g) ⇄ 2HI(g)
Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with ∆n_g = 0.

Question 20.
Draw the lewis structure of PCl₅ and SF₆
Answer:

Question 21.
How are naphthalene and camphor purified?
Answer:

1. Naphthalene, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non volatile impurities.

2. Substances to be purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resulting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 22.
How will you convert ethyl chloride into
(I) Ethane
(II) n – butane
Answer:
(I) Conversion of ethyl chloride into ethane:

(II) Conversion of ethyl chloride into n – butane:
Wurtz reaction:

Question 23.
Chloroform is kept with a little ethyl alcohol in a dark coloured bottle, why?
Answer:
(I) Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carboxyl chloride (phosgene), it is therefore stored in closed dark coloured bottles completely filled so that air is kept out.

(II) With the use of 1 % ethanol we can stabilise chloroform, because ethanol can convert the poisonous COCl₂ gas into non poisonous diethyl carbonate.
COCl₂ + 2C₂H₅OH → CO(OC₂H₅)₂ + 2HCl.

Question 24.
How does classical smog differ from photochemical smog?
Answer:
Classical smog:

1. Classical smog is caused by coal-smoke and fog.
2. It occurs in cold humid climate.
3. The chemical composition is the mixture of SO₂, SO₃ gases and humidity.
4. Chemically it is reducing in nature because of high concentration of SO₂ and so it is also called reducing smog.
5. It is primarily responsible for acid rain.
6. It also causes bronchial irritation.

Photochemical smog:

1. Photochemical smog is cause by photochemical oxidants.
2. It occurs in warm, dry and sunny climate.
3. The chemical composition is the mixture of NO₂ and O₃ gases.
4. Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃ and so it is also called oxidising smog.
5. It causes irritation to eyes, skin and lungs and increase the chances of asthma.
6. It causes corrosion of metals, stones and painted surfaces.

PART – III

Answer any six questions in which question No. 28 is compulsory. [6 × 3 = 18]

Question 25.
An ice cube at 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?
Answer:

1. In an ice cube, each atom is surrounded tetrahedrally by four water molecules through hydrogen bond and its density is low.
2. Liquid water at 0°C has the density as 999.82 kg/cm³. Maximum density is attained by water only at 4°C as 1000 kg/cm³.
3. When the temperature changed from 4°C to 0°C, the density of water decreases rather than increases. This is called anomalous expansion of water.
4. The reason for this phenomenon lies in the structure of ice lattice and hydrogen bonding in water,
5. At 0°C, ice cube sinks in liquid water at 0°C because of the lesser density and greater volume of water.

Question 26.
Write the chemical equations for the reactions involved in Solvay process of preparation of sodium carbonate?
Answer:
Solvay process:
The Solvay process is represented by the below chemical equations:

Question 27.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if
(a) It Is compressed to a smaller volume at constant temperature
(b) The temperature is raised while keeping the volume constant
(c) More gas is introduced into the same volume and at the same temperature
Answer:
(a) If a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) If more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. If the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Question 28.
Calculate \(\Delta \mathrm{H}_{\mathrm{r}}^{0}\) for the reaction
CO₂(g) + H₂(g) → CO(g) + H₂O (g)
given that \(\Delta \mathrm{H}_{\mathrm{f}}^{0}\) for CO₂(g), CO(g) and H₂O(g) are – 393.5, – 111.31 and – 242 kJ mol^-1 respectively.
Answer:
Given:
\(\Delta \mathrm{H}_{\mathrm{f}}^{0}\) CO₂ = -393.5 KJ mol^-1

Question 29.
Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O₂ is paramagnetic?
Answer:
(I) Electronic configuration of O atom is 1s² 2s² 2p⁴

(II) Electronic configuration of O₂ molecule is

(III) Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \) = \(\frac{10-6}{2}\) = 2

(IV) Molecule has two unpaired electrons, hence it is paramagnetic.

Question 30.
Give the principle involved in the estimation of halogen in an organic compound by Carius method?
Answer:
Estimation of halogens: Carius method
(I) A known mass of the organic compound is heated with fuming HNO₃ and AgNO₃.

(II) C, H and S gets oxidised to CO₂, H₂O and SO₂ and halogen combines with AgNO₃ to form a precipitate of silver halide.

(III) The precipate AgX is filtered, washed, dried and weighted.

(IV) From the mass of AgX and the mass of organic compound taken, the percentage of halogen are caluculated.
(V)

Question 31.
What polymerisation? Explain with suitable example?
Answer:
A polymer is a larga molecule formed by the combination of large number of small molecules (monomers). This process is known as polymerisation, a few examples are:

Question 32.
Compare \(\mathbf{S}_{\mathrm{N}^{1}}\) and \(\mathbf{S}_{\mathrm{N}^{2}}\) reaction mechanisms?
Answer:

Question 33.
From where does ozone come in the photochemical smog?
Answer:
(I) Photochemical smog is formed by the combination of smoke, dust and fog with air pollutants in the presence of sunlight.

(II) Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃. So it is also called oxidising smog.

(III) Photochemical smog is formed by following reactions:
N₂ + O₂ 2NO
2NO + O₂ 2NO₂

(O) + O₂ O₃
O₃ + NO NO₂ + (O)

(IV) NO and O₃ are strong oxidising agents and they can react with unbumt hydrocarbons in polluted air to form formaldehyde, acrolein and PAN.

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) An atom of an element contains 35 electrons and 45 neutrons. Deduce

1. The number of protons
2. The electronic configuration for the element
3. All the four quantum numbers for the last electron

(II) How many unpaired electrons are present in the ground state of Fe²⁺ (z = 26), Mn²⁺ (z = 25) and argon (z=18)?

[OR]

(b)
(I) Explain why hydrogen is not placed with the halogen in the periodic table.
(II) Complete the following reactions.
Al₄C₃ + D₂O → ?
CaC₂ + D₂O → ?
Mg₃N₂, + D₂O → ?
Ca₃P₂ + D₂O → ?
Answer:
(a) (I) An element X contains 35 electrons and 45 neutrons

1. The number of protons must be equal to the number of electrons. So the number of protons = 35.
2. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.
3. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. n = 4, l = 1, m = +1, s = –\(\frac{1}{2}\)

(II) Fe → Fe²⁺ + 3e^–
Fe (Z = 26) Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

[OR]

(b) (I)

1. Hydrogen resembles alkali metals as well as halogens.
2. Hydrogen resembles more alkali metals than halogens.
3. Electron affinity of hydrogen is much less than that of halogen atom. Hence the tendency to form hydride ion is low compared to that of halogens.
4. In most of its compounds hydrogen exists in +1 oxidation state. Therefore it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

(II)

Question 35 (a).
(I) Why alkafr metals have high chemical reactivity? How this changes along the group?
(II) Distinguish between alkali metals and alkaline earth metals?

[OR]

(b)
(I) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude?
(II) Explain the graphical representation of Charles’ law?
Answer:
(a) (I) Alkali metals exhibit high chemical reactivity because of their low ionization enthalpy and their larger size.
The reactivity of alkali metals increases from Li to Cs, since the value of ionization energy decreases down the group (Li to Cs). All the alkali metals are highly reactive towards the more electronegative elements such as oxygen and halogens.

(II)
Alkali Metals:

1. Alkali metals are soft.
2. They have a single electron in the valence shell and their electronic configuration is [noble gas] ns¹.
3. They have low melting points.
4. Hydroxides are strongly basic.
5. Carbonates do not decompose.
6. Nitrates give corresponding nitrites and oxygen as products.
7. They show +1 oxidation states.
8. Their carbonates are soluble in water except Li₂CO₃.
9. Except Li, alkali metals do not form complex compounds.

Alkaline earth metals:

1. Alkaline earth metals are hard.
2. They have two electrons in the valence shell and their electronic configuration is [noble gas] ns².
3. They have relatively high melting points.
4. Hydroxides are less basic.
5. Carbonates decompose to form oxide, when heated to high temperatures.
6. Nitrates give corresponding oxides, nitrogen dioxide and oxygen as products.
7. They show +2 oxidation states.
8. Their carbonates are insoluble in water.
9. They can form complex compounds.

[OR]

(b)
(I) The volume of the gas is inversely proportional to pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease.
As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

(II)

1. Variation of volume of the gas sample with temperature at constant pressure.
2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P₁ to P₅.
3. i.e. P₁ < P₂ < P₃ < P₄ < P₅. When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.
5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Question 36 (a).
(I) Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?
(II) Derive the relationship between standard free energy (∆G°) and equilibrium constant (K_eq)

[OR]

(b)
(I) 2.56g of Sulphur is dissolved in 100 g of carbon disulphide. The solution boils at 319. 692 K. What is the molecular formula of Sulphur in solution. The boiling point of CS₂ is 319. 450K. Given that K_b for CS₂ = 2.42 K kg mol^-1
(II) Show that the sum of mole fraction of a solution is equal to one?
Answer:
(a) (I) A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from its constituents means no heat change.

(II)

1. In a reversible process, system is at all times in perfect equilibrium with its surroundings.

2. A reversible chemical reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up.

3. This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible.

4. It is possible only if at equilibrium, the free energy of a systepi is minimum.

5. Lets consider a general equilibrium reaction,
A + B ⇄ C + D
The free energy change of the above reaction in any state (∆G) is related to the standard free energy change of the reaction (∆G°) according to the following equation.
∆G = ∆G° +RTIn Q ………………. (1)
where Q is reaction quotient and is defined as the ratio of concentrajion of the products to the concentration of the reactants under non-equilibrium condition.

6. When equilibrium is attained, there is no further free energy change i.e. ∆G = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes, ∆G° =-RTln K_eq ……………… (2)
This equation is known as Van’t Hoff equation.
∆G° = -2.303 RTlogKeq ………………. (3)
We also know that,
∆G° = ∆H° -T∆S° = – RT In K …………….. (4)

[OR]

W₂ = 2.56 g; W₁ = 100 g
T = 319.692; K_b = 2.42 K kg mol^-1
∆T_b = (319.692 – 319.450) K = 0.242 K
M₂ = image 28
M₂ = 256 g mol^-1
Molecular mass of sulphur in solution = 256 g mol^-1
Atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \(\frac{256}{2}\) = 8
Hence, molecular formula of sulphur is S₈.

(II) Consider a solution containing two components A and B whose mole fractions are x_A and x_B respectively. Let the number of moles of two components A and B are n_A
and n_B respectively.

Question 37 (a).
(I) Explain about the procedure and calculation behind the carius method of estimation of sulphur?
(II) What is the difference between distillation, distillation under reduced pressure and steam distillation?

[OR]

(b)
(I) An organic compound (A) of a molecular formula C₂H₄ which is a simple alkene. A reacts with dil H₂SO₄ to give B. A again reacts with Cl₂ to give C. Identify A, B and C and write the equations.
(II) Why chloro acetic acid is stronger acid than acetic acid?
Answer:
(a) (I) Carius method

• Procedure: A known mass of the organic compound is taken in a clean carius tube and few mL of fuming HNO₃ is added and then the tube is sealed. It is then placed in an iron tube and heated for 5 hours. The tube is allowed to cool and a hole is made to allow gases to escape.
• The carius tube is broken and the content collected in a beaker. Excess of BaCl₂ is added to the beaker. H₂SO₄ formed is converted to BaSO₄ (white ppt.) The precipitate is filtered, washed, dried and weight. From the mass of BaSO₄, percentage of S is calculated.

(II) Calculation:
Mass of organic compound = Wg
233 g of BaSO₄ contains 32 g of sulphur
Percentage of sulphur = (\(\frac{32}{233}\) × \(\frac{x}{w}\) × 100)%

(II) Distillation is used in case of volatile liquid mixed with a non-volatile impurities.
Distillation under reduced pressure:
This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.
Steam distillation is used to purify steam volatile liquids associated with water immiscible impurities.

[OR]

(b) (I)

1. C₂H₄ is CH₂ = CH₂ is a simple alkene. A is ethylene.
2. Ethylene (A) reacts with dil H₂SO₄ to give ethanol (B)

3. Ethylene (A) reacts with Cl₂ to give 1, 2 dichloro ethane (C)

(II) Chloro acetic acid: image 31
Chloro acetic acid has Cl – group and it has high electronegativity and shows -I effect. Therefore Cl – atom to facilitate the dissociation of O – H bond very fastly. Whereas in the case of acetic acid, has CH₃ group and it shows +1 effect, therefore dissociation of O – H bond will be more difficult. Thus chloro acetic acid is stronger acid than acetic acid.

Question 38 (a).
(I) Write a chemical reaction useful to prepare the following:

1. Freon – 12 from carbon tetrachloride.
2. Carbon tetrachloride from carbon disulphide.

(II) What are ambident nucleophiles? Explain with an example.

[OR]

(I) Write about hydrosphere (or) Why Earth is called as Blue planet?
(II) Even though the use of pesticides increases the crop production, they adversely affect the living organisms. Explain the function and the adverse effects of the pesticides.
Answer:
1. Freon-12 from carbon tetrachloride:
Freon-12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride. This reaction is called “Swarts reaction.”

2. Carbon tetrachloride from carbon disulphide:
Carbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl₃ as catalyst to give carbon tetrachloride.

(II) Nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide group is a resonance hybrid of two contributing structures and therefore it can act as a nucleophile in two different ways:

It can attack through carbon to form cyanides and through nitrogen to form isocyanides or carbylamines.

[OR]

(b)

1. Hydrosphere include all types of water sources like oceans, seas, rivers, lakes, streams, underground water, polar ice – caps, clouds etc.
2. It covers about 75% of the earth’s surface. Hence earth is called as Blue planet:

(II) Pesticides are the chemicals that are used to kill or stop the growth of unwanted organims. But these pesticides can affect the health of human beings. Pesticides are classified as

1. Insecticides
2. Fungicides and
3. Herbicides.

1. Insecticides:
Insecticides like DDT, BHC, Aldrin can stay in soil for a long period of time and are absorbed by soil. They contaminate root crops like carrot, radish..

2. Fungicides:
Organomercury compounds dissociate in soil to produce mercury which is highly toxic.

3. Herbicides:
They are used to control unwanted plants and are also known as weed killers. Eg, Sodium chlorate, sodium nitrate. They are toxic to mammals.

Students can Download Tamil Nadu 11th Maths Model Question Paper 5 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 5 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
If n((A × B) ∩ (A × C)) = 8 and n(B ∩ C) = 2 then n(A) = …………………
(a) 6
(b) 4
(c) 8
(d) 16
Answer:
(b) 4

Question 2.
The value of log₃ \(\frac{1}{81}\) is ……………….
(a) -2
(b) -8
(c) -4
(d) -9
Answer:
(c) -4

Question 3.
The value of log₃ 11 log₁₁ 13 log₁₃ 15 log₁₅ 27 log₂₇ 81 is ……………………
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 4.
The value of sin(45° + θ) – cos (45° – θ) is …………………..
(a) 2 cos θ
(b) 1
(c) 0
(d) 2 sin θ
Answer:
(c) 0

Question 5.
If tanα and tan β are the roots of x² + ax + b = 0 then \(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}\) is equal to ……………………..
(a) \(\frac{b}{a}\)
(b) \(\frac{a}{b}\)
(c) –\(\frac{a}{b}\)
(d) –\(\frac{b}{a}\)
Answer:
(c) –\(\frac{a}{b}\)

Question 6.
If a² – aC₂ = a² – aC₄ then the value of a is …………………….
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 7.
If ⁿP_r = 840, ⁿC_r= 35 then n = …………………..
(a) 1
(b) 6
(c) 5
(d) 4
Answer:
(a) 1

Question 8.
If 2x² + 3xy – cy² = 0 represents a pair of perpendicular lines then c = …………………….
(a) -2
(b) \(\frac{1}{2}\)
(c) – \(\frac{1}{2}\)
(d) 2
Answer:
(d) 2

Question 9.
The number of terms in the expansion of [(a + b)²]¹⁸ = …………………..
(a) 19
(b) 18
(c) 36
(d) 37
Answer:
(d) 37

Question 10.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …………………..
(a) (7, 3)
(b) (4, 1)
(c) (1,-1)
(d) (3, 4)
Answer:
(b) (4, 1)

Question 11.
Let A and B be two symmetric matrices of same order. Then which one of the following statement is not true?
(a) A + B is a symmetric matrix
(b) AB is a symmetric matrix
(c) (AB) = (BA)^T
(d) A^TB = AB^T
Answer:
(d) A^TB = AB^T

Question 12.
If [3 -1 2] B = [5, 6] then the order of B is ………………….
(a) 3 × 2
(b) 2 × 3
(c) 3 × 1
(d) 1 × 1
Answer:
(a) 3 × 2

Question 13.
1f \(\underset { x\rightarrow 0 }{ lim } \) \(\frac{sin px}{tan 3x}\) = 4 then the value of p is …………………….
(a) 6
(b) 9
(c) 12
(d) 4
Answer:
(c) 12

Question 14.
For \(\vec { a } \) = \(\vec { i } \) + \(\vec { j } \) – 2\(\vec { k } \), \(\vec { b } \) = \(\vec { i } \) + 2\(\vec { j } \) + \(\vec { k } \) and \(\vec { c } \) = \(\vec { i } \) – 2\(\vec { j } \) + 2\(\vec { k } \) the unit vector parallal to is \(\vec { a } \)
+ \(\vec { b } \) + \(\vec { c } \) is ……………………….
(a) \(\frac{\vec{i}+\vec{j}-\vec{k}}{\sqrt{3}}\)
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)
(c) \(\frac{\vec{i}+\vec{j}+\vec{k}}{3}\)
(d) \(\frac{\vec{i}-\vec{j}+\vec{k}}{\sqrt{6}}\)
Answer:
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)

Question 15.
The differential co-efficient of log₁₀ x with respect to log_x 10 is …………………….
(a) 1
(b) -(log₁₀x)²
(c) (log_x10)²
(d) \(\frac { x^{ 2 } }{ 100 } \)
Answer:
(b) -(log₁₀x)²

Question 16.
\(\frac{d}{dx}\)(e^x+5logx) is …………………..
(a) e^xx⁴(x + 5)
(b) e^xx(x + 5)
(c) e^x + \(\frac{5}{x}\)
(d) e^x – \(\frac{5}{x}\)
Answer:
(a) e^xx⁴(x + 5)

Question 17.
If f(x) = x tan^-1x thenf'(1) = ……………………..
(a) 1 + \(\frac { \pi }{ 4 } \)
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)
(c) \(\frac{1}{2}\) – \(\frac { \pi }{ 4 } \)
(d) 2
Answer:
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)

Question 18.
∫cosec xdx = …………………….
(a) log tan \(\frac{x}{2}\) + c
(b) -log (cosec x + cot x) + c
(c) log (cosecx – cot x) + c
(d) all of them
Answer:
(d) all of them

Question 19.
Ten coins are tossed; The probability of getting atleast 8 heads is …………………..
(a) \(\frac{7}{64}\)
(b) \(\frac{7}{32}\)
(c) \(\frac{7}{128}\)
(d) \(\frac{7}{16}\)
Answer:
(c) \(\frac{7}{128}\)

Question 20.
Two items are chosen from a lot containing twelve items of which four are defective. Then the probability that atleast one of the item is defective is …………………..
(a) \(\frac{19}{33}\)
(b) \(\frac{17}{33}\)
(c) \(\frac{23}{33}\)
(d) \(\frac{13}{34}\)
Answer:
(a) \(\frac{19}{33}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A?
Answer:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

Question 22.
Prove that \(\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}\) = \(\frac{1+sinθ}{cosθ}\)
Answer:

Question 23.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Answer:
No. of non-collincar points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in ¹⁵C₃ ways.
∴ No. of Tnangle formed ¹⁵C₃
= \(\frac{15 \times 14 \times 13}{3 \times 2 \times 1}\) = 455

Question 24.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1?
Answer:
Givc sum of the intercepts = 1
⇒ when x intercept a then y intercept = 1 – a
Equation of the line is \(\frac{x}{a}\) + \(\frac{y}{1-a}\) = 1
The line passes through (8, 3) ⇒ \(\frac{8}{a}\) + \(\frac{3}{1-a}\) = 1
(i.e) 8 (1 – a) + 3a = a(1 – a)
8 – 8a + 3a = a – a²
a² – 6a + 8 =0
(a – 2)(a – 4) = 0 ⇒a = 2 or 4
1. When a = 2 equation of the line is \(\frac{x}{2}\) + \(\frac{y}{-2}\) = 1 (i.e) \(\frac{x}{2}\) – y = 1 ⇒ x – 2y = 2
2.When a = 4 equation of the line is \(\frac{x}{4}\) + \(\frac{y}{1-4}\) = 1 (i.e) \(\frac{x}{4}\) – \(\frac{y}{3}\) = 1 ⇒ 3x – 4y = 12

Question 25.
Find the values of p, q, r, & s if \(\left[\begin{array}{ccc}
p^{2}-1 & 0 & -31-q^{3} \\
7 & r+1 & 9 \\
-2 & 8 & s-1
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & -4 \\
7 & \frac{3}{2} & 9 \\
-2 & 8 & -\pi
\end{array}\right]\)
Answer:
When two matrices (of some order) are equal then their correspondings entries are equal.

Question 26.
Find |\(\vec { a } \) × \(\vec { b } \)| where \(\vec { a } \) = 3\(\vec { i } \) + 4\(\vec { j } \) and \(\vec { b } \) = \(\vec { i } \) + \(\vec { j } \) + \(\vec { k } \)
Answer:

Question 27.
At the given point x₀ discover whether the given function is continous or discontinous citing the reasons for your answer?
Answer:

Question 28.
Evaluate y = xe^-x2
Answer:

Question 29.
Evaluate ∫\(\frac{1}{x logx}\) dx?
Answer:

Question 30.
Evaluate [((256)^-1/2)^-1/4]³
Answer:

PART – III

III. Answer any seven questions. Question No. 40 Is compulsory. [7 × 3 = 21]

Question 31.
Find the largest possible domain for the real valued functionsf defined by f(x) = \(\sqrt { x^{ 2 }-5x+6 } \)

Question 32.
Show that tan 75° + cot 75° = 4?

Question 33.
There are 10 bulbs in a room. Each one of them can be operated independently. Find the number of ways in which the room can be illuminated?

Question 34.
Find the \(\sqrt [ 3 ]{ 126 } \) approximately to two decimal places?

Question 35.
Find the equation of the line passing through the point of intersection 2x + y = 5 and x + 3y + 8 = O and parallel to the line 3x + 4y = 7?

Question 36.
If \(\left|\begin{array}{ccc}
a & b & a \alpha+b \\
b & c & b \alpha+c \\
a \alpha+b & b \alpha+c & 0
\end{array}\right|\) = 0 prove that a, b, c are in G.P. or a is a root of ax² + 2bx + c =0

Question 37.
Evaluate \(\underset { x\rightarrow 3 }{ lim } \)\(\frac { x^{ 2 }-9 }{ x-3 } \) ¡f it exists by finding f(3^–) and f(3⁺)

Question 38.
Find the derivative of tan^-1 (1 + x²) with respect x² + x + 1?

Question 39.
Evaluate: ∫x⁵e^{x2 dx}

Question 40.
Prove that the line segment joining the mid points of the adjacent sides of a quadrilateral from parllelogram?

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) Graph the functions f(x) = x³ and g(x) = \(\sqrt [ 3 ]{ x } \) on the same coordinate plane. Find fog and graph it on the plane as well. Explain your results.

[OR]

(b) 1f x = -2 is one root of x³ – x² – 17x = 22 then find the other roots of the equation?

Question 42.
(a) lf A + B + C= it prove that cosA + cos B + cosC = 1 + 4 sin (\(\frac{A}{2}\)) sin (\(\frac{B}{2}\) sin (\(\frac{C}{2}\))

[OR]

(b) If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) show that A² – 4A – 5I = O

Question 43.
(a) If ⁿ⁺¹C₈ : ^(n-3)P₄ = 57 : 16, find the value of n?

[OR]

(b) If the letters of the word IITJEE arc permuted in all possible ways and the strings thus formed are arranged in the lexicographic order, find the rank of the word IITJEE?

Question 44.
(a) The line \(\frac{x}{a}\) + \(\frac{y}{b}\)= 1 moves in such a way that \(\frac { 1 }{ a^{ 2 } } \) + \(\frac { 1 }{ b^{ 2 } } \) = \(\frac { 1 }{ c^{ 2 } } \) where c is a constant. Find the locus of the foot of the perpendicular from the origin on the given line?

[OR]

(b) Show that the equation 4x² + 4xy + y² – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them?

Question 45 (a).
Prove that \(\left|\begin{array}{lll}
1 & x^{2} & x^{3} \\
1 & y^{2} & y^{3} \\
1 & z^{2} & z^{3}
\end{array}\right|\) = (x – y) (y – z) (z – x) (xy + yz + zx)

(b) Evaluate \(\underset { x\rightarrow \infty }{ lim } \) \(\frac{3}{x-2}\) – \(\frac{2 x+11}{x^{2}+x-6}\)

Question 46.
(a) Evaluate \(\frac{1}{6 x-7-x^{2}}\)

(b) Evaluate ∫e^tan-1x (\(\frac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \)) dx

Question 47 (a).
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?

[OR]

(b) Firm manufactures PVC pipes in three plants viz. X, Y and Z. The daily production volumes from the three firms X, Y and Z are respectively 2000 units, 3000 units and 5000 units. It is known from the past experience that 3% of the output from plant X, 4% from plant Y and 2% from plant Z are defective. A pipe is selected at random from a day’s total production,

1. find the probability that the selected pipe is a defective one.
2. if the selected pipe ¡s a defective, then what is the probability that it was produced by plant Y?

Students can Download Tamil Nadu 11th Physics Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 sec average velocity is ………………..
(a) 0
(b) 2 m/s
(c) 4 m/s
(d) 2π m/s
Hint:
Displacement of the cyclist in half revolution is
d = diameter of the circular track
i.e., d= 80 m
Time taken, t = 40 s
Average velocity, V =  = \(\frac{80}{40}\)
V = 2 m/s
Answer:
(b) 2 m/s

Question 2.
A wheel has angular acceleration of 3.0 rad/s² and an initial angular speed of 2.00 rad/s. In a time of 2 seconds it has rotated through an angle of (in radian) ………………..
(a) 10
(b) 12
(c) 4
(d) 6
Answer:
(a) 10

Question 3.
If the origin of co-ordinate system lies at the centre of mass. The sum of the moments of the masses of the system about the centre of mass is …………………
(a) May be greater than zero
(b) May be less than zero
(c) May be equal to zero
(d) Always zero
Answer:
(d) Always zero

Question 4.
Dimensional formula for co-efficient of viscousity
(a) ML^-2 T^-2
(b) ML^-2 T^-1
(c) ML^-1 T^-1
(d) M^-1 L^-1 T^-1
Answer:
(c) ML^-1 T^-1

Question 5.
Action and reaction …………………
(a) Acts on same object
(b) Acts on two different objects
(c) Have resultant not zero
(d) Acts on the same direction
Answer:
(b) Acts on two different objects

Question 6.
A spring is stretched by applying load to its free end. The strain produced in the spring is …………………
(a) Volumetric
(b) Shear
(c) Longitudinal
(d) Longitudinal and shear
Answer:
(d) Longitudinal and shear

Question 7.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s^-2
(b) 25 rad s^-2
(c) 5 ms^-2
(d) 25 ms^-2

Hint:
τ = F × r
Iα = F × r
MR² × α = 30 × \(\frac{40}{100}\); \(\frac { 3\times 40\times 40\times \alpha }{ 100\times 100 } \) = 12
\(\frac { 3\times 16\times \alpha }{ 100 } \) = 12; α = 25 rad/s²
Answer:
(b) 25 rad s^-2

Question 8.
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (E is total energy) ………………
(a) \(\frac{2}{3}\) E
(b) \(\frac{1}{3}\) E
(c) \(\frac{1}{4}\) E
(d) \(\frac{1}{2}\) E
Hint:
PE = \(\frac{1}{2}\) kx²
⇒ \(P E_{V_{2}}=\frac{1}{2} K\left(\frac{A}{2}\right)^{2}=\frac{1}{4}\left(\frac{1}{2} K A^{2}\right)\)
Answer:
(c) \(\frac{1}{4}\) E

Question 9.
A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5 rad/s and 7.5 m/s² respectively. Amplitude of the oscillation is ………………
(a) 0.36
(b) 0.28
(c) 0.61
(d) 0.53
Hint:
x = A sin ωt
∴ a = \(\frac { d^{ 2 }x }{ dt^{ 2 } } \) = -Aω² sinωt
∴Maximum acceleration |a_max| = Aω²
Now Aω² = 7.5
A = \(\frac { 7.5 }{ \omega ^{ 2 } } \) = \(\frac { 7.5 }{ (3.5)^{ 2 } } \) = 0.61
Answer:
(c) 0.61

Question 10.
If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved, then its fundamental frequency will become ………………….
(a) \(\frac{n}{4}\)
(b) \(\sqrt{2n}\)
(c) n
(d) \(\frac { n }{ \sqrt { 2 } } \)
Hint:

Answer:
(c) n

Question 11.
The theory of refrigerator is based on …………….
(a) Joule-Thomson effect
(b) Newton’s particle theory
(c) Joule’s effect
(d) None of the above
Answer:
(d) None of the above

Question 12.
Work done by 0.1 mole of a gas at 27°C to double its volume at constant pressure is ………………..
(a) 54 cal
(b) 60 cal
(c) 546 cal
(d) 600 cal
Hint:
Workdone (W) = – p.dv = nRT
= 0.1 × (0.2 cal) × (273 + 27) = 0.1 × 2 × 300
W = 60 cal
Answer:
(b) 60 cal

Question 13.
When a lift is moving upwards with acceleration a, then time period of simple pendulum in it will ………………..
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)
(b) 2π\(\sqrt { \frac { g+a }{ l } } \)
(c) \(\frac{1}{2π}\)\(\sqrt { \frac { 1 }{ g+a } } \)
(d) \(\frac{1}{2π}\)\(\sqrt { \frac { g+a }{ l } } \)
Answer:
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)

Question 14.
A disc is rotating with angular speed ω. If a child sits on it, what is conserved?
(a) Linear momentum
(b) Angular momentum
(c) Kinetic energy
(d) Potential energy
Answer:
(b) Angular momentum

Question 15.
The vectors \(\vec { A } \) and \(\vec { B } \) are such that |\(\vec { A } \) + \(\vec { B } \)| = |\(\vec { A } \) – \(\vec { B } \)|. The angle between the two vector is ………………..
(a) 45°
(b) 60°
(c) 75°
(d) 90°
Hint:
The angle between two vector is always 90°.
Answer:
(d) 90°

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
Velocity – time graph of a moving object is shown below. What is the acceleration of the object? Also draw displacement – time graph for the motion of the object?
Answer:
The given graph shows that the velocity of the object is constant. That is, the velocity of the object is not changing, so the acceleration of the object is zero. Since the acceleration of an object is given by


Displacement – time graph for the motion of the object is shown in the figure above.

Question 17.
Can a body subjected to a uniform acceleration always move in a straight line?
Answer:
It will be a straight line in one dimensional motion but not applicable for two dimensional motion because the projectile has a parabolic path but it has a uniform acceleration.

Question 18.
Calculate the viscous force on a ball of radius 1mm moving through a liquid of viscosity 0.2 Nsm^-2 at a speed of 0.07 ms^–
Answer:
Radius of the ball (a) = 1mm = 1 × 10^-3m
Co-effecient of viscosity of liquid (η) = 0.2 Nsm^-2
Speed of the ball (v) = 0.07 ms^-1
According to Stoke’s law
Viscous force F = 6 π η av
= 6 × 3.14 × 1 × 10^-3 × 0.2 × 0.07
= 0.26376 × 10^-3 = 2.64 × 10^-4N

Question 19.
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10ms^-2)
Answer:
Given data: F = 30 N, load (m) 2 kg; height = 10m, g = 10 ms^-2
Gravitational forcc F = mg = 30 N
The distance moved h = 10 m
Work done on the object W = Fh = 30 × 10 = 300 J

Question 20.
Why are shockers used in automobiles like car?
Answer:
In the event of jump or jerk, the lime of action of force increases. Since the product of force aid time is constant in a given situation. therefore the force decreases.

Question 21.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighted 250 N on the surface?
Answer:
As g_d = g (1 – \(\frac{d}{R}\)) ⇒ mg_d = mg(1 – \(\frac{d}{R}\))
Here d = \(\frac{R}{2}\)
∴mg_d = (250) × (1 – \(\frac { R/2 }{ R } \)) = 250 × \(\frac{1}{2}\) = 125N

Question 22.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors \(\vec { A } \) and \(\vec { B } \) are perpendicular to each other than their scalar product \(\vec { A } \).\(\vec { B } \) = O because cos 90° = 0. Then the vectors \(\vec { A } \) and \(\vec { B } \) are said to be mutually orthogonal.

Question 23.
An air bubble of radius r in water is at a depth of h below the water surface at some instant. If P is atmospheric pressure and d and T are the density and surface tension of ater respectively. Calculate the pressure P inside the bubble?
Answer:
Excess ot pressure inside the air bubble in water = \(\frac{2T}{r}\)
∴ Total pressure inside the air bubble
= atmospheric pressure + pressure due to liquid column + Excess pressure due to surface tension
= P + hρg + \(\frac{2T}{r}\)

Question 24.
Define beats?
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second.
n = |f₁ – f₂| per second.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Define centripetal acceleration and give any two examples?
Answer:
The acceleration that is directed towards the centre of the circle along the radius and perpendicular to the velocity of the particle is known as centripetal or radial or normal acceleration.
Example:-

1. In the case of planets revolving round the Sun or the moon revolving round the earth, the centripetal force is provided by the gravitational force of attraction between them.
2. For an electron revolving round the nucleus in a circular path, the electrostatic force of attraction between the electron and the nucleus provides the necessary centripetal force.

Question 26.
Write any six properties of vector product of two vectors?
Answer:
(I) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\vec { A } \) and \(\vec { B } \), even though the vectors \(\vec { A } \) and \(\vec { B } \) may or may not be mutually orthogonal.

(II) The vector product of two vectors is not commutative, i.e., \(\vec { A } \) × \(\vec { B } \) ≠ \(\vec { B } \) × \(\vec { A } \). But, \(\vec { A } \) × \(\vec { B } \) = – \(\vec { B } \) × \(\vec { A } \) .
Here it is worthwhile to note that |\(\vec { A } \) × \(\vec { B } \)| = |\(\vec { B } \) × \(\vec { A } \)| = AB sin θ i.e., in the case of the product vectors \(\vec { A } \) × \(\vec { B } \) and \(\vec { B } \) × \(\vec { A } \), the magnitudes are equal but directions are opposite to each other.

(III) The vector product of two vectors will have maximum magnitude when sin θ = 1, i.e., θ = 90° i.e., when the vectors \(\vec { A } \) and \(\vec { B } \) are orthogonal to each other.
(\(\vec { A } \) × \(\vec { B } \))_max = AB\(\hat { n } \)

(IV) The vector product of two non-zero vectors will be minimum when sin θ = 0, i.e., θ = 0° or θ = 180°
(\(\vec { A } \) × \(\vec { B } \))_min = 0
i.e., the vector product of two non-zero vectors vanishes, if the vectors are either parallel or antiparallel.

(V) The self-cross product, i.e., product of a vector with itself is the null vector
\(\vec { A } \) × \(\vec { A } \) = AA sin 0° \(\hat { n } \) = \(\vec { 0 } \)
In physics the null vector \(\vec { 0 } \) is simply denoted as zero.

(VI) The self-vector products of unit vectors are thus zero.
\(\hat { i } \) × \(\hat { i } \) = \(\hat { j } \) × \(\hat { j } \) = \(\hat { k } \) × \(\hat { k } \) = 0

Question 27.
Show that the pressure of the gas is equal to two third of mean kinetic energy per unit volume?
Answer:
The internal energy of the gas is given by
U = \(\frac{3}{2}\) NkT
The above equation can also be written as
U = \(\frac{3}{2}\) PV
Since PV = NkT
P = \(\frac{2}{3}\) \(\frac{U}{V}\) = \(\frac{2}{3}\) u
From the equation (1), we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density (u = \(\frac{U}{V}\))
Writing pressure in terms of mean kinetic energy density using equation.
P = \(\frac{1}{3}\) nm\(\overline { V^{ 2 } } \) = \(\frac{1}{3}\) ρ\(\overline { V^{ 2 } } \)
where ρ = nm = mass density (Note n is number density)
Multiply and divide R.H.S of equation (2) by 2, we get
P = \(\frac{2}{3}\)(\(\frac{ρ}{2}\) \(\overline { V^{ 2 } } \))
P = \(\frac{2}{3}\) \(\overline { KE } \)
From the equation (3), pressure is equal to \(\frac{2}{3}\) of mean kinetic energy per unit volume.

Question 28.
Derive an expression for gravitational potential energy?
Answer:
The gravitational tbrce is a conservative force and hence we can define a gravitational potential energy associated with this conservative force field.
Two masses m₁ and m₂, are initially separated by a distance r’. Assuming m₁ to be fixed in its position. work must be done on m₂ to move the distance from r’ to r.


To move the mass m₂, through an infinitesimal displacement d\(\vec { r } \) from \(\vec { r } \) to \(\vec { r } \) + d\(\vec { r } \) , work has to be done externally. This infinitesimal work is given by
dW = \(\vec { F } \)_ext . d\(\vec { r } \) ……………….. (1)
The work is done against the gravitational force, therefore,
\(\vec { F } \)_ext = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \) ………………. (2)
Substituting equation (2) in (1), we get
dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).d\(\vec { r } \) ………………… (3)
d\(\vec { r } \) = dr \(\hat { r } \) ⇒ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).(dr \(\hat { r } \))
\(\hat { r } \).\(\hat { r } \) = 1 (Since both are unit vectors)
∴ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \) dr …………….. (4)
Thus the total work done for displacing the particle from r’ to r is

This work done W gives the gravitational potential energy difference of the system of masses and m₁ and m₂ when the separation between them are r and r’ respectively.

Question 29.
A satellite orbiting the Earth in a circular orbit of radius 1600 km above the surface of the Earth. What is the acceleration experienced by satellite due to Earth’s gravitational force?
Answer:
g’ = g(1 – \(\frac { 2h }{ R_{ e } } \))
= g(\(\frac { 1-2\times 1600\times 10^{ 3 } }{ 6400\times 10^{ 3 } } \)) = g(1 – \(\frac{2}{4}\))
g’ = g (1 – \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = g (1- \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = \(\frac{8}{2}\) (or) g’ = \(\frac{9.8}{2}\) = 4.9ms^-2

Question 30.
Explain the v ariation of a g with latitude?
Answer:
When an object is on the surface fo the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
This centrifugal force is given by mωR’.
\(\mathrm{OP}_{z}, \cos \lambda=\frac{\mathrm{PZ}}{\mathrm{OP}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\)
R’ = R cos λ
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
\(a_{\mathrm{PQ}}=\omega^{2} \mathrm{R} \cos \lambda=\omega^{2} \mathrm{R} \cos ^{2} \lambda\)
Since R’ = R cos λ

Therefore, g = g – ω²R cos² λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g’ = g, it is maximum. At the equator, g’ is minimum.

Question 31.
A bullet of mass 50g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g = 10 ms^-2
Answer:
m₁ = 50 g = 0.05 kg; m₂ = 450 g = 0.45 kg
The speed of the bullet is u₁ The second body is at rest (u₂ = 0). Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\)
v = \(\frac{0.05 u_{1}+(0.45 \times 0)}{(0.05+0.45)}\) = \(\frac{0.05}{0.50}\)u₁
The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,
v = \(\sqrt{2gh}\)
v = \(\sqrt{2 \times 10 \times 1.8}\) = \(\sqrt{36}\)
v = 6 ms^-1
Substituting this in the above equation, the value of u₁ is
6 = \(\frac{0.05}{0.50}\)u₁ or u₁ = \(\frac{0.05}{0.50}\) × 6 = 10 × 6
u₁ = 60ms^-1

Question 32.
If the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? If not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 33.
Calculate how many times more intense is 90 dB sound compared to 40 dB sound?
Answer:
Given Data:
L = log \(\frac { I }{ I_{ 0 } } \) = log I – log I₀
We get 90 dB = 9 B = log I₁ – log I₀ ……………….. (1)
40 dB = 4 B = logI₂ – logI₀ ………………….. (2)
Subtract (2) from (1)
50 dB = 5B = log I₁ – logI₂
5 = log₁₀ (\(\frac{I_{1}}{I_{2}}\))
\(\frac{I_{1}}{I_{2}}\) = 10⁵

PART – IV

Answer all the questions. [ 5 × 5 = 25]

Question 34 (a)
Obtain an expression for the time period T of a simple pendulum. The time period T depends on

1. Mass of the bob(m)
2. Length of the pendulum (l)
3. Acceleration due to gravity (g) at the place where the pendulum is suspended, (constant k = 2π)

Answer:
Example:
An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon

1. Mass m of the bob
2. Length l of the pendulum and
3. Acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is k = 2π.

Solution:

Here k is the dimensionless constant. Rewriting the above equation with dimensions.

Comparing the powers of M, L and T on both sides, a = 0, b + c = 0, -2c = 1
Solving for a, b and c ⇒ a = 0, b = 1/2, and c = -1/2
From the above equation
T = \(\mathrm{k} \mathrm{m}^{0} l^{1}=g^{-1}-2\)
T = \(k\left(\frac{1}{g}\right)^{1}\) = \(k \sqrt{1 / g}\)
Experimentally k = 2π, hence
T = \(2 \pi \sqrt{1 / g}\)

[OR]

(b) Obtain an expression for the escape speed in detail?
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v_i the initial total energy of the object is
\(\mathrm{E}_{i}=\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) ………………. (1)
where, M_E is the mass of the Earth and R_E the radius of the Earth.
The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_F = 0
E_i = E_f ……………….. (2)
Substituting (1) in (2) we get,

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e i.e..

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity.

Question 35 (a).
Derive an expression for loss of kinetic energy in perfect inelastic collision?
Answer:
In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KE_i be the total kinetic energy before collision and KE_f be the total kinetic energy after collision.
Total kinetic energy before collision,
\(\mathrm{KE}_{i}=\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\) …………….. (1)
Total kinetic energy after collision
KE_f = \(\frac{1}{2}\) (m₁ + m₂)v² …………….. (2)
Then the loss of kinetic energy is
Loss of KE, ∆Q = KE_f – KE_i = \(\frac{1}{2}\) (m₁ + m₂)v² – \(\frac{1}{2}\) m₁ u₁² – \(\frac{1}{2}\) m₂ u₂² ………………. (3)
Substituting equation v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\) in equation (3), and on simplying (expand v by using the algebra) (a + b)² = a² + b² + 2ab, we get
Loss of KE, ∆Q = \(\frac{1}{2}\) \(\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\) (u₁ – u₂)²

(b) A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. Calculate the initial velocity of the shell?
Answer:
Given Data :
m = 200 gm = 0.2 kg; M = 4 kg.
Energy generated = 1.05 KJ = 1.05 × 10³ J
According to law of Conservation of linear momentum
mv = Mv’
∴v’ = (\(\frac{m}{M}\)) v
Total K.E of the gun and bullet

[OR]

(c) State parallel axis theorem?
Answer:
Parallel axis theorem: Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²

(d) Calculate the moment of inertia of uniform circular disc of mass 500 G radius 10 cm about

1. The diameter of the disc
2. The axis, tangent to the disc and parallel to its diameter
3. The axis through the centre of the disc and perpendicular to its plane

Answer:
1. Given Data: M = 500 g = 0.5 kg. R = 10 cm = 10 × 10^-2 m
Moment of inertia of disc about diameter = Id = \(\frac{1}{4}\) MR²
Id = \(\frac{1}{4}\) × 0.5 × 0.1 kg m² = 0.0125 kg m²

2. Apply a parallel axes theorem, moment of inertia of the disc about a tangent to the disc and parallel to the diameter of the disc
= \(\frac{1}{4}\) MR² + MR² = \(\frac{5}{4}\) MR² = \(\frac{5}{4}\) × 0.5 × 1
= 0.0625 kgm²

3. Moment of inertia of the disc about an axis passing through the centre of disc and perpendicular to the plane of the disc
= \(\frac{1}{2}\) MR² = \(\frac{1}{2}\) × 0.5 × 0.1 = 0.025 kgm²

Question 36 (a).
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it?
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum.

When two particles interact with each other, they exert equal and opposite forces on each other. The particle 1 exerts force \(\vec { F } \)₁₂ on particle 2 and particle 2 exerts an exactly equal and opposite force \(\vec { F } \)₁₂ on particle 1 according to Newton’s third law.
\(\vec { F } \)₂₁ = –\(\vec { F } \)₁₂ …………… (1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as
\(\vec { F } \)₁₂ = \(\frac{d \bar{p}_{1}}{d t}\) and \(\vec { F } \)₂₁ = \(\frac{d \vec{p}_{2}}{d t}\) ……………… (2)

Here \(\vec { P } \)₁ is the momentum of particle 1 which changes due to the force \(\vec { F } \)₁₂ exerted by particle 2. Further \(\vec { P } \)₂ is the momentum of particle 2. This changes due to \(\vec { F } \)₂₁ exerted by particle 1.
Substitute equation (2) in equation (1)

It implies that \(\vec { P } \)₁ + \(\vec { P } \)₂ = (constant vector always)

\(\vec { P } \)₁ + \(\vec { P } \)₂ is the total linear momentum of the two particles ( \(\vec { P } \)_tot = \(\vec { P } \)₁ + \(\vec { P } \)₂). It is also called as total linear momentum of the system. Flere, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system ( \(\vec { P } \)_tot) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec { P } \)₁ and \(\vec { P } \)₂ can vary,
in such a way that \(\vec { P } \)₁ + \(\vec { P } \)₂ is a constant vector.

The forces \(\vec { F } \)₁₂ and \(\vec { F } \)₁₂ are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

To find the recoil velocity of a gun when a bullet is fired from it:
Consider the firing of a gun. Here the system is Gun + bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec { P } \)₁ be the momentum of the bullet and \(\vec { P } \)₂ the momentum of the gun before firing. Since initially both are at rest.

\(\vec { P } \)₁ = 0, \(\vec { P } \)₂ = 0.

Total momentum before firing the gun is zero, \(\vec { P } \)₁ + \(\vec { P } \)₂ = 0

According to the law of conservation of linear momentum, total linear momemtum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec { P } \)₁ + \(\vec { P } \)₂. To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec { P } \)₂ to \(\vec { P } \)₂. Due to the conservation of linear momentum, \(\vec { P } \)₁+ \(\vec { P } \)₂‘= 0. It implies that \(\vec { P } \)₁‘ = –\(\vec { P } \)₂ the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum \(\vec { P } \)₂. It is called ‘recoil momentum’. This is an example of conservation of total linear momentum.

[OR]

(b) Derive an expression for escape speed?
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v;, the initial total energy of the object is
\(E_{i}=\frac{1}{2} M v_{i}^{2}-\frac{G M M_{E}}{R_{E}}\) ………………. (1)
where, M_E is the mass of the Earth and RE the radius of the Earth. The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.

When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_f = 0

According to the law of energy conservation,
E_i – E_f = 0 …………….. (2)

Substituting (1) in (2) we get,
\(\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{F}}}=0\)
\(\frac{1}{2} \mathrm{M} v_{i}^{2}=\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) …………….. (3)

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e, i.e.,

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity.

Question 37 (a).
Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dt}\) ∝ (T – T_s) …………….. (1)

The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,

T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.

Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount of T in time dt, then the amount of heat lost is,
dQ = msdT ………………. (2)
Dividing both sides of equation (2) by dt
\(\frac{dQ}{dt}\) = \(\frac{msdT}{dt}\) ……………….. (3)
From Newton’s of cooling
\(\frac{dQ}{dt}\) ∝ (T – T_s)
\(\frac{dQ}{dt}\) = -a(T – T_s) ……………….. (4)

Where a is some positive constant.
From equation (3) and (4)

Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides, we get
\(\mathrm{T}=\mathrm{T}_{\mathrm{s}}+b_{2} e^{\frac{-a}{m s}}\) ……………. (6)
Here b₂ = _e^b1 = Constant

[OR]

(b) Derive an expression for pressure exerted by the gas on the wall of the container?
Answer:
Expression for pressure exerted by a gas:
Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision. the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

A molecule of mass in moving with a velocity \(\vec { v } \) having components (v_x v_y, v_z) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with sanie speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (-v_x, v_y, v_z)
The x-component of momentum of the molecule bêfore collision = mv_x
The x-component of momentum of the molecule after collision = -mv_x
The change in momentum of the molecule in x direction
= Final momentum – initial momentum = -mv_x – mv_x = -2mv_x
According to law of conservation of linear momentum, the change in momentum of the wall = 2mv_x

The number of molecules hitting the right side wall in a small interval of time ∆t.

The molecules within the distance of v_x ∆t from the right side wall and moving towards the right will hit the wall in the time interval &. The number of molecules that will hit the right side wall in a time interval ∆t is equal to the product of volume (Av_x∆t) and number density of the molecules n). Here A is area of the wall and ii is number of molecules per
unit volume \(\frac{N}{V}\) We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.

Te no.of molecules that hit the right side wall in a time interval ∆t

= \(\frac{n}{2} \mathrm{A} v_{x} \Delta t\) ……………….. (1)
In the same interval of time ∆t, the total momentum transferred by the molecules
\(\Delta \mathrm{P}=\frac{n}{2} \mathrm{A} v_{x} \Delta t \times 2 m v_{x}=\mathrm{A} v_{x}^{2} m n \Delta t\) ……………….. (2)

From Newton’s second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)

F = \(\frac{\Delta p}{\Delta t}=n m \mathrm{A} v_{x}^{2}\) ……………….. (3)
Pressure P = force divided by the area of the wall

P = \(\frac{F}{A}\) = nmv²_x …………………. (4)
Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term v²_x by the average \(\bar{v}_{x}^{2}\) in equation (4)

P = nm\(\bar{v}_{x}^{2}\) ………………… (5)

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\bar{v}_{x}^{2}\) = \(\bar{v}_{y}^{2}\) = \(\bar{v}_{z}^{2}\). The mean square speed is written as

\(\bar{v}^{2}\) = \(\bar{v}_{x}^{2}\) + \(\bar{v}_{y}^{2}\) + \(\bar{v}_{z}^{2}\) = 3\(\bar{v}_{x}^{2}\)
\(\bar{v}_{x}^{2}\) = \(\frac{1}{3}\) \(\bar{v}^{2}\)
Using this in equation (5), we get
P = \(\frac{1}{3}\)nm \(\bar{v}^{2}\) or P = \(\frac{1}{3}\) \(\frac{N}{V}\) m\(\bar{v}^{2}\) as [n = \(\frac{N}{V}\)] ……………… (6)

Question 38 (a).
Explain with graphs the difference between work done by a constant force and by a variable force. Arrive at an expression for power and velocity. Give some examples for the same?
Answer:
Work done by a constant force: When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW = (F cos θ) dr ………………. (1)
The total Work done in producing a displacement from initial position r_i to final position r_f is,

The graphical representation of the work done by a constant force is shown in figure given below. The area under the graph shows the work done by the constant force.

Work done by a variable force:
When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation.

dW = F cos θ dr [F cos θ is the component of the variable force F]
where, F and θ are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation,
W = \(\int_{r_{i}}^{r_{f}} d \mathrm{W}=\int_{r_{i}}^{r_{f}} \mathrm{F} \cos \theta d r\) ………………. (4)

A graphical representation of the work done by a variable force is shown in figure given below. The area under the graph is the work done by the variable force.

Expression for power and velocity

The work done by a force \(\vec { F } \) for a displacement \(\bar { dr } \) is
W = ∫\(\vec { F } \).\(\vec { dr } \) ……………. (1)
Left hand side of the equation (1) can be written as
W = ∫dW = ∫\(\frac{dW}{dt}\) (multiplied and divided by dt) ………………… (2)
Since, velocity is \(\vec { v } \) = \(\frac{d \vec{r}}{d t}\); \(\vec { dr } \) = \(\vec { v } \) dt. Right hand side of the equation (I) can be written as

Substituting equation (2) and equation (3) in equation (1), we get

This relation is true for any arbitrary value of di. This implies that the term within the bracket must be equal to zero, i.e.,
\(\frac{dW}{dt}\) – \(\vec { F } \).\(\vec { v } \) = 0 Or \(\frac{dW}{dt}\) = \(\vec { F } \).\(\vec { v } \)
Hence power P = \(\vec { F } \).\(\vec { v } \)

[OR]

(b) Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dt}\) ∝ (T – T_s)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T_s = Temperature of the surrounding
From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in
time dt, then the amount of heat lost is,
dQ = msdT ………………. (2)
Dividing both sides of equation (2) by dt

\(\frac{dQ}{dt}\) = \(\frac{msdT}{dt}\) ……………….. (3)
From Newton’s law of cooling
\(\frac{dQ}{dt}\) ∝ (T – T_s)
\(\frac{dQ}{dt}\) -a(T – T_s) …………………. (4)
Where a is a positive constant.
From equation (3) and (4)
– a(T – T_s) = ms \(\frac{dT}{dt}\)
\(\frac{d T}{T-T_{s}}\) = -a\(\frac{a}{ms}\) dt ………………. (5)
Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides we get,
\(\mathrm{T}=\mathrm{T}_{s}+b_{2} e^{\frac{-a}{m s}}\) ………………… (6)
Here b₂ = e^b1 = constant

Students can download 6th Social Science Term 3 Civics Chapter 2 Local Bodies: Rural and Urban Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science Civics Solutions Term 3 Chapter 2 Local Bodies: Rural and Urban

Samacheer Kalvi 6th Social Science Local Bodies: Rural and Urban Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
……………… is set up with several village panchayats
(a) Panchayat Union
(b) District Panchayat
(c) Taluk
(d) Revenue village
Answer:
(a) Panchayat Union

Question 2.
_______ is National panchayat Raj day
a) January 24
(b) July 24
(c) November 24
(d) April 24
Answer:
(d) April 24

Question 3.
The oldest urban local body in India is ………………
(a) Delhi
(b) Chennai
(c) Kolkata
(d) Mumbai
Answer:
(b) Chennai

Question 4.
_________ District has the highest number of Panchayat Unions.
(a) Vellore
(b) Thiruvallore
(c) Villupuram
(d) Kanchipuram
Answer:
(c) Villupuram

Question 5.
The head of a corporation is called a ………………
(a) Mayor
(b) Commissioner
(c) Chair Person
(d) President
Answer:
(a) Mayor

II. Fill in the blanks

1. ……………… is the first state in India to introduce town Panchayat.
2. The Panchayat Raj Act was enacted in the year ………………
3. The tenure of the local body representative is ……………… years.
4. ……………… is the first municipality in Tamil Nadu.

Answer:

1. Tamil Nadu
2. 1992
3. 5
4. Walajahphet Municipality

III. Match the following Answer

Answer:
1. – d
2. – c
3. – a
4. – b

IV. Answer the following

Question 1.
Is there any corporation in your district? Name it.
Answer:
Yes, Tirunelveli Corporation.

Question 2.
What is the need for local bodies?
Answer:
In order to fulfill the requirements of the people and to involve them directly in governance, there is a need for an effective system of local bodies.

Question 3.
What are the divisions of a rural local body?
Answer:
The rural bodies are categorized into village panchayats, Panchayat Unions, and District panchayats.

Question 4.
What are the divisions of an Urban local body?
Answer:
The Urban local bodies are Categorized into City Municipal Corporations, Municipalities and Town Panchayats.

Question 5.
Who are the representatives elected in a Village Panchayat?
Answer:
The Elected Representatives in a Village Panchayat

1. Panchayat president
2. Ward members
3. Councillor
4. District Panchayat Ward Councillor

Question 6.
List out a few functions of corporations.
Answer:

1. Drinking-Water Supply
2. Street Light
3. Maintenance of Clean Environment
4. Primary Health facilities
5. Corporation Schools
6. Birth and Death registration etc.

Question 7.
List out a few means of revenue of village Panchayats.
Answer:

1. House tax
2. Professional tax
3. Tax on shops
4. Water charges
5. Specific fees for property tax Funds from Central and State Governments etc.

Question 8.
When are Grama Sabha meetings convened? What are the special on those days?
Answer:

1. The Grama Sabha meetings are Convened on January 26, May 1, August 15, and October 2.
2. These days are celebrated as National festival days every year.

Question 9.
What are the special features of the Panchayat Raj system?
Answer:

1. Special features of Panchayat Raj
2. Grama Sabha
3. Three-tier local body governance
4. Reservations
5. Panchayat Elections
6. Tenure
7. Finance Commission
8. Account and Audit etc.

Question 10.
What is the importance of Grama Sabha?
Answer:

1. Grama Sabha is essential for the effective functioning of Village Panchayat.
2. It enhances public participation in the planning and implementation of schemes for social benefit.

V. HOTS

Question 1.
Local bodies play an important role in the development of villages and cities. How?
Answer:

1. India is a vast nation. It is very difficult for a single government to run the entire Country.
2. Our Constitution has provided for three separate levels of government.
3. Union government
4. State government
5. Local government
6. The Local government takes care of the local administration of cities and villages.
7. The main jobs of these bodies are
8. Keeping an area clean
9. Construction of roads and schools
10. arrangements for water and electricity etc.

VI. Activities

1. Prepare a questionnaire to interview a local body representative.
2. Discuss; If there is a contribution to the improvement of your school by local body representatives.
3. If I were a local body representative, I would
4. Find out the number of local bodies in your district and list them.
   

Question 1.
Prepare …………… representative
1. What is your plan to bring in a good drainage system?
2. When will the bridgework Connecting Maharaja Nagar and Thyagaraja Nagar be completed?
3. How many months it will take to bring street lights in our area?

Question 2.
Discuss: If there representatives:
1. Local body members of our area have Contributed much to the upliftment of our school.
2. They met the VIPS, the Common public, and the old Students to meet out the needs of the school.
3. From the Sponsorship they supplied things like Computer Laboratory equipment and books for the library.

Question 3.
If I were …………… them.
1. I would take steps to eradicate Dengu and all sorts of infectious diseases.
2. I would try to maintain a proper drainage system.
3. I would bring in proper roads and lightings etc.

Samacheer Kalvi 6th Social Science Local Bodies: Rural and Urban Additional Important Questions and Answers

I. Fill in the blanks Answer

1. There are ………………. Corporations in Tamil Nadu.
2. The Chennai Corporation was founded in ……………….
3. ………………. District has the most number of municipalities
4. A ………………. Panchayat is between a Village and a city
5. The ………………. Panchayats are the local bodies of Villages.
6. The ………………. and ………………. Districts have the lowest number of Panchayat Unions.
7. The Constituencies are also called……………….

Answer:

1. Twelve
2. 1688
3. Kanchipuram
4. Town
5. Village
6. Nilgris and Perambalur
7. Wards

II. Choose the Correct answer

Question 1.
advocated Panchayat Raj as the foundation of India’s Political System ……………….
(a) Jawaharlal Nehru
(b) Mahatma Gandhi
(c) Rajendra Prasad
Answer:
(b) Mahatma Gandhi

Question 2.
A Bio element Officer (BDQ) is the administrative head of a __________
(a) Village Panchayat
(b) District Panchayat
(c) Panchayat Union
(d) Town Panchayat
Answer:
(c) Panchayat Union

Question 3.
In the 2011 Local Bodies election, percent Seats were won by women ……………….
(a) 38
(b) 28
(c) 48
Answer:
(a) 38

Question 4.
Discretionary function of a Village Panchayat __________
(a) Cleaning roads
(b) Libraries
(c) Water supply
(d) Street lighting
Answer:
(b) Libraries

Question 5.
The Tamil Nadu State Election Commission is situated in ……………….
(a) Chennai
(b) Coimbatore
(c) Trichy
Answer:
(a) Chennai

III. Answer the following briefly

Question 1.
Write about the officials of Municipal Corporation.
Answer:

1. A City Municipal Corporation has a Commissioner, who is an IAS Officer.
2. Government Officials are deputed as Commissioners for municipalities.
3. The administrative officer of a Municipality is an Executive officer (EO)

Question 2.
Explain briefly about Panchayat Union.
Answer:

1. Many Village Panchayats join to form a Panchayat Union.
2. A Councillor is elected from each Panchayat.
3. The Councillors will elect a Panchayat union Chairperson among themselves.
4. A Vice-Chairperson is also elected.
5. A Block Development Officer (BDO) is the administrative head, of a Panchayat Union.
6. The Services are provided on the Panchayat Union level.

Question 3.
List out the Strength of local bodies in Tamil Nadu.
Answer:
Tamil Nadu:

1. Village Panchayats – 12, 524
2. Panchayat Unions – 385
3. District Panchayats – 31
4. Town Panchayats – 561
5. Municipalities – 125
6. City Municipal Corporation – 12

IV. Answer the following in detail

Question 1.
Discuss the District Panchayat
Answer:

1. A District Panchayat is formed in every district.
2. A district is divided into wards on the basis of a 50,000 population.
3. The ward members are elected by the village panchayats.
4. The members of the District Panchayat elect the District Panchayat Committee Chairperson.
5. They provide essential services and facilities to the rural population.
6. They also provide planning and execution of development programmes for the district.

Question 2.
List out the Works carried out by local bodies during natural disasters and outbreaks of diseases.
Answer:

1. Rescuing the public and settle them in a safer places.
2. Arranging food packets and pure drinking water.
3. Assisting them with medical Aids.
4. Creating awareness about the clean environment to the public.
5. Keeping the medicines in the upto date conditions.
6. Preventing them from getting panic against the diseases and make the situation, calm.

V. Mind map

Students can download 6th Social Science Term 3 Geography Chapter 1 Asia and Europe Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science Geography Solutions Term 3 Chapter 1 Asia and Europe

Samacheer Kalvi 6th Social Science Asia and Europe Text Book Back Questions and Answers

I. Choose the correct Answer

Question 1.
Which is not the Western margin of Asia?
(a) Black Sea
(b) Mediterranean Sea
(c) Red Sea
(d) Arabian Sea
Answer:
(d) Arabian Sea

Question 2.
The Intermontane …………… plateau is found between Elbruz and Zagros.
(a) Tibet
(b) Iran
(c) Deccan
(d) The Yunnan
Answer:
(b) Iran

Question 3.
Equatorial climate:
(i) Uniform throughout the year.
(ii) The average / mean rainfall is 200 mm.
(iii) The average temperature is 10°C.
(iv) Of the statements give above.
(a) i alone is correct
(b) ii and iii are correct
(c) i and iii are correct
(d) i and ii are correct
Answer:
(a) i alone is correct

Question 4.
Match list I correctly with list II and select your answer from the codes given below.

Codes:
(a) 2, 3, 4, 1
(b) 4, 3 , 2, 1
(c) 4, 3, 1, 2
(d) 2, 3, 1, 4
Answer:
(a) 2, 3, 4, 1

Question 5.
India is the leading producer of ______
(a) Zinc
(b) Mica
(c) Manganese
(d) Coal
Answer:
(b) Mica

Question 6.
The natural boundary between Spain and France is ……………
(a) The Alps
(b) The Pyrenees
(c) The Carpathian
(d) The Caucasus
Answer:
(b) The Pyrenees

Question 7.
The western and north-western Europe enjoys mild and humid climate.
Choose the correct option:
(a) These regions are found near the equator
(b) It is influenced by the North Atlantic Drift
(c) It is surrounded by mountains
(d) All of the above
Answer:
(b) It is influenced by the North Atlantic Drift

Question 8.
Which of the following statements is incorrect?
(a) Europe produces electricity from hydel power.
(b) All the rivers of Europe originate in the Alps.
(c) Most of the rivers in Europe are used for inland navigation.
(d) The fivers of Europe are perennial in nature.
Answer:
(b) All the rivers of Europe originate in the Alps.

Question 9.
Choose the incorrect pair.
(a) The Meseta – Spain
(b) The Jura – France
(c) The Pennine – Italy
(d) The Black Forest – Germany
Answer:
(c) The Pennines – Italy

Question 10.
Which country in Europe has a very low density of population?
(a) Iceland
(b) The Netherlands
(c) Poland
(d) Switzerland
Answer:
(a) Iceland

II. Fill in the blanks

1. The Taurus and the Pontine ranges radiate from the …………….
2. The wettest place in the world is …………….
3. Iran is the largest producer of ……………. in the world.
4. Europe connected with south and south east Asia by …………….
5. The national dance of Philippines is …………….
6. The second highest peak in Europe is …………….
7. The type of climate that prevails in the central and eastern parts of Europe is …………….
8. The important fishing ground in North Sea is ……………..
9. The density of population in Europe is ……………..
10. The river ……………. passes through nine countries of Europe.

Answer:

1. Armenian
2. Mawsynram
3. dates
4. the Suez canal
5. Tinikling
6. the Mont Blanc
7. Continental type
8. Dogger Bank
9. 34 persons/Km²
10. Danube

III. Match the following

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

IV. Let us learn

Question 1.
Assertion (A) : Italy has dry summers and rainy winters
Reason (R) : It is located in the Mediterranean region
(a) Both A and R are individually true and R is the correct explanation for A
(b) Both A and R are individually true but R is not the correct explanation for A
(c) A is true, but R is false
(d) A is false, but R is true
Answer:
(d) A is false, but R is true

Question 2.
Places marked as 1, 2, 3 and 4 in the given map are noted for the following plains.
(a) Indo – Gangetic plain
(b) Manchurian plain
(c) Mesopotamian
(d) Great plains of China
Match the plains with the notation on the map and select the correct answer using the codes given below.
Codes:

Answer:
b) 2 1 3 4

Question 3.
In the given outline map of Asia, the shaded areas indicate the cultivation of ……………..

(a) Sugarcane
(b) Dates
(c) Rubber
(d) Jute
Answer:
(b) Dates

V. Answer in Brief

Question 1.
Name the important intermontane plateaus found in Asia.
Answer:
The plateau of Anatolia, The plateau of Iran, and the plateau of Tibet are the important intermontane plateaus found in Asia.

Question 2.
Write a short note on the monsoon climate.
Answer:

1. The south, southeast and eastern parts of Asia are strongly influenced by monsoon winds.
2. Summer is hot and humid while winter is cool and dry.
3. The summer monsoon winds bring heavy rainfall to India, Bangladesh, Indo-China, Philippines and Southern China.

Question 3.
How does physiography play a vital role in determining the population of Asia?
Answer:

1. In Asia, the population is unevenly distributed because of various physical features.
2. China and India alone cover three-fifth of Asia’s population.
3. River plains and industrial regions have a high density of population, whereas low density is found in the interior parts of Asia.

Question 4.
Name the ports found is Asia.
Answer:
Tokyo, Shanghai, Singapore, Hong Kong, Chennai, Mumbai, Karachi and Dubai are the important seaports in Asia.

Question 5.
Asia is called the ‘Land of Contrasts – Justify.
Answer:
The biggest continent Asia is called “the land of contrasts”.
Because:

1. Asia is the biggest continent has different types of land features such as mountain, plateau, plain, valley, bay, island etc.
2. It has different climatic conditions from the equator to polar region.
3. Apart from this many races, languages, religions and cultures are followed by people who live in Asia. Therefore, Asia is called ‘the land of contrasts’.

Question 6.
Name the important mountains found in the Alpine system.
Answer:
The important mountain ranges in the Alpine system are the Sierra Nevada, the Pyrenees, the Alps, the Apennines, the Dinaric Alps, the Caucasus and the Carpathian.

Question 7.
What are the important rivers of Europe?
Answer:
The important rivers of Europe are Volga, Danube, Dnieper, Rhine, Rhone, Po, and Thames.

Question 8.
Name a few countries which enjoy the Mediterranean type of climate.
Answer:
The Mediterranean and sub-mediterranean climate regions in Europe are found in much of southern Europe, mainly in southern Portugal, most of Spain, the southern coast of France, Italy, the Croatian Coast, much of Bosnia, Montenegro, Kosovo, Serbia, Albania, Macedonia, Greece and the Mediterranean Islands.

Question 9.
Give a short note on the population of Europe.
Answer:

1. Europe is the third-most populous continent, after Asia and Africa. The population density in Europe is.34 persons / km2.
2. High population density is often associated with the coalfields of Europe.
3. Other populous areas are sustained by mining, manufacturing, commerce, offering large market, labour forces and productive agriculture.
4. Monaco, Malta, San Marino, and the Netherlands are the most densely populated countries; Iceland and Norway have a very low density of population.

Question 10.
Name the important festivals celebrated in Europe.
Answer:

1. The Europeans celebrate both religious and holiday festivals.
2. Christmas, Easter, Good Friday, the Saint Day, Redentore, Tomatina and Carnival are the important festivals of Europe.

VI. Distinguish

Question 1.
Intermontane plateaus and southern plateaus.
Answer:
Intermontane plateaus:

1. Intermontane plateaus are found in the mountain ranges.
2. Eg, The plateau of Anatolia, The plateau of Iran, and the plateau of Tibet

Southern plateaus:

1. The southern plateaus are relatively lower than the northern plateaus.
2. Eg. Arabian Plateau, Deccan Plateau, Shan Plateau, and the Yunnan Plateau.

Question 2.
Cold desert and hot desert.
Answer:
Cold desert:

1. A barren or desolate area especially sandy region of little rainfall, featuring cold dry winter.
2. Cold deserts are found in the antarctic, Green land, Western China, Turkartan the Gopi crodi desert in Mongolia.

Hot desert:

1. A barren or desolate area often the sandy region of little rainfall.
2. The largest hot desert is the Sahara desert.

Question 3.
Tundra and Taiga.
Answer:

Thunder:

1. The Arctic and northern Scandinavian highland have Tundra type of vegetation.
2. The winters are very long and severe summers are very short and warm.
3. It is the land with few animals like polar bear, reindeer and walrus
4. No trees. Lowest form of vegetation like Mosses and Lichen only available.

Taiga:

1. They are found in the south of the Tundra region in Norway, Sweden, Finland, Germany, Poland and Austria.
2. The winters are long and cold. Summers are short and warm.
3. It is the land of fur bearing animals. Eg. Mink, silver fox, squirrel etc.
4. Pine, fir, spruce and larch are the important tree varieties.

Question 4.
The North-Western highlands and Alpine mountain range
Answer:
The North-Western Highlands:

1. This region includes the mountains and plateaus of Norway, Sweden, Finland, Scotland, and Iceland.
2. This region has a fjord coast. It was created by glaciations.
3. A lot of lakes here serve as reservoirs for producing hydroelectricity.

The Alpine mountain range:

1. The Alpine mountain system consists of a chain of young fold mountains found in the southern part of Europe.
2. The Sierra Nevada, the Pyrenees, the Alps, the Apennines, the Dinaric Alps, the Caucasus and the
3. Carpathian is the important mountain ranges.
4. The Pyrenees is a natural boundary between Spain and France.

VII. Give Reasons

Question 1.
Asia is the leading producer of rice.
Answer:
Because:

1. In Asia, India has the largest area of arable lands.
2. Agriculture is intensively practiced in the riverine plains of Asia
3. China and India are the leading producers of rice in the world.

Question 2.
Asia is the largest and most populous continent in the world.
Answer:
Because:

1. Most of the land of Asia in the northern hemisphere has different physical and cultural features.
2. Lofty mountains, plateaus, plains, islands and peninsulas are the major physiographic features.
3. Many perennial rivers flow through different parts. These river valleys are the cradles of ancient civilizations.
4. River plains and industrial regions have high density of population. Population density 143 persons/Km².
5. Asia covers about 30 percent of the world’s population.

Question 3.
Although Western Europe is located in the high latitudes, it has a moderate climate.
Answer:
Because:

1. The western part has a mild, generally humid climate, influenced by the North Atlantic Drift.
2. North Atlantic Drift is a Warm ocean current which brings warmth to the Western Europe.

VIII. Answer in Paragraph

Question 1.
Give an account of the drainage system in Asia.
Answer:

1. The rivers of Asia originate mostly from the central highlands.
2. The Ob, Yenise and Lena are the major rivers flow towards the north and drain into the Arctic Ocean. They remain frozen during winter.
3. Many perennial rivers like Brahmaputra, Indus, Ganga and Irrawaddy originate from the high mountains.
4. They do not freeze during winter.
5. The Euphrates and Tigris flow in west Asia.
6. The Amur, Huang He, Yangtze and Mekong flow in the south and southeastern parts. The Yangtze is the longest river in Asia.

Question 2.
Describe the mineral sources found in Asia.
Answer:

1. Asia has a variety of mineral deposits like Iron, Coal, Manganese, Bauxite, Zinc, Tungsten, Petroleum, Tin etc. Oil and Natural Gas found in the west Asian countries.
2. Iron Ore: Asia has the largest deposits of iron ore in the world. China and India are the important iron ore deposit countries of Asia.
3. Coal: Coal is a fossil fuel. China and India are the largest producers of coal in Asia.
4. Petroleum: Petroleum is a mineral oil. The largest petroleum reserves are found in South West Asia. The important petroleum-producing countries are Saudi Arabia, Kuwait, Iran, Bahrain, Qatar and UAE.
5. Bauxite is found in India and Indonesia.
6. Mica: India is the largest producer in the world.
7. Tin is found in Myanmar, Thailand, Malaysia, and Indonesia.

Question 3.
What are fjords? How do they protect harbours from bad weather conditions?
Answer:

1. A fjord is a narrow and deep-sea inlet between steep clifts.
2. Fjords reduce the speed of wind irrespective of their direction.
3. The force of sea waves is also controlled.
4. Fjord coast was created by glaciations in the past.
5. Areas with fjords are best suited for natural harbours. Example : Norway.

Question 4.
Describe the climatic divisions of Europe.
Answer:

1. The climate of Europe varies from the subtropical to the polar climate.
2. The Mediterranean climate of the south has warm summer and rainy winter.
3. The western and northwestern parts have a mild, generally humid climate, influenced by the North Atlantic Drift.
4. In central and eastern Europe, the climate is humid continental-type.
5. In the northeast, subarctic and tundra climates are found.
6. The whole of Europe is subject to the moderating influence of prevailing westerly winds from the Atlantic Ocean.
7. North Atlantic Drift is a warm ocean current which brings warmth to Western Europe.
8. The westerly wind further transports warmth across Europe.

X. Activity

Question 1.
Complete the following
I belong to ………… district. My district is famous for the following : …………, ………… and ………… . The boundaries of my districts are in the north, in the east, in the south and in the west. It spreads for an area of km². There are ………… taluks and ………… villages in my district. …………, …………, ………… are the important mountain / plain / plateaus (If all, mention all features). The rivers …………, ………… ………… criss – cross my district. …………, …………, ………… are common trees and wildlife such as, are found here. …………, …………, ………… are important minerals available in my district. Based on this …………, ………… industries are located here. The major crops are …………, …………, ………… . (Coastal districts may write the variety of fish). The total population is …………. We celebrate …………, …………, ………… festivals.
Answer:
Tirunelveli, Nellaiyappar Temple, Courttalam, Halwa, Virudhunagar, Tuticorin, Kanyakumari, Western Ghats, 6823sq.kms, 16 taluks, 559 Villages, Thamirabarani, Chittar, Manimuthar, Palm, Neem, Cocount, Monkeys, Tigers, Elephants, Bluemetal, Lime stone, Thorium, Cement, Ginning, Vessels, Paddy, Cotton, Sugarcane, 33, 22, 644, Pongal, Deepavali, Christmas.

Question 2.
If you get a chance to settle in Europe, which country would you choose? List out the reasons why?

Question 3.
Choose any region is Asia. In the map of Asia, mark its distribution of natural vegetation and wildlife. Paste related pictures.

Samacheer Kalvi 6th Social Science Asia and Europe Additional Important Questions and Answers

I. Choose the Correct Answer

Question 1.
……………. separates Asia from Africa
(a) Suez Canal
(b) Bering Strait
(c) Mediterranean Sea
(d) Palk Striat
Answer:
(a) Suez Canal

Question 2.
Shan plateau is located in ______
(a) Saudi Arabia
(b) Myanmar
(c) India
(d) China

Question 3.
The South Asian rivers
(i) remain frozen during winter
(ii) flow towards the north
(iii) are perennial
Of the statements given above
(a) i alone is correct
(b) iii alone is correct
(c) All the three are correct
(d) All the three are wrong
Answer:
(b) iii alone is correct

Question 4.
Find out the wrong pair.
(a) Coal – China
(b) Iron ore – India
(c) Bauxite – Iran
(d) Tin – Myanmar
Answer:
(c) Bauxite – Iran

Question 5.
River yangtze flows in ……………
(a) Inida
(b) Japan
(c) Myanmar
(d) China
Answer:
(d) China

Question 6.
The light house of the Mediterranean is ______
(a) Mt. Stromboli
(b) Mt. Etana
(c) Mt. Vesuvius
(d) None of the above

Question 7.
Choose the incorrect pair:
(a) Siberian Plain – Ob, Yenisey
(b) Manchurian Plain – Amur
(c) Greet plain of China – Yangtze, Sikiang
(d) Mesopotamian Plain – Irrawaddy
Answer:
(d) Mesopotamian Plain – Irrawaddy

II. Match the following

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

III. True/False

1. There are two knots found in Asia.
2. Ob, Yensie, Lena remain frozen during winter
3. Teak, Sandal wood are coniferous trees
4. The fjord region has a lot of lakes
5. Wheat is the dominant crop throughout Europe.

Answer:

1. True
2. True
3. False
4. True
5. True

IV. Answer in Brief

Question 1.
What are the cradles of ancient civilization? Why?
Answer:

1. The river valleys are the cradles of civilizations.
2. Because the ancient civilizations Indus valley, Mesopotamian and Chinese civilizations were born in Asian river valleys.

Question 2.
Mention the Physical divisions of Asia.
Answer:
The physical divisions of Asia are

1. The Northern lowlands
2. The Central High Mountains
3. The Southern Plateaus
4. The Great Plains and
5. The Island Groups.

Question 3.
What are the rare species found in Asia?
Answer:

1. Orang – Utan
2. Komodo Dragon
3. Giant Panda

Question 4.
Mention the Great Plains of Asia.
Answer:
The great plains of Asia are

1. West Siberian plain (Ob and Yenisey)
2. Manchurian Plain (Amur)
3. Great Plain of China (Yangtze and Sikiang)
4. Indo-Gangetic Plain (Indus and Ganga)
5. Mesopotamian plain (Tigris and Euphrates) and
6. The Irrawaddy plain (Irrawaddy)

Question 5.
Name the countries where fishing is a large industry.
Answer:

1. Norway
2. Iceland
3. Russia
4. Denmark
5. The United Kingdom
6. The Netherlands

V. Give Reasons

Question 1.
Europe is a modern and economically developed continent.
Answer:
Avilability of sources, efficient educated work force, research, contact with other nations and innovations are the factors that transformed like this.

Question 2.
Varied patterns of agricultural activities are in use in Europe.
Answer:

1. Europe is an industrially developed continent. It has great diversity in its topography, climate and soil.
2. These interact to produce varied patterns of agriculture activities.

[The varied patterns of agriculture activities in use: Mediterranean agriculture, Dairy farming, mixed livestock, crop farming, horticulture]

Question 3.
There is no winter in the equatorial region.
Answer:
The areas found in and around the equator of Asia have uniform climate throughout the year. There is no winter. Average temperature 27°c. The mean rainfall 1270 mm.

VI. Answer in Paragraph

Question 1.
Describe Industries in Europe.
Answer:

1. Large scale Industries : Steel and Iron are, Ship building, Motor vehicle, Aircraft construction, Pharmaceutical drugs.
2. Small scale Industries that produce nondurable goods are found throughout Europe.
3. Some countries have a reputation for speciality goods.
   • Bicycles – English, Italian and Dutch
   • Glass – Swedish and Finnish
   • Perfumes and fashion goods – Parisian
   • Precision instruments – Swiss

VII. Mind map

Students can download 6th Social Science Term 3 Civics Chapter 1 Democracy Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science Civics Solutions Term 3 Chapter 1 Democracy

Samacheer Kalvi 6th Social Science Democracy Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
The early man settled near ……………….. and practiced agriculture.
(a) plains
(b) bank of rivers
(c) mountains
(d) hills
Answer:
(b) bank of rivers

Question 2.
The birth place of democracy is ________
(a) China
(b) America
(c) Greece
(d) Rome
Answer:
(c) Greece

Question 3.
……………….. is celebrated as the International Democracy Day.
(a) September 15
(b) October 15
(c) November 15
(d) December 15
Answer:
(a) September 15

Question 4.
Who has the right to work in a direct Democracy?
(a) Men
(b) Women
(c) Representatives
(d) All eligible voters
Answer:
(d) All eligible voters

II. Fill in the blanks 

1. Direct Democracy is practiced in ………………
2. The definition of democracy is defined by ………………
3. People choose their representatives by giving their ………………
4. In our country ……………… democracy is in practice.

Answer:

1. Switzerland
2. Abraham Lincoln
3. Votes
4. Parliamentary

III. Answer the following

Question 1.
What is Democracy?
Answer:

1. The citizens of a country select their representatives through elections.
2. Thus they take part in the direct governance of a country. This is termed democracy.

Question 2.
What are the types of democracy?
Answer:

1. There are various types of democracy in practice around the world.
2. Among those, direct democracy and representative democracy are the most popular forms of government.

Question 3.
Define: Direct Democracy
Answer:

1. In a direct democracy, only the citizens can make laws.
2. All changes have to be approved by the citizens.
3. The politicians only rule over parliamentary procedure. Eg. Switzerland.

Question 4.
Define Representative Democracy.
Answer:

1. In a representative democracy, all the members should be represented by a group of representatives.
2. To select their representative’s elections are held.
3. On behalf of people, these representatives obtain the power to take decisions in a democratic manner.
4. This is termed Representative Democracy.

Question 5.
What are the salient features of our constitution that you have understood?
Answer:

1. The Constitution of India guides the Indians in all aspects and maintains law and order.
2. It ensures freedom, equality and justice to everyone.
3. It defines the political principles, the structure of government, the powers and responsibilities.
4. It fixes the Rights and Duties and Directive Principles of the Citizens.
5. It is the longest written constitution in the world.

IV. HOTS

Question 1.
Compare and contrast direct democracy and representative democracy.
Answer:

V. Activity

Question 1.
Find out your area’s representative’s names and write down

1. MP
2. MLA
3. Local body member

Answer:

1. MP – KRPPrabakara
2. MLA – TPM Mohideenkhan
3. Local body member – A. Radhakrishnan

Question 2.
Discuss about the merits and demerits of democracy.
Answer:
The merits of democracy are :

1. A democratic government is better form of government because it is more accountable form of government.
2. Democracy improves the quality of decision making,
3. Democracy enhances the dignity of citizens.
4. Poor and least educated has the same status as the rich and educated.
5. Democracy allows us to correct own mistake.

Demerits:

1. Leaders keep on changing leading to instability.
2. Democracy in all about political competition and power play and there is no scope for mortality.
3. So many people have to be consulted in a democracy that it lead to delays.
4. Democracy leads to corruption for it is based on electoral competition.
5. Ordinary people do not know what is good for them, they should not decide anything

Samacheer Kalvi 6th Social Science Democracy Additional Important Questions and Answers

I. Fill in the blanks Answer

1. The UNO General Assembly resolved to observe 15th September as the International Day of Democracy in ………………
2. ……………… constitution is the longest written constitution in the world.
3. The Drafting committee of the Constituent Assembly was headed by ………………
4. In India, all the people above ……………… years of age enjoy universal Adult Franchise.
5. The oldest and longest functioning parliament in the world is ………………

Answer:

1. 2007
2. Indian
3. Dr. B.R. Ambedkar
4. 18
5. The Iceland Democracy

II. Choose the Correct answer

Question 1.
The Chief Architect of our constitution is ………………
(a) Dr. Rajendra Prasad
(b) Dr. B.R. Ambedkar
(c) Dr. S. Radhakrishnan
Answer:
(b) Dr. B.R. Ambedkar

Question 2.
USA follows ______
(a) Direct democracy
(b) Representative democracy
(c) Monarchy
(d) Dictatorship
Answer:
(b) Representative democracy

Question 3.
Presidential Democracy is practised in ………………
(a) USA
(b) Canada
(c) (a) and (b)
Answer:
(c) (a) and (b)

Question 4.
Presidential democracy is followed in
(a) USA
(b) India
(c) England
(d) Switzerland
Answer:
(a) USA

Question 5.
The Constitution of India guarantees ……………… fundamental rights to its citizens.
(a) 6
(b) 9
(c) 8
Answer:
(a) 6

III. Match the following

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

IV. Answer the following questions

Question 1.
What is a Government
Answer:
A group of people with the authority to govern a country is called government.

Question 2.
How did Abraham Lincoln define democracy?
Answer:
Abraham Lincoln defined democracy as “Government of the people, by the people, and for the people”

Question 3.
What is meant by democratic decision making?
Answer:

1. In the system of democracy, the power to take decisions does not lie with the head.
2. All the members of the group hold open discussions and take final decisions only when everyone is convinced.
3. This is called democratic way of decision making.

IV. Answer the following in detail

Question 1.
What are the Aims of Democracy?
Answer:

1. To preserve and promote the dignity and fundamental rights of the individual
2. To achieve Social justice and Social development of the Community.
3. To establish the rule of law.
4. To enable the People to choose their government.
5. To work towards the development of the country with the help of People’s Participation.

V. Mind map

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Tamilnadu Samacheer Kalvi 6th Social Science Geography Solutions Term 3 Chapter 3 Understanding Disaster

Samacheer Kalvi 6th Social Science Understanding Disaster Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
……………… was the founder of Gupta dynasty.
(a) Chandragupta I
(b) Sri Gupta
(c) Vishnu Gopa
(d) Vishnugupta
Answer:
(b) Sri Gupta

Question 2.
Prayog prashasti was composed by ………………
(a) Kalidasa
(b) Amarasimha
(c) Harisena
(d) Dhanvantri
Answer:
(c) Harisena

Question 3.
The monolithic iron pillar of Chandragupta is at ………………
(a) Mehrauli
(b) Bhitari
(c) Gadhva
(d) Mathura
Answer:
(a) Mehrauli

Question 4.
……………… was the first Indian to explain the process of surgery.
(a) Charaka
(b) Sushruta
(c) Dhanvantri
(d) Agnivasa
Answer:
(b) Sushruta

Question 5.
……………… was the Gauda ruler of Bengal.
(a) Sasanka
(b) Maitraka
(c) Rajavardhana
(d) Pulikesin II
Answer:
(a) Sasanka

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : Chandragupta I crowned himself as a monarch of a large kingdom after eliminating various small states in Northern India.
Reason (R) : Chandragupta I married Kumaradevi of Lichchavi family.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is not correct.
(d) A is not correct but R is correct.
Answer:
(a) Both A and R are true and R is the correct explanation of A

Question 2.
Statement I : Chandragupta II did not have cordial relationship with the rules of South India.
Statement II : The divine theory of kingship was practised by the Gupta rulers.
(a) Statement I is wrong but statement II is correct.
(b) Statement II is wrong but statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
(a) Statement I is wrong but statement II is correct.

Question 3.
Which of the following is arranged in chronological order?
(a) Srigupta – Chandragupta I – Samudragupta – Vikramaditya
(b) Chandragupta I – Vikramaditya – Srigupta – Samudragupta
(c) Srigupta – Samudragupta – Vikramaditya – Chandragupta I
(d) Vikramaditya – Srigupta – Samudragupta – Chandragupta I
Answer:
(a) Srigupta – Chandragupta I – Samudragupta – Vikramaditya

Question 4.
Consider the following statements and find out which of the following statements (s) is/are correct.
(1) Lending money at high rate of interest was practised.
(2) Pottery and mining were the most flourishing industries,
(a) 1. is correct
(b) 2. is correct
(c) Both 1 and 2 are correct
(d) Both 1 and 2 are wrong
Answer:
(a) 1. is correct

Question 5.
Circle the odd one
(1)
Answer:
Samudragupta

(2)
Answer:
Harshacharita

III. Fill in the blanks Answer

1. …………….., the king of Ceylon, was a contemporary of Samudragupta
2. Buddhist monk from China …………….., visited India during the reign of Chandragupta II.
3. …………….. invasion led to the downfall of Gupta Empire.
4. …………….. was the main revenue to the Government.
5. The official language of the Guptas was ……………..
6. …………….., the Pallava king was defeated by Samudragupta.
7. …………….. was the popular king of Vardhana dynasty.
8. Harsha shifted his capital from …………….. to Kanauj.

Answer:

1. reign of
2. Fahien
3. Huns
4. Land tax
5. Sanskrit
6. Vishnugopa
7. Harsha Vardhana
8. Thaneswar

IV. State whether True of False

1. Dhanvantri was a famous scholar in the field of medicine.
2. The structural temples built during the Gupta period resemble the Indo – Aryan style.
3. Sati was not in practice in the Gupta Empire.
4. Harsha belonged to Hinayana school of thought
5. Harsha was noted for his religious intolerance.

Answer:

1. True
2. False
3. False
4. False
5. False

V. Match the following

A.

Answer:
b) 2,4,1,3,5

B.

Answer:
c) 3, 5,1, 2,4

VI. Answer in one or two sentences

Question 1.
Who was given the title Kaviraja? Why?
Answer:

1. Samudragupta was given the title Kaviraja.
2. Because he was a lover of poetry and music.

Question 2.
What are the two types of disasters? Give examples.
Answer:

1. Disaster can be classified as natural and man-made disaster.
2. Natural disaster: Earthquakes, Volcanoes, Tsunami, Cyclones, Floods, Landslides, Avalanches, Thunder and lightning.
3. Man made disaster: Fire, Destruction of building, Accidents in industries, Accident in transport, Terrorism, Stampede.

Question 3.
Explain the Divine Theory of Kingship.
Answer:

1. The Divine Theory of Kingship was practised by the Gupta rulers.
2. The king is the representative of God on earth. He is answerable only to God and not to anyone else.

Question 4.
Chennai, Cuddalore and Cauvery delta are frequently affected by floods. Give reason.
Answer:

1. In our State, Northeast Monsoon season starts from October. It will continue till December.
2. Every year, during this Northeast Monsoon season, low pressure depressions are formed in the Bay of Bengal.
3. The low pressure depressions are then transformed into cyclones and hit the coastal districts.
4. Heavy rain follows the depressions and cyclones.
5. Hence, Chennai, Cuddalore and Cauvery delta are often affected by floods

Question 5.
Who were the Huns?
Answer:

1. Huns were the nomadic tribes.
2. They were terrorising Rome and Constantinople.
3. The white Huns came to India through Central Asia.
4. They were giving trouble to all Indian frontier states.

Question 6.
Differentiate Landslide – Avalanches.
Answer:

Landslide

1. The movement of a mass of rocks, debris, soil etc., down slope is called landslide.

Avalanches

1. A large amount of ice, snow and rock falling quickly down the side of a mountain is called an Avalanches.

Question 7.
Name the books authored by Harsha.
Answer:

1. Ratnavali
2. Nagananda
3. Priyadharshika

VII. Answer the following briefly

Question 1.
Write a note on Prashasti.
Answer:

1. Prashasti is a Sanskrit word, meaning communication or in praise of.
2. Court poets flattered their kings listing out their achievements.
3. These accounts were later engraved on pillars so that the people could read them.

Question 2.
Give an account of Samudragupta’s military conquests.
Answer:

1. Samudragupta was a great general and he carried on a vigorous campaign all over the country.
2. He defeated the Pallava king Vishnugopa.
3. He conquered nine kingdoms in northern India.
4. He reduced 12 rulers of southern India to the status of feudatories and to pay tribute.
5. He received homage from the rulers of East Bengal, Assam, Nepal, the eastern part of Punjab and various tribes of Rajasthan.

Question 3.
Describe the land classification during the Gupta period.
Answer:
Classification of land during Gupta period.

1. Kshetra – Cultivable land
2. Khila – Wasteland
3. Aprahata – Jungle (or) Forest land
4. Vasti – Habitable land
5. Gapata saraha – Pastoral land

Question 4.
Write about Sresti and Sarthavaha traders.
Answer:
Sresti:
Sresti traders were usually settled at a standard place.

Sarthavaha:
Sarthavaha traders caravan traders who carried their goods to different places.

Question 5.
Highlight the contribution of Guptas to architecture.
Answer:

1. From the earlier tradition of rock-out shrines, the Guptas were the first to contruct temples.
2. These temples, adorned with towers and elaborate carvings, were dedicated ‘ to all Hindu deities.
3. The most notable rock – cut caves are found at Ajanta and Ellora, Bagh and Udaygiri.
4. The structural temples built during this period resemble the Dravidian style.

Question 6.
Name the works of Kalidasa.
Answer:

1. Kalidasa’s famous dramas were Sakunthala, Malavikagnimitra and Vikramaoorvashiyam.
2. Other sigrificant works were Meghaduta, Raghuvamsa, Kumarasambava and Ritusamhara

Question 7.
Estimate Harshvardhana as a poet and a dramatist.
Answer:

1. Harsha himself was a poet and dramatist.
2. Around him gathered a best of poets and artists.
3. His popular works are Ratnavali, Nagananda and Priyadharshika
4. His royal court was adorned by Banabhatta, Mayura, Hardatta and Jayasena.

VIII. HOTS

Question 1.
The gold coins issued by Gupta kings indicate.
Answer:
(a) the availability of gold mines in the kingdom
(b) the ability of the people to work with gold
(c) the prosperity of the kingdom
(d) the extravagant nature of kings.
Answer:
(c) the prosperity of the kingdom

Question 2.
The famous ancient paintings at Ajanta were painted on.
(a) walls of caves
(b) ceilings of temples
(c) rocks
(d) papyrus
Answer:
(a) walls of caves

Question 3.
Gupta period is remembered for.
(a) renaissance in literature and art
(b) expeditions to southern India
(c) invasion of Huns
(d) religious tolerance
Answer:
(a) renaissance in literature and art

Question 4.
What did Indian scientists achieve in astronomy and mathematics during the Gupta period?
Answer:

1. Invention of Zero and the cosequent evolution of the decimal system to the modern world.
2. Aryabhatta explained the true causes of solar and lunar eclipses. He was the, first Indian astronomer to declare that the earth revolves around its own axis.

IX. Student activity (For Students)

1. Stage any one of the dramas of Kalidasa in the classroom.
2. Compare and contrast the society of Guptas with that of Mauryas.

X. Life Skills (For Students)

1. Collect information about the contribution of Aryabhatta, Varahamihira and Brahmagupta to astronomy.
2. Visit a nearby ISRO centre to know more about satellite launching.

XI. Answer Grid

Question 1.
Who was Toromana?
Answer:
Answer:
Toromana was the chief of white Huns.

Question 2.
Name the high ranking officials of Gupta Empire.
Answer:
Dandanayakas and Maha dandanayakas

Question 3.
Name the Gupta kings who performed AsVamedha yagna.
Answer:
Samudragupta and Kumaragupta I

Question 4.
Name the book which explained the causes for the lunar and solar eclipses.
Answer:
Surya Siddhanta

Question 5.
Name the first Gupta king to find a place on coins.
Answer:
Samudragupta

Question 6.
Which was the main source of information to know about the Samudragupta’s reign?
Answer:
Allahabad Pillar

Question 7.
Harsha was the worshipper of in the beginning.
Answer:
Shiva

Question 8.
Universitv reached its fame during Harsha period.
Answer:
The Nalanda

Samacheer Kalvi 6th Social Science Understanding Disaster Additional Important Questions and Answers

I. Choose the Correct Answer

Question 1.
The successor of Sri Gupta …………….
(a) Kumaragupta I
(b) Skandagupta
(c) Vishnugupta
(d) Ghatotkacha
Answer:
(d) Ghatotkacha

Question 2.
‘Nitisara’ emphasises the importance of …………….
(a) Trade
(b) Military
(c) Agriculture
(d) Treasury
Answer:
(d) Treasury

Question 3.
The Huhs chief who crowned himself as king.
(a) Yasodharman
(b) Attila
(c) Mihirakula
(d) Toromana
Answer:
(d) Toromana

Question 4.
The Gupta coins were known as Dinara …………….
(a) Copper
(b) Silver
(c) Bronze
(d) Gold
Answer:

Question 5.
The place Harsha went to participate in the great Kumbhamela held.
(a) Allahabad
(b) Kasi
(c) Ayodhya
(d) Prayag
Answer:
(d) Prayag

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : The last of the great Guptas Narasimha Gupta I was paying tribute to Mihirakula.
Reason (R) : He stopped paying tribute as Mihirakula’s hostility towards Buddhism.
(a) Both A and R are true and R is the correct explantion of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is not correct
(d) A is not correct but R is correct
Answer:
(b) Both A and R are correct but R is no correct explanation of A

Question 2.
Statement I : Criminal law was not more severe than that of the Gupta age.
Statement II : Death punishment was the punishment for violation of the laws and for plotting against the king.
(a) Statement I is wrong but statement II is correct
(b) Statement II is wrong but statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
(d) Both the statements are wrong

III. Fill in the blanks

1. In the assembly at ……………. Harsha distributed his weath.
2. The capital of China ……………. was a great centre of art and learning.
3. ……………. was wife of chandragupta I.
4. The military campaigns of kings were financed through revenue.
5. The peasants were required to pay various taxes and were reduced to the position of ……………..

Answer:

1. Prayag
2. Xi’an
3. Kumaradevi
4. surpluses revenue
5. serfs

IV. Match the following

Answer:
b) 4, 5, 2,1, 3

V. Answer in one or two sentences

Question 1.
Write a note on ‘Lichchhavi’.
Answer:

1. Lichchhavi was an old gana – Sanga and its territory lay between the Ganges and the Nepal Terai.
2. Chandragupta I married Kumaradevi of the famous and powerful lichchhavi family.

Question 2.
What do you know about ‘Kaviraja’?
Answer:

1. In one of the gold coins issued by Samudragupta he is portrayed playing harp (Veena).
2. He was a lover of poetry and music and for this, he earned the title ‘Kaviraja’.

Question 3.
What did the travel accounts of Fahien provide information about the condi¬tions of the people of Magadha?
Answer:

1. According to Fahien the people of Magadha were happy and prosperous.
2. Gaya was desolated. Kapilvasthu had become a jungle, but at Pataliputra people were rich and prosperous.

VII. Answer the following briefly

Question 1.
Name the officials employed by the Gupta rulers.
Answer:

1. High – ranking officials were called dandanayakas and mahadandnayakas.
2. The provinces known as deshas or bhuktis were administered by the governors designated as Uparikas. The districts such as vishyas, were controlled by vishyapatis. At the village level gramika and gramadhyaksha were the functionaries.
3. The military designations.
   Baladhikrita (Commander of infantry)
   Mahabaladhikrita (Commander of cavalry)
   Dutakas (spies)

Question 2.
Mention the importance of Forecasting and Early warning.
Answer:
(i) Weather forecasting, Tsunami early warning system, cyclonic forecasting and warning provide necessary information. This information help in reducing risks during disasters.

(ii) School Disaster Management Committee, Village Disaster Management Committee, State and Central government institutions take mitigation measures, together during disaster.

(iii) Newspaper, Radio, Television and social media bring updated information and give alerts on the vulnerable area, risk preparatory measures and relief measures including medicine.

VIII. Mind map