Category: Class 11

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TN State Board 11th Chemistry 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: 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.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is …………………….
(a) 0%
(b) 4.4%
(c) 16%
(d) 8.4%
MgCO₃ → MgO + CO₂↑
MgCO₃: (1 × 24) + (1 × 12) + (3 × 16) = 84g
CO₂: (1 × 12) + (2 × 16) = 44g
100% pure 84 g MgCO₃ on heating gives 44 g CO₂
Given that 1 g of MgCO₃ on heating gives 0.44 g CO₂
Therefore, 84 g MgCO₃ sample on heating gives 36.96 g CO₂
Percentage of purity of the sample = \(\frac { 100% }{ 44gCO_{ 2 } } \) × 36.96 g CO₂ = 84%
Percentage of impurity = 16%
Answer:
(c) 16%

Question 2.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and Tb (atomic no. 65) are …………………….. [NEET – Phase II]
(a) [Xe] 4f⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²
(c) [Xe] 4f⁷ 6s² [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
(d) [Xe] 4f ⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Solution:
Eu : [Xe] 4f⁷, 5d⁰, 6s²
Gd : [Xe] 4f⁷, 5d¹, 6s²
Tb : [Xe] 4f⁹, 5d⁰, 6s²
Answer:
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²

Question 3.
Match the following:

Answer:

Question 4.
Various successive ionization enthalpies (in kJ mol^-1) of an element are given below.

The element is ……………………….
(a) Phosphorus
(b) Sodium
(c) Aluminium
(d) Silicon
Answer:
(c) Aluminium

Question 5.
Non-stoichiometric hydrides are formed by ……………………..
(a) Palladium, vanadium
(b) Carbon, nickel
(c) Manganese, lithium
(d) Nitrogen, chlorine
Answer:
(a) Palladium, vanadium

Question 6.
Which is the function of sodium – potassium pump?
(a) Maintenance of ion balance
(b) Used in nerve impulse conduction
(c) Transmitting nerve signals
(d) Regulates the blood level
Answer:
(c) Transmitting nerve signals

Question 7.
∆S is expected to be maximum for the reaction ………………………
(a) Ca_(s) + 1/2O_2(g) → CaO_(s)
(b) C_(s) + O_2(g) → CO_2(g)
(c) N_2(g) + O_2(g) → 2NO_(g)
(d) CaCO_3(s) → CaO_(s) + CO_2(g)
Solution:
In CaCO_3(s) → CaO_(s) + CO_2(g) entropy change is positive. In (a) and (b) entropy change is negative; in (c) entropy change is zero.
Answer:
(d) CaCO_3(s) → CaO_(s) + CO_2(g)

Question 8.
Which one of the following is a reversible reaction?
(a) Ripening of a banana
(b) Rusting of iron
(c) Tarnishing of silver
(d) Transport of oxygen by Hemoglobin in our body
Solution:
All the other three reactions are irreversible reactions. But the hemoglobin combines with O₂ in lungs to form oxyhemoglobin. The oxyhemoglobin has a tendency to form hemoglobin by releasing O₂. So it is a reversible reaction.
Answer:
(d) Transport of oxygen by Hemoglobin in our body

Question 9.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………………………….
Solution:
(a) P₁ + x₁ (P₂ – P₁)
(b) P₂ – x₁ (P₂ + P₁)
(c) P₁ – x₂ (P₁ – P₂)
(d) P₁ + x₂ (P₁ – P₂)
P_total = P₁ + P₂
= P₁x₁ + P₂ x₂
= P₁(1 – x₂) + P₂x₂ [∵x₁ + x₂ = 1, x₁ = 1 – x₂]
= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂ = P₁ – x₂ (P₁ – P₂)
Answer:
(c) P₁ – x₂ (P₁ – P₂)

Question 10.
Which of the following molecule contain no n bond?
(a) SO₂

(b) NO₂

(c) CO₂

(d) H₂O

Solution:
Water (H₂O) contains only σ bonds and no π bonds.
Answer:
(d) H₂O

Question 11.
IUPAC name of  is …………………….
(a) Trimethylheptane
(b) 2 -Ethyl -3, 3- dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 – Butyl -2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 12.
Which one of the following is an example for free radical initiators?
(a) Benzoyl peroxide
(b) Benzyl alcohol
(c) Benzyl acetate
(d) Benzaldehyde.
Answer:
(a) Benzoyl peroxide

Question 13.
Some meta-directing substituents in aromatic substitution are given. Which one is most – deactivating?
(a) – COOH
(b) – NO₂
(C) – C ≡ N
(d) – SO₃H
Answer:
(b) – NO₂

Question 14.
Consider the following statements.
(I) S_N² reaction is a bimolecular nucleophilic first order reaction.
(II) S_N² reaction take place in one step.
(III) S_N² reaction involves the formation of a carbocation. Which of the above statements is/are not correct?
(a) (II)
(b) (I) only
(c) (I) & (III)
(d) (III)
Answer:
(c) (I) & (III)

Question 15.
The questions given below consists of an assertion and the reason. Choose the correct option out of the choices given below each question.
Assertion (A): If BOD level of water in a reservoir is more than 5 ppm it is highly polluted.
Reason(R): High biological oxygen demand means high activity of bacteria in water.
(a) Both (A) and R are correct and (R) is the correct explanation of (A)
(b) Both (A) and R are correct and (R) is not the correct explanation of (A)
(c) Both (A) and R are not correct
(d) (A) is correct but( R) is not correct
Answer:
(d) (A) is correct but( R) is not correct

PART – II

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

Question 16.
What is the actual configuration of copper (Z = 29)? Explain about its stability?
Answer:
Copper (Z = 29)
Expected configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
Actual configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
The reason is that fully filled orbitals have been found to have extra stability.

Copper has the electronic configuration [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s² due the symmetrical distribution and exchange energies of d electrons.

Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

Question 17.
What is screening effect?
Answer:
Screening effect:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell elections act as a shield between the nucleus and the valence electrons. This effect is called shielding effect (or) screening effect.

Question 18.
Ice is less dense than water at 0°C. Justify this statement?
Answer:
In ice, each oxygen atom is surrounded tetrahedrally by through hydrogen bonds to four water molecules. That is, the presence of two hydrogen atoms and two lone electron pairs (on oxygen) in each water molecule results in a three-dimensional structure. The arrangements creates an open structure, which in turn accounts for the fact that ice is less dense than water at 0°C.

Question 19.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions.

So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire, or leave it in the direct sunlight, even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

1. The gas pressure increases.
2. More of the liquefied propellant turns into a gas.

Question 20.
One mole of PCl₅ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant?
Answer:
Given that [PCl₅]_initial = \(\frac { 1mol }{ dm^{ 3 } } \)
[Cl₂]_eq = 0.6 mol dm^-3
PCl₅⇄ PCl₃ + Cl₂
[PCl₅]_eq = 0.6 mole dm^-3
[PCl5]_eq = 0.4 mole dm^-3
∴ K_C = \(\frac{0.6×0.6}{0.4}\)
K_C = 0.9

Question 21.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. It is because the fact that this process involves decrease of entropy.

Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.
(Gas + Solvent ⇄ Solution + Heat)

Question 22.
Give the general formula for the following classes of organic compounds?
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol

(b) Aliphatic ketones

(c) Aliphatic amines

Question 23.
Identify which of the following shows +1 and -I effect?

(I) -NO₂
(II) -SO₃H
(III) -I
(IV) -OH
(V) CH₃O^–
(VI) CH_3-

Answer:

Question 24.
Write the products A & B for the following reaction?

Answer:

PART – III

Answer any six questions in which question No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Balance by oxidation number method: Mg + HNO₃ → Mg(N0₃)₂ + NO₂ + H₂O.
Answer:
Step 1:

Step 2:
Mg + 2HNO₃ → Mg(NO₃)₂ + NG₂ + H₂O

Step 3:
To balance the number of oxygen atoms and hydrogen atoms 2HNO₃ is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + H₂O

Step 4:
To balance the number of hydrogen atoms, the H₂O molecule is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + 2H₂O

Question 26.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals?
Answer:

1. n = 4, l = 2
2. n = 5, l = 3
3. n = 7, l = 0

1. n = 4, l = 2
If l = 2, ‘m’values are -2,-1, 0, +1, +2. So, 5 orbitals such as d_xy, d_yz ,d_xz, \(d_{ x^{ 2 }-y^{ 2 } }\) and \(d_{ z^{ 2 } }\)

2. n = 5, l = 3
If l = 3, ‘m’ values are -3, -2, -1, 0, +1, +2, +3
So, 7 orbitals such as \(\mathrm{f}_{\mathrm{z}}^{3}, \mathrm{f}_{\mathrm{xz}}^{2}, \mathrm{f}_{\mathrm{yz}}^{2}, \mathrm{f}_{\mathrm{xyz}}, \mathrm{f}_{\mathrm{z}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{x}\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{z}^{2}\right)}\)

3. n = 7, l = 0
If l = 0, ‘m’ values are 0. Only one value. So, 1 orbital such as 7s orbital.

Question 27.
Prove that ionization energy is a periodic property?
Answer:
Variation in a period:
On moving across a period from left to right, the ionization enthalpy value increases.
This is due to the following reasons:

1. Increase of nuclear charge in a period
2. Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus, thus ionization enthalpy increases. Hence, ionization energy is a periodic property.

Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons:

1. A gradual increase in atomic size
2. Increase of screening effect on the outermost electrons due. to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 28.
Explain how heat absorbed at constant pressure is measured using coffee cup calorimeter with neat diagram?
Answer:

1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2. We know that ∆H = qp (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, ∆H_r
3. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and ∆H_r will Reaction also be negative. Mixture
4. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ∆H_r will also be positive.

Question 29.
Consider the following reactions,
(a) H₂(g) + I₂(g) ⇄ 2 HI(g)
(b) CaCO₂(s) ⇄ CaO(s) + CO₂(g)
(c) S(s) + 3F₂(g) ⇄ SF₆(g)
In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product?
Answer:
(a) H₂(g) + I₂2(g) ⇄ 2HI(g)
In the above equilibrium reaction, volume of gaseous mdlecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

(b)
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product CO₂, the pressure should be decreased or volume should be increased.

(c)
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 30.
You are provided with a solid ‘A’ and three solutions of A dissolved in water one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
(I) Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

(II) Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

(III) Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCl in 1 litre of water at 25°C.
2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCl in 1 litre of water at 25°C.
3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCl in 1 litre of water at 25°C.

Question 31.
Explain about the salient features of molecular orbital theory?
Answer:

• When atoms combine to fonn molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.
• The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.
• The number of molecular orbitals formed is the same as the number of combining atomic orbitals. Half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.
• The bonding molecular orbitals are represented as σ (sigma), π (pi), δ (delta) and the corresponding anti-bonding orbitals are called C*, 7t* and 5*.
• The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Aufbau’s Principle, Pauli’s exclusion principle and Hund’s rule as in the case of filling of electrons in the atomic orbitals.
• Bond order gives the number of covalent bonds between the two combining atoms.

The bond order of a molecule can be calculated using the following equation:
Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \)
N_b = Number of electrons in bonding molecular orbitals.
N_a = Number of electrons in anti-bonding molecular orbitals.
(vii) A bond order of zero value indicates that the molecule does not exist.

Question 32.
Explain the types of addition reactions?
Answer:
Addition reactions are classified into three types they are,

1. Electrophilic addition reaction
2. Nucleophilic addition reaction
3. Free radical addition reaction

1. Electrophilic addition reaction:
An electrophilic addition reaction can be described as an addition reaction in whifth a reactant with multiple bonds as in a double or triple bond undergoes has its n bond broken and two new a bond are formed.

(ethane) Br Bf

2. Nucleophilic addition reaction:
A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron deficient or electrophilic double or triple bond, a n bond, reacts with a nucleophilic which is an electron rich reactant with the disappearance of the double bond and creation of two new single or a bonds.

3. Free radical addition reaction:
It is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non radical or between two radicals.

Question 33.
Complete the following reaction and identify A, B and C?

Answer:

PART – IV

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

Question 34 (a).
(I) How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
(II) How many unpaired electrons are present in the ground state of

(a) Cr³⁺ (Z = 24)
(b) Ne (Z = 10)

[OR]

(b) (I) The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain?
(II) Explain why cation are smaller and anions are larger in radii than their parent atoms?
Answer:
(a) (I) Formula for total number of nodes = n – 1
For 2s orbital: Number of radial nodes = 1.
For 4p orbital: Number of radial nodes = n – l – 1.
= 4 – 1 – 1 = 2
Number of angular nodes = l
∴Number of angular nodes = l.
So, 4p orbital has 2 radial nodes and 1 angular node.

For 5d orbital:
Total number of nodes = n – 1
= 5 – 1 =4 nodes
Number of radial nodes = n – l – 1
= 5 – 2 – 1 = 2 radial nodes.
Number of angular nodes = l = 2
∴5d orbital have 2 radial nodes and 2 angular nodes.

For 4f orbital:
Total number of nodes = n – 1
= 4 – 1 = 3 nodes
Number of radial nodes = n – l – 1
= 4 – 3 – 1 = 0 node.
Number of angular nodes = l
= 3 nodes
∴ 5d orbital have 0 radial node and 3 angular nodes.

(II) (a) Cr (Z = 24) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Cr³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴.
It contains 4 unpaired electrons.
(b) Ne(Z=10) 1s² 2s² 2p⁶. No unpaired electrons in it.

[OR]

(b) (I)

• Electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
• Half filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
• For e.g. Beryllium (Z = 4) 1s² 2s² (completely filled electronic configuration)
  Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ (half filled electronic configuration)
  Both beryllium and nitrogen have high ionization energy due to more stable nature.
• In the case of beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.
• Noble gases have stable ns² np⁶ configuration and the addition of further electron is unfavorable and they have zero electron affinity.

(II) A cation is smaller than the parent atom because it has fewer electrons while its nuclear charge remains the same. The size of anion will be larger than that of parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 35 (a).
(I) Arrange NH₃, H₂O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement?
(II) Can we use concentrated sulphuric acid and pure zinc in the preparation of dihydrogen?

[OR]

(b) (I) Beryllium halides are covalent whereas magnesium halides are ionic. Why?
(II) What happens when
Sodium metal is dropped in water?

1. Sodium metal is heated in free supply of air?
2. Sodium peroxide dissolves in water?

Answer:
(a) 1. Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is
HF > H₂O > NH₃

2. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding.

3. Among N, F and O the increasing order of their electronegativities are
N < O < F 4. Hence the expected order of the extent of hydrogen bonding is HF > H₂O > NH₃

(II) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂.
Zn + 2H₂SO₄ (Conc) → ZnSO₄ + 2H₂O + SO₂ Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

[OR]

(b)
(I) Beryllium ion (Be²⁺) is smaller in size and it is involved in equal sharing of electrons with halogens to form covalent bond, whereas magnesium ion (Mg²⁺) is bigger and it is involved in transfer of electrons to form ionic bond.

(II)

1. 2Na + 2H₂O → 2NaOH + H₂
2. 2Na + O₂ → Na₂O₂
3. Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

Question 36 (a).
(I) Define Gibb’s free energy?
(II) You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below?
Answer:

[OR]

(b) (I) Define mole fraction.
(II) Differentiate between ideal solution and non-ideal solution.
Answer:
(a) (I) Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work.
G = H – TS, where G = Gibb’s free energy
H – enthalpy;
T = temperature;
S = entropy

(II) For ethanol:
Given: T_b = 78.4°C = (78.4 + 273) = 351.4 K

[OR]

(b) (I) Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all components present in the solution.

(II)
Ideal Solution:

1. An ideal solution is a solution in which each component obeys the Raoult’s law over entire range of concentration.
2. For an ideal solution,
   ΔH_missing = 0, ΔV_mixing = 0
3. Example: Benzene and toulene n – Hexane and n – Heptane.

Non – Ideal Solution:

1. The solution which do not obey Raults’s law over entire range of concentrations are called non-ideal solution.
2. For an ideal solution,
   ΔH_mixing ≠ 0, ΔV_mixing ≠ 0
3. Example: Ethyl alcohol and Cyclo hexane, Benzene and aceton.

Question 37 (a).
(I) What is dipole moment?
(II) Describe Fajan’s rule?

[OR]

(b) (I) How does Huckel rule help to decide the aromatic character of a compound?
(II) Draw cis-trans isomers for the following compounds
(a) 2- chloro-2-butene
(b) CH₃-CCl=CH-CH₂CH₃
Answer:
(a)
(I)

1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: m = q × 2d, where m is the dipole moment, q is the charge, 2d is the distance between the two charges.
2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.
3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).
4. 1 Debye = 3.336 × 10^-3o C m

(II)

1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of ah anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the cation greater will be the attraction on the electron cloud of the anion. Similarly higher the magnitude of negative charge on anion, greater is its polarisability. For example, Na⁺ < Mg²⁺ < Al³⁺, the covalent character also follows the order:
NaCl < MgCl₂ < AlCl₃.

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation, e.g., LiCl is more covalent than NaCl.

4. Cation having ns² np⁶ nd⁰ configuration shows greater polarising power than the cations with ns² np⁶ configuration, e.g., CuCl is more covalent than NaCl.

[OR]

(b) (I) A compound is said to be aromatic, if it obeys the following rules:

1. The molecule must be cyclic.
2. The molecule must be co-planar.
3. Complete delocalisation of rc-electrons in the ring.
4. Presence of (4n + 2)π electrons in the ring where n is an integer (n = 0, 1, 2 …)

This is known as Huckel’s rule.
Example –  – Benzene

1. It is cyclic one.
2. It is a co-planar molecule.
3. It has six delocalised n electrons.
4. 4n + 2 = 6

4n = 6 – 2
4n = 4
⇒ n = 1
It obey Huckel’s rule, with n = 1, hence benzene is aromatic in nature.

(II) (a) 2-Chloro-2-butene:

(b)

Question 38 (a).
(I) Reagents and the conditions used in the reactions are given below. Complete the table by writing down the product and the name of the reaction?
Answer:

(II) What is the IUPAC name of the insecticide DDT? Why is their use banned in most of the countries?

[OR]

(b) (I) Explain about green chemistry in day-to-day life?
(II) How acetaldehyde is commercially prepared by green chemistry?
Answer:
(a)
(I)

(II)

•  The IUPAC name of the insecticide DDT is p, p’-dichloro-diphenyl trichloroethane.
• Even DDT is an effective insecticide. Now-adays it is banned because of its long term toxic effects.
• DDT is very persistent in the environment and it has a high affinity for fatty tissues. As a result, DDT gets accumulated in animal tissue fat, in particular that of birds of prey with subsequent thinning of their eggs shells and impacting their rate of reproduction. That is why DDT is banned in most of the countries.

[OR]

(b)
(I)
1. Dry cleaning of clothes:
Solvents like tetrachloroethylene used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloro ethylene, liquefied CO₂ with suitable detergent is an alternate solvent used. Liquefied CO₂ is not harmful to the ground water. Nowadays H₂O₂ is used for bleaching clothes in laundry, gives better result and utilises less water.

2. Bleaching of paper:
Conventional method of bleaching was done with chlorine. Nowadays H₂O₂ can be used for bleaching paper in the presence of catalyst.

1. Instead of petrol, methaftol is used as a fuel in automobiles.
2. Neem based pesticides have been synthesised, which are more safer than the chlorinated hydrocarbons.

(II) Acetaldehyde is commericially prepared by one step oxidation of ethene in the presence of ionic catalyst in aqueous medium with 90% yield.

Students can Download Tamil Nadu 11th Maths Model Question Paper 1 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 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 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 – 1

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

Question 1.
For non empty sets A and B if A ⊂ B then (A × B) n (B × A) is equal to ………………….
(a) A ∩ B
(b) A × A
(c) B × B
(d) none of these
Answer:
(b) A × A

Question 2.
The solution set of the inequality |x – 1| ≥ |x – 3| is ………………..
(a) [0, 2]
(b) [2, ∞)
(c) (0, 2)
(d) (-∞, 2)
Answer:
(b) [2, ∞)

Question 3.
The numer of solutions of x² + |x – 1| = 1 is …………………
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

Question 4.
Which of the following is not true?
(a) sin θ = – \(\frac{3}{4}\)
(b) cos θ = -1
(c) tan θ = 25
(d) sec θ = \(\frac{1}{4}\)
Answer:
(d) sec θ = \(\frac{1}{4}\)

Question 5.
Let f_k(x) = \(\frac{1}{k}\) [sin^k x + cos² x] where x ∈ R and k ≥ 1. Then f₄(x) – f₆(x) = …………………
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{12}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{3}\)
Answer:
(b) \(\frac{1}{12}\)

Question 6.
If A and B are coefficients of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)^2n-1 respectively then \(\frac{A}{B}\) =
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{n}\)
(c) 1
(d) 2
Answer:
(d) 2

Question 7.
The value of 15C₈ + 15C₉ – 15C₆ – 15C₇ is …………………
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(a) 0

Question 8.
The slope of the line which makes an angle 45° with the line 3x -y = – 5 are …………………
(a) 1, -1
(b) \(\frac{1}{2}\), -2
(c) 1, \(\frac{1}{2}\)
(d) 2, \(\frac{-1}{2}\)
Answer:
(b) \(\frac{1}{2}\), -2

Question 9.
The sum of the binomial co-efficients is ………………….
(a) 2n
(b) 2ⁿ
(c) n²
(d) 1
Answer:
(b) 2ⁿ

Question 10.
If the square of the matrix \(\begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix}\) satisfy the relation
(a) 1 + α² + βγ = 0
(b) 1 – α² – βγ = 0
(c) 1 – α² + βγ = 0
(d) 1 + α² – βγ = 0
Answer:
(b) 1 – α² – βγ = 0

Question 11.
The value of x for which the matrix A = \(\begin{bmatrix} e^{ x-2 } & e^{ 7+x } \\ e^{ 2+x } & e^{ 2x+3 } \end{bmatrix}\) is singular is ………………….
(a) 9
(b) 8
(c) 7
(d) 6
Answer:
(b) 8

Question 12.
If A = \(\begin{pmatrix} \lambda & 1 \\ -1 & -\lambda \end{pmatrix}\) then for what value of λ, A² = 0?
(a) 0
(b) ±1
(c) -1
(d) 1
Answer:
(b) ±1

Question 13.
lim_x→1 \(\frac { xe^{ x }-sinx }{ x } \) is ……………….
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(d) 0

Question 14.
If the points whose position vectors 10\({ \vec { i } }\) + 3\({ \vec { j } }\) + 12\({ \vec { i } }\) – 5\({ \vec { j } }\) and a\({ \vec { i } }\) + 11\({ \vec { j } }\) are collinear then the value of a is ……………….
(a) 3
(b) 5
(c) 6
(d) 8
Answer:
(d) 8

Question 15.
If y = mx + c and f(0) = f'(0) = 1 then f(2) = ………………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
lim_x→1 \(\frac { e^{ x }-e }{ x-1 } \) = ……………….
(a) 1
(b) e
(c) ∞
(d) 0
Answer:
(b) e

Question 17.
∫2^3x+5 dx is ………………..
(a) \(\frac { 3(2^{ 3x+5 }) }{ log2 } \) + c
(b) \(\frac { 2^{ 3x+5 } }{ 2log(3x+5) } \) + c
(c) \(\frac { 2^{ 3x+5 } }{ 2log3 } \) + c
(d) \(\frac { 2^{ 3x+5 } }{ 3log2 } \) + c
Answer:
(d) \(\frac { 2^{ 3x+5 } }{ 3log2 } \) + c

Question 18.
∫x² cos xdx is ………………..
(a) x²sinx + 2xcosx – 2 sinx + c
(b) x² – 2xcosx – 2sinx + c
(c) -x² + 2xcosx + 2sinx + c
(d) -x² – 2xcosx + 2 sinx + c
Answer:
(a) x²sinx + 2xcosx – 2 sinx + c

Question 19.
∫\(\frac { x }{ 1+x^{ 2 } } \) dx = ………………
(a) tan^-1 x + c
(b) log (1 + x²) + c
(c) log x + c
(d) \(\frac{1}{2}\) log (1 + x²) + c
Answer:
(d) \(\frac{1}{2}\) log (1 + x²) + c

Question 20.
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) \(\frac{7}{128}\)

PART – II

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

Question 21.
To secure an A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If A scored 84, 87, 95, 91 in first four subjects, what is the minimum mark be scored in the fifth subject to get an A grade in the course?
Answer:
Required marks = 5 × 90 = 450
Total marks obtained in 4 subjects = 84 + 87 + 95 + 91 = 357
So required marks in the fifth subject = 450 – 357 = 93.

Question 22.
Prove that (1 + tan1°) (1 + tan2°) (1 + tan 3°) …. (1 + tan 44°) is a multiple of 4?
Answer:
45°= 1 + 44 (or) 2 + 43 (or) 3 + 42 (or) 22 + 23
So we have 22 possible pairs
sa the product is (2) (22) = 44
which is ÷ by 4

Question 23.
Find the distinct permutations of the letters of the word MISSISSIPPI?
Answer:
MISSISSIPPI
Number of letters = 11
Here M-1 time, I-4timcs, S-4times, P-2timcs
So total number of arrangement is of this word = \(\frac { 11! }{ 4!4!2! } \)
\(\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 1\times 2\times 1\times 4! } \) = 34650

Question 24.
Compare |A| using Sarrus rule if A = \(\left[\begin{array}{ccc}
3 & 4 & 1 \\
0 & -1 & 2 \\
5 & -2 & 6
\end{array}\right]\)
Answer:

|A| = [3(-1) (6) + 4(2)(5) + 1(0)(-2)] -[5(-1)(1) + (-2)(2)3 + 6(0)(4)] = [-18 + 40 + 0] – [-5 – 12 + 0] = 22 + 17 = 39.

Question 25.
If |\({ \vec { a } }\)|, |\({ \vec { b } }\)| = 6, |\({ \vec { c } }\)| =7 and \({ \vec { a } }\) + \({ \vec { b } }\) + \({ \vec { c } }\) = 0 find \({ \vec { a } }\).\({ \vec { b } }\) + \({ \vec { b } }.\) \({ \vec { c } }\) + \({ \vec { c } }\).\({ \vec { a } }\)
Answer:
Given \(\bar { a } \)+\(\bar { b } \)+\(\bar { c } \) = 0
⇒ (\(\bar { a } \)+\(\bar { b } \)+\(\bar { c } \))² = 0
(i.e;) |\(\bar { a } \)|² + |\(\bar { b } \)|² + |\(\bar { c } \)|² + 2 [\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = 0
⇒ 5²+6²+7²+2(\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = 0
⇒2(\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = – 25 – 36 – 49 = -110
⇒\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \) = \(\frac{-110}{2}\) = -55

Question 26.
Find \(\lim _{ t\rightarrow 0 }{ \frac { \sqrt { t^{ 2 }+9-3 } }{ t^{ 2 } } } \)
Answer:
We can’t apply the quotient Iieorem it.:iiediatcly. Use the algebra technique of rationalising the numerator.

Question 27.
Evaluate y = e^-x.log x?
Answer:
y = e^-x log x = uv(say)
Here u = e^-x and v = log x
⇒ u’ = -e^-x and v’ = uv’ + vu’
Now y = uv ⇒y’ = uv’ + vu’
(i.e). \(\frac{dy}{dx}\) = e^-x(\(\frac{1}{x}\)) + log x(- e^-x)
= e^-x(\(\frac{1}{x}\) – log x)

Question 28.
Evaluate ∫\(\frac { 1 }{ \sqrt { 1+(4x)^{ 2 } } } \) dx
Answer:
Let I = ∫\(\frac { 1 }{ \sqrt { 1+4x^{ 2 } } } \)dx = ∫\(\frac { 1 }{ \sqrt { 1+(2x^{ 2 }) } } \) dx
Putting 2x = t ⇒ 2 dx = dt ⇒ dx = \(\frac{1}{2}\) dt
Thus, I = \(\frac{1}{2}\)∫\({ \frac { 1 }{ \sqrt { 1^{ 2 }+t^{ 2 } } } }\) dt
I = \(\frac{1}{2}\) log |t+\(\sqrt { t^{ 2 }+1 } \) + c = \(\frac{1}{2}\) log |2x + \(\sqrt { (2x)^{ 2 }+1 } \)| + c
I = \(\frac{1}{2}\) log |2x+\(\sqrt { 4^{ 2 }+1 } \)| + c.

Question 29.
Nine coins are tossed once. Find the probablity to get atleast 2 heads?
Answer:
Let S be the sample and A be the event of getting at least two heads.
Therefore, the event \(\bar { A } \) denotes, getting at most one head.
n(S) = 2⁹ = 521, n(\(\bar { A } \)) = 9C₀ + 9C₁ = 1 + 9 = 10
P(\(\bar { A } \)) = \(\frac{10}{512}\) = \(\frac{5}{256}\)
P(A) = 1- P(\(\bar { A } \)) = 1 – \(\frac{5}{256}\) = \(\frac{251}{256}\)

Question 30.
If P(A) denotes the power set of A, then find n(P(P(ϕ))))?
Answer:
Since P(∅) contains 1 element, P(P(∅)) contains 2¹ elements and hence P(P(P(∅))) contains 2² elements. That is, 4 elements.

PART – III

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

Question 31.
A simple cipher takes a number and codes it, using the function f(x) = 3x – 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines)?

Question 32.
Show that 4 sin A sin (60° + A). sin(60° – A) = sin 3A?

Question 33.
How many strings of length 6 can be formed using letters of the word FLOWER if (i) either starts with F or ends with R?

Question 34.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle 135° with the positive x- axis?

Question 35.
Using factor theorem prove that \(\left|\begin{array}{lll}
x+1 & 3 & 5 \\
2 & x+2 & 5 \\
2 & 3 & x+4
\end{array}\right|\) = (x – 1)² + ( x+ 9)

Question 36.
Find the unit vectors perpendicular to each of the vectors \({ \vec { a } }\) + \({ \vec { b } }\) and \({ \vec { a } }\) – \({ \vec { b } }\). Where \({ \vec { a } }\) = \({ \vec { i } }\) + \({ \vec { j } }\) + \({ \vec { k } }\) and b = \({ \vec { i } }\) + 2\({ \vec { j } }\) + 3\({ \vec { k } }\)?

Question 37.
If y = tan^-1 \(\left(\frac{1+x}{1-x}\right)\), find y’?

Question 38.
Evaluate ∫\(\frac { x }{ \sqrt { 1+x^{ 2 } } } \) dx?

Question 39.
If A and B are two events such that P(A∪B) = 0.7, P(A∩B) = 0.2 and P(B) = 0.5 then show that A and B are independent?

Question 40.
Resolve into partial fractions \(\frac { 2x^{ 2 }+5x-11 }{ x^{ 2 }+2x-3 } \)

PART – IV

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

Question 41.
(a) A photocopy store charges Rs. 1.50 per copy for the first 10 copies and Rs. 1.00 per copy after the 10th copy. Let x be the number of copies, and let y be the total cost of photocopying.
(a) Draw graph of the cost as x goes from 0 to 50 copies?
(b) Find the cost of making 40 copies

[OR]

(b) If the difference of the roots of the equation 2x² – (a + 1)x + a – 1 = 0 is equal to their product then prove that a =2?

Question 42.
(a) Prove that tan 70° – tan 20° – 2 tan 40° = 4 tan 10°?

[OR]

(b) In a ∆ABC, if a = 2\(\sqrt{3}\), b = 2\(\sqrt{2}\) and C = 75° find the other side and the angles?

Question 43.
(a) Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n?

[OR]

(b) Evaluate y = (x² + 1) \(\sqrt [ 3 ]{ x^{ 2 }+2 } \)?

Question 44.
(a) Show that f(x) f(y) = f(x + y), where f(x) = \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\)

[OR]

(b) The chances of X, Y and Z becoming managers of a certain company are 4 : 2 : 3. The probabilities that bonus scheme will be introduced if X, Y and Z become managers are 0.3, 0.5 and 0.4 respectively. 1f the bonus scheme has been introduced, what is the probability that Z was appointed as the manager?

Question 45.
(a) If the equation λx² – 10xy + 12y² + 5x – 16y – 3 = O represents a pair of straight lines, find

1. The value of λ and the separate equations of the lines
2. Angle between the lines

[OR]

(b) Show that

1. \(\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{3 n^{2}+7 n+2}=\frac{1}{6}\)
2. \(\lim _{n \rightarrow \infty} \frac{1^{2}+2^{2}+\ldots+(3 n)^{2}}{(1+2+\ldots+5 n)(2 n+3)}=\frac{9}{25}\)
3. \(\lim _{n \rightarrow \infty} \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots .+\frac{1}{n(n+1)}=1\)

Question 46.
(a) A man repays an amount of ₹3250 by paying ₹20 in the first month and then increases the payment by ₹15 per month. How long will it take him to clear the amount?

[OR]

(b) Find the area of the triangle whose vertices are A (3, -1, 2), B (1, -1, -3) and C (4, -3, 1)?

Question 47.
(a) Evaluate (x + 1) \(\sqrt { 2x+3 } \)?
[OR]

(b) Evaluate ∫cosec²x dx?

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

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TN State Board 11th Maths Model Question Paper 2 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 – 1

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

Question 1.
Let R be the set of all real numbers. Consider the following subsets of the plane R × R:
S= {(x, y) : y = x + 1 and 0 < x < 2} and T = {(x, y): x – y is an integer}
Then which of the following is true?
(a) T is an equivalence relation but S is not an equivalence relation.
(b) Neither S nor T is an equivalence relation
(c) Both S and T are equivalence relation
(d) S is an equivalence relation but T is not an equivalence relation.
Answer:
(a) T is an equivalence relation but S is not an equivalence relation.

Question 2.
If the set A has m elements the set B has n elements and the number of elements in A × B is …………………
(a) m + n
(b) mn
(c) \(\frac{m}{n}\)
(d) m²
Answer:
(b) mn

Question 3.
If \(\frac{ax}{(x+2)(2x-3)}\) = \(\frac{2}{x+2}\) + \(\frac{3}{2x-3}\) then a = ……………….
(a) 8
(b) 7
(c) 5
(d) 4
Answer:
(b) 7

Question 4.
The number of solutions of x² + |x – 1| = 1 is ………………….
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

Question 5.
If a, 8, b are in A.P. a, 4, b are in G.P. and a, x, b are in H.P then x = ………………..
(a) 2
(b) 1
(c) 4
(d) 16
Answer:
(a) 2

Question 6.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are ……………….
(a) 45
(b) 40
(c) 10!
(d) 2¹⁰
Answer:
(a) 45

Question 7.
The value of e^2logx = …………………..
(a) 2x
(b) x²
(c) \(\sqrt{x}\)
(d) \(\frac{x}{2}\)
Answer:
(b) x²

Question 8.
The n^th term of the sequence 1, 2, 4, 7, 11 …. is …………………
(a) n³ + 3n² + 2n
(b) n³ – 3n² + 3n
(c) n\(\frac{(n+1)(n+2)}{3}\)
(d) \(\frac { n^{ 2 }-n+2 }{ 2 } \)
Answer:
(d) \(\frac { n^{ 2 }-n+2 }{ 2 } \)

Question 9.
The last term in the expansion (2+\(\sqrt{3}\))⁸ is ………………
(a) 81
(b) 27
(c) 9
(d) 3
Answer:
(a) 81

Question 10.
A line perpendicular to the line 5x -y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5sq.units, then its equation is …………………..
(a) x + 5y ± 5\(\sqrt{2}\) = 0
(b) x – 5y ± 5\(\sqrt{2}\) = 0
(c) 5x + y ± 5\(\sqrt{2}\) = 0
(d) 5x – y ± 5\(\sqrt{2}\) = 0
Answer:
(a) x + 5y ± 5\(\sqrt{2}\) = 0

Question 11.
A factor of the determinant \(\left|\begin{array}{ccc}
x & -6 & -1 \\
2 & -3 x & x-3 \\
-3 & 2 x & x+2
\end{array}\right|\) is ……………….
(a) x + 3
(b) 2x – 1
(c) x – 2
(d) x – 3
Answer:
(a) x + 3

Question 12.
If λ\(\vec { a } \) + 2λ\(\vec { j } \) + 2λ\(\vec { k } \) is a unit vector then the value of λ is ………………
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{1}{3}\)

Question 13.
One of the diagonals of parallelogram ABCD with \(\vec { a } \) and \(\vec { b } \) are adjacent sides is \(\vec { a } \) + \(\vec { b } \). The other diagonal BD is ………………….
(a) \(\vec { a } \) – \(\vec { b } \)
(b) \(\vec { a } \) – \(\vec { b } \)
(c) \(\vec { a } \) + \(\vec { b } \)
(d) \(\frac{\vec{a}+\vec{b}}{2}\)
Answer:
(b) \(\vec { a } \) – \(\vec { b } \)

Question 14.
If (1, 2, 4) and (2, -3λ, -3) are the initial and terminal points of the vector \(\vec { i } \) + 5\(\vec { j } \) – 7\(\vec { k } \) then the value of λ …………………..
(a) \(\frac{7}{3}\)
(b) –\(\frac{7}{3}\)
(c) \(\frac{5}{3}\)
(d) \(\frac{-5}{3}\)
Answer:
(b) –\(\frac{7}{3}\)

Question 15.
If y = mx + c and f(0) =f'(0) = 1 then f(2) = …………………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
The derivative of (x + \(\frac{1}{x}\))² w.r.to. x is ………………..
(a) 2x – \(\frac { 2 }{ x^{ 3 } } \)
(b) 2x + \(\frac { 2 }{ x^{ 3 } } \)
(c) 2(x + \(\frac{1}{x}\))
(d) 0
Answer:
(a) 2x – \(\frac { 2 }{ x^{ 3 } } \)

Question 17.
If f(x) is \(\left\{\begin{array}{cc}
a x^{2}-b, & -1<x<1 \\
\frac{1}{|x|}, & \text { elsewhere }
\end{array}\right.\) is differentiable at x = 1, then …………………
(a) a = \(\frac{1}{2}\), b = \(\frac{-3}{2}\)
(b) a = \(\frac{-1}{2}\), b = \(\frac{3}{2}\)
(c) a = \(\frac{-1}{2}\), b = \(\frac{-3}{2}\)
(d) a = \(\frac{1}{2}\), b = \(\frac{3}{2}\)
Answer:
(c) a = \(\frac{-1}{2}\), b = \(\frac{-3}{2}\)

Question 18.
∫\(\frac { \sqrt { tanx } }{ sin2x } \) dx is ………………
(a) \(\sqrt{tanx}\) + c
(b) 2\(\sqrt{tanx}\) + c
(c) \(\frac{1}{2}\) \(\sqrt{tanx}\) + c
(d) \(\frac{1}{4}\) \(\sqrt{tanx}\) + c
Answer:
(a) \(\sqrt{tanx}\) + c

Question 19.
An urn contains 5 red and 5 black balls. A balls is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. The probability that the second ball drawn is red will be ………………….
(a) \(\frac{5}{12}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{7}{12}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 20.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\), and P(B/A) = \(\frac{2}{3}\) then
P(B) = ………………….
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(b) \(\frac{1}{3}\)

PART – II

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

Question 21.
If n(P(A)) = 1024, n(A∪B) = 15 and n(P(B)) = 32 then find n(A∩B)
Answer:
n(P(A)) = 1024 = 2¹⁰ ⇒ n(A) = 10
n(A∪B) = 15
n(P(B)) = 32 = 2⁵ ⇒ n(B) = 5
We know n(A∪B) = n(A) + n(B) – n(A∩B)
(i.e) 15 = 10 + 5 – n(A∩B)
⇒ n(A∩B) = 15 – 15 = 0

Question 22.
Simplify (343)^2/3
Answer:
(343)^2/3 = (7³)^2/3 = 7^3×2/3 = 7² = 49

Question 23.
Show that cos36° cos 72° cos 108° cos 144° = \(\frac{1}{16}\)
Answer:
LHS = cos36° cos(90° – 18°) cos(90° – 18°) cos(90° + 18°) cos(180° – 36°)
= sin² 18° cos² 36°
= (\(\frac { \sqrt { 5-1 } }{ 4 } \))² (\(\frac { \sqrt { 5+1 } }{ 4 } \))² = \(\frac{1}{16}\) = RHS

Question 24.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour?
Answer:
There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour.
∴Number of ways of selection = ⁶C₃ × ⁵C₃ × ⁵C₃
= \(\frac { 6! }{ 3!3! } \) × \(\frac { 5! }{ 3!2! } \) × \(\frac { 5! }{ 3!2! } \)
= 20 × 10 × 10 = 2000

Question 25.
Find |A| if A = \(\left[\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right]\)
Answer:
\(\left[\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right]\)
= 0M₁₁ – sin αM₁₂ + cos αM₁₃
= 0 – sin α(0 – cos α sin β) + cos α (-sin α sin β – 0) = 0

Question 26.
For any vector prove that \(\vec { r } \) = [\(\vec { r } \).\(\vec { i } \)) + (\(\vec { r } \).\(\vec { j } \))j + [\(\vec { r } \).\(\vec { k } \)}k
Answer:
Let \(\vec { r } \) = x\(\hat { i } \) + y\(\hat { j } \) + z\(\hat { k } \)

Question 27.
Calculate \(\lim _{x \rightarrow-2}\) (x³ – 3x + 6) (-x² + 15)
Answer:

Question 28.
Evaluate y = e^x sin x
Answer:

Question 29.
Integrate the following with respect to x
\(\frac{4}{3+4x}\) + (10x + 3)⁹ – 3 cosec(2x + 3) cot (2x + 3)
Answer:

Question 30.
P(A) = 0.6, P (B) = 0.5 and P(A∩B) = 0.2 find P(A/B)
Answer:
Given that P(A) = 0.6, P(B) = 0.5, and P(A∩B) = 0.2
P(A/B) = \(\frac { p(A∩B) }{ p(B) } \) = \(\frac{0.2}{0.5}\) = \(\frac{2}{5}\)

PART – III

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

Question 31.
A quadratic polynomial has one of its zeros 1 + \(\sqrt{5}\) and it satisfies p(1) = 2. Find the quadratic polynomial?

Question 32.
Prove that

1. tan^-1 (\(\frac{1}{7}\)) + tan^-1(\(\frac{1}{13}\)) = tan^-1(\(\frac{2}{9}\))
2. cos^-1\(\frac{4}{5}\) + tan^-1\(\frac{3}{5}\) = tan^-1\(\frac{27}{11}\)

Question 33.
The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP?

Question 34.
Find the equation of the line passing through the point (5, 2) and perpendiular to the line joining the points (2, 3) and (3, -1)?

Question 35.
Find the area of the triangle whose vertices are (0,0), (1,2) and (4,3)?

Question 36.
If \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) are three vectors such that \(\vec { a } \) + 2\(\vec { b } \) + \(\vec { c } \) = 0 and |\(\vec { a } \)| = 3, |\(\vec { b } \)| = 4, |\(\vec { c } \)| = 7 fimd the angle between \(\vec { a } \) and \(\vec { b } \)

Question 37.
Evaluate: \({ \underset { x\rightarrow 0 }{ lim } }\) \(\frac { 3^{ x }-1 }{ \sqrt { 1+x-1 } } \)

Question 38.
Find \(\frac{dy}{dx}\) for y = tan^-1 \((\frac { cosx+sinx }{ cosx-sinx } )\)

Question 39.
Evaluate: ∫x⁵ e^x2

Question 40.
How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?

PART – IV

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

Question 41.
(a) If f : R – {-1, 1} → R is defined by f(x) = \(\frac { x }{ x^{ 2 }-1 } \), verify whether f is one-to-one or not?

[OR]

(b) Solve: log₂ x + log₄ x + log₈ x = 11

Question 42.
(a) Prove that \(\frac{\sin x+\sin 3 x+\sin 5 x+\sin 7 x}{\cos x+\cos 3 x+\cos 5 x+\cos 7 x}\) = tan 4x

[OR]

(b) If x + y + z = xyz, then prove that \(\frac { 2x }{ 1-x^{ 2 } } \) + \(\frac { 2y }{ 1-y^{ 2 } } \) + \(\frac { 2z }{ 1-z^{ 2 } } \) = \(\frac { 2x }{ 1-x^{ 2 } } \) \(\frac { 2y }{ 1-y^{ 2 } } \) \(\frac { 2z }{ 1-z^{ 2 } } \)

Question 43.
(a) If the letters of the word GARDEN are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, then and the ranks of the words

1. GARDEN
2. DANGER

[OR]

(b) \(\underset { x\rightarrow a }{ lim } \) \(\frac{\sqrt{x-b}-\sqrt{a-b}}{x^{2}-a^{2}}\) (a>b)

Question 44.
(a) If the binomial coefficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42 then find n?

[OR]

(b) Evalute \(\sqrt { x^{ 2 }+y^{ 2 } } \) = tan^-1(\(\frac{y}{x}\))

Question 45.
Let \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) be three vectors such that |\(\vec { a } \)| = 3, |\(\vec { b } \)| = 4, |\(\vec { c } \)| = 5 and each one of them being perpendicular to the sum of the other two, find |\(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \)|.

[OR]

(b) Evaluate ∫sec³ 2xdx

Question 46.
(a) Find all the equations of the straight lines in the family of the lines y = mx – 3, for which m and the x-coordinate of the point of intersection of the lines with x – y = 6 are integers?

[OR]

(b) There are two identical boxes containing respectively 5 white and 3 red balls, 4 white and 6 red balls. A box is chosen at random and a ball is drawn from it

1. Find the probability that the ball is white
2. If the ball is white, what is the probability that it from the first box?

Question 47 (a).
If A_i B_i, C_i are the cofactors of a_i, b_i, c_i, respectively, i = 1 to 3 in

[OR]

(b) Express the matrix \(\left(\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right)\) as the sum of symmetric martix and a skew-symmetric martix?

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Tamil Nadu 11th Commerce Model Question Papers Tamil Medium 2020-2021

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11th Commerce Model Question Paper Design 2020-2021 Tamil Nadu

Types of Questions	Marks	No. of Questions to be Answered	Total Marks
Part-I Objective Type	1	20	20
Part-II Very Short Answers (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily)	2	7	14
Part-Ill Short Answers (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily)	3	7	21
Part-IV Essay Type	5	7	35
Total Marks			90

Tamil Nadu 11th Commerce Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

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11th Chemistry Model Question Paper Design 2020-2021 Tamil Nadu

Types of Questions	Marks	No. of Questions to be Answered	Total Marks
Part-I Objective Type	1	15	15
Part-II Very Short Answers (Totally 9 questions will be given. Answer any Six. Any one question should be answered compulsorily)	2	6	12
Part-Ill Short Answers (Totally 9 questions will be given. Answer any Six. Any one question should be answered compulsorily)	3	6	18
Part-IV Essay Type	5	5	25
Total			70
Practical Marks + Internal Assessment (20+10)			30
Total Marks			100

Tamil Nadu 11th Chemistry Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

It is necessary that students will understand the new pattern and style of Model Question Papers of 11th Standard Chemistry Tamilnadu State Board Syllabus according to the latest exam pattern. These Tamil Nadu Plus One 11th Chemistry Model Question Papers State Board Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN HSLC Board Exams and Score More marks.

We hope the given Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern Class 11th Chemistry Model Question Papers 2020 2021 with Answers Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding the Tamil Nadu Government 11th Chemistry State Board Model Question Papers with Answers 2020 21, TN 11th Std Chemistry Public Exam Question Papers with Answer Key, New Paper Pattern of HSC Class 11th Chemistry Previous Year Question Papers, Plus One +1 Chemistry Model Sample Papers, drop a comment below and we will get back to you as soon as possible.

Subject Matter Experts at SamacheerKalvi.Guide have created Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern 11th Model Question Papers 2020-2021 with Answers Pdf Free Download in English Medium and Tamil Medium of TN 11th Standard Public Exam Question Papers Answer Key, New Paper Pattern of HSC 11th Class Previous Year Question Papers, Plus One +1 Model Sample Papers, Tamil Nadu 11th Quarterly Half Yearly Model Question Papers are part of Samacheer Kalvi.

Let us look at these Tamil Nadu Government 11th State Board Model Question Papers with Answers for All Subjects 2020-21 Tamil Medium Pdf. Students can view or download the Class 11th New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming Tamil Nadu HSC Board Exams. Students can also read Tamilnadu Samcheer Kalvi 11th Books Solutions.

The higher officials of Tamil Nadu Directorate of Government Examinations conducts Public examination to Class 11 Students in the month of March. Recently Government of Tamil Nadu has made changes in the examination pattern to reduce the pressure and tension in students. Here we have provided the Tamil Nadu 11th Model Question Papers PDF, students can view and download the latest pdf of 11th Model Question Papers. Before preparing for the public examinations students need to know about TN 11th Model Question Papers. These Model Papers helps students to prepare well. In this page, we have provided all subject wise Tamil Nadu Plus One Model Question Papers. Before appearing for HSC examinations students must practice all the TN Model Question Papers to score high marks in TN Plus One Examinations.

11th New Public Exam Model Question Papers Tamil Nadu 2020 2021 English Tamil Medium

11th New Model Question Papers 2020 2021 Tamil Nadu

• Tamil Nadu 11th Maths Model Question Papers
• Tamil Nadu 11th Physics Model Question Papers
• Tamil Nadu 11th Chemistry Model Question Papers
• Tamil Nadu 11th Biology Model Question Papers
• Tamil Nadu 11th English Model Question Papers
• Tamil Nadu 11th Tamil Model Question Papers
• Tamil Nadu 11th Commerce Model Question Papers
• Tamil Nadu 11th Economics Model Question Papers
• Tamil Nadu 11th Accountancy Model Question Papers
• Tamil Nadu 11th Computer Science Model Question Papers

Here we have given TN 11th Model Question Papers with Answers for all subjects in English Tamil Medium. Previous Year Question Papers and Model Papers will help students to know about their preparation level not only that these papers will also help Students to analyze about the repeated questions and important questions that are asked in previous year public examinations. As per the new exam pattern, there will be no change in the syllabus. Now students can get their Tamil Nadu 11th Model Question Papers of all Subjects in English Tamil Medium from the above list.

It is necessary that students will understand the new pattern and style of Model Question Papers of 11th Standard Tamilnadu State Board Syllabus according to the latest exam pattern. These Tamilnadu Plus One 11th Model Question Papers State Board Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN HSLC Board Exams and Score More marks.

We hope the given Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern Class 11th Model Question Papers 2020 2021 with Answers Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding the Tamil Nadu Government 11th State Board Model Question Papers with Answers 2020 21 for All Subjects, TN 11th Std Public Exam Question Papers with Answer Key, New Paper Pattern of HSC Class 11th Previous Year Question Papers, Plus One +1 Model Sample Papers, Tamil Nadu 11th Quarterly Half Yearly Model Question Papers, drop a comment below and we will get back to you as soon as possible.

Subject Matter Experts at SamacheerKalvi.Guide have created Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern 11th Maths Model Question Papers 2020-2021 with Answers Pdf Free Download in English Medium and Tamil Medium of TN 11th Standard Maths Public Exam Question Papers Answer Key, New Paper Pattern of HSC 11th Class Maths Previous Year Question Papers, Plus One +1 Maths Model Sample Papers are part of Tamil Nadu 11th Model Question Papers.

Let us look at these Government of Tamil Nadu State Board 11th Maths Model Question Papers Tamil Medium with Answers 2020-21 Pdf. Students can view or download the Class 11th Maths New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming Tamil Nadu HSC Board Exams. Students can also read Tamilnadu Samcheer Kalvi 11th Maths Guide.

TN State Board 11th Maths Model Question Papers 2020 2021 English Tamil Medium

Tamil Nadu 11th Maths Model Question Papers English Medium 2020-2021

• Tamil Nadu 11th Maths Model Question Paper 1 English Medium
• Tamil Nadu 11th Maths Model Question Paper 2 English Medium
• Tamil Nadu 11th Maths Model Question Paper 3 English Medium
• Tamil Nadu 11th Maths Model Question Paper 4 English Medium
• Tamil Nadu 11th Maths Model Question Paper 5 English Medium

Tamil Nadu 11th Maths Model Question Papers Tamil Medium 2019-2020

• Tamil Nadu 11th Maths Model Question Paper 1 Tamil Medium
• Tamil Nadu 11th Maths Model Question Paper 2 Tamil Medium
• Tamil Nadu 11th Maths Model Question Paper 3 Tamil Medium
• Tamil Nadu 11th Maths Model Question Paper 4 Tamil Medium
• Tamil Nadu 11th Maths Model Question Paper 5 Tamil Medium

11th Maths Model Question Paper Design 2020-2021 Tamil Nadu

Types of Questions	Marks	No. of Questions to be Answered	Total Marks
Part-I Objective Type	1	20	20
Part-II Very Short Answers (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily)	2	7	14
Part-Ill Short Answers (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily)	3	7	27
Part-IV Essay Type	5	7	35
Total			90
Internal Assesment			10
Total Marks			100

Tamil Nadu 11th Maths Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

It is necessary that students will understand the new pattern and style of Model Question Papers of 11th Standard Maths Tamilnadu State Board Syllabus according to the latest exam pattern. These Tamil Nadu Plus One 11th Maths Model Question Papers State Board Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN HSLC Board Exams and Score More marks.

We hope the given Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern Class 11th Maths Model Question Papers 2020 2021 with Answers Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding the Tamil Nadu Government 11th Maths State Board Model Question Papers with Answers 2020 21, TN 11th Std Maths Public Exam Question Papers with Answer Key, New Paper Pattern of HSC Class 11th Maths Previous Year Question Papers, Plus One +1 Maths Model Sample Papers, drop a comment below and we will get back to you as soon as possible.

Subject Matter Experts at SamacheerKalvi.Guide have created Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern 11th Biology Model Question Papers 2020-2021 with Answers Pdf Free Download in English Medium and Tamil Medium of TN 11th Standard Biology (Bio Botany Bio Zoology) Public Exam Question Papers Answer Key, New Paper Pattern of HSC 11th Class Biology Previous Year Question Papers, Plus One +1 Biology Model Sample Papers are part of Tamil Nadu 11th Model Question Papers.

Let us look at these Government of Tamil Nadu State Board 11th Biology Model Question Papers Tamil Medium with Answers 2020-21 Pdf. Students can view or download the Class 11th Biology New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming Tamil Nadu HSC Board Exams. Students can also read Tamilnadu Samcheer Kalvi 11th Biology Guide.

NEET Biology MCQ

TN State Board 11th Biology Model Question Papers 2020 2021 English Tamil Medium

Tamil Nadu 11th Biology Model Question Papers English Medium 2020-2021

• Tamil Nadu 11th Biology Model Question Paper 1 English Medium
• Tamil Nadu 11th Biology Model Question Paper 2 English Medium
• Tamil Nadu 11th Biology Model Question Paper 3 English Medium
• Tamil Nadu 11th Biology Model Question Paper 4 English Medium
• Tamil Nadu 11th Biology Model Question Paper 5 English Medium

Tamil Nadu 11th Biology Model Question Papers Tamil Medium 2020-2021

• Tamil Nadu 11th Biology Model Question Paper 1 Tamil Medium
• Tamil Nadu 11th Biology Model Question Paper 2 Tamil Medium
• Tamil Nadu 11th Biology Model Question Paper 3 Tamil Medium
• Tamil Nadu 11th Biology Model Question Paper 4 Tamil Medium
• Tamil Nadu 11th Biology Model Question Paper 5 Tamil Medium

11th Biology Model Question Paper Design 2020-2021 Tamil Nadu

Tamil Nadu 11th Biology Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

It is necessary that students will understand the new pattern and style of Model Question Papers of 11th Standard Biology Tamilnadu State Board Syllabus (Bio Botany Bio Zoology) according to the latest exam pattern. These Tamil Nadu Plus One 11th Biology Model Question Papers State Board Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN HSLC Board Exams and Score More marks.

We hope the given Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern Class 11th Biology Model Question Papers 2020 2021 with Answers Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding the Tamil Nadu Government 11th Biology State Board Model Question Papers with Answers 2020 21 (Bio Botany Bio Zoology), TN 11th Std Biology Public Exam Question Papers with Answer Key, New Paper Pattern of HSC Class 11th Biology Previous Year Question Papers, Plus One +1 Biology Model Sample Papers, drop a comment below and we will get back to you as soon as possible.

Subject Matter Experts at SamacheerKalvi.Guide have created Tamil Nadu State Board Samacheer Kalvi 11th Computer Science Book Answers Solutions Guide Pdf Free Download of Volume 1 and Volume 2 in English Medium and Tamil Medium are part of Samacheer Kalvi 11th Books Solutions.

Let us look at these TN State Board New Syllabus Samacheer Kalvi 11th Std Computer Science Guide Pdf of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank and revise our understanding of the subject.

Students can also read Tamil Nadu 11th Computer Science Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 11th Computer Science Book Solutions Answers Guide

Tamilnadu State Board Samacheer Kalvi 11th Computer Science Book Back Answers Solutions Guide Volume 1 and Volume 2.

Samacheer Kalvi 11th Computer Science Book Back Answers

Unit 1 Fundamentals of Computer and Working with a Typical Operating Systems (Windows & Linux)

• Chapter 1 Introduction to Computers
• Chapter 2 Number Systems (Part 1)
• Chapter 2 Boolean Algebra (Part 2)
• Chapter 3 Computer Organization
• Chapter 4 Theoretical Concepts of Operating System
• Chapter 5 Working with Typical Operating System (Windows & Linux)

Unit 2 Algorithmic Problem Solving

• Chapter 6 Specification and Abstraction
• Chapter 7 Composition and Decomposition
• Chapter 8 Iteration and Recursion

Unit 3 Introduction to C++

• Chapter 9 Introduction to C++
• Chapter 10 Flow of Control
• Chapter 11 Functions
• Chapter 12 Arrays and Structures

Unit 4 Object Oriented Programming with C++

• Chapter 13 Introduction to Object-Oriented Programming Techniques
• Chapter 14 Classes and Objects
• Chapter 15 Polymorphism
• Chapter 16 Inheritance

Unit 5  Computer Ethics and Cyber Security

• Chapter 17 Computer Ethics and Cyber Security
• Chapter 18 Tamil Computing

We hope these Tamilnadu State Board Samacheer Kalvi Class 11th Computer Science Book Solutions Answers Guide Volume 1 and Volume 2 Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 11th Standard Computer Science Guide Pdf of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, drop a comment below and we will get back to you as soon as possible.