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Karnataka 1st PUC Chemistry Model Question Paper 5 with Answers (Old Pattern)

Time: 3.15 Hours
Max Marks: 70

Instructions:

1. The questions paper has four parts A, B, C and D.
2. In part A each question carries ONE Marks. In part B, each question carries TWO marks, in part C each question carries FIVE marks. In Part D-1 carries TEN Marks and each question in D -2 carries FIVE marks.
3. Write balanced chemical equations and draw diagram wherever necessary.

Part – A

I. Answer of all the following questions: ( 1 × 9 = 9 )

Question 1.
What is a limiting reagent in a reaction ?
Answer:
The reactant which gets completely consumed or limits the amount of product formed in a reaction

Question 2.
State the law of conservation of mass.
Answer:
It states that matter can neither be created nor be destroyed

Question 3.
Select an isoelectronic pair among the following Na⁺,Cl^–,F^–,Li⁺
Answer:
Na⁺ ,F^–

Question 4.
What is the effect of temperature on Viscosity of liquids ?
Answer:
Viscosity of liquids decrease as temperature rises.

Question 5.
A closed thermos flask containing hot coffee represents what type of a system ?
Answer:
Isolated.

Question 6.
Write the expression for Kc for the equilibirium, N_2(g) + 3H_2(g) ⇌ 2NH_2(g)
Answer:
\(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{N}_{2}\right]^{1}}\)

Question 7.
H₂⁺and H₂^–have same bond order but H₂⁺is more stable than H₂^–, why ?
Answer:
H₂^–has en electron in antibonding MO which makes it less stable than H₂⁺

Question 8.
What volume of O₂ is produced by 1 litre of “10 volume” H₂O₂ at STP ?
Answer:
10 litres.

Question 9.
Name the alkali metal which is radioactive ?
Answer:
Francium (Fr)

Question 10.
Which element is diagonally related to lithium ?
Answer:
Magnesium (Mg)

Question 11.
Give an example for a Heterocyclic compound.
Answer:
Pyridine, Furan, Thiophene.

Part – B

II. Answer any Five questions: ( 5 × 2 = 10 )

Question 12.
Write balanced chemical equation for the combustion of methane. How many mole of methane are required to produced 88g Carbon dioxide by the combustion of Methene ?
Answer:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
\(\begin{array}{cc}{12+4 \times 1} & {12+2 \times 16} \\{16 g} & {44 g}\end{array}\)
44g of CO₂ requires 16g of CH₄.
88g of CO₂ requires ?
= \(\frac{88 \times 16}{44}\) = 32g of CH₄ = 2 mole of methane.

Question 13.
Size of Na⁺ ion is smaller than Na. Give reason.
Answer:
During the formation of Na⁺ ions effective nuclear charge increases thus electron cloud shrinks.

Question 14.
What causes temporary and permanent hardness of water ?
Answer:
Temporary Hardness – presence of dissolved Bicarbonates of Mg and Ca.
Permanent Hardness – Presence of dissolved Chlorides and sulphates of Mg and Ca.

Question 15.
Derive Ideal gas equation using gas laws.
Answer:
From Boyle’s law V ∝ \(\frac{1}{p}\) ………..(1)
From Charle’s law V ∝ T  ………..(2)
From(1) and (2)
V ∝ \(\frac{T}{p}\)
PV ∝ T
Or PV – RT

Question 16.
(i). Chemical equilibrium is dynamic. Give reason.
Answer:
Both forward and backward reactions are occurring at the same rate.

(ii). What is the effect of a Catalyst on the equilibrium of a reversible reaction.    2
Answer:
A catalyst has no effect on the position of equilibrium but it helps the reaction to attain equilibrium quickly.

Question 17.
Diamond is a bad conductor of electricity but Graphite is a good conductor. Justify the statement.
Answer:
Due to sp³ hybridization in Diamond no free electrons are present.
In Graphite due to sp³ hybridization, there are free electrons to conduct electricity.

Question 18.
Write any two common chemicals of photochemical smog.
Answer:
Ozone, Nitric oxide, Acrolein, Frmaldehyde and peroxy acetyl nitrate.

Part – C

Note : Answer any Four questions: ( 4 × 4 = 16 )

Question 19.
a. Define mole fraction.
Answer:
It is ratio of number of moles of a particular component to the total number of moles
of the solution. Mole fraction = \(\frac{\mathbf{n}_{\mathrm{B}}}{\mathbf{n}_{\mathrm{A}}+\mathbf{n}_{\mathrm{B}}}\)

b. Calculate the mass percent of the solute, if 4 g of a solute is dissolved in 18g of water.
Answer:

Question 20.
a. Explain the formation of line spectrum of Hydrogen based on Bohr’s atomic theory. Write the energy level diagram for Lyman and Balmer series of the hydrogen spectrum.
Answer:
According to Bohr’s Atomic model:
A sample of H₂ gas contains large number of hydrogen molecules. When electric discharge is passed through hydrogen gas the molecules absorbs energy and split up into atoms. Now the electron present in different atoms absorbs different energy and jump to different high energy state. The electrons in the higher energy is unstable thus returns back by emitting radiations having different wavelength. Hydrogen spectrum consists of many spectral lines event through the hydrogen only one electron.

Lymon series : The lines obtained when and electron jumps from 2nd, 3rd, 4th…. i.e., from any higher energy level to first energy level (n₁ = 1) are called Lyman series. These lines are found in ultraviolet region.

Balmer series : The lines obtained when electrons jumps from 3rd, 4th 5th… from any higher energy level to second energy level are called Balmer series. This is found in visible region.

Or (Internal choice)

b. (i) Calculate the wavelength of a ball of mass 0.05 kg moving with a velocity of 5ms^-1. Given : h = 6.626 × 10^-34 Js.               2+2
Answer:

Question 21.
a. Classify the following physical quantities having α Extensive and Intensive properties Mass, Density, Temperature and Heat capacity.
Answer:
Extensive property – Mass, Heat capacity
Intensive property – Density, Temperature

b. State Hess’s law of constant summation.
Answer:
Hess’s law : .Whether a chemical reaction takes place in a single step or in several steps, the total change in enthalpy remains the same. Consider the formation of CO₂ from C.

Question 22.
Using Le-Chatelier’s principle explain the effect of
(i) Addition of CH₄ (ii) Addition of CS₂ (iii) Removal of S₂ (iv) Removal of H₂S on the equilibrium,
CH_4(g) + 2S_2(g) ⇌ CS_2(g) + 2H₂S_{(g)                          4}
Answer:
i) Increase in concentration of CH₄ , favours forward reaction.
ii) Increase in concentration of S₂, favours backward reaction.
iii) Decrease in concentration S₂, favours backward reaction.
iv) Decrease in concentration of H₂S , favours forward reaction.

Question 23.
a. Give the composition of Washing soda. An aquous solution of washing soda is alkaline, Give reason.
b. What happens when Cl₂ (Chlorine) gas is passed into milk of Lime ? Write the equation.   2+2
Answer:
Bleaching powder is formed.
2Ca(OH)₂ + 2Cl₂ → CaCl₂ + Ca(OCl)₂ + 2H₂O

Question 24.
Draw a neat labeled diagram and give the calculations involved in the estimation of Nitrogen in an Organic compound by Duma’s method.
Answer:
Principle : The organic compound containing nitrogen when heated with excess of copper oxide in the atmosphere of carbon dioxide, yields nitrogen in addition to carbon dioxide and water.

Traces of nitrogen oxides formed during combustion of organic compound are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The percentage of nitrogen present in a given organic compound is calculated from the volume of nitrogen collected over potassium hydroxide solution from a known mass of organic compound.

Procedure : The apparatus used for the estimation of nitrogen by this method is shown in the figure.

A known mass of organic compound is mixed with copper oxide and placed in the combustion tube. The carbon dioxide gas is passed through the combustion tube to displace air present in the tube. The combustion tube is now heated in the furnace. the nitrogen evolved collects in the nitrometer. The volume of the nitrogen collected is recorded after adjusting the levels of potassium hydroxide solution in the two limbs are equal. Room temperature and atmosphere pressure are recorded.

Calculation:
Mass of organic compound = mg
Volume of nitrogen in nitrometer = V cm³
Room temperature = t° C = (273 + t) K
Atmosphere pressure = P₁ mm
Aqueous tension at room temperature = P′ mm
Pressure of dry nitrogen gas formed = P = (P – P′) mm
Volume of nitrogen at STP (V₀) = \(\frac{\mathrm{PV} \times 273}{760 \times(273+\mathrm{t})} \mathrm{cm}^{3}\)
22,400 cm³ of nitrogen of STP = 28 g of nitrogen
Mass of V₀ cm³ of nitrogen = \(\frac{28 \times V}{22,400} g\)
Percentage of nitrogen =\(\frac{28 \times V_{0} \times 100}{22,400 \times m}\)

Part – D

Note : Answer any Four questions : ( 2 × 5 = 10 )

Question 25.
a. (i) Name the isotope of Hydrogen containing two neutrons.
(ii) Mention any two limitations of Bohr’s atomic theory.
b. Atomic number and Mass number of an element are 29 & 64. How many protons and Neutrons are present in it? Why the Electronic configuration of the element.
Answer:
a) i) Tritium (₁T³)
ii) It fails to account for the finer details of the hydrogen atom spectrum observed by using sophisticated spectroscopic technique, i.e., stark effect and zeeman effect.

b) It could not explain the ability of atoms to form molecules by chemical bonds.

Question 26.
a. Explain the shape of Ammonia molecule using VSEPR theory.
Answer:
Ammonia (NH₃)
Electronic configuration of N: 1s² 2s² 2p³
Valance orbital representation

1. when 2s² orbital hybridised with 2p_x, 2p_y, and 2p_z forms the following sp³ hybridised structure.
2. In 2s² orbital, the electrons are paired up hence they are not involved in bond formation called loan paired of electron (L.P).
3. When sp³ hybridized N combines with 3 hydrogen atom, forms pyramidal structure of ammonia.
   Pyramidal structure angle = 107°

Or (Internal Choice)

Mention any three postulates of Molecular Orbital Theory.
Answer:

1. The electrons in an atom are found in atomic orbitals, the electrons in a molecule are found in molecular orbitals.
2. The molecular orbitals are formed by the combination of atomic orbitals of comparable energies and proper symmetry.
3. The BMO has lower energy and hence greater stability than the corresponding ABMO.
4. The molecular orbitals are filled by electrons in accordance with Aufbau principle, Pauli’s exclusion principle and the Hund’s rule.

b. Lithium molecule has 4 electrons in bonding molecular orbital and 2 electrons in anti bonding molecular orbital. Write the electronic configuration and calculate the bond order of Lithium molecule.
Answer:

Or (Internal choice)

Calculate the Bond order, show that Helium Molecule (He₂) does not exist.
Answer:

Reason: since bond order is 0 the molecule does not exist.

Question 27.
a. For the oxidation of Iron : 4Fe_(s) + 3O_2(g) → 2Fe₂O_3(g)
Entropy change and Enthalpy change at 298 K are – 549.4 J/K/mol and -1648 × 10³ J/mol respectively. Calculate the free energy change for the reaction. Is the reaction spontaneous at 298 K.
Answer:
∆G = ∆H – T∆S
∆G = -1648000 + 163721.2
= -1484278.8 J/mol
=-1484.2788 kjmol
The reaction is spontaneous.

b. Define Enthalpy of solution.
Answer:
The enthalpy of solution at infinite dilution is the enthalpy change observed an dissolving the substance in an infinite amount of solvent when the interations between the ions are negligible.

Question 28.
a. In the Galvanic cell, the cell reaction is Zn_(s) + 2Ag_(aq)⁺ → Zn²⁺+ 2Ag_(a)
(i) Which of the electrode is the Anode ?
(ii) Write the Cathodic and Anodic reactions.
(iii) Identify the oxidizing and reducing agents.
Answer:
i) Zn elelctrode.

ii) Cathodic reaction : Ag⁺+e^– → Ag
Anodic rection : Zn – 2e^– → Zn²⁺

iii) Oxidising agent: Ag⁺
Reducing agent: Zn

b. Show that the reaction CaCO₃ → CaO + CO₂ is not a Redox reaction.    3+2
Answer:

Since there is not change in oxidation state of any element before and after the reaction. It is not a redox reaction.

Question 29.
a. Explain the Ampoteric nature of Aluminium, give equations.
Answer:
Aluminium reacts both with acid and base so it behaves as amphoteric nature.
2Al(s) + 6HCl(aq)  → 2Al³⁺(aq) + 6Cl^–(aq) + 3H₂(g) behaving as acid.
2Al(s) + 2NaOH(aq) + 6H₂O(l) → 2Na⁺[Al(OH)₄]^– (aq) + 3H₂(g) behaving as base.

b. Name the type of hybridization of
(i) Boron in Diborane
(ii) Carbon in Graphite
Answer:
i) sp³ hybridization
ii) sp² hybridisation

Question 30.
a. Explain the mechanism of chlorination of Methane, with equations.
Answer:
Mechanism of chlorination of methane involves three types.
Step 1 : Initiation : Chlorine absorbs energy and undergoes homolysis to give chlorine free radicals.

Step 2 : Propagation : Chlorine free radical reacts with methane to give methyl free radical.
Cl^• + CH₄ → CH₃^•+ HCl
The methyl free radical reacts with chlorine to form methyl chloride and chlorine free radical.
CH₃^•+ Cl₂ → CH₃Cl + Cl^•

Step 3 : Termination : Free radicals combine to form stable products.
Cl^• + Cl^• → Cl₂ (Chlorine)
CH₃^•+ Cl^• → C₂H₆ (Ethane)
CH₃^• + Cl^• → CH₃Cl (Methyl Chloride)

b. How do you convert Benzene to Benzene hexachloride (BHC) ? 3+2
Answer:

Under ultra violet light, three chlorine molecules add to benzene to produce benzene hexa chloride, C₆H₆Cl₂ which is also called gammaxane.

Part – E

Answer any Three of the following questions: ( 3 5 = 15 )

Question 31.
a. (i) Explain why BF₃ molecule has zero dipole moment although the B-F bonds are polar ?
(ii) Alkali metals and Halogens form Ionic compounds. Give reason.
Answer:
i) Shape of BF₃ is Trigonal planar.
The net sum of bond moment is equal to zero because of cancellation of bond moments due to symmetry in the molecule.

ii) Alkali metals have low ionization enthalpy hence they easily form cations.
Halogens have high electron gain enthalpy hence they easily form Anions.

b. What is Hydrogen bonding ? Sketch the Hydrogen bonding in liquid Hydrogen fluoride ?   3+2
Answer:
It is an electrostatic force of attraction between an electropositive element and electronegative element Hs⁺-Fs.

Question 32.
a. Write the expression for the compressibility factor (Z) for one mole of a gas and name the terms in it. What is the value of Z for an ideal gas ?
Answer:
z = \(\frac{p v}{R T}\)    z = compressibility, P = pressure, V = Volume, T = Temperature in K, Z = 1 for ideal gas, R = Universal gas constant.

b.A balloon is filled with Hydrogen gas at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 liter volume, up to what volume can the balloon be inflated ?
Answer:

Question 33.
a. What is a Bronsted – Lowery Acids ? Identify the two Conjugate acid-base pairs in the following.
NH₃ + HCl ⇌ NH₄⁺+ Cl^–
Answer:
Bronsted – lowery theory
An acid is substance which donates proton base is a substance which accepts a portion so this theory is also called as proton theory.

Conjugate acid base pair

b. Calculate the solubility of Lead Chloride at 298. K, if its solubility product is 1.6 × 10^-5.
Ans: For PbCl₂k_sp = 4s³

Question 34.
a. Write any two differences between Resonance effect and Electromeric effect.
Answer:

Resonance Effect	Electromeric Effect
1. It takes place in conjugated double bond system	1. It takes place in the molecule having c = c
2. It is a permanent effect takes place in absence of attacking reagent.	2. It is temporary effect takes place in presence of attacking reagent.

b. What type of structural isomerism is exhibited by the Propanone and Propanal pair ?
Write their structural and identify the functional groups present in them.      2+3
Answer:
Functional isomerism
Structure : CH₃ – CO – CH₃, CH₃ – CH₂ – CHO
Functional groups : Ketone (C = O); Aldehyde (-CHO)

Question 35.
a. (i) Name the chief product obtained when Benzene is heated with concentrated Sulphuric acid. Give equation ?
Answer:
Benzene Sulphonic Acid
Suiphonation : Benzene reacts with concentrated sulphuric acid at 80°C to form benzene suiphoric acid.

Mechanism : This involves the following steps.
Step 1 : Generation of electrophile SO₃.
2H₂SO₄ ⇌ SO₃ + H₃O⁺ + HSO₄^–

Step 2 : The electrophile SO₃ attacks tihe benzene ring to form a carbocation

Step 3 : Loss of a proton to form a sulphonate ion. The proton is removed by HSO₄^–

Step 4 : Addition of proton to give benzene sulphonic acid.

(ii)Write the resonance hybrid structured of Benzene.
Answer:

b. How is Ethane prepared from sodium acetate by electrolytic method ? 2+1+2
Answer:

Karnataka 1st PUC Chemistry Model Question Paper 1 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
2. Write balanced chemical equations and draw labeled diagram wherever allowed.
3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

Note : Answer any nine questions: ( 9 × 1 = 9 )

Question 1.
Define Molarity of a solution.
Answer:
Number of moles of solute in one litre of the solution.

Question 2.
Name the element having highest value for its Electron gain enthiaply.
Answer:
Chlorine

Question 3.
On what parameter do the Elements are classified in the modern periodic table?
Answer:
Atomic number

Question 4.
What type of Van der waals force exists between HCI molecules?
Answer:
Dipole – dipole interaction

Question 5.
What is the relationship between ∆H and ∆U for the reaction PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) ?
Answer:
∆H = ∆U + RT

Question 6.
The value of Ionic Product of water at 298 K is 1 × 10^-14, what is its [H⁺] ?
Answer:
1 × 10^-7 or 10^-7 Mole/dm³

Question 7.
Name any two gases responsible for Green house effect.
Answer:
Any two gases (CH₄, CO₂, NO₂ etc.)

Question 8.
What is the role of Heavy water in a nuclear reactor?
Answer:
As a moderator Or To slow down fast moving neutrons.

Question 9.
What is Catenation?
Answer:
The tendency of Carbon atoms to link with one another through covalent bonds to form chains and rings.

Question 10.
What type of structural isomerism is shown by the pair : But-1-yne and But-2-yn?
Answer:
Position isomerism.

Question 11.
Which one of the following acids is not present in Acid rain?
Answer:
H – COOH

Part – B

Note : Answer any Five questions: ( 5 × 2 = 10 )

Question 12.
Express i) 5 litre of milk in cubic meter (m³) ii) 25° C in Kelvin.   ( 1 + 1 )
Answer:
i) 5  10^-3 m^-3 …………1   ii) 298 K ………..1

Question 13.
Which of the following species are Isoelectronic with Ne ?  ( 2 )
(Atomic no. of Ne = 10). N^3-,Na⁺,Al³⁺,Ar,Rb⁺ & F^–
Answer:
All elements contains total 10 electrons.

Question 14.
Mention the type of Hydrogen bond in the following compounds.  ( 1 + 1 )
i) Water ii) ortho-Nitrophenol
Answer:
i) Inter molecular hydrogen bonding …………1
ii) Intra molecular hydrogen bonding ………..1

Question 15.
Write Vander Walls equation for one mole of a gas and name any two terms in it.   ( 1 + 1 )
Answer:

Question 16.
Calculate the pH of a solution whose Hydrogen ion conc. is 1.5 × 10³ M.    ( 2 )
Answer:
pH = – log[H⁺]
pH = – log(1.5 × 10^-3)
pH = 3 – log 1.5 = 3 – 0.1761
pH = 2.8239

Question 17.
What is the composition of Permutit (Zeolite) ? How does it works in softening of hard water?  ( 1 + 1 )
Answer:
Hydrated sodium aluminium silicate / Sodium aluminium silicate ………..(1) It exchanges Mg²⁺ and Ca²⁺ ions for its Na⁺ hence soften the hard water …. (1)

Question 18.
Explain the action of heat on Orthoboric acid. Give example.  ( 1 + 1 )
Answer:
When hydro basic acid is heated two moles of water is liberated as follows.

Part – C

Note : Answer any Four questions:  ( 4 × 4 = 16 )

Question 19.
Mention one significance for each of the four quantum numbers.   ( 4 )
Answer:
n = size of the atom
l = orientation, shape
m = subduell (s, p, d, f)
s = Spin of electrons

Question 20.
Write the Electronic configuration, Energy level diagram for the molecular orbitals of Oxygen molecule (O₂). Calculate its bond order and give reason as to why O₂ is paramagnetic.    ( 4 )
Answer:

Atomic number of oxygen = 8. Electronic configuration of oxygen = 1s²2s²2p⁴. When two oxygen atoms combines, the molecular orbital energy level diagram is as shown in the figure.
From the diagram, the molecular electronic configuration of oxygen is

Magnetic property: There are two unpaired electrons {\(\pi_{2 \mathrm{p}_{\mathrm{x}}^{1}}^{*}=\pi_{2 \mathrm{p}_{\mathrm{y}}^{1}}^{*}\)} therefore, oxygen molecule is a paramagnetic in nature.

Question 21.
Calculate the heat of reaction of the following reaction :
C₆H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H₂O_(g) ; ΔH = ?
C_(graphite) + O_2(g) → CO_2(g);  ΔH = -395.0 kJ … (1)
H_2(g) + \(\frac{1}{2}\)O(g) → H₂O(l); ΔH = -269.5 kJ … (2)
6C_(graphite) + 6H_2(g) + O_2(g) → C₆H₁₂O₆; ΔH = -1169.8 kJ … (3)
Answer:
Multiplying equation (1) and (2) each by 6 reversing (3), we get,
6C_(graphite) + 6O_2(g) → 6CO_2(g); ΔH = -2370 kJ …(4)
6H_2(g) + 3O_2(g) → 6H₂O(l); ΔH = -1616.4 kJ … (5)
C₆H₁₂O_6(s) → 6C_(graphite) + 3O_{2(g) ;} ΔH = +1169.8 kJ …..(6)

Adding (4), (5) and (6),
C₆H₁₂O_6(s) +6O_2(g) → 6CO_2(g) +6H₂O_(g) ;
ΔHg (C₆H₁₂O₆) = -2370.0 – 1616.4 + 1169.8 = -2816.6 kJ

Question 22.
A buffer solution of pH 8.3 is prepared from ammonium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8 × 10^-5. What is the mole proportion of ammonium chloride and ammonium hydroxide?
Answer:
pOH= 14 – pH =14 – 8.3 = 5.7

Question 23
a. A coulourless liquid ‘A’ contains H and O elements only. It decomposes slowly on exposure to light. It is stabilized by mixing urea to store in the presence of light, (i) Suggest possible structure of A. (ii) Write chemical equations for its decomposition reaction in light.   ( 2 + 2 )
Answer:
The liquid A is hydrogen peroxide (H₂O₂)

b. Give three uses of different allotropic forms of carbon.
Answer:

Forms of carbon	Uses
Diamond	Gemstone, cutting, drilling, grinding, polishing, industry.
Graphite	Reducing agent, refractories, pencils, high temperature, crucibles, electrode making, moderator in nuclear reactors, high strength composite materials.
Activated carbon	Rubber industry, pigments in ink, paints and plastics
Coke	Fuel, strut manufacture
Charcoal	Fuel, reducing agent, Adsorption.

Question 24
a. Explain the mechanism of Halogenation or chlorination of benzene.   ( 3 + 1 )
Answer:
Halogenation: Benzene reacts with chlorine in the presence of FeCl₃ or AlCl₃ to form chlorobenzene.

Machanism : This involves the following steps.
Step 1: Generation of electrohile Cl + Cl – Cl + FeCl₃ → Cl⁺+ FeCl₄^–
Step 2 : The electrophile Cl+ attacks benzene ring to form a carbon cation which is resonance stabilised.

Step 3: Loss of a proton to give chiorobenzene. The proton is removed by FeCl₄.

b. Name the product obtained when Benzene is hydrogenated in presence of heated Nickel catalyst.
Answer:
Cyclohexane  (1)

Part – D

Note : Answer any Four questions: ( 4 × 5 = 20 )

Question 25
a. i) What is the mass percentage of Carbon in Methane ( CH₄ ) (Molecular mass of CH₄ =16)
Answer:
\(\frac{12}{16}\) × 100 = 75% or 75% …….. (1)

ii) Expess 9.8 g H₂SO₄ in mole (Molecular mass of H₂SO₄ = 98)
Answer:
9.8 g H₂SO₄ = 0.1 mole or 0.1 mole …. (1)

iii) What mass of Calcium carbonate is to be decomposed to obtain 4.4 of CO₂ in the following reaction
CaCO₃ → CaO + CO₂ (Molecular mass of CaCO₃ = 100)
Answer:
100 × \(\frac{4.4}{44}\) = 10g or g …… (1)

b. State Avegadro’s law.
How many atoms of Hydrogen are present in 1 mole of water?
Answer:
Under the same temperature & pressure equal volume of gases contains equal number of molecules.
= 2 × 6.022 × 1023 atoms …….. (1)

Question 26
a. The average kinetic energy of a gas molecule at O °C is 5.621 × 10^-27 J. Calculate Boltzmann constant. Also calculate the number of molecules present in one mole of the gas.
Answer:
Average kinetic energy KE = \(\frac{3}{2}\)KT

No. of molecules in 1 mole of the gas (Avagadro’s no.) = \(\frac{R}{k}\)

b. How vapour pressure of liquid is related to (i) temperature, (ii) nature of liquid, (iii) boiling point, (iv) atmospheric pressure?
Answer:
(i) Vapour pressure is directly proportional to temperature.
(ii) If intermolecular force of attraction is less in the liquid, its vapour pressure will be high.
(iii) Higher the vapour pressure, lower will be boiling point.
(iv) If atmospheric pressure is low, boiling point will be less and higher will be the vapour pressure.

Question 27
a. What happens to the solubility of BaSO₄ when a few drops of BaCl₂ solution is added to its saturated solution ? Give reason.
Answer:
Solubility decreases, Due to common ion effect

b. What is meant by Conjugate acid-base pair ?
Write the conjugate acids for CN^– & H₂O.
Answer:
Acid base pair which differ by one photon.
Conjugate acid of CN^– is HCN ; Conjugate acid of H₂O is H₃O⁺

c. An aqueous solution of NH₄Cl is acidic. Give reason.
Answer:
Due to hydrolysis produce HCl, reaction

Question 28
a. Give reason for diagonal relationship of lithium with magnesium.
Answer:
Both Lithium and magnesium have small size and high charge density. The electronegativities of Li is 1.0 and Mg is 1.2. They are low and almost same. Their ionic radii are similar. Hence they show similarities which is known as diagonal relationship between first element of a group with the second element in the next higher group.

b. What is the effect of heat on the following compounds ?
(a) Magnesium chloride hexahydrate
(b) Gypsum
(c) Magnesium sulphate heptahydrate
Answer:

c. Between LiCl and NaCl, LiCl is soluble in Ethanol but NaCl does not give reason.
Answer:
Because LiCl is covalent compound hence dissolves in organic solvent (Ethanol) but NaCl is ionic compound hence insoluble in ethanol.

Question 29
a. How is the estimation of halogens by Carius method ?   ( 4 )

b. Which method is employed for the estimation of carbon and hydrogen organic compounds ?
Answer:
Liebig’s method.

Question 30
a. What is Wurtz reaction ? Give example.   ( 2 )
Answer:
When alkyl halides are heated with sodium metal in ether medium higher alkanes are formed. This reaction is known as Wurtz reaction and employed for the synthesis of higher alkanes containing even number of carbon atoms.

b. How is Methane prepared from sodium acetate ?  ( 2 )
Answer:
Decarboxylation : When sodium salt of carboxylic acid is heated with soda lime (mixture of sodium hydroxide and calcium oxide), an alkane containing one carbon atom less them parent carboxylic acid is formed. This reaction is called decarboxylation.

c. What is the compostion of Zeigler Natta Catalyst?  ( 1 )
Answer:
Triethyl aluminium and titanium tetrachloride inert solvent.

Part – E

Answer any Three of the following questions:  ( 3 × 5 = 15 )

Question 31
a. Draw the structure of d-orbital (Orbital whose Azimuthal quantum no = 2).
Answer:
d – orbital has 5 orientations.

b. What is Isobars ?
Answer:
Atoms of different elements having the same mass number but different atomic numbers are called isobars.

Question 32
a. Define hybridization ? Explain the hybridization in Methane molecule.
(Refer Q. No. 41)
Answer:
CH₄ – Methane Molecule.
The Molecular formula of Methane is CH₄
Electronic configuration of C is ground state – Is² 2s² 2p²
Elecronic configuration of C in excited state – Is² 2s¹ 2p³
Valance orbital representation –

The valence orbital contains unpaired electron. Hence sp³ hybridized carbon combine with 4 hydrogen atom forms methane molecule.

b. What type of orbital over lapping results in the formation of a n-bond ?
Answer:
P-P parallel / sideway / lateral overlapping.

c. Between H₂S & H₂O, H₂O) is more polar. Why ?     3 + 1 + 1
Answer:
Because Oxygen is more electronegative than sulphur.

Question 33
a. Explain the steps involved in Born Haber cycle for the formation of NaCl.
Answer:
Step 1 : Sublimation of metallic sodium of gaseous sodium atom with enthalpy of sublimation = ∆H_s.

Step – 2 : Dissociation of molecule of chlorine to chlorine atoms with enthalpy of dissociation = ∆H_d

Step – 3 : Ionization of gaseous sodium with enthalpy of ionization = ∆H_i

Step – 4 : Addition of electron to gaseous chlorine atom with enthalpy of electron affinity = ∆H_Ea

Step – 5 : Close packing of gaseous sodium ion and chloride’ ion to form lattice structure of NaCl, with lattice chloride ion to form lattice structure of NaCl, with Lattice energy = U ;
NaCl⁺_(g) + Cl^–_(g) → NaCl∆; ∆H = U

Step – 6 : But sum of all the energies will be equal to the heat of formation of one mole of sodium chloride form its reluctant i.e., ∆Hf
Na_(s) + 1/2 Cl_2(g) → NaCl_(s); ∆H = ∆Hf

b. Define the term Entropy. What happens to Entropy when ice melts to liquid water ?  ( 3 + 2 )
Answer:
It is disorderness of the system. Entropy increases.

Question 34
a. Balance MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O
Answer:
Step – 1:
Assign oxidation number :

Step – 2:
Write the oxidation number changes.
Change in oxidation number of Mn + 4 to + 2 = 2
Change in oxidation number of Cl – 1 to 0 = +1

Step – 3: Cross multiply the numbers. Co-efficient 2 is multiplied to HCl and 1 is multiplied to MnO₂.
MnO₂ + 2HCl → MnCl₂ + H₂O + Cl₂

Step – 4: Check the oxidation number, and 2 molecules of HCl on left hand side to balance the chlorine atoms of MnCl₂. In order to balance oxygen and hydrogen atom 2 molecule of H₂O has to be added on the right hand side.
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

b. Calculate the oxidation number of the element underline in each of the following cases, (a) Al in Al₂O₃ (b) P in P₂O₅
Answer:
The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

Question 35
a. Describe with a neat diagram the estimation of carbon and hydrogen by Leibig’s method.
Answer:
Principle : A known mass of an organic compound is strongly heated with dry cupric oxide (CuO), when carbon and hydrogen are quantitatively oxidized to CO₂ and H₂O respectively. The masses of CO₂ and H₂O thus formed are determined. From this, the percentages of carbon and hydrogen can be calculated.

Procedure : Pure and dry oxygen is passed through the entire assembly of the apparatus (Fig) till the CO₂ and moisture is completely removed.
sample in platinum boat

A boat containing weighed organic substances is introduced inside from one end of the combustion tube by opening it for a while. The tube is now strongly heated till the whole of the organic compound is burnt up. The flow of oxygen is continued to drive CO₂ and water vapours completely to the U-tubes. The apparatus is cooled and the U-tubes are weighed separately.

Observed and Calculations.

1. Mass of organic compound taken = w.g.
2. Mass of water produced = x g (Increase in mass of CaCk tube)
3. Mass of carbon dioxide produced = y g. (Increase in mass of KOH tube)

To determine % of carbon :
Molar mass of CO₂ = 44g mol^-1
Now, 44g of CO₂ = contains 12 g of C.
∴ y g of CO₂ will contain of \(\frac{12 y}{44}\)g of C.
This amount of carbon was present in w. g. of the substance
∴ % C = \(\frac{12 y}{44} \times \frac{100}{w}\)
To determine % of Hydrogen
Molar mass of water = 18 g mol^-1
Now, 18g of H₂O contains 2 g of H₂
∴ x g of H₂O will contain \(\frac{2 x}{18} g\) of H₂
This amount of hydrogen was present in weight of substance.
∴ % H₂ = \(\frac{2 x}{18} \times \frac{100}{w}\)

b. Write the functional grop for i) Aldehyde ii) Ketone.     ( 4 + 1 )
Answer:
i) Aldehyde : (-CHO)
ii) Ketone : (- CO -)

Karnataka 1st PUC Chemistry Model Question Paper 2 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
2. Write balanced chemical equations and draw labeled diagram wherever allowed.
3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

Note : Answer any nine questions: ( 9 × 1 = 9 )

Question 1.
Write the value of Avagadro’s number.
Answer:
6.023 × 10²³

Question 2.
Name the fundamental particles in an atom.
Answer:
Electron, Proton, Neutrons.

Question 3.
Write the general electronic configuration of transition elements.
Answer:
(Noble Gas) (n – 1)d^1–10 ns^1–2

Question 4.
State Law of Conservation of Energy of mass.
Answer:
The total mass of the products in a reaction is equal to mass of the reactants ‘or’ the mass neither be created nor be destroyed.

Question5.
When is ∆H=∆u inareaction?
Answer:
H_2(g) + I_2(g) ⇌ 2HI_g

Question 6.
Define solubility product of weak electrolyte.
Answer:
Solubility product of weak electrolyte is defined as the product of molar concentration of ions of electrolyte raised to appropriate power within saturated solution of that electrolyte.

Question 7.
What is the valency of hydrogen in hydride ion?
Answer:
– 1

Question 8.
Name the isotope of hydrogen which used In nuclear reactor.
Answer:
\(_{1}^{2} \mathrm{H}\) (deuterium)

Question 9.
Name any four alkali metals.
Answer:
Lithium, Sodium, Potassium and Rubedium.

Question 10.
What is the self linking property of an atom known as?
Answer:
Catenation.

Question 11.
Write the IUPAC name of

Answer:
2 Methyl 2 Butene.

Part – B

Note : Answer any Five questions:  ( 5 × 2 = 10 )

Question 12.
Calculate the wave number and wavelength of the first spectral line of Lyman series of hydrogen spectrum.
Rydberg constant R = 10.97 x 106 m^–1
Answer:

8.2275 × 10⁶ m^–1 Wave number v̄ = 8.2275 × 10⁶ m^–1
Wavelength λ = \(\frac{1}{8.2275 \times 10^{6}}\)m = 121.5 × 10^–9 m = 121.5 nm

Question 13.
What is electro negativity? How does it change in a period as well as in a group?
Answer:
The ability of an atom to attract the shared electron pair (of a covalent bond) in a molecule towards itself is called electro negativity. In a period from left to right the electro negativity increases. Down a group electro negativity value decreases.

Question 14.
Mention any two characteristics of ionic compounds.
Answer:

1. They have high Melting point.
2. Many of it are solid in nature.
3. Does not conduct in solid state but good conductors in fused or aqueous state.
4. Soluble in polar solvent & insoluble in non-polar solvent.
5. Ionic bond is non-directional.
6. They do not exhibit Isomerism.

Question 15.
Write the acidity and calculate equivalent weight of following base.
a) Ca(OH)₂ b) Fe(OH)₃
Answer:

Question 16.
Among NH₃, H₂O and HF, which would you except to have highest magnitude of hydrogen bonding and why ?
Answer:
HF, because Flourine is more electronegative.

Question 17.
Calculate the [OH^–] of a solution whose pH is 9.62.
Answer:
pH + pOH = 14 pOH = 14 – pH = 14 – 9.62 = 4.38
pOH = 4.38 = – log₁₀ [OH^– ]; 4.38 = – log₁₀ [OH^– ]
log₁₀[ OH^– ] = – 4.38 ; [OH^–] = antilog (–4.38) = antilog (5 – 4.38 – 5)
= antilog (0.62 – 5) = antilog (0.62) × 10^–5 = 4.169 × 10^–5 mol dm^–3.

Question 18.
Name the oxides of nitrogen. What are the sources ?
Answer:
Nitrogen monoxide (NO), Nitrogen dioxide (NO₂), Dinitrogen trioxide (N₂O₃),
Dinitrogen tetroxide (N₂O₄), Dinitrogen pentoxide (N₂O₅)
The source of oxides of nitrogen are combustion of fossil fuel, especially, petroleum. They are also formed by reaction of N₂ and O₂ in presence of lightening.

Part – C

Note : Answer any Four questions:  ( 4 × 4 = 16 )

Question 19.
4g of copper chloride on analysis was founded to contain 1.890 g of copper (Cu) and 2.110 g of chlorine (Cl). What is the empirical formula of copper chloride? [At. mass of Cu = 63.5 u, Cl = 35.5 u].
Answer:

Empirical formula = CuCl₂.

Question 20.
Based on VSEPR theory explain the structure of water.
Answer:
Electronic configuration of O: Is² 2s² 2p⁴Valence orbited representation :

1. When 2s² orbital hybridised with 2p_x, 2p_y and 2p_z forms the following sp³ hybridisation structure.
2. In 2s² and 2p_x ² the electrons are paired up hence they are not involved in bond formation called loan paired of electron (L.P.)
3. When sp³ hybridised O combines with 2 hydrogen atom forms Bond structure water.

Question 21.
State and illustrate Hess’s law.
Answer:
Hess’s law : Whether a chemical reaction takes place in a single step or in several steps, the total change in enthalpy remains the same.
Consider the formation of CO₂ from C.

Question 22
a. Define oxidation and reduction according to electronic concept.
Answer:
Oxidation is a process in which loss of electrons take place. Reduction is a process in which gain of electrons takes place.

b. Define cathode and anode.
Answer:
Cathode is electrode towards which cations are attracted. Anode is electrode which attracts anions.

Question 23
a. Give the IUPAC name of the following compounds.

Answer:
(i) Propyle Benzene
(ii) 3 Chloro Propenal

b. In Leibig’s method. 0.24 g of organic compound on combustion with dry oxygen produced of 0.62 g of CO₂ and 0.11 g of H₂O. Determine the percentage composition of the compound.
Answer:
Mass of organic compound = m = 0.24 g
Mass of carbon dioxide formed = 0.62 g
Mass of water formed = 0.11 g
Percentage of carbon = \(\frac{12}{4} \times \frac{0.62}{0.24} \times 100\) = 70.45
Percentage of hydrogen = \(\frac{2}{4} \times \frac{0.11}{0.24} \times 100\) = 5.09
Percentage of oxygen = [100 – (70.5 + 5.0)] = 24.46

Question 24
a. Why does benzene undergo electrophilic substitution easily and nucleophilic substitution with difficulty?
Answer:
Because of de-localisation of 671 electrons in the benzene ring.

b. Write the balanced chemical equations for the combination of the following hydrocarbons 2.
(i) Butane (ii) Toulene
Answer:
(i) C₄H₁₀ + \(\frac{3}{2}\)O₂ → 5H₂O + 4CO₂

(ii) C₆H₅CH₃ + 9O₂ → 7CO₂ + 4H₂O

Part – D

Note : Answer any Four questions:  ( 2 × 5 = 10 )

Question 25
a. Summarize the Bohr’s Model of an atom.
Answer:
Bohr’s Model of an atom, the postulates are

1. Electrons revolve around the nucleus of an atom in a certain definite path called Orbit or stationary state of shell.
2. The shells are having different energy levels denoted as K, L, M, N
3. As long as the electron remains in an orbit, they neither absorb nor emit energy.
4. The electron can move only in that orbit in which angular momentum is quantized, i.e., the angular momentum of the electron is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\).

b. Mention the Merits of Bohr’s theory.
Answer:

1. Explains the formation of hydrogen spectrums
2. It explains the stability of atom
3. Ryd berg constant is calculated by it.

Question 26
a. Explain the structure of CH₄ on the basis of VSEPR theory.

b. Define electronegativity.
Answer:
The ability of an atom to attract the shred electron pair (of a covalent bond) in a molecule towards itself is called electro negativity. In a period from left to right the electronegativity increases. Down a group electronegativity value decreases.

Question 27
a. State First law of thermodynamics and write its mathematical form.
Answer:
The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is ΔU = Q − W.

b. 1 m³ of C₃H₄ at STP is burnt in oxygen, according to the thermochemical reaction : C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l); ΔH = – 1410 kJ mol^–1 Assuming 70% efficiency, determine how much of useful heat is evolved in the reaction.
Answer:
22.4 L of C₂H₄ at STP produces 1410 kJ of energy.
1000 L of C₂H₄ at STP produces \(\frac{1410}{22.4}\) × 1000 (efficiency)

Question 28
a. Explain Arrhenius concept of acids and bases with example.
Answer:
An electrolyte which when dissolved in water, produces H ion is called acid.
An electrolyte which when dissolved in water, produces OH^– ion is called base.

b. Explain the buffer of acidic buffer.
Answer:
Consider acidic buffer mixture,
CH₃COOH ⇌ CH₃COO^– + H⁺ ; CH₃COONa → CH₃COO^– + Na⁺
Here, acetic acid is a weak electrolyte, in it’s solution there exists equilibrium between it’s ions and molecules. Where as sodium acetate completely dissociates into it’s ions.

Therefore, the buffer mixture contains large number of CH₃COO^– ions followed by Na⁺, H⁺, CH₃COOH.
Case i) When an acid added to this solution :
H⁺ ion of the acid combines with CH₃COO^– ion in buffer solution and makes equillibrium with acetic acid.
CH₃COO^– + H⁺ ⇌ CH₃COOH
As a result pH remains constant.

Case ii) When a base is added to this solution :
OH^– ion of the base combines with H⁺ ion in buffer solution forms water molecule.
H⁺+ OH^– → H₂O.
As a result pH remains constant.

b. The pOH of a solution is 5.725. Calculate the [H⁺].
Answer:
pH + pOH = 14   pH = 14 – pOH = 14 – 5.725 = 8.275
pH = –log₁₀ [H⁺]  ⇒ 8.275 = log₁₀ [H⁺] ⇒
[H⁺] = antilog (- 8.275) = antilog(9 – 8.275 – 9).
[H⁺] = antilog (0.275 – 9) = 5.309 × 10^-9 mol dm^-3.

Question 29
a. Describe an experiment to determine the percentage of nitrogen in an organic compound by Kjeldahl’s method.
Answer:
Principle : When a nitrogenous organic compound is heated with cone. H₂SO₄ using (CUSO₄K₂SO₄) as a catalyst, the nitrogen from the compound is quantitatively converted to ammonium sulphate.

This ammonium sulphate is decomposed by heating with excess of alkali and the ammonia evolved is absorbed in known excess of a standard solution of H₂SO₄. Part of acid is neutralized by ammonia. The excess of acid left behind after neutralization with ammonia is estimated by back titration with standard alkali. From this, the amount of acid actually consumed by ammonia can be obtained which can be used to determine the percentage of nitrogen in the compound.

Procedure : A known exact mass of organic compound (about 0.5g) ix mixed with 10g K₂SO₄, Ig CuSO₄ and 25 ml of conc. H₂SO₄. The mixture is heated strongly in a Kjeldahl’s flask. Till the contents become clear. This step is known as digestion.

The Kjeldahl’s flask is now cooled and the liquid is heated in a round-bottomed flask with excess of caustic soda solution The ammonia evolved is absorbed in a known volume of a standard acid.

The amount of unreacted acid is determined by titrating it against a standard alkali. (NH₄ )₂ SO₄ + 2NaOH → 2NH₃ ↑+Na₂SO₄ + 2H₂O

Observation :

• Mass of organic compound taken = Weight
• Normality of standard acid = N
• Volume of standard acid taken = V₁ ml
• Volume of alkali (Normality = N) required for back titration = V₂ ml

Calculation: Volume of acid used up by ammonia = Volume of ammonia produced = (V₂ – V₁) = V ml of normality N.
Now,
1000 ml of 1 normal NH₃ = 17g NH₃ = 14g N₂
∴ V ml of N – normal ammonia will contain \(\frac{14 \times \mathrm{N} \times \mathrm{V}}{1000}\)gN₂
This amount of nitrogen was present in w g of the compound
∴ %N₂ = \(\frac{14 \times N \times V}{1000} \times \frac{100}{w}\) or %N₂ = \(\frac{1.4 \mathrm{NV}}{\mathrm{w}}\)
Where, N and V are the normality and volume respectively of the acid used up by ammonia.

b. What are isomorphous salts ? Give two examples.
Answer:
Isomorphous salts are those which have same crystalline structure , e.g., MgSO₄, 7H₂O and ZnSO₄, 7H₂O are isomorphous.

Question 30
a. Discuss the molecular orbital theory of benzene.
Answer:
The electronic configuration of carbon in the ground state is
                                                         ’
In the excited state of carbon, one of the 2s electrons gets promoted to 2p orbital. So the electronic configuration of carbon in the excited state is

• All the six carbon atoms in benzene undergoes sp² hybridization to form three sp² hybrid orbitals.
• Each carbon atom in benzene possesses an unhybridised p-orbital containing one electron. They overlap sideways forming three π bonds. The π electrons are delocalised.
• Benzene is a planar molecule with a π electron cloud above and below the ring as shown in above structure. The bond angles are 120°.
• Bond length is 0.139 nm and dipole moment is zero.
• It has twelve a bonds and three π bonds.
• Two carbon atoms to form two sigma bonds and one hydrogen atom to form another sigma bonds.

Part – E

Answer any Three of the following questions:  ( 3 × 5 = 15 )

Question 31
a. Calculate the mass of hydrochloric acid in 200 cm³ of 0.2 N solution of it.
What volume of this acid solution will react exactly with 25 cm³ of 0.14 N solution of sodium hydroxide?
Answer:
Normality of hydrochloric acid = 0.2;
Eq. mass of hydrochloric acid = 36.5
Mass of hydrochloric acid in one dm³ of the solution is given by
W = N × E = 0.2 × 36.5 g
∴ mass of hydrochloric acid in 200 cm3 of the solution = \(\frac{0.2 \times 36.5}{5}\) = 1.46 g

Substituting we have V₁ × 0.2 = 25 × 0.14
∴V₁ = \(\frac{25 \times 0.14}{0.2}\) = 17.5 cm³

b. Write the difference between isotope & isobars.
Answer:

Isotopes	Isobars
Different atoms of the same elements having same atomic number but different mass number are called Isotopes. Ex: 1H1 (Protium), 1H2 (Deuterium), 1H3 (Tritium) 17Cl35 , 17CS37	Element having different atomic number but same mass number are called Isobars. 18Ar40 , 19K40 , 20Ca40

Question 32
a. 50 ml of oxygen were collected at 10° C under 750 mm pressure. Calculate volume at STP.
Answer:
V₁ = 50 ml    V = ?
P₁ = 750 mm   P₂ = 760 mm
T₁ = 10 + 273=283K   T₂ = 273.15K

b. Give any two differences between ideal and real gas.
Answer:

Ideal Gas	Real Gas
1. Obeys ideal gas equation PV = nRT	1. Does not obey the ideal gas equation but obeys Van der walls equation.
2. Ideal gas does not exist in nature.	2. All gases which exist in nature are real gases.

Question 33
a. A buffer solution of pH 8.3 is prepared from ammonium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8 × 10⁵. What is the mole proportion of ammonium chloride and ammonium hydroxide?
Answer:
pOH= 14 – pH = 14 – 8.3 = 5.7

b. Write the conjugate acid for the following.
(i) C₂H₅NH₂ (ii) CH₃COO^–
Answer:
(i) C₆H₅NH₃⁺ (ii) CH₃COOH

Question 34
a. Describe in brief the manufacture of caustic soda using the Castner-Kellner cell.
Answer:
The Castner-Kellner cell consists of large rectangular trough divided into three compartments with partition short of reaching the bottom of the tank Thus mercury in one compartment can flow into another but solution cannot mix. Graphite anodes are used in outer compartments filled with NaCl solution. The middle compartment contains very dilute solution of caustic soda and filled with iron rods as cathode.

On passing electric current Cl₂ is liberated in outer compartments and sodium liberated at cathode. Mercury forms amalgam which is passed into middle compartment in which mercury acts as anode (having induced +ve potential).
At anode Na⁺ + e^– → Na
At cathode Na + Hg → Na – Hg
2 (Na – Hg) + 2H₂O  → 2NaOH + Hg + H₂
The concentration of NaOH goes on increasing in the middle compartment. When the concentration of NaOH reaches 20% the solution is replaced by dilute solution.

b. Give reason for diagonal relationship of lithium with magnesium.
Answer:
Both Lithium and magnesium have small size and high charge density. The electro negativities of Li is 1.0 and Mg is 1.2. They are low and almost same. Their ionic radii are similar. Hence they show similarities which is known as diagonal relationship between first element of a group with the second element in the next higher group.

Question 35
a. Describe the following in brief:
(i) Ozone depletion over Antarctica (do not write reaction)
(ii) BOD and COD
(iii) Eutrophication
Answer:

(i) Ozone depletion over Antarctica : In summer season, nitrogen dioxide and methane react with chlorine monoxide preventing ozone depletion whereas in winter, special type of clouds called polar stratosphereic could s are formed over Antartica. These clouds provide surface on which chlorine nitrate molecules formed gets hydrolysed to form HOCl. It also reacts with HCls to give Cl₂. When sunlight returns to Antarctica again ozone depletion starts by free radicals.

(ii) BOD : It is amount of oxygen required by bacteria to decompose organize wastes present in water.

COD : It is the amount of oxygen (in ppm) required to oxidize the contaminants. COD is determined by using chemical oxidizing agent K₂Cr₂O₇.

(iii) Eutrophication : The process in which nutrient enriched water bodies support a dense plant polulation which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as Eutrophication.

b. Distinguish between photochemical smog and classical smog.
Answer:

Photochemical smog	Classical smog
1. It is formed by oxides of nitrogen, hydrocarbons, etc.,	1. It is formed by oxides of sulphur, carbon and particulate matter from combustion.
2. It is oxidizing in nature.	2. It is reducing in nature.
3. Photochemical smog occurs in cities having large number of vehicles. It is not harmful.	3. It occurs in both urban and rural areas. It is harmful.

Karnataka 1st PUC Chemistry Model Question Paper 3 with Answers

Time: 3.15 Hours
Max Marks: 70

Instructions:

1. The questions paper has four parts A, B, C and D.
2. In part A each question carries ONE Marks. In part B, each question carries TWO marks, in part C each question carries FIVE marks. In Part D-1 carries TEN Marks and each question in D-2 carries FIVE marks
3. Write balanced chemical equations and draw diagram wherever necessary.

Part – A

I. Answer of all the following questions: (1 × 9 = 9)

Question 1.
State “Law of definite proportions”.
Answer:
The given compound always contains exactly the same proportion of elements by weight.

Question 2.
Name the fundamental particle of an atom that has highest value for its e/m
Answer:
Electron.

Question 3.
Write the two resonance (canonical/contributing) structures of Ozone.
Answer:

Question 4.
A molecule XY₄ has four bond pairs and 2 lone pairs of electrons for its central atom. Predict the shape of the molecule.
Answer:

Question 5.
How many valence electrons are present around phosphorous in PCls?
Answer:
P – 15 = 1s² 2s² 2p⁶ / 3s² 3p³.

Question 6.
What is the change in internal energy of a system, if 10J of heat is supplied to it and 15J of work is done by it?
Answer:
Since it is a closed system.
∆u = q – w
∆u = 10J – 15J   ∆u= -5J

Question 7.
H^– is a Lewis base. Give reason.
Answer:
H^– can donate a pair of electron. Hence it is a lewis base.

Question 8.
What is the composition of water gas ?
Answer:
CO(g) + H₂(g) or a mixture of carbon monoxide and dihydrogen gas is called as water gas.

Question 9.
Which alkali metal is the strongest reducing agent ?
Answer:
Lithium of Li.

Question 10.
Draw the staggered conformation of Ethane.
Answer:

Question 11.
Mention one use of Chromatography.
Answer:
Chromatography is used to:

• Separate mixtures into their components
• purify components
• test the purity of compounds

Part – B

II. Answer any Five questions: ( 5 × 2 = 10 )

Question 12.
(i) How many significant figures are in 0.2500 g?
(ii) If the mass of one molecule of water is 18 amu, what is the mass of one mole of water molecules ?
Answer:
(i) Four
(ii) 18g.

Question 13.
(i) How does the atomic radius very down a group in the periodic table ?
(ii) Arrange the following in the decreasing order of their ionic radius : N^-3, Mg⁺², Na⁺¹, O^-2
Answer:
(i) Atomic radii increases down a group in the periodic table.
(ii) N^-3 > O^-2 > Na⁺ > Mg²⁺

Question 14.
Among N, Cu, Rn and U, identify the element that: (i) belongs to d-block, (ii) is an actionid
Answer:
(i) Cu,
(ii) U

Question 15.
(i) State Charles’ Law.
Answer:
Charle’s law states that volume of fixed mass of gas is directly proportional to absolute temperature at constant pressure.
\(\left(\frac{\mathrm{V}_{1}}{\mathrm{T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{T}_{2}}\right)_{\mathrm{P}}\)

(ii) Give the relationship between molecular mass and density of a gas.
Answer:
(ii) d = \(\frac{P M}{R T}\)

Question 16.
What happens when (i) Sodium hydride is treated with water ?
(ii) Hydrogen peroxide is treated with lead sulphide ?
Answer:
(i) Sodium hydride when treated with water liberate dihydrogen gas
NaH(s) + H₂O(I) → NaOH(l) + H₂(g)

(ii) Hydrogen peroxide on reacting with lead sulphide produces lead sulphate.
4H₂O₂ + PbS → PbSO₄ + 4H₂O

Question 17.
What is the repeating unit in Organo silicon polymer? Name the starting (raw) material used in the manufacture of Organo silicon polymer.
Answer:
R₂ SiO. The starting material is alyl or aryl substituted silicon chlorides.

Question 18.
How is Ozone layer formed in the stratosphere? Name a chief chemical that causes its depletion.
Answer:
In stratosphere the free oxygen atoms combine with the molecular oxygen is presence of UV radiations to form ozone.

The chemical that causes its depletion is chlorofluorocarbon compounds or CFCs or fireons.

Part – C

I. Answer any Four questions: (4  × 4 = 16)

Question 19.
(i) State Heisenberg’s uncertainty principle. Give its mathematical equation.
Answer:
It is impossible to determine both the momentum (particle nature) and position (wave nature) of a moving sub atomic particle simultaneously with absolute accuracy.
Mathematically ∆x × ∆p = h / 4π where ∆x = uncertainty in position:
∆p = uncertainty in momentum ; h = Plank’s constant = 6.626 × 1O^-34 Js.

(ii) Calculate the wave number of the spectral line of shortest wave length appearing in Given, R = 1.09 × 107 m1
Answer:
\(\overline{\mathbf{v}}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right) \mathrm{m}^{-1}\)
v̄ = 1.09 × 10⁷ m^-1 \(\left(\frac{1}{2^{2}}-\frac{1}{0}\right)\)
v̄ = 1.09 × 10⁷ (\(\frac{1}{4}\))
v̄ = 0.2725 × 10⁷   v̄ = 2.725 × 10⁶ m^-1

Question 20.
(i) Mention two conditions for the linear combination of atomic orbitals.
(ii) Write the electronic configuration of C₂ molecular. What is its magnetic property ?
Answer:
(i) (a) The combining atomic orbitals must have the same or nearly the same energy
(b) The combining atomic orbitals must have the same symmetry about the molecular axis.
(c) The combining atomic orbitals must overlap to the maximum extent.
It is diamagnetic. E.C. of C₂ molecule is \(\sigma 1 \mathrm{s}^{2} \sigma^{*} 1 \mathrm{s}^{2} \sigma 2 \mathrm{s}^{2} \sigma^{*} 2 \mathrm{s}^{2} \pi 2 \mathrm{px}^{2}=\pi 2 \mathrm{py}^{2}\)

Question 21.
(i) Define “Standard Enthalpy of Vapourisation”
(ii) Write thermo chemical equation for vaporization of Ethanol (C₂H₃OH)
(iii) Calculate the enthalpy of vapourisation of ethanol, given enthalpies of formation of liquid Ethanol and gaseous Ethanol as -277.6 kJ and -235.4kJ respectively.
Answer:
(i) Amount of heat required to vaporize one mole of a liquid at constant temperature under standard pressure (1 bar) is called standard enthalpy of vaporization.
(ii) C₂H₃OH(l) → C₂H₃OH(g) ; ∆H = +QkJ
(iii) ΔH_vap= Hp-Hr .
ΔH_vap =-235.4kJ-(-277.6kJ)
ΔH_vap = 42.2kJ

Question 22.
(a) For
(i) What is the oxidation number of Oxygen in (2) ?
(ii) What type of Redox reaction is it?
(b) Balance the Redox reaction using oxidation number method.
SO₂ + H₂S → S + H₂O
Answer:
(i) -2
(ii) Disproportionation or Disproportionate reaction.

(b)

Multiply oxidation equation by 2 and reduction equation by 1.
SO₂ → S
2H₂S → 2S
Add both the equations
SO₂ + 2H₂S → 3S
Balancing the atoms of hydrogen and oxygen by introducing suitable number of H₂O molecules to the product.

Question 23.
(a) How is sulphur detected using the ‘Sodium fusion extract’ of the given organic compound ?
(b) Give the IUPAC name of

Write the bond line diagrame of 2-methyl pentane.
Answer:
(a) Te sodium fusion extract is acidified with acetic acid and lead acetate. A black precipitate of lead sulphite is formed which indicates the presence of sulphur.

(b) Cyclohexanol

Question 24.
(a) Give equation for each of the following reactions.
(i) Water is dropped on Calcium carbide
(ii) Hydrogen bromide is added to propene in presence of peroxide
(iii) Phenol is heated with Zinc dust.
(iv) Benzene is treated with Chlorine in presence of Ferric chloride.
Answer:
(a) (i) CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

(b) (i) Write the equation for the steps involved in the mechanism of nitration of Benzene between Toluene and Nitrobenzene which one is more reactive towards Nitration ?
Answer:
Nitration benzene reacts with a mixture of concentrated nitric acid and concentrated sulphuric acid at 50°C to form nitrobenzene.

Mechanism: This involves the following steps
Step 1: Generation of electrophile nitronium ion NO₂⁺
HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺+ 2HSO₄^–

Step 2: The electrophile NO₂⁺ attacks the benzene ring to form a carbocation which is resonance stabilized.

Step 3: Loss of proton to give nitrobenzene. The proton is removed by HSO₄^–

Part – D

Answer any Four questions: ( 4 × 5 = 20 )

Question 25
a. Calculate the mas of Magnesium required to completely react with 250 cm³ of O.1 M HCl.
Given : Mg + 2HCl – MgCl₂ + H₂    Atomic mass of Mg = 24
Or (Internal choice)
100cm³ of a solution of HCl completely neutralizes 25cm³ of 0.1 M NaOH.
Calculate the mass of HCl present in 100cm³.
(b) Mention two postulates of ‘Dalton’s atomic theory’.
Answer:
(a) 250 × 0.1 M = 25 millimoles of HCl
From the equation it is clear that 2HCl is reacting with one Mg.
∴ 25 millimoles of HCl rect with 12.5 millimoles of Mg
∴ Mass of magnesium required = volume of Mg × atomic weight = 12.5 × lO^-3 × 24 = 0.3 g of Mg.
OR
(a) V_Hcl = 100cm³, V_NaOH = 25cm³, MHCl = ?, M_NaOH = 0.1M
aHcl V_Hcl × M_Hcl = V_NaOH × M_NaOH × aNaOH
here aHCl = basicity of HCl = 1, aNaOH = acidity of NaOH = 1
M_Hcl = \(\frac{25 \times 0.1}{100}\) = 0.025m
Mass of HCl present in 100cm³

b. Mention two postulates of ‘Dalton’s Atomic theory’
Answer:
The postulates of Dalton’s atomic theory are :

1. The matter consists of indivisible atoms.
2. All the atoms of a given elements have identical properties including identical mass. Atoms of different element differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
   OR (Internal choice)

What is Empirical formula ? Give an example for a compound whose Empirical formula and molecular formula are the same
Answer:
Emperical formula represents the simplest whole number ratio of various atoms present in a compound.
Example : C₂H₈.

Question 26.
(a) Define the terms:
(i) Bond order
(ii) Bond length
(iii) Bond enthalpy
Answer:
(i) It is the number of covalent bonds holding the atoms in the molecule.
Example: If the bond is formed by the sharing of two electron pairs, then the bond order is 2. O=O or C = C bond in alkenes.
(ii) Bond length: It is the equilibrium distance between nuclei of two bonded atoms in a molecule.
(iii) Bond enthalpy: It is the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.

b. With respect to the formation of Ethane molecule mention.
(i) hybridation of Carbon. (ii) number of sigma bonds in the molecule.
Answer:
(i) sp²
(ii) 5

Question 27.
(a) Write three postulates of ‘Kinetic theory of gases’.
Answer:

1. Gases are made up of large number of the minute particles.
2. Pressure is exerted by a gas
3. There is no loss of kinetic energy.
4. Molecules of gas attract on one another.
5. Kinetic energy of the molecule in directly proportional to absolute temperature.
6. Actual volume of the gaseous molecule very small.
7. Gaseous molecules are at always in motion.
8. There is more influence of gravity on movement of gaseous molecule.

(b) Two gases A & B have critical temperatures as 250 and 125 K respectively. Which one of these can be liquefied easily and why ?
Answer:
Gas a can be easily liquefied. Because higher critical temperature, greater is the inter molecular force of attraction.

Question 28.
(a) Calculate the pOH of a solution obtained when 0.05 mol NH₄Cl is added and dissolved in 0.025M ammonia solution. Kb for ammonia is 1.77 × 10^-5
(b) For the equilibrium : BaCO₃(s) ⇌ BaO(s) + CO₂(g)
(i) Write the expression of Kp.
(ii) What is the effect of ‘increase in pressure’ on the above equilibrium ?
Answer:
(a) pK_b = – log[Kb]
= – log [1.77 × 10^-3]
= 5 – 0.2480 = 4.752

pOH= 4.752 + log2 = 4.752+ 0.3010 = 5.053

(b) (i) K_p = P_CO2
(ii) Increase in pressure shifts the equilibrium to left or backward reaction is favoured.

Question 29.
(a) Give reason (i) Coodination number of Be is 4, but that of Mg is six
(ii) Lithium iodide is covalent but potassium iodide is ionic.
Answer:
(a) (i) Beryllium does not have ‘d’ subshell in n = 2 level. It has only 4 valence orbital.
Hence it can have a coordination number 4.
Mg has ‘d’ subshells in n = 3 level. Hence it has more valence orbital. Therefore it can have a coordination number 6.

(ii) Lithium iodide is covalent because of the smaller atomic size of lithium atom, it distorts to the electron to the electron cloud of iodine towards it. Or K has less polarizing power than Li⁺.

(b) Compared the 2^nd Ionisation enthalpies and Hydration enthalpies of Alkali and Alkaline earth metals/ions.
Answer:
Alkali metals have higher value of 2^nd ionization enthalpy than alkaline earth metals.
since they achieve a stable noble gas configuration after loosing 1 electron from ns¹ shell.

Alkaline earth metal ions have higher value of hydration enthalpy because of their smaller ionic size than alkali metals.

(c) What is the chemical formula of plaster of paris ?
Answer:
CaSO₄. 1/2H₂O

Question 30
(a) Name the method by which Halogen present in an organic compound is estimated ? O.1 g of an organic compound gives 0.08g of silver bromide, Calculate the percentage of bromine.
Atomic masses : Ag = 108, Br = 80
(b) What is inductive effect ?
Which one of the following shows maximum hyper conjugation effect ?
CH₃CH = CH₂, (CH₃)₂ = C = CH₂        CH₂ = CH₂
Answer:
(a) Carius method

molecular mass of AgBr = 108 + 80 = 188
Mass of silver bromide = 0.09g
Mass of organic compound = O.1 g
% of bromine = \(\frac{80 \times 0.08 \times 100}{188 \times 0.1}\) = 34.04%

(b) The percentage displacement of bonded sigma electrons along the chain of carbon atoms due to the presence of a substitutent group of different electro negativity is called inductive effect.
(CH₃)₂ C = CH₂

Part – E

Answer any three questions: (3 × 5 = 15)

Question 31.
(a) For the Element with atomic number 24:
(i) Write the electron configuration
(ii) Write the value of u & 1 for its electron in the valence shel.
(ii) How many unpaired electrons are present in it ?
Answer:
(a) (i) 1s² 2s² 2p⁶ 3s² 3p⁶, 4s¹ 3d⁵
(b) n = 4, l = 0
(c) Six or 6.

(b) What is photo electric effect ? Does the effect support particle nature or wave nature of light ?
Answer:
The effect in which the electrons gets ejected when certain metals are exposed to a beam of light is called as photo electric effect.
It supports the particle nature of light.

Question 32.
(a) What is a spontaneous process ?
For the equilibrium A + 2B ⇌ C
∆H is + 400 kJ and ∆S is + 200 JK^-1
Calculate the temperature above which the reaction becomes spontaneous ?
Answer:
The process that takes place on its won without the assistance of external agency is called spontaneous process.
∆G = ∆H – T∆S
At equilibrium, ∆G = 0
T∆S = ∆H

T = 2000 K.

(b) For Cl₂ → 2Cl Assign the sigans for ∆G and ∆S.
Answer:
∆H = +ve (endothermic reaction)
∆S = -ve (entropy increases)

Question 33.
(a) Calculate the solubility of Ag₂CrO₄4 in 0.1M AgNO₃K_sp of Ag₂ CrO₄ = 1 × 10^-12
Ans:

K_sp = [Ag⁺]² [CrO₄^2- ]
1 × 10^-12 = (2 x 0.1)² [CrO₄^2- ]

b. An aqueous solution of sodium acetate has pH greater than 7.
Explain with equation.
Answer:
Sodium acetate is a salt of a strong base and a weak acid which undergoes hydrolysis to form strong alkaline (basic) solution. Hence the pH is greater than 7.
CH₃COONa + H₂O ⇌ CH₃COOH + NaOH

Question 34.
(a) Give reasons :
(i) The stability of +3 oxidation state of 13 group elements decreases down the group.
(ii) Boron is used as control rods in nuclear reactors.
(iii) Graphite is soft and slippery.
Answer:
(a) (i) The stability of +3 oxidation state of group 13 elements decreases down the group due to the inert pair effect on ns2 electrons. This is because of the poor shielding effect of electron in ‘d’ and ‘f’ orbitals.
(ii) Because it absorbs neutrons
(iii) Due to the weak vander waal’s force of attraction between the layers of hexagonal rings.

b. Complete the following equation :
(i) 2Al + 2 NaOH + → 6H₂O
(ii)
Answer:
(i) 2Al + 2NaOH + 6H₂O → 2Na + [Al(OH)₄ ] + 3H₂
(ii)

Question 35.
(a) Give two tests to distinguish between alkanes and alkenes.
(b) Naphthalene is an aromatic compound justify the statement using Huckle rule.
(c) Draw cis and trans structures of CHBr = CHBr
Answer:
(a) (i) Test with bromine water or CCl₄
Alkenes react with bromine water of CCl₄ to form dibromo alkane, Where as alkanes do not show such rection.
(ii) Test with alkaline. KMnO₄ or Baeyer’s reagent test:
Alkenes react with Baeyer’s reagent to produce glycol, Where as alkenes do not show such reaction.

(b) According to Hucle rule, if naphthalein is an aromic compound it should possess (4n + 2) π electrons delocalized in the ring.
In naphthalein, n = 2
Number of delocalized p electrons = (4n + 2) = 4 × 2 + 2
= 10π electrons
Hence naphthalein is an aromatic compound.

(c)

Karnataka 1st PUC Chemistry Model Question Paper 4 with Answers

Time: 3.15 Hours
Max Marks: 70

Instructions:

1. The questions paper has four parts A, B, C and D.
2. In part A each question carries ONE Marks. In part B, each question carries TWO marks, in part C each question carries FIVE marks. In Part D-1 carries TEN Marks and each question in D-2 carries FIVE marks
3. Write balanced chemical equations and draw diagram wherever necessary.

Part – A

I. Answer of all the following questions: (1 × 9 = 9)

Question 1.
What is the S.I. Unit of density ?
Answer:
Kg/m² or kgm^-3

Question 2.
How do isotopes of an element differ from one another ?
Answer:
Atoms having same atomic number but different atomic masses.
OR
Isotopes of an element differ only in the number of protons.

Question 3.
What are representative elements ?
Answer:
Elements of s-block and p-block are representative elements.

Question 4.
Between CO & CO₂, which diffuses faster ?
Answer:
CO diffuses faster.

Question 5.
Give a chemical reaction for which ∆H = ∆U.
Answer:
2H_(g) ⇌ H_2(g) + I_2(g)

Question 6.
Give the relation between K_p and K_c for the equilibrium N₂(g) + 3H₂(g) ↔ 2NH₃(g).
Answer:
K_p = K_c = (RT)²

Question 7.
What is the chemical used in Clarke’s process to remove temporary hardness of water ?
Answer:
Lime or Ca(OH)₂ or Calcium hydroxide.

Question 8.
Write the General outer electronic configuration of s-block elements.
Answer:
General electronic configuration of s-block elements is ns^1-2

Question 9.
Complete the following equation 2Fe⁺² + 2H⁺ + H₂O₂ → ……. + H₂O
Answer:
2Fe⁺²+ 2H⁺ + H₂O₂ → 2Fe⁺³+ 2H₂O

Question 10.
Suggest a suitable method to separate sugar and salt from an aqueous solution containing them.
Answer:
Fractional crystallization is a suitable method to separate sugar and salt from an aqueous solution containing them due to difference in their solubilities in water.

Question 11.
Trans-2 Butene has higher melting point than cis-2-butene, why ?
Answer:
Because more close packing of molecules is possible in trans-isomer due to more symmetry.

Part – B

II. Answer any Five questions: (5 × 2 = 10)

Question 12.
Name the quantum number that specifies
(i) Size of an orbital
(ii) Shape of an orbital.
Answer:
(i) Principal quantum number specifies size of an orbital
(ii) Azimuthal quantum number specifies shape of an orbital.

Question 13.
(i) Distinguish between Sigma and Pi bonds.
Answer:

Sigma (σ) Bond	Pi (π) Bond
1. The bond is formed by the axial overlap of the atomic orbitais.	1. The bond is formed by the sidewise overlap of the atomic orbitais.
2. s-orbitais can take part in the sigma bond formation	2. s-orbitais do not take part in the pi-bond formation.

Question 14.
Give the reasons for the deviation of real gases from ideal behaviour ?
Answer:

1. There is no intermolecular force of attraction within gaseous molecules (at high pressure and low temperature there is appreciable intermolecular force of attraction).
2. The actual volume of the molecule is negligible as compared to total volume of the gas. (At high pressure the gaseous molecule occupy minimum volume, which is not negligible).

Note: Because of this deviation Vander waal modified the ideal gas pressure and ideal gas volume respectively.

Question 15.
Why do group 1 metals have lower first ionisation Enthalpy than corresponding group 2 metals.
Answer:

1. Size of 1^st group metals is greater than size of 2^nd group metals, hence effective nuclear charge increase of 2^nd group metal.
2. Upon losing one electron, 1^st group metals attain stable octet structure.

Question 16.
Justify the position of Hydrogen in the periodic table.
Answer:
Hydrogen has been placed at the top of the alkali metal family.
Hydrogen and alkali metals resemble in the following aspects.

1. Electronic configuration : Hydrogen has one electron in its valance shell like the alkali metals H – 1s¹.
2. Electropositive character : As both hydrogen and alkali metals form monovalent cations by losing the electron in the valance shell. H → H⁺ + e^–

Question 17.
What happens when formic acid is heated with Cone. H₂SO₄? Give the equation.
Answer:
Dehydration takes place or CO is formed

Question 18.
What is meant by Biochemical Oxygen Demand (BOD)? What is its significants?
Answer:
The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water is called Biochemical Oxygen Demand (BOD) The amount of BOD in the water is a measure of the amount of organic material in the water, in terms of how much oxygen will be required to break it down biologically.

Part – C

I. Answer any Four questions: (4 × 4 = 16)

Question 19.
A compound contains 4.07% hydrogen, 24.27% Carbon and 71.65% Chlorine. Its molar mass is 98.96. Calculate is Empirical and Molecular formulae.
Answer:

Empirical formula = CH₂Cl
Molecular formula = n × Empirical formula

Molecular formula = 2 × CH₂Cl = C₂H₄Cl₂.

Question 20.
(a) Explain the formation of methane molecule on the basis of hybridization.
Answer:
The electronic configuration of the central carbon atom (z = 6) in methane is 2, 4. Each of the four electrons present in the valence shell forms shared pairs with the electrons of the hydrogen atoms as shown below :

Methane has a tetrahedral structure which is multiplanar, in which carbon atom lies at the centre and the four hydrogen atoms lie at the four comers of regular tetrahedran. All H – C – H bond angles are 109.5°

(b) Between O₂ & O₂^–  , which one has higher bond order ?
Answer:
O₂ has higher bond order than O₂^–

Question 21.
(a) What is Oxidation Number ? What is the oxidation Number of Cl is KClO₃?
Answer:
Oxidation number is the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that the electrons in a co-valent bond belong entirely to the more electro negative elements.
Oxidation number of Cl in KClO₃ is +5.

(b) Write separate equations for the oxidation and reduction reactions occurring in the following redox reaction.
2Fe + 2HCl → FeCl₂ + H₂
Answer:
2Fe + 2HCl → FeCl₂ + H₂
Fe → Fe²⁺ +2e^–
2H⁺ + 2e^– → H₂

Question 22.
(a) What are Isothermal and Adiabatic process ?
Answer:
Isotherman process : A process is said to be isothermal in case the temperature of the system remains fixed or constant i.e., operation is carried at constant temperature.

Adiabatic process : In an adiabatic process, no exchange of heat between the system and the surroundings is possible i.e., the system is completely isolated or insulated from the surroundings.

(b) State the II law of Thermodynamics. Give the equation that relates Gibbs energy with Entropy and Enthalpy of a system ?
Answer:
Second law of thermodynamics can be stated in a number of forms:

• All spontaneous processes or naturally occurring processes are thermo dynamically irreversible.
• Without the help of an external agency, a spontaneous process cannot to reversed.
• The entropy of an isolated system increase if it is to be spontaneous in a particular direction.
• In a non-isolated system, the total entropy of both the system and surroundings must increase or must be positive.
• The total entropy of the universe must tend to increase in a spontaneous process.
  G = H – TS

Question 23.
(a) Giving justification, catogorise the following species as Nuclophile or Electrophile.
(i)SO₃, (ii) C₂H₅O
Answer:
(i) SO₃ – Electoplite, because it is an electron deficient and accepts a pair of electrons.
(ii) C₂H₂O – Nucleophile because it is an electron rich and donates electrons.

OR (Internal Choice)

Mention the differences between Fractional distillation and Steam Distillation.
Answer:

Fractional distillation	Steam distillation
1. Liquid. and impurities must process different boiling points.	1. Liquid must be steam voilatile impurities present must be no volatice.
2. Liquid may or may not be mixable with water	2. Liquid must not mixable with water.

(b) Write the IUPAC names of the following compounds.
Answer:

OR (Internal Choice)

Explain position isomerism with suitable example.
Answer:
The isomers differ with respect to the position of the multiple bonds, substituents or functional groups
Ex .

Question 24.
(a) Give equations for the reaction that occurs when
(i) Ehyene is treated with Baeyer’s reagent.
(ii) Enthyene is treated with sodium metal in the ratio I : 2.
Answer:

(b) Explain Friedel – Craft’s acylation with an example.
Answer:
Friedel-Craft’s acylation : Reaction between benzene and acetyl chloride in presence of anhydrous AlCl₃, is called Friedel-craft’s acylation.

Part – D

Answer any Four questions: (4 × 5 = 20)

Question 25.
(a) Write any three postulates of Bohr’s atomic model.
Answer:
Bohr’s Model of an atom, the postulates are

1. Electrons revolve around the nucleus of an atom in a certain definite path called Orbit or stationary state of shell.
2. The shells are having different energy levels denoted as K, L, M, N, ……
3. As long as the electron remains in an orbit, they neither absorb nor emit energy.
4. The electron can move only in that orbit in which angular momentum is quantized, i.e., the angular momentum of the electron is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\).

b. Calculate the energy with 5th orbit of hydrogen atom.
Answer:

Question 26.
Define enthalpy of formation.
Calculate Enthalpy of formation of C₂H_6(g) given the enthalpy of combustion of C₂H_4(g) C₂H_6(g) and H_2(g) as – 140 kJ/mol, -1550 kJ/mol and -286 kJ/mol respectively.
Answer:
(a) Heat of formation of a compound is the change in enthalpy produced when one mole of the compound is formed from its elements denoted as ∆H_f.

(b) C₂H_4(g) + CO_2(g) → 2CO_3(g) + 2H₂O_(g)           ∆_cH° = — 140kJ/mol … (1)
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(g)                                 ∆_cH° =-286kJ/mol …(2)
C₂H_6(g) + \(\frac{7}{2}\) O_2(g) → 2CO_2(g) + 3H₂O_(g)  ∆_cH° = -1550kJ/mol … (3)
Required equation
C₂H_4(g) + H_2(g) → C₂H_6(g)      ∆_fH° = ?
Reverse equation (3)
2CO_2(g) + 3H₂O_(g) → C₂H_6(g) + \(\frac{7}{2}\) O_2(g)   ∆_cH° = -1550kJ / mol …(4)
Add equation (1), (2) and(4)
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(g)
C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(g)
2CO_2(g) + 3H₂O_(g) → C₂H_6(g) + \(\frac{7}{2}\) O_2(g)
C₂H_4(g) + H_2(g) → C₂H_6(g)
∆_fH° = 15550 kJ/mol – 140kJ/mol – 286 kJ/mil
∆_fH° = 1124kJ/mol.

Question 27.
(a) (i) State Le-Chatelier’s principle,
(ii) What is common ion effect. Give an example.
Answer:
(i) Definition : “when a constraint applied to a system at equilibrium in a reversible reaction, the equilibrium shifts so as to nullify the constraint”.
[Constraint is change in temperature or pressure, addition of reactant or product]

(ii) Supression in degree of dissociation of a weak electrolyte by addition of a common ion is called common ion effect.
Example: CH₃COOH and CH₃COONa

OR (Internal Choice)

(i) What is the pH of 1M NaCl at 298K.
(ii) What happens to the pH of a solution of Acetic acid when some sodium acetate is dissolved in it? Explain
Answer:
(i) pH = 7
(ii) pH of the acid solution increases
(iii) Because of common ion effect dissociation of acetic acid decreased and hence [H⁺] decreases.

(b) Acetic acid has a dissociation constant of 1.8 × 10^-5, Calculate the pH value of the decinormal solution of Acetic acid.
Answer:

pH = -log[H⁺] = -log 01.34 × 10^-1 = 2.872

OR (Internal Choice)

Calculate the ionization constant of 0.05M weak acid if it’s degree of dissociation is 0.018.
Answer:
\(\mathrm{K}_{\mathrm{a}}=\frac{\alpha^{2}}{\mathrm{C}}=\frac{(0.018)^{2}}{0.05}=6.48 \times 10^{-3}\)

Question 28.
(a) Describe LCAO method for the formation molecular orbitals of Hydrogen. Write the energy level diagram for these orbitals.
Answer:
The formation of bonding molecular orbital when two atomic orbitals of hydrogen atoms (Is orbital) combine by the addition of their wave functions is shown below.

The atomic orbitals of these atoms may be represented by the wave function TA, TB Mathematically, the formation of molecular orbitals may be described by the linear combination of atomic orbitals that can take place by addition and by substruction of wave functions of individual atomic orbitals as shown below :

Ψ_mo = Ψ_A ± Ψ_B
Therefore, the two molecular orbitals σ and are σ’ formed as
σ = Ψ_A = ’Ψ_B
σ’ = Ψ_A = Ψ_B
The MO σ formed by the addition of atomic orbitals is called the bonding molecular orbital while the molecular orbital σ’ formed by the substraction of atomic orbital is known as antibonding molecular orbitals.

(b) Give two difference Bonding and Antibonding molecular orbitals.
Answer:

Bonding molecular orbital	Antibonding molecular orbital
1. A bonding molecular orbital is formed by the linear combination of two atomic orbitals when their wave functions are added.	1. An antibonding molecular orbital formed by the linear combination of two atomic orbitals when their wave functions are added.
2. It is formed when the lobes of the combining atomic orbitals have same sign	2. It is formed when the lobes of the combining atomic orbitals have opposite signs.

Question 29.
(a) What is the formulae of inorganic benzene ? Write the structure of Diborane and explain the nature of bonding in it.
Answer:
Formulae of inorganic benzene is B₃N₃H₆.
Structure of Diborane

Bonding in diborane : Each B atom uses sp3 hybrides for bonding. Out of the four sp3 hybrids one each B atom, one is without electrons shown in shaded region. The terminal B – H bonds are normal 2 – centre – 2 electron bonds but the two bridge bonds are 3 – centre – 2 – electron bridge bonds are also referred to as balance bonds.

(b) What are Fullurenes ? How are they prepared ?
Answer:
Carbon exhibits many allotropic forms: both crystallins as well as amorphous. Diamond and graphite are two well known crystalline forms of carbon, third form of corbon known as fullerences.

Fullerenes are made by heating of graphite in an electric are in the presence of inert gases such as helium or argon. The sooty material formed by condensation of vapourised Ca small molecular consists of-mainly C60 with smaller quantity of C70 and traces of fallerences consiting of even number of carbon atoms upto 350 or above.

Question 30 a.
(a) Writ the principles involved in the estimation of (i) Halogens (ii) Sulphur present in an organic compound.
Answer:
(i) Halogens : Halogens reacts with silver nitrate to for precipitate of silver halides.
X + Ag⁺ → AgX .
X represents a halogen -Cl, -Br or -1

(ii) Sulphur
S⁺+ [Fe(CN)₅NO]² → [Fe(CN)₅NOS]^4-
Sulphur reacts with nitropresside to form a violet colour complex.

(b) Name the element estimated by Kjeldahl’s method.
Answer:
Nitrogen is estimate by Kjeldahl’s method.

Part – E

Answer any Three questions: (3 × 5 = 15)

Question 31.
(a) Calculate the mole fraction of Ehtanol (C₂H₅OH) in the solution containing 20g of Ethanol and 100g of water.
Answer:
Weight of ethanol = 20g
Weight of water = 100g
Molecular mass of C₂H₅ – OH = – 46
No of moles of ethanol = \(\frac{20}{46}\) = 0.4348 moles.
n₈ = No of moles of water = \(\frac{100}{18}\) = 5.555 moles

X_A = 0.07259

(b) How are 2 mole NaOH and 2 Molar NaOH different ?
Answer:
2 mole NaOH
1 mole of NaOH = 40g of NaOH = 6.023 × 10²³ molecules.
2 mole of NaOH 2 × 40g = 80g of NaOH
= 2 × 6.023 × 10²³ molecules.
2 molar NaOH
1 molar NaOH = 40g of NaOH in 1000 ml of water
2 molar NaOH = 80g of NaOH in 100 ml of water
= 2 moles of NaOH dissolved in 1000 ml of its solution.

Question 32.
(a) Sate Gay Lussac’s law.
Calculate the pressure exerted by 4 mole of a gas occupying a volume of 1.5m³ at 100 K. Given R = 8.314 J/k/mol.
Answer:
(i) Gay Lussac’s law (Pressure – temperature relationship)
Gay Lussac’s law states that at constant volume, pressure of a fixed amount of a gas varies directly with the temperature.
(ii) Given n = 4 moles
V = 1.5m³, T = 100k, R = 8.314 J/k/mol,    P = ?

(b) Draw z vs P graph for an Ideal gas, and CO₂ gas
Answer:

Question 33.
(a) What are buffer solutions ? Given an example for an acidic buffer.
Answer:
The solution which resist the pH or pOH when an acid or base is added to it. Buffer solution is a mixture of weak electrolyte and its salt.
A solution containing a weak acid and its salt with a strong base.
(i) CH₃COOH + CH₃COONa
(ii) HCOOH + HCOOK

(b) Calculate the solubility of A2 x 3 pure water, Assuming that neither king of ion reacts with water. The solubility product of A2X3 Ksp = 1.1 × 10^-23 .Explain with equation.
Answer:
k_sp =[A³⁺]² [X^-2]³ = [2s]² [3s]³ =108s⁵
S⁵ = \(\frac{1.1 \times 10^{-23}}{108}\) = 1 × 10^-25
S⁵ = 1 × 10^-5 mol dm^-3

Question 34.
(a) Give equations for the reactions that occur when
(i) Sodium metal is dropped into water
(ii) Magnesium metal is heated in air.
(iii) Sodium peroxide dissolves in water.
Answer:

Hydrogen gas is liberated.

(ii) NO₂ gas is liberated.
2Mg(NO₃)₂ → 2MgO + 4NO₂ + O₂

(iii) H₂O₂ is formed.
Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

b. What happens when limited amount of CO₂ gas is passed into milk of lime ? Give equation.
Answer:
Lime water turns milky or white Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Question 35.
(a) Between -NO₂ and -Br, which one of these is a in eta directing group? Write equation for the conversion of Benzene into p-Bromonitrobenzene.
Answer:
-NO₂ is meta directing group.

(b) State anti Merkovnikoff s rule and explain with an example.
Answer:
In the presence of peroxide, addition of HBr to un symmetrical alkenes like propene takes place contrary to the Merkovnikoff s rule. This happens only with HBr but not with HCl or HI. This reaction is know as peroxide or Kharash effect or anti- Merkovnikoff s rule

Students can Download 1st PUC Chemistry Model Question Paper 3 with Answers, Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Model Question Paper 3 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
2. Write balanced chemical equations and draw labeled diagram wherever allowed.
3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

I. Answer ALL of the following (each question carries one mark): ( 10 × 1 = 10 )

Question 1.
‘Cis platin’ a medicine used in the treatment of which disease?
Answer:
Cancer/Cancer Tumour.

Question 2.
Write the mathematical expression for Boyle’s law.
Answer:
V ∝ \(\frac{1}{P}\) at constant temperature.

Question 3.
Give the example for gaseous reversible reaction for which Kp = Kc
Answer:
H_2(g) + I_2(g) ⇌ 2HI_(g)

Question 4.
Which group elements in the periodic table are called Noble gases?
Answer:
18th group.

Question 5.
What is the oxidation number of the element in its free state?
Answer:
0 or Zero.

Question 6.
Write the general electronic configuration of First group elements.
Answer:
ns¹

Question 7.
Why is boric acid considered as weak acid?
Answer:
Due to small size of B and presence of only six electrons in the valence shell i.e. protonic acid.

Question 8.
Mention the structure of SiO₄^4-.
Ans.
Tetrahedral structure.

Question 9.
Give an example for non-benzenoid compound.
Answer:
Pyridine

Question 10.
Write the expanded form of ‘CNG
Answer:

Part – B

II. Answer any FIVE of the following questions carrying TWO marks ( 5 X 2 = 10 )

Question 11.
Convert 37°C to °F.
Answer:
°F = \(\frac{9}{5}\) + °C +32 = \(\frac{9}{5}\) X 37 + 32 = 98.6°

Question 12.
Under what conditions of temperature and pressure real gases tend to behave ideally?
Answer:
Low pressure and high temperature.

Question 13.
What is dipole moment? What is its SI unit.
Answer:
It is the product of +ve or negative charge and the distance between the centre of them, μ = e x d
SI unit coulomb meter or cm.

Question 14.
Write any two diagonal relationship between Beryllium and Aluminium.
Answer:
(a) Both forms covalent compounds and soluble in organic compounds.
(b) Both BeCl₂ and AlCl₃ act as Lewis acids.

Question 15.
Give the reaction for the synthesis of water gas and producer gas.
Answer:

Question 16.
How is chloromethane converted to methane?
Answer:
Chloromethane heated with zinc and NaOH to methane.

Question 17.
Illustrate Markovnikov’s rule with an example.
Answer:

The Br^– of HBr is added to double bonded carbon as containing less of OH atoms to give 2–bromopropanol.

Question 18.
What is BOD? What is its significance?
Answer:
The amount of oxygen consumed by the microorganisms in decomposing the organic matter present in water is called BOD. It is a water quality parameter to know the amount of organic matter available for bacteria.

Part – C

III. Answer any FIVE of the following questions; carrying THREE marks: ( 5 x 3 = 15 )

Question 19.
(a) Give reason : Ionic radius of F^– is more than atomic radius of F.
Answer:
Due to more number of electrons in F^– than F.

(b) How does ionization enthalpy varies down the group?
Answer:
Ionisation enthalpy decreases down the group due to increase in atomic size of atoms.

(c) Ionization enthalpy of nitrogen is more than that of oxygen. Give reason.
Answer:
Due to completely half filled p-orbitals (1s² 2s² 2p³) in nitrogen but in oxygen atom p-orbital is more than half-filled.

Question 20.
With the help of MOT write the energy level diagram of hydrogen molecule. What is its bond order and predict magnetic property.
Answer:
BO = \(\frac{1}{2}\) (N_BMO N_ABMO)
= \(\frac{1}{2}\)(2 – 0 ) = 1
It is diamagnetic due to absence of unpaired electrons in the molecule.

Question 21.
Calculate the formal charge of each oxygen atom of ozone molecule
Answer:

Question 22.
(a) Give any two differences between sigma and pi bonds.
Answer:
Sigma Bond

1. Head to Head overlapping of AO.
2. It is strong bond due to large overlapping.
3. Free rotation is possible.

Pi Bond

1. Lateral (sideways) overlapping of AO.
2. It is weak bond due to small overlapping.
3. Free rotation about n bond is not possible.

(b) What is the magnetic nature of oxygen molecule?
Answer:
It is paramagnetic due to presence of unpaired electrons in π_px π_py antibonding molecular orbitals.

Question 23.
Balance the following redox reaction using oxidation number method.
MnO₂ + Br^– → Mn²⁺ +Br₂ + H₂O (In Acidic medium)
Answer:

Step II – Oxidation Reaction x 2
Reduction Reaction x 1

Balance the O atoms (add to RHS) and balance the H+ ions (To LHS)
MnO₂ + 2B^– + 4H⁺ → Mn⁺² +2Br₂ +2H₂O

Question 24.
(i) Explain the laboratory preparation of dihydrogen.
Answer:
In the laboratory, dihydrogen is prepared by adding dil. H2S04 to granulated zinc.
i.e. Zn + H₂SO₄ (dil) → ZnSO₄ + H₂
The H₂ gas liberated is collected by downward displacement of water.

(ii) Give an example of ionic hydride.
Answer:
LiH/NaH/CaH₂/BaH₂ etc.

Question 25.
Give any three points of differences between lithium and other alkali metals.
Answer:

1. Small size of lithium and its ion.
2. High charge to size ratio (polarising power).
3. High ionization enthalpy and low electropositive character of lithium.
4. Absence of d-orbitals.

Question 26.
(i) Write the molecular formula of silica.
Answer:
SiO₂

(ii) Write the partial structure of silicone.
Answer:

(iii) Name the catalyst used in gasoline production.
Answer:
ZSM-5.

Part – D

IV. Answer any FIV E of the following questions’carrying FIVE marks: ( 5 x 5 = 25 )

Question 27.
(a) An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 gram of this substance is subjected to complete combustion.
Answer:
Given % carbon = 69%, H = 4.8%, w = 0.20g

(b) Give an example each for element and compound.
Answer:
Element = sodium/Na.
Compound = NaOH/sodium hydroxide.

Question 28.
(a) Write the significance of quantum numbers n, I and m.
Answer:
(i) Principal quantum number (n): Determines energy and size of the orbits (electrons).
(ii) Azimuthal Q.N. (1): Determines the shape of the suborbitals (s,p,d,f).
(iii) Magnetic Q.N. (m): Determines the orientation of the orbitals.

(b) State Pauli’s exclusion principle.
Answer:
For two electrons of an atom the values of n, 1 and m but s is different.
OR
No two electrons of an atom can have the same values of n, 1 and m are same 3 is different.

(c) Write the electronic configuration of copper (Z = 29).
Ans. Cu29= 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

Question 29.
(a) The FM station of All India Radio, Hassan, broadcast on a frequency of 1020 kilohertz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter.
Answer:
Given γ = 1020 kilohertz = 1020 x 1000 hertz
C = 3 x10⁸ ms^-11 λ = ?

(b) What is the maximum number of electrons present in third main energy level?
Answer:
Third Main energy level = M r /.
∴ No. of electrons in M shell = 18e.

Question 30.
(a) Give any three postulates of kinetic molecular theory of gases.
Answer:

1. ll gases are made up of with many number of tiny particles called molecules.
2. Gas molecules are separated from each other by large distance. Hence the volume of the gas molecules is negligible compared to total volume of the gas.
3. There is no force of attraction between the gas molecules.
4. The average kinetic energy of gas molecules is directly proportional to Kelvin temperature.

(b) On a ship sailing in Pacific Ocean where temperature is 23.4°C, a balloon is filled with 2L air. What will be the volume of the balloon when the ship reaches 1 the Indian Ocean where temperature is 26.1°C?
Answer:
V₁= 2L V₂ = ?
T₁= 23.4°C + 273 = 296.4 K
T₂ = 26.1°C + 273K = 299.1K

Question 31.
(a) The combustion of one mole benzene takes place at 298K and 1 atm. After combustion, CO₂(g) and H₂O (l) are produced and 3267.0 kJ of beat is liberated. Calculate the standard enthaipy of formation of benzene.
Given : Standard enthalpy of formation of CO₂(g) and H₂O (1) are – 393.5 kJ mol^-1 and -285.0 kJ mol^-1 respectively.
Answer:
Required equation : 6C (s) + 3H₂(g) → C₆H₆ (l); ∆H_f = ?
(i) C₆H₆(l) + \(\frac{15}{2}\)O₂(g) → 6CO₂(g) + 3H₂O(l); ∆H_c = -3267KJ
(ii) C(s) + O₂(g) → CO₂ (g); ∆H_f = -393.5 KJ
(iii) H₂(s) + \(\frac{1}{2}\)O₂(g) → H₂O(l); ∆H_f= -285 KJ
reverse the equation (i) + (ii) 6 + (iii) * 3

(b) Write the mathematical expression for First law of thermodynamics.
Answer:
∆U = q +w or ∆U = q – P∆V

(c) Give an example of isolated system.
Answer:
Tea placed in a thermos flask.

Question 32.
(a) What are the values for ArH° and ArS° for reaction to be spontaneous at all
temperatures?
Answer:
When ∆_rH° = -ve and ∆_rS° = +ve the process is spontaneous at all temperture.

(b) Write the relationship between
(i) Enthalpy (H) and infernal energy (U) (ii) C_p and C_v.
(iii) Free energy (G) Enthalpy (H) and Entropy (S).
Answer:
(i) ∆H = ∆U + P∆V
(ii) C_p – C_v= R for 1 mole of an ideal gas
C_p – C_v =nR for n moles of an ideal gas
(iii) ∆G = ∆H – T∆S

Question 33.
(a) The following concentrations were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium 500K. [Ni] = 1.5 x 10^-2 M, [H₂] = 3.0 x 10^-2 M and [NHJ = 1.2 x 10^-2 m. Calculate the equilibrium constant.
Answer:
Given [N₂] = 1.5 x 10^-2 M, [H₂] = 3.0 x 10^-2 M and [NH₃] = 1.2 x 10^-2m
For N₂(g) + 3H₂ (g) → 2NH₃ (g)

b) Define common ion effect
Answer:
It is the process of decreasing (or suppression) the dissociation of a weak electrolyte by adding a strong electrolyte having a common ion is called common ion effect.

(c) What is the relationship between [HsO+] and [OH-] for neutral solution?
Answer:
[H₃O₊] = [OH^–].

Question 34.
(a) What is acid and bases according to Arrhenius concept?
Answer:
Arrhenius Acid : Substances which give hydrogen ion in water are called Arrhenius Acid. Example: HCl. HNO₃
Arrhenius base : Substances which gives hydroxyl ion (OH-) in water are called Arrhenius bases.
Example: NaOH, KQH, NH₄OH, etc.

(b) Give an example for (i) so.M-vapoisr equilibrium (ii) liquid-vapour equilibrium,
(i) I₂(s) → I₂ (vapour) / NH₄Cl(s)⇌ NH₄Cl(vapour) (ii) H₂O(l) → H₂O(v).

(c) Write the equilibrium K. expression for H2 + I2
Answer:
K_c = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}\)

Part – E

V. Answer any TWO of the following questions carrying FIVE marks: ( 2 x 5 = 10 )

Question 35.
(a) For the molecule CH₃CH₂CH₂OH
(i) Identify functional group
Answer:
Functional group = – OH

(ii) Write the bond line forarulbu
Answer:
Bond line formula =

(iii) Write the succeeding homologue.
Succeeding homologue = CH₃CH₂CH₂CH₂OH

(b) Explain inductive effect with a suitable example.
Answer:
The permanent displacement of a (sigma) electrons along the saturated carbon chain away/ towards the group/atom attached at the end of the chain.

Question 36.
(a) An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Answer:
Given % carbon = 69%, H = 4.8%, w = 0.20g

(b) What are nucleophiles? Give an example.
Answer:
Chemical species having -ve charge reacts with nucleus of an atom/molecule is called nucleophile, e.g: H^–, CN^–, H₂O, NH₃, CP^–, Br^– etc.

Question 37.
(a) Explain the mechanism of chlorination of methane.
Answer:
Mechanism of chlorination of methane :
Step I : Chain initiation: Methane and Cl₂ heated in UV light, Cl₂ undergoes homolytic fission giving chlorine free radicals,

Step II: Chain propagation: Cl reacts with methane gives methyl free radical,

Methyl free radical reacts with Cl2 molecule gives C1‘ radical, i.e. CH₃ + Cl₂ → CH₃Cl + Cl’ These are repeated several times.

Step III: Chain termination:
The reaction goes to the end when the free radicals reacts with each other.

(b) Write the Newman’s projections of ethane.
Answer:

Students can Download 1st PUC Chemistry Model Question Paper 4 with Answers, Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Model Question Paper 4 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
2. Write balanced chemical equations and draw labeled diagram wherever allowed.
3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

I. Answer ALL of the following (each question carries one mark): ( 10 × 1 = 10 )

Question 1.
Give an example for a homogeneous mixture.
Answer:
Air, Gasoline, Sea water, etc.

Question 2.
Write the ideal gas equation for ‘n’ moles of a gas.
Answer:

Question 3.
What is hydronium ion?
Answer:
H₃O⁺ is called hydronium ion

Question 4.
Write the IUPAC name of an element with atomic number 107.
Answer:
Ununnillium.

Question 5.
Identify the element in the compound K₂Cr₂O₇ showing negative oxidation number.
Answer:
Oxygen.

Question 6.
Name the gas liberated when a piece of sodium metal is added into the water.
Answer:
Hydrogen (H₂).

Question 7.
What is Zeolite?
Answer:
Zeolites are alumino silicates having a three-dimensional cage like structure with a general
formula Mⁿ⁺_x/n [(AlO₂)_x(SiO₂)_y]H₂O, where M is a cation ion like K⁺Na⁺,Ca²⁺.
These are obtained by replacing some of the silicon atoms by aluminium atoms. The aluminosilicate thus formed acquires negative charge and it is balanced by the cations like
Na⁺,K⁺ or Ca²⁺

Question 8.
Give the composition of water gas.
Answer:
Water gas [CO + H₂] is a mixture of carbon monoxide and hydrogen.

Question 9.
Write the bond line formula for 3-chloro octane.
Answer:

Question 10.
Alkanes are not soluble in water. Give reason.
Answer:
Non-polar and saturated.

Part – B

II. Answer any FIVE of the following questions carrying TWO marks: (5 x 2 = 10 )

Question 11.
Calculate the molarity of NaOH in the solution, prepared by dissolving lOg in enough water to form 500mL of the solution. Molar mass of NaOH is 40 g mol^-1
Answer:

Question 12.
Give any two differences between ideal and non-ideal (real) gas.
Answer:
Ideal Gas

1. Obeys ideal gas equation PV = nRT.
2. Ideal gas does not exist in nature.

Real Gas:

1. Does not obeys the ideal gas equation but obeys van der Waals equation.
2. All gases which exist in nature are real gases.

Question 13.
Mention any two conditions for the combination of atomic orbitals.
Answer:
The conditions for the combination of atomic orbitals are:
(a) The combining atomic orbitals should have comparable energy.
1s orbital can combine with another Is orbital but not with 2s orbital.

(b) The combining atomic orbitals should have the same symmetry about the molecular axis. For example: 2p orbital can combine with 2s orbital or another 2p_z orbital but not with 2p_x or 2p_y.
2s cannot combine with 2p_x or 2p_y .
(c) The combining atomic orbitals must overlap to the maximum extent.

Question 14.
Complete the following equations:
(i) 4Li + O₂ → _____
Answer:

(ii) 2Na + O₂ → _____
Answer:
2Na + O₂ → Na₂O₂ (Sodium peroxide)

Question 15.
(a) What is dry ice?
Answer:
Solid CO₂ is called dry ice. It is called dry ice because it directly changes into gaseous state without becoming liquid.

(b) Write the formula of basic structural unit of silicates.
Answer:
Silicates are the compounds in which the anions are present as either discrete SiO^- tetrahedral or a number of such units joined together through comers.

Silicates are basically made up SiO⁴_4- tetrahedra (as shown in the figure) in which silicon atom is bonded to four oxygen atoms in a tetrahedral manner. In the figure, big circles indicate oxygen atoms and small circle indicates silicon atoms.

Question 16.
Explain Wurtz reaction with a suitable example.
Answer:
When alkyl halides are heated with sodium metal in ether medium higher alkanes are formed. This reaction is known as Wurtz reaction and employed for the synthesis of higher alkanes containing even number of carbon atoms.

Question 17.
What are benzenoids? Give an example.
Answer:
Aromatic compounds with palanar structure and delocalized n e A above and below the plane of the ring, eg: Benzene, Toluene.

Question 18.
How ozone is formed in the atmosphere? Give equations.
Answer:
The UV radiations of very short wavelength (< 250 nm) have sufficient energy to cause photo – dissociation of oxygen molecules into oxygen atoms. Highly reactive oxygen atoms combine with oxygen molecules to form ozon

Part – C

III. Answer any FIVE of the following questions carrying THREE marks: (5 x 3 = 15 )

Question 19.
(a) Write the general electronic configuration of p-block elements.
Answer:
General electronic configuration of p-block elements is ns² np^1_6.
Both s and p block together is called representative or normal elements.

(b) How does ionisation enthalpy varies along a period and down the group?
Answer:
The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom. Ionization energy increases along the period and decreases down the group.

Question 20.
Define the terms (a) Bond angle (b) Bond order (c) Co-valent radius.
Answer:
(a) Bond angle: It is defined as the “angle between the oribitals containing bonding electron pairs around the central atom in a molecule or in a complex ion”. It is expressed in degrees.
Example: In CO₂ the bond angle is 180°. So CO₂ has linear shape.
(b) Bond order: It is the number of covalent bonds holding the atoms in the molecule. Example: If the bond is formed by the sharing of two electron pairs, then the bond order is 2. O = O or C = C bond in alkenes.
(c) Co-valent radius: A chemical bond formed by the sharing up of one or more electrons between the atoms.

Question 21.
Give any three important postulates of VSEPR theory.
Answer:
Bonding : The bond formation takes place if there exists impaired electron in the valance shell.
Shape : The geometry or the shape of a molecule depends on the number of valence shell electrons surroundings the central atom.
Repulsion : Electron pair tend to repell one another because electron clouds have similar charge.
Stability : As a result of electron pair repulsion these electron pairs try to stay apart as possible in order to attain minimum energy and maximum stability.
Repulsive interaction : Lane pair with lane pair electrons are having maximum repulsive interaction, bond pair-bond pair electrons are having minimum repulsive interaction. When angle of repulsion decreases angle between the electron pair increases (109°. 28′ less repulsion, 104° more repulsion).

Question 22.
Explain bonding in H₂ molecule based on the basis of Molecular orbital theory.
Answer:
Atomic number of hydrogen is 1 = Is’

(a) Electronic configuration of hydrogen molecule is σ1s²σ*1s⁰
(b) Bond order – \(\frac{2-0}{2}\) = 1
(c) Magnet property = diamagnetic.

Question 23.
(a) What is the oxidation number of Cl in KClO,sub>3 ?
Answer:
+5

(b) Write the separate equations for the oxidation and reduction reactions occuring in the given redox reaction. 2Fe + HCI → FeCL₂ + H₂
Answer:

Question 24.
Explain the process of softening of temporary hardness by Clark’s method.
Answer:
Clark’s process: Temporary hardness can be removed by the addition of a calculated amount of lime, where magnesium carbonate or calcium carbonate is precipitated.
Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ +2H₂O
Mg(HCO₃)₂ + Ca(OH)₂ → CaCO₃ + MgCO₃ + 2H₃O

Question 25.
(a) Give the composition of plaster of paris and mention one use of it.
Answer:
Plaster of Paris is CaSO₄ + \(\frac{1}{2}\)H₂O It is prepared by heating gypsum at 373 K.

Uses:
(i) It is used for manufacture of statues,
(ii) It is used for filling gaps before white washing.

(b) Write any one biological importance of Magnesium.
Answer:
Biological importance of Mg and Ca are:
(a) Mg ion is present in chlorophyll which is responsible for photosynthesis.
(b) Ca²⁺ ions are required for maintaining heart beat.
(c) Ca²⁺ and Mg²⁺ ions are required for clotting of blood.

Question 26.
(a) Boron shows anamolous behaviour with other elements of the same group. Give reasons.
Answer:

1. Due to smaller size
2. High ionisation enthalpy
3. High electronegativity.

(b) Name the gas liberated when formic acid is heated with concentrated H₂SO₄.
Answer:
Pure carbon monoxide is formed by the dehydration of formic acid with cone. H₂SO₄.

Part – D

IV. Answer any FIVE of the following questions carrying FIVE marks: ( 5 x 5 = 25 )

Question 27.
(a) The percentage composition of an organic compound found to contain 39.9% carbon, 6.7% hydrogen and the rest is oxygen. If the molecular mass of the compound is 60 gmol^-1. Determine the molecular formula of the compound.
Answer:
Percentage of oxygen = 100 – (39.9 + 6.7) = 100 – 46.6 = 53.4

∴ Empirical formula = CH₂O
Molecular mass = Eq. mass x basicity = 60 x 1 = 60
Empirical formula weight = CH₂O = 12 + 2 + 16 = 30
Nows 30n = 60 => n = 2
∴ Molecular formula = (Empirical formula) x n = (CH₂O)₂ = C₂H₄O₂

(b) State Avagadro law.
Answer:
Avogadro’s law states that “Equal volumes of all gas under similar conditions of temperature and pressure contain equal number of mole cols.”

Question 28.
(a) Give any three postulates of Bohr’s theory of atomic model.
Answer:
Bohr’s model of an atom, the postulates are:

1. Electrons revolve around the nucleus of an atom in a certain definite path called orbit or stationary state of shell.
2. The shells are having different energy levels denoted as K, L, M, N…
3. As long as the electron remains in an orbit, they neither absorb nor emit energy.
4. The electron can move only in that orbit in which angular momentum is quantized,
   i.e., the angular momentum of the electron is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

(b) State Hund’s rule of maximum multiplicity.
Answer:
Hund’s rule or maximum multiplicity: Electron pairing does not take place until orbitals of same energy are singly occupied.

Question 29.
(a) Name the four quantum numbers and mention what they indicate?
Answer:
In order to define state energy and location of electron a set of four numbers are used. These numbers are called quantum numbers, (i) Principal quantum number (n). (ii) Azimuthal quantum number (1). (iii) Magnetic quantum number (m). (iv) Spin quantum number (s).

(b) What is node?
Answer:
The region where probability of finding the electron is zero is called a node.

Question 30.
(a) Write the important postulates of Kinetic theory of gases.
Answer:

1. Gases are made up of large number of the minute particles.
2. Pressure is exerted by a gas.
3. There is no loss of kinetic theory.
4. Molecules of gas attract on one another.
5. Kinetic energy of the molecule is directly proportional to absolute temperature.
6. Actual volume of the gaseous molecule is very small.
7. Gaseous molecules are always in motion.
8. There is more influence of gravity in the movement of gaseous molecule.

(b) What is Boyle’s temperature?
Answer:
The temperature at which real gases obey ideal behaviour for appreciable range of pressure is called Boyle’s temperature (or) Boyle’s point.

(c) Falling liquid droplets are spherical. Give reason.
Answer:
Due to surface tension.

Question 31.
(a) Calculate the standard enthalpy of formation of benzene from the following data.
Answer:
C₆H₆ + O₂ + \(\frac{15}{2}\)O₂  → 6CO₂ + 3H₂O_(l); ∆H= – 3264KJ
C_(s) + O_2(g)  → CO_2(g); ∆H = -393.5KJ
H₂ + \(\frac{1}{2}\)O₂ → H₂O_(l); ∆H = -285.9KJ

(b) What is entropy?
Answer:
Measure of disordemess of a system.

Question 32.
(a) Calculate the total work done when one mole of a gas expands isothermally and reversibly from an intial volume of 10dm³ to a final volume of 20dm³ at 298K.
Answer:
T = 27°C = 27 + 273 = 300K , V₁ = 10 dm³ V₂ = 20 dm³ , R = 8.314 J/K/mol
w = -2.303 nRTlog \(\left(\frac{v_{2}}{v_{1}}\right)\)
w = -2.303 x 1 x 8.314 x 300 x log \(\left(\frac{20}{10}\right)\)= -2.303 x 8.314 x 300 x log2
= -2.303×8.314x300x0.3010 = -1729 joule = -1.723 kJ.

(b) What is an intensive property? Give an example.
Answer:
Intensive property of a system is that property of the system which does not depend on the quantity of the substance present in the system. E.g; Density, Viscosity.

Question 33.
(a) For the hydrolysis of sucrose the equilibrium constant K_c is x x 10^-3 at 300K calculate ∆G° at 300K?
Answer:
Given K_c = 2 x 10^-3
T = 300K; ∆G° = ?
We know that ∆G° = -2.303 RT log K_c = -2.303 x 8.314 x 300 x 2 x 10^-3
= -2.303 x 8.314 x 3 x 2 = -11.488 J/mol.

(b) Explain common ion effect with suitable example.
Answer:
Suppression in the degree of dissociation of a weak electrolyte by the addition of a common ion is called common ion effect.

Question 34.
(a) Calculate the pOH of a solution obtained when 0.05 mol NH₄Cl is added and dissolved , in 0.025M ammonia solution. (Given Kb for ammonia = 1.77 x 10^-5).
Answer:
pK_b = -logK_b = -logl.8 x 10^-5 = -(5.2553) = 4.7447
p^OH = pK_b + log \(\frac{[\text { salt }]}{[\text { base }]}\) = 4.7447 + log \(\frac{0.05}{0.12}\) = 4.74 – 0.38 = 4.36
∴ pH = 14 – p^OH= 14 – 4.36 = 9.64

(b) What is solubility product?
Answer:
It is the product of the molar concentration of ions of the saturated solution of a sparingly soluble solution of the salt and the coefficients are raised to their powers at constant temperature.

Part – E

V. Answer any TWO of the following questions carrying FIVE marks: ( 2 x 5 = 10 )

Question 35.
(a) Write the principles involved in the estimation of
(i) Halogens (ii) Sulphur present in an organic compound by Carius method.
Answer:
(i) Halogens: When an organic compound containing halogen (Cl, Br or 1) is heated in a sealed tube with fuming nitric acid and excess of silver chloride, silver halide is formed from the mass of silver halide obtained, the percentage of the halogen can be calculated.

Procedure: In a hard glass tube (Carius tube), 5 ml of fuming HNO₃ and 2 to 2.5 g AgNO₃ are taken. A small narrow weighing tube, containing a small amount (nearly 0.15 -0.2g) of accurately weighed organic compound, is introduced in the Carius tube in such a way that nitric acid does not enter the weighing tube. The Carius tube is now sealed and heated in a furnace at 300°C for about six hours.

The tube is then cooled and its narrow end is cut off and the contents are completely transferred to a beaker by washing with water. The precipitate of silver halide formed is filtered through a weighed sintered glass crucible. It is washed, dried and weighed.

Observation and calculation:
(i) Mass of organic compound taken = W₁g
(ii) Mass of silver halide obtained = w₂g
(a) For chlorine:

143.5g of AgCl contains 35.5 g of chlorine.
w₂g of AgCl will contain \(\frac{35 \cdot 5 \times w_{2}}{143 \cdot 5}\) of chlorine
∴ % Cl₂ = \(\frac{35 \cdot 5 \times w_{2}}{143 \cdot 5} \times \frac{100}{w_{1}}\)

(b) For bromine:

188 g of AgBr contains 80 g of bromine
w₂ of AgBr will contain \(\frac{980 \times w_{2}}{188}\) g of bromine

(ii) Sulphur Present in an organic compound by carius method:
Principle: An organic compound containing sulphur to heated in Carius tube with fuming nitric acid or sodium peroxide. Sulphur present is oxidised to sulphuric acid. It is then precipitated as burium sulphate by adding excess of barium chloride solution. The precipitate formed is filtred, washed, dried and weighed. The % of S is calculated from the mass of barium sulphate.

Calculations: Mass of organic substance = ‘m’
Mass of barium sulphate formed = ‘x’ g.
Molecular mass of barium sulphate = 233 g.

(b) Name the element estimated by Kjeldahl’s method.
Answer:
Nitrogen.

Question 36.
(a) Describe the estimation of Carbon and Hydrogen by Liebig’s method.
Answer:

Principle: A known mass of an organic compound is strongly heated with dry cupric oxide (CuO), when carbon and hydrogen are quantitatively oxidized to CO₂ and H₂O respectively. The masses of CO₂ and H₂O thus formed are determined. From this, the percentages of carbon and hydrogen can be calculated.
Procedure : Pure and dry oxygen is passed through the entire assembly of the apparatus (Figure) till the CO₂ and moisture is completely removed.

A boat containing weighed organic substances is introduced inside from one end of the combustion tube by opening it for a while. The tube is now strongly heated till the whole of the organic compound is burnt up. The flow of oxygen is continued to drive CO₂ and water vapours completely to the U-tubes. The apparatus is cooled and the U-tubes are weighed separately.

Observation and calculations:

1. Mass of organic compound taken = w.g.
2. Mass of water produced = x g (Increase in mass of CaCl₂ tube).
3. Mass of carbon dioxide produced = y g (Increase in mass of KOH tube).

To determine % of carbon:
Molar mass of CO₂ = 44g mol^-1
Now, 44g of CO₂ = contains 12 g of C
y g of CO₂ will contain of \(\frac{12 y}{44}\) f of C
This amount of carbon was present in w. g. of the substance
∴ % C = \(\frac{12 y}{44} \times \frac{100}{w}\)

To determine % of Hydrogen
Molar mass of water = 18 g mol^-1
Now, 18 g of H₂O contains 2 g of H₂O
∴ x g of H₂O will contain \(\frac{2 x}{18}\) g of H₂O
This amount of hydrogen was present in weight of substance.
∴ H₂ = \(\frac{2 x}{18} \times \frac{100}{w}\)

(b) What is Cracking?
Answer:
The decomposition of higher alkane into a mixture of lower alkanes alkenes, etc. by the application

Question 37.
(a) Explain the mechanism of chlorination of Methane.
Answer:
Mechanism of chlorination of methane involves three types:

Step 1: Initiation: Chlorine absorbs energy and undergoes homolysis to give chlorine free radicals.

Step 2: Propagation: Chlorine free radical reacts with methane to give methyl free radical.

The methyl free radical reacts with chlorine to form methyl chloride and chlorine free radical.

Step 3: Termination: Free radials combine to form stable products.

(b) Name the product formed when phenol vapours are passed over heated zinc dust. Write the chemical equation.
Answer:
Benzene

Students can Download 1st PUC Chemistry Model Question Paper 2 with Answers, Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Model Question Paper 2 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
2. Write balanced chemical equations and draw labeled diagram wherever allowed.
3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

I. Answer ALL of the following (each question carries one mark): ( 10 × 1 = 1 )

Question 1.
Express 0.001023 into scientific notation.
Answer:
1.023 x 10^-3.

Question 2.
Define critical temperature.
Answer:
The temperature above which gas cannot be liquified.

Question 3.
Give the example which acts as Lewis base as well as Bronsted base.
Answer:
NH₃ or Ammonia.

Question 4.
How does electronegativity related to atomic size?
Answer:
Atomic size ∝ \(\frac{1}{\text { Electronegativity }}\)
Electronegativity

Question 5.
What is the oxidation state of oxygen in peroxide?
Answer:
-1 or minus 1

Question 6.
Which alkali metal is act as strong reducing agent?
Answer:
Lithium.

Question 7.
Write the formula of inorganic benzene.
Answer:
B₃N₃H₃

Question 8.
What is dry ice?
Answer:
Solid CO₂.

Question 9.
Complete the reaction NH₄CNO

Answer:

Question 10.
Write the name of the chain isomer of n-Butane.
Answer:
CH₃-CH₂-CH₂-CH₃

Part – B

II. Answer any FIVE of the following questions carrying TWO marks: ( 5 × 2 = 10 )

Question 11.
Calculate the average atomic mass of chlorine using the following data.

Answer:

Question 12.
Draw the graph of pressure versus volume of a gas at a different temperatures to illustrate the Boyle’s law.\
Answer:

Question 13.
Give any two conditions for hybridization of atomic orbitals.
Answer:
(i) Should contain half filled atomic orbitals.
(ii) Energy of the combining atomic orbitals should be equivalent.

Question 14.
Explain the reactivity of second group elements towards hydrogen.
Answer:
Hydrogen combines with second group elements to form metallic hydrides,
i.e. M + H₂ → MH₂
OR
Ca + H₂ → CaH₂
when M = Mg, Ca, Ba, etc.

Question 15.
How many sigma and pi bonds are in carbon monoxide?
Answer:
2σ bonds and one 7t-bond.

Question 16.
Write the structural isomers of an alkene with molecular formula C₂H₈.
Answer:
H₃C-H₂C-HC=CH₂ , CH₃-CH=CH-CH₃

Question 17.
What are the characteristics for any ring system to be called as aromatic compound?
(i) It should be planar.
(ii) Obeys Huckel rule i.e. (4n + 2)π electrons.
(iii) Should contain delocalized π – electrons above and below the plane of the molecule.

Question 18.
(a) Which oxide of nitrogen in higher concentration will retard the rate of photosynthesis in plants?
NO₂ or Nitrogen dioxide.
(b) Name the compound formed when carbon monoxide binds to haemoglobin.
Carboxyhaemoglobin (Hb + CO → HbCO).

Part – C

III. Answer any FIVE of the following questions carrying THREE marks: ( 5 × 3 = 15 )

Question 19.
Why Beryllium exhibit anomalous behaviour from the rest of the elements in the group.
Answer:
Due to

1. Smaller size compared to other elements in the group.
2. High ionisation enthalpy in the ground state.

Question 20.
With the help of hybridization explain the structure of methane.
Answer:
CH₄
EC of C = 1s²2s²2p_x¹2p_y¹2p_z¹
The four half filled orbitals overlaps with each other giving 4sp³ hybrid orbitals. These combines with s-orbital of hydrogen along the axis giving 4σ bonds with bond angle 109°28′ and tetraheral geometry.

Question 21.
(a) What is bond enthalpy? How it is related to the bond order?
Answer:

1. The amount of energy required to break one mole of bonds of same type to separate them into gaseous atoms.
2. Bond enthalpy oc Bond order.

(b) Write the resonance structure of CO₂.
Answer:

Question 22.
(a) Give any two differences between bonding and anti-bonding molecular orbitals.
Answer:
BMO :

1. Formed by addition overlapping Ψ_AB = Ψ_A + Ψ_B
2. Less energy than ABMO.
3. More stable.

ABMO :

1. Formed by the subtraction overlapping of atomic orbitals. of atomic orbitals i.e. Ψ_AB = Ψ_A -Ψ_B
2. More energy than BMO.
3. Less stable.

(b) What is the dipole moment of BeF2?
Answer:
Dipole moment is 0 (zero)
F ⇌ Be ⇌ F

Question 23.
Balance the following redox reaction by half reaction method.
MnO₄^– (aq) + 1^– (aq) → MnO₂ (s) + I₂(s) : Basic medium.
Answer:

Question 24.
Give the reactions to show amphoteric nature of water.
Answer:

(ii) Mention any one method of removal of temporary hardness of water.
Answer:
By boiling.

Question 25.
How caustic soda is commercially prepared from brine by Castner-Kellner cell.
Answer:
Caustic soda by Castner-Kellner cell: NaOH is manufactured by the electrolysis of aqueous solution of NaCl (Brine).
i.e. 2NaCl → 2Na⁺ + 2Cl^–
Sodium ions are discharged at mercury cathode.
Sodium deposited at mercury forms sodium amalgam.
Chlorine liberated at anode removed from the cell.
At cathode : 2Na⁺ + 2e → Na, Na + Hg → Na – Hg
At anode : 2Cl^– – 2e → Cl₂
Na-Hg is treated with water to form NaOH
i.e. Na/Hg + 2H₂O → 2NaOH + H₂ + Hg.

Question 26.
Give the example of element of group 14
(i) Shows maximum catenation capacity.
Answer:
Carbon

(ii) Used as semiconductor.
Answer:
Silicon

(iii) Which reacts with water.
Answer:
Tin

Part – D

IV. Answer any FIVE of the following questions carry ing FIVE marks: ( 5 x 5 = 25 )

Question 27.
(a) M atch the following :

Answer:
(i)-(c),
(ii)-(d),
(iii)-(a)

(b) Define molarity. Write the expression to calculate the molarity of the solution for the given mass and volume.
Answer:
It is the number of moles of solute present in 1000 ml solvent or 1dm3.
M = \(\frac{\mathrm{W}}{\mathrm{GMM}} \times \frac{1000}{\mathrm{V}}\)

Question 28.
(a) Give any three posulates of Bohr’s model for hydrogen atom.
Answer:

• Electrons are revolving around nucleus in a closed circular path called orbits or main shells or energy levels.
• When an electron jumped from higher energy level to lower energy level, the difference of energy emitted as radiation, i.e. E₂ – E₁ = ∆E= hγ
• The angular momentum of an electron has discrete values. It is given by the equation
  mvr = \(\frac{n h}{2 \pi}\)

(b) Calculate the mass of a photon with wavelength 5,OA°
Answer:

Question 29.
(a) Sketch the shapes of P_x and dz².
Answer:

(b) Identify the property exhibited by ₁₉K⁴⁰ and ₁₉Ca⁴⁰.
Answer:
Isobars (∵ mass number are equal).

(c) Write the orbital (box) type electronic configuration of p⁴ and d⁴ according to Hund’s rule of maximum of multiplicity.
Answer:

Question 30.
a) Derive ideal gas equation using gas laws.
Answer:
Ideal gas equation:
According to Boyle’s law V ∝ \(\frac{1}{P}\) T at constant T
According to Charle’s law V ∝ at constant P
According to Avogadro’s law V ∝ n at constant T and P
On coming V ∝ \(\frac{1}{P}\) V ∝ \(\frac{1}{P}\) x T x n
or PV=nRT
For ‘n’ moles, PV = RT for 1 mole.
R = gas constant, T = Kelvin Temp, P = Pressure, V = Volume of gas
n = Number of moles of gas.

(b) At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 ml.
Initial conditions
P₁ = 760 mm
V₁ = 600 mm
T₁ = 250C = 298K

Final conditions
P ₂= ?
v₂ = 640 mL
T₂= 10°C + 273k = 283K

Question 31.
(a) If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourization of 1 mole of water at 1 bar ad 100°C is 41kJ mole^-1. Calculate the internal energy change when 1 mole of water is vapourized at 1 bar pressure and 100°C.
H₂O (l) → H₂0 (g)
∆n_g = 1 – 0 = 1
W.K.T. LH = ∆U + ∆n_gRT
OR ∆U = ∆H – ∆n_gRT = 41 x 10³ – 1 x 8.314 x 373
= 41000— 3101,122 = 37898.878 J/mol
OR = 37.898878 K/J/mol.

(b) State Hess’s law of constant heat summation.
Answer:
The total amount of heat liberated or absorbed is same whether the reaction takes place in one step or more than one step / (or several steps).

(c) What is the value for standard enthalpy of formation of an element.
Answer:
Zero/0

Question 32.
(a) Calculate ∆G° for conversion of oxygen to ozone, -O₂(g) —> O₃(g) at 298K. If
K_p for this conversation is 2.47 x 10^-29.
Answer:
∆rG° = ? R = 8.314J/k/mol T = 298K K_p = 2.47 x 10^-29.
W.K.T. ∆rG° = -2.303RTlogK_p = 2.303x 8.314 x 298 x 2.47 x 10^-29.
= 163229 J/mol or 163.2 KJ/mol.

(b) What is thermochemical equation? Write the thermochemical equation for the molar combustion of ethanol (Given ∆_rH° = -1367 kJ mol^-1).

1. It is a balanced reaction, gives the information about physical state of reactants and product as well as heat liberated or absorbed.
2. C₂H₅OH_(l)+ 3O₂(g) → 2CO₂(g) + 3H₂0(/) ; ∆H° = -1367 KJ/mol

(c) What is the value of ∆G in a spontaneous process?
Answer:
∆G = -ve

Question 33.
For the equilibrium 2NOCl (g) ⇌ 2NO(g) + Cl₂(g) the value of the equilibrium constant Kc is 3.75 x 10^-6 at 1069K. Calculate the K_p for the reaction at this temperature?
Answer:
For the reaction 2NOCl (g) ⇌ 2NO(g) + Cl₂(g)
∆n = 3 – 2 = 1. ∴ K_p = K_c (RT)^∆n
K_p = (3.75 x 10^-6 x 0.0831 x 1069) x 1 = 3.75 x 0.0831 x 1069 x 10^-6 = 333.13 x 10^-6 OR K_p= 3.33 x 10^-4

(b) Write any two general characteristics of equilibria involving physical process.
Answer:

1. It is dynamic in nature.
2. It can be achieved in a closed vessel.
3. It depends only on temperature but not on concentration or pressure.

Question 34.
(a) The p^H of the blood is 7.4. Calculate the [H⁺].
p^H = 7.4 [H⁺] = ?
W.K.T [H⁺] = Antilog (-p^H)
[H⁺]= Antilog ₁₀ (-7.4)
Add-1 to-7 and +1 to-0.4 ∴ [H⁺] = Antilog \(\overline{8.6}\)
[H⁺] = 10^-8 x Antilog of (+0.6).

(b) Derive the Henderson-Hasselbalch equation for acid buffer.
Answer:
Henderson-Hasselbalch equation:
Consider a weak acid and its salt BA,

Part – E

V. Answer any TWO of the following questions carrying FIVE marks: ( 2 x 5 = 10 )

Question 35.
(a) For the following bond cleavage, use curved arrows to show electron flow, mention the type of bond cleavage, and reactive intermediate formed.
CH₃CH₂O + OCH₂CH₃ → CH₃CH₂O + OCH₂CH₃

1. Homolytic cleavage
2. Ethoxide free radicals.

(b) Give the hybridisation and geometry of carbocation.
Answer:
sp² and planar.

Question 36.
(a) 0.2033 g of an organic compound on combustion gave 0.3780 g of CO₂ and 0.1288 g H₂O. In a separate experiment 0.1877 g of the compound on analysis by Dumas method produced 31.7 ml. of nitrogen collected over water at 14°C and 758 mm pressure. Determine the percentage composition of the compound. (Aqueous tension of water at 14°C = 12 mm pressure).
Answer:
Given mass of organic compound = 0.2033 g |
Volume of nitrogen (V₁) = 31.7 mL
T₁ = 287K P₁ = (P – f) = 758 – 12 = 746 mm;
P₂ = 760mm U₂ = ? T₂ = 273K

To convert the volume of of N₂ at STP into mass
22400 mL of nitrogen at STP weighs = 28 g


% age of oxygen = 100 – (18.1985 + 50.71 + 7.03) = 24.06%

(b) What are nucleophiles? Give an example.
Answer:
The negatively charged species reacts at nucleus or +ve centre are called nucleophiles.
Example: Cl^–, Br^–, I^–, \(\mathrm{C} \overline{\mathrm{N}}\), NH₃, etc.

Question 37.
(a) Explain the mechanism of addition of HBr to propene in the presence of peroxide catalyst.
Answer:
It is a free radical mechanism.
It has the following steps:
(a) Initiation

(b) Propagation

(c) Termination

(b) Write the structures of cis and trans isomers of But-2-ene.
Answer:

Students can Download 1st PUC Maths Model Question Paper 5 for Practice, Karnataka 1st PUC Maths Model Question Paper with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Model Question Paper 5 for Practice

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. Part-A carries 10 marks. Part-B carries 20 marks. Part-C carries 30 marks. Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

I. Answer all the questions ( 1 × 10 = 10 )

Question 1.
If set A = {1, 3, 5}, then find the number of elements in P(A)1

Question 2.
Find the value of tan \(\frac{19 \pi}{3}\).

Question 3.
Express (5t)( \(-\frac{3}{5}\) i ) in the form a + ib,. where a, b ∈ R

Question 4.
If \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\) find x.

Question 5.
Find the tenth term of G.P : 5, 25, 125, ……………..

Question 6.
Find the slope of the line \(\frac{x}{3}+\frac{y}{2}\) = 1.

Question 7.
Find the derivative of x² – 2 at x = 0.

Question 8.
Write the contrapositive of “if a number is divisible by 9 then it is divisible by 3”.

Question 9.
Define mutually exclusive events.

Question 10.
If for some non empty sets A and B containing 3 elements A × B – {(3,4), (5, -3), (6,1)} . Find the set A.

Part – B

II. Answer any Ten questions ( 10 × 2 = 20 )

Question 11.
If A = {1, 2, 3, 4}, B = {2, 3, 5} and C = {3, 5, 6}, find A ∪ (B ∩C).

Question 12.
If X and Y are the two sets such that n(X) = 17, n(Y) 23 and n(X ∪ F) = 38. Find n(X ∩ Y)

Question 13.
Find the range and domain of the real function f(x) = \(\sqrt{9-x^{2}}\)

Question 14.
The minute hand of a clock is 2.1 cm long. How far does its tip move in 20 minute? ( use π = \(\frac{22}{7}\)

Question 15.
If sin A = \(\frac{3}{5}\) and A is in I quadrant, then find sin 2A.

Question 16.
Evaluate: \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}\)

Question 17.
A die is thrown. Write the sample space. Also find the probability of the event “A number greater than or equal to 3 will appear”.

Question 18.
Write the component statement of the following compound statement and check whether the compound statement is true or false; “Zero is less than every positive integer and every negative integer “.

Question 19.
The co-efficient of variation and standard deviation are 60 and 21 respectively. What is the arithmetic mean of the distribution?

Question 20.
Find the equation of the line perpendicular to the line x + y + 2 = 0 and passing through the point (-1, 0).

Question 21.
Represent the complex number z = -1 + i in polar form.

Question 22.
Solve 3x + 2y > 6 graphically.

Question 23.
Find the distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0

Question 24.
Show that the points A(l, 2, 3), B(-l, – 2, -3), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram.

Part – C

III. Answer any Ten questions ( 10 × 3 = 30 )

Question 25.
In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee.
Find how many students were taking neither tea nor coffee?

Question 26.
Write the relationR defined as R = {(x,; x + 5):x∈ {0,1, 2, 3, 4}} in roster system. Write down its range and domain.

Question 27.
Provethat (cosx + cosy)² + (sinx – siny)² = 4cos² \(\left(\frac{x+y}{2}\right)\)

Question 28.
Solve the equation x² + 3x + 9 = 0.

Question 29.
Find the real 0 such that \(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\) is purely real.

Question 30.
In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(i) The student has opted for NCC or NSS
(ii) The student has opted for NCC but not NSS

Question 31.
Find the coefficient of x⁶y³ in the expansion of (x + 2y)⁶

Question 32.
Find the sum of the sequence : 7, 77, 777, 7777……………..  to n terms.

Question 33.
Find the value of n so that \(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) may be geometric mean between a and b.

Question 34.
Find the derivative of the function ‘sin x’ with respect to ‘x’ from first principle.

Question 35.
Find the center and radius of the circle 2x² + 2y² + 8x + 10y – 8 = 0

Question 36.
Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that
(i) all vowels occur together
(ii) all vowels do not occur together.

Question 37.
Verify by the method of contradiction that √2 is irrational,

Question 38.
Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains
(i) 3 kings
(ii) At least 3 kings.

Part – D

IV. Answer any Six questions ( 6 × 5 = 30 )

Question 39.
The real-valued function f is defined by

Draw the graph of f(x) and hence find the domain and range

Question 40.
Prove that lim \(\lim _{\theta \rightarrow 0}\left(\frac{\sin \theta}{\theta}\right)=1\) (θ being in radians) and hence show that \(\lim _{\theta \rightarrow 0}\left(\frac{\tan \theta}{\theta}\right)=1\)

Question 41.
Prove by mathematical induction that

Question 42.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls ?
(ii) atleast 3 girls ?
(iii) at most 3 girls ?

Question 43.
The second, third and fourth terms in the binomial expansion (x + a)ⁿ are 240, 720 and 1080, respectively. Find x, a and n.

Question 44.
Derive an expression for the coordinates of a point that divides the line joining the points A(x₁, y₁, z₁) and B (x₂, y₂, z₂) internally in the ratio m:n. Hence, find the coordinates of the midpoint of A body of mass where A(2, -3, 4) and B(-1, 2, 1).

Question 45.
Derive the equation of the line in the form x cos α + y sin α = p where p is the length of perpendicular from origin to straight line and α is the angle made by the perpendicular with positive direction of x axis. Using this find the equation of the straight line with p = 4 and α = 120°.

Question 46.
Prove that

Question 47.
Solve the following linear inequalities 2x + y ≥ 6 and 3x + 4y ≤ 12 graphically.

Question 48.
Find the mean deviation about the median age for the age distribution of 100 persons given below:

Part – E

V. Answer any ONE question ( 1 × 10 = 10 )

Question 49.
(a) Prove geometrically that cos (x + y) = cos x cos y – sin x sin y using unit circle method and hence write the formula for cos (x – y). (6)
(b) Find the sum to n terms of the series: 1.2.3 + 2.3. 4 + 3.4.5 ……………………  (4)

Question 50.
(a) Define Hyperbola as a set of points and
Derive its equation in the form \(\frac{x^{2}}{a^{2}}-\frac{x^{2}}{a^{2}}=1\)
(b)

Karnataka 1st PUC Maths Model Question Paper 1 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. Part-A carries 10 marks. Part-B carries 20 marks. Part-C carries 30 marks. Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

I. Answer all the questions: ( 10 × 1 =10 )

Question 1.
If A= {1,2}, B = {x : x ∈ A and x² – 9 = 0}. Find A x B.

Question 2.
Define subset of a set.

Question 3.
Convert \(\frac{2 \pi}{3}\) radians into degree measure

Question 4.
Express \(\frac{5+i \sqrt{2}}{2 i}\) in the form x + iy.

Question 5.
Find n if (n – l)p₃: np₄ = 1: 9

Question 6.
Find the tenth term of G.P : 5, 25, 125………….

Question 7.
Find the slope of the line joining the points (3, -2) and (-1,4).

Question 8.
Evaluate: \(\lim _{x \rightarrow 0}\left(\frac{\cos x}{\pi-x}\right)\)

Question 9.
Write the contrapositive of “if a number is divisible by 9 then it is divisible by 3”.

Question 10.
Write the sample space for the experiment “a.coin is tossed repeatedly three times”.

Part – B

II. Answer Any Ten Questions ( 10 x 2 = 20 )

Question 11.
If the universal set U = { 1, 2, 3, 4, 5, 6, 7 } A = { 1, 2, 5, 7 } , B = { 3,4, 5, 6}. Verify (A ∪ B)’= A’ ∩ B’

Question 12.
In a class of 35 students, 24 likes to play cricket, 5 likes to play both cricket and football. Find how many students like to play football?

Question 13.
If A = { 1,2,3 }, B = { 3,4 },C={ 4,5,6 }. Find A x (B∪C).

Question 14.
A wheel makes 360 revolutions in one minute, through how many radians does it turn in one second ?

Question 15.
Find the value of (sin(15°))

Question 16.
Find the value of x and y, if (x + 2y) + i(2x – 3y) is the conjugate of 5 + 4i.

Question 17.
Solve 7x +1 ≤ 4x + 5 and represent the solution graphically on the number line.

Question 18.
Find the equation of the line passing through (-1, 1) and (2, -4)

Question 19.
Find the equation of the line passing through (-4, 3) with slope ( \(\frac{1}{2}\) )

Question 20.
Find the ratio in which the line segment joining the points (4, 8,10) and (6,10, -8) is divided by YZ-Plane.

Question 21.
Evaluate : \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x}\)

Question 22.
Write the component statement of the following compound statement and check whether the compound statement is true or false; “ Zero is less than every positive integer and every negative integer”.

Question 23.
The co-efficient of variation and standard deviation are 60 and 21 respectively. What is the arithmetic mean of the distribution.

Question 24. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will be “not an ace”.

Part – C

III. Answer Any Ten Questions ( 10 x 3 = 30 )

Question 25. Let A = {1, 2, 3, 14} Define a relation R from A to A by R = {(.x, y): 3x -y = 0, x, y ∈ A} write its domain
and range.

Question 26.
Find the general solution of 2 cos² x + 3 sin x = 0.

Question 27.
Express √3 + i in the polar form.

Question 28.
Solve: 3x² – 4x + \(\frac{20}{3}\) = 0

Question 29.
How many numbers greater than 10,00000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4.

Question 30.
Using the Binomial theorem, which number is among (1.1)¹⁰⁰⁰⁰ and 1000.

Question 31.
In an A.P, if m^th term is n and n^th term is m. Then find p^th term (m # n).

Question 32.
Find the sum of n terms of an A.P whose k^th term is (5k + 1).

Question 33.
Find the co-ordinates of the foci and latus rectum of the hyperbola 3x² – y² = 3.

Question 34.
Find the derivative of sin x from first principle.

Question 35.
Given p : 25 is a multiple of 5 : q: 25 is a multiple of 8. Write the compound statement connecting these two statements with “and”, “or”. In 60^th cases check the validity of the statement.

Question 36.
The student Anil and Ashima appeared in the examination, the probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination 0.10. The probability that the both will qualify the examination is 0.02. Find the probability that only one of them qualify the examination.

Question 37.
A letter is chosen at random from the ward ‘ASSASSINATION’, Find the probability that the letter is (i) an vowel (ii) consonant.

Question 38.
In a survey of 600 students in a school, 150 students were found to be taking tea and 250 taking coffee, 100 were taking both tea and coffee. Find how many student were taking neither tea nor coffee.

Part -D

IV. Answer any Six questions ( 6 x 5 = 30 )

Question 39.
Define a polynomial function. If the function from f: R → R is defined as f(x) = x² then draw the graph of/ and find the domain and range.

Question 40.
Prove that

Question 41.
Prove by mathematical induction that 1³ + 2³ + …………….n³ = \(\left[\frac{n(n+1)}{2}\right]^{2}\)

Question 42.
Solve graphically 2x + y ≥ 4,x + y ≤ 3, 2x – 3y ≤ 6

Question 43.
A group consists of 4 girls and 7 boys, In how many ways can a team of 5 members be selected if the team has
(i) No girl
(ii) At least one boy and one girl
(iii) At least three girls

Question 44.
State and prove Binomial theorem for all natural numbers.

Question 45.
Derive the formula for the angle between two straight lines with slopes m₁ and m₂ hence find the slope of the line which makes an angle \(\frac{\pi}{4}\) with the line x – 2y + 5.

Question 46.
Derive the formula for the distance between two points x – 2y + 5 and x – 2y + 5 And hence find the distance between (2, -1, 3) and (-2, 1, 3).

Question 47. Prove geometrically: \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\) where θ is in radian and hence deduce that \(\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\)

Question 48.
Find the mean deviation about the mean for the following data

Part – E

V .Answer any ONE ( 1 x 10 = 10 )

Question 49.
a) Prove geometrically that cos (x + y) = cos x cos y – sin x. sin y and hence prove that cos (x – y) = cos x. cos y + sin x. sin y. 6
b) Find the sum of first n terms of the series 1² + 2² + …. + n².

Question 50.
a) Define ellipse and derive the equation of the ellipse in standard form as \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 (a > b)
b) Find the derivation of \(\frac{x^{5}-\cos x}{\sin x}\) with respect to x.

Karnataka 1st PUC Maths Model Question Paper 2 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. Part-A carries 10 marks. Part-B carries 20 marks. Part-C carries 30 marks. Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

I. Answer all the questions: ( 10 × 1 =10 )

Question 1.
Define an empty set.

Question 2.
If ( \(\frac{x+1}{2}\) ,7 ) find ‘x’

Question 3.
Convert ( \(\frac{7 \pi}{6}\) ) into degrees .

Question 4.
Find the real number x if (x – 2i) (1 + i) is purely imaginary.

Question 5.
Given 5 flags of different colours how many different signals can be made if each signal requires the use of 2 flags, one below the other .

Question 6.
For what value of x the numbers \(\frac{-2}{7}\), x, \(\frac{-7}{2}\) are in G. P.

Question 7.
Find the slope of the line \(\frac{x}{3}+\frac{y}{2}\) = 1

Question 8.
Find the derivative of x² – 2 at x = 0 .

Question 9.
Write the negation of ‘For all a, b ∈ I, a – b∈ I’.

Question 10.
Define sure event.

Part – B

II. Answer any Ten questions ( 10 × 2 = 20 )

Question 11.
In a school, there are 20 teachers who teach mathematics or physics. Of these 12 teach mathematics and 4 teach both physics and mathematics. How many teach physics?

Question 12.
If A ={1,2}, form the set A × A × A.

Question 13.
Taking the set of natural numbers as the universal set. If A = {x: x ∈ N, and 2x +1 > 10} and B = {x : x ∈ N, and 3x – 1 > 8 find A’ and B’.

Question 14.
Find the value of cos (- 1710°).

Question 15.
Prove that sin 2x = \(\frac{2 \tan x}{1+\tan ^{2} x}\)

Question 16.
Find the least positive integer m such that \(\left(\frac{1+i}{1-i}\right)^{4 m}\) =1

Question 17.
Solve {3 (2x-5 ) -7} ≥ 9(x-5).

Question 18.
Find the distance of a point (3,-5) from the line 3x – 4y – 5 =0.

Question 19.
Find the angle between the lines y – y√3 x – 5 = 0 and √3 y – x + 6 = 0.

Question 20.
Evaluate \(\lim _{x \rightarrow-2} \frac{\frac{2}{x}+\frac{1}{2}}{x+2}\) .

Question 21.
Show that the points P(-2, 3, 5), Q (1, 2, 3) and R (7,0,-1) are collinear.

Question 22.
Write the converse and contrapositive of ‘If a parallelogram is a square, then it is a rhombus’.

Question 23.
Write the mean of the given data 6,7,10,12,13,4,6,12 .

Question 24.
Given P(A) = \(\frac{3}{5}\) and P (B) = \(\frac{1}{5}\) find P (A or B)

Part – C

III. Answer any Ten questions ( 10 × 3 = 30 )

Question 25.
There are 200 individuals with a skin disorder. 120 has been exposed to the chemical A, 50 to chemical B and 30 to both chemical A and B. Find the number of individuals exposed to
(i) chemical A but not to chemical B
(ii) Chemical A or chemical B.

Question 26.
Let A = {1, 2}, B = {1, 2, 3, 4}, and C = {5, 6}. Verify that A × (B ∩C) = (A × B) ∩ (A × C)
sin A sin B sin C

Question 27.
Prove that in any triangle ABC, \(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)

Question 28.
Express \(\frac{-1+i}{\sqrt{2}}\) in the polar form.

Question 29.
Solve the equation x² + \(\frac{x}{\sqrt{2}}\) + 2 = 0

Question 30.
In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together.

Question 31.
Find the middle term in the expansion of \(\left(\frac{x}{3}+9 y\right)^{10}\)

Question 32.
The number of bacteria in a certain time double every hour. If there are 30 bacteria present in the culture originally. How many bacteria will be present at the end of 2^nd hour, 4^th hour, and n^th hour.

Question 33.
The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°. Find the number of sides of the polygon.

Question 34.
Find the equation of the ellipse whose center at origin, major axis on the X axis and passes through the point (4, 3) and (6, 2).

Question 35.
Find the derivative of tan x w. r. t x from first principle.

Question 36.
Verify by the method of contradiction that √2 is irrational

Question 37.
One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be
(i) diamond
(ii) not an ace
(iii) a black card.

Question 38.
A fair coin 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.

Part – D

IV. Answer any Six questions ( 6 × 5 = 30 )

Question 39.
Define modulus function. Draw the graph of modulus function, Write down its domain and range.

Question 40.
Prove that cos² 2x – cos² 6x = sin 4x . sin 8x

Question 41.
Prove by mathematical induction that

Question 42.
Solve the following system of inequalities graphically: 5x + 4y ≤ 40, x ≥ 2, y ≥ 3;

Question 43.
What is the number of ways choosing four cards from a pack of 52 playing cards. In how many of these
(i) Four cards of the same suit
(ii) are face cards
(iii) two red and two black card
(iv) cards are of the same colour.

Question 44.
For all real numbers a, b and positive integer V prove that,
(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁ a^n-1b + ⁿC₂ a^n-2b² + …………….+ⁿC_n-1 ab^n-1 + ⁿC_nbⁿ
Hence prove that C₂ + C₂ + C₂ + ………….C_n = 2ⁿ

Question 45.
Derive a formula for the perpendicular distance of a point (x₁, y₁) from the line Ax + By + C = 0.

Question 46.
Derive the section formula in 3-D for internal division. Also find the coordinates of the midpoint of the line joining the points A (1,-2,3) and B (3,4,8).

Question 47.
Prove that \(\lim _{\theta \rightarrow 0} \frac{\sin x}{x}\) = 1 (x being in radians) and hence evaluate \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\)

Question 48.
The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that on observation 8 was incorrect. Calculate the correct mean and the standard deviation in each of the following cases
(i) if wrong item is omitted
(ii) if it is replaced by 12

Part – E

V. Answer any One question ( 1 × 10 = 10 )

Question 49.
(a) Prove geometrically that cos(A + B) = cos A cos S-sin A sin B
(b) Find the derivative of f(x) = 2x² + 3x – 5, also prove that F'(0) + 3f'(-1) = 0

Question 50.
(a) Define parabola as a set of all points in the plane and derive its equation in the form y² = 4ax, a > 0 and hence also find the focus and vertex.
(b) Find the sum to ‘n’ terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + …………..

Karnataka 1st PUC Maths Model Question Paper 3 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. Part-A carries 10 marks. Part-B carries 20 marks. Part-C carries 30 marks. Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

I. Answer all the questions ( 1 × 10 = 10 )

Question 1.
Write the set {x : x e R and – 4 < x ≤ 6} as an interval.

Question 2.
IfA = {1,2}, B = {3,4) then show that A x (B ∩ Φ) = Φ

Question 3.
If cos x = -3/5, x lies in the HI quadrant then find the value of sin x.

Question 4.
Evaluate: \(i^{24}+\left(\frac{1}{i}\right)^{26}\)

Question 5.
Find the number of 4 digits that can be formed using the digits 1, 2, 3, 4, 5. If no digit is repeated.

Question 6.
Which term of 2, 2√2, 4 is 128.

Question 7.
Reduce 6x + 3y – 5 = 0 into slope-intercept form.

Question 8.
Find \(\lim _{x \rightarrow 5}|x|-5\)

Question 9.
Write the negation of “Every natural number is greater than zero”.

Question 10.
If \(\frac{2}{11}\) is the probability of an event A then what is the probability of the event ‘not A’?

Part – B

II. Answer any ten questions ( 2 × 10 = 20 )

Question 11.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2,4, 6, 8} and B = {2, 3, 5, 7} verify (A ∩B)’ = A’ ∪ B’.

Question 12.
If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements. How many elements does X ∩ Y have?

Question 13.
If A = {1, 2} and B = {3, 4} write A × B. How many subsets will A × B have?

Question 14.
The minute hand of a clock is 1.5 cm long. How far does its tip move in 40 Minute? (Use π = 3.142)

Question 15.
Prove that sin 3x = 3 sin x – 4 sin³x.

Question 16.
If x + iy = \(\frac{p+i q}{p-i q}\) prove that x² + y² = 1.

Question 17.
Solve the inequality (2x – 5) > (1 – 5x) and represent the solution graphically on the number line.

Question 18.
By using the concept of equation of the line prove that the three points (3, 0), (-2, -2) and (8, 2) are collinear.

Question 19.
Find the equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).

Question 20.
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).

Question 21.
Compute the derivative of sin² x.

Question 22.
By giving a counter example, show that the following statements is false: “If n is an odd integer then n is a prime”.

Question 23.
The mean and variance of heights of XI students are 162.6cm and 127.69cm² respectively. Find the C.V.

Question 24.
A card is selected from a pack of 52 parts calculate the probability that the card is (i) an Ace (ii) a black card.

Part – C

III. Answer Any Ten Questions ( 3 x 10 = 30 )

Question 25.
In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple and orange juices. Find how many students were taking neither apple juice nor orange juice.

Question 26.
If f(x) = x² and g(x) = 2x + 1 be two real functions find (i) (f + g) (x) (ii) (f – g) (x) (iii) (fg) (x)

Question 27.
Find the general solution of sec²2x = 1 – tan2x.

Question 28.
Express \(\frac{1-i}{1+i}\) into polar form

Question 29.
Solve: 2x² + √3x – 1 = 0

Question 30.
If 5.⁴P_r = 6.⁵P_r-1, then find r.

Question 31.
Find the coefficient of x⁵ of (x + 3 )⁸.

Question 32.
Insert five number between 8 and 26 such that resulting sequence is an A.P.

Question 33.
The sum of first three terms of a G.P. is \(\frac{13}{12}\) and their product if -1. Find the common ratio and the terms.

Question 34.
Find the equation of parabola with vertex at the origin, axis along x-axis and passing through the point (2, 3) also find its focus.

Question 35.
Differentiate \(\frac{x+1}{x}\) from first principle

Part – D

IV. Answer any six questions ( 5 × 6 = 30 )

Question 36.
Verify by the method of contradiction that √7 is irrational.

Question 37.
A bag contains 9 discs of which 4 are red 3 are blue and 2 are yellow. The discs are similar in shape and size.
The disc is drawn at random from the bag. Calculate the probability that will be ’
(i) red
(ii) no blue
(iii) either red or blue.

Question 38.
A and B are events such that P(A) = \(\frac{1}{4}\), P(B) = \(\frac{1}{2}\) and P(A and B) = \(\frac{1}{8}\). Find
(i) P(A or B)
(ii) P(not A and not B)

Question 39.
Define Signum function. Draw the graph of it and write down it Domain and Range.

Question 40.
Prove that cos2x cos \(\frac{x}{2}\) – cos3x cos \(\frac{9x}{2}\) = sin5x sin \(\frac{5x}{2}\)

Question 41.
Prove that 10^2n-1 + 1 is divisible by 11, ∀ n ∈ N by the principle of mathematical Induction.

Question 42.
Solve the following system of linear inequalities graphically. x + y ≥ 5, x – y ≤ 3.

Question 43.
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.

Question 44.
Prove Binomial Theorem for positive integers with real numbers. Hence prove that
ⁿC₀ + ⁿC₂ + ⁿC₄+ …………………. = ⁿC₁ + ⁿC₃ + ⁿC₅ + ………………..

Question 45.
P(a. b) is the midpoint of the line segment between axes. Show that the equation of the line is \(\frac{x}{a}+\frac{y}{b}=2\)

Question 46.
Derive the formulae for distance between two points (x₁, y₁, z₁) and I(x₁, y₂, z₂) and hence find the distance between P(1, -3, 4) and Q(-4, 1,2).

Question 47.
Prove that \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\) where x is in radian and hence evaluate: \(\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x}\)

Question 48.
The mean and standard deviation of 100 observations were evaluated as 40 and 5.1 respectively. By a student who took by mistake, 50 instead of 40 for one observation. What are correct mean and standard deviation?

Part – E

V. Answer Any One Question: ( 1 × 10 = 10 )

Question 49.
a) Define ellipse as a set of all points in the plane and derive its equation in the standard form as \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
a > b
b) Find the derivate \(\frac{2}{x+1}-\frac{x^{2}}{3 x-1}\)

Question 50.
a) Prove geometrically that cos(x+y) = cosxcosy-sinxsiny and hence show that cos ( \(\frac{\pi}{2}\) + x) = – sin x.
b) Find the sum to n terms of series:
\(\frac{1}{1.2}+\frac{1}{2.3}+\ldots \ldots \ldots \ldots\)

Karnataka 1st PUC Maths Model Question Paper 4 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. Part-A carries 10 marks. Part-B carries 20 marks. Part-C carries 30 marks. Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

I. Answer all the questions ( 1 × 10 = 10 )

Question 1.
If A has 4 elements. How many subsets does A has?

Question 2.
Convert 520° in to radian measure.

Question 3.
Find the conjugate of √3i -1

Question 4.
If \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\) then find x

Question 5.
Find the 20th term of the G.P. \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \cdots \cdots \cdots \cdots\)

Question 6.
Find the slope of the line making inclination of 60° with positive direction of x-axis.

Question 7.
Find the derivative of 2 × – \(\frac{3}{4}\)

Question 8.
Write the negation of ‘For all a, b ∈ I, a – b ∈ I’.

Question 9.
Define mutually exclusive events.

Question 10.
If for some non empty sets A and B containing 3 elements A × B = {(3,4), (5,-3), (6,1)}. Find the set A.

Part – B

II. Answer any Ten questions ( 10 × 2 = 20 )

Question 11.
If A = {1,2, 3, 4}, B = {2, 3, 5} and C = {3, 5, 6}, find A ∪ (b ∩ C).

Question 12.
If X and Y are the two sets such that n (X) = 17, n (Y) = 23 and n (X∪ Y) = 38. Find n (X ∩ Y).

Question 13.
Find the range and domain of the real function f(x) = \(\sqrt{9-x^{2}}\)

Question 14.
The minute hand of a clock is 2.1 cm long. How far does its tip move in 20 minute? ( use π = \(\frac{22}{7}\) )
tan A:-tan y

Question 15.
Prove that tan(x – y) = \(\frac{\tan x-\tan y}{1+\tan x \tan y}\)

Question 16.
Evaluate:
\(\lim _{x \rightarrow-1}\left[1+x+x^{2}+x^{2}+\cdots \cdots x^{10}\right]\)

Question 17.
A die is thrown. Write the sample space. Also find the probability of the event “A number greater than or equal to 3 will appear”.

Question 18.
Write the converse and contrapositive of ‘If a parallelogram is a square, then it is a rhombus’.

Question 19.
Two series A and B with equal means have standard deviations 9 and 10 respectively. Which series is more consistent?

Question 20.
Find the equation of the line perpendicular to the line x + y + 2 = 0 and passing through the point (-1, 0).

Question 21.
Represent the complex number z = -1 +i in polar form.

Question 22.
Solve 3x + 2y > 6 graphically.

Question 23.
Find the distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0.

Question 24.
Show that the points A(1, 2, 3), B(-1, -2, -3), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram.

Part – C

III. Answer any TEN questions ( 10 × 3 = 30 )

Question 25.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis? How many like tennis only and not cricket?

Question 26.
Write the relation R defined as R = {(x,x + 5)}: x ∈ {0,1,2,3,4} in roster system. Write down its range and domain.

Question 27.
Prove that (cosx + cosy)² + (sinx – sin y)² =4cos² ( \(\frac{x+y}{2}\) )

Questioin 28.
Solve the equation x² + 3x + 9 = 0.

Question 29.
Find the real θ such that \(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\) is purely real.

Question 30.
In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) the student has opted for NCC or NSS (ii) The student has opted for NCC but not NSS.

Question 31.
Find the coefficient of x⁶y³ in the expansion of (x + 2y)⁶

Question 32.
Find the sum of the sequence : 7,77,777,7777, ……………….

Question 33.
In If \(\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}\) is the A.M. between a and b, then find the value of n.

Question 34.
Find the derivative of the function ‘-x’ with respect to ‘x’ from first principle.

Question 35.
Find the centre and radius of the circle x² + y² + 8x + 10y – 8 = 0.

Question 36.
How many words with or without meaning can be made from the letters of the word MONDAY assuming that no letter is repeated, if (i) 4 letters are used at a time, (ii) All letters are used at a time (iii) All letters are used but first letters a vowel.

Question 37.
Verify by the method of contradiction that √2 is irrational.

Question 38.
Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (i) 3 kings (ii) At least 3 kings.

Part – D

IV. Answer any Six questions ( 6 × 5 = 30 )

Question 39.
Define greatest integer function. Draw the graph of greatest integer function, Write the domain and range of the function.

Question 40.
Prove that \(\lim _{\theta \rightarrow 0}\left(\frac{\sin \theta}{\theta}\right)=1\) (0 being in radians) and hence show that \(\lim _{\theta \rightarrow 0}\left(\frac{\tan \theta}{\theta}\right)=1\)

Question 41.
Prove by mathematical induction that

Question 42.
How many words with or without meaning each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

Question 43.
Find “a” if 17^th and 18^th terms of the expansion (2 + a)⁵⁰ are equal.

Question 44.
Derive an expression for the coordinates of a point that divides the line joining the points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) internally in the ratio m : n. Hence, find the coordinates of the midpoint of AB where A(2,-3,4) and B(-1,2,1)

Question 45.
Derive the equation of the line with slope m and y-intercept c. Also find the equation of the line for which tan θ = \(\frac{1}{2}\) and y-intercept is \(\frac{-3}{2}\)

Question 46.
Prove that

Question 47.
A manufacture has 600 Utters of a 12% solution of acid. How many liters of a 30% acid solution must be added to it so that the acid content in the resulting mixture be more that 15% but less than 18%.

Question 48.
Find the mean deviation about the median age for the age distribution of 100 persons given below

Part – E

V. Answer any One question ( 1 × 10 = 10 )

Question 49.
(a) Prove geometrically that cos(x + y) = cosx cosy – sinx siny using unit circle method and hence find the value of cos ( \(\frac{\pi}{2}\) + x ) = – sin x. (6)
(b) Find the sum to n terms of the series: 1.2.3 + 2.3.4 + 3.4.5 …………….. (4)

Question 50.
(a) Define Hyperbola as a set of points. Derive its equation in the form \(\frac{x^{2}}{a^{2}}-\frac{x^{2}}{a^{2}}=1\)
(b) Suppose

What are the possible values of a and b?

Students can Download 1st PUC Maths Model Question Paper 4 with Answers, Karnataka 1st PUC Maths Model Question Paper with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Model Question Paper 4 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. Part-A carries 10 marks. Part-B carries 20 marks. Part-C carries 30 marks. Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

Answer ALL the questions. Each question carries one mark. (10 × 1 = 10)

Question 1.
Define power set of a set.
Answer:
All possible subset of a given set are called powerset.

Question 2.
If G = {7,8}, H = {5,4,2}, Find H × G.
Answer:
H × G = {(5.7),(5.8),(4,7),(4,8),(2,7),(2,8)}

Question 3,
Convert 225° into radian measure
Answer:

Question 4.
Express (2 – i) – (6 + 3i) in a + ib form.
Answer:
(2 – i) – (6 + 3i) = -4 -4i in the form (a + ib)

Question 5.
Find n, if nc₉ = ⁿc₈
Answer:
n = 9 + 8 = 17

Question 6.
Find 17th term of sequence whos: nth term is given by a = 4n – 3. term of sequence whos
Answer:
a₁₇ = 4(17) – 3 = 68 – 3 = 65

Question 7.
Define the slope of a straight line
Answer:
Slope = m = gradient or inclination of a line with respect to positive x-axis i.e., m = tan θ

Question 8.
Evalaute \(\lim _{x \rightarrow 0} \frac{(x+1)^{2}-1}{x}\)
Answer:

Question 9.
Write the negation of ” √2 is not a complex number.”
Answer:
~(√2 is not a complex no.) = √2 is a complex no.”

Question 10.
Two coins are tossed. Find a sample space.
Answer:
Sample space ={HH, HT,TT,TH}

Part – B

Answer any TEN Questions. (10 × 2 = 20 )

Question 11.
Let A = {1,2,3,4,5,6} and B = {2,4,6,8}, Find A – B and B – A.
Answer:
A – B = {1,3,5), (B – A) = {8}

Question 12.
If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements, Y has 32 elements, how many elements does X ∩ Y have?
Answer:
n(x ∪ y) = n(x) + (n ∪ y) – n(x ∩ y)
50 = 28 + 32 (x ∩ y) ⇒ [∴ n(X ∩ Y) = 10]

Question 13.
A function f is defined by f(x) = 2x – 5 write the value of (i) f(7) (ii) f(-3)
Answer:
Given f(x) = 2x – 5
∴ (i) f(7) = 2(7) – 5 = 9. (ii) f(-3) = 2(-3) – 5 = -11.

Question 14.
Find the angle in radian through which a pendulum swings, if its length is 75 cm and the tip describes an arc of length 10 cm.
Answer:
θ = ? r = 75 cm, s = 10 cm
S = rθ, (formula) 10 = 75 x θ.
∴ θ = \(\frac{10}{75}=\frac{2}{15}\) ⇒ ∴ θ = [0.133 radian.]

Question 15.
Find the solution of sin x = \(-\frac{\sqrt{3}}{2}\)
Answer:
α = \(-\frac{\pi}{3}\) ∴ G.S. x = 2nπ ± α
∴ [x = -2nπ ± \(-\frac{\pi}{3}\)]

Question 16
Write the Multplicative inverse of 4 – 3i.
Answer:

Question 17.
Solve the inequality 3(1 – x) <2 (x + 4) and represent the solution graphically on the number line.
Answer:
Given 3(1 – x) <2 (x + 4)
3 – 3x < 2x + 8 = 3 – 8 < 2x + 3x
-5 < 5x or 5x > -5 i.e., [x > -1]

Question 18.
Find the equation of straight line passing through the point (-4, 3) and having slope \(\frac { 1 }{ 2 }\)
Answer:
Equation of a straight line is givenby the formula
y – y₁ = m(x – x₁)
y = 3 = \(\frac { 1 }{ 2 }\)(x + 4) = 2y – 6 = x + 4 ⇒ ∴ [x – 2y + 10 = 0]

Question 19.
Find the distance between parallel lines 15x + 8y – 34 = 0, 15x + 8y + 31 = 0
Answer:

Question 20.
Find the equation of set of the points P such that ies distances from the points A(3, -4, 5) and B(-2, 1, 4) are equal.
Answer:
Given PA= PB where P =(x, y ,z)
∴ flash distance formula
\(\sqrt{(x-3)^{2}+(y-4)^{2}+(2+5)^{2}}=\sqrt{(x+2)^{2}+(y+1)^{2}+(z-4)^{2}}\)
(squaring on both sides)
∴ (x – 3)² + (y –  4)² + (2 + 5)² = (x + 2)² + (y – 1)² + (z – 4)²

∴(-6x – 4x) + (-8y + 2y) + (102 + 82) + (50 + 21) = 0 = 10x – 6y + 182 + 71 = 0 or
[10x + 6y – 182 – 71 = 0] equation of the plane.

Question 21.
Evaluate: \(\lim _{x \rightarrow 0}\left[\frac{1-\cos x}{x}\right]\)
Answer:

Question 22.
Write converse and contrapositive of “something is cold implies that it has low temperature”
Answer:
Converse: q → p = “Something has low temperature implies it is cold”.
Contrapositive: = q → -p= something does not have low temperature implies it is not cold.

Question 23.
Coefficient of ‘variation of distribution arc 60 and the standard deviat,on is 21. What is the arithmetic mean of the distribution?
Answer:
Given variation = 60
σ = 21, X̄ = ?
Take coefficient of variation = \(\frac{\sigma}{\bar{X}} \times \omega_{0}\)

X̄ = Arithmatic mean = 35.

Question 24.
A card is selected from a pack of 52 cards. Calculate the probability that the card is (i) an arc (ii) black card.
Answer:

Part – C

Answer any TEN of the following questions. Each question carries THREE marks. (10 × 3 = 30)

Question 25.
In a group of 65 people, 40 like cricket, 10 like cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Answer:
n(C) = 40, n(T) = ? n( C ∩ T) = 10
n(C ∪ T) = ? n(T – C) = n(C ∪ T) = n(C) + A(T) – n(C ∩ T)
40 = 40 + n(T) -10 – 40 – 30 = n(T)
∴ [N(T) = 10] ⇒ n(T – C) = n(T)n(C ∩ T) = 10 – 10 = 0

Question 26.
Let A = {1,2,3,…14). Define a relation R from A to A by R = {{(x,y)}}: 3x – 4 = 0, x,y ∈ A}. Write its domain, co-domain and Range.
Answer:
R={(x,y): x ∈ A, y ∈ B, 3x – y = 0}
3(1) – 3 = 0, 3(2) – 6 = 0, 3(3) – 9 = 0, 3(4) – 12 = 0
∴ R = {(1,3), (2,6),(3,9),(4,12)}
Domain of R = A ⇒ Co-domain of R = A
Range of R = {3,6,9,12}

Question 27.
Prove that cos3x = 4cos³x – 3cosx.
Answer:
Take cos(A+B) =cos Acos B – sin Asin B
Put A = x, B = 2x
∴ cos(x + 2x) = cos x.cos2x – sin x sin 2x
cos(3x) = cos x(2cos²x – 1) – sin x(2 sin cos x)
= 2cos³x – cosx – 2cosx(sin²x) = 2cos³x – cosx – 2cosx(1 – cos²x).
= 2 cos³x – cos x – 2 cos x + 2cos³x [4cos³r – 3 cosx.]

Question 28.
Convert the complex number -1 -i into polar form.
Answer:
-1 -i = r(cosθ + isinθ) polar form

∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+1}=\sqrt{2}\)
θ = tan^-1y/x = tan^-1\(\left(\frac{-1}{-1}\right)\) = -(π – standard) = \(-\left(\pi-\frac{\pi}{4}\right)=-\frac{3 \pi}{4}\)

Question 29.
If x + iy = \(\frac{a+i b}{a-i b}\) prove that x² + y²
Answer:

Question 30.
How many words with or with out meaning can be made from the letters of the word “Monday” assuming that no letters is repeated if
(i) 4 letters are used at a time.
(ii) All letters are used at a time.
(iii) All letters are used but first letter is a vowel.
Answer:
Monday : Given number of letters = 6
∴ total number of permutation = 6P₆ = 6!= 720
(i) It 4 letters are together = 3! × 4!= 6 × 24 = 144
(ii) All letters are used at a time. = 6! = 720
(iii) All letters are used but 1st letter is a vowel = 5!= 120 .

Question 31.
Find the middle term in the expansion of \(\left(3-\frac{x^{3}}{6}\right)^{7}\)
Answer:
x = 3, a = \(-\frac{x^{3}}{6}\), n = 7, T_n = ? r = 3
T_n = ? r = 4
n = 7 (add) no. there are two middle terms.

Question 32.
Insert Five numbers between 8 and 26 such that the resulting sequence is an arithmetic progression.
Answer:
Let the five AM’s are A₁, A₂, A₃, A₄, A₅ ∴ 5 AM’s are
∴ 8, A₁, A₂, A₂₃, A₄, A₅, A₆ ∴ A₁ = 11
n = 7, a = 8, d = ?, a_n = 26. A₂ = 14
an = a + (n – 1) d. A₃ = 17.
2b = 8 + (7-1)d A₄ = 20
∴ [d = 3] A₅ = 23

Question 33.
Find sum t0 n terms of sequence 8, 88, 888, 88888,..
Answer:
Let S_n = 8 + 88 + …… + n terms
S_n = 8{1 + 11 +……… + n terms}
X^ly by 9 on both side
∴ \(\frac{95 n}{8}\) = 9 + 99 + 999 +…. + to n term
GP

Question 34.
Find the coordinates of focus, directrix and latus rectum of parabola y² = 8x.
Answer:
Equation of parabola y² = 8x compare with the standard form
y² = 4ax ∴ 4a = 8 ∴ a = 2
∴ focus = (a,0) = (2,0)
Equation of directrix : x = -a.
x = -2 or [x + 2 = 0]
Latus rectum = 4a = 8.

Question 35.
Find the derivative of sin x with respect to x from 1st principle.
Answer:
y = sin x as x → x + Δx, y + Δy
∴ (y + Δy) = sin(x + Δx)
Δy = sin(x + Δx) – y
Δy = sin(x + Δx) – sin x
÷ Δx and apply lt Δx → 0 on both side

Question 36.
Using method of contradiction verify “√2 is irrational.”
Answer:
Let us assume that √2 is rational i.e.. the gives statement is false.
∴ √2 = \(\frac { 1 }{ 2 }\) ≠ 0 a and b have no conunon factor.
Squaring ∴ 2 = \(\frac{a^{2}}{b^{2}}\) ⇒ [a² = 2b²] ⇒ 2 divides a
Again put a = 2C(C ∈ Z) ⇒ ∴ (2c)² = 2b²
∴ 4c² = 2b² or [b² = 2c²] 2 divide b
i.e., 2 divides both a and b hence our assumption is i.e.. a and b does flot have common contradict out statement √2 is rational is false.
∴ √2 is irrational.

Question 37.
If A and B are events such that P(A) = 0.42, P(B)=0.48, P(A and B) = 0.16 determine
(i) P (not A)
(ii) P (not B)
(iii) P (A or B).
Answer:
P(A) = 0.42 P(8) = 0.48, P(A ∩B) = 0.16
(i) P( A¹ = 1 – 0.42 = 0.58 (ii) P(B¹) = 1 – 0.48 = 0.52
(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.42 + 0.48 – 0.16 = 0.50 – 0.16 = 0.74.

Question 38.
A committee of two persons is selected from two men and two women. What is the probability that the committee will have
(i) no man ?
(ii) one man?
(iii) two men?
Answer:

(i) Probability (no man) = women × men = 2C₂, × 2C₀ = 1
(ii) Probability (one man) = 2C₁ × 2C₁ = 2 × 2 = 4
(iii) Probability (two even) = 2C₀ × 2C₂ = 1

Part – D

Answer any SIX Questions. (6 × 5 = 30)

Question 39.
Define signum function. Draw the graph. Also write its Domain and Range.

Question 40.
Prove that \(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Answer:

Question 41.
Prove by Mathematical induction that for all n ≥ 1
1² + 2² + 3² +…+ n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Answer:

Question 42.
Solve the system of inequalities graphically x + 2y ≤ 10, x + y ≥1, x – y ≤ 0, x ≥ 0, y ≥ 0.
Answer:

Question 43.
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected, if the team has (i) no girls (ii) atleast one boy and one girl? (iii) atleast 3 girls
Ans.
Given 4 girls, 7 boys
Select 5 members

(ii) atleast 1 B and 1 G
(a) 4C₁ × 7C₄ = 140
(b) 4C₂ × 7C₃ = 210
(e) 4C₃ × 7C₂ = 21
(d) 4C₂ × 7C₁ = 7
Total number of selection: 378ways.

Question 44.
State and prove Binomial theorem for positive integer n.
Answer:

Question 45.
Derive the equation of straight line in the form xcosω + ysin ω = P, where P is length of normal to the line from origin, w is the inclination of normal with positive x-axis.
Answer:
Carried a line cutting X-axsis at A and Y axis at B.
Let OA = a, OB = b are x and y interscept.
∴ By intercept form of equation of line.
\(\frac{x}{a}+\frac{y}{b}=1\) is the equation of the line AB.
Draw OM ⊥ AB. i.e., normal line
Let OM = P. ∴ In O OAM
cosω = \(=\frac{O M}{O A}=\frac{P}{\cos w}\)  ⇒ ∴ a = \(\frac{O M}{O B}=\frac{P}{b}\)

Question 46.
Derive an expression for the co-ordinates of a point that divides the line joining points A(x¹, y¹, z¹) and B(x², y², z²) internally in the ratio m : n. Hence find the coordinates of the mid point of AB where A = (1,2,3) and B = (5,6,7).
Answer:
Proof: Let p(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be the given points
Let R(x, y, z) divide PQ internally in the ratio m : n
Draw PL, QM, RN perpendicular to xy-plane.
∴ PL ∥ RN ∥ QM
∴ PL, RN, QM lie in one plane
So that the points L, N, M lie in a straight line which is the intersection of the plane and XY plane.
Through the point R draw a line AB It to the line LM. The line AB intersect the line LP externally at A and the line MQ at B.

Triangle APR and LIQR are similar.

∴ n(z – z₁) = m(z₂ – z)
∴ nz = nz₁ = mz₂ – m₂
∴ (nz + mz) = (mz₂ + nz₁)
∴ z(m + n) = mz₂ + nz₁

Question 47.
Prove that \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\), where x measured in radians. Also evaluate \(\lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x}\right)\)

Question 48.
Find mean deviation about mean for the data.

Answer:

Part – E

Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) Prove geometrically that cos(x + y) = cos x cos y – sin xsin y. and hence deduce that cos 2x cos²x – sin²x.
Answer:
Prove that cos(x + y) = cos x cos y.sin x sin y

Proof Consider a unit circle (radius = 1 unit) with centre is (0, 0).
Consider 4 point P₁, P₂, P₃ and P₄

The co-ordinate of P₁, P₂, P₃ and P₄ are given by
P₁ = (cosx,sinx) P₂ = [cos(x + y). sìn(x + y)]
P₃ = [cos(-y) sin(-y)] P₄ = [1,0]
From the figure OP₁OP₃ is congruent to ∆P₂OP₄
∴ From distance formula
P₁p₃ = P₂P₄ …… (1)
Takè the distance (P₁P₃)² = [cosx – cos(-y)]² + [sin x – sin (-y)]²
= (cosx – cosy)² + (sin x + sin y)²
= cos²x + cos²y = cosxcosy + sin²x + sin²y + 2sin cosy = 1 + 1 + 2(cosxcosy – sinxsiny)
(p₁, P₃)² = 2 – 2 cos (x + y)
Again (P₂P₄)² = [1^{(a-b)2-cos(x+y)}]² + |q – sin (x + y)|²
= 1 + cos²(x + y) -2cos(x + y) + sin²(x + y) = 1 + 1 – 2cos(x + y) = 2 – 2cos(x + y)

⇒ LHS = RHS
∴ [cos(x + y) = cosx cosy – sin x sin y]

(ii) Show that cos 2x = cos²x – sin x²x
Take cos(x + y) = cos xcos y – sin xsin y
Put y = x
∴ cos(x + x) = cosx cosx – sinx sinx
cos2x = cos² x – sin²x

(b) Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 +…
Answer:
(1 × 2) + (2 × 3) + (3 × 4) +……+ n term nth term.

n^2th term, 9n = n(n + 1), an = n² + n
to find the sum apply Σ on both side
∴ Σan = Σn² + n

Question 50.
(a) Define hyperbola as a set of points. Derive its equation in the standard form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Answer:
Definition: “Hyperbola is the set of all points in a plane the difference of where distances from the fixed point in the plane is constant.”
Derivain: prove that \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) (standard hyperbola equation)
Let F₁, and F₂, are the two fixed point called focii of the hyperbola. The unit point of the sine segment. Joining the Foci is called the centre of the hyperbola.

A line through the Foci is called the transverse axis. And the line through the centre of perpendicular to the transverse axis is called conjugate axis.
Assume the distances between the two focii is “2c’ the distance between two and define the quantity ‘b’ as b = \(\sqrt{c^{2}-a^{2}}\) and 2b is the length of the conjugaton, (PF₂ – PF₁ = 2a = transverse axis)

The coordinate of F₁ = (-0,0) and F₂, =(0,0)
Let P(x,y) be a junction the hyperbola such that the distance of the distance from P to the farter point be “2a’ i.e., PF₁ – PF₂ = 2a ∴ from the distance formula


Again squaring on both side. \(\therefore\left(\frac{x c}{a}-a\right)^{2}=(x-c)^{2}+y^{2}\)
On simplification further to get \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{c^{2}-a^{2}}=1\)
put c² – a² = b²
∴ the standard form of hyperbola equation is
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

(b) Find the derivative of \(\frac{x^{5}-\cos x}{\sin x}\) with respect to x
Answer:

Students can Download 1st PUC Maths Model Question Paper 3 with Answers, Karnataka 1st PUC Maths Model Question Paper with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Model Question Paper 3 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. Part-A carries 10 marks. Part-B carries 20 marks. Part-C carries 30 marks. Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

Answer ALL the questions. Each question carries one mark. (10 × 1 = 10)

Question 1.
Write the set {x : x ∈ R and -4< x ≤ 6) as an interval.
Answer:
{-3, -2, -1,0,1,2,3,4,5,6} ∴ x ∈ (-3,6)]

Question 2.
If (x + 1, y – 2) = (3,1), find the value of x and y.
Answer:
x + 1 = 3, ∴ x = 2
y – 2 = 1, ∴ y = 3.

Question 3.
Convert \(\frac{2 \pi}{3}\) radians into degree measue.
Answer:

Question 4.
Evaluate 7! – 5!
Answer:
7! – 5! = (7 × 6 × 5 × 4 × 3 × 2 × 1) – (5 × 4 × 3 × 2 × 1) = 5040 – 120 = 4920

Question 5.
Find the 20^th term of the GP. \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots \ldots\)
Answer:

Question 6.
Find n if nC₇ = nC₆
Answer:
If nC₇ = nC₆
∴ n = 7 + 6 = 13 (∵ nC_r = nC_r-1).

Question 7.
Find the slope of the joining the points (3, 2) and (-1, 4)
Answer:
Slope of the line = m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{(4)-(2)}{(-1)-(3)}=\frac{2}{-4}=\left[\frac{-1}{2}\right]\)

Question 8.
Evalaute \(\lim _{x \rightarrow 0} \frac{\cos x}{x-x}\)
Answer:

Question 9.
Write the negation of the statement “√2 is a complex number. ”
Answer:
“√2 is not a complex no.”

Question 10.
If \(\frac { 2 }{ 11 }\) is the probability of an event, then what is the probability of the event ‘not
Answer:
P(not A) = p(Ā) = 1 – \(\frac { 2 }{ 11 }\) = \(\left[\frac{9}{11}\right]\)

Part – B

Answer any TEN Questions. (10 × 2 = 20)

Question 11.
If A and B are two disjoint sets and n(A) = 15 and n(B) = 10. Find n(A ∪ B) and n(A ∩ B).
Answer:
n(A ∩ B) = 0
∵ A and B are disjoint net
∴ n(A ∪ B) = n(A) + n(B) = 15 + 10 = 25.

Question 12.
If U = {x : X ≤ 10 and x ∈ N} is the universal set and A = {x : X ∈ N and x is prime}, B = {x : X ∈ N and x is even} are the subsets of U, find A ∩ B’ in Roster form.
Answer:
u = {1,2,3,4,5,6,7,8,9,10}, A = {2,3,5,7), B = {2,4,6,8,10,….}
∴ B’ = {1,3,5,7,9} ⇒ ∴ A ∩ B’ ={3,5,7}.

Question 13.
If A = {1,2} B = (3,4), write A × B How many subsets will A × B have?
Answer:
A × B = {(1,3), (1,4),(2,3),(2,4)} and the number of subset = 2^m×n = 2^2×2 = 16

Question 14.
Find the value of sin 75°.
Answer:
sin 75° = sin (45° +30°) = sin Acos B + cos Asin B = sin 45o cos 30° + cos 45osin 30°

Question 15.
Find the radius of the circle in which a central angle of \(\frac{\pi}{3}\) radians intercepts an arc of length 37.4 cm (use π = \(\frac{22}{7}\))
Answer:

Question 16.
Express \(\frac{1+3 i}{1-2 i}\) in the form of a + ib.
Answer:

Question 17.
Solve 7x + 3 < 5x + 9. Show the graph of the solution on number line.
Answer:
7x + 3 < 5x + 9. ⇒ ∴ 7x – 5x < 9 – 3 ⇒ 2x < 6 ⇒ ∴[x < 3] x ∈ (-∞,3).

Question 18.
Find the equation of the line parallel to the line 3x – 4y + 2=0 and passing through the point (–2, 3)
Answer:
Line parallel to 3x – 4y + 2 = 0 can be assumed on
3x – 4y + k = 0 ∴ at (-2,3)
3(-2) – 4(3) + k = 0
∴ -6 -12 + k = 0
∴ k =18
∴ the line parallel to a given line is
3x – 4y + 18 = 0.

Question 19.
Find the distance between the parallel line 15x + 8y – 34= 0 and 15x + 8y + 31 = 0
Answer:
Distances between the parallel line is given by

Question 20.
Find the distance between the points (-3,7,2) and (2,4,-1)
Answer:
d = \(\sqrt{(-3-2)^{2}+(7-4)^{2}+(2+1)^{2}}=\sqrt{25+9+9}=\sqrt{43}\)

Question 21.
Evaluate \(\lim _{x \rightarrow 2}\left(\frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}\right)\)
Answer:

Question 22.
Write the converse and contrapositive of the statement “If the two lines are parallel then they do not intersect in the same plane.
Answer:
P : the two lines are paraallel V : line do not internet in the some plane.
∴ converse = q → p = If the two lines do not internet in the same plane then they are parallel.
Contrapositive: If the two lines intersect in the same plane they are parallel.

Question 23.
If the coefficient of variation and standard deviation of the distributions are 60 and 21 respectively, find the arithmetic mean of the distribution.
Answer:
C.V. = 60, σ = 21, X̄ = 2, Formula: C.V.= \(\frac{\sigma}{\bar{X}} \times 100\)
60 = \(\frac{21}{\bar{X}} \times 100\) ⇒ ∴ X̄ = \(\frac{21 \times 100}{60}\) = 35

Question 24.
Three coins are tossed at once find the probability of getting atleast two heads.
Answer:
3 coins tossed the sample space.

Part – C

Answer any TEN of the following questions. Each question carries THREE marks. (10 × 3 = 30)

Question 25.
If U = (1,2,3,4,5,6), A = {2,3}, B = {3,4,5} show that (A ∪ B)’ = A’ ∩ B’
Answer:
LHS = (A ∪ B) = {2,3,4,5}’ = {1,6}
RHS = A’ ∩ B’ = {1,4,5,6} ∩ {1,2,6} = {1,6} ∴ LHS = RHS.

Question 26.
Let f : R → R and g : R → R be defined by f(x) = x + 1, g(x) = 2x – 3, find (f + g)(x),(f – g)(x) and \(\left(\frac{f}{g}\right)\)(x)
Answer:
(i) (f + g)(x) = f(x) + g(x) = (x + 1) + (2x – 3) = 13x – 21
(ii) (f – g)(x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = 4 – x
(iii) (f/g)(x) = \(\left[\frac{x+1}{2 x-3}\right]\)

Question 27.
Find the general solution of the equation sin 2x + cos x = 0
Answer:
Given sin 2x + cos x = 0, sin 2x = -cosx

sin 2x + cos x = 0, ∴ cos x=-sin 2x
cosx = cos \(\left(\frac{\pi}{2}+2 x\right)\)
∴ GS
x = 2nπ ± α (n ∈ z)
Put x = 2x ⇒ α = \(\frac{\pi}{2}\) + 2x ⇒ x = 2nπ ± (\(\frac{\pi}{2}\) + 2x)
(x + 2x) = (2nπ ± \(\frac{\pi}{2}\)) ⇒ 3x = 2nπ ±\(\frac{\pi}{2}\)
∴ [x = 2nπ ± \(\frac{\pi}{6}\)]

Question 28.
Express √3 + i in polar form.
Answer:
(√3 + i) = [cose + isin) ….. (1)

r = \(\sqrt{x^{2}+y^{2}}=\sqrt{3+1}=2\) ⇒ θ = tan^-1 \(\frac{y}{x}\) = tan^-1 \(\frac{y}{x}\) = tan^-1 \(\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\)

Question 29.
Solve the equation x² + 3x + 9 = 0.
Answer:
Given

Question 30.
In how many ways can the letters of the word Permutations be arranged if (i) the words starts with P and end with S (ii) Vowels are all together. Answer:
n = 12
T repeats = 2 times ,
(i) the word starts with P and ends with S
(P ……… (5)
Total words = \(\frac{9 !}{2 !}\) = 9 × 8 × 7 × 6 × 5 × 4 × 3 = 1,81,440 ways

(ii) Vowels are together: vowels = E,U,A,I,O = S together
∴ they form 1 group remaining \(\frac{+7}{8}\) letters
∴ Total words = \(\frac{8 !}{2 !}\) = 8 × 7 × 6 × 5 × 4 × 3 = 20,160 ways.

Question 31.
Find the middle term in the expansion of \(\left(\frac{x}{3}+9 y\right)^{10}\)
Answer:
x = \(\frac{x}{3}\), a = 9y, n = 10, r = 5
n = 10 (even) there is only one middle term is \(\left(\frac{n}{2}+1\right)=\left(\frac{10}{2}+1\right)\) = 6^th term
General term T_r+1 = ⁿC_r x^n-r a^r ⇒ T_s+1 = ¹⁰C₅ \(\left(\frac{x}{3}\right)^{10-5}\) (9y)⁵
T₆ = ¹⁰C₅\(\left(\frac{x}{3}\right)^{5}\) g⁵ y⁵ ⇒ T₆ = ¹⁰C₅ . \(\frac{x^{5}}{3^{5}}\) × 9⁵ y⁵ = ¹⁰C₅ (xy)⁵ × \(\left(\frac{3^{10}}{3^{5}}\right)\)
T₆ = [¹⁰C₅ (xy)⁵ 3⁵] is the middle term.

Question 32.
Insert Five Arithmetic means between 8 and 26 such that resulting sequence is an A.P.
Answer:
a = 8, a_n, = 26, n = 7, d = ?
Let the 5 AM’s are A₁, A₂, A₃, A₄, and A₅
∴ (8), A₁, A₂, A₃, A₄, A₅ (26) ⇒ Take a_n, = a + (n – 1) d
26 = 8 + (7 – 1)d ⇒ 18 = 6d ∴ d = 3
∴ A_r, = a + d = 8 + 3 = 11 ⇒ A₂ = a + 2 = 8 + 6 =14 ⇒ A₃ = 17, A₄ = 20, A₅ = 23.

Question 33.
In an A.P. if m^th term is “n” and n^th term is “m” where m ≠ n, find p^th term..
Answer:
Given a_m = n, a_m = m a_p = ? a = ? d = ?
n^th term 2,
a_n = a + (n – 1)d

Question 34.
Find the co-ordinate of the focus, equation of the directrix and length of the Latus Rectum of the parabola y² = 8x.
Answer:
Equation of parabola y² = 8x compare with the standard form
y² = 4ax ∴ 4a = 8 ∴ a = 2
∴ focus = (a,0) = (2,0)
Equation of directrix : x = -a.
x = -2 or [x + 2 = 0]
Latus rectum = 4a = 8.

Question 35.
Find the derivative of sin x with respect of x from first principle.
Answer:
y = sin x as x → x + Δx, y + Δy
∴ (y + Δy) = sin(x + Δx)
Δy = sin(x + Δx) – y
Δy = sin(x + Δx) – sin x
÷ Δx and apply lt Δx → 0 on both side

Question 36.
Verify by the method of contradiction √2 is irrational
Answer:
Let us assume that √2 is rational i.e.. the gives statement is false.
∴ √2 = \(\frac { 1 }{ 2 }\) ≠ 0 a and b have no conunon factor.
Squaring ∴ 2 = \(\frac{a^{2}}{b^{2}}\) ⇒ [a² = 2b²] ⇒ 2 divides a
Again put a = 2C(C ∈ Z) ⇒ ∴ (2c)² = 2b²
∴ 4c² = 2b² or [b² = 2c²] 2 divide b
i.e., 2 divides both a and b hence our assumption is i.e.. a and b does flot have common contradict out statement √2 is rational is false.
∴ √2 is irrational.

Question 37.
If E and F are the evens such that P(E) = \(\frac { 1 }{ 4 }\) , P(F) = \(\frac { 1 }{ 2 }\) and P(E and F) = \(\frac { 1 }{ 8 }\) . Find (i) P (E or F) (ii) P (not E and not F)
Answer:
(i) P(E or F) = P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{7}=\frac{7+14-4}{28}=\left[\frac{17}{28}\right]\)
(ii) P(E’ ∩ F’) = P(E ∪ F)’ = 1 – P(E ∪ F) = 1 – \(1-\frac{17}{28}=\frac{11}{28}\)

Question 38.
A bag contain 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the Probability that it will be (i) red (ii) yellow (iii) blue
Answer:
Total ball = 4 Red + 3 blue + 2 yellow = 9 ball
(i) P (red ball) = \(\frac{4}{9}\)
(ii) P (Yellow) = \(\frac{2}{9}\)

Part – D

Answer any SIX questions. (6 × 5 = 30)

Question 39.
Define a modulus function. Draw the graph and write down its domain and range.
Answer:
Definition of modulus function: The function f: R → R defined by f(x) = |x|. For each x ∈ R is called modulus function. For each non-negative value of n. f(x) = x. But for negative value

Question 40.
Prove that cos²x + cos² \(\left(x-\frac{\pi}{3}\right)=\frac{3}{2}\)
Answer:

Question 41.
Prove by Mathematical induction 1³ + 2³ + 3³ + ………. + n³ \(\frac{n^{2}(n+1)^{2}}{4}\), n ∈ N
Answer:
at n = 1, LHS = RHS, \(\mathrm{l}^{3}=\frac{\mathrm{l}^{2}(1+1)^{2}}{4}\) [1 = 1]
at n = k(+)
1³ + 2³ + 3³ + ………. + k³ \(\frac{k^{2}(k+1)^{2}}{4}\) ….. (1)
Add (k+1) on both side.



Hence the given series is true for n = 1, 2, …….., k, k + 1,….. for all positive integer of n.

Question 42.
Solve the following system of inequalities graphically 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6.
Answer:

Question 43.
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (1) no girl (ii) atleast one boy and one girl.
Answer:
Given 4 girls and 7 boy’s
(i) no girls:

(ii) atleast 1 boy and 1 girls
Girls × boys
(a) 4C₁ × 7C₄ = 7 (b) 4C₂ × 7C₃ = 84.
(c) 4C₃ × 7C₂ = 210 (d) 4C₄ × 7C₁ = 140
Total number of selection 441 ways .

Question 44.
State and prove Binomial theorem for all positive integers.
Answer:
Statement: (a + b)ⁿ =nC₀aⁿ + nc₁a^n-1 b + nC₂ a^n-2 b² + nC₃ a^n-3 b³ + …….. + nC_n-1 a.b^n-1 + nC_nbⁿ
Proof: By aplying principle of mathemaical induction
Let P(n) : (a + b)ⁿ = nC₀aⁿ + nc₁a^n-1b¹ + ……. + ⁿC _nbⁿ
For n = 1 p(1) = (a + b)¹ = 1C₀a¹ + 1C₁a^1-1b¹
[(a + b) = (a – b)] is true
For n = K(f). p(k) = (a + b)k = kC₀a^k + kC₁a^k-1b +….+ KC_kb^k …..(l)
is true for n = k
We shall prove that p(k + 1) is also true i.e..


Thus it has been proved that R(k + 1) is tense for p(K) is true and it is that for all positive integer n.

Question 45.
Derive an expression for the co-ordinate of a point that divides the line joining the points A(x₁, y₁, z₁) and B(x₂, y₂, z₂)internally in the m : n. Hence find the co-ordinate of the midpoint of AB where A = (1,2,3) and B = (5,6,7)
Answer:
Proof: Let p(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be the given points
Let R(x, y, z) divide PQ internally in the ratio m : n
Draw PL, QM, RN perpendicular to xy-plane.
∴ PL ∥ RN ∥ QM
∴ PL, RN, QM lie in one plane
So that the points L, N, M lie in a straight line which is the intersection of the plane and XY plane.
Through the point R draw a line AB It to the line LM. The line AB intersect the line LP externally at A and the line MQ at B.

Triangle APR and LIQR are similar.

∴ n(z – z₁) = m(z₂ – z)
∴ nz = nz₁ = mz₂ – m₂
∴ (nz + mz) = (mz₂ + nz₁)
∴ z(m + n) = mz₂ + nz₁

Question 46.
Prove that \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\) (being in radians) and hence show that \(\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\)
Answer:
\(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\)
Proof: Consider a circle with centre ‘O’ and radius ‘r’. Mark two point A and B on the circumference of the circle so that

At ‘A’ draw a tangent to the circle produce OB to cut the tangent at C. Joint AB.
Draw BM ⊥ OA,
Here OA = OB = r
From the figure

Area of triangle OAB < area of the sector AOB < area of triangle OAC

∴ BM < rθ < AC … (1)
In triangle OBM : sinθ = \(\frac{B M}{O B}\)
∴ BM = OB sinθ = rsinθ
In triangle OAC: tan θ = \(\frac{AC}{O A}\) ∴ AC = OA tan θ = r tan θ

sin θ < 0< tan θ
÷ sin θ
\(1<\frac{\theta}{\sin \theta}<\frac{1}{\cos \theta}\)
apply lt θ → 0

Question 47.
Derive the expression for the length of the perpendicular drawn from the point (x₁ y₁) of the line ax + by + c = 0.
Answer:
Given ax + by + c = 0
∴ ax + by = -c
÷ C .



P(x, y)

Question 48.
Find the mean deviation about mean for the following data:

Answer:

Part – E

Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) Prove Geometrically that cos(x + y) = cos x cos y.sin x sin y
Answer:
Prove that cos(x + y) = cos x cos y.sin x sin y

Proof Consider a unit circle (radius = 1 unit) with centre is (0, 0).
Consider 4 point P₁, P₂, P₃ and P₄

The co-ordinate of P₁, P₂, P₃ and P₄ are given by
P₁ = (cosx,sinx) P₂ = [cos(x + y). sìn(x + y)]
P₃ = [cos(-y) sin(-y)] P₄ = [1,0]
From the figure OP₁OP₃ is congruent to ∆P₂OP₄
∴ From distance formula
P₁p₃ = P₂P₄ …… (1)
Takè the distance (P₁P₃)² = [cosx – cos(-y)]² + [sin x – sin (-y)]²
= (cosx – cosy)² + (sin x + sin y)²
= cos²x + cos²y = cosxcosy + sin²x + sin²y + 2sin cosy = 1 + 1 + 2(cosxcosy – sinxsiny)
(p₁, P₃)² = 2 – 2 cos (x + y)
Again (P₂P₄)² = [1^{(a-b)2-cos(x+y)}]² + |q – sin (x + y)|²
= 1 + cos²(x + y) -2cos(x + y) + sin²(x + y) = 1 + 1 – 2cos(x + y) = 2 – 2cos(x + y)

⇒ LHS = RHS
∴ [cos(x + y) = cosx cosy – sin x sin y]

(ii) Show that cos 2x = cos²x – sin x²x
Take cos(x + y) = cos xcos y – sin xsin y
Put y = x
∴ cos(x + x) = cosx cosx – sinx sinx
cos2x = cos² x – sin²x

(b) Find the sum of the series 5, 55, 555, 5555, … to n terms.
Answer:
S_n = 5 + 55 + 555 + ……… to n term
S_n = 5[1 + 11 + 111 + ………to n term]
\(\frac{S_{n}}{5}\) = 1 + 11 + 111 +……. to n term.
Multiply both side by 9
∴ \(\frac{9 S_{n}}{5}\) = 9 + 99 + 999+ ………. n term
\(\frac{9 S_{n}}{5}\) = (90 – 1) + (100 – 1) + (1000 – 1) +
\(\frac{9 S_{n}}{5}\) = (10 + 100 + 1000 + … nth) GP (-1 – 1 – 1 … + n term)

Question 50.
(a) Define ellipse as a set of all points in the plane and derive its equation as \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:
Let F₂ and F₂, be the focii. ‘O’ be the mid point of the line segment F₁F₂. ‘0 be the origin. And a line from O through F₂, be + ve and F₁ be -ve ∴ the co-ordinate of F₁(C₁ .0) and F₂ (c₂ o)

Let p(x,y) be a locus on the ellipse.
∴ PF₁ + PF₂ = 2a
Using distance formula

Squaring on both side

(b) Find the derivative of \(\frac{x^{3}-\cos x}{\sin x}\) with respect to X.
Answer:

Students can Download 1st PUC Maths Model Question Paper 2 with Answers, Karnataka 1st PUC Maths Model Question Paper with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Model Question Paper 2 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
2. Part-A carries 10 marks. Part-B carries 20 marks. Part-C carries 30 marks. Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

Answer ALL the questions. Each question carries one mark. (10 × 1 = 10)

Question 1.
If n(A) = 3, then find n[P(A)].
Answer:
n(A) = 3 then n[p(A)] = 2³ = 8 elements.

Question 2.
If (x – 2, y + 3) = (5, -6), then find the values x and y.
Answer:
x – 2 = 5 ∴ x=7
y + 3 = -6 ∴ y=-9.

Question 3.
Convert 750° into radians.
Answer:

Question 4.
Find the conjugate of i(2 + i).
Answer:
Conjugate i(2 + i) = 2i + i² = -1 + 2i
∴ conjugate Z̄ = -1 -2i.

Question 5.
If nC₅ = nC₁₀, then find n
Answer:
n = 5 + 10 = 15

Question 6.
Find the 5^th term of the GP. \(1, \frac{1}{5}, \frac{1}{25},\)
Answer:
a₅ = (1)\(\left(\frac{1}{5}\right)^{5-1}\) = \(\left(\frac{1}{5}\right)^{4}=\left[\frac{1}{625}\right]\)

Question 7.
Find the slope of the straight line 3x – 4y + 6 = 0
Answer:
Stage = m = \(-\frac{(3)}{-4}=\left[\frac{3}{4}\right]\)

Question 8.
Evaluate \(\lim _{x \rightarrow 4} \frac{x^{3}-64}{x-4}\)
Answer:
= n : a^n-1 = 3 (4)^3-1 = 3(4)² = 48

Question 9.
Write the negation of “Every natural number is greater than 0″.
Answer:
“Every natural number is not greated than 0.”

Question 10.
Define “sample space”.
Answer:
Sample space: Set of all possible outcome of an event is called a sample space.

Part – B

Answer any TEN Questions. (10 × 2 = 20)

Question 11.
If U = {x : x ≤ 6 and x ∈ N}, A = {3,5}, B = {2,5,6} then find (A ∪ B)’.
Answer:
A ∪ B = {3,5} ∪ [2;5,6] = [2,3,5,6]
∴ (A ∪ B)’ = {1,4}.

Question 12.
If X and Y are two sets such that n(X ∩ Y) = 10, n(X) = 38, n(Y) = 25 then find – n(X ∪ Y).
Answer:
If n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y) = 38 + 25 – 10 = 53

Question 13.
Finid the domain and range of the real function f(x) = \(\sqrt{9-x^{2}}\)
Answer:
Domain of f(x) = {x : x ∈ 2, -3 ≤ x ≤ 3}, Range of f(x)=[x : x ∈ R]

Question 14.
If in two circles, arcs of the same length subtend angles 60° and 75o at the centre, then find the ratio of their radii.
Answer:
S₁, = 5, S₂ = 5, θ₁ = 60° = \(\frac{\pi^{c}}{3}\) r₁ : r₂ = ?

∴ r₁ : r₂ = 5 : 4

Question 15.
Find the value of sin 75°
Answer:
sin 75° = sin ( 45° +30°)
[sin (x + y)= sin x cos y + cosxsin y] = sin 45° . cos 30° + cos 45°. sin 30°

Question 16.
Express the following in a + ib form \(\left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)\)
Answer:

Question 17.
Solve thr inequality 3(1 – x) <2 (x + 4) and represent it on number line
Answer:
3(1 – x) <2 (x + 4) ⇒ ∴ 3 – 3x < 2x + 8
⇒ ∴ 3 – 8 < 2x + 3x
-5 < 5x ⇒ -1 < x or x > -1.

Question 18.
Find the acute angle between the straight lines 5x + 6y – 1 = 0 and x – 11y + 8 = 0.
Answer:

Question 19.
Find the distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0.
Answer:

Question 20.
Tind the ratio in which the YZ – plane divides the line segment formed by joining the points (-2,6,7) and (2,-5,6) internally.
Answer:

Let the ratio be m : n = k : 1 (Assume)
(x₁, y₁, z₁) = (-2,6,7) ⇒ (x₂, y₂,z₂) = (2,-5,6)
yz plane divides the x coordinate is = 0

∴ 0 = \(\frac{2 k-2}{k+1}\) ⇒ ∴ 0 = 2k – 2 ∴ k = 1 ∴ ratio is 1 : 1

Question 21.
Evaluate \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x}\)
Answer:

Question 22.
Write the converse and contrapositive of “If two lines are parallel then they do not intereset in the same plane.”
Answer:
Converse: (q → p) If two line donot interest in the same plane then they are parallel,
Contrapositive (~q → -p): If two line intersect in the same plane then they are not paralle..

Question 23.
The coefficient of variation of a distribution is 60 and the standard deviatio is 21.What is the arithmetic mean of the distribution?
Answer:
C.V.= 60, σ = 21, X̄ = ? .

Question 24.
A letter is choosen at random from the word “ASSASSINATION”. Find the probability that the letter is (i) a vowel (ii) a consonant.
Answer:
Total letters = 13
Vowels = AAAIIO = 6
Consonant = SSSSNNT = 7
‘A’ repeas = 3 times. I repeats = 2
‘S’ repeats. 4 times. N repeats = 2
Probability of choosen a letter it a vowel

Probability of choosen a consonant = \(\frac{7 C_{1}}{13 C_{1}}=\left[\frac{7}{13}\right]\)

Part – C

Answer any TEN of the following questions. Each question carries THREE marks. (10 × 3 = 30)

Question 25.
In a class of 35 students, 24 like to play cricket and 16 like to play football. Also each student likes to play atleast one of the two grams. How many students like to play both cricket and football?
Answer:
Cricket = C, Football = F, n(C ∪ F) = 35, n(C ∩ F) = ?
n(C) = 24 2(F) = 16
n(C ∩ F) = n(C) + n(F) – n(C ∪ F) = 24 + 16 – 25 = 40 – 35 = 5

Question 26.
Let f = {1,1),(2,3)(0,-1),(-1-3}} be a linear function from Z into Z. Find f (x).
Answer:
Let the function be f (x) = ax + b
f (1) = a(1) + b = a + b = 1 ⇒ f(2) = a(2) + b = 2a + b = 2
f(0) = a(0) + b = [b = -1] …..(1)
f(-1) = a(-1) + b = -3
∴ b – a = -3 … (2)

(i) in (2) (-1) -a = -3
∴ -1 + 3 = a ⇒ ∴ a = 2 ⇒ ∴ f(x) = 2x – 1.

Question 27.
Find the general solution of the euation cos 4x = cos2x.
Answer:
cos 4x = cos 2x , cos 4x – cos 2x = 0
By transformation formula cosC – cos D = -2 sin \(\left(\frac{C+D}{2}\right)\) sin² (-1)
-2 sin\(\left(\frac{6 x}{2}\right)\) × sin \(\left(\frac{2 x}{2}\right)\) = 0
∴ 2 sin 3x × sin x = 0 or sin 3x = 0 or sin x = 0
2 = 0°
∴ General Solution for sin x is
x = nπ + (-1)ⁿα ⇒ 3x = nπ + 0 ∴[x = \(\frac{n \pi}{3}\) or nπ]
or
cos 4x = cos 2x ⇒ assume α = 2x
∴ General solution
x = 2nπ ± 2 ⇒ 4x = 2nπ ± 2x ⇒ ∴ 2x = 2nπ and 6x = 2nπ
x = nπ, x = \(\frac{n \pi}{3}\)

Question 28.
Solve the equation √2x² + x + √2 = 0. .
Answer:
a = √2, b = 1, c = √2

Question 29.
Express the complex number -1 + i√3 in polar form.
Answer:
-1 + i√3 = r(cos θ + isin θ) polar form

∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+3}=\sqrt{4}=2\)
θ = tan^-1(\(\left(\frac{y}{x}\right)\)) = tan^-1√3 – 1 = -(π – Standard angle) = \(-\left(\pi-\frac{\pi}{3}\right)=\frac{(-2 \pi)}{3}\)
∴ -1 + i√3 = 2cis\(\left(\frac{-2 \pi}{3}\right)\)

Question 30.
How much words.with or without meaning can be made from the letters of the word “MONDAY” assuming that no letter is repeated if
(i) Four letters are used at a time
(ii) All letters are used at a time
(iii) All letters are used but first letter is vowel?
Answer:
Monday: Given 6 letters
∴ number of arrangement = 6P₆ = 6!
(i) Taken 4 letters at a time
= 6P₄ = \(\frac{6 !}{2 !}\) = 6 × 5 × 4 × 3 = 360 ways

(ii) all letters at a limt = 6P₆ = 6! = 720 ways.

(iii) start with 0 ……… = 5P₅ = 120
Start with A ……… = 5P₅ = 120
240 ways.

Question 31.
Find the middle terms in the expansion of \(\left(\frac{n+1}{2}\right)\) and \(\left(\frac{n+1}{2}\right)\) + 1
Answer:
x = 3 , a = \(-\frac{x^{3}}{6}\), n = 9, T₅ = ?, r = 4
Formula T_r+1 = nC_r x^n-r, a^r

Question 32.
Insert five numbers between 8 and 26 such that the resulting sequence is an arithmetic progression.
Answer:
8A₁, A₂, A₃, A₄, A₅,26
a = 8, n = 7, a_n = 26, d = ?
a_n = a + (n-1)d ⇒ 26 = 8 + (7 – 1)d = 26 – 8 = 6d
18 = 6d ∴ d=3
Common difference
∴ A₁ = a + d = 8+3 =11, A₂ =11 + 3 = 14, A₃ = 17, A₄ = 20, A₅ = 23 are the 5 AM’s.

Question 33.
The sum of first three terms of a geometric progression is \(\frac { 39 }{ 10 }\) and their product is 1. Find the common ratio and the terms.
Answer:
Let the 3 term in GP are \(\frac { a }{ r }\) a and ar

Qudratic equation
∴ 10r² – 25r – 4r + 10 ⇒ 50(2r – 5) -2 (2r – 5) = 0
(2r – 5) = 0 or 5r – 2 = 0 ⇒ ∴ r = \(\frac{5}{2}\) or r = \(\frac{2}{5}\)
∴ the three terms are
at a = 1, r = \(\frac{5}{2}\)
I term \(\frac{a}{r}=\frac{1}{\frac{5}{2}}=\frac{2}{5}\)
II term a = 1
III term a_r = \(\frac{5}{2}\)

at a = 1, r = \(\frac{2}{5}\)
I term a_r = \(\frac{5}{2}\)
II term a = 1
III term = \(\frac{2}{5}\)

Question 34.
Find the equation of the circle with radius 5, whose centre lies on x-axis and passes through the point (2, 3).
Answer:
r = 5 .
Centre lies on X – axis ⇒ pt(2,3) = (x, y)
Let the circle equation be x² + y² + 2fx + 2gx + C =0 … (1)
∵ centre lies on x-axis ∴ f = 0
∴ (1) becomes
x² + y² +2 gx + c = 0) at (2,3)
4 + 9 + 2g(2) + c = 0 ∴ 48 + c = -13 …(2)
Given radius = 5
∴ \(\sqrt{g^{2}+f^{2}-c}=5 \mathrm{SBS}\)
Put f = 0. :g² – c = 25 …(3)

∴ g² + 4g – 12 = 0 ⇒ g² + 5g – 2y = 12 = 0 ⇒ g(g – 16) – 2(g + 6)=0
∴ g = -6 or g = -2
Substitute in (2) 4g + C = -13
at g = -6 4(-6) + C = -13 ⇒ C = -13 + 24 = 11
at g = -2 4(-2) + 1 = -13 ⇒ C =-13 + 8 = -5
∴ at g = -6, C = 11, f = 0 (1) becomes
x² + y² – 12x + 11 = 0
at g = -2, c = -5, f = 0
x² + y² – 4x – 5 = 0

Question 35.
Find the derivative of sin x with respect to x fron first principles.
Answer:
y = sin x as x → x + Δx, y + Δy
∴ (y + Δy) = sin(x + Δx)
Δy = sin(x + Δx) – y
Δy = sin(x + Δx) – sin x
÷ Δx and apply lt Δx → 0 on both side

Question 36.
Verify by the method of contradiction that “√7 is irrational.”
Answer:
Prove that √7 is irrational
Assume √7. is an rational :: √7 = \(\frac{p}{q}\)
Squaring on b.s.
∴ 7 = \(\frac{p^{2}}{q^{2}}\)
∴ 7q² = p² i.e., p in a multiple of again put p = 7k (k+1) :
∴ 7q² = (7k)² ⇒ 7q² = 49k² ⇒ q² = 7k²
⇒ q is also a multiple of q ⇒ ∴ both “p’ and ‘q’ are the factor of 7

∴ our assumption that √7 is an partial is wrong. ∴ √7 is an irrational.

Question 37.
A committee of two pers.ns ¡s selected from 2 men and 2 women. What is the probability that the committee will have
(i) no man (ii) one man.
Answer:
Given 2 mens × 2 women → select 2 person.

Question 38.
If E and F are two events such that P(E) = \(\frac { 1 }{ 4 }\) and P(F) = \(\frac { 1 }{ 2 }\) and P(E and F) = \(\frac { 1 }{ 8 }\) then find p(not E and not F)
Answer:
P(E) = \(\frac { 1 }{ 4 }\), P(F) = \(\frac { 1 }{ 2 }\), and p(E ∪ F) = \(\frac { 1 }{ 8 }\) P(A ∩ F) = 1
From addition rule
p(E ∩ F) = p(F) + p(F) – p(E ∪ F) = \(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2+4-1}{8}=\frac{5}{8}\)

Part – D

Answer any SIX Questions. (6 × 5 = 30)

Question 39.
Define signum function Draw its garph. Write its domain, co – domain and range
Answer:
Signum function: The function f:R → R defined by

a signum function: The domain of the signum function is R (Real nos) and the bugs and the signum function is the set {-1,0,1}

Question 40.
Prove that \(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Answer:

Question 41.
Prove that 1.2 + 2.3 + 3.4 + ……….. +n(n – 1) = \(\frac{n(n+1)(n+2)}{3}\) ∀n ∈ N, by using principal of mathematical induction.
Answer:
at n = 1
LHS = RHS


Hence the given statement is true for n =1,2,…..k, k +1…. i.e., for all (+) integral value of n.

Question 42.
Solve the system of linear inequalities: 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, y ≥ 0 graphically.
Answer:

Question 43.
What is the number of ways of choosing 4 cards from a pack In how many of these (i) four cards are of the same suit? (ii) four card belong to four different suits?
Answer:
Given 52 cards → select 4 cards.
(i) 4 cards are of same smit.
∴ total number of selection \(=\left[\frac{13 C_{4}+13 C_{4}+13 C_{4}+13 C_{4}}{4}\right]\)

(ii) 4 cards are of different suits
Total number of selection = 13C₁ × 13C₁ × 13C₁ × 13C₁ = (134)⁴ = (13)⁴ = 28.561

Question 44.
Prove binomial theorem for any positive integer n.
Answer:
Prove binomial theorem for any positive integral.
Statement: (a + b)ⁿ =nC₀aⁿ + nc₁a^n-1 b + nC₂ a^n-2 b² + nC₃ a^n-3 b³ + …….. + nC_n-1 a.b^n-1 + nC_nbⁿ
Proof: By aplying principle of mathemaical induction
Let P(n) : (a + b)ⁿ = nC₀aⁿ + nc₁a^n-1b¹ + ……. + ⁿC _nbⁿ
For n = 1 p(1) = (a + b)¹ = 1C₀a¹ + 1C₁a^1-1b¹
[(a + b) = (a – b)] is true
For n = K(f). p(k) = (a + b)k = kC₀a^k + kC₁a^k-1b +….+ KC_kb^k …..(l)
is true for n = k
We shall prove that p(k + 1) is also true i.e..


Thus it has been proved that R(k + 1) is tense for p(K) is true and it is that for all positive integer n.

Question 45.
Derive the equation of straight line in the normal form x,cos θ + y,sinω = P, where Po is the length of normal to the line from the origin and ‘ω’ is inclination of normal with positive x-axis.
Answer:
Derive the eqn. of straight line in the following
x cos ω + y sin ω = p (normal form)
consider a line cutting x-axis at A and y-axis at B.
Let the x-intercept = OA = a y-intercept OB = b

Drawn OM ⊥ to AB from 0
OM is called the length of the normal = P. It makes an angle of the norma = P. It makes an angle w (omega) w.r.t. X-axis.
In the angle AOM

In the angle ROM

Equation of the line in intercept is given by \(\frac{x}{a}+\frac{y}{b}=1\) substitute ‘a’ and ‘b’ then

∴ [xcosα w + ysin α = p] in the equation of line AB is normal form.

Question 46.
Derive the expression for the co-ordinates of a point that divides the line joining points A(x₁,y₁,z₁) and B(x₂, y₂,z₂) internally in the ratio m:n, also find the midpoint of line AB, where A = (2,8,1) and B = (2,-2,5).
Answer:
Proof: Let p(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be the given points
Let R(x, y, z) divide PQ internally in the ratio m : n
Draw PL, QM, RN perpendicular to xy-plane.
∴ PL ∥ RN ∥ QM
∴ PL, RN, QM lie in one plane
So that the points L, N, M lie in a straight line which is the intersection of the plane and XY plane.
Through the point R draw a line AB It to the line LM. The line AB intersect the line LP externally at A and the line MQ at B.

Triangle APR and LIQR are similar.

∴ n(z – z₁) = m(z₂ – z)
∴ nz = nz₁ = mz₂ – m₂
∴ (nz + mz) = (mz₂ + nz₁)
∴ z(m + n) = mz₂ + nz₁

Mid point of AB = \(=\frac{2+2}{2}, \frac{F-2}{2}, \frac{1+5}{2}=(2,3,3)\)

Question 47.
Prove that \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\) Where “x” being measured in radians.
Answer:
\(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\)
Proof: Consider a circle with centre ‘O’ and radius ‘r’. Mark two point A and B on the circumference of the circle so that

At ‘A’ draw a tangent to the circle produce OB to cut the tangent at C. Joint AB.
Draw BM ⊥ OA,
Here OA = OB = r
From the figure

Area of triangle OAB < area of the sector AOB < area of triangle OAC

∴ BM < rθ < AC … (1)
In triangle OBM : sinθ = \(\frac{B M}{O B}\)
∴ BM = OB sinθ = rsinθ
In triangle OAC: tan θ = \(\frac{AC}{O A}\) ∴ AC = OA tan θ = r tan θ

sin θ < 0< tan θ
÷ sin θ
\(1<\frac{\theta}{\sin \theta}<\frac{1}{\cos \theta}\)
apply lt θ → 0

Question 48.
Find the mean deviation about mean for the following data:

Answer:

Part – E

Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) Defind Hyperbola as a set of points and derive the equation of hyperbola in standard form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Answer:
Definition hyperbola: A Hyperbola is the act of all points in a plane the difference of whose distances.
From two fixed points in the plane is constant.
The two fixed points are called Focii and the mid point ‘O’ is called the centre of the hyperbola.
Distance between two focii F₁ and F₂ = 2c
Distance between two velocities = AB = 2a
(length of transverse axis) and b = \(\sqrt{c^{2}-a^{2}}\)
∴ 2b = length of conjugate axis and \(e=\frac{c}{a}\) is called the eccentricity of the hyperbola.
Derivation : Equation of hypergola in standard form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

From the definition PF₁ – PF₂ = 2a
Using the distane formula
\(\sqrt{(x+1)^{2}+y^{2}}-\sqrt{(x-c)^{2}+y^{2}}=2 a\)
∴ \(\sqrt{(x+c)^{2}+y^{2}}-\sqrt{(x-c)^{2}+y^{2}}=2 a\)
Squaring on both side

(b) Find the deviation of \(\frac{x^{2}-\cos x}{\sin x}\) with repect to x.
Answer:
y = \(\left[\frac{x^{5}-\cos x}{\sin x}\right]\) by using quotient rule

Question 50.
(a) Prove geometrically that cos(x + y) = cos x, cos y – sin x.sin y.
Answer:
Prove that cos(x + y) = cos x cos y.sin x sin y

w
Proof Consider a unit circle (radius = 1 unit) with centre is (0, 0).
Consider 4 point P₁, P₂, P₃ and P₄

The co-ordinate of P₁, P₂, P₃ and P₄ are given by
P₁ = (cosx,sinx) P₂ = [cos(x + y). sìn(x + y)]
P₃ = [cos(-y) sin(-y)] P₄ = [1,0]
From the figure OP₁OP₃ is congruent to ∆P₂OP₄
∴ From distance formula
P₁p₃ = P₂P₄ …… (1)
Takè the distance (P₁P₃)² = [cosx – cos(-y)]² + [sin x – sin (-y)]²
= (cosx – cosy)² + (sin x + sin y)²
= cos²x + cos²y = cosxcosy + sin²x + sin²y + 2sin cosy = 1 + 1 + 2(cosxcosy – sinxsiny)
(p₁, P₃)² = 2 – 2 cos (x + y)
Again (P₂P₄)² = [1^{(a-b)2-cos(x+y)}]² + |q – sin (x + y)|²
= 1 + cos²(x + y) -2cos(x + y) + sin²(x + y) = 1 + 1 – 2cos(x + y) = 2 – 2cos(x + y)

⇒ LHS = RHS
∴ [cos(x + y) = cosx cosy – sin x sin y]

(ii) Show that cos 2x = cos²x – sin x²x
Take cos(x + y) = cos xcos y – sin xsin y
Put y = x
∴ cos(x + x) = cosx cosx – sinx sinx
cos2x = cos² x – sin²x

(b) Find the sum to n terms of the
Answer:
Let F₂ and F₂, be the focii. ‘O’ be the mid point of the line segment F₁F₂. ‘0 be the origin. And a line from O through F₂, be + ve and F₁ be -ve ∴ the co-ordinate of F₁(C₁ .0) and F₂ (c₂ o)

Let p(x,y) be a locus on the ellipse.
∴ PF₁ + PF₂ = 2a
Using distance formula

Squaring on both side

Students can Download 1st PUC Maths Model Question Paper 1 with Answers, Karnataka 1st PUC Maths Model Question Paper with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Model Question Paper 1 with Answers

Time : 3 hrs 15 min
Max. Marks : 100

Instructions:
1. The question paper has five parts A, B, C, D and E and answer all parts,
2. Part-A carries 10 marks, Part-B carries 20 marks, Part-C carries 30 marks, Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

Answer any TEN questions : (10 × 1 = 10)

Question 1.
Given that the numbers of subsets of a set A is 16. Find the number of elements of A.
Answer:
Let n(A) = m
n [P(A)] = 16 ⇒ 2^m = 16 ⇒ m= 4
n(A)= 4

Question 2.
If tan x = \(\frac { 3 }{ 4 }\) and x lies in the third quadrant, find sin X.
Answer:
tan x = \(\frac{3}{4}=\frac{-3}{-4}\)

ordinate = -3, abscissa = 4, distance = 5 sin x = \(\frac { -3 }{ 5 }\)

Question 3.
Find the modulus of \(\frac{1+i}{1-i}\)
Answer:
\(\frac{1+i}{1-i}=\frac{(1+i)^{2}}{1-i}=\frac{2 i}{2}\) = i Modulus of i= 1

Question 4.
Find ‘n’ if ⁿC₇ = ⁿC₆
Answer:
ⁿC_n-7 = ⁿC₆ ⇒ n – 7 = 6 ⇒ n = 13.

Question 5.
Find 20th term of GP. \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}—-\)
Answer:

Question 6.
Find the distance between 3r + 4y +5= 0 and 6x + 8y +2=0.
Answer:
3x + 4y + 5 = 0) ………….. (1)
3x + 4y + 1 = 0 …………. (2)
Required distance \(\left|\frac{5-1}{\sqrt{9+16}}\right|=\frac{4}{5}\) units

Question 7.

Answer:
\(\lim _{x \rightarrow 0+}\) f(x) = \(\lim _{x \rightarrow 0} \frac{x}{x}=\frac{1}{1}=2\) ∴ f(x) in Discontinuous

Question 8.
Write the negation of For all a, b ∈ I, a – b ∈ I’.
Answer:
“There exists a, b ∈ I, such that a – b € I’
or ‘∃ a, b ∈ I, a – b ∉ I’.

Question 9.
A letter is chosen at random from the word “ASSASINATION”. Find the probability that letter is vowel.
Answer:
No of ways of selecting one vowel out of six vowels (3A’S, 21’s, 10’s) = ⁶C₁ = 6.
P(1 vowel) = \(\frac{^{6} \mathrm{C}_{1}}{^{13} \mathrm{C}_{1}}=\frac{6}{13}\)

Question 10.
Let A = {2,3,4} and R be a relation on A defined by
R={(x,y)|x,y ∈ A,x divides y}, find ‘R’.
Answer:
R= {(2, 2), (2, 4), (3, 3), (4,4)}.

Part – B

Answer any TEN questions : (10 × 2 = 20)

Question 11.
If A and B are two disjoimt sets and n(A) = 15 and n(B) = 10 find n(A ∪ B), n(A ∩ B)
Answer:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∩ B) = 0, n(A ∪ B) = 15 + 10 = 25.

Question 12.
If U = {x:r s≤10, r ∈ N} A = {x : x ∈ N, x is prime) B = {x : x ∈ N, x is even} write A ∩ B’ in roster form.
Answer:
U = {1, 2, 3 … 10), A = {2, 3, 5, 7}
B = {2, 4, 6, 8, 10} B¹ = {3, 5, 7,9}
[A ∩ B¹ = {3,5,7)]

Question 13.
If A × B = {(a,1) (a,2) (a,3) (b,1) (b,2) (b,3)}, find the sets A and B and hence fimnnd B × A
Answer:
A = {a, b}, B = {1, 2, 3}
B × A= {(1, a) (1, b) (2, a) (2, b) (3, a) (3, b)}

Question 14.
The difference between two acute angles of a right angled triangle is to \(\frac{3 \pi}{10}\) radiAnswer: Express the angles in degrees.
Answer:
Let A and B be acute angles
Given A + B = \(\frac{\pi}{2}\) = 90° : A – B = \(\frac{3 \pi}{10}\) = 54° ⇒ A = \(\frac{2 \pi}{5}\) and B = \(\frac{\pi}{5} \) ⇒ A = 72 and B = 18°.

Question 15.
Find sin \(\frac{x}{2}\) if tan x = \(-\frac{4}{3}\)and x lies in second quadrant,
Answer:
tan x = \(\frac{-4}{3}=\frac{4}{-3}\) ⇒ ∴ cos x = \(\frac{-3}{5}\)

Question 16.
\(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-5 x+6}\)
Answer:

Question 17.
CV = 60 σ = 21
Answer:

Question 18.
Write the inverse, converse of ‘If a parallelogram is a square, then it is rhombus’.
Answer:
Inverse: If a parallelogram is not a square then it is not a rhombus
Converse: If a parallelogram is a rhombus then it is a square.

Question 19.
On her vacations Veena visits four cities A, B, C and D in random order. What is the probability that she visits A before B?
Answer:
n (S) = 24
P(visiting A before B) = \(\frac{12}{24}\) = 1/2

Question 20.
In a triangle ABC with vertex A(2,3), B(4, -1) and C(1,2). Find the Length of the alitude from vertex A.
Answer:
Equation BC = x + y – 3 = 0
Length of the alitude from A (2, 3) = \(\left|\frac{2+3-3}{\sqrt{2}}\right|=\left|\frac{2}{\sqrt{2}}\right|=\sqrt{2}\) = √2.

Question 21.
Represent the complex number z = 1 + i in polar form.
Answer:
r = √2 and θ = \(\frac{\pi}{4}\) Polar form is √2 \(\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\)

Question 22.
Obtain all pairs of consecutive odd natural numbers such that in each pair both are more than 50 and their sum is less than 120.
Answer:
Taking the pair as x, x + 2,
we get x > 50, 2x + 2 < 120 ⇒ x < 59 and writing the required pairs are (51, 53), (53, 55), (55, 57), (57, 59).

Question 23.
A line cuts off equal intercepts on the co – ordinate axes. Find the angle made by the line with the positive x-axis.
Answer:
Slope =-1 ⇒ tanθ = -1
angle made = 135°.

Question 24.
If the origin is the centroid of the triangle PQR with vertices P (2a, 4, 6) Q(-4, 3b, -10) and R(8, 14, 2c) then find the values of a, b, c.
Answer:

Part-C

Answer any EIGHT of the following questions : (10 × 3 = 30)

Question 25.
Out of a group of 200 students (who know at least one language), 100 students know English, 80 students know Kannada, 70 students know Hindi. If 40 students know all the three languages, find the number of students who know exactly two languages.
Answer:
writing n(E)= 100, n(K)= 80, n(H)= 70 .
n(E ∪ K ∪ H) = 200, n (E ∩ K N ∩ H) = 40
We know that n(E ∪ K ∪ H) = n(E) + n (K) + n(H) –n (E ∩ K)
– n (K ∩ H) -n (H ∩ E) + n (E ∩ K ∩ H)
∴ n (E ∩ K) + n (K ∩ H) + n (H ∩ E) = 90

Question 26.
Let R: Z → Z be a relation defined by R = {(a,b) : a, b, ∈ Z, a – b ∈ z}. Show that
(i) ∀ a ∈ Z, (a, a) ∈ R.
(ii) (a, b) ∈ R ⇒ (b,a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R
Answer:
∀ a ∈ Z, (a, a) ∈ R since a – a = 0 ∈ Z
(a, b) ∈ R ⇒ a – b ∈ Z ⇒ ∴ b – a ∈ Z ⇒ (b, a) ∈ R
(a, b) ∈ R, (b, c) ∈ R ⇒ a – b ∈ Z, b – C ∈ Z
∴ a – b + b – c ∈ Z ⇒ (a, c) ∈ Z

Question 27.
Prove that (cos x + cos y)² + (sin x – sin y)= 4cos² \(\left(\frac{x+y}{2}\right)\)
Answer:
LHS = cos²x + cos² y + 2 cos x cos y + sin²x + sin²y – 2sin x sin y
= 1 + 1 + 2 (cos x cos y – sin x sin y) = 2 [1 + cos (x + y)] = 4cos² \(\left(\frac{x+y}{2}\right)\)

Question 28.
Solve: √2x² + x + √2 = 0.
Answer:
√2x² + x + √2 = 0; Here a = √2, b = 1,c = √2
∴ D = 5² – 4ac = (1)² – 4(√2)(√2) = 1 – 8 = -7

Question 29.
How many letters words with or with out meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ü) all letters are used at a time
(iii) all letters are used but first letter is a vowel?
Answer:
There are 6 letters in the word MONDAY. So, the total number of words is equal to the number of arrangements of these letters taken four at a time.
The number of such arrangements = ⁶P₄ = \(\frac{6 !}{(6-4) !}=\frac{6 !}{2 !}=\frac{6.5 .4 .3 .2 .1}{2.1}\) = 360
Hence, total number of words = 360
(ii) Total number of arrangements in this case = ⁶P₆ = 6! = 720
(ii) Total number of arrangements when all letters are used but the first letter is a vowel
= 2 × ⁵P₅ = 2 × 5! = 2 × 5 × 4 × 3 × 2 × 1 = 240.

Question 30.
If x + iy = \(\frac{2+i}{2-i}\) then prove that x² + y²
Answer:

Question 31.
Find the term independent of x in the expansion of \(\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}\)
Answer:

The term will be independent of x if the index of x is zero, i.e., 12 – 3r= 0. Thus, r=4
Hence 5th term is independent of x and is given by (-1)^{4 6}C₄ \(\frac{(3)^{6-8}}{(2)^{6-4}}=\frac{5}{12}\)

Question 32.
8, A₁, A₂, A₃, 24
Answer:
T_n = a + (n − 1)d
24 = 8 + (5 – 1)d
16 = 4d ⇒ d = 4
3 Am’s are 12, 16, 20.

Question 33.
(i) At least one man.
Answer:
1 man and I woman
or 2
men and 0 women
= ²C₁ × ²C₁ + ²C₂ × ²C₀ = 2 × 2 + 1 × 1 = 4 + 1 = 5.

(ii) At most one man
Answer:
1 men and 1 women
or
0 men and 2 women
= ²C₁ × ²C₁ + ²C₀ × ²C₂ = 2 × 2 + 1 × 1 = 4 + 1 = 5.

Question 34.
Find the derivative of the function ‘cos x’ w.r.t ‘x’ from first principle
Answer:
Let f ‘(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

Question 35.
A parabola with vertex at origin has its focus at the centre of x² + y² – 10x + 9 = 0 Find the direction its directrix and latus rectum.
Answer:
The centre of the circle = (5,0).
The equation y² = 20x ⇒ y² = 4ax ⇒ a = 5
directrix is x = -5, LR = 20.

Question 36.
If an A.P. if m^th term is n and the n^th term is m, where m ≠ n, find the p^th term
Answer:
We have
a_m = a + (m – 1) d = n, …… (1)
and a_n = a + (n – 1) d = m …….. (2)
Solving (1) and (2), we get
(m, n) d = n – m, or d = -1,
and a = n + m – 1 ……. (3)
Therefore a_p = a + (p – 1)d ……. (4)
= n + m – 1 + (p – 1) (-1)= n + m – p
Hence, the p^th term is n + m – p.

Question 37.
Verify by the method of contradiction that √2 is irrational
Answer:
If possible Let √2 is rational
∴ √2 = \(\frac{p}{q}\), p, q ∈ z, q ≠ 0
We assume that p and q donot have any common factor
p = √2q ⇒ p = 2q²
p² is a multiple of 2 ⇒ ∴ p is a multiple of 2
∴ p = 2k, k ∈ z ⇒ p² = 4k
2q² = 4k² ⇒ q² = 2k²
q² = 2k²
q² is a multiple of 2 ⇒ ∴ q is a multiple of z
∴ p and q are both multiple of 2 and hence has a common factor z which is a contradiction
∴ our assumption is wron
∴ √2 is irrational

Question 38.
Two students Anil and Sunil a appear in the examination The prpbability that Anil will qualify in the examination is 0.05 and that Sunil will qualify is 0.10. The probability that both will qualify in the examination is 0.02. Find the probability that Anil and Sunil will not qualify in the examination.
Answer:
Let A, B denote the events that Anil, Sunil qualify in the exam
P (A) = 0.05, P(B) = 0.1, P (A ∩ B) = 0.02
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.05 + 0.1 – 0.02 = 0.13
P(A’ ∩ B’) = 1 – P (A ∪ B) = 1 – 0.13 = 0.87.

Part – D

Answer any Six questions : (6 × 5 = 30)

Question 39.
Define signum function. Draw the graph of the signum function. Also write its domain and range.
Answer:
Singum function :
The function f:R → R defined by


is called the signum function. The domain of the signum function is R and the range is the set {-1, 0, 1}.

Question 40.
Prove that \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\) (q being in radians) and hence show that \(\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\)
Answer:
Consider a circle with centre O and radius ‘r’ Mark two points A and B on the circle so that

Join AB. Draw BM ⊥ OA
From the figure it is clear that.
Area of ∆OAB < Area of sector OAB < sector of ∆OAC …. (1)
Area of ∆OAB = \(\frac { 1 }{ 2 }\) OA.BM
[In ∆OBM. Sin θ \(\frac{\mathrm{BM}}{r}\) ⇒ BM = r Sin θ ]

∴ Area of ∆OAB = \(\frac { 1 }{ 2 }\) r r. sin θ = \(\frac { 1 }{ 2 }\)r² sin θ
Area of sector OAB = \(\frac { 1 }{ 2 }\) r²θ
Area of ∆OAC = \(\frac { 1 }{ 2 }\) OA.AC
= \(\frac { 1 }{ 2 }\)r² tan θ [In ∆OAC, tan θ = \(\frac{A C}{O A}=\frac{A C}{r}\) ⇒ AC = r tan θ]
∴ (1) becomes

Question 41.
1² + 2² + 3² + ………… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Answer:
Let P(n) : 1² + 2² + 3² + ………… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)

step 1 : Prove that P(1) is true
when n= 1; L.H.S = 1² = 1
R.H.S = \(\frac{1(1+1)(2+1)}{6}\)
∴ L.H.S = R.H.S ⇒ ∴ P(1) is true

step 2: Assume that P(m) is true
i.e., 1² + 2² + 3² + ………… + m² = \(\frac{m(m+1)(2 m+1)}{6}\) …… (i)

step 3: Prove that P(m + 1) is true
i.e., 1² + 2² + 3² + ………… + m² + (m+1)² = \(\frac{(m+1)(m+2)(2 m+3)}{6}\)
L.H.S = [1² + 2² + 3² + ………… + m²] + (m+1)²

= R.H.S ⇒ ∴ P(m+1) is true
Conclusion: P(1) is true ⇒ P(m) is true ⇒ P(m+1) is true
∴ By principle of mathematical induction, the result is true for all natural numbers ‘n’.

Question 42.
A group consists of 7 boys and 5 girls. Find the number of ways in which a team of 5 members can be selected so as to have atleast one boy and one girl.
Answer:
Number of ways of selecting
1 B and 4 G = ⁷C₁ × ⁵C₄ = 35
2 B and 3G = ⁷C₂ × ⁵C₃ = 210
3 B and 2 G = ⁷C₃ × ⁵C₂ = 350
4. B and 1 G = ⁷C₄ ×⁵C₁ = 175
Total number of selections = 770.

Question 43.
State and prove binomial theorem for positive integers.
Answer:
Statement: “If’n’ is a +ve integer then
(x + a)ⁿ = ⁿC₀ xⁿ a⁰ + ⁿC₁ x^n-1 a¹ + ⁿC₂ x^n-2 a² + ………….. + ⁿC_n x⁰ aⁿ.

Proof : (By Mathematical induction):
Let p(n) : (x + a)ⁿ = ⁿC₀ xⁿ a⁰ + ⁿC₁ x^n-1 a¹ + ⁿC₂ x^n-2a² + ………….. + ⁿC_n x⁰ aⁿ.

Step1 : Prove that P(1) is true
when n = 1, L.H.S = (x + a)¹ = x + a; RHS = ¹c₀ x¹ a⁰ + ¹c₁ x⁰ a¹ = 1.x.1 + 1.1.a = x+a
L.H.S. = R.H.S. ⇒ ∴ P(1) is true

Step 2: Assume that P(m) is true
i.e., (x + a)^m = ^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc₂  x^m-2 a² + ………….+ ^mc^mx⁰a^m …….(i)

Step 3 : Prove that P(m + 1)^m+1 is true
i.e. (x + a)^m+1 = ^m+1c₀ x^m+1 a⁰ + ^m+1c₁ x^m a¹ + ^m+1c₂ x^m-1 a² + …….. + ^m+1c_m+1 x⁰ a^m+1
Multiply both sides of equation (i) by (x+a)
∴ (x + a)^m (x + a) = (x + a) [^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc₂ x^m-2 a² + ………….+ ^mc^mx⁰a^m]
(x + a)^m+1 = x [^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc₂ x^m-2 a² + ………….+ ^mc^mx⁰a^m] + a [^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc² x^m-2 a² + ………….+ ^mc^mx⁰a^m]
(x + a)^{m + 1} = ^mc₀x^m+1 a⁰ + ^mc₁x^ma¹ + ^mc₂ x^m-1 a² + ……. + ^mc_m x⁰ a^m+1
= ^mc₀x^m+1a⁰ + (^mc₁ + ^mc₀)x^m a¹ + (^mc₂ + ^mc₁) x^m-1 a² + ………… + (^mc_m + ^mc_m-1) x¹ a^m + ^mc^m x⁰ a^m+1

(x + a)^m+1 = ^m+1c₀ x^m+1 a⁰ + ^m+1c₁ x^m a¹ + ^m+1c₂ x^m-1 a² + …….. + ^m+1c_m+1 x⁰ a^m+1
[∴ ^mc_m = ^m+1c₀
^mc₁ + ^mc₀ = ^m+1c₁
^mc₂ + ^mc₁ = ^m+1c₂
^m+1c_m]
⇒ ∴ P(m + 1) is true
Conclusion: P(1) is true, P(m) is true ⇒ P(m+1) is true
∴ By principle of mathematical induction the result is true for all natural numbers n.

Notes: 1. The number of terms in the expansion of (x + a)ⁿ is n + 1.
2. The Gen term of binomial expansion is T_r+1 = ⁿc_r x^n-r a^r.

Question 44.
Derive an expression for the coordinates of a point that divides the line joining the points A(x₁, y₁, z₁,) and B (x₂, y₂, z₂.) internally in the ratio m : n. Hence, find the coordinates of the midpoint of AB where A = (1, 2, 3) and B = (5, 6, 7).
Answer:

Let the two given points be P (x₁, y₁, z₁,) and Q (x₂, y₂, z₂). Let the point R (x, y, z) divide PQ in the given ratio m : n internally, Draw PL,
QM and RN perpendicular to the XY-plane. Obviously PL ∥ RN ∥ QM and feet of these perpendiculars lie in a XY-plane. The points L, Mand N will lie on a line which is the intersection of the plane containing PL, RN and QM with the XY-Plane. Through the point R draw a line ST parallel to the line LM. Line ST will intersect the line LP externally at the point S and the line MQ at T, as shown in Fig 12.5.

Also note that quadrilaterals LNRS and NMTR are parallelograms.
The triangles PSR and QTR are similar. Therefore,


Similarly, by drawing perpendiculars to the XZ and YZ-planes, we get

Hence, the coordinates of the point R which divides the line segment joining two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂,) internally in the ratio m: n are

Question 45.
Derive a formula for the angle between two lines with slopes m¹ and m². Hence find the slopes of the lines which make an angle \(\frac{\pi}{4}\) with the line x – 2y + 5 = 0.
Answer:
m₁ = tan θ₁, m₂ = tan θ₂

Question 46.
Prove that \(\frac{\cos 4 x+\cos 2 x+\cos 3 x}{\sin 4 x+\sin 2 x+\sin 3 x}=\cot 3 x\)
Answer:

Question 47.
Solve graphically 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Answer:
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6 …..(1)
Points A(2, 0) and B(0,4) lie on 2x + y = 4
Plot the points and join them to get line AB
(0,0) does not satisfy 2x + y ≥ 4
⇒ Half plane given by 2x + y ≥ 4 is away from origin …..(2)
Points C(3, 0) and D(0, 3) lie on x + y= 3
Plot the points and join them to get line CD.
(0,0) satisfies x + y ≤ 3 ….. (3)
⇒ Half plane given by x + y ≤ 3 is towards origin
Points C(3, 0) and E(0, -2) lie on 2x – 3y = 6

Plot the points and join them to get line CE.
(0,0) satisfies 2x – 3y ≤ 6
⇒ Half plane given by 2x – 3y ≤ 6 is towards origin
From (2), (3), (4) common region shown shaded in figure represents solution of (1).

Question 48.
Find the mean deviation about the mean for the following data.

Answer:

Part – E

Answer any ONE questions : (1 × 10 = 10)

Question 49.
(a) To cos (A + B) = cos x. cos y – sin x sin y and hence find cos 75°.
Answer:

(i) becomes
∴ cos (A + B)= cos A.cos B – sin B
cos 75° = cos (45 +30)
= cos 45 cos 30 – sin 45 sin 30

(b) Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + ……..
Answer:
nth term T_n = 1² + (1² + 2²) + (1² + 2² + 3²) + ……. + n² = \(\frac{n(n+1)(2 n+1)}{6}=\frac{2 n^{3}+3 n^{2}+n}{6}\)
∴ T_n = \(\frac{2 n^{3}+3 n^{2}+n}{6}\)
Sum to n terms, S_n = ΣT_n
= \(\frac{1}{6}\) [2Σn³ +3Σn² + Σn]

∴ S_n = \(\frac{n(n+1)^{2}(n+2)}{12}\)

Question 50.
(a) An elipse is the set of all ninte in a plane tha e of whose distance from two fived points in the need of all points in a plane the sun of whose distance points in the plane is a constant.
Answer:

Let F₁ and F₂ be the foci and O be the mid point of the line segment F₁F₂ Let O be the origin and the line from 0 through F₂ be the positive x-axis and that through F₁ as the negative x-axis. Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F₁ be (-c, 0) and F₂ be (c, 0).

Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a
i.e., PF₁ + PF₂ = 2a. …(1)
Using the distance formula, we have
\(\sqrt{(x+c)^{2}+y^{2}}+\sqrt{(x-c)^{2}+y^{2}}=2 a\)
i.e., \(\sqrt{(x+c)^{2}+y^{2}}=2 a-\sqrt{(x-c)^{2}+y^{2}}\)
Squaring both sides, we get
(x + c)² + y² = 4a² – 4a \(\sqrt{(x-c)^{2}+y^{2}}\) + (x – c)² + y²
which on simplification gives
\(\sqrt{(x-c)^{2}+y^{2}}=a-\frac{c}{a} x\)
Squaring again and simplifying, we get

(b) \(y=\frac{x^{5}-\cos x}{\sin x}\)
Answer: