Kerala SSLC Chemistry Model Question Paper 2 English Medium

Students can Download Kerala SSLC Chemistry Model Question Paper 2 English Medium Pdf, Kerala SSLC Chemistry Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Chemistry Model Question Paper 2 English Medium

General Instructions:

  1. The first 15 minutes is the cool off time. You may use the time to read and plan your answers.
  2. Answer the questions only after reading the instructions and questions thoroughly.
  3. Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
  4. Five questions are given in each section. Answer any four from each section.
  5. Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks

Section  – A

(Answer any 4 questions from 1 to 5. Each question carries 1 score) (4 × 1 = 4)

Question 1.
Which one of the following is not possible?
a) 2s 2
b) 3p3
c) 2d5
d) 4F2
Answer:
NH4OH

Kerala SSLC Chemistry Model Question Paper 2 English Medium

Question 2.
Which metal cannot displace copper from CuSO4. solution.
a) Magnessium
b) Silver
c) Zinc
d) Iron
Answer:
Silver

Question 3.
The substance obtained at the cathode when NaCl solution is electrolysed is …………..
a) Sodium
b) Chlorine
c) Hydrogen
d) Sodium hydroxide
Answer:
Hydrogen

Question 4.
The chemical formula of liquor ammonia is ………..
a) NH4OH
b) NH4Cl
c) (NH2)SO4
d) NH3
Answer:
NH4OH

Question 5.
The symbol of carboxylase functional group is ……………
a) -OH
b) – COOH
c) -CH3
d) – O –
Answer:
-COOH

Kerala SSLC Chemistry Model Question Paper 2 English Medium

Section – B

(Answer any 4 questions from 6 to 10. Each question carries 2 scores)(4 × 2 = 8)

Question 6.
The subshell electron Configuration of certain elements are given below.
Find out the wrong ones and correct them (symbols are not real)
a) 12P = 1s22s22p63s2
b) 6Q = 1s22s23s2
c) 178R = 1s22s22p63s23p6
d) 20S = 1s22s22p63s23p63d2
Answer:
Q – 1s22s22p2
S – 1s22s22p63s23p63d64s2

Question 7.
Calculate the number of molecules in 90g of water.
(Hint -Atomic mass H = 1 O = 16)
Answer:
1 GMM of water = 18 g.
90 g water = \(\frac{90}{18}\) = 5 mol
No. of molecules in 5 mol = 5 x 6.022 x 1023

Kerala SSLC Chemistry Model Question Paper 2 English Medium

Question 8.
Observe the solutions and the metals placed in three test tubes.
Kerala SSLC Chemistry Model Question Paper 2 English Medium - 1
a) In which test tube a chemical reaction is likely to take place? Why. (1)
b) Write the name of the reaction. (1)
c) Write the equation of the redox reaction taking place here. (1)
Answer:
a) In test tube 2. More reactive iron displaces copper from CuSO4 solution.
b) Displacement reactions
c) Fe + Cu2+ → Fe2+ + Cu

Question 9.
Metallurgy involves all the process leading to the separation of a pure metal from its ore.
a) Distinguish a mineral from an ore. (1)
b) What type of ores are usually concentrated by froth floatation process. Given an example. (2)
Answer:
a) Elements or their compounds, occurring naturally and obtained by mining are called minerals, Ore is the mineral from which a metal is extracted economically, easily and quickly.
b) Sulphide ores eg. Copper pyrites.

Kerala SSLC Chemistry Model Question Paper 2 English Medium

Question 10.
Analyse the chemical equations given below and answer the questions that follow.
Kerala SSLC Chemistry Model Question Paper 2 English Medium - 2
a) Identify A, B and C. (2)
b) Write what type of chemical reactions (i) and (ii) are (1)
Answer:
a) A = CH3-CH = CH2 or C3H6
B = CH3 – CH2 – CH3 or C3H8
C = CH3 – CH2 – CH2 – Cl

b) i) Addition reaction
ii) Substitution reaction

Section – C

(Answer any 4 questions from 11 to 15. Each question carries 3 scores) (4 × 3 = 12)

Question 11.
The outermost electronic configuration of an element , X is 3s2 3p1. Answer the questions given below related
to this element. (Symbol of the element is not actual).
a) What is the atomic number of this element? (1)
b) Write the group number and period number of X? (1)
c) Write the formula of the oxide of this element?(1)
Answer:
a) 13
b) Group-13, Period-3
c) X2O3

Kerala SSLC Chemistry Model Question Paper 2 English Medium

Question 12.
The atomic masses of certain elements are S = 32, O = 16, H = 1,C = 12
a) Find out 1 GMM of H2SO4 (1)
b) Find out the number of molecules in 220 gCO2(2)
c) What will be the volume of 1 mole SO2 at STP?(1)
Answer:
a) 98g
b) 220g CO2 = \(\frac{220}{44}\) = 5 mol
No. of molecules in 5 mol = 5 x 6.022 x 1023
c) 22.4 L

Question 13.
Analyse the picture of an electro chemical cell given below, and answer the questions that follow.
Kerala SSLC Chemistry Model Question Paper 2 English Medium - 3
a) Which are the anode and cathode in this cell?(1)
b) Write the equation of chemical reaction taking
place at both of the electrodes. (2)
c) Write the equation of the redox reaction.
Answer:
a) Anode – Mg
Cathode – Cu
b) At anode Mg →Mg2+ + 2e
At cathode Ca2+ + 2e → Cu
c) Mg + Cu2+ → Mg2+ + Cu

Kerala SSLC Chemistry Model Question Paper 2 English Medium

Question 14.
Certain metals their characteristics and the method of refining are tabled below. Match them suitably.(3)
Kerala SSLC Chemistry Model Question Paper 2 English Medium - 4
Answer:
Tin-Low melting point – Liquation
Mercury – Low boiling point – Distillation
Copper – Low electropositivity – Electrolysis

Question 15.
The structural formula of an organic compound is given below.
CH3-CH2-CH2-CH2-OH
a) Write the IUPAC name of the compound. (1)
b) Write the structure of a position isomer of this compound. (1)
c) Write the structure of a functional group isomer of this compound. (1)
Answer:
a) Butan -1 – ol
Kerala SSLC Chemistry Model Question Paper 2 English Medium - 8
(c) CH3 – CH2 – O – CH2 – CH3 (OR)
CH3 – O – CH2 – CH2 – CH3

Kerala SSLC Chemistry Model Question Paper 2 English Medium

Section – D

(Answer any 4 questions from 16 to 20. Each question carries 4 scores)(4 × 4 = 16)

Question 16.
The chemical equation showing the preparation of 8 -10% of ethnol is given below
Kerala SSLC Chemistry Model Question Paper 2 English Medium - 5
a) Write the products A and B. (1)
b) How the product obtained is converted in to rectified spirit? (1)
c) What is denatured spirit? (1)
Answer:
a) A = C6H12O6
A = CO2
b) By fractional distillation
c) Ethanol added with poisonous substances to avoid its misuse as a beverage.

Question 17.
Analyse the compounds given below
Kerala SSLC Chemistry Model Question Paper 2 English Medium - 6
a) Which among these compounds is an ester?(1)
b) Choose the compounds required to form this ester and write the chemical equation. (1)
c) Write any two uses of esters. (1)
Answer:
a) CH3-COO-CH3
b) CH3-COOH-CH3-OH →CH3-COO-CH3
c) Used as a flavouring agent for food. Used for making perfumes.

Question 18.
Two reversible reactions are given below:
i) H2(g) + l2(g)⇌ 2Hl(g)
ii) N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat
a) In which reaction change of pressure has no effect.
b) Suggest two methods to increase the amount of product in the second reaction.
c) Even though the forward reaction is exothermic the temperature is maintained at 450°C. Why?
Answer:
a) Pressure has no effect in reaction -1. Because there is no change in number of molecules due to forward and backward reactions.
b) Increase pressure Increase the concentration of the reactants Remove the products continuously from the system.
c) 450°C is the optimum temperature of this reaction. At this temperature maximum product is obtained. At low temperature NH3 decomposes back in the reactants.

Kerala SSLC Chemistry Model Question Paper 2 English Medium

Question 19.
At one end of a glass tube cotton soaked in cone.
HCl and at the other end of it cotton soaked in cone.
NH4OH solution is inserted.
Kerala SSLC Chemistry Model Question Paper 2 English Medium - 7
a) What will be the observation after sometime?
b) What will happen if the glass tube is heated?
c) Write equations of the chemical reactions in a and b
Answer:
a) A white precipitate is formed inside the glass tube. NH3 gas and HCI gas combines to form NH4Cl.
b) The precipitate disappears on heating. On heating NH4Cl decomposes into NH3 and HCl

c) NH3 + HCI → NH4Cl
NH4Cl → NH3 + HCl

Question 20.
Match those given in columns A, B and C suitably.

A B C
Bayles Law Reversible Reaction V&T
Charles Law Volume & number of molecules V ∝ 1/p
Avagadros Law Volume & pressure V ∝ η
Le-Chatelers principle Volume and temperature Equillibrium state

Answer:

A B C
Bayles Law Volume & pressure V&T
Charles Law Volume and temperature V ∝ 1/p
Avagadros Law  Volume & number of molecules V ∝ η
Le-Chatelers principle  Reversible Reaction Equillibrium state

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Students can Download Kerala SSLC Chemistry Model Question Paper 1 English Medium Pdf, Kerala SSLC Chemistry Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Chemistry Model Question Paper 1 English Medium

General Instructions:

  1. The first 15 minutes is the cool off time. You may use the time to read and plan your answers.
  2. Answer the questions only after reading the instructions and questions thoroughly.
  3. Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
  4. Five questions are given in each section. Answer any four from each section.
  5. Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks

Section – A

(Answer any 4 questions from 1 to 5. Each question carries 1 score) (1 × 5 = 5)

Question 1.
The element with 1s2 2s2 2p6 3s2 3p6 is placed in the group of the periodic table.
a) 1
b) 6
c) 8
d) 18
Answer:
18

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Question 2.
The reaction taking place at the anode of a zinc – copper galvanic cell …………
a) Zn → Zn2+ + 2e
b) Zn2+ + 2e → Zn
c) Cu → Cu2+ + 2e
d) Cu2+ + 2e → Cu
2.
Answer:
Zn → Zn<>sup>2+ + 2e

Question 3.
The metal that liberate hydrogen gas when react with steam is ………….
a) Magnesium
b) Zinc
c) Iron
d) Sodium
Answer:
Iron

Question 4.
The substance used to remove moisture from ammonia is ……………….
a) Vanadium pentoxide
b) Sulphuric acid
c) Calcium oxide
d) Silicon dioxide
Answer:
Calcium oxide

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Question 5.
The functional group present in the compound CH3 – O-CH3 is …………
a) Hydroxyl
b) Alkoxy
c) Methyl
d) Carboxylic
Answer:
Alkoxy

Section – B

(Answer any 4 questions from 6 to 10. Each question carries 2 scores)(4 × 2 = 8)

Question 6.
Calculate the number of molecules in 90g of water. (Hint-Atomicmass H = 1 O = 16)
Answer:
1 GMM of water = 18 g.
90 g water = \(\frac{90}{18}\) = 5 mol
No. of molecules in 5 mol = 5 x 6.022 x 1023

Question 7.
The molecular mass of CO2 is 44.
a) Find out the number of molecules in 220g CO2.
Answer:
a) 220g CO2 = \(\frac{224}{44}\) = 5 mol
∴ No. of molecules = 5 x 6.022 x 1023
b) 5×22.4 = 112 L

b) What will be the volume of this much CO2 at STP?
(Hint: The molar volume of gases at STP = 22.4 L)

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Question 8.
The equation of the industrial production of ammonia is given below.
N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat
a) By which name the industrial production of ammonia is known?
b) What will be the effect of increased pressure in ‘ this reaction?
c) What will happen if ammonia is removed from the system at regular intervals?
Answer:
a) Heber’s process
b) Due to forward reaction number of molecules decreases. So increase in pressure is favourable to forward reaction.
c) Forward reaction increases, more product will be formed.

Question 9.
Certain elements are arranged in the order as they appear in the electrochemical series.
Answer the questions that follows based on this Na > Mg > Al > Zn > Cu > Ag
a) Which of these element is likely to be found in free state in nature?
b) Which of these metal will produce the stablest compound?
c) Which of these metal is produced by reducing its ore with CO?
Answer:
a) Ag
b) Na
c) Zn

Question 10.
The structural formula of an organic compound is given below.
Kerala SSLC Chemistry Model Question Paper 1 English Medium - 1
a) What is the total number of carbon atom in the longest carbon chain in this compound?
b) Write the name and position of the branch.
c) Write the IUPAC name of the compound.
Answer:
a) 5
b) Methyl, 3
c) 3 – methyl pentane

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Section – C

(Answer any 4 questions from 11 to 15. Each question carries 3 scores) (4 × 3 = 12)

Question 11.
The subshell electron configuration of certain elements are given below. (Symbols are not real)
A = [Ar] 4s2 3d1 B = [Ne]3s1
C = [Ar] 4s1 D = [Ne]3s23p5
a) Write any characteristic of the element A.
b) Which among these belong to the same group?
c) Which among these belong to the same period?
Answer:
a) Shows variable oxidation states or forms coloured compounds.
b) B and C
c) C and D

Question 12.
Certain samples are given below.
Kerala SSLC Chemistry Model Question Paper 1 English Medium - 2
a) Find out the number of moles in each sample.
b) Arrange them in the increasing order of number of molecules.
(Hint: Atomic mass of H = 1, He = 4, Ca = 20)
Answer:
200g H2 = \(\frac{200}{2}\) = 100 mol
200 g He = \(\frac{200}{4}\) = 50 mol
200 g Ga = \(\frac{200}{20}\) = 10 mol
200g Ca < 200g He < 200g H2

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Question 13.
a) Which substances are used as the anode,
cathode and electrolyte when copper is refined by electrolysis.
b) It is said that electrolytic refining of copper is economic. Why?
Answer:
a) Anode – Impure copper (Copper to be purified)
Cathode – Pure Copper rod
Electrolyte – CuSO4 solution
b) When copper is purified by electrolysis the residue left behind at the anode, called anode mud, often contains precious metals like gold.

Question 14.
Match those given columns suitably.

Metal Name of ore Chemical formula
Aluminium Calamine Al2O3.2H2O
Iron Bauxite ZnCO3
Zinc Haematite Fe2O3

Answer:
Aluminium – Bauxite – Al2O3-2H2O
Iron – Haematite – Fe2O3
Zinc – Calamine – ZnCO3

Question 15.
There are compounds with same molecular formula but different structural formula,
a) Write two compounds with different structural formula and the molecularformulaC3H8O.
b) By what name these compounds are known?(1)
Answer:
a) CH3-O-CH2-CH3
CH3-CH2-CH2-OH
b) Isomers

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Section – D

(Answer any 4 questions from 16 to 20. Each question carries 4 scores)(4 × 4 = 16)

Question 16.
In the table given below, under A certain chemical reactions and under B types of chemical reactions are given. Match them suitably.
Kerala SSLC Chemistry Model Question Paper 1 English Medium - 3
Answer:
CH4 + Cl2 → CH3 – Cl + HCl
– Substitution reaction

CH2 = CH2 + H2 → CH3 – CH3
-Addition reaction

CH3-CH2-CH2-CH3 → CH4-C3H6
– Thermal cracking

nCH2 = CH2 → [-CH2-CH2-]n
– Polymerisation

Question 17.
The chemical reactions taking place inside blast furnance during the manufacture of ‘iron is given below. Analyse it and answer the questions given below.
CaCO3(s) → CaO(s) + CO2(g)
CaO(s) + SiO2(s) → CaSiO3(s)
C(s) + O2(g) → CO2(g) + Heat
CO(g) + C(s) + Heat → 2CO(g)
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
a) Write the chemical formula of the ore of iron.
b) Which compound is the actual reducing agent in the reaction?
c) Which equation shows slag formation?
Answer:
a) Fe2O3
b) CO
C) CaO(s) + SiO2(s) →CaSiO3(s)
d) FeO3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Question 18.
The equation of a reversible reaction is given below.
2SO2(g) + O2(g) ⇌ 2SO3(g) + Heat
a) Write the equations of endothermic and exothermic reactions.
b) Write the mole ratio of the reactants and products
c) Suggest methods to get more product.
Answer:
a) Endothermic
2SO3(g)+Heat → 2SO2(g +O2(g)
Exothermic
2SO2(g) + O2(g) → 2SO3(g)+Heat

b) 2:1:2
c) Increase the concentration of the reactants
Remove the product from the system
Decrease the temperature
Increase the pressure.

Question 19.
a) Sulphuric acid is known as ‘the king of chemicals’.
What is the basis for this?
b) How concentrated sulphuric acid is diluted in the laboratory?
c) The wooden cupboards in which concentrated sulphuric acid is often seen charred. Why?
Answer:
a) Due to the wide range of uses sulphuriç acid is known as the king of chemicals eg: Used to make explosives, petroleum refining dehydrating agent, manufacture of chemical fertilizers, etc.
b) Take water in a beaker and add concentrated acid slowly along the side and stir continuously. Otherwise, the add will spurt over as the dissolution is exothermic.
c) Sulphuric absorbs hydrogen and oxygen present in substances in the ratio as that in water. So carbon only remains (dehydrating property)

Kerala SSLC Chemistry Model Question Paper 1 English Medium

Question 20.
a) How do galvanic cells and electrolytic cells differ each other?
b) Write any two practical uses of electrolysis.
Answer:
a)

  • Galvanic cells converts chemical energy in to electrical energy.
  • Electrolytic cell converts electrical energy into chemical energy
  • The anode in a galvanic cell is negative and cathode is positive.
  • The anode in an electrolytic cell is positive and cathode is negative

b)

  • Production of metal .
  • Production of non metals
  • Production of compounds
  • Refining of metals
  • Electroplating

 

Kerala SSLC Biology Question Paper March 2019 English Medium

Students can read Kerala SSLC Biology Question Paper March 2019 with Answers English Medium and Kerala SSLC Biology Previous Year Question Papers with Answers helps you to score more marks in your examinations.

Kerala Syllabus Class 10 Biology Question Paper March 2019 with Answers English Medium

Instructions :

  • The first 15 minutes Is the Cool-off time.
  • You may use the time to read the questions and plan your answers.
  • Answer only on the basis of instructions and questions given.
  • Consider score and time while answering.

Time: 1½ Hours
Total Score: 40 Marks

Section – A

Answer any five questions from Q.No. 1 to 6. Each carries on score. (5 × 1 = 5)

Question 1.
Identify the word pair relation and fill the blanks.
(a) Monkeys: Cercopithecoidea::
Chimpanzee: ………………..
(b) A.I.Oparin: Theory of chemical evolution::
Hugo de Vries: ………………….
Answer:
a) Hominoidea
b) Mutation theory

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Question 2.
Find out the parts that are not related to retina from the following.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 1
Answer:
Conjuctiva
Iris

Question 3.
“Myelin sheath accelerates the speed of impulses through axon and provides nutrition to it.”
(a) How does myelin sheath form?
Answer:
Myelin sheath is formed due to the repeated encir- * cling of Schwann cells around the axone.

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Question 4.
Identify the relation in the Indicator (A) and complete (B) accordingly.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 2
Answer:
i. Ribose Sugar
ii. AUGC.

Question 5.
Find out the fungal diseases from the following: Malaria, Ringworm, Filariasis, Athelete’s foot
Answer:
Ring worm Athlete’s foot

Question 6.
Complete the statement suitably:
“In …….(a)……………. the specialised part in pancreas two types of cells are found. Of these (b)……. cells produce insulin.”
Answer:
a) Islets of Langerhans
b) Beta cells

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Answer any six questions from Q.No. 7 to 13. Each carries on score. (6 × 2 = 12)

Question 7.
Make suitable pairs of different white blood cells and the function they perform.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 3
Answer:
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 15

Question 8.
Write the name of pathogens and symptoms of the given diseases:
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 4
Answer:
A: Malaria
Pathogens: Protozoa – plasmodium

Symptoms

  • High fever with shivering
  • Profuse sweating, severe headache
  • Vomitting, diarrhoea, anemia

B: Tuberculosis
Pathogens: bacteria: Mycobacterium tuberculosis

Symptoms

  • Loss of body weight
  • Fatigue
  • Continuous dry cough

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Question 9.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 5

  • Excess blood is lost even though minor, injuries.
  • Loss of body balance
  • Accumulation of insoluble proteins in the neutral tissues of brain
  • Production of dopamine reduces
  • Irregular flow of electric charges in brain.

Answer:
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 16

Question 10.
Observe the illustration and answer the questions.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 6
(a) Name the Scientists who devised the experi-mental set up shown above.
(b) Which theory of evolution is substantiated by this experiment?
Answer:
a) Stanley L.Miller and Harold C.Urey
b) Theory of chemical evolution

Question 11.
Observe the illustration and answer the questions.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 7
Answer:
a. A: Genetic scissors: Restriction endonuclease
B: Genetic glue : Ligase

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

b. Yes

  • They will have the ability to prouce insulin.
  • They change that occured in the genetic constitution will be transferred to the next generation too. In case of mutation the gene might loss the capacity for the production.

Question 12.
List out the four major concepts to be included in a blood donation campaign.
Answer:

  • Blood cannot be made artificially. So we can save life of a person by donating blood. So do-nate blood and save life.
  • A healthy person can donate 300 ml of blood within a period of 6 months.
  • If the level of blood decreases beyond a certain level, it may cause the death of the individual. During such situation for the sustenance of the life the blood donation by a person becomes inevitable.
  • Before transfusion, blood group matching should be ensured.

Question 13.
Mutation cause variations in organisms. It leads to evolution of species:
(a) What is mutation?
(b) Explain two other factors that cause variations in organisms.
Answer:
a) A sudden heritable change in the genetic con-stitution of an organism is called mutation.
b) Crossing over in chromosomes

Answer any five from Q.No. 14 to 20. Each carries 3 score. (5 x 3 = 15)

Question 14.
Observe the illustration and answer the questions.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 8
(a) Why do the forelimbs of these organisms show differences in external appearance?
(b) What inferences regarding evolution can be drawn from the anatomy of these organs?
(c) Write any two other scientific evidences which proves evolution.
Answer:
a) Difference in their external appearances are their adaptations to live in their own habitats.
b) Anatomical resemblances justify the interference that all organisms evolved from a common ancestor.
c) Biochemistry and Physiology, Molecular Biology and Evidence from fossils

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Question 15.
Observe the illustration.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 9
(a) What does R, r denote in the illustration
(b) Which character is expressed in first generation. Why?
Answer:
a) Gametes / Allele
b) Red Flower
Hybridization experiment, the allele that controls the dominant character (Red) that is expressed, and other character remains hidden (recessive character-white) in the offsprings of the first generation.

Question 16.
“Smoking harmfully affects internal organs.”
This is a general statement.
Explain how smoking affects brain, heart and lungs.
Answer:
a) Brain: Stroke, Addiction to nicotine
b) Lungs: Lung cancer, Bronchitis, Emphysema
c) Heart: Hypertension, loss of elasticity of arteries, Decrease in functional efficiency.

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Question 17.
There are certain mistakes in the given chart. Find out and correct it.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 10
Answer:
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 17

Question 18.
Analyse the statement and answer the questions. “Antibiotics, the miraculous medicines of 20th century helped a lot to bring many diseases under control. But the use of antibiotics without consulting doctor is not advisable.”
(a) Why antibiotics considered as miraculous medi-cines?
(b) Write two side effects of antibiotics.
Answer:
a) Antibiotics are drugs obtained from microorgan isms that are used to destroy the growth of other microorganisms that cause diseases. Antibiotics are biochemical substances extracted from living things like bacteria and fungi which can or prevent the spreading of germs. Antibiotics target microorganisms such as bacteria, fungi and parasites.

b) Side effects of antibiotics:

  • Regular use develops immunity in pathogens against antibiotics.
  • Destroys useful bacteria in body.
  • Reduces the quantity of some vitamins, in body.

Question 19.
Correct mistakes if any in the underlined part of the given statements.
(a) Curvature of lens increases when viewing near objects.
(b) Vitreous humor is formed from blood, and is re-absorbed by blood.
(c) Membraneous labyrinth in the inner ear is filled with Perilymph.
(d) Eustachian tube amplifies and transmits the vi-brations of tympanum to the internal ear.
Answer:
(a) Curvature of lens increases when viewing near objects.
(b) Vitreous humor is formed from blood,and is re-absorbed by blood.
(c) Membraneous labyrinth in the inner ear is filled with endolvmph.
(d) Ear ossicles amplifies and transmits the vibration of tympanum to the internal ear.

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Question 20.
Observe the figure and answer the questions:
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 11
(a) Identify the partsAand B.
(b) What is the function of ‘A’?
(c) Explain the process that takes place in ‘B’.
Answer:
a) A: mRNA B: Ribosome
b) mRNA:mRNA carries information from DNA to ribosomes and controls protein synthesis.
c) mRNA molecule that carries information from DNA to ribosomes.
mRNA reaches ribosomes. tRNA carry different kinds of amino acids to ribosomes.
Based on the information in mRNA protein is synthesized in ribosomes adding amino acids.

Answer any two from Q.No. 21 to 23. Each careries 4 score. (2 x 4 = 8)

Question 21.
Analyse the given informations related to plant hor mones and answer the questions.
(a) to increase the size of apple
(b) to prevnt dropping of premature fruits.
(c) to increase the production of latex.
(d) to enable flowering of pineapple plants at a time.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 12
(i) Match hormones and their functions properly.
(ii) Write the name and function of any other two hormones occur naturally in plants.
Answer:
i) a) Gibberellins
b) Auxins
c) Ethyphon
d) Ethylene

ii)

Plant hormones Functions
Cytokinins Promotes cell division, cell growth and differentiation along with auxin.
Abscisic acid Control the dormancy of embryo in the seed.
Control flowenng
Helps to sustair the plant in adverse conditions

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Question 22.
Observe the illustration and answer the following questions.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 13
(a) Name the cells (A) and (B).
(b) Explain the role of these cells in making vision possible.
(c) How impulses are generated in these cells when light rays fall on it?
Answer:
a) A. Rod cells B. Cone cells

Photo receptors Functions
Rod cells Vision in dim light, black and white vision
Cone cells Bright light vision, colour vision

b) Working of the cone
When the light falls on cone cells, the photopsin, in them dissociate into retinal and opsin. This chemical change creates impulses.
Working of rod cells :
d cells the pigment rhodopsin in them dissociate into retinal and opsin. This chemical change creates impulses.

c) The sense of vision :
When the pigment photoreceptors dissociate, impulses are forrmed. When get the sense of vision, when these impulses reach the brain through optic nerve.

Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium

Question 23.
(a) Redraw the diagram.
Name and label the parts that perform the given functions.
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 14
(a) Relay station of impulses.
(b) Controls heartbeat, breathing etc.
(c) Maintains equilibrium of the body.
Answer:
Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium - 18
a) Thalamus
b) Medulla oblongata
c) Cerebellum

Kerala SSLC Social Science Question Paper March 2019 with Answers Malayalam Medium

Students can read Kerala SSLC Social Science Question Paper March 2019 with Answers Malayalam Medium and Kerala SSLC Social Science Model Question Papers helps you to score more marks in your examinations.

Kerala Syllabus Class 10 Social Science Question Paper March 2019 with Answers Malayalam Medium

Instructions:

  • The first 15 minutes is the cool-off time. You may use the time to read the questions and plan your answers.
  • Answer all questions in PART – A. Answer any one from the questions given under each question number in PART – B.

Time: 2½ Hours
Total Score: 80 Marks

Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 1
Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 2

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Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 10
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Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 13
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Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 17
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Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 22

Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 23
Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 24
Kerala SSLC Social Science Previous Year Question Paper March 2019 Malayalam Medium 25

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Students can Download Chapter 2 Accounting for Partnership – Basic Concepts Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Plus Two Accountancy Accounting for Partnership – Basic Concepts One Mark Questions and Answers

Question 1.
A partner is entitled to get 6% per annum as
(a) Profit
(b) Interest on capital
(c) Interest on loan
(d) Remuneration
Answer:
(c) Interest on loan

Question 2.
Profit and Loss Appropriation is an extension of
(a) Capital Account
(b) Current Account
(c) Trading Account
(d) Profit and Loss Account
Answer:
(d) Profit and Loss Account

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 3.
Find odd one and state the reason
(a) Interest on capital
(b) Interest on drawings
(c) Salary
(d) Commission
Answer:
(b) Interest on drawings

Question 4.
Complete the following

  • Interest on loan – Charge against profit.
  • Interest on Partners capital – _______.

Answer:

  • Appropriation of profit
  • Reason: All others are increase to capital A/c.

Question 5.
Find the odd one and state reason.
(a) Interest on partner’s capital
(b) Interest on partner’s loan.
(c) Interest on partner’s drawings
(d) Borrowings from the firm
Answer:
(d) Borrowings from the firm. Others are P/L appropriation A/c items.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 6.
Rugma, Neha and Lekshmi are partners in a firm sharing profits and loses in the ratio of 3:3:4. Their fixed capitals were Rs. 1,00,000, Rs. 2,00,000 and Rs. 3,00,000 respectively. For the year 2005, interest on capital was credited to them @ 10% instead of 9% per annum. You are required to rectify the mistake by passing an adjustment entry.
Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 1

Plus Two Accountancy Accounting for Partnership – Basic Concepts Two Mark Questions and Answers

Question 1.
Define Partnership.
Answer:
According to Section 4 of the Indian Partnership Act 1932, a partnership is “the relation between persons who have agreed to share the profits of a business carried on by all or any of them acting for all”. The persons entered into agreement are individually known as ‘partners’ and collectively as ‘firm’.

Question 2.
What is Partnership Deed?
Answer:
Partnership agreement may be oral or written. When the agreement is written, it is called Partnership Deed. It is also called ‘Articles of Partnership’.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 3.
What is meant by Profit and Loss Appropriation Account?
Answer:
Profit and Loss Appropriation Account is an extension of the Profit and Loss Account. It is prepared to show the appropriation (distribution) of profit among partners.

Question 4.
What are the circumstances in which goodwill is need to be valued?
Answer:

  1. When a new partner is admitted.
  2. When a partner is retired ordied.
  3. When two or more firms are amalgamated.
  4. When a firm is dissolved or its business is sold.

Question 5.
What is divisible profit?
Answer:
Divisible profit is the balance net profit that remains after making all adjustments to net profit regarding interest on capital, salary to partners, interest on partners loan, interest on drawings, etc. and which is distributed among partners in their profit sharing ratio.

Question 6.
Edwin and Abel are partners sharing profits and losses in the ratio of 1:1. Edwin drew Rs. 1000/- at the beginning of every month for the year ending 31st Dec. 2004. Calculate interest on drawings at 6% perannum.
Answer:
Total amount withdrawn by Edwin = 1,000 × 12 = 12,000
1,000 × 12 = 12,000
\(Average period =\frac{\text { Total period (months) }+1}{2}\)
\(=\frac{12+1}{2}\) = 6 : 5 months
Interest on drawings = 12,000 × 6/100 × 6.5/12 = Rs. 390.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 7.
Christy and Fiyona are partners in a firm with equal profit sharing ratio. Christy drew regularly Rs. 15000 at the end of every month.
Answer:
Total amount withdrawn by Christy = Rs. 1,500 × 2 = 18,000
\(Average period =\frac{\text { Total period (months) }-1}{2}\)
\(=\frac{12-1}{2}\) = 5.5 months
Interest ondrawings= 18, 000 × 5/100 × 5.5/12 = Rs. 412.5

Question 8.
After closing the books of accounts, it was discovered that an item, interest on capital was omitted to be recorded in the books of accounts. Even then, there was no difference in the closing balance of capital account, before and after the treatment of the item. What do you infer from this?
Answer:
Partner’s Share of Capital and their profit sharing ratio are in accordance with their capital account balances.

Question 9.
Paul, Kumar, and Lakshman are partners in a firm, sharing profits and losses in the ratio of 3:2:1. After the preparation of final accounts, it was discovered that interest on drawing had not been taken into consideration. The interest on drawings of partners amounted to Rs. 600, Rs. 400 and Rs. 200. Give necessary adjustment journal entry.
Answer:
Items which are omitted while preparing P&L Appropriation A/c can be brought into accounts through P&L adjustment a/c by passing the following entry.

Paul’s capital Dr. 600
Kumar’s capital Dr. 400
Lakshman’s capital Dr. 200
To P&L Adjustment A/c 1,200

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 10.
Give your suggestions to the following arguments. “Under capitalisation method, the firm will have good will only if the value of net tangible assets are more than the capitalised value of profit.”
Answer:
Under capitalisation method
Value of goodwill = Total value of business – Net as-sets
Total value of business
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 2
Net assets = Assets – Liabilities The above equation proves that, a firm will have goodwill only if the value of net tangible assets are less than the capitalised value of profit.

Question 11.
Sanu and Manu are in partnership, who have not made any written agreement. Sanu has given a loan of Rs. 12,000/- to the firm in addition to his capital contribution. During the year the firm made a net loss of Rs. 40,000. Regarding the interest on loan, Manu is of the opinion that no interest be paid being the loan was not external one. Is Mr.Manu right in his stand? State your views.
Answer:
Normally partnership deed contains rules and regulations regarding the conduct of partnership business. In such cases partnership may not have a written agreement. In some other firms, partnership deed may be silent on some matter.

Then relevant discussion in the IPA 1932 becomes applicable. As per IPA 1932, interest on loan is payable at 6% p.a. on Partner’s loan. So Manu’s opinion that interest on loan is not payable, is wrong.

Question 12.
A business has been purchased by a firm for Rs. 1,00,000. But its net tangible assets were worth Rs. 92,000.

  1. What does this difference in value indicate?
  2. Where is it shown in the Balance sheet of the firm?

Answer:
Total value of business-Net Assets = Value of goodwill 1,00,000 – 92,000 = 8,000
So ‘Rs. 8000’ implies the value of goodwill of the firm. Goodwill refers to the value of reputation of a business. It is an intangible asset. So it is shown on the asset side of B/S.

Plus Two Accountancy Accounting for Partnership – Basic Concepts Three Mark Questions and Answers

Question 1.
Match the following.

a. Partnership deed Maximum 10 partners
b. Banking business If no partnership deed
c. Profit and losses shared equally Written agreement of partners
d. Registration of partnership Fixed capital Method
e. Current A/c Not compulsory

Answer:

a. Partnership deed Written agreement of partners
b. Banking business Maximum 10 partners
c. Profit and losses shared equally If no partnership deed
d. Registration of partnership Not compulsory
e. Current A/c Fixed capital Method

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 2.
Ameer and Sudheer are partners sharing profits equally. They are entitled to salaries as follows, Ameer Rs. 6000, Sudheer Rs.4000. The partnership has made a profit of Rs. 15,000. How much is the increase in capital of Mr.Ameer?
(a) 3500
(b) 3900
(c) 8500
(d) 9300
Answer:
(c) 8500
Notes: A:S = 1:1
A’s salary 6000
S’s salary 4000
Out of Net profit of Rs. 15,000.
P/L Appropriation A/c:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 3
A will get salary 6000 and share of profit 2500. So increase in capital is Rs. 8500.

Question 3.
Gomez and Arun Gomez are partners sharing profits and losses in the ratio of 2:1. They are allowed interest at 10% per annum on capitals and loans to the partnership.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 4
The partnership has made a net profit of Rs. 40,000 for the year. How much is the total increase in the net worth of A.Gomez?
(a) 24,800
(b) 25,000
(c) 26,800
(d) 27,100
Answer:
(c) 26,800
Notes:
Capital A/c of A.Gomez:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 5
P/L Appropriation A/c:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 6
A Gomez’s opening capital (capital in the begining of the year) is Rs. 20,000 and closing capital (capital at the end of the year) is Rs. 46,800. So total increase in the net worth of A.Gomez is Rs. 26,800. (46,800-20,000).

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 4.
Lalu and Beena are in partnership. He is also entitled to a salary of Rs. 12,000 per annum. Profits and losses are shared equally. The partnership has made a net profit of Rs. 30,000. How much is Lalu’s total increase in his Capital A/c?
(a) 18,000
(b) 15,000
(c) 21,000
(d) 42,000
Answer:
(c) 21,000
Notes: P&L Appropriation A/c:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 7
Lalu’s capital A/c:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 8
Increase in capital = 12000 + 9000 = 21000.

Question 5.
“Partnership deed must be in writing.” Do you agree with the statement? Give reasons in favour of having partnership deed in writing.
Answer:
Partnership is the result of agreement between two or more persons. The agreement may be oral or written. The written agreement is called Partnership Deed. It is always advisable to put the partnership agreement in writing because of the reasons given below:

  1. To avoid disputes, quarrels, and misunderstanding among the partners.
  2. To remind the partners about their rights, duties, and liabilities.
  3. To maintain healthy atmosphere to carry on business smoothly.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 6.
X, Y, and Z are partners in a firm sharing profits and losses in the ratio of 4:3:2. During 2005, their fixed capital and drawings were as follows:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 9
Partners are entitled to a salary of Rs. 12,000 p.a. and interest on capital @ 5% p.a. You are required to prepare the Current Accounts of partners.
Answer:
Current Account:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 10

Plus Two Accountancy Accounting for Partnership – Basic Concepts Five Mark Questions and Answers

Question 1.
Define Partnership Deed. Mention some of its contents.
Answer:
Partnership deed is a written document which contains the rules and regulations regarding the conduct of business.
Contents of Partnership deed:

  1. Name and address of the firm.
  2. Name and Address of partners
  3. Nature of business
  4. Duration of partnership
  5. Capital introduced by partners
  6. Interest on capital
  7. Drawing made by partner
  8. Interest on partner drawings
  9. Salary, commission and other remunerations payable to partners
  10. Rights, Duties, and Liabilities of Partners.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 2.
What are the rules applicable as per the Partnership Act in the absence of an agreement?
Answer:
In India Partnerships are governed by the Indian Partnership Act 1932. The following are the rules applicable as per the Partnership Act in the absence of an agreement.

  1. Profits and Losses – Profits and losses are to be shared equally among partners.
  2. Salary or remuneration – Partners are not entitled to salary or any other remuneration.
  3. Interest on capital- Partners are not entitled to interest on capital.
  4. Interest on drawings – No interest is charged on drawings made by partners.
  5. Interest on loan – partners are entitled to get an interest at 6% per annum for any loans they have given to the firm.

Question 3.
Distinguish between Fluctuating and Fixed Capital methods.
Answer:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 11

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 4.
What are the general characteristics/ features of a partnership? Explain.
Answer:
1. Number of members:
The minimum member of persons required for a partnership is two (2). Maximum number is ten (10) in case of banking business and twenty (20) in other partnerships.

2. Business Purpose:
The purpose of forming a partnership should be to carry out some business. The business must not be illegal.

3. Agreement:
For the formation of a partnership an agreement is must. The agreement may be oral or written. Only competent persons can enter into a partnership agreement.

4. Profit sharing:
The profits and losses of a partnership business must be shared among the partners. Profits must be shared in an agreed ratio or equally.

5. Mutual agency:
Mutual agency is there in partnership. Every partner is an agent and a principal at a time. He is an agent when he acts for others and a principal when the others act for him.

6. Unlimited liability:
The liability of the partners . in a firm is unlimited. Every partner is individually and jointly liable for all the debts of the firm.

7. No legal existence:
A partnership has no legal existence. It has no existence different from its members.

8. No transfer of share:
A partner cannot transfer his share in the firm to outsiders without the consent of the other partners.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 5.
Sabu and Sekhar commenced business in partnership on 1st January, 2005. No written agreement was in force between them. They contributed Rs. 40,000 and Rs. 10,000 respectively as capital. In addition, Sabu advanced Rs. 20,000 on 1st July, 2005 as loan to the firm. Sabu met with an accident on 1st April, 2005 and could not attend the partnership business upto 30th June, 2005. The . profits for the year ended on 31st December, 2005 amounted to Rs. 50,600. Dispute arise between them for sharing profits.
Sabu claims:

  • He should get an interest @ 10% p.a. on capital. Sekhar claims:
  • Net Profit should be shared equally
  • He should be allowed remuneration of Rs. 1,000 p.m. during the period of Sabu’s illness.

You are required to:

  1. In your opinion how much profit will each partner get?
  2. State your reason.

Answer:
In the absence of agreement, partners are not entitled to interest on capital contributed by them. So Sabu’s claim is not admitted.

  1. In the absense of agreement, partners are not entitled to any salary or other remuneration.
  2. In the absence of written agreement, partners are entitled to share profits equally, Here, net profit is divided equally among Sabu and Sekhar.

Profit for the year = 50,600
Less:Interest on Sabus loan (20000 × 6/100 × 6/12) = 600
The actual profit = 50,000
In the absence of agreement, partners are entitled to interest on loan (to the firm) at the rate of 6% p.a. Sabu’s share of profit = 50,000 × 1/2 = 25000
Stephen’s share of profit = 50000 × 1/2 = 25000.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 6.
Sona and Jerin started a partnership business on 1st January 2004. Sona contributed Rs. 50,000 and Jerin Rs. 25,000 as capital. They decided to share profits and losses in the ratio of 2:1. Sona was entitled to a salary of Rs. 2000 per month. Partners are entitled to interest on their capitals at 5% per annum. The drawings of Sona and Jerin during the year are Rs. 9,000 and Rs. 6,000 respectively. The profit of the firm after making all the adjustment was Rs. 15,000. Prepare the capital accounts of the partners under fluctuating capital method.
Answer:
Partners Capital Account:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 12

Question 7.
A and B are partners sharing profits in the ratio of 2: 3. On 1st January 2001, they admitted C into the firm for a sixth share of profits with a guaranteed minimum of Rs. 25000. A & B continue to share profits as before but agrees to suffer any excess over 1/6 of profit going to C equally. The profits of the firm forthe year was Rs. 75,000. Prepare Profit and Loss appropriation account.
Answer:
Notes:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 13
The ratio in which this difference is to be borne 1: 1 (equally)
Dr. Profit and Loss Appropriation Account Cr.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 14

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 8.
From the following Balance Sheet of Aneesh and Jaya, calculate interest on capital @ 5% per annum, payable to Jaya for the year ending 31.12.05.
Balance Sheet as on 31.12.2005:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 15
During the year, Jaya’s drawings were Rs. 3,000 and the firm made a profit of Rs. 4,000.
Answer:
Interest on capital is payable onthe opening capital (ie. capital on 1.1.05)
Opening capital = Closing capital + drawings – Net Profit
Closing capital = Rs. 6000
Drawings = 3000
Net Profit during the year = 4000
P&L appropriation shown in B/S = 2000
∴ Net profit credited to partners capital = 4000 – 2000 = 2000
Net profit credited to Jaya = 2000 × 1/2 (ratio being 1:1) = 1000
Opening capital = 6000 + 3000 – 1000 = 8000
Interest on capital = 8000 × 5/100 = 400.

Question 9.
Aby and Anu are partners sharing profits in the ratio of 4:1. Their capital a/c balances are
Aby : 4,00,000
Anu : 5,00,000
Profit made during the year was Rs. 1,00,000.
Anu is of the opinion that their agreement must include a provision for interest on capital @10% p.a. Otherwise the profit sharing ratio must be made equal. Why did Anu put forward such an opinion? Will it be worthwhile to her if such changes are made. Which of the above condition is more advantageous to her. Give your advice.
Answer:
Profit sharing ratio of Aby and Anu is 4:1.
Net Profit = 1,00,000
As per this ratio Aby will get Rs. 80,000 (1,00,000 × 4/5) and Anu will get 20,000 (1,00,000 × 1/5) as their share of profit. But Anu has contributed Rs. 1,00,000 more than Aby’s capital. Now Anu is in a disadvantageous position.
Conditions:
1. If there is a provision for interest on capital @10%.
Anu’s interest on capital = 5,00,000 × \(\frac{10}{100}\) = 50,000
Her share of profit = (Net Profit – Interest on capital of Aby and Anu) × 1/5
=[1,00,000 – (40,000 + 50,000)] × \(\frac{1}{5}\)
= 1,00,000 – 90,000 × 1/5
=10,000 × \(\frac{1}{5}\) = 2,000
Anu will get interest on capital = 50,000+ share of profit 2000 = 52,000.

2. Profit sharing ratio made equal:
Abu’s share of profit = 1,00,000 × 1/2 = 50,000
Anu’s share of profit = 1,00,000 × 1/2 = 50,000
So First condition is more advantageous to Anu.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 10.
The partner’s capital account prepared by Mr.Jose, an accountant in Gokul and Co, where Mr. Raman and Mrs.Seetha are partners, is given below. Rectify the errors, if any in the capital accounts prepared by him and show the partners capital accounts under fixed capital method. What should have been the profit of the firm as per profit and loss account?
Partners Capital A/c
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 16
Answer:
Calculation of NP:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 17
Capital A/c:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 18

Question 11.
T and S are partners with equal profit sharing ratio. T withdrew the following amounts during the year 2005.

31st January 2,000
31st March 1,500
30th April 2,000
31st May 2,500
30th Sept 1,000
30th Nov 1,500

The interest on drawings charted is at 6% p.a. Assuming that the accounting year ends on 31st December. Calculate the interest on drawings under product method.
Answer:
Calculation of Interest on Drawings:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 19
Interest on drawings = 73,500 × 6/100 = Rs. 4,410
Interest for one month= 4,410 × 1/12 = Rs. 367.5.

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts

Question 12.
On 1st January 2006, X&Y entered into partnership contributing Rs. 60,000 and Rs. 40,000 respectively. They agreed to share profits & losses in the ratio of 3 : 2. Y is allowed a salary of Rs. 15,000 per year. Interest on capital is to be allowed at 10% per annum. During the year X withdrew Rs. 9,000 and Y Rs. 8,000 as drawings. The interest on drawings paid by X and Y was Rs.150 and Rs. 130 respectively. Profits as on 31st December, 2006 before the above mentioned adjustments were Rs. 65,000. Show the distribution of Profits by preparing Profit and Loss Appropriation A/c & Prepare Partner’s Capital Accounts.
Answer:
Profit and Loss Appropriation Account:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 20
Capital A/c:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 21

Plus Two Accountancy Accounting for Partnership – Basic Concepts Eight Mark Questions and Answers

Question 1.
On 1st January 2005 Sneha and Surya started partnership business by contributing capitals of Rs. 50,000 and Rs. 60,000 respectively. They share profits in the ratio of 2 : 3. Sneha is entitled to a salary of Rs. 12,000 per annum. Interest on capital allowed is at 6% p.a. Surya is entitled to a commission of Rs. 3,000. During this year Sneha withdrew Rs. 3,000 and Surya Rs. 2,000. Interest on drawings charged is Rs. 100 and Rs. 150 respectively. Profit in the year before making the adjustments was Rs. 25,000. Pass necessary journal entries, Prepare Profit and Loss Appropriation Account and Partners Capital Accounts.
Answer:
Journal:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 22
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 23
Dr. Profit & Loss Appropriation Account Cr.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 24

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts
Dr. Partners Capital Account Cr.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Accounting for Partnership – Basic Concepts - 25

Plus One Physics Notes Chapter 8 Gravitation

Students can Download Chapter 8 Gravitation Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 8 Gravitation

Summary
Introduction
Gravitational force is one of the fundamental forces exist in nature. All celestial bodies in the universe are moving under the influence of the gravitational force.

Kepler’S Laws
The three laws of Kepler are Known as

  1. Law of orbits
  2. Law of areas
  3. Law of periods

1. Law of Orbits: All planets move in elliptical orbits with the sun situated at one of the foci of the ellipse.
Ellipse:
An ellipse is the path traced out by a planet around the sun. The closest point ‘P’ is called the perihelion and Athe aphelion. The semimajor axis is half the distance AP.
Plus One Physics Notes Chapter 8 Gravitation 1
Drawing an ellipse:
Plus One Physics Notes Chapter 8 Gravitation 2
Select two points F1 and F2. Take a length of a string and fix its ends at F1 and F2 by pins. With the tip of a pencil stretch the string and then draw a closed curve. F1 and F2 are called the focii.

Plus One Physics Notes Chapter 8 Gravitation

2. Law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of time, ie. aerial velocity is constant
Proof:
Plus One Physics Notes Chapter 8 Gravitation 3
Let the sun be at the origin and let the position and momentum of the planet be denoted by \(\overrightarrow{\mathrm{r}}\) and \(\overrightarrow{\mathrm{p}}\). Let ∆\(\overrightarrow{\mathrm{A}}\) be the area swept out by the planet of mass m in time interval A t.
From the figure:
Plus One Physics Notes Chapter 8 Gravitation 4
But we know angular momentum of a planet is constant because the gravitational force is central force. Hence we get \(\frac{\Delta \overline{\mathrm{A}}}{\Delta \mathrm{t}}\) = a constant.
Note:

  1. Gravitational force is a central force. If direction of force on the planet is along the vector joining sun and planet, it is called central force.
  2. The law of area is the consequence of conservation of angular momentum.

3. Law of periods (Kepler’s third law): The square of the time period of revolution of a planet is proportional to the cube of the semi major axis of the ellipse traced out by the planet.
T2 α a3
Where ‘a’ is semi major axis.

Plus One Physics Notes Chapter 8 Gravitation

Universal Law Of Gravitation
On the basis of Keplers laws and observations of motion of planets, Newton gave a universal law of force acting between any two particles of matter.

Statement:
Every body in the universe attracts each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Explanation
Plus One Physics Notes Chapter 8 Gravitation 5
Consider two masses m1 and m2 separated by a distance r. The force of attraction between two particles can be written as
Plus One Physics Notes Chapter 8 Gravitation 6
The above equation can be written in vector form as
Plus One Physics Notes Chapter 8 Gravitation 7
\(\hat{\mathbf{r}}\) is the unit vector from m1 to m2. G is a constant called the gravitational constant. In S.l. Unit G = 6.67 × 10-11 Nm2 × \(\bar{g}^{2}\).
Note:

  1. The negative sign of gravitational force shows it is an attractive force.
  2. \(\vec{F}_{12}=-\vec{F}_{21}\) ie; force on Ist particle due to IInd particle = – force on IInd particle due to Ist particle.

Deduction of Kepler’s third law:
Planet of mass m is orbiting around the sun of mass M in circular orbit of radius r, with constant angular velocity w, then.
Centripetal force = Gravitational force.
Plus One Physics Notes Chapter 8 Gravitation 8

Plus One Physics Notes Chapter 8 Gravitation
The above result holds equally good for elliptical orbit provided we replace r with a. (the semi major axis of the elli pse)

The Gravitational Constant
The value of the gravitational constant G can be determined experimentally. This was first done by English scientist Henry Cavendish in 1798.
Cavendish experiment to determine gravitational constant:
Plus One Physics Notes Chapter 8 Gravitation 9
Two equal masses (m) are attached at two ends of the rod of length L. The bar is suspended from a rigid support by a fine wire. When two large lead spheres are brought close to small ones from opposite sides as shown, a deflecting torque is produced.
deflecting torque τ = F × L
= \(\frac{G M m}{d^{2}}\) × L
Where d is the distance between m and M. Due to this deflecting torque a restoring couple is produced on the wire.
For equilibrium,
Deflecting couple = Restoring couple
Plus One Physics Notes Chapter 8 Gravitation 10
Where ‘C’ is the restoring couple per unit twist and θ is the angle of rotation. Knowing C, θ, d, M, m and L, we can calculate G. The currently accepted
value is G = 6.67 × 10-11 Nm2/kg2

Question 1.
is it possible to shield a body from gravitational effect?
Answer:
It is not possible to shield a body from gravitational effect because the gravitational force does not depend on the nature of the intervening medium.

Plus One Physics Notes Chapter 8 Gravitation

Acceleration Due To Gravity Of The Earth
Acceleration due to gravity is the acceleration experienced by a body falling freely towards the earth.
Relation between acceleration due to gravity and gravitational constant:
Consider a body of mass ‘m’ at a distance ‘r’ from centre of earth. According to Newton law of
Gravitation, F = \(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\) ______(1)
Force acting on mass ‘m’
F = mg ______(2)
comparing eq(1) and eq(2)
g = \(\frac{G M}{r^{2}}\)
On the surface of earth r = R.
∴ acceleration due to gravity on the surface of earth
Plus One Physics Notes Chapter 8 Gravitation 11
The value of g is 9.8 m/s2. There is a slight variation for g from place to place depending on the height or depth of the place.

Question 2.
Where is the maximum value of g on the surface of earth. Why?
Answer:
It is on the poles. The distance between the pole and centre of earth is minimum at the poles. Hence
g = \(\frac{G M}{R^{2}}\) is maximum at the poles.

Question 3.
There is a popular statement regarding Convendish: “Cavendish weighed the earth”. Comment on this statement.
Answer:
Acceleration due to gravity g = \(\frac{G M}{R^{2}}\)
∴ M = \(\frac{\mathrm{gR}^{2}}{\mathrm{G}}\)
Knowing g, R, and G, we can find the mass of earth. The value of G is experimentally found out by Covendish. This is the reason for the above statement.

Plus One Physics Notes Chapter 8 Gravitation

Acceleration Due To Gravity Below And Above The Surface Of Earth
1. Variation of ‘g’ with altitude:
The acceleration due to gravity on the surface of earth,
g = \(\frac{G M}{R^{2}}\) _____(1)
Plus One Physics Notes Chapter 8 Gravitation 12
At a height h, the acceleration due to gravity can be written as,
gh = \(\frac{\mathrm{Gm}}{(\mathrm{R}+\mathrm{h})^{2}}\) _____(2)
eq(1)/eq(2)
Plus One Physics Notes Chapter 8 Gravitation 13
This equation shows that acceleration due to gravity decreases as height increases. The above equation is valid when h<<R.

2. Variation of g with depth:
Plus One Physics Notes Chapter 8 Gravitation 14
If we assume the earth as a sphere of radius R with uniform density r,
mass of earth = volume × density
M = \(\frac{4}{3}\) πR3 ρ _____(1)
We know acceleration due to gravity on the surface, GM
g = \(\frac{G M}{R^{2}}\) ______(2)
Substituting eq(1) in eq(2), we get
Plus One Physics Notes Chapter 8 Gravitation 15

Plus One Physics Notes Chapter 8 Gravitation
g = \(\frac{4}{3}\) πGRρ ______(3)
Therefore the acceleration due to gravity at a depth d is given by
gd = \(\frac{4}{3}\) πG(R – d)ρ _______(4)
eq(4)/eq(3)
Plus One Physics Notes Chapter 8 Gravitation 16
The above equatiqn shows that, when depth increases g decreases.

Gravitational Potential Energy
Gravitational potential energy:
The gravitational potential energy of a body at a point is defined as the amount of workdone in bringing the body from infinity to that point without acceleration.
Expression for gravitational P.E:
Plus One Physics Notes Chapter 8 Gravitation 17
Consider the earth as a uniform sphere of radius R and mass M. Consider a point A at distance ‘r’ from the centre of earth. P is another point at a distance ‘x’ from O. Q lies at distance dx from P.

By definition, the gravitational potential energy of the body at point A, is the work done in bringing the body of mass ‘m’ from infinity to that point A. The gravitational force on the body at the point P is given by F = \(\frac{\mathrm{GMm}}{\mathrm{x}^{2}}\).
If the body is displaced from P to Q Work done,
dw = F.dx
= \(\frac{\mathrm{GMm}}{\mathrm{x}^{2}}\). Therefore, workdone in bringing the body from infinity to the point A,
Plus One Physics Notes Chapter 8 Gravitation 18

Plus One Physics Notes Chapter 8 Gravitation
Since this workdone is stored inside the body as its gravitational potential energy, the gravitational potential energy (U) of a body of mass m at distance r from the centre of the earth is given by,
Plus One Physics Notes Chapter 8 Gravitation 19
Question 4.
Gravitational potential energy at a point, U = \(\frac{-\mathrm{GMm}}{\mathrm{r}}\)

  1. What is meant by negative sign in the above equation.
  2. What is the value of U at r = ∞

Answer:

  1. Negative sign shows that the potential energy is due to attractive gravitational force. It means, to bring a body from infinity to point P, work has been done by the gravitational field of earth, ie; by attractive force.
  2. When r = ∞, the gravitational potential energy becomes zero.

Note: Maximum value of gravitational potential energy is zero.
Gravitational Potential:
The gravitational potential at a point is the amount of workdone in bringing a unit mass from infinity to that point without acceleration.
Explanation
Gravitational potential energy of mass m at a point is v = \(\frac{-\mathrm{GMm}}{\mathrm{r}}\)
If we take, m = 1 we get gravitational potential.
∴ Gravitational potential v = \(\frac{-G M}{r}\)
Note: The gravitational potential at infinity is taken to be zero.

Question 5.
What is the difference between gravitational potential energy and gravitational potential?
Answer:
Gravitational potential energy of mass m at a distance ‘r’ from centre of earth.
w = \(\frac{-\mathrm{GMm}}{\mathrm{r}}\)
But gravitational potential at a distance r from the centre of earth, w = \(\frac{-G M}{r}\)
From the equation we can understand that, gravitational potential energy depends on the mass of the body. But gravitational potential at point is independent of the mass of the body.

Plus One Physics Notes Chapter 8 Gravitation

ESCAPE SPEED
The minimum speed with which a body is projected so that it never returns to the earth is called escape
speed or escape velocity.
Expression for escape speed:
Force on a mass m at a distance r from the centre of earth = \(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)
Work done for giving small displacement dr,
Plus One Physics Notes Chapter 8 Gravitation 20
Work done in taking the body to infinity from surface of earth,
Plus One Physics Notes Chapter 8 Gravitation 21
This energy is given in the form of K.E. = \(\frac{1}{2}\)mv2e
where ve is the escape speed
Plus One Physics Notes Chapter 8 Gravitation 22

Plus One Physics Notes Chapter 8 Gravitation
This escape velocity \(\sqrt{2 \mathrm{Rg}}\) estimated to be 11.2 km/s on the earth. But escape velocity of moon is 2.3 km/s.
Note: Escape velocity is independent of mass of the escaping body.

Question 6.
Moon has no atmosphere. Why?
Answer:
Escape velocity on moon is 2.3 km/s. The r.m.s. velocity of the gas molecules is greater than this value. Gas molecules easily escape from the surface of moon. Hence there is no atmosphere on moon.

Earth Satellites
Satellites:
The moon is the natural satellite of the earth. It goes round the earth in about 27.3 days almost in a circular. Orbit of radius 3.84 × 105 km. The gravitational force of attraction between the earth and the moon supplies the necessary centripetal force to keep the moon in its circular orbit.
Artificial Satellites:
A man made satellite is called an artificial satellite.
eg: INSAT, EDUSATetc.
Orbital velocity of a satellite:
Orbital velocity of a satellite is the velocity required for a satellite to revolve round the earth in a fixed orbit.
Expression for orbital velocity:
Consider a satellite of mass m revolving with orbital velocity ‘v’ around the earth at a height ‘h’ from the surface of earth. Let M be the mass of earth and R be radius of earth.
Plus One Physics Notes Chapter 8 Gravitation 23
The gravitational force of attraction between earth and satellite.
Plus One Physics Notes Chapter 8 Gravitation 24
Centripetal force required forthe satellite
Plus One Physics Notes Chapter 8 Gravitation 25
For stable rotation,
Centripetal force = Gravitational force
Plus One Physics Notes Chapter 8 Gravitation 26

Plus One Physics Notes Chapter 8 Gravitation
The above equation shows that orbital velocity decreases as h increases.
Period of satellite:
Period of satellite is time taken by the satellite to revolve once around the planet in a fixed orbit.
Plus One Physics Notes Chapter 8 Gravitation 27
Plus One Physics Notes Chapter 8 Gravitation 28
Case -1
For a minimum orbit
For a satellite very close to the surface of earth, h can be neglected in comparison to R.
ie; R + h ≈ R
Plus One Physics Notes Chapter 8 Gravitation 29
Where T0 is called period of minimum orbit. If we substitute the value g = 9.8 m/s2 and R = 6400 km,
We get
Plus One Physics Notes Chapter 8 Gravitation 30
T0 = 85 minutes.

Plus One Physics Notes Chapter 8 Gravitation

Energy Of An Orbiting Satellite
The kinetic energy of the satellite in a circular orbit at height h from surface of earth with speed vis k.E
= \(\frac{1}{2}\) mv2
Plus One Physics Notes Chapter 8 Gravitation 31
The potential energy at distance (R+h) from the centre of the earth is P.E = \(\frac{-G M m}{(R+h)}\)
∴ Total energy E = k.E. + p.E
Plus One Physics Notes Chapter 8 Gravitation 32

Question 7.
What is the physical meaning of -ve energy?
Answer:
If total energy of satellite is negative, it moves in a closed path. This satellite can’t escape from earth’s gravitational field.

Question 8.
What is the condition fortotal energy to escape from earth’s gravitational field.
Answer:
If total energy of satellite is zero or positive, it can escapes to infinity.

Geostationary And Polar Satellites
A geostationary satellite is a satellite which appears to be stationary in the sky. The period of this satellite must be the same as the period of the rotation of the earth about its own axis. Its direction of rotation is from west to east.

Question 9.
Calculate the height of geostationary satellite.
Answer:
If h is the height of satellite, period
Plus One Physics Notes Chapter 8 Gravitation 33
Substituting the value of
T = 24 × 60 × 60 sec., and
R= 6400 × 1000 m g = 9.8 m/s2
we get
Plus One Physics Notes Chapter 8 Gravitation 34

Plus One Physics Notes Chapter 8 Gravitation
R + h = 42648.54 km.
h = 42648.54 – R = 42648.54 – 6380
h = 36268.54 km.
Example: INSAT
Polar satellite:
Plus One Physics Notes Chapter 8 Gravitation 35
Polar satellite orbits in the north-south direction. The height of polar satellite is 500 to 800 km. Since its time period is around 100 minutes, it crosses any altitude many times a day. The camera fixed on the satellite can view small strips of the earth in one orbit.

Adjacent strips are viewed in the next orbit. So that in effect the whole earth can be viewed strip by strip during the entire day. This type of satellite is used for remote sensing, meteorology and environmental studies.

Plus One Physics Notes Chapter 8 Gravitation

Weightlessness
A satellite can be considered to be a continuously falling body. Since a freely falling body experiences weightlessness, the satellite and object inside it feel weightlessness.

Plus One Business Studies Notes Chapter 12 International Business – II

Students can Download Chapter 12 International Business – II Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 12 International Business – II

Contents

  • Export procedure – Export documents
  • Import procedure – import documents
  • Foreign Trade Promotion Measures
  • International trade Institutions
  • World Bank – IDA – IFC – MIGA – IMF – WTO

Plus One Business Studies Notes Chapter 12 International Business - II

Export Procedure:
1. Receipt of enquiry and sending quotations:
The prospective buyer of a product sends an enquiry to different exporters requesting them to send information about price, quality, terms of payment etc. The exporter sends a reply to the enquiry in the form of a quotation referred to as proforma invoice.

2. Receipt of order or indent:
If the buyer is satisfied with the export price and other terms and conditions, he places the order or indent for the goods.

3. Assessing importer’s creditworthiness and securing a guarantee for payments:
After receipt of the indent, the exporter makes necessary enquiry about the creditworthiness of the importer. To mimimise the risk, most exporters demand a letter of credit from the importer.

4. Obtaining export license:
The exporter must apply for export license to the appropriate authority. The following procedure is followed for obtaining the export license.

  1. Opening a bank account in any bank authorised by the Reserve Bank of India
  2. Obtaining Import Export Code (IEC) number
  3. Registration cum Membership Certificate (RCMC) from appropriate export promotion council
  4. Registering with Export Credit and Guarantee Corporation (ECGC) in order to safeguard against risks of non payments.

5. Obtaining pre-shipment finance:
After obtaining the export license, the exporter approaches his banker in order to obtain pre-shipment finance for carrying out production.

6. Production or procurement of goods:
Exporter, after obtaining the pre-shipment finance from the bank, proceeds to get the goods ready as per the orders of the importer.

7. Pre-shipment inspection:
Quality control and pre shipment inspection is compulsory in India as per Quality Control and Inspection Act. Inspection certificate is provided by Export Inspection Council. The customs authority permits the shipment of goods only if there is inspection certificate.

8. Excise clearance:
According to Central Excise Tariff Act, excise duty on the material used in manufacturing goods is to be paid. For this purpose exporter apply to the concerned Excise Commissioner in the region with an invoice.

But in many cases the government exempts payment of excise duty or later on refunds it if the goods so manufactured are meant for exports. The refund of excise duty is known as duty drawback.

9. Obtaining certificate of origin:
In order to obtain Tariff concessions or other exemptions, the importer may ask the exporter to send certificate of origin.

10. Reservation of shipping space:
The exporting firm applies to the shipping company for provision of shipping space. Then the shipping company issues a shipping order. A shipping order is an instruction to the captain of the ship that the specified goods after their customs clearance at a designated port be received on board.

11. Packing and forwarding:
The goods are then properly packed and marked with necessary details such as name and address of the importer, gross and net weight, port of shipment and destination, country of origin, etc. The exporter then makes necessary arrangement for transportation of goods to the port,

12.Insurance of goods:
In order to protect the goods against the risk of loss or damage the exporter gets the goods insured with an insurance company.

Plus One Business Studies Notes Chapter 12 International Business - II

13.Customs clearance:
The goods must be cleared from the customs before these can be loaded on the ship. For obtaining customs clearance, the exporter prepares the shipping bill. Shipping bill contains particulars of the goods being exported, the name of the ship, the port at which goods are to be discharged, exporter’s name and address, etc. Five copies of the shipping bill along with the following documents are then submitted to the Customs Appraiser at the Customs House.

  • Export Contract or Export Order
  • Letter of Credit
  • Commercial Invoice
  • Certificate of Origin
  • Certificate of Inspection, where necessary
  • Marine Insurance Policy

14. Obtaining mates receipt:
A mate receipt is a receipt issued by the commanding officer of the ship when the cargo is loaded on board, and contains the information about the name of the vessel, berth, date of shipment, description of packages, condition of the cargo at the time of receipt on board the ship etc.

15. Payment of freight and issuance of bill of lading:
The clearing & forwarding agent (C&F agent) hands overthe mate s receipt to the shipping company for the computation of freight. After receipt of the freight, the shipping company issues a bill of lading which serves as an evidence that the shipping company has accepted the goods for carrying to the designated destination.

16. Preparation of invoice:
The exporter prepares an invoice for the dispatched goods. Invoice contains information regarding the quantity of goods sent & the amount to be paid by the importer. It is duly attested by the customs.

17.Securing payment:
After the shipment of goods, the exporter informs the importer about the shipment of goods. Various documents like certified copy of invoice, bill of lading, packing list, insurance policy, certificate of origin and letter of credit are sent by the exporter through his bank.

These documents are required by the importer for getting the goods cleared from customs. The exporter get payment from his bank on the submission of necessary documents called negotiations of the documents.

Documents for Export Transaction:
A. Documents related to goods:
1. Export Invoice:
It is a seller’s bill which contains information about goods like quantity, number of packages, marks on packing, name of ship, port of destination, terms of delivery &, payments etc.

2. Packing List:
A packing list is a statement of the number of cases or packs and the details of the goods contained in these packs.

3. Certificate of Origin:
This is a certificate which specifies the country in which the goods are being produced. This certificate enables the importer to claim tariff concessions or other exemptions.

Plus One Business Studies Notes Chapter 12 International Business - II

4. Certificate of Inspection:
For ensuring quality, the government has made inspection of certain goods compulsory by some authorised agency like export Inspection council of India (EICI) etc. After inspecting the goods, the agency issues a certificate of inspection that the consignment has been inspected as required under the export (Quality Control & Inspection) Act, 1963.

B. Documents Related to Shipment:
1. Mate’s Receipt:
A mate receipt is a receipt Issued by the commanding officer of the ship when the cargo is loaded on board, and contains the information about the name of the vessel, berth, date of shipment, description of packages, condition of the cargo at the time of receipt on board the ship etc.

2. Shipping Bill:
The shipping bill is the main document on the basis of which customs office grants permission for the export. It contains full details regarding the goods being exported name of the vessel, exporter s name & address, country of final destination etc.

3. Bill of lading:
It is a document issued by the shipping company after the cargo is loaded on the ship. It is an acknowledgement of goods by the shipping company. It is a document which shows the title to the goods and is freely transferable by endorsement and delivery.

4. Airway Bill:
It is a document issued by the airline company acknowledging the receipt of goods for transshipment through air carriage.

5. Marine Insurance Policy:
It is a certificate of insurance contract whereby the insurance company agrees in consideration of premium to indemnify the insured against loss incurred by perils of the sea.

6. Cart Ticket:
It is also known as cart chit or gate pass. It is prepared by the exporter and contains details regarding export cargo like number of packages, shipping bill number, port of destination etc.

C. Document related to payment:
1. Letter of credit:
A letter of credit is a guarantee issued by the importer’s bank that it will honour up to a certain amount the payment of export bills to the bank of the exporter.

2. Bill of Exchange:
In export & import transaction, exporter draws the bill on the importer asking him to pay a specified amount to a certain person or the bearer of the instrument. The documents required by the importer for claiming title of exported goods are passed on to him only when the importer accepts this bill.

3. Bank Certificate of Payment:
It is a certificate that the necessary documents relating to the particular export consignment have been negotiated and payment has been received in accordance with the exchange control regulations.

Plus One Business Studies Notes Chapter 12 International Business - II

Import Procedure:
1. Trade enquiry:
It is a written request by the importer to the exporters for supply of relevant information regarding the price, quality, quantity and various terms and conditions of export etc. Then, the exporter prepares the quotation and sends it to the importer. The quotation is known as proforma invoice.

2. Obtaining import license:
In India it is compulsory to get the IEC number from the Directorate General Foreign Trade ( DGFT).

3. Obtaining foreign exchange:
As foreign exchange transactions are controlled by Reserve Bank of India, the importer has to submit an application along with necessary documents to the RBI to issue foreign exchange.

4. Placing order or indent:
After obtaining the import license, the importer places an import order or indent with the exporter for supply of the specified products. It contains information about the price, quality, quantity of goods ordered, ports of shipment and destination etc..

5. Obtaining letter of credit:
The importer must obtain letter of credit from his bank and send it to the exporter.

6. Arranging for finance:
The importer should make arrangements in advance to pay to the exporter on arrival of goods at the port.

7. Receipt of shipment advice:
After loading the goods on the ship, exporter dispatches the shipment advice to the importer. A shipment advice contains information about the shipment of goods.

8. Retirement of import documents:
After shipping the goods, the overseas supplier prepares necessary documents and hands overto his banker for their onward transmission to the importer when he accepts the bill of exchange. The acceptance of bill of exchange for the purpose of getting delivery of the documents is known as retirement of import documents.

9. Arrival of goods:
When the goods arrive in the importer’s country, the person in charge of the carrier informs the officer in charge at the dock or the airport about it. He provides a document called import general manifest for unloading of cargo. Import general manifest is a document that contains the details of the imported goods.

10. Customs clearance and release of goods:
All the goods imported into India have to pass through customs clearance after they cross the Indian borders. The importer must fulfill the following formalities for clearing the goods.

  • Endorsement for delivery
  • Payment of dock charges
  • Bill of entry

Plus One Business Studies Notes Chapter 12 International Business - II

Documents used in an Import Transaction:

  1. Trade enquiry:
  2. Proforma invoice
  3. Import order or indent
  4. Letter of credit
  5. Shipment advice
  6. Bill of lading
  7. Airway Bill
  8. Bill of entry

Bill of entry:
It is a form supplied by the customs office to the importer who filled it at the time of receiving the goods. It has to be in triplicate and is to be submitted to the customs office. It contains information such as name and address of the importer, name of the ship, number of packages, description of goods, quantity and value of goods, name and address of the exporter, port of destinations, customs duty payable etc.

Foreign Trade Promotion Measures and Schemes:
1. Duty drawback scheme: it refers to the refund of customs and excise duties paid on imported inputs used in the manufacture of export goods.

2. Export manufacturing under bond scheme:
Under this facility firms can produce goods without payment of excise and other duties. The firms can avail this facility after giving an undertaking (i.e. bond) that they are manufacturing goods for export purposes.

3. Exemption from payment of sales taxes:
Goods meant for export purposes are not subject to sales tax. Income derived from export operations had been exempt from payment of income tax. Now this benefit of exemption from income tax is available only to 100 per cent Export Oriented Units (100 per cent EOUs) and units set up in Export Processing Zones (EPZs) or Special Economic Zones (SEZs) for selected years.

4. Advance license scheme:
It is a scheme under which an exporter is allowed duty free supply of domestic as well as imported inputs required for the manufacture of export goods.

5. Export Promotion Capital Goods Scheme (EPCG):
The main objective of this scheme is to encourage the import of capital goods for export production. This scheme allows export firms to import capital goods at zero or concessional rate of import duty.

6. Scheme of recognising export firms as export house, trading house and superstar trading house:
The registered exporters having a record of export performance over a number of years are granted the status of export houses/ trading houses/star trading houses/super star trading houses subject to the fulfillment of annual average export performance.

7. Export of Services:
In order to boost the export of services, various categories of service houses have been recognised, i.e. Service Export House, International Service Export House, International Star Service Export House.

8. Export finance:
Two types of export finances are made available to the exporters by authorised banks. They are termed as pre-shipment finance and post shipment finance. Pre shipment finance is provided to an exporter for financing the purchase, processing, manufacturing or packaging of goods for export purpose. Post-shipment finance is provided to the exporter from the date of extending the credit after the shipment of goods to the export country.

9. Export Processing Zones (EPZs):
They are industrial estates, which form special enclaves separated from the domestic tariff areas. These are usually situated near seaports or airports. They are intended to provide an internationally competitive duty free environment for export production at low cost. They can import capital goods and raw materials for production of export goods without license.

10. 100 percent Export Oriented Units (100 percent EOUs):
These units are established with the main purpose of exporting their entire production except those which are specifically permitted to be sold in the domestic market.

11 .Special Economic Zone (SEZ):
It is specifically a delineated duty free enclave, which is deemed to be a foreign territory for the purpose of trading and imposing duties. It provided an internationally competitive and duty free environment for the production of export of goods.

Plus One Business Studies Notes Chapter 12 International Business - II

International Trade Institutions:
Government has set up from time to time various institutions in order to facilitate the process of foreign trade. The Department of Commerce in the Ministry of Commerce is the apex body responsible for regulation and administration of the country’s external trade.
Other organizations are:

  1. Export Promotion Councils
  2. Commodity Boards
  3. Export Inspection Council
  4. Indian Institute of foreign trade
  5. State Trading Organisation.

All these organizations help exporters by way of promotion of specific export products, quality inspection, participation in trade fairs and exhibitions, conducting training programmes, carrying out overseas researches, disseminating product and market information, and providing packaging consultancy and testing.

Trade Agreements:
At the international level, there are various international organisations such as the World Bank, IMF and WTO for fostering economic cooperation, trade and investments . among the countries.
1. World Bank:
The International Bank for reconstruction and development i.e. World Bank was established in 1945, whose main objectives were to aid the task of reconstruction of the war affected economies of Europe and assist in the development of the underdeveloped nations of the world.
Functions:

  • To bring about economic growth and widen the scope of international trade.
  • To develop infrastructural facilities such as ‘ energy, transportation etc.
  • To help different countries for raising cash crops so that their income rises.
  • In addition to providing financial assistance for infrastructure development, agriculture, industry, health etc.

2. International Development Association (IDA):
International Development Association (IDA) was set up in 1960 as a part of the World Bank. IDA was established soft loans primarily to provide soft loans to the poorest countries. So it is called the Soft Loan Window of the IBRD.
Objectives of IDA:

  1. To provide development finance on easy terms to the less developed countries
  2. To provide assistance for poverty eradication in the poorest countries
  3. To provide finance at concessional interest rates in order to promote economic development, raise productivity and living standards in less developed nations
  4. To extend macro economic management services relating to health, education, nutrition, human resource development and population control.

3. International Finance Corporation (IFC):
IFC was established in July 1956 in orderto provide finance to the private sector of developing countries. All the members of the World Bank are eligible to become members of IFC.

Plus One Business Studies Notes Chapter 12 International Business - II

4. The Multinational Investment Guarantee Agency (MIGA)
The Multinational Investment Guarantee Agency was established in April 1988 to supplement the functions of the World Bank and IFC.
Objectives of MIGA:

  • To encourage direct foreign investment into the developing countries
  • To. provide insurance cover to investors against political risks
  • To provide guarantee against noncommercial risks
  • To insure new investments and expansion of existing investments
  • To provide promotional and advisory services
  • To establish credibility.

5. International Monetary Fund (IMF) International Monetary Fund:
it is the second international organization next to the World Bank. It came into existence in 1945 and its objective is the maintenance of exchange rates and providing short term loans to the countries facing short term foreign exchange problems.
Objectives of IMF:

  1. To provide international monetary cooperation through a permanent institution.
  2. To facilitate expansion of balanced growth of international trade and to promote and maintain high levels of employment.
  3. To promote exchange rate stability among member countries.
  4. To help in the establishment of International Payment System.

World Trade Organisation:
The General Agreement for Tariffs and Trade (GATT) was transformed into World Trade Organisation (WTO) with effect from 1 st January 1995. The headquarters of WTO are situated at Geneva, Switzerland.

The World Trade Organization deals with the global rules of trade between nations. Its main function is to ensure that trade flows as smoothly, predictably and freely as possible.

It is concerned with solving trade problems between countries and providing a forum for multilateral trade negotiations. It has a global status similar to that of the IMF and the World Bank. India is a founding member of WTO.
Objectives of WTO:

  1. To ensure reduction of tariffs and other trade barriers imposed by different countries
  2. To engage in activities which improve the standards of living, create employment, increase Income, facilitate higher production and trade
  3. To facilitate the optimal use of the world’s resources
  4. To promote an integrated, more viable and durable trading system.

Functions of WTO:

  1. To remove barriers of International trade.
  2. To Act as a dispute settlement body.
  3. To ensure that all the rules regulations prescribed in the Act are duly followed by the member countries for the settlement of their disputes.
  4. Laying down a commonly accepted code of conduct for international trade.
  5. To consult with IMF, IBRD and its affiliated agencies to bring better understanding and cooperation in global economic policy making.

Plus One Business Studies Notes Chapter 12 International Business - II

Benefits of WTO:

  1. It helps to promote international peace and facilitates international business
  2. All disputes between member nations are settled with mutual consultations
  3. It makes international trade and relations very smooth and predictable.
  4. Free trade improves the standard of living of the people by increasing the income level
  5. Free trade provides an opportunity for getting varieties of qualitative products
  6. Economic growth has been increased
  7. The system encourages good government.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Students can Download Chapter 11 Structuring Database for Accounting Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Plus One Accountancy Structuring Database for Accounting One Mark Questions and Answers

Question 1.
_______are known facts that can be recorded and which have implicit meaning.
Answer:
Data.

Question 2.
Data transformation is called_______.
Answer:
Information

Question 3.
Anything in the real world with independent existence is called______.
Answer:
Entity

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 4.
______represents the database at collection of tables comprising different volumes.
Answer:
Relational Data Model (RDM)

Question 5.
_______is a collection of concepts used to describe the structure of a database.
Answer:
Data Model

Question 6.
A______does not have key attributes of its own.
Answer:
Weak entity

Question 7.
Each simple attribute of an entity type is associated with a value set called______of values.
Answer:
Time ware

Question 8.
When structure of AIS is based on both human and computer resources, it is called______AIS.
Answer:
Liveware

Question 9.
SQL stands for_______.
Answer:
Structured Query Language

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 10.
_______is a collection of all entity instances of a particular entity type.
Answer:
Entity Set

Question 11.
Qualification of a person is a______attribute.
Answer:
Multi-valued

Question 12.
Height of a person is______attribute.
Answer:
Single valued

Plus One Accountancy Structuring Database for Accounting Two Mark Questions and Answers

Question 1.
What is Database?
Answer:
Database can be defined as “any organised collection of information. It is a centrally controlled integrated collection of logically organised data.
For example: a data regarding purchase order may be required by accountants, financial manager, and purchase managers. In such a case, interrelated data should be properly organised and stored. Such a collection of interrelated data are called ‘database’.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 2.
What do you mean by DBMS?
Answer:
A Database Management system is a software system that manages the creation and use of databse. It is a general purpose software system which helps the users in the process of defining, creating, updating, reading, maintaining and protecting the database. All access to the database is through the database management system. It helps the users to obtain data in a more orderly and logical fashion.

Question 3.
What are the functions of DBMS in a supermarket?
Answer:
With the help of DBMS the present volume of sales is known and this information together with past records and perhaps statistics obtained from market research, forms some of the data needed for sales forecasting which in turn influence stock control. Computers are widely used in everyday cash transactions. Computer help to minimise excess stock and release money for other purpose.

Question 4.
What is meant by Null value? What-are the reasons that lead to their occurrence in database relations?
Answer:
Absence of a data item is represented by a special value called null value. There are three situation which may require the use of null values.

  • When a particular attributes doesnot apply to an entity.
  • Value of an attribute is unknown, although it exists.
  • Unknown because it does not exist.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 5.
What do you mean by Multi-valued attribute? How is it different from complex and composite attribute?
Answer:
An attribute with multiple values for an entity is called multivalued attributes. For example the qualifications of a person.
The composite attributes can be divided into smaller sub-parts to represent some more basic attributes with independent meanings. For example – Name of a person that is normally subdivided into First name, Middle name, and Last name is a composite attribute.

The composite and Multi-valued attributes may be grouped into constitute “Complex ones”. The paren-thesis () are used for showing grouping of components of composite attributes. The braces {} are used for showing the multi-valued attributes.

Question 6.
State main categories of data model.
Answer:

  1. Relational Data Model
  2. Hierarchical model
  3. Network model

Question 7.
What is a participation role? State the circumstances under which the use of role names becomes necessary in description of relationship types.
Answer:
Each entity type that participates in a relationship type plays a particular role in the relationship. The role name signifies the role that a participating entity of an entity type plays in each relationship instance.

In PREPARED BY relationship type, EMPLOYEE plays the role of document creator and voucher plays the role of document created. Entity TEACHER and Entity STUDENT are related with a relationship TEACHER – teach – STUDENT. The teaches is a participating role in the entity set TEACHER and STUDENT.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 8.
Distinguish between stored and derived attribute.
Answer:
Two or more attributes may be related in such a way that one or more becomes basic while other becomes dependent on that basic attribute.
E.g.:- Date of birth of a person is a stored attribute while age of that person is derived attribute.

Plus One Accountancy Structuring Database for Accounting Three Mark Questions and Answers

Question 1.
What are the functions of DBMS in the following organisations?

  1. Insurance
  2. Railway
  3. Bank
  4. Telecom department
  5. Payroll department

Answer:

  1. Insurance – To enter the data relating to the amounts of insurance premium paid, date of payment, date of maturity, etc. and to retrieve the same.
  2. Railway – For reservation of railway tickets.
  3. Bank – Money can be deposited in one bank and be withdrawn from another bank.
  4. Telecom department – Enter customers details, amount of bill, date of payment, etc.
  5. Payroll department – To enter the name of the employee, amount and date of payment and type of employees.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 2.
What is meant by Key attributes of an Entity type?
Answer:
Every entity has one of its attributes, which contains unique values for identifying the entity instances is called identifier or key attributes of an entity type.
For example, “RollNo” as attribute of Entity type ‘Student’ has unique values through which a student instance can be identified. Similarly, code is a key attribute of entity type ‘Accounts’ because its data values are required to be unique.

Question 3.
What do you understand by relationship type? How is it different from relationship instance and relationship set?
Answer:
Whenever entities from different entity types are related to one another in a particular manner, they constitute a relationship type. The relationship prepared by between the two entity types vouchers and employees associates each voucher with the employees who prepared it.

Similarly, the relationship authorised by between the two entity types, vouchers and employees associate each voucher with the employee who authorises it.

The degree of a relationship type is the number of participating entity types. A relationship type of de-gree two is called binary and that of degree three is called ternary. AVOUCHER (entity), Authorised by (relationship) and EMPLOYEES (entity) together signify a binary relationship.

ASUPPLIER (entity) SUPPLY (relationship) PARTS (entity) to PROJECT (entity) signify a ternary relationship because three entities, namely supplier, parts, and projects are participating in supply relationship in any transaction.
Relationship sets

  • 1: 1 (one to one relationship set)
  • 1: M (one to many relationship set)
  • M: M (Many to many relationship set)

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 4.
What do you understand by the concept of weak entity used in data modelling? Explain the relevance of owner entity type, partial key and identifying relationship in the context of such modelling.
Answer:
Weak Entity Types:
Entity Types, which do not have identifier (or key attributes) of their own are, called weak entity types. Such entity types are identified by some of their attribute values. These other entity types are called identifying or owner entity type. Accordingly, the relationship type that relates a weak entity type to its owner is called identifying relationship of the weak entity.

A weak entity type always has a total participation constraint (existence dependency) with respect to its identifying relationship because it cannot be identified without its owner entity, eg. a voucher may be accompanied by a set of support documents such as bills, issued by other parties to the transaction, details of which need to be stored. Such SUPPORT DOCUMENT entity type which is used to keep track of support documents attached to each voucher via 1:N relationship is a weak entity.

This is because they are identified as distinct entities only after determining the particular voucher. A weak entity type normally has a partial key, which is a set to attribute that can uniquely identify weak entities that are related to the same owner entity.

Assuming that two support documents of a voucher do not have the same document Id, the said id can be a good partial key. Otherwise a composite attribute of all the weak entity’s attributes will be the partial key.

Question 5.
Define foreign key. How is this concept useful in relational data model? Illustrate with suitable example.
Answer:
While key and entity constraints are specified on individual relation, the referential integrity constraints is specified between two or more relations. This constraint is specified to maintain consistency among the typles of such relations.

Accordingly, a tuple in one relation that refers to another relation must refer to an existing tuple in that other relation. In referencing Accounts type, Accounts relation uses its attributes Type, which acts as foreign key to reference the tuples of relation Accounts type through its primary key Catld.

The value of Type cannot be null because of total participation of Accounts in classify relationship.
Say e.g. there are two tables EMP (EMPNO, ENM, SAL, DPTNO (Foreign key)
DEPT (DPTNO (Primary key), DNM, LOC)
The foreign key DPTNO of EMP table takes its value as a reference from DEPT TABLE DPTNO.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 6.
What integrity constraints are specified on database scheme? Why each is considered important?
Answer:
While key and entity constraints are specified on invidividual relation, the referential integrity is specified between two or more relations. This constraint is specified to maintain consistency among the tuples of such relations.

Accordingly, a tuple in one relation that refers to another relation must refer to an existing tuple in that other relation. In referencing Accounts Type, Accounts relation uses its attribute Type, which acts as foreign key to reference the tuples of relation Accounts Type through its primary key Catld. The value of Type cannot be null because of total participation of Accounts in classify relationship.

For example, there are two tables EMP (EMPNO, ENM, SAL, DPTNO (Foreign Key) DEPT (DPTNO {Primary key}, DNM, LOC). The foreign key DPTNO of EMP table takes its value as a reference from DEPT TABLE DPTNO.

Question 7.
In relational data model, a row is called______,a column header is called on_____and the table as
such is called a______.
Answer:

  • a row is called a tuple
  • a column header is called an attribute
  • The table is called a relation.

Plus One Accountancy Structuring Database for Accounting Four Mark Questions and Answers

Question 1.
Explain Entities and Attributes.
Answer:
Entities
Anything in the real world with independent existence is called entity such as an object with physical existence (example – car, person, house) or conceptual existence (example – job, university, account, voucher, a company). In accounting reality, there exist five entities – Accounts, vouchers, employees, accounts type and support documents.

Attributes
Attributes are some properties of interest or characteristics that further describe the entity such as height, weight and date of birth in case of a person and code and name in case of accounts. An entity has a value for each of its attributes, which is the data stored in the database.

In the context of the accounting reality, the following, attributes specific to each entity types have been stated below as:

Entity type List of attributes
1. Accounts type → Catld, Category
2. Accounts  → Code, Name, Type
3. Employees → Empld, Fname, Minit, Lname, Superld
4. Vouchers  → Vno, Date, Debit, Credit, Amount, Narration,  PrepBy
5. Support Documents → Sno, dDate, Name

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 2.
Discuss the process of designing a Database for Accounting.
Answer:
The process of designing a database for accounting is described below:
1. Reality:
It refers to some aspect of real world situation, for which database is to be designed. In the case of accounting, it is accounting reality that is to be expressed with complete description.

2. ER Design:
This is formal blue print, a pictorial presentation, in which Entity Relations Model concepts are used to represent description of reality.

3. Relational Data Model:
It represents a collection of related data values and hence typically corresponds to real world entity or relationship. It consists of rows and columns.

The table name and column name are used to help in interpreting the meaning of volumes of each rows. Each row of a table is called a data record. All values in a column, which belongs to a particular domain, are of same data type.

4. Normalisation:
This is the process of refining a database design through which the possibility of duplicate or redundant data items is reduced or eliminated.

5. Refinement:
This is the outcome of the process of normalisation. The final database design is arrived at after the process of normalisation is completed. The following chart showing the process of designing a Database for Accounting:
Plus One Accountancy Structuring Database for Accounting Four Mark Questions and Answers 1

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 3.
What do you mean by Entity Type? How it is different from entity set?
Answer:
Entity Types and Entity Sets
An entity type is defined as a collection of entities, which share a common definition in terms of their attributes. Each entity type is assigned a name for its subsequent identification. The attributes of entity type are used to describe it in the database. The value of attributes of an entity belonging to entity type are known as “Entity Instance”.

An entity set is a collection of all entity instances of a particular entity type. An entity type is described by a set of attributes called “Schema”. The set of entities pertaining to a particular entity type share the same set of attributes. The collection of entities of a particular entity type is grouped into entity set, called the extension of the entity type.
For example: Entity type: Accounts
Plus One Accountancy Structuring Database for Accounting Four Mark Questions and Answers 2
Entity set: Collection of entity instances of an entity
Extension or Instances of entity type:
Plus One Accountancy Structuring Database for Accounting Four Mark Questions and Answers 3

Question 4.
Discuss the basic concepts of Entity Relationship (ER) Model. Illustrate as to how an ER model is diagrammed.
Answer:
It is a popular conceptual data model, which is mostly used in database-oriented applications. The major elements of ER model are entities, attributes, identifiers and relationships that are used to express a reality for which a database is to be designed.

The model is best depected with the help of ER symbols, the list and description of which is shown in figure below. The formal blue print with a pictorial presentation of ER model is called as ER Diagram. While preparing an ER Diagram, the following symbols are used to represent different types of entities, attributes, identifiers, and relationships.

The elements of ER model that are meant to describe and display the reality are discussed in the context of an accounting reality given below.
Plus One Accountancy Structuring Database for Accounting Four Mark Questions and Answers 4

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Question 5.
Discuss the different types of update operations in relation to the integrity constraints which must be satisfied in a relational database model.
Answer:
The following are the update operations in a relation base model. The three basic types of updates are as given below:
1. insert:
This operation is performed to add a new tuple in a relation, eg. an attempt to add another record of an account with data values corresponding to Code, Name and its Type to Accounts relation shall be made by performing insert operation. The insert operation is capable of violating any of the four constraints.

2. Delete:
This operation is carried out to remove a tuple from a relation. A particular data record from a table can be removed by performing such a operation. The delete operation can violate only referential integrity, if tuple being removed is referred by foreign key from other tuples in the database.

3. Modify:
The operation aims at causing a change in the values of some attribute in existing tuples. This is useful in modifying existing values of an accounting record in a data table. Usually, this operation does not cause problems provided the modification is directed on neither primary key or foreign key.

Whenever applied, these operations must enforce integrity constraints specified on relational database schema. Retrieval operation on Relational Data Model does not cause violation of any integrity constraints.

Plus One Accountancy Structuring Database for Accounting Eight Mark Questions and Answers

Question 1.
Discuss the steps you would take to transform an ER Model into various relations of Relational Data Model. Give suitable examples.
Answer:
The rules or guidelines required to design the relational database scheme attempt to provide a step by step procedure that transforms ER design into Relational Data Model design to constitute the desired database.
The following specific steps are required to cause its transformation into relational data model.
1. Create a Relation for Every Strong Entity:
For each strong entity type (which has primary key) in ER schema, a separate relation that includes all the simple attributes of that entity is created.

Either choose bn of the key attributes of such an entity as the primary key for this relation, or choose a set of simple attributes that uniquely identify this entity as the primary key of the relation so created,

eg. employee entity is strong because its finds its primary key in Empld which is one of its unique attribute. Therefore, a separate relation for employee has been created as shown below.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

Employee (Empld, Fname, Minit, Lname, Address, PhoneNo, Superld). Similarly, separate relations need be created for the following strong entities whose Primary key attribute have been underlined.

  • Accounts (Code, Name, Type)
  • Vouchers (VNo, vDate, amount, narration)
  • Accounts Type (Catld, Category)

2. Create a separate relation for each weak entity type:
Every weak entity has an owner entity and an identifying relationship through which such weak entity type is identified. For every weak entity type, a separate relation is created by including its attributes. The primary key of this new relation is the combination of its unique attribute(s) for a particular tuple of the owner relation along with primary key attribute of such owner relation.

Furthermore, the primary key of owner entity is included as foriegn key in such a relation key of owner entity and the partial key of weak entity, eg. Support entity, with Vouchers as its owner Entity, does not have a primary key of its own.

It has partial key which is the Sno assigned to each document. Therefore, the Primary key of Vouchers, Vno along with Sno is designed as composite key for support entity and the relation so formed is shown below as
Support (vNo, Sno, dName, sDate)

3. Identify Entity Types Participating in Binary 1: N Relationship Type:
Identify the first relation on n-side of relationship and second on 1-side of such relationship. The primary key of second relation should be included in first relation as its foreign key. eg. An employee can authorize a number of vouchers. It implies that Vouchers entity participates in Auth.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 11 Structuring Database for Accounting

By relationship on n-side while Employees entity participates in same relationship on 1-side. Therefore, the vouchers relation as already formed above in step 1, must also include as foreign key the primary key of Employees, which is Empld.

4. Identify Entity Types Participating in Binary M: N Relationship Type:
For each binary M: N relationship type, create a new relation to represent such relationship. This new relation should include as foreign keys, the primary keys of the relations that represent the participating entity types.

Plus One Business Studies Notes Chapter 11 International Business – I

Students can Download Chapter 11 International Business – I Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 11 International Business – I

Contents

  • Meaning – reasons for international business Differences between Foreign business and Domestic business
  • Scope and Benefits of international business
  • Modes of entry into international business – Exporting and Importing – advantages – Limitations
  • Contract Manufacturing – advantages – limitations
  • Licensing and Franchising – advantages – limitations
  • Joint Venture – advantages – limitations
  • Wholly Owned Subsidiaries – advantages – limitations

International or external business can be defined as those business activities that take place beyond the geographical boundaries of a country. In other words buying and selling of goods and services between two countries are called external trade.

It involves not only the international movements of goods and services, but also of capital, personnel, technology and intellectual property like patents, trademarks, know how and copyrights.

Plus One Business Studies Notes Chapter 11 International Business - I

Reasons for international business:
1. Because of unequal distribution of natural resources and differences in productivity levels, a country cannot produce all that they need

2. Labour productivity and production costs differ among nations due to various socio-economic, geographical and political reasons.

3. Availability of various factors of production such as labour, capital and raw materials differ among nations.

Differences between International Business and Domestic Business:

Domestic Business International Business.
Both the buyers and sellers are from the same country Buyers and sellers are from different Countries
Various stakeholders such as suppliers, employees, middlemen .shareholders and partners are usually citizens of the same country Various stakeholders such as suppliers, employees, middlemen, shareholders and partners are from different nations
The factors of production Like capital, labour and raw material can move freely within the country There are, restrictions on free mobility of factors of production across countries
Domestic markets are relative more homogeneous in nature International markets lack homogeneity due to differences, in languages, preferences customs etc across markets.
Business systems and practices are relatively more homogeneous within a country Business systems and practices vary considerably across countries.
It has to face the political system and risk of only one country Different countries have different forms of political systems and risk.
Business laws, regulations and economic policies are uniformly applicable within a country. Business laws, regulations and economic policies are differ widely among nations.
Currency of domestic country is used. They use different currencies for business transactions.

Scope of International Business:
1. Merchandise exports and imports:
merchandise exports means sending tangible goods abroad, merchandise imports means bringing tangible goods from a foreign country to one’s own country. It is also known as trade in goods (visible trade), include only tangible goods and exclude services.

2. Service exports and imports:
It involves trade in intangibles. It is also known as invisible trade which includes services received from other countries or services rendered to other countries. eg: Tourism and travel, communication, marketing, transportation services etc.

3. Licensing and franchising:
Permitting another party in a foreign country to produce and sell goods under their trademarks, patents or copyright in lieu of some fee is called licensing. Franchising is similar to licensing, but it is a term used in connection with the provision of services.

4. Foreign investments:
Foreign investment involves investments of funds abroad in exchange for financial return. Foreign investment can be of two types.
(a) Direct investments:
Direct investment takes place when a company directly invests in properties such as plant and machinery in foreign countries with a view to undertaking production and marketing of goods and services in those countries. This is also known as Foreign Direct Investment, i.e., FDI.

(b)Portfolio investment:
Under portfolio investment, a company makes investment by acquiring shares or providing loans to a foreign company and earns income by way of dividends or interest on loans.

In this investor does not get directly involved in production or marketing of goods. It simply earns an income by investing in shares, bonds, bills, or notes in a foreign country or providing loans to foreign business firms.

Plus One Business Studies Notes Chapter 11 International Business - I

Benefits of International Business:
The benefits of international business to the nations and business firms are.
Benefits to Nations:
1. Earning of foreign exchange:
It helps a country team foreign exchange which can be used for importing capital goods, technology, petroleum products and fertilisers, pharmaceutical products, etc.

2. More efficient use of resources:
External trade enables a country to utilize the available resources in the best possible manner.

3. Improving growth prospects and employment potentials:
External trade helps to accelerate the economic growth and employment opportunities of a country.

4. Increased standard of living:
Foreign trade helps in raising the standard of living of a country.

5. International relation:
External trade helps to promote harmonious and cordial relationship among the nations.

Benefits to Firms:
1. Prospects for higher profits:
When the domestic prices are lower, business firms can earn more profits by selling their products in countries where prices are high.

2. Increased capacity utilisation:
It help firms in using their surplus production capacities and improving the profitability of their operations. Large scale production helps to reduce the cost of production.

3. Prospects for growth:
It helps firms in improving their growth prospects by creating demands for their products in foreign countries.

4. Enhances competition:
External trade enhances competition, which compels the domestic firms to improve technology of production, production process and quality of the products.

5. Improved business vision:
It improves business vision as it makes firms to grow, more competitive and diversified.

Mode of Entry into International Business:
1. Exporting and Importing:
When goods are sold to a foreign country, it is called export trade. When goods are purchasing from a foreign country, it is called import trade.
Advantages:

  • It is the easiest way of gaining entry into international markets.
  • Business firms are not required to invest that much time and money in host countries.
  • It is less risky as compared to other modes of entry into international business

Limitations:

  • It involves additional packaging, transportation and insurance costs.
  • Exporting is not possible in case the foreign country restricts imports.
  • The export firms do not have much contact with the foreign markets.

2. Contract Manufacturing (Outsourcing):
When a firm enters into a contract with one or a few local manufacturers in foreign countries to get certain goods produced as per its specifications it is called contract manufacturing. It is also known as outsourcing and it can take place in following forms.

  • Production of certain components
  • Assembly of components into final products
  • Complete manufacture of the products

Advantages:

  • It Permits the international firms to get the goods produced on a large scale without requiring investment in setting up production facilities.
  • There is no investment risk involved in the foreign countries.
  • It helps to get the products at lower cost
  • Local producers in foreign countries can ensure greater utilization of their idle production capacities.

Limitations:
(a) It may affect the quality of the products.

(b) Local manufacturer in the foreign country loses his control over the manufacturing process because goods are produced strictly as per the terms and specifications of the contract.

(c) The local firm cannot sell the contracted output as per their will.

3. Licensing and Franchising:
Licensing is a contractual arrangement in which one firm grants access to its patents, trade secrets or technology to another firm in a foreign country for a fee called royalty. The firm that grants permission is known as licensor and the firm that receives the rights to use technology or patents is called the licensee.

Plus One Business Studies Notes Chapter 11 International Business - I

Franchising is similar to licensing. But it is used in connection with the provision of services. The parent company is called the franchiser and the other party to the agreement is called franchisee.
Advantages:

1. It is a less expensive mode of entering into international business.

2. There is no investment risk

3. Since the business in the foreign country is managed by the licensee/franchisee who is a local person, there are lower risks of business takeovers or government interventions.

4. Since licensee/franchisee is a local person, he has greater market knowledge and customer contacts. It helps the licensor/franchiser in successfully conducting its marketing operations.

Limitations:
(a) The licensee can start marketing an identical product under a slightly different brand name.

(b) Trade secrets may lose in the foreign markets.

(c) Conflicts often develop between the licensor/franchiser and licensee/franchisee oyer issues such as maintenance of accounts, payment of royalty, etc.

4. Joint Ventures:
Joint venture means establishing a firm that is jointly owned by two or more independent firms. It can be brought into existence in three major ways.

  • Foreign investor buying an interest in a local firm.
  • Local firm acquiring an interest in an existing foreign firm.
  • Both the foreign and local entrepreneurs jointly forming a new enterprise

Plus One Business Studies Notes Chapter 11 International Business - I

Advantages:

  1. Since the local partner also contributes to the equity capital, the international firm has less financial burden to expand the business globally.
  2. It helps to execute large projects requiring huge capital outlays and manpower.
  3. The foreign business firm benefits from local partner’s knowledge of the host countries.
  4. The foreign business firm shares cost and risks with local partner. So they can enter into foreign market very easily and without high risk.

Limitations:

  1. Foreign firms entering into joint ventures share the technology and trade secrets with local firms. It leads to leakage of technology and secrets to others.
  2. The dual ownership arrangement may lead to conflicts

5. Wholly Owned Subsidiaries:
The parent company (holding company) acquires full control over the foreign company by making 100% investment in its equity capital. It is called wholly owned subsidiaries. It can be established in either of the two ways. i.e.

  • Setting up a new firm altogether to start operations in a foreign country
  • Acquiring an existing firm in the foreign country

Advantages:

  1. The parent firm is able to exercise full control over its operations in foreign countries.
  2. It is not required to disclose its technology or trade secrets to others.

Limitations:

  1. It is not suitable for small and medium size firms which do not have enough funds to invest abroad.
  2. The parent company alone has to bear the entire losses.
  3. It is subject to higher political risk.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

Students can Download Chapter 10 Mechanical Properties of Fluids Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

Summary
Introduction
Unlike a solid, a fluid has no definite shape. Fluid is that which can flow. Liquids and gases are fluids. A fluid has no definite shape. It assumes the shape of the container.
Difference between the liquids and gases:

  • Liquid is incompressible, while a gas is compressible.
  • Liquid has a free surface, while a gas no free surface.

Common properties of liquids and gases:

  • Liquids and gases can flow
  • Liquids and gases can exert pressure on the walls of the container.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

Pressure
In fluids, the molecules are free to move. When they move, they collide with the walls of the container and exert a force on the walls. The force is acting normal to the surface. The force acting per unit area normal to the surface is called pressure.

ie; P = F/A. Its unit is N/m2 or Pascal (Pa), Another common unit of pressure is atmosphere. The pressure exerted by the atmosphere at sea level is called the atmospheric pressure.
I atm = 1.0313 × 105Pa
Properties of Fluid Pressure:

  • The pressure will be equal at any point on the same horizontal level of a liquid at rest.
  • The pressure will be the same in all directions at any point in a liquid at rest.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 1

Question 1.
When a surface is immersed in a liquid, what is the direction of pressure on the surface?
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 2
Answer:
The direction of pressure is always normal to the surface.
Density:
The density r for a fluid of mass m occupying volume V
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 3
The density of water at 4°C (277k) is 1 × 103kg m-3
Relative density:
The relative density of a substance is the ratio of its density to the density of water at 4°C. The relative density of aluminium is 2.7. Its density is 2.7 × 103kg m-3.
1. Variation of pressure with Depth:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 4

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
Consider a fluid at rest in a container. Consider a cylindrical element of fluid having base area ‘A’ and height ‘h’. Let P1 and P2 be pressures at points 1 and 2 respectively.
The force at top of cylindrical element F1 = P1A
This force is acting vertically down ward.
Similarly, the force at bottom F2 = P2A
This force is acting vertically upward.
∴ Total force acting on the cylindrical element of fluid = P2A – P1A
The fluid is at rest. Hence, Net force on the cylindrical element of fluid = Weight of cylindrical element of fluid.
ie.: (P2 – P1)A = mg
(P2 – P1)A = vρg (∴ m = ρv)
(P2 – P1)A = Ahρg (∴ v = Ah)
P2 – P1 = hρg _____(1)
The above equation shows that pressure difference depends on the vertical distance h between the points (1 and 2), density of the fluid, acceleration due to gravity.
Gauge Pressure:
If the point 1 is shifted to the top of fluid, P1 may be replaced by atmospheric pressure ‘Pa’. Let pressure at bottom of cylindrical element be ‘P\ then from eq (1), we can write
P – Pa = hρg
P = Pa + hρg
The above equation shows that, the pressure P (below the surface of a liquid open to the atmosphere) is greater than atmospheric pressure by an amount rgh. The excess pressure (P – Pa) at depth h is called guage pressure.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

Hydrostatic paradox:
Question 2.
In which points (see figure) pressure is maximum?
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 5
Answer:
The liquid pressure is the same at all points (A, B, C). Because all the points lie at same horizontal level. The pressure at A, B and C do not depend upon base area or shape of the container. This paradox is called hydrostatic paradox.

2.  Atmospheric pressure and Gauge pressure:
Atmospheric pressure:
The pressure exerted by the atmosphere at any point is due to the weight of air above that point. At sea level the atmospheric pressure is maximum and is taken as 1 atmosphere.
1 atm = 1.013 × 105 Pa
Mercury Barometer:
Italian scientist Evangelista Torricelli (1608 -1647) devised a method for measuring atmospheric pressure. This device is known as mercury barometer.
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 6

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
A long glass tube closed at one end is filled with mercury. Then it is inverted in a mercury reservoir. The mercury level inside the tube falls up to a height 76 cm.

Consider two points A and B in the reservoir. Since the reservoir is open to the atmosphere, the pressure at,
P0 = Pa (atmospheric pressure).
The pressure at B is PB = Hrg
The points A and B lie at same level, hence
pressure at A = pressure at B
Pa = PB
ie; Pa = Hrg
In the experiment, it is found that the mercury column in the barometer has a height of about 76 cm at sea level. 76 cm of mercury = 1 atm.
Different unit of pressure:
1 mm of mercury = 1 torr
1 torr= 133 Pa
1 bar= 105 Pa
Open – Tube Manometer:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 7
open-tube manometer is a useful instrument for measuring pressure differences. This device consists of a U shaped tube containing liquid of density r. One end is open to the atmosphere and other end is connected to the system whose pressure is to be measured.

The pressure ‘P’ at A is equal to the pressure at B.
ie: P = Pa + hρg
(or) P – Pa = hρg
P is called absolute pressure.

3. Hydraulic Machines:
Hydraulic machines are based on Pascal’s law. Examples of hydraulic machines:

  • Hydraulic Lift
  • Hydraulic press
  • Hydraulic brake.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

Pascal’s Law:
Whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions.

Hydraulic lift:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 8
A hydraulic lift is used to lift heavy load. Consider a liquid enclosed in a vessel with two cylinders C1 and C2 attached as shown in the figure. The cylinders are provided with two pistons having areas A1 and A2 respectively.
If F1 is the force exerted on the area A1.
pressure P1 = \(\frac{F_{1}}{A_{1}}\)
If F2 is the force exerted on the area A2,
pressure P2 = \(\frac{F_{2}}{A_{2}}\)
According to pascal’s law P1 = P2 ie;
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 9
Using this method we can lift heavy load by applying small force.
Hydraulic Brake:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 10
Hydraulic brake is based on Pascals law. When the brake pedal (F) is pressed, a pressure is produced on liquid. This pressure will reach at A1 and A2. Due to large area of A1 and A2, a large force is produced. This large force will push brake shoes (S1 ans S2) in to the inner rim of wheel. Hence the motion of wheel will be retarded.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

Streamline Flow
Fluid dynamics:
The study of the fluids in motion is known as fluid dynamics.
Steady Flow:
The flow of the fluid is said to be steady if at any given point, the velocity of each fluid particle remains constant in time. The path taken by a fluid particle under a steady flow is a streamline.
Turbulent flow:
As the speed of flow increase, at a certain speed the stream line flow stops and irregular flow starts. This irregular flow is called turbulent.
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 11
The speed at which the turbulent flow starts is called critical speed.
Equation of continuity:
Flow rate of a incompressible fluid is constant
ie Av = cont.
Where A is the area of cross section of flow and v is the velocity of flow. ‘Av’ is also called volume flux.
Proof:
Consider a flow of incompressible liquid. Let P, R and Q be three points of a flow. Consider three planes perpendicular to direction of flow at P, R and Q.
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 12
Let AP, AR and AQ be the area of cross section and VP, VR and VQ be the corresponding speeds at P, R, Q respectively.
The mass of liquid passing through area AP in a time Dt, MP = AP VP rP ∆t
Similarly, MR = AR VR rR ∆t
MQ = AQ VQ rQ ∆t
But we know rP = rR = rQ = r
Hence we get MP = AP VP r ∆t
MR = AR VRr ∆t
MQ = AQ VQ r ∆t
the mass of liquid flowing out = the mass flowing in
ie; MP = MR = MQ
∴ APVP = ARVR = AQVQ
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 13

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

Bernoulli’S Principle
Bernoulli’s theorem:
As we move along a streamline the sum of the pressure (p), the kinetic energy per unit volume \(\frac{\rho v^{2}}{2}\) and the potential energy per unit volume (ρgh) remains a constant.

(OR)

The total energy of an incompressible non-viscous liquid flowing from one place to another without friction is a constant. Mathematically Bernoulli’s theorem can be written as
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 14
Proof:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 15
Consider an incompressible liquid flowing through a tube of non uniform cross section from region 1 to region 2. Let P1 be the pressure, A1 the area of cross section and V1 the speed of flow at the region 1.

The corresponding values at region 2 are P2, A2, and V2 respectively. Region 1 is at a height h1 and region 2 is at a height h2.
The workdone on the liquid in a time ∆t at the region 1 is given by
W1 = force × distance
= P1A1∆ x1
= P1∆V1 (∵ A1∆x1 = ∆V)
Where ∆x1 is the displacement produced at region 1, during the time interval ∆t.
Similarly, the workdone in a time ∆t at the region 2 is given by,
W2 = -P2 A2 ∆ x2
W2 = -P2 ∆ V2
[Here -ve sign appears as the direction of \(\vec{p}\) and ∆x are in opposite directions.]
Net workdone
DW = P1 ∆ V1 – P2 ∆ V2

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
According the equation of continuity
∆V1 = ∆V2 = ∆V
∆W= P1 ∆ V – P2 ∆ V
∆W = (P1 – P2) ∆ V _____(1)
This work done changes the kinetic energy, pressure energy and potential energy of the fluid.
If ∆m is the mass of liquid passing through the pipe in a time ∆t. the change in Kinetic energy is
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 16
Change in gravitational potential energy is given by
∆p.E = ∆mgh2 – ∆mgh1
∆p.E = ∆mg(h2 – h1) ______(3)
According to work-energy theorem work done is equal to the change in kinetic energy plus the change in potential energy.
ie; ∆w = ∆kE + ∆PE ______(4)
Substituting eq. 1, 2 and 3 in eq. 4, we get
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 17
Dividing by ∆V , we get
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 18

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
Case – 1
When the fluid at rest, its velocity is zero everywhere. Hence Bernoullis equation becomes
p1 + ρgh1 = P2 + ρgh2
(P1 – P2) = ρg(h2 – h1 )
Note: Bernoullis law is conservation law of energy for a flowing liquid.

1. Speed of Afflux: Torricellis law:
Torricellis law may be stated as the velocity of afflux through a hole at a depth ‘h’ will be equal to the velocity gained by a freely falling body when it travels a distance ‘h’. ie; v = \(\sqrt{2 \mathrm{gh}}\).
Proof
Consider a tank containing a liquid of density r with a small hole in its side. Let Y1 be the height of the hole, and y2 be height of water in the tank. Applying Bernoullis equation at points (1) and (2) we get
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 19
(∵ P1 = Pa, the atmospheric pressure).
If the cross sectional area of the tank A2 is much larger than that of the hole (ie; A2 >> A1 ), we take
v2 ≈ 0.
∴ eq(1) can be written as.
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 20
Note: The equation (2) shows that the speed of afflux is determined by the container pressure. Such a situation occurs in rocket propulsion.

2. Venturi – meter:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 21

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
A venturimeter is used to measure the flow of a liquid. It is based on Bernoullis theorem. Venturimeter consists of a horizontal tube of nonuniform cross section. A manometer (U tube) is attached to it as shown in the figure.

When a liquid flows through the tube, the manometer level falls in the limb (1) and rises in the limb (2). Let ‘h’ be the height difference of the manometer liquid in the two limbs. Let A be the cross-sectional area and V1 the velocity at the wide limb(1) and V2 be the corresponding values at the narrow limb (2).
According to Bernoullis theorem
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 22
Where ρm is the density of manometer liquid.
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 23
Using the above equation, the rate of flow can be measured.
Atomiser (application of Bernoulli’s theorem):
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 24

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
Atomiser is used for getting a fine spray of perfumer or insecticide. It consists of a cylinder with a piston. A small vessel containing liquid to be sprayed is attached to the cylinder. When the piston is moved forward air is blown out through a small opening of the cylinder.

As the velocity of flow of air increases, the pressure at the opening decreases. Due to the lower pressure at the opening, the liquid rises through the narrow tube and gets sprayed out along with air.

Blood flow and heart attack:
Bernoulli’s principle helps in explaining blood flow in artery. The artery may get constricted due to the accumulation of plaque on its inner walls. The speed of blood flow increases at which plaque is deposited. The increase in speed will lowers the pressure inside the artery and it may collapse due to the external pressure. This may result in heart attack.

4. Dynamic lift:
Dynamic lift is the force that acts on a body by virtue of its motion through a fluid (air). (This dynamic lift can be partly explained on the basis of Bernoulli’s principle)
Example:

  1. Ball moving without spin
  2. Ball moving with spin
  3. Aerofoil or lift on aircraft wing.

1. Ball moving without spin:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 25
When a ball move in air without spinning, the velocity of fluid (air) above and below the ball is the same. Hence pressure difference between above and below is zero. Therefore there is no upward or downward force on the ball.

2. Ball moving with spin:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 26

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
When a ball is thrown in a air with spin, air will be dragged along with it. Therefore the velocity of air above (larger) and below (smaller) are different. This difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called Magnus effect.

3. Aerofoil or lift on aircraft wing:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 27
An aerofoil is a solid piece, shaped to provide an upward vertical force as it moves horizontally through air. Due to the curved shape at the top, the velocity of the air on the top side increases. Due to increase in velocity, the pressure at this region decreases. Thus an upward force develops on the aerofoil. This principle is applied in the design of the wings of aeroplanes.

Viscosity
Viscosity is liquid friction. When liquid layer moves over another liquid layer, there is a force of friction between the liquid layers, opposing the motion of layers.

Laminar flow:
When a liquid flows on a surface, the layer of liquid in contact with the surface will be rest. The velocity of the layer increases as the distance of the layer from the surface increases. The velocity is maximum on the top most layer. This type of flow is known as laminar.

Coefficient of viscosity:
The coefficient of viscosity for a fluid is defined as the ratio of shearing stress to the strain rate. Explanation:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 28
Consider a portion of liquid having shape ABCD. When this liquid flows, it take the shape AEFD after a short interval of time Dt
During this time interval the liquid has undergone a shear strain of \(\frac{\Delta \mathbf{x}}{\ell}\). Since, the stress in a flowing fluid increases with time, we take rate.
ie. strain rate = \(\frac{\Delta \mathbf{x}}{\ell \Delta t}\)
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 29

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
Unit of viscosity:
The SI unit of viscosity is poiseiulle (PI). Its another units are NSm-2 or PaS.
Note:

  • Blood is thicker (more viscous) than water. The relative viscocity (n/nwater) of blood remains constant between 0° C and 37° C.
  • The viscocity of liquid decreases with temperature while it increases in the case of gases.

1. Stokes law:
If a body is moving through a fluid, its velocity is influenced by the viscosity of the medium. Consider a small sphere of radius ‘a’ falling under gravity in air: The viscocity of air opposes the motion of the sphere. This opposing force is called viscous force and can be written as.
F = 6πaηv
This is called stokes, formula.
Where η is the coefficient of viscocity of the medium and v is the velocity of the body.

(i) Terminal velocity:
The constant velocity attained by a body as it falls down through a fluid medium is called the terminal velocity.
Expression for terminal velocity:
Considers sphere of radius ‘a’ densitity r falling through a liquid of density s and viscocity h. The viscous force acting on the sphere can be written as
F = 6πaηv
Where v is the velocity of sphere. This force is acting in upward direction. When the body comes down its velocity will increase. Hence the viscous force will also increases.

When the viscous force is equal to the weight of the body in the medium. The net force on the body is zero. Now the body falls without acceleration. It moves with a constant velocity called the terminal velocity.
The weight of a body in a medium,
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 30
When body has terminal velocity, we can write.
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 31

Reynolds Number
The turbulence in a fluid is determined by a dimension parameter called Reynolds number.
Reynolds number Re = \(\frac{\rho \mathbf{v d}}{\eta}\)
Where ‘r is the density of a fluid, V- the speed of flow of the liquid, ‘d’- the dimension of pipe, h- viscocity of the liquid. Re – the dimensionless number. The flow is streamline when the Re is below 1000, unsteady when Re is in between 1000 and 2000 and turbulent when Re > 2000.
Critical velocity:
The maximum velocity of a fluid in a tube for which the flow remains streamlined is called its critical velocity.
Critical velocity v = \(\frac{R_{e} \eta}{\rho d}\).

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

Surface Tension
Liquids acquire a free surface when poured in a container. These surfaces possess some additional. energy. This phenomenon is known as surface tension.

1. Surface Energy:
Question 4.
What is the reason for surface tension?
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 32
Answer:
Consider a liquid enclosed in a vessel as shown in figure. Let A, B, and C be three molecules in the liquid. For a molecule A, the forces of attraction on the molecule are equal in all directions. Hence net force on it is zero. Consider a molecule B near the surface. In this case the molecule is partially outside the liquid. So there is a net downward force.

Consider another molecule Con the surface of liquid. Half of the sphere is outside the liquid. Hence the downward force on molecule is maximum. This, when a molecule comes to the surface, it is doing work to overcome this downward pull. The work done resides as RE. in the molecule on the surface. Thus molecule on the surface of liquid have P.E. This energy is called the surface energy.
Note:
1. To increase the area of surface of a liquid, work has to be done. This energy is required to increase the surface area.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

2. Surface energy and surface tension:
Question 5.
Show that Surface tension is equal to surface energy per unit area.
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 33
Answer:
Consider a horizontal liquid film as shown in figure. The bar at one end of film is free to move. Move the bar a small distance as shown in figure. Since the area of the surface increases, some work has to be done against an internal force.
workdone W = Fd _____(1)
This workdone is stored as additional energy in the film. If the surface energy of the film is S per unit area, the workdone to increase the surface area.
W = S × increase in area
W = S × 2/d ______(2)
Where 2/d is the increase in area due to two sides of film.
equating eq(1) and eq(2),we get
S 2/d = Fd
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 34
This quantity S is the magnitude of surface tension. It is equal to the surface energy per unit area.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

3. Angle of contact:
Angle of contact is the angle between the solid surface and the tangent drawn to the liquid surface at the point of contact inside the liquid.

If angle of contact of liquid is acute(<90°), the meniscus in the tube will be concave. The liquid level in the tube will rise to a certain height. This is called capillary rise. If the angle of contact is obtuse (>90°), the liquid level falls in the tube. This is called capillary fall or capillary depression.

4. Drops and bubbles:
Due to surface tension, the liquid surface always tends to have the minimum surface area. For a given volume, a sphere has a minimum surface area. Hence, srnail drops and bubbles of a liquid assume spherical shape.

On the other hand, for bigger drops the effect of gravity predominates over surface tension and the drop gets flattened.

(i) Excess of pressure inside a spherical drop The molecules near the surface of a drop experience a resultant pull inwards due to surface tension. Because of this inward pull, the pressure inside is greater than the outside.

Due to the forces of surface tension, the drop tends to contract. But due to excess inside pressure, the drop tends to expand. When the drop is in equilibrium, these two forces will be equal and opposite.

Expression for excess pressure inside a liquid-drop:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 35

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
Consider a drop of liquid of radius r. Let pi and po be the values of pressure inside and outside the drop. Let the radius of liquid of drop increases by a small amount Dr under the pressure difference.
The outward force acting on the surface of the drop,
f = pressure difference × surface area,
ie. f = (pi – po)4pr2
If the radius of liquid drop is increased by Dr due to the above force
The work done (DW) = f.Dr
= (pi – po)4pr2Dr _____(1)
The increase in surface area of the drop
= 4p(r + Dr)2 – 4pr2
= 4p(r2 + 2rDr + Dr2) – 4pr2 = 8prDr [ Neglecting Dr2]
If S is the surface tension of the liquid, the work done to increase the surface area,
DW = Increase in surface area × surface tension
DW= 8prDr.S _____(2)
From equation (1) and (2), we get 2S
pi – po = \(\frac{2 \mathrm{S}}{\mathrm{r}}\).

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

(ii) Expression for excess pressure inside a spherical bubble:
A liquid bubble has air both inside and outside it. Hence, it has two surfaces;
Hence the total increase in the surface area, = 16prDr
The workdone to increase the surface area,
= 16prDrS ______(3)
From equation (1) and (3) we get,
pi – po = \(\frac{4 \mathrm{S}}{\mathrm{r}}\)

5. Capillary rise:
A tube of very fine core is called a capillary tube. When a capillary tube is dipped in a liquid, the liquid immediately rises in the tube. This phenomena is called capillarity or capillary rise. In the case of mercury, capillary depression is observed.
Expression for capillary rise:
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 36
Consider a capillary tube of radius ‘a’ dipped in a liquid of density rand surface tension S. If the liquid has a concave meniscus it will rise in the capillary tube. Let h be the rise of the liquid in the tube. Let p1 be the pressure on the concave Side of the meniscus and p0, that on the other side.
The excess pressure on the concave side of the meniscus can be written as
pi – po = \(\frac{2 \mathrm{S}}{\mathrm{r}}\) _____(1)
Where R is the radius of the concave meniscus. The tangent to the meniscus at the point A makes an angle q with the wall of the tube.
In the right angled triangle ACO
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 37

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids
substituting the values of R in the equation (1)
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 38
Considering two points M and N in the same horizontal level of a liquid at rest,
pressure at N = pressure at M
But pressure at M = pi, the pressure over the concave meniscus and pressure at N = po + hrg
∴ po = po + hrg
or pi – po = hrg ______(3)
From equations (2) and (3), we get
hρg = \(\frac{2 \mathrm{S} \cos \theta}{\mathrm{a}}\)
Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids 39
case-1
For water q = 0, ie. cosq = 1
hence S = \(\frac{1}{2}\) haρg

6. Detergents and Surface Tension:
Dirty clothes containing grease and oil stains sticking on them cannot be cleaned by soaking them in water. – This is because water will not wet greasy dirt. If water could wet grease, then the flow of water would carry away some greasy dirt.

Plus One Physics Notes Chapter 10 Mechanical Properties of Fluids

But, by adding detergent or soap to water, the greasy dirt can be removed. The molecules of the detergent are hairpin shaped. When the detergent is added to water, one end of hair pin shaped molecule get attached to the water surface and other end to stain.

This results in the formation of water – grease interface. Thus surface tension between water and grease reduces. Hence dirt can be removed by the moving water.