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Students can Download Chapter 2 Data Representation and Boolean Algebra Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 2 Data Representation and Boolean Algebra

Number System:
It is a systematic way to represent numbers in different ways. Each number system has its own Base, that is a number and that number of symbols or digits used.

1. Most Significant Digit (MSD): The digit with most weight is called MSD. MSD is also called Left Most Digit(LMD)
2. Least Significant Digit (LSD): The digit with least weight is called LSD. LSD is also called Right Most Digit(RMD)
   • eg: 106 : Here MSD : 1 and LSD : 6
   • 345.78: Here MSD : 3 and LSD : 8
3. A Binary Digit is also called a bit.
4. The weight of each digit of a number can be represented by the power of its base.

Number conversions:
In general, to convert a decimal number into another number system(binary or octal or hexadecimal) do the following. Divide the number successively by the base of the number system do you want to convert and write down the remainders from bottom to top.

To convert a decimal fraction into another number system .multiply the number by the base of the number system do you want to convert then integer part and fractional part are separated again multiply the fractional part by the base and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Decimal to Binary:
Divide the number by the base 2 successively and write down the remainders from bottom to top.

Decimal fraction to binary:
multiply the number by the base 2 then integer part and fractional part are separated again multiply the fractional part by the base 2 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Decimal to Octal:
Divide the number by the base 8 successively and write down the remainders from bottom to top.

Decimal fraction to octal:
multiply the number by the base 8 then integer part and fractional part are separated again multiply the fractional part by the base 8 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.
eg: (55)₁₀ = ()₈

(0.140625)₁₀ = (0.11)₈

Decimal to Hexadecimal:
Divide the number by the base 16 successively and write down the remainders from bottom to top.

Decimal fraction to hexadecimal:
multiply the number by the base 16 then integer part and fractional part are separated again multiply the fractional part by the base 16 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Converting a number from any number system into decimal: For this multiply each digit by its corresponding weight and sum it up.

Binary to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 2.

Converting binary fraction to decimal

101.101 = 1 × 2² + 0 × 2¹ + 1 × 2⁰ + 1 × 2^-1 + 0 × 2^-2 + 1 × 2^-3
= 4 + 0 + 1 + 1/2 + 0 + 1/8
= 5 + 0.5 + 0.125
(101.101)₂ = (5.625)₁₀

Octal to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 8.
Eg: (1007)₈ =()₁₀?

1 × 8³ + 0 × 8² + 0 × 8¹ + 7 × 8⁰
= 512 + 0 + 0 + 7
=(519)₁₀
Converting octal fraction to decimal (600.005)₈ =()₁₀?

= 6 × 8² + 0 × 8¹ + 0 × 8⁰ + 0 × 8^-1 + 0 × 8^-2 + 5 × 8^-3
= 384 + 0 + 0 + 0 + 0 + 0.009765625
= (384.009765625)₁₀

Hexadecimal to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 16.
Eg: (100)₁₆ = ()₁₀?

= 1 × 16² + 0 × 16¹ + 0 × 16⁰
= 256 + 0 + 0
= (256)₁₀
Converting Hexadecimal fraction to decimal (60A.4)₈ =()₁₀?

= 6 x 16² + 0 x 16¹ + 10 x 16⁰ + 4 x 16^-1
= 1536 + 0 + 0 + .25
= (1536.25)₁₀

Octal to binary conversion:
Convert each octal digit into its 3 bit binary equivalent. Consider the following table

Hexadecimal to binary conversion:
Convert each Hexadecimal digit into its 4 bit binary equivalent. Consider the following table

1010 1011 1100 (ABC)₁₆=(101010111100)₂

Binary to octal conversion
Divide the binary number into groups of 3 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding octal equivalent.
eg: (10100110)₂= ()₈?
Insert one zero in the left side to form 3 bits group

(10100110)₂= (246)₈

Binary to Hexadecimal conversion:
Divide the binary number into groups of 4 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding Hexadecimal equivalent.
eg: (100100110)₂ = ()₁₆?
Insert 3 zeros in the left side to form 4 bits group

(100100110)₂ = (126)₁₆

Octal to Hexadecimal conversion:
First convert octal number into binary(see 1.6.7), then convert this binary into hexadecimal(also see 1.6.10)
eg: Convert (67)₈ = ( )₁₆
Step I: First convert this number into binary equivalent for this do the following:

Step II: Next convert this number into hexadecimal equivalent for this do the following.

So the answer is (67)₈ = ( 37)₁₆

Hexadecimal to octal conversion:
First convert Hexadecimal to binary(see 1.6.8), then covert this binary into octal(also see 1.6.9)
eg: Convert (A1)₁₆ = ( )₈?
Step I: First convert this number into binary equivalent. For this do the following

Step II. Next convert this number into octal equivalent. For this do the following.
So the answer is (A1)₁₆ = (241)₈

Data representation:
The data stored in the computer memory is in the form of binary.

Representation of integers
There are three ways to represent integers in computer. They are as follows:

1. Sign and Magnitude Representation (SMR)
2. 1’s Complement Representation
3. 2’s Complement Representation

1. SMR:
Normally a number has two parts sign and magnitude, eg: Consider a number+5. Here + is the sign and 5 is the magnitude. In SMR the most significant Bit (MSB) is used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is -ve. eg: If a computer has word size is 1 byte then

Here MSB is used for sign then the remaining 7 bits are used to represent magnitude. So we can , represent 2⁷ = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to +127 and 0 to -127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.

2. 1’s Complement Representation:
To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but +ve number has no 1 ’s complement,
eg:
(i) To find the 1’s complement of -21
+21 = 00010101
To get the 1’s complement change all 0 to 1 and.all 1 to 0.
-21 = 11101010
1’s complement of-21 is 11101010

(ii) Find the 1’s complement of +21. Positive numbers are represented by using SMR.
+21 = 00010101 (No need to take the 1’s complement)

3. 2’s Complement Representation:
To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg: To find the 2’s complement of -21
+21 = 00010101
First take the 1’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of -21 is 1110 1011

Representation of floating point numbers:
A real number consists of an integer part and fractional part and represent by using Exponent and Mantissa method. This method is also used to represent too big numbers as well as too small numbers.
Eg: .0000000000000000000000001 can be represented easily as 1 × 10^-25. Here T is the Mantissa and -25 is the exponent.

A computer with 32 bit word length is used 24 bits for mantissa and the remaining 8 bits used to store exponent.

Representation of characters:
1. ASCII(American Standard Code for Information Interchange):
It is 7 bits code used to represent alphanumeric and some special characters in computer memory. It is introduced by U.S. government. Each character in the keyboard has a unique number.

Eg: ASCII code of ‘a’ is 97, when you press ‘a’ in the keyboard , a signal equivalent to 1100001 (Binary equivalent of 97 is 1100001) is passed to the computer memory. 2⁷ = 128, hence we can represent only 128 characters by using ASCII. It is not enough to represent all the characters of a standard keyboard.

2. EBCDIC(Extended Binary Coded Decimal Interchange Code):
It is an 8 bit code introduced by IBM(lnternational Business Machine). 2⁸ = 256 characters can be represented by using this.

3. ISCII(lndian Standard Code for Information Interchange):
It uses 8 bits to represent data and introduced by standardization committee and adopted by Bureau of Indian Standards(BIS).

4. Unicode:
The limitations to store more characters is solved by the introduction of Unicode. It uses 16 bits so 2¹⁶ = 65536 characters (i.e, world’s all written language characters) can store by using this.

Binary arithmetic:
Binary addition:
The rules for adding two bits

eg: Find the sum of binary numbers 110011 and 100001.

Binary subtraction:
The rules for subtracting a binary digit from another digit.

Subtraction using 1’s complement:
The steps are given below:

Step 1: Add 0s to the left of the subtrahend, to make two numbers with same number of bits.
Step 2: Find 1’s complement of subtrahend.
Step 3: Add the complement with minuend.
Step 4: If there is a carry, ignore the carry, the result is positive then add the carry 1 to the result.
eg: Subtract 1101 from 111100 using 1’s complement method.
Step 1: Insert two Os to the left of 1101. Hence the subtrahend is 001101.
Step 2: 1’s complement of 001101 is 110010
Step 3: Add this to the minuend.

Step 4: Ignore the carry the result is positive and add add the carry 1 to 101110

Hence the result is 101111.

Subtraction using 2’s complement:
The steps are given below :
Step 1: Add 0s to the left of the subtrahend, to make two numbers with same number of bits.
Step 2: Find 2’s complement of subtrahend.
Step 3: Add the complement with minuend.
Step 4: If there is a carry, ignore the carry, the result is positive.
eg: Subtract 1101 from 111100 using 2’s complement method.
Step 1: Insert two 0s to the left of 1101. Hence the subtrahend is 001101.
Step 2: Find the 2’s complement of 001101.
1’s complement is 110010.
2’s complement is 110010 + 1 = 110011
Step 3: Add this to the minuend.

Step 4: Ignore the carry the result is positive. Hence the result is 101111.

Introduction to Boolean algebra:
The name Boolean Algebra is given to honour the British mathematician George Boole. Boolean algebra deals with two states true or false otherwise Yes or No and numerically either 0 or 1.

Binary valued quantities:
A logical decision which gives YES or No values is a binary decision, A statement which gives YES or NO values(TRUE or FALSE) is a logical statement or truth function. A variable which can assign TRUE or FALSE (1 or 0) values is a logical variable

Boolean operators and logic gates:
Logical Operators are AND, OR and NOT. A logical gate is a physical device (electronic circuit)that can perform logical operations on one or more logical inputs and produce a single logical output. A table represents the set f all possible values and the corresponding results in a statement is called truth table.
1. The OR operator and OR gate:
The OR operator gives a 1 either one of the operands is 1. If both operands are 0, it produces 0. The truth table of X OR Y is

The logical OR gate is given below.

The truth table and the gate for the Boolean expression Y = A + B + C

2. The AND operator and AND gate:
The AND operator gives a 1 if and only if both operands are 1. If either one of the operands is 0, it produces 0 The truth table of X AND Y is

The logical AND gate is given below.

The truth table and the gate for the Boolean expression Y = A . B . C

3. The NOT operator and NOT gate:
It produces the vice versa. NOT gate is also called inverter. It is a unary operator that means it has only one input and one output. The truth table of NOT X is

Basic postulates of Boolean algebra:
Boolean algebra consists of some fundamental laws. These laws are called postulates.
Postulate 1: Principles of 0 and 1
If A ≠ 0 , then A = 1 and A 1, then A = 0

Principle of Duality:
When changing the OR(+) to AND(.), AND (.) to OR(+), 0 to 1 and 1 to 0 in a Boolean expression we will get another Boolean relation which is the dual of the first, this is the principle of duality.

Basic theorems of Boolean algebra:
There are some standard and accepted rules in every theory, these rules are known as axioms of the theory.

Identity law:
If X is a Boolean variable, the law states that

1. 0 + X = X
2. 1 + X = 1 (these are additive identity law)
3. 0 . X = 0
4. 1 . X = X (these are multiplicative identity law)

Following are the truth tables

Idempotent law:
This law states that

1. X + X = X
2. X . X = X

Involution law:
This states that
\(\overline{\overline{\mathrm{X}}}=\mathrm{x}\)
The compliment of compliment of a number is the number itself.

Complimentary law:
This law states that

1. \(x+\bar{x}=1\)
2. \(x \cdot \bar{x}=0\)

The truth table is given below

Commutative law:
This law allows to change the position of variable in OR and AND

1. X + Y = Y + X
2. X . Y = Y . X

The truth table is given below

Associative law:
It allows grouping of variables differently

1. X + (Y + Z) = (X + Y) + Z
2. X . (Y . Z) = (X . Y) . Z

The truth table is given below

Distributive law:
This law allows expansion of multiplication over addition and also allows addition operation over multiplication.

1. X . (Y + Z) = X . Y + X . Z
2. X + Y . Z = (X + Y) . (X + Z)

The truth table is given below

Absorption law:
It is a kind of distributive law in which two variables are used and result will be one of them

1. X + (X . Y) = X
2. X . (X + Y) = X

The truth table is given below

De Morgan’s theorem:
Demorgan’s first theorem states that
\(\overline{\mathrm{X}+\mathrm{Y}}=\overline{\mathrm{X}} \cdot \overline{\mathrm{Y}}\)
ie. the compliment of sum of two variables equals product of their compliments.

The second theorem states that
\(\overline{\mathrm{X} . {\mathrm{Y}}}=\overline{\mathrm{X}}+\overline{\mathrm{Y}}\)
ie. The compliment of the product of two variables equals the sum of the compliment of that variables.

Circuit designing for simple Boolean expressions:
By using basic gates such as AND, OR and NOT gates we can create logic circuits.

Universal gates:
By using NAND and NOR gates only we can create other gate hence these gates are called Universal gate.

NAND gate:
The output of AND gate is inverted by NOT gate is the NAND gate

NOR gate:
The output of OR gate is inverted by NOT gate is the NOR gate.

Students can Download Chapter 1 Some Basic Concepts of Chemistry Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

INTRODUCTION
Chemistry is the branch of science which deals with the composition, properties and transformation of matter. These aspects can be best understood in terms of basic constituents of matter: atoms and molecules. That is why chemistry is called the sci-ence of atoms and molecules.

IMPORTANCE OF CHEMISTRY
Chemistry plays an important role in almost all walks of life. In recent years chemistry has tackled with a fair degree of success some of the pressing aspects of environmental degradation. Safer alternatives to environmentally hazardous refrigerants like CFCs, responsible for ozone depletion in the stratosphere, have been successfully synthesised.

NATURE OF MATTER
Matter is anything which has mass and occupies space. Matter can exist in three physical states: solid, liquid and gas.

At the macroscopic or bulk level, matter can be clas-sified as mixtures or pure substances.
A material containing only one substance is called a pure substance. Materials containing more than one substance are called mixtures. Pure substances are further classified into two types: elements and com¬pounds. Mixtures are also of two types: homoge¬neous mixtures and heterogeneous mixtures. A mix¬ture is said to be homogeneous if it has same com¬position throughout. Some examples of homoge¬neous mixtures are air, gasoline, kerosene, milk, alloys, etc. Heterogeneous mixtures are the mixtures which have different composition in different parts. Some examples of heterogeneous mixtures are iron and sulphur, muddy water, etc.

PROPERTIES OF MATTER AND THEIR MEASUREMENT
Every substances has characteristic properties. These are classified into two categories – physical properties and chemical properties.
Physical properties are those properties which can be measured or observed without changing the iden¬tity or the composition of the substance. Some ex¬amples of physical properties are colour, odour, melt¬ing point, boiling point, density, etc. chemical properties are characteristic reactions of different substances; these include acidity or basic¬ity, combustibility, etc.

THE INTERNATIONAL SYSTEM OF UNITS (SI UNITS)
A unit may be defined as the standard of reference chosen to measure any physical quantity. There are many different systems of units.
The improved metric system of units accepted inter-nationally is called International System of Units or SI units. The SI system has seven base units from which all other units are derived.

UNCERTAINTY IN MEASUREMENT
Scientific measurements involving some measuring devices have some degree of uncertainty. The magnitude of uncertainty depends on the accuracy of the measuring device and also on the skill of its operator.
The closeness of a set of values obtained from identical measurements of a quantity is known as precision of the measurement.
The term accuracy is defined as the closeness of a measurement or a set of measurements to its true value.

SIGNIFICANT FIGURES
The total number of digits in a measurement is called the number of significant figures. It includes the num¬ber of figures that are known with certainty plus the last uncertain digit, beginning with the first non-zero digit.
1) All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.

2) Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.
3) Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.

4) Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example,0.200 g has three significant figures. But, if otherwise, the zeros are not significant. For example, 100 has only one significant figure.

5) Exact numbers have an infinite number of significant figures. For example, in 2 balls or 20 eggs, there are infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e.,2 = 2.000000 or 20 = 20.000000

LAWS OF CHEMICAL COMBINATIONS
Law of Conservation of Mass
It states that matter can neither be created nor destroyed.
This law was put forth by Antoine Lavoisier in 1789.

Law of Definite Proportions
This law was given by, a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight. It is sometimes also referred to as Law of definite composition.

Law of Multiple Proportions
This law was proposed by Dalton in 1803. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

Gay Lussac’s Law of Gaseous Volumes
This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Avogadro Law
In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules Avogadro made a distinction between atoms and molecules which is quite understandable in the present times.

DALTON’S ATOMIC THEORY
The main postulates of the theory are:-

1. Matter is made up of extremely small, indivisible particles called atoms.
2. Atoms of the same element are identical in all respects i.e. size and mass.
3. Atoms of different elements are different, i.e., they possess different sizes, shapes, masses, and chemical properties.
4. Atoms of different elements may combine with each other in a simple whole number ratio to form compound atoms or molecules.
5. Atoms can neither be created nor destroyed, i.e., atoms are indestructible.

ATOMIC AND MOLECULAR MASSES
ATOMIC MASS
The mass of an atom is extremely small. These are very inconvenient for calculations. This difficulty was overcome by expressing atomic masses as relative masses, i.e., with respect to the mass of an atom of a standard substance. The scale in which the relative masses are expressed is called atomic mass unit scale or amu scale. One atomic mass unit (amu) is equal to one-twelfth (1/ 12) the mass of an atom of carbon-12. Recently the symbol ‘u’ (unified mass) is used in place of amu.

AVERAGE ATOMIC MASS
The atomic masses of many elements have fractional values because they exist as mixture of isotopes.
In the case of such elements, the atomic mass is taken as the average of the atomic masses of the various isotopes. For example, ordinary chlorine is a mixture of two isotopes with atomic masses 35 u and 37 u and they are present in the ration 3:1. Therefore,
Atomic mass of Chlorine = \(\frac{35 \times 3+37 \times 1}{3+1}\) = 35.5u

MOLECULAR MASS
Molecular mass is the sum of atomic masses of the elements present in a molecule.
For example, molecular mass of water
= 2 x atomic mass of hydrogen + 1 x atomic mass of oxygen
= 2 × (1.008u) + 1 × 16.00 u
= 18.02 u

FORMULA MASS
In ionic compounds, we use formula mass instead of molecular mass. Formula mass of an ionic compound is the sum of the atomic masses of all atoms in a formula unit of the compound.

MOLE CONCEPT AND MOLAR MASSES
‘Mole’ was introduced as the seventh base quantity for the amount of substance in SI system. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the ¹²C isotope. This number is known as ‘Avogadro constant’ (N_A = 6.022 × 10²³). The mass of one mole of a substance in grams is called its molar mass. The molar mass is numerically equal to atomic/ molecular/ formula mass in u.

PERCENTAGE COMPOSITION
One can check the purity of a sample by analysing its mass percentage
Mass % of an element in a compound =

EMPIRICAL FORMULA FOR MOLECULAR FORMULA
An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

Problem 1.2
A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulae?
Solution:
Step 1. Conversion of mass per cent to grams.
Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen is present, 24.27g carbon is present and 71.65 g chlorine is present.

Step 2. Convert into number moles of each element
Divide the masses obtained above by respective atomic masses of various elements.

Step 3. Divide the mole value obtained above by the smallest number
Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:CI.
In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4. Write empirical formula by mentioning the numbers after writing the symbols of respective elements.
CH₂Cl is, thus, the empirical formula of the above compound.

Step 5. Writing molecular formula
a) Determine empirical formula mass. Add the atomic masses of various atoms present in the empirical formula.
ForCH₂Cl, empirical formula mass is 12.01 + 2 1.008 + 35.453 = 49.48 g

b) Divide Molar mass by empirical formula mass

c) Multiply empirical formula by ‘n’ obtained above to get the molecular formula
Empirical formula = CH₂Cl, n = 2.
Molecular formula = 2 × [CH₂Cl]=C₂H₄Cl₂

STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
‘Stoichiometry’ deals with the calculation of masses (sometimes volumes also) of the reactants and prod¬ucts involved in a chemical reaction. The coefficients of reactants and products in a balanced chemical equation is called the stoichiometric coefficients.

LIMITING REAGENT
The amount of the product obtained in a chemical reaction is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is com¬pletely consumed in a reaction containing two or more reactants.

REACTIONS IN SOLUTIONS
1. Mass per cent :
It is obtained by the following relation:

2.Mole Fraction :
It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB respectively; then the mole fractions of A and B are given as

3. Molarity :
It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution.

4. Molality :
It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by ‘m’.

Note :
Molarity of a solution changes with temeprature. But molality of a solution does not change with temperature since mass remains unaffected with temperature.

Students can Download Chapter 14 Oscillations Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 14 Oscillations

Plus One Physics Oscillations One Mark Questions and Answers

Question 1.
Fill in the blanks :
A girl is swinging on a swing in a sitting position. When she stands up, the period of the swing will______.
Answer:
Decreases

Question 2.
A particle executes a simple harmonic motion with a frequency f. What is the frequency with which its kinetic energy oscillates?
Answer:
Frequency of oscillation of kinetic energy is 2f.

Question 3.

Answer:
a. 7/2R
b. ω = \(\sqrt{k / m}\)

Question 4.
Can a simple pendulum vibrate at centre of earth?
Answer:
No. Because ‘g’ at centre of earth is zero.

Question 5.
A glass window may be broken by a distant explosion. Why?
Answer:
The sound waves can cause forced vibrations in glass due to difference between frequency of sound wave and natural frequency of glass. This can break the glass window.

Plus One Physics Oscillations Two Mark Questions and Answers

Question 1.
A simple pendulum is transferred from earth to moon. Will it go faster or slower?
Answer:
The value of g at moon is low compared to earth. The decrease in g will increase time period of simple pendulum. Hence pendulum will vibrate slower.

Question 2.
Here five examples of accelerated motion are given in first column. Match each examples given in the second column.

Answer:

Question 3.
A girl is swinging a swing in sitting position. What shall be the effect of frequency of oscillation if

1. if she stands up
2. if another girl sits gently by her side

Answer:
1. If the system is considered as simple pendulum, length of pendulum is reduced as girl stands up.

So frequency of oscillation is increased.

2. The time period and hence frequency of simple pendulum is independent of mass. Hence there is no change in frequency.

Plus One Physics Oscillations Three Mark Questions and Answers

Question 1.
A student is advised to study the variation of period of oscillation with the length of a simple pendulum in the laboratory. According he recorded the period of oscillation for different lengths of the pendulum.

1. If he plots a graph between the length and period of oscillation, what will be the shape of the graph?
2. How would you determine the value of acceleration due to gravity using l – T² graph?

Answer:
1.

This is the equation parabola. Hence the shape of graph between period(T) and length (l) will be parabola.

2. Find slope of l – T² graph. The acceleration due to gravity can be found using formula g = 4P² × slope of (l – T²) graph.

Question 2.
A simple pendulum has a bob of mass m is suspended from the ceiling of a lift which is lying at the ground floor of a multistoried building.

1. Find the period of oscillation of pendulum when the lift is stationary.
2. What is the tension of the string of the pendulum when it is ascending with an acceleration ‘a’?
3. What is the period of oscillation of the pendulum while the lift is ascending?

Answer:

1. T = \(2 \pi \sqrt{l /g}\)
2. Tension, T= m (g + a)
3. T = 2π\(\sqrt{\frac{\ell}{g+a}}\)

Question 3.
A body tied a spherical pot with a string and suspended it on a clamp. He then filled it with water. Length of the string if 90 cm and diameter of the pot is 20 cm. The pot is slightly displaced to one side and leave it to oscillate. Considering the above example as a simple pendulum (g = 9.8 ms^-2)

1. What is the length of the Pendulum
2. Calculate the period of oscillation of the pendulum.

Answer:
1. Length of pendulum l = 90 + 10 = 100 cm

2.

Question 4.

1. Motion repeated at regular intervals of time is called periodic. Explain the simple harmonic motion with a figure.
2. A particle executes a simple harmonic motion with a period 2 seconds, starting from its equilibrium at time t = 0. Find the minimum time in which it is displaced by half the amplitude.

Answer:
1. A periodic motion in which acceleration is directly proportional to displacement but opposite in direction is called SHM.

The graphical variation of simple harmonic motion with time given above.

2. y = a sin wt

Sin π/6 = sin wt
wt = π/6
\(\frac{2 \pi}{T}\)t = π/6
But T = 2s. Hence we get

Question 5.
A spring of spring constant ‘k’ is used to suspend a mass ‘m’ at its free end while the other end of the spring is rigidly fixed.

1. If the mass is slightly depressed and released, then name the motion of the mass.
2. Write down the expression for the period of oscillation of the mass.
3. If this system is taken into outer space then what happens to its period? Why?

Answer:

1. Simple Harmonic Motion
2. T = \(2 \pi \sqrt{m / k}\)
3. Period of oscillation does not change.

Plus One Physics Oscillations Four Mark Questions and Answers

Question 1.
A simple harmonic motion is represented by x(t) = Acosωt.

1. What do you mean by simple harmonic motion.
2. An SHM has amplitude A and time period T, What is the time taken to travel from x = A to x = A/2

Answer:
1. Simple harmonic motion is the simplest form of oscillatory motion. The oscillatory motion is said to be simple harmonic motion if the displacement ‘x’ of the partide from the origin varies with time as

2. x = Acosωt

Question 2.

1. Define simple harmonic motion for a particle moving in a straight line.
2. Use your definition to explain how simple harmonic motion can be represented by the equation.
3. Show that the above equation is dimensionally correct.
4. A mechanical system is known to perform simple harmonic motion. What quantity must be measured in order to determine frequency for the system?

Answer:

1. A periodic motion in which acceleration is directly proportional to displacement and opposite in direction is called simple harmonic oscillation.
2. Mathematically, a simple harmonic oscillation can be expressed as
   y = a sin wt (or) y = a cos wt
3. y = a sin w t
   Sin wt has no dimension. Hence we need to consider dimension of ‘a’ only, ie, y = a, L = L
4. Its period is determined.

Question 3.
A particle executes simple harmonic motion according to the equation x = 5sin\(\left(\frac{2 \pi}{3} t\right)\)

1. find the period of the oscillation
2. What is the minimum time required for the particle to move between two points 2.5cm on either side of the mean position?

Answer:
1. x = 5sin\(\left(\frac{2 \pi}{3} t\right)\), when we compare this equation
with standard S.H.M, x = a sin wt.
We get wt = \(\frac{2 \pi}{3} t\)

2. y = a sin wt
2.5 = 5 sin w × t

Time taken to travel 2.5 from the mean position is 0.25 sec. Hence time taken to travel 2.5 cm on either side of the mean position is 0.5 sec.

Question 4.
A mass m is suspended at one end of a spring and the other end of the spring is firmly fixed on the ceiling. If the mass is slightly depressed and released it will execute oscillation.

1. Write down the expression for the frequency of oscillation of the mass.
2. If the spring is cut into two equal halves and one half of the spring is used to suspend the same mass then obtain an expression for the ratio of periods of oscillation in two cases.
3. If this system is completely immersed in water then what happens to the oscillation?

Answer:
1. f = \(\frac{1}{2 \pi} \sqrt{k / m}\)

2. f₁ = [atex]\frac{1}{2 \pi} \sqrt{k / m}[/latex] ____(1)

3. When spring is cut in to half,spring constant becomes 2k. Hence frequency,

4. If this system is immersed in water, the amplitude of oscillation decrease quickly. Hence system comes to rest quickly.

Question 5.
Starting from the origin, a body oscillates simple harmonically with an amplitude of 2m and a period of 2s.

1. What do you mean simple harmonic motion.
2. Draw the variation of displacement with time for the above motion.
3. After what time, will its kinetic energy be 75% of the total energy?

Answer:
1. The oscillatory motion is said to be simple harmonic motion if the displacement ‘x’ of the particle from the origin varies with time as

where
x(t) = displacement x as a function of time t
A = amplitude
ω = angular frequency
(ω t + Φ) = phase (time-dependent)
Φ = phase constant or initial phase

2.

3. y = a sin ω t
ν = aω cos ωt
Kinetic energy,

Question 6.
A body of mass 16 kg is oscillating on a spring of force constant 100 N/m.

1. What do you mean by spring constant.
2. Derive a general expression for period of oscillating spring.

Answer:
1. Spring constant is the force required for unit extension.

2.

Consider a body of mass m attached to a massless spring of spring constant K. The other end of spring is connected to a rigid support as shown in figure. The body is placed on a frictionless horizontal surface.

If the body be displaced towards right through a small distance Y, a restoring force will be developed.

Comparing this equation with standard differential equation \(\frac{d^{2} x}{d t^{2}}\) + ω²x = 0
We get ω² = k/m

Question 7.
A particle execute simple harmonic motion according to the equation, x = 5sin \(\left(\frac{2 \pi}{3}\right) t\)

1. What is the period of the oscillation?
2. Write an expression for velocity and acceleration of the above particle.

Answer:
1. When x = 5sin\(\left(\frac{2 \pi}{3}\right) t\) with standard equation.
x = asin ω t we get

2. Velocity V = \(\frac{d}{d t}\) 5sin\(\left(\frac{2 \pi}{3}\right) t\)

Question 8.

1. Arrive the differential equation of SHM.
2. What do you mean by seconds pendulum.

Answer:
1. The force acting simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion.

Comparing this equation with standard differential equation \(\frac{d^{2} x}{d t^{2}}\) + ω²x = 0,
We get ω² = k/m.

2. A pendulum having period 2 sec is called seconds pendulum.

Question 9.
SHM is a type of motion in which both speed and acceleration change continuously.

1. Which of the following condition is sufficient for SHM?
   • a = ky,
   • a = ky
   • a = ky²
2. Draw a graph of SHM between
   • displacement-time
   • speed – time
   • acceleration -time

Answer:
1. a = ky

2. Variation of displacement Y with time t

Variation of velocity Y with time t

Variation of acceleration with time

Question 10.

1. Is oscillation of a mass suspended by a spring is simple harmonic?
2. Write period of oscillation of the spring?
3. There are two springs, one delicate and another stout. For which spring, the frequency of oscillation will be more?
4. Two un equal springs of same material are loaded with same load, which one will have large value of time period?

Answer:
1. Yes

2. T = \(2 \pi \sqrt{\frac{m}{k}}\)

3. When a stout spring loaded with mass ‘m’, the extension (l) produced is large.
∴ T is large, because T = \(2 \pi \sqrt{\frac{m}{k}}\),
T is small, i.e, frequency is large. Stout spring oscillate with larger frequency.

4. When a longer spring is locked with weight mg, the extension T is more
∴ T is large, because T = \(2 \pi \sqrt{\frac{l}{g}}\)
So longer spring will have a large value of period.

Question 11.
A simple pendulum consists of a metallic bob suspended from a long straight thread whose one
end is fixed to a rigid support.

1. What is the time period of second’s pendulum?
2. Derive an expression for period of simple pendulum.

Answer:
1. 2 sec

2.

Consider a mass m suspended from one end of a string of length L fixed at the other end as shown in figure. Suppose P is the instantaneous position of the pendulum. At this instant its string makes an angle θ with the vertical.

The forces acting on the bob are (1) weight of bob F_g(mg) acting vertically downward. (2) Tension T in the string.

The gravitational force F_g can be divided into a radial component F_gCosθ and tangential component F_gSinθ. The tangential component F_gSinθ produces a restoring torque.
Restoring torque τ = – F_g sinθ . L
τ = – m_g sinθ . L _____(1)
-ve sign shown that the torque and angular displacement θ are oppositely directed. For rotational motion of bob,
τ = Iα ______(2)
Where I is moment of inertia about the point of suspension and α is angular acceleration. From eq (1) and eq (2).
Iα = – mgsinθ . L
If we assume that the displacement θ is small, sinθ ≈ θ
∴ Iα = -mgθ . L
Iα + mgθ L = 0

Comparirig eq (3) with standard differential equation


∴ period of simple pendulum,

for simple pendulum I = mL²
Substituting I = mL² we get

The above equation gives that T is indepent of mass.

Plus One Physics Oscillations Five Mark Questions and Answers

Question 1.

1. Which of the following condition is sufficient for the simple harmonic motion?
   • a = ky
   • a = ky²
   • a = -ky
   • a = -ky²
     Where ‘a’ – acceleration, y – displacement
2. Prove that simple harmonic motion is the projection of uniform circular motion on any diameter of the circle.
3. Represent graphically the variations of potential energy, kinetic energy and total energy as a function of position ‘x’ for a linear harmonic oscillator. Explain the graph.

Answer:
1. a = -ky

2.

Consider a particle moving along the circumference of a circle of radius ‘a’ and centre O, with uniform angular velocity w. AB and CD are two mutually perpendicular diameters along X and Y axis. At time t = 0.

let the particle be at P₀ so that ∠P₀OB = Φ. After time ‘t’ second, let the particle reach P so that ∠POP₀ = ω t. N is the foot of the perpendicular drawn from P on the diameter CD.

Similarly M is the foot of the perpendicular drawn from P to the diameter AB. When the particle moves along the circumference of the circle, the foot of the perpendicular executes to and fro motion along the diameter CD or AB with O as the mean position.

From the right angle triangle O MP, we get
Cos (ωt + Φ) = \(\frac{O M}{O P}\)
∴ OM = OPcos(ωt + Φ)
X= a cos (ωt + Φ) _______(1)
Similarly, we get
Sin (ωt + Φ) = \(\frac{y}{a}\) (or)
Y = a sin (ωt + Φ) _______(2)
Equation (1) and (2) are similar to equations of S.H.M. The equation(1) and (2) shows that the projection of uniform circular motion on any diameter is S.H.M.

3. KE = PE
\(\frac{1}{2}\)mω²(a² – x²) = \(\frac{1}{2}\) = mω²x²
Solving we get, x = \(\frac{a}{\sqrt{2}}\)
where a is the amplitude of oscillation.

Question 2.
The spring has a scale that reads from zero to 30 kg. The length of the scale is 30cm.

1. Calculate the force constant K
2. If the period of oscillation is 1 sec. Calculate mass of the body attached to the spring.
3. If the spring is cut into two halves, What is the force constant of each half?

Answer:
1. 30 cm = 30kg
1 cm = \(\frac{30}{30}\) = 1kg
∴ Spring constant K = 1 kg/cm = 10N/cm =1000N/m.

2.

3. If spring of spring constant K is cut in to half, spring constant of each half became, 2K

Plus One Physics Oscillations NCERT Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Answer:
(b) A freely suspended bar magnet displaced from its N-S direction and released and
(c) A hydrogen molecule rotating about its centre of mass

Question 2.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = 200x²
(c) a = 10x
(d) a = 100x²
Answer:
Only (c) ie. a = -10x represent SHM. This is because acceleration is proportional and opposite to displacement (x).

Question 3.
A spring having spring constant 1200Nm^-1 is mounted on a horizontal table as shown in figure. A mass of 3 kg is attached to the free end of the spring. The mass is the pulled sideways to a distance of 2.0cm and released.

Determine

1. the frequency of oscillations,
2. maximum acceleration of the mass and
3. the maximum speed of the mass.

Answer:
k = 1200Nm^-1, m = 3kg, a = 2.0cm = 2 × 10^-2m
1.

2. Maximum acceleration = ω² a = \(\frac{k}{m}\)a

3. Maximum speed = aω = a\(\sqrt{\frac{K}{m}}\)

Question 4.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0m. If the position moves with simple harmonic motion with an angular frequency of 200 rev/min, what is its maximum speed?
Answer:
a = \(\frac{1}{2}\)m, ω = 200 rev/min
U_max = aω = \(\frac{1}{2}\) × 200m/min = 100 m/min.

Question 5.
A circular disc of mass 10kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = -αθ, where J is the restoring couple and θ the angle of twist).
Answer:

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Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Plus One Chemistry Environmental Chemistry One Mark Questions and Answers

Question 1.
The greatest affinity for haemoglobin is shown by
a) NO
b) CO
c) O₂
d) CO₂
Answer:
b) CO

Question 2.
London smog is ___________ in nature.
Answer:
reducing

Question 3.
Addition of phosphate fertilizers into water leads to
a) Increased growth of decomposers
b) Reduced algal growth
c) Increased algal growth
d) Nutrient enrichment
Answer:
d) Nutrient enrichment

Question 4.
Which of the following is a viable particulate?
a) Algae
b) Smoke
c) Mist
d) Fumes
Answer:
a) Algae

Question 5.
Which of the following is not a greenhouse gas?
a) CO₂
b) CH₄
c) O₂
d) Water vapour
Answer:
c) O₂

Question 6.
Methyl isocyanate is prepared by the action of ___________ on methyl amine
Answer:
COCl₂

Question 7.
In a photo chemical smog the gas that causes eye irritation is ___________ .
a) CO₂
b) CH₄
c) PAN
d) Acrolein
e) both (c) and (d)
Answer:
PAN

Question 8.
Super sonic jet planes can contribute to zone depletion by introduction of the gas straight to the stratosphere
Answer:
NO

Question 9.
The difference between the amounts of dissolved oxygen in a sample of water saturated with oxygen and that after incubation for a period of 5 days is known as.
Answer:
BOD

Question 10.
Excess of nitrate ions in drinking water causes.
Answer:
Methenoglobinemia

Plus One Chemistry Environmental Chemistry Two Mark Questions and Answers

Question 1.
Write the equation for the combustion of ethane.
a) Complete combustion
b) Incomplete combustion
Answer:
a) 2C₂H₆(g) +7O₂ → 4CO₂(g) + 6H₂O(v)
b) 2C₂H₆(g)+5O₂ → 4CO(g) + 6H₂O(v)

Question 2.
Write the three common pollutants.
Answer:
1. Gases such as CO and SO₂
2. Compounds of metals like lead, mercury, zinc.
3. Radioactive substance

Question 3.
Three equations are given below. After studying it, write about Acid rain.
(1) CO₂(g) + H₂O(l) → 2H+ (aq) + CO₃^2-(aq)
(2) 2SO₂(g) + O₂(g) +2H₂O(l) → 2H₂SO₄(aq)
(3) 4NO₂(g) + O₂(g) + 2H₂O(l) → 2HNO₃ (aq)
Answer:
Oxides of Sulphur and Nitrogen, mist of HCI and Phosphoric acid, etc. present in polluted airdissolve in rain water making it more acidic. This is known as acidic rain. SO₂ and NO₂ present in polluted air are the major contributors of acid rain.
2SO₂(g) +O₂(g) + 2H₂O(l) → 2H₂SO₄(aq)
4NO₂(g) +O₂(g) + 2H₂O(l) → 2HNO₃(aq)

Question 4.
Marble of Taj Mahal reacts with acid rain. What is its result?
Answer:
The marble with which Taj Mahal is made is continuously eaten away by acid rain. It is due to the presence of chemical factories in Agra. The acids present in acid rain react with marble and making the marble dull and rough.
CaCO₃ + H₂SO₄ → CaSO₄ +H₂O + CO₂

Question 5.
Why usage of chlorofluorocarbons being discouraged? or Explain ozone layer depletion?
Answer:
The decomposition of CFC’s destroying ozone. CF₂Cl₂(g) + hv → Cl(h) + CF₂Cl(g)
The reactive Cl atom reacts with O₃ to form ClO radical. Cl(g) + O₃ → ClO(g) + O₂(g)
Thus each Cl atom produced, can destroy many O₃ molecules. This leads to ozone depletion.

Question 6.
What is meant by BOD?
Answer:
It is biochemical oxygen demand. It is the amount of dissolved oxygen required by micro-organisms to oxidise organic and inorganic matter present in polluted water.

Question 7.
1. How we can control air pollution?
2. How does the soil pollution occur?
Answer:
1. Air pollution can be controlled by the following ways.
• Exhaust to oxidize CO to CO₂.
• CO₂ level can be maintained by planting trees.
• Hydrogen gas is looked upon as pollution less future fuel.
• The large amount of nitrogen oxides emitted from power plant can be removed by scrubbing it with H₂SO₄.
2. It occurs due to
• Indiscriminate use of fertilizers, pesticides etc.
• Dumping of waste materials.
• Deforestation

Question 8.
What do you mean by greenhouse effect?
Answer:
Global warming is caused by the excess amount of CO₂ in the atmosphere. When CO₂ accumulate due to the decrease of trees it will increase the temperature of earth. This leads to melting of ice. So the sea level rises.

Question 9.
What do you mean by global warming?
Answer:
As more and more infrared radiations are trapped, the atmosphere becomes hotter and the global temperature rises up. This is known as global warming. There has been a marked increase in the levels of CO₂ in the atmosphere due to severe deforestation and burning of fossil fuels.

Question 10.
Give two examples in which green chemistry has been applied.
Answer:
1) Oxidative cracking process in the formation of ethylene is a significant step.
2) Fuel cells for cellular phones have been developed. This cell last for the full life time of the phone.

Question 11.
Oxygen play a key role in the troposphere while ozone in the stratosphere.
i) How is ozone formed in the atmosphere?
ii) What are the causes of depletion of ozone layer?
Answer:
i) By electric discharge of O₂ during lightning.
ii) UV radiations enter to the earth surface and it causes skin diseases.

Question 12.
No new industries are allowed in thickly populated cities by Government order. Why such an order is issued?
Answer:
The city will be destroyed due to acid rain caused by industrial pollution. The Govt, order was to protect the environment (orthe city).

Question 13.
Discuss the importance of dissolved oxygen in water.
Answer:
It helps living organisms in water. They inhale dissolved oxygen in water.

Question 14.
Define environmental chemistry.
Answer:
Environmental chemistry is defined as the branch of science which deals with the chemical processes occurring in the environment. It involves the study of origin, transport, reactions, effects and the fates of chemical species in the environment.

Question 15.
Explain tropospheric pollution.
Answer:
Tropospheric pollution occurs due to the presence of gaseous and the particulate pollutants.
• Gaseous air pollutants. These include mainly oxides of sulphur (SO₂, SO₃), oxides of nitrogen (NO, NO₂) and oxides of carbon (CO, CO₂) in addition , to hydrogen sulphide (H₂S), Hydrocarbons, ozone and other oxidants.
• Particulate pollutants. These include dust, mist, fumes, smoke, smog, etc.

Question 16.
Which gases are responsible for greenhouse effect? List some of them.
Answer:
The main gas responsible for greenhouse effect is CO₂. Other greenhouse gases are methane, nitrous oxide, water vapours, chlorofluorocarbons (CFC’s) and ozone.

Question 17.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.
Answer:
Excessive phytoplankton (organic pollutants such as leaves, grass, trash, etc) present in water is biodegradable. A large population of bacteria decomposes this organic matter in water. During this process they consume the oxygen dissolved in water. Water has already limited dissolved oxygen (≈10 ppm) which gets is further depleted. When the level of dissolved oxygen falls below 6 ppm, the fish cannot survive. Hence, they die and float dead in water.

Question 18.
How can domestic waste be used as mannure?
Answer:
Domestic waste comprises two types of materials, biodegradable such as leaves, rotten food, etc, and non-biodegradable such as plastics, glass, metal scrap, etc. The non-biodegradable waste is sent to industry for recycling. The biodegradable waste should be deposited in the land fills. With the passage of time, it is converted into compost manure.

Question 19.
For your agricultural field or garden, you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Answer:
The compost producing pit should be set up at a suitable place or in a tin to protect ourselves from bad odour and flies. It should be kept covered so that flies cannot make entry into it and the bad odour is minimized. The recyclable material like plastics, glass, newspapers, etc., should be sold to the vendor who further sells it to the dealer. The dealer further supplies it to the industry involved in recycling process.

Plus One Chemistry Environmental Chemistry Three Mark Questions and Answers

Question 1.
Fish grow in cold water as well as in warm water.
1. Do you agree? What is the reason?
2. pH of rain water is 5.6. Is it true? What is the reason?
Answer:
1. No. Fish grow in cold water. Warm water contains less dissolved O₂ than cold water,

2. Normally rain water contains dissolved CO₂ and hence it shows acidic pH of 5.6.
H₂O + CO₂ \(\rightleftarrows \) H₂CO₃ or 2H⁺ + CO₂
When pH of rain water became below 5.6, it becomes acid rain.

Question 2.
Write the mechanism of formation of photochemical smog.
Answer:
At high temp, the petrol and diesel engines, N₂ & O₂ combine to form NO which is emitted into atmosphere. NO is then oxidised in airto form NO₂ which absorbs sunlight and form NO and free O atom.

The O atoms being reactive and combine with O₂ to form O₃.
O₂ (g) + O(g) → O₃(g)
The O₃ react with NO formed by the photochemical decomposition of NO₂.
NO(g) + O₃(g) → NO₂(g) + O₂(g)
NO₂ and O₃ are good oxidising agents and they react with unburnt hydrocarbons in the polluted airto form substances such as acrolein and formaldehyde. These are the main substances of photochemical smog.

Question 3.
1. What are the major pollutants of water?
2. What is meant by eutrophication?
Answer:
1. Micro-organism present in domestic sewages, organic wastes, plant nutrients, toxic metals, sediments, pesticides and radioactive substances,

2. Addition of P to water as PO₄^3- ion encourages the formation of algae which reduces the dissolved oxygen content of water. This is called eutrophication.

Question 4.
1. Acid rain causes extensive damage to vegetation and aquatic life. )
i) What do you mean by acid rain?
ii) Name the chemicals responsible for acid rain.
2. List gases which are responsible for greenhouse effect.
Answer:
1. i) When the pH of the rain water drops below 5.6, it is called acid rain.
ii) Oxides of nitrogen and sulphur mist of hydrochloric acid and phosphoric acid etc.

2. Such as carbon dioxide, methane, ozone, chlorofluorocarbon compounds (CFC).

Question 5.
Statues and monuments in India such as Tajmahal are affected by acid rain. How?
Answer:
The statues of monuments in India are affected by acid rain. For example, the air in the vicinity of Taj Mahal contains very high levels of oxides of sulphur and nitrogen. This results in acid rain which reacts with marble of Taj Mahal causing pitting.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
CaCO₃ + 2HNO₃ → Ca(NO₃)2 +H₂O + CO₃.
As a result, the monument is being slowly eaten away and the marble is getting decolourised and lustreless. Thus, acid rain is considered as a threat to Taj Mahal.

Question 6.
1. Writethree major consequences of air pollution.
2. Write any two suitable methods to control air pollution.
Answer:
1. We cannot get pure oxygen for inspire.
Air pollution causes global warming.
It leads to acid rain.
It leads to diseases.

2. Plant trees.
Reduce the use of motor vehicles.
Do not burn plastics.

Question 7.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
CO combines with haemoglobin to forms a complex entity, carboxyhaemoglobin which is about 300 times more stable than oxy-haemoglobin. In blood, when the concentration of carboxyhaemoglobin reaches 3 -4%, the oxygen-carrying capacity of the blood is significantly reduced. In other words, the body becomes oxygen-starved. This results into headache, nervousness, cardiovascular disorder, weak eyesight etc.

Plus One Chemistry Environmental Chemistry Four Mark Questions and Answers

Question 1.
Classify the following pollutants:
(a) Carbon monoxide (CO)
(b) Detergents
(c) Plastic
(d) DDT
(e) Sewage
(f) Cigarette smoke
Answer:

Question 2.
As a result of stratospheric pollution, the ozone layer is destructed.
a) What is ozone layer?
b) How is it useful to us?
c) Write a harmful effect of ozone layer depletion.
Answer:
a) The layer of ozone seen in stratosphere is called ozone layer.
b) Ozone layer plays a significant role in protecting earth from UV rays.
c) Due to ozone layer depletion agricultural crops were found to give reduced yields. Small aquatic organisms which are very sensitive are destroyed due to the increase in the level of UV radiation.

Question 3.
a) What is smog?
b) What are the adverse effects of photochemical smog?
c) Write any two methods to control photochemical smog.
Answer:
a) Smog is a mixture of smoke and fog. This is the most common example of air pollution that occurs in many cities throughout the world. There are two types of smog:
1) Classical smog
2) Photochemical smog

b) Adverse effects of photochemical smog:
• Eye irritants
• Nose irritation
• Head ache
• Chest pain
• Dryness of throat
• Cough
• Difficulty in breathing

c) 1) Use catalytic converters in automobiles.
2) Plant certain plants (e.g. Pinus) which can metabolise nitrogen oxide.

Students can Download Chapter 13 Hydrocarbons Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Plus One Chemistry Hydrocarbons One Mark Questions and Answers

Question 1.
Which of the following cannot be prepared by Wurtz reaction?
a) CH₄
b) C₂H₆
c) C₃H₈
d) C₄H₈
Answer:
a) CH₄

Question 2.
The cyclic polymerization of propane produces __________ .
Answer:
1, 3, 5-trimethylbenzene

Question 3.
The reaction

a) Hydration
b) Dehydration
c) Dehydrogenation
d) Dehalogenation
Answer:
b) Dehydration

Question 4.
Say TRUE or FALSE.
Calcium carbide on hydrolysis gives ethylene.
Answer:
False

Question 5.
3-Hexyne reacts with Na/liquid NH₃ to produce
a) cis-3-Hexene
b) trans-3-Hexene
c) 3-Hexylamine
d) mm2-Hexylamine
Answer:
b) trans-3-Hexene

Question 6.
Fill in the blanks after finding the correct relationship
CH₃ – O – CH₃ : Ether, CH₃-CH₂-OH: …………….
Answer:
Alcohol

Question 7.
Choose the correct answer from the brackets given below:
1) General formula of alkene (C_nH_2n, C_nH_4n-2)
2) The 2 different forms of elemental carbon (Bitumen, Diamond, Charcoal, Coke, Led, Graphite)
3) Find the odd man out. Give reason (C₂H₄, C₃H₆, C₄H₈, C₂H₆)
Answer:
1) C_nH_2n
2) Diamond, graphite
3) C₂H₆. It is an alkane. All others are alkenes.

Question 8.
The structural formula of a compound and the name given by a student to it is given below:

Answer:
The name given by the student is wrong. The correct name of the compound is 3-Ethyl -2-methyl nonane.

Question 9.
Which is not the isomer of CH₃-CH₂-O-CH₂-CH₃? (1)
a) CH₃-O-CH₂-CH₂-CH₃
b) CH₃-CH₂-CH₂-CH₂-OH
c) CH₃-CH₂-CO-CH₃
d) CH₃-CH₂(OH)-CH-CH₃
Answer:
c) CH₃CH₂COCH₃

Question 10.
Gammexane has the formula
Answer:
C₆H₆Cl₆

Question 11.
Bayer’s reagent is __________ .
Answer:
dil. alkaline KMnO₄

Question 12.
The electrophile attacking benzene during nitration is __________ .
Answer:
NO₂⁺

Question 13.
The compound that is least readily nitrated is __________ .
a) phenol
b) Toluene
c) Ethylbenzene
d) Benzoic acid
e) Xylene
Answer:
d) Benzoic acid

Question 14.
The hydrocarbon formed when Beryllium carbide is treated with water is __________ .
Answer:
Methane

Plus One Chemistry Hydrocarbons Two Mark Questions and Answers

Question 1.
The lUPAC names of 2 compounds with their structural formulae is given below. Make out the errors and correct them.
A. 2, 2-Dimethyl-3-hexyne

B. 1-Butyne
CH₃-CH₂-CH=CH₃
Answer:
A is wrong. The strucutral formula of 2,2-Dimethyl -3- hexyne is

Question 2.
Is there an organic compound named 2-Ethylpentane? Why? If no, write the correct answer.
Answer:
No. When an ethyl group comes with the second carbon atom, the longest chain will have 6 carbon atoms. Hence its correct name will be 3-Methyl hexane.

Question 3.
Write down 2 similarities and 2 difference of the hydrocarbons given below:

Answer:

Question 4.
An equation for combined Chemical reaction of acetylene is given below:

Find the products X’ and ‘Y’?
Answer:

Question 5.
In a Chemistry class, teacher asked students to write the geometrical form of dicarboxylic acid with the formula C₄H₄O₄. For this question, one student wrote

Both of them argued fortheir answers. Hearing this argument teacher told that both of them are right and she also explained the reason for it. Can you write the answer given by the teacher?
Answer:
Two geometrical isomers are possible for dicarboxylic acid. They are cis and transform. In the cis form, similar groups are present on the same side of the double bond whereas in transform, identical groups are present on different side of the double bond.

Question 6.
Match the following:

Answer:

Question 7.
Addition of HBr to propene yields 2-bromopropane, what happens if benzoyl peroxide is added to the above reaction.
Answer:
When propene is allowed to react with HBr in the presence of peroxide, 2 bromo propane is obtained as the minor product and this is called peroxide effect. Anti markonikov’s rule.

Plus One Chemistry Hydrocarbons Three Mark Questions and Answers

Question 1.
Reactions
CH₃-C ≡ C-CH₃ + H₂
Reaction: 2
X + HCl → Y
a) What type of reaction is reaction 1 ?
b) Find X and Y.
c) Which are the different products obtained from the reaction of X with oxygen?
Answer:
a) Addition reaction

Question 2.

Here are some functional groups. You are supposed to form 3 structures of hydrocarbons using 3 different functional groups and try to find their names.
Answer:

Question 3
CH₃ – CH₂ – CH₂ – CH₂ – OH, CH₃ – CH₂ – CH – CH₃
i)For which isomerism can the example above be considered?
ii) Define it.
Answer:
i)Position isomerism
ii) Position isomerism arises as a result of the difference in the position of double bond, triple bond or functional group.

Question 4.
Explain the following with necessary chemicals equations.
i) Wurtz reaction
ii) Kolbe’s reaction
iii) Ozonolysisof alkenes
Answer:
i) Wurtz reaction – when alkyl halide is allowed to react with metallic sodium in presence of dry ether an alkane is obtained.

ii) Kolbes reaction – When a solution of sodium acetate is electrolized, ethane is obtained.

iii) Ozonolysis ofalkene: When an alkene is allowed to react with ozone, an ozonide is obtained. This on hydrolysis gives Aldehyde or ketone. The whole process is called ozonolysis.

Question 5.
b) Complete the reaction:

Answer:

Question 6
1. Explain the reaction between sodium metal and bromoethane in dry ether. (3)
2. Draw Sawhorse and Newman’s projections of the different conformers of ethane.
Answer:
1. 2CH₃CH₂Br + 2Na → CH₃-CH₂-CH₂-CH₃ + 2NaBr
2.

Question 7.
Analyse the given reactions, give the major products. Justify your answer.
a) HBris added to 1-Butene two products are obtained.
b) Action of excess chlorine with benzene in dark.
c) Addition of chlorine to benzene in uv light.
Answer:

c) When Benzene is allowed to react with chlorine in presence of sunlight. Benzene hexa chloride (BHC) is obtained.

Question 8.
Predict the product in the following reactions and identify the rules:-

Answer:
a) When propene is allowed to react with HBr in the presence of peroxide, 2-Bromo propane is obtained as the minor product. (Peroxide effect, Kharasch effect or Anti Markownikoffs rule of addition).

Question 9.
Write any two necessary condition for a compound to be aromatic. Convert Acetylene to benzene.
Answer:
Cyclic, Planar and should contain (4n+2)π electrons.

Plus One Chemistry Hydrocarbons Four Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2.
Fill in the blanks:

Answer:

Question 3.
Given below are the structures of some hydrocarbons. Pick out the correct IUPAC names for them from the box.

Answer:
1) 4 – Ethyl -2, 3-dimethyl heptane
2) 3, 4-Dimethyl hex – 3-ene
3) 4, 5, 5- triethyl 3 methyl 2 heptyne
4) 3ethyl-2, 5, 5, trimethylheptane

Question 4.
a) I am an unsaturated hydrocarbon.
b) My wordroot is pent.
c) My suffix is ene. The double bond lies between 2nd and 3rd carbon atoms.
d) l have a branch of methyl group on my second carbon atom. Who am I?
Answer:

Question 5.
From the given table find out the isomer pairs and which type of isomerism they have?

Answer:
Isomer pairs
a – ii – functional group isomerism
b – i – chain isomerism
c-iv-position isomerism
d – iii – metamerism Question 6

Qn 6.
Given below is the structural formula of a compound written by a student.

i) Draw all the possible conformers of the compound?
ii) Arrange them in the order of stability?
Answer:
i)

ii) Eclipsed

Question 7.
A cross word puzzle.
Down
1. Two or more compounds having the same
molecular formula but different physical or chemical properties are known as isomers and the phenomenon is known as
2. Hydrocarbons having the general formula C_nH_2n.
3. IUPAC name of C₆H₅CH₃.
4. ………….. is added to the word root to show whether the hydrocarbon is saturated or unsaturated.
5. Hydrocarbons contain carbon-carbon triple bond.

Cross
6. From where does the inorganic compounds mainly originate from?
7. Name the alkene with the moelcular formula C₁₀H₂₀
8. Prefix of the functional group carboxylic acid.
9. Compounds with two rings.
10. Compounds having same molecular formula but different arrangements of carbon atoms on either side of the same functional group are called ……..

Answer:

Question 8.
1. Name the product obtained when HBr is added to propene. Why?
2. Acetylene is more acidic than ethylene or ethane. Why?
Answer:
1. When propene is treated with HBr, both 2-bromopnopane and 1-bromopropane are formed as products. The major product is2-Bromo-propane. This is due to the rule known as Markonikoffs rule of addition. It states that when a hydrogen halide is added to an unsymmetrical alkene, the halogen atom will goes to the doubly bonded carbon containing lesser number of hydrogen atoms.

2. In Acetylene, the hybridisation of carbon is sp and hence 50% s-character is present.

Question 9.
a) How will you prepare butane in the laboratory using ethyl bromide (CH₃CH₂Br) as one of the raw materials. Write relevant equation.

Identify the product ‘X’. Statethe law that explains the formation of X.
Answer:

The law of Anti-Markownikoff’s rule of addition explains the formation of 1 -Bromopropane.

Question 10.
1. Addition of HBrto propene yields 2-Bromopropane, while in the presence of Benzoyl peroxide. The same reaction yields 1-Bromopropane. Give reason. Justify your answer.
2. Three compounds are given. Benzene, m- dinitrobenzene and toluene. Identify the compound which will undergo nitration most easily and why?
Answer:
1. When propene is allowed to react with HBr in the presence of “peroxide” 2-Bromopropane is obtained as the minor product. (Peroxide effect or Kharach effect or Anti-Markownikoff’s role of addition) Peroxide effect is applicable only in the case of HBr.

2. Toluene. This is because the -CH₃ group, being an activating group activates the benzene ring towards electrophilic substitution in toluene.

Question 11.
a) Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene:
i) In the absence of peroxide.
ii) In the presence of peroxide.
b) Howwill you convert:
i) Benzene to toluene
ii) Benzene to nitrobenzene
Answer:

Question 12.
The equations for two chemical reactions are given below:
i) CH ≡ CH + HCl → A → B
ii) OH₄ + O₂ → C + D
Which are the products A, B, C, and D?
Answer:

Question 13.
The 2 conformations shown below belongs to the compound cyclohexane. Infinite number of conformations are possible for cyclo hexane. But these 2 conformations given below has a peculiarity. Try to find it. Also, define the terms conformers and the phenomenon conformation?

Answer:
Infinite number of conformations are possible for cyclohexane. Out of it chair conformation is the most stable one and the boat conformation is the least stable form.

The different arrangement of a compound which arises as a result of rotation about carbon single bond are called conformers and the phenomenon is called conformation.

Plus One Chemistry Hydrocarbons NCERT Questions and Answers

Question 1.
How do you account for formation of ethane during chlorination of methane? (3)
Answer:
Chlorination of methane takes place through a free radical chain mechanism as given below:

From the above mechanism, it is evident that during chain propagation step, CH3free radicals areproduced. In the chain termination step, the two free CH₃ radicals may combine together to form ethane (CH₃-CH₃) molecule.

Question 2.
For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: (4)
a) C₄H₈ (one double bond)
b) C₃H₈ (one triple bond)
Answer:
a) Isomers of C₄H₈ having one double bond are:

Question 3.
What are the necessary conditions for any compound to show aromaticaly? (3)
Answer:
The conditions for a compound to show aromaticity are:
i) The molecule must be cyclic
ii) It must have a conjugated system of (4n+2) π- electrones
iii) The molecule must be planar so that delocalization of π-electrones can take place.

Question 4
Explain why the following system are not aromatic?

Answer:

It dose not contain all the π- electrons in the ring. Therefore, it is not an aromatic compound.

It contains only four electrons, therefore, the system is not aromatic because it dose not contain (4n + 2) π- electrones.

Cyclootatetraene is a conjugated system having 8 π-electrons.
Therefore, the molecule is not contain (4n+2) π-electrons.

Question 5.
In the alkane, H₂CCH₂C(CH₃)₂ CH₂CH(CH₃)₂, identify 1°, 2°, 3° carbon atoms and give the total number of atoms bonded to each one of these.
Answer:

1° Carban atoms = 5
Hydrogen atoms attached to 1° carbon atoms = 15 2° Carbon atoms = 2
Hydrogen atoms attached to 2° carbon atoms = 4 3° Carbon atom = 1
Hydrogen atoms attached to 3° carbon atom = 1

Question 6.
What effect does branching of an alkane chain has on its melting point?
Answer:
As the branching increases melting point increases.

Question 7.
Why does benzene undergo electrophilic subsitution easily and nucleophilic substitutions with difficulty? (2)
Answer:
Benzene molecule has two n cloud rings, one above and the other below the plane of atoms. Therefore, it is likely to be attached by electrophiles which subsequently brings about substitution.

The nucleophiles would be repelled by the π-electron rings and hence benzene reacts with nucleophiles with difficulty.

Students can Download Chapter 1 The Discipline of Computing Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 1 The Discipline of Computing

Summary
Computing milestones and machine evolution:
People used pebbles and stones for counting earlier days. They draw lines to record information. eg: 1 line for one, 2 lines for two, 3 lines for three, etc. In this number system the value will not change if the lines are interchange. This type of number system is called non positional number system.

Counting and the evolution of the positional number system:
In positional number system, each and every number has a weight. Earlier sticks are used to count items such as animals or objects. Around 3000 BC the Egyptians use number systems with radix 10(base-the number of symbols or digits used in the number system) and they write from right to left.

Later Sumerian/Babylonian use number system with largest base 60 and were written from left to right. They use space for zero instead of a symbol, 0. In 2500 BC, the Chinese use simple and efficient number system with base 10 very close to number system used in nowadays.

In 500 BC, the Greek number system known as Ionian, it is a decimal number system and used no symbols for zero. The Roman numerals consists of 7 letters such as l, V, X, L, C, D, M. The Mayans used number system with base 20 because of the sum of the number of fingers and toes is 10 + 10 = 20.

It is called vigesimal positional number system. The numerals are made up of three symbols; zero (shell shape, with the plastron uppermost), one (a dot) and five (a bar or a horizontal line). To represent 1 they used one dot, two dots for 2, and so on

The Hindu – Arabic number system had a symbol(0)for zero originated in India 1500 years ago. Consider the table to compare the number system

Evolution of the computing machine:
(a) Abacus:
In 3000 BC Mesopotamians introduced this and it means calculating board or frame. It is considered as the first computer for basic arithmetical calculations and consists of beads on movable rods divided into two parts. The Chinese improved the Abacus with seven beads on each wire. Different Abacus are given below.

(b) Napier’s bones:
A Mathematician John Napier introduced this in AD 1617.

(c) Pascaline:
A French mathematician Blaise Pascal developed this machine that can perform arithmetical operations.

(d) Leibniz’s calculator:
In 1673, a German mathematician and Philosopher Gottfried Wilhelm Von Leibniz introduced this calculating machine.

(e) Jacquard’s loom:
In 1801, Joseph Marie Jacquard invented a mechanical loom that simplifies the process of manufacturing textiles with complex pattern. A stored program in punched cards was used to control the machine with the help of human labour. This punched card concept was adopted by Charles Babbage to control his Analytical engine and later by Hollerith.

(f) Difference engine:
The intervention of human beings was eliminated by Charles Babbage in calculations by using Difference engine in 1822. It could perform arithmetic operations and print results automatically

(g) Analytical engine:
In 1833. Charles Babbage introduced this. Charles Babbage is considered as the “Father of computer It is considered as the predecessor of today’s computer. This engine was controlled by programs stored in punched cards. These programs were written by Babbage’s assistant, Augusta Ada King, who was considered as the first programmer in the World.

(h) Hollerith’s machine:
In 1887, Herman Hollerith an American made first electromechanical punched cards with instructions for input and output. The.card contained holes in a particular pattern with special meaning. The Us Census Bureau had large amount of data to tabulate, that will take nearly 10 years.

By this machine this work was completed in one year. In 1896, Hollerith started a company Tabulating Machine Corporation. Now it is called International Business Machines(IBM).

(i) Mark-1:
In 1944 Howard Aiken manufactured automatic electromechanical computer in collaboration with engineers at IBM that handled 23 decimal place numbers and can perform addition, subtraction, multiplication and subtraction.

Generations of computers:
There are five generations of computers from 16^th century to till date.

First generation computers (1940 – 1956):
Vacuum tubes were used in first generation computers. The input was based on punched cards and paper tapes and output was displayed on printouts. The Electronic Numerical Integrator and Calculator(ENIAC) belongs to first generation was the first general purpose programmable electronic computer built by J. Presper Eckert and John V. Mauchly.

It was 30-50 feet long, weight 30 tons, 18,000 vacuum tubes, 70,000 registers, 10,000 capacitors and required 1,50,000 watts of electricity. It requires Air Conditioner. They later developed the first commercially successful computer, the Universal Automatic Computer(UNIVAC) in 1952. Von Neumann architecture

The mathematician John Von Neumann designed a computer structure that structure is in use nowadays. Von Neumann structure consists of a central processing unit(CPU), Memory unit, Input and Output unit. The CPU consists of arithmetic logical unit(ALU) and control unit(CU).

The instructions are stored in the memory and follows the “Stored Program Concept”. Colossus is the secret code breaking computer developed by a British engineer Tommy Flowers in 1943 to decode German messages.

Second generation computers (1956 -1963):
Transistors, instead of Vacuum tubes, were used in 2^nd generation computers hence size became smaller, less expensive, less electricity consumption and heat emission and more powerful and faster.

A team contained John Bardeen, Walter Brattain and William Shockley developed this computer at Bell Laboratories. In this generation onwards the concept of programming language was developed and used magnetic core (primary) memory and magnetic disk(secondary) memory.

These computers used high level languages(high level language means English like statements are used)like FORTRAN (Formula translation) and COBOL(Common Business Oriented Language). The popular computers were IBM 1401 and 1620.

Third generation computers (1964 – 1971):
Integrated Circuits(IC’s) were used. IC’s or silicon chips were developed by Jack Kilby, an engineer in Texas Instruments. It reduced the size again and increased the speed and efficiency. The high level language BASIC(Beginners All purpose Symbolic Instruction Code) was developed during this period.

The popular computers were IBM 360 and 370. Due to its simplicity and cheapness more people were used. The number of transistors on IC’s doubles approximately every two years. This law is called Moore’s Law, it is named after Gordon E Moore. It is an observation and not a physical or natural law.

Fourth generation computers (1971 onwards):
Microprocessors are used hence computers are called microcomputers. Microprocessor is a single chip which contains Large Scale of IC’s(LSI) like transistors, capacitors, resistors,etc due to this a CPU can place on a single chip. Later LSI were replaced by Very Large Scale Integrated Circuits(VLSI). The popular computers are IBM PC and Apple II.

Fifth generation computers (future):
Fifth generation computers are based on Artificial Intelligence(AI). Al is the ability to act as human intelligence like speech recognition, face recognition, robotic vision and movement etc. The most common Al programming language are LISP and Prolog.

Evolution of computing:
Computing machines are used for processing or calculating data, storing and displaying information. In 1940’s computer were used only for single tasks like a calculator. But nowadays computer is capable of doing multiple tasks at a time.

The “Stored Program Concept” is the revolutionary innovation by John Von Neumann helped storing data and information in memory. A program is a collection of instructions for executing a specific job or task.

Augusta Ada Lowelace: She was the Countess of Lowelace and she was also a mathematician and writer. She is considered as the first lady computer programmer.

Programming languages:
The instructions to the computer are written in different languages. They are Low Level Language(Machine language), Assembly Language(Middle level language) and High Level Language(HLL).

In Machine Language 0’s and 1 ’s are used to write program. It is very difficult but this is the only language which is understood by the computer. In assembly language mnemonics (codes) are used to write programs

Electronic Delay Storage Automatic Calculator(EDSAC) built during 1949 was the first to use assembly language. In HLL English like statements are used to write programs. A-0 programming language developed by Dr. Grace Hopper, in 1952, for UNIVAC-I is the first HLL.

A team lead by John Backus developed FORTRAN @IBM for IBM 704 computer and ‘Lisp’ developed by Tim Hart and Mike Levin at Massachusetts Institute of Technology. The other HLLs are C, C++, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

Usually programmers prefer HLL to write programs because of its simplicity. But computer understands only machine language. So there is a translation needed. The program which perform this job are language processors.

Algorithm and computer programs:
The step-by-step procedure to solve a problem is known as algorithm. It comes from the name of a famous Arab mathematician Abu Jafer Mohammed Ibn Musaa Al-Khowarizmi, The last part of his name Al-Khowarizmi was corrected to algorithm.

Theory of computing:
It deals with how efficiently problems can be solved by algorithm and computation. The study of the effectiveness of computation is based upon a mathematical abstraction of computers is called a model of computation, the most commonly used model is Turing Machine named after the famous computer scientist Alan Turing.

1. Contribution of Alan Turing:
He was a British mathematician, logician, cryptographer and computer scientist. He introduced the concept of algorithm and computing with the help of his invention Turing Machine.

He asked the question Can machines think’ led the foundation for the studies related to the computing machinery and intelligence. Because of these contributions he is considered as the Father of Modern Computer Science as well as Artificial Intelligence.

2. Turing Machine:
In 1936 Alan Turing introduced a machine, called Turing Machine. A Turing machine is a hypothetical device that manipulates symbols on a strip of tape according to a table of rules. This tape acts like the memory in a computer. The tape contains cells which starts with blank and may contain 0 or 1.

So it is called a 3 Symbol Turing Machine. The machine can read and write, one cell at a time, using a tape head and move the tape left or right by one cell so that the machine can read and edit the symbol in the neighbouring cells. The action of a Turing machine is determined by

1. the current state of the machine
2. the symbol in the cell currently being scanned by the head and
3. a table of transition rules, which acts as the program.

3. Turing Test:
The Turing test is a test of a machine’s ability to exhibit intelligent behaviour equivalent to, or indistinguishable from, that of a human. The test involves a human judge engages in natural language conversations with a human and a machine designed to generate performance indistinguishable from that of a human being.

All participants are separated from one another. If the judge cannot reliably tell the machine from the human, the machine is said to have passed the test. The test does not check the ability to give the correct answer to questions; it checks how closely the answer resembles typical human answers. Turing predicted that by 2000 computer would pass the test.

Students can Download Chapter 13 Kinetic Theory Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 13 Kinetic Theory

Plus One Physics Kinetic Theory One Mark Questions and Answers

Question 1.
The value of \(\frac{P V}{T}\) for one mole of an ideal gas is nearly equal to
(a) 2 Jmol^-1K^-1
(b) 8.3 Jmol^-1K^-1
(c) 4.2 Jmol^-1K^-1
(d) 2 cal mol^-1K^-1
Answer:
(d) 2 cal mol^-1K^-1
The value of \(\frac{P V}{T}\) for one mole of an ideal gas = gas constant = 2 cal mol^-1K^-1.

Question 2.
Mean free path of a gas molecule is
(a) inversely proportional to number of molecules per unit volume
(b) inversely proportional to diameter of the molecule
(c) directly proportional to the square root of the absolute temperature
(d) directly proportional to the molecular mas
Answer:
(a) inversely proportional to number of molecules per unit volume.

Question 3.
If for a gas \(\frac{R}{C_{v}}\) = 0.67, this gas is made up of molecules which are
(a) monoatomic
(b) diatomic
(c) Polyatomic
(d) mixture of diatomic and polyatomic molecules
Answer:
(a) monoatomic
For a gas, we know \(\frac{R}{C_{v}}\) = γ – 1
or, 0.67 = γ – 1, or γ = 1.67
Hence the gas is monoatomic.

Question 4.
According to kinetic theory of gases, molecules of a gas behave like
(a) inelastic spheres
(b) perfectly elastic rigid spheres
(c) perfectly elastic non-rigid spheres
(d) inelastic non-rigid spheres
Answer:
(b) According to kinetic theory of gases, gas molecules behave as a perfectly elastic rigid spheres.

Question 5.
Which one of the following is not an assumption of kinetic theory of gases?
(a) The volume occupied by the molecules of the gas is negligible.
(b) The force of attraction between the molecules is negligible.
(c) the collision between the molecules are elastic.
(d) All molecules have same speed.
Answer:
(d) Molecules of an ideal gas moves randomly with different speeds.

Question 6.
What is the shape of graph between volume and temperature, if pressure is kept constant?
Answer:
PV = nRT. Hence graph will be straight line.

Question 7.
What is the shape of graph between pressure p and I/V for a perfect gas at constant temperature?
Answer:
Straight line

Question 8.
Identify the minimum possible temperature at which all molecular motion ceases.
Answer:
Absolute temperature (OK or – 273.15°C).

Question 9.
What is the formula for average translational kinetic energy of a gas molecule?
Answer:
3/2 K_BT

Plus One Physics Kinetic Theory Two Mark Questions and Answers

Question 1.
Mention the conditions under which the real gases obey ideal gas equation.
Answer:
Low pressure and high temperature.

Question 2.
Why the temperature rises when gas is suddenly compressed?
Answer:
The work done on gas during compression increases the kinetic energy of molecules and hence temperature of gas rises.

Question 3.
Why evaporation causes cooking?
Answer:
During evaporation, fast moving molecules escape from liquid. Hence average kinetic energy of molecules left behind is decreaesd. This will reduce temperature & causes cooling.

Question 4.
When automobile travels long distance air pressure in tyres increases slightly. Why?
Answer:
As automobile moves, work is being done against force of friction. This work is converted in to heat and it increases the temperature. As P a T, increase in temperature will increase pressure.

Question 5.
PV = µ RT is the ideal gas equation. Real gas obeys ideal behaviour at high temperature and at low pressure.

1. Give an example for ideal gas
2. Why real gases obey ideal gas equation at high temperature and at low pressure.

Answer:

1. Hydrogen
2. The interaction between molecules can be neglected at high T and at low temperature.

Question 6.
A vessel of volume V contains a gas of µ moles at a temperature T.

1. What is the ideal gas equation if gas is considered to be ideal one?
2. The variation of pressure P with number of moles per unit volume in a vessel is shown in the graph. Analyse the graph and choose the correct one and justify your answer.

(i) The temperature inside the vessel decreases.
(ii) The temperature inside the vessel increases.
Answer:
1. PV = µRT

2. P = \(\frac{1}{3}\)nmc^-2
In this case, when n increases, (mc^-2) decreases to maintain P as constant. Temperature is directly proportional to mc^-2. Hence we can say that temperature inside the vessel decreases.

Plus One Physics Kinetic Theory Three Mark Questions and Answers

Question 1.
1 mole of ideal gas is taken in vessel.

1. State the following statements as true or false.
   • In gas equation R is constant.
   • All real gas obeys gas equation at all temperature and pressures.
2. Draw the variation of R with pressure for the above ideal gas.
3. Draw the variation of R with volume for this ideal gas.

Answer:
1. Following statements as true or false:

• True
• False

2.

3.

Question 2.

1. Air pressure in a car tyre increase during driving. Why?
2. Air is filled in a vessel at 60°C. To what temperature should it be heated in order that 1/3^rd of air may escape out of the vessel? (Expansion of air may be neglected).

Answer:
1. During driving the temperature of air inside the tyre increases due to motion.

2. T₁ = 60 + 273 = 333K
V₁ = V; T₂ = ? V₂ = V+ V/3
V₂ = \(\frac{4}{3}\)V

Question 3.
Find the degrees of freedom of the following.

1. A body is confined to move in a straight line
2. A body moves in a plane
3. A body moves in a space

Answer:

1. 1
2. 2
3. 3

Plus One Physics Kinetic Theory Four Mark Questions and Answers

Question 1.
An enclosed vessel contains many number of molecules moving in random direction.

1. Explain the term pressure in terms molecular concept.
2. Derive an expression for the pressure exerted by the gas molecules by assuming postulates of kinetic theory of gases.

Answer:
1. Pressure P = \(\frac{2}{3} n \overline{K E}\)
Where ‘n’ is the number of gas molecules per unit volume. \(\overline{\mathrm{KE}}\) is the average kinetic energy of a gas molecules moving in random direction.

2.

Consider molecules of gas in a container. The molecules are moving in random directions with a velocity V. This is the velocity of a molecule in any direction.

The velocity V can be resolved along x, y and z directions as V_x, V_y, and V_z respectively. If we assume a molecule hits the area A of container with velocity V_x and rebounds back with -V_x.

The change in momentum imparted to the area A by the molecule = 2mV_x. The molecules covers a distance V_xt along the x-direction in a time t. All the molecules within the volume AV_xt will collide with the area in a time t.

If ‘n’ is the number of molecules per unit volume, the total number of molecules hitting the area A, N = AV_xt n.
But on an average, only half of those molecules will be hitting the area, and the remaining molecules will be moving away from the area. Hence the momentum imported to the area in a time t.

Q = 2mv_x × \(\frac{1}{2}\) AV_xt n.
= nmV_x2 At
The rate of change of momentum,
\(\frac{Q}{t}\) = nmV_x² A
But rate of change of momentum is called force, ie. force F = nmV_x2A
∴ pressure P =nmV_x 2 (P = \(\frac{F}{A}\))
Different molecules move with different velocities. Therefore, the average value V2_x is to be taken. If \(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) isthe average value then the pressure.
p = nm\(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) ………(1)
\(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) is known as the mean square velocity.
Since the gas is isotropic (having the same properties in all directions), we can write

Hence the eq (1) can be written as

But nm = ρ, the density of gas
∴ P = \(\frac{F}{A}\) ρ\(\overline{\mathbf{V}}^{2}\).

Question 2.
1. Fill in the blanks

2. What happens to the value of ratio of specific heat capacity, if we consider all rotational degrees of freedom of a 1-mole diatomic molecule?
Answer:
1.

2. Total degrees of freedom = 3 (trans) + 3 (Rot) = 6
∴ C_V = 3R, C_P = 4R
Ratio of specific heat γ = \(\frac{4}{3}\)
Ratio of specific heat capacity decreases.

Plus One Physics Kinetic Theory NCERT Questions and Answers

Question 1.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer:
PV= µRT or V = \(\frac{\mu \mathrm{RT}}{\mathrm{P}}\)

= 22.4 × 10^-3 m³ = 22.4 litre.

Question 2.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m³ at a temperature of 27°C and 1-atmosphere pressure.
Answer:
V = 25.0m³, T = (27 + 273), K = 300 K, k = 1.38 × 10^-23JK^-1
PV= nRT = n(Nk)T = (nN)kT = N_tkT
Here N_t represents the total number of air molecules in the given gas.

Question 3.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³s^-1. Identify the gas.
Answer:
According to Graham’s law of diffusion of gases, the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass. If R₁ and R₂ be the rates of diffusion of two gases having molecular masses M₁ and M₂ respectively, then

Question 4.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.
Answer:
Consider one mole of oxygen gas at STP. It occupies 22.4 litre of volume which will contain 6.023 × 10²³. (ie. Avogadro number) molecules. Considering spherical shape of molecule, volume of oxygen molecule

Volume of 6.023 × 10²³ molecules
= \(\frac{4}{3}\) × 3.142(1.5)³ × 10^-30 × 6.02 × 10²³ m³
= 85.1 × 10^-7m³
= 8.51 × 10^-6m³ = 8.51 × 10^-3
litre Molecular volume of one mole of oxygen (∵ 1m³ = 10³ litre)
∴ Molecular volume of one mole of oxygen = 8.51 × 10^-3 litre
Actual volume occupied by one mole of oxygen at STP = 22.4 litre
Fraction of molecular volume to actual volume

Students can Download Chapter 7 Control Statements Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 7 Control Statements

Plus One Control Statements One Mark Questions and Answers

Question 1.
An if statement contains another if statement completely. Then it is known as _________.
Answer:
Nested if

Question 2.
From the following which is not optional with switch statement.
(a) break
(b) default
(c) case
(d) none of these
Answer:
(c) case.

Question 3.
To exit from a switch statement _______ is used.
(a) quit
(b) exit
(c) break
(d) none of these
Answer:
(c) break

Question 4.
From the following which statement is true for switch statement.
(a) switch is used to test the equality
(b) switch is used to test relational or logical expression
(c) switch can handle real numbers case data
(d) none of these
Answer:
(a) switch is used to test the equality

Question 5.
Sonet wants to execute a statement more than once. From the following which is exactly suitable.
(a) if
(b) loop
(c) switch
(d) if else if ladder
Answer:
(b) loop

Question 6.
Odd one out
(a) for
(b) if
(c) switch
(d) if-else if ladder
Answer:
(a) for. It is a loop the others are branching statement.

Question 7.
Odd one out
(a) for
(b) if
(c) while
(d) do-while
Answer:
(b) if. It is a branching statement and the others are loops.

Question 8.
From the following which loop does the three things, initialisation, checking and updation.
(a) while
(b) do-while
(c) for
(d) none of these
Answer:
(c) for

Question 9.
Predict the output Output
(a) 10
(b) 1 to 10
(c) 11
(d) none of these
Answer:
(c) 11.

Question 10.
From the following which is exit controlled loop
(a) for
(b) while
(c) do-while
(d) None of these
Answer:
(c) do-while

Question 11.
_____________ statement is used for unconditional jump from one location to another.
Answer:
goto.

Question 12.
Sunitha wants to skip one iteration. From the following which will help her?
(a)continue
(b) break
(c) for
(d) case
Answer:
(a) continue

Question 13.
To terminate a program, from the following which is used.
(a) break
(b) continue
(c) end()
(d) exit()
Answer:
(d) exit()

Question 14.
Which header file is needed to use exit() function in a program?
(a) iostream
(b) cstdlib
(c) math
(d) iomanip
Answer:
(b) cstdlib

Question 15.
In while loop, the loop variable should be updated?
(a) along with while statement
(b) after the while statement
(c) before the while statement
(d) inside the body of while
Answer:
(d) Inside the body of while

Question 16.
How many times the following loop will execute?
int S = 0, i = 0;
do
{
S + = i;
i++;
} while(i < 5);
Answer:
5 times

Question 17.
1. statement takes the program control out of the loop even though the test expression is true.

2. Consider the following code fragment. How many times will the character be printed on the screen?
for (i=0; i< 10; i =i+2);
cout <<“*”;
}
Answer:

1. break or goto
2. Only one time because of semicolon(;) in the end of the for(i=0;i<10;i=i+2);

Question 18.
Which selection statement tests the value of a variable or an expression against a list of integers or character constants? (SAY-2015) (1)
(a) For
(b) If
(c) Switch
(d) Conditional expression
Answer:
(c) switch

Question 19.
How many times the following loop will execute? (MARCH-2016) (1)
int m = 2
do
{
cout<<“Welcome”; m++ ;
} while (m>10);
Answer:
Only one time

Question 20.
________ statement takes the program control outside a loop even though the test expression is true. (SCERT SAMPLE – II) (1)
Answer:
break

Question 21.
Read the following C++ statement for (int n = 1; n<10; n+=2); cout<<n
Now, choose the correct output from the following options. (SCERT SAMPLE – II)(1)
(a) 1
(b) 13579
(c) 11
(d) 10
Answer:
(c) 11. This is because of for statement is end with; (semi colon). Here cout<<n; executes only once.

Question 22.
____________ search method is an example for ‘divide and conquer method’.
Answer:
goto.

Question 23.
1. Name the type or loop which can be used to ensure that the body of the loop will surely be executed at least once.

2. Consider the code given below and predict the output. (MARCH-2017) (1)
for (int i=1; i<=9;i=i+2)
{
if (i==5) continue;
cout<<i<< ” “;
}
Answer:

1. do while loop(Exit controlled loop)
2. 1 3 7 9. It bypasses one iteration of the loop when i = 5.

Plus One Control Statements Two Mark Questions and Answers

Question 1.
Your friend Arun asked you that is there any loop that will do three things, initialization, testing and updation. What is your answer. Explain?
Answer:
Yes. There is only one loop namely for loop that will do this three things. The other loops will do the checking only, initialisation must be do before the loop and updation must be inside the loop.
The syntax of for loop is given below For(initialisation; testing; updation)
{
Body of the for loop;
}

Question 2.
While writing a program Geo uses while loop but forgets to update the loop variable. What will happen?
Answer:
The loop variable inside the while loop must be updated otherwise the loop will not be terminated. The loop will be work infinitely.

Question 3.
Draw the flow chart of if statement.
Answer:

Question 4.
Draw the flow chart of if else statement
Answer:

Question 5.
Write a while loop that display numbers from 500 to 550.
Answer:
int i = 500
while (i<=550)
{
cout<<i;
i = i + 1;
}

Question 6.
Distinguish between exit(0) function and return statement
Answer:
Both are used to terminate the program but both are different. Return is a keyword and exit(0) is a function. The difference is, we can use more than one exit(0) function but we can use only one return statement in a scope. To use exit(0), the header file cstdlib should be used.

Question 7.
Draw the flowchart of for loop
Answer:

Question 8.
How many time the following for loop will execute? Justify.
for(i = 0; ; i ++)
{
if(i > 5)
cout<<“continue”;
else
cout<<“over”;
}
Answer:
Here the loop becomes infinite because the check condition is missing.

Question 9.
Predict the output.
#include<iostream.h>
int main()
{
int a = 0;
start:
cout<<endl<< ++a;
if(a < 5)
goto start;
}
Answer:
1
2
3
4
3

Question 10.
for(int i=2, sum=0; i <= 20; i=i+2)
sum += i;
Rewrite the above code using while loop.
Answer:
int i = 2; sum=0;
while (i<=20)
{
sum += i;
i = i + 2;
}

Question 11,.
Rewrite the following code using switch case statement.
if(day == 1)
cout<<“Sunday”;
else if(day == 2)
cout<<“Monday”;
else if(day == 7)
cout<<“Saturday”;
else
cout <<“Wednesday”;
Answer:
switch (day)
{
case 1: cout<<“Sunday”;break;
case 2: cout<<“Monday”;break;
case 7: cout<<“Saturday”;break;
default : cout<<“Wednesday”;
}

Question 12.
Pick the odd one out from the following. Give reason.

1. for, while, do….while
2. if, switch, for

Answer:

1. do…..while. It is an exit controlled loop while others are entry controlled loop
2. for. It is a loop while others are branching statements.

Question 13.
State whether the following statements are True or False. In either case justify your answer.

1. Break statement is essential in switch
2. For loop is an entry controlled loop
3. Do…while loop is an entry controlled loop
4. Switch is a selection statement

Answer:

1. False. It is not essential in single case statement
2. True. Because it will first check the condition. If it is true then only the body will be executed.
3. False. It is an exit controlled loop.
4. True.

Question 14.
Write the equivalent code for the following statement. R = (P<Q?P:Q)
Answer:
if(P<Q)
R = P;
else
R = Q;

Question 15.
Examine the following code snippet and find out the output? What will happen if the statement int ch; is replaced by char ch;
int ch;
for(ch=’A’;ch<=’Z’;++ch)
cout<<ch<<”;
Answer:
This code snippet will print 65, 66, 67,……., 90. If the statement int ch; is replaced by char ch; it prints A, B, C, ……., Z.

Question 16.

1. _______ is an entry control loop.
2. _______ Explain the memory allocation for the following declaration statement. int A[10] [10];

Answer:

1. while or for loop
2. To store an integer 4 bytes is used in Geany Editor. int A[10] [10]; → It needs 10 × 10 × 4 = 400 bytes

Question 17.
Differentiate between break and continue statements in C++. (SAY -2016) (2)
Answer:
break statement:
It is used to skip over a part of the code i.e. we can premature exit from a loop such as while, do-while, for or switch.
Syntax:
while (expression)
{
if (condition)
break;
}

continue statement:
It bypasses one iteration of the loop. That is it skips one iteration and continue the loop with next iteration value.
Syntax :
while (expression)
{
if (condition)
continue;
}

Plus One Control Statements Three Mark Questions and Answers

Question 1.
Compare if else and conditional operator?
Answer:
We can use conditional operator as an alternative of if-else statement. The conditional operator is a ternary operator.
The syntax of if-else
if (expression 1)
expression 2;
else
expression 3;

First expression 1 is evaluated if it is true expression 2 will be executed otherwise expression 3 will be executed. Instead of this, we can be written as follows using conditional operator Expression 1 ? expression 2: expression 3;

Question 2.

Answer:

Question 3.
Rewrite the program following program using if else.
#include<iostream>
using namespace std;
int main()
{
int a,b,big;
cout<<“Enter two integers”;
cin>>a>>b;
big = (a>b)?a:b;
cout<<“Biggest number is “<<big<<endl;
return 0;
}
Answer:

Question 4.


Is it possible to rewrite the above program using switch statement? Distinguish between switch and if else if ladder.
Answer:
No. It is not possible to write the above code using switch statement. Following are the difference between switch and if else if ladder.

1. Switch can test only for equality but if can evaluate a relational or logical expression
2. If else is more versatile
3. If else can handle floating values but switch cannot
4. If the test expression contains more variable if-else is used
5. Testing a value against a set of constants switch is more efficient than if-else.

Question 5.
Rewrite the following using nested switch construct.

Answer:

Question 6.
Consider the following output and write down the code for the same.
*
* *
* * *
* * * *
Answer:
#include<iostream>
using namespace std;
int main()
{
int i,j;
for(i=1;i<5;i++)
{
for(j=1;j<=i;j++)
cout<<n*”;
cout<<endl;
}
}

Question 7.
Consider the following output and write down the code for the same.
1
1 2
1 2 3
1 2 3 4
Answer:
#include<iostream>
using namespace std;
int main()
{
int ij;
for(i=1;i<5;i++)
{
for(j=l;j<=i;j++)
cout<<j<<“”;
cout<<endl;
}
}

Question 8.
Consider the following output and write down the code for the same.
1
2 2
3 3 3
4 4 4 4
Answer:
#include<iostream>
using namespace std;
int main()
{
int ij;
for(i=1;i<5;i++)
{
for(j=1;j<=i;j++)
cout<<i<<””;
cout<<endl;
}
}

Question 9.
Consider the following output and write down the code for the same.
1
2 3
4 5 6
7 8 9 10
Answer:
#include<iostream>
using namespace std;
int main()
{
int ij,k=0;
for(i=1;i<5;i++)
{
for(j=l;j<=i;j++)
cout<<++k<<“”;
cout<<endl;
}
}

Question 10.
Consider the following output and write down the code for the same.
1
1 3
1 3 5
1 3 5 7
Answer:
#include<iostream>
using namespace std;
int main()
{
int i,j;
for(i=1;i<5;i++)
{
for(j=1;j<=i;j++)
cout<<j × 2-1<<”
cout<<endl;
}
}

Question 11.
Consider the following output and write down the code for the same.
2
4 4
6 6 6
8 8 8 8
Answer:
#include<iostream>
using namespace std;
int main()
{
int i,j;
for(i=1;i<5;i++)
{
for(j=1;j<=i;j++)
cout<<i × 2<<“”;
cout<<endl;
}
}

Question 12.
Write a program to print the sum of first n natural numbers.
Answer:
#include<iostream>
using namespace std;
int main()
{
int n,i,sum=0;
cout<<“Enter a value for n”;
cin>>n;
for(i=1;i<=n;i++)
{
sum = sum + i;
}
cout<<“The sum of first ” <<n<<” numbers is “<<sum;
}

Question 13.
Write a program to read a number and check whether it is palindrome or not.
Answer:

Question 14.
Write a program to print the factorial of a number.
Answer:
#include<iostream>
using namespace std;
int main()
{
int n,i;
long fact = 1;
cout<<“Enter a number”;
cin>>n;
for(i=1;i<=n;i++)
fact = fact × i;
cout<<“The factorial of “<<n<<” is “<<fact;
}

Question 15.
Write a program to print the Fibonacci series.
Answer:
#include<iostream>
using namespace std;
int main()

Question 16.
Write a program to read a number and check whether the given number is Armstrong or not.
Answer:

Question 17.
Write down the code for the following output using while loop.
*
* *
* * *
* * * *
Answer:

Question 18.
Distinguish between entry controlled loop and exit controlled loop.
Answer:
An entry controlled loop first checks the condition and execute(or enters in to) the body of loop only if it is true. But exit control loop first execute the body of the loop once even if the condition is false then check the condition. The for loop and while loop are entry controlled loops but do-while loop is an exit controlled loop.

Question 19.
Write a program to find the largest of 3 numbers.
Answer:

Question 20.
Check whether a given number is prime or not.
Answer:

Question 21.
Write a program to print the prime numbers less than 100.
Answer:

Question 22.
Write a program to read number and display its factors.
Answer:
#include<iostream>
using namespace std;
int main()
{
int n,i;
cout<<“Enter a number greater than zero”;
cin>>n;
cout<<“The factors are”;
for(i=1;i<=n;i++)
if(n%i==0)
cout<<i<<“,”;
}

Question 23.
Write a program to print the Armstrong numbers less than 1000.
Answer:

Question 24.
Char result;
float marks;
cin>>marks;
if (marks >= 50)
result = ’P’;
else
result = ’F’;
cout<<result;
Rewrite the above code without using if statement.
Answer:
result=(marks>=50) ? ’P’: ’F’;

Question 25.
The output of a program is given below.
1
3
5
7
9
The sum is 25
Write a C++ program for obtaining the above output.
Answer:
#include<iostream>
using namespace std;
int main()
{
int sum=0,i;
for (i=1; i<=9; i+=2)
{
cout<<i<<endl;
sum = sum + i;
}
cout<<“The sum is”<<sum;
}

Question 26.
Find out the error in syntax if any and correct it?
Answer:
1. while (test condition);
{
}

2. do (condition)
{
}while

3. switch (condition)
{
Case 1:
Case 2:
Case 3:
Case 4:
}
Answer:
1. No need of semi colon. The corrected toop is given below
while (test condition)
{
}

2. In do … while loop the while must be end with semicolon.
do (condition)
{
}while;

3. switch contains expression instead of condition switch(expression)
{
Case 1:
Case 2:
Case 3:
Case 4:
}

Question 27.
Given the total mark of each student in SSLC examination. Write a C++ code fragment to find the grades.
Answer:

Question 28.
You are given the heights of 3 students. Write the relevant code segment to find the maximum height?
Answer:

Question 29.
Write the easiest code snippet for printing your name 1000 times. Explain.
Answer:
#include<iostream>
using namespace std;
int main()
{
int i;
char name[20];
cout<<“Enter your name:
cin>>name;
for(i=0;i<1000;i++)
cout<<name<<endl;
}

Question 30.
Given a code segment for(i=1; i<10; i++)
cout<<i;

1. Rewrite the code using do….while loop
2. What will be the output when i = 0? Give reason.

Answer:
1. i = 1;
do{
cout<<i; i++;
}while(i<10);

2. When i = 0, it will execute one more time. ie. the for loop execute 9 times but here this loop executes 10 times.

Question 31.
Whenever a string is entered the inverse of that string is displayed( eg: if we enter ‘CAR’ the output is ‘RAC’). Write a suitable programme for the output.
Answer:

Question 32.
Write a C++ program to display as follows
A
A B
A B C
A B C D
A B C D E
Answer:

Question 33.
Write C++ program forgetting the following output. (SAY-2016) (3)
1
1 2
1 2 3
1 2 3 4

OR

Consider the following C++ program and answer the following questions.
#include<iostream.h>
int main()
{
int a, p = 1;
for(a=1;a<=5;a+=2)
p = p × a;
cout<<p;
}

(a) Predict the output of the above code.
(b) Rewrite the above program using while loop.

Answer:

Plus One Control Statements Five Mark Questions and Answers

Question 1.
“We know that the execution of a program is sequential”. Is it possible to change this sequential manner and explain different jump statements in detail.
Answer:
The execution of a program is sequential but we can change this sequential manner by using jump statements. The jump statements are
1. goto statement:
By using goto we can transfer the control anywhere in the program without any condition. The syntax is goto label;
Example:

2. break statement:
It is used to skip over a part of the code i.e. we can premature exit from a loop such as while, do-while, for or switch.

3. continue statement:
It bypasses one iteration of the loop.

4. exit(0) function:
It is used to terminate the program. For this the header file cstdlib must be included.

Question 2.
Mr. X wants to get an output 9 when inputting 342 and he also wants to get 12 when inputting 651. Write the program and draw a suitable flowchart for X?
Answer:

Question 3.
Explain conditional statements in detail?
Answer:
1. Simple if:
The syntax is given below if(expression)
statement;

or

if(expression)
{
Block of statements
}
First expression evaluates if it is true then only statement will be executed.
eg: if (n>0)
cout<<n<<” is positive”;

2. if else:
The syntax is given below, if (expression)
statement 1;
else
statement 2;

or

if (expression)
{
statement block 1;
}
else
{
statement block 2;
}
First expression evaluates if it is true statement block 1 will be executed otherwise statement block 2 will be executed. Only one block will be executed at a time so it is called branching statement.
eg:
if (n>0)
cout<<n<<” is positive”;
else
cout<<n<<” is negative”;

3. if else if ladder:
The syntax will be given below
if (expression!)
{
statement block 1;
}
else if (expression 2)
{
statement block 2;
}
else if (expression 3)
{
statement block 3;
}
else
{
statement block n;
}
Here first expression 1 will be evaluated if it is true only the statement blockl will be executed otherwise expression 2 will be executed if it is true only the statement block 2 will be executed and so on. If all the expression evaluated is false then only statement block n will be evaluated .
eg:
If (mark>=90)
cout<<“Your grade is A+”;
else if (mark>=80)
cout<<“Your grade is A”;
else if (mark>=70)
cout<<“Your grade is B+”;
else if (mark>=60)
cout<<“Your grade is B”;
else if (mark>=50)
cout<<”Your grade is C+”;
else if (mark>=40)
cout<<“Your grade is C”;
else if (mark>=30)
cout<<“Your grade is D+”;
else
cout<<“Your grade is D”;

4. conditional operator:
It is a ternary operator and it is an alternative for if else construct. The syntax is given below.
expression 1? expression 2: expression 3;

or

expression 1? Value if true : value if false;
Here expression 1 will be evaluated if it true expression 2 will be executed otherwise expression 3 will be executed.
eg:
n>0?cout<<n<<” is positive”:cout<<n<<” is negative”;

5. Switch:
It is a multiple branch statement. Its syntax is given below.
switch(expression)
{
case value: statements;break;
case value: statements;break;
case value: statements;break;
case value: statements;break;
case value: statements;break;
……..
default : statements;
}
First expression evaluated and selects the statements with matched case value.
eg:
switch (n)
{
case 1: cout<< “Sunday”;break;
case 2: cout<< “Monday”;break;
case 3: cout<< “Tuesday”;break;
case 4: cout<< “Wedesday”;break;
case 5: cout<< “Thursday”;break;
case 6: cout<< “Friday”;break;
case 7: cout<< “Saturday”;break;
default : cout<< “lnvalid”
}

Question 4.
Explain different loops in detail?
1. For loop:
The syntax of for loop is for(initialization; checking ; update loop variable)
{
Body of loop;
}
First part, initialization is executed once, then checking is carried out if it is true the body of the for loop is executed. Then loop variable is updated and again checking is carried out this process continues until the checking becomes false. It is an entry controlled loop.
eg: for(i=1,j=1;i<=10;i++,j++)
cout<<i<<” × “<<j<<” = “<<i × j;

2. While loop:
It is also an entry controlled loop The syntax is given below
Loop variable initialised while(expression)
{
Body of the loop;
Update loop variable;
}
Here the loop variable must be initialised outside the while loop. Then the expression is evaluated if it is true then only the body of the loop will be executed and the loop variable must be updated inside the body. The body of the loop will be executed until the expression becomes false.
eg:
i = 1;
j = 1;
while(i<=10)
{
cout<<i<<” × “<<j<<” = “<<i × j; i++;
j++;
}

3. do While loop:
It is an exit controlled loop. The syntax is given below
do
{
Statements
}while(expression);
Here the body executes atleast once even if the condition is false. After executing the body it checks the expression if it false it quits the body otherwise the process will be continue.

Question 5.
Write a program to do the following:

1. Inputs the values for variables ‘n’ and ‘m’.
2. Prints the numbers between ‘m’ and ‘n’ which are exactly divisible by ‘m’.
3. Checks whether the numbers divisible by ‘m’ are odd or even.

OR

Write a program using nested loop that inputs a number ‘n’ which generates an output as follows. Hint: if the value of ‘n’ is 5, the output will be as ‘n’
Answer:
1.
25
25 16
25 16 9
25 16 9 4
25 16 9 4 1

2.
#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int i,n,m;
cout<<“Enter values for n and m”;
cin>>n>>m;
for(i=1;i<=n;i++)
if(i%m == 0)
cout<<i<<“,”;
getch();
}

3.
#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int i,n,m;
cout<<“Enter values for n and m”;
cin>>n>>m;
for(i=1;i<=n;i++)
if(i%m == 0)
{
cout<<i<<“\t”;
if(i%2 == 0)
cout<<“even”<<endl;
else
cout<<“odd”<<endl;
}
getch();
}

OR

#include<iostream.h>
#include<conio.h>
#include<string.h>//for strlen()
main()
{
clrscr();
int n,i,j;
cout<<“enter a value for n:”;
cin>>n;
for(i=n;i>0;i- -)
{
for(j=n;j>=i;j~)
cout<<j × j<<“\t”;
cout<<endl;
}
getch();
}

Question 6.
Write a C++ program to display Fibonacci series. (SAY-2015)
Answer:

Question 7.
1. Write a C++ program to accept an integer number and check whether it is an Armstrong number or not. (Hint: Sum of the cubes of the digits of an Armstrong number is equal to that number itself)

OR

2. rite a C++ program to accept an integer number and print its reverse (Hint: If 234 is given, the output must be 432).
Answer:
1.

2. #include<iostream>
void main()

Question 8.
1. Write a menu driven program which accepts 3 numbers and show options to find and display.

• the biggest number
• the smallest number
• the sun of the numbers
• the product of the numbers

OR

2. Write a C++ program to check whether a number is palindrome or not.
Answer:
1.

OR
2.

Question 9.
Answer any one question from 22(a) and 22(b).
1. Write a C++ program to display all leap years between 1000 and 2000 excluding all century years.

OR

2. Write a C++ program to find the sum of the first 10 numbers of Fibonacci series. (Fibonacci series is 0, 1, 1,2, 3, 5, 8, 15 where 0 and 1 are the first two terms and reamaining terms are obtained by the sum of the two preceding terms.)
Answer:
1.

OR
2.

Question 10 (MARCH-2017)
Write a program to check whether the given number is palindrome or not. (MARCH-2017) (5)

OR

Write a program to print the leap years between 2000 and 3000.
(A century year is leap year only if it is divided by 400 and a noncentury year is leap year only if it is divided by 4).
Answer:

OR

Students can Download Chapter 12 Organic Chemistry: Some Basic Principles and Techniques Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques One Mark Questions and Answers

Question 1.
Which of the following does not contain fused benzene ring?
a) Naphthalene
b) Anthracene
c) Diphenyl
d) Phenanthrene
Answer:
c) Diphenyl

Question 2.
Hemolytic fission of a covalent bond results in the formation of ___________ .
CH₂=CH – C ≡ CH₂OH
Answer:
Free radicals

Question 3.
The IUPAC name of is
a) 2-pentyl-4-en-1-ol
b) 4-penten-2-yn-1-ol
c) 1-pentene-3-yn-5-ol
d) 5-Hydroxy-1-pentene-3-yne
Answer:
b) 4-penten-2-yn-1-ol

Question 4.
Lassigne’s solution on treating with sodium nitro prusside solution gives a violet colour indication the presence of ___________ in the organic compound.
Answer:
sulphur

Question 5.
In Kjeldahl’s method, nitrogen present is estimated as ___________ .
Answer:
NH₃

Question 6.
In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
a) Na₄[Fe(CN)₆]
b) Fe₄[Fe(CN)₆]₃
c) Fe₂[Fe(CN)₆]
d) Fe₃[Fe(CN)₆]₄
Answer:
The Prussian blue colour i§ due to the formation Fe₄[Fe(CN)₆]₂. Thus, option (b) is correct,

Question 7.
Which of the following carbocation is most stable?

Answer:
The order of stability of carbocation is : 3º>2º>1º

Thus, option (b) is correct.

Question 8.
The best and latest technique for isolation, purification and separation of organic compounds is:
a) Crystallisation
b) Distillation
c) Sublimation
d) Chromatography
Answer:
Chromatography. Thus, option (d) is correct.

Question 9.
The following reaction is classified as:
CH₂CH₂l + KOH (aq) → CH₃CH₂OH + Kl
a) Electrophilic Substitution
b) Nucleophilic substitution
c) Elimination
d) Addition.
Answer:
This is an example of nucleophilic substitution reaction since the nucleophile l – is replaced by the nucleophile OH^– ion. Thus, option (b) is correct.

Question 10.
Which of the carbocation is more stable?

Answer:

Question 11.
Absolute alcohol cannot be obtained by fractional distillation because
Answer:
constant boiling azeotrope mixture is formed with water

Question 12.
Lassaigne’s test fails in
a) NH₂NH₂
b) H₂NCONH₂
c) C₆H₂NHNH₂
d) NH₂OH
Answer:
a) Hydrazine NH₂NH₂

Question 13.
Beilstein test is for the detection of __________ .
Answer:
Halogens

Question 14.
Glycerine can be purified by __________ .
Answer:
Distillation under reduced pressure.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Two Mark Questions and Answers

Question 1.
2-Butene exhibits geometrical isomerism.
1. Represent the cis-trans isomers of 2-Butene.
2. Explain ozonolysis with a suitable example.
Answer:
1.

2. When an alkene is allow to react with ozone, an ozonide is obtained. This on hydrolysis gives aldehyde/ ketone. The whole process is called ozonolysis. Eg:

Question 2.
Draw the structures of the molecules represented by the IUPAC names, Pent-3-en-1-ol and 2- Nitrocyclohexene.
Answer:
a) CH₃ – CH = CH – CH₂ – CH₂ – OH

Question 3.
Addition of HBr to propene yields 2-bromopropane, what happens if benzoyl peroxide is added to the above reaction.
Answer:
When propene is allowed to react with HBr in the presence of peroxide, 2 bromo propane is obtained as the minor product and this is called peroxide effect.

Question 4.
Write the IUPAC name of

Answer:
i) 2-Pentanone
ii) 3-Methyl-1-pentanal

Question 5.
Write any two necessary condition for a compound to be aromatic. Convert Acetylene to benzene.
Answer:
Cyclic, Planar and should contain (4n+2) n electrons.

Question 6.
What are hybridisation states of each carbon atom in the following compounds?
CH₂ = C = O, CH₃CH = CH₂,(CH₃)₂CO,
CH₂ = CHCN, C₆H₆
Answer:
The hybridisation of each carbon is written as superscript on the carbon atom in the molecule.

Question 7.
Which is expected to be more stable, O₂ NCH₂CH₂O or
Answer:

has -l- effect and hence it tends to disperse the -ve charge on the O-atom. In contrast, CH₃CH₂ exerts + l – effect. It, therefore, tends to intensity the -ve change and hence destabilizes it.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Three Mark Questions and Answers

Question 1.
a) Write the structure of the following compounds.
i) Hexane-2, 4-dione
ii) 3-Bromo-4 methyl hexane-2-ol
iii) 2-Bromo-4-methyl hept-5-enal
Answer:
a) i) Hexane 2, 4-dione

Question 2.
Isomers are compounds with same formula and different properties.
a) Write all the possible structural isomers of C₅H₁₂.
b) Give the IUPAC name of above isomers.
Answer:

b) 1) Pentane
2) 2-Methyl butane
3) 2,2-Dimethyl propane

Question 3.
a) Give the IUPAC names of:

b) How is nitrogen detected by Lassaignes test?
c) Name a suitable technique for separation of the components from a mixture of benzene (b.p.353 K) and aniline (b.p.-457 K)
Answer:
a) (i) 2-Chloro 3-methyl hexane
(ii) 2-Methylbutanal

b) About 2 mL of the extract is boiled with about 1 mL of freshly prepared ferrous sulphate solution. One or two drops of con.H₂SO₄ are added to the solution. Presence of nitrogen is indicated by the appearance of a blue colouration.

c) Distillation

Question 4.
i) Draw the structure of propanone. Write the hybridisation of each carbon in propanone.
ii) Arrange the following carbocations in the increasing order of their stability. Justify.
CH₃⁺, CH₃CH₂⁺, (CH₃)₂CH⁺
iii) What is the method used to separate a mixture of o-Nitrophenol from p-Nitrophenol? Which property is utilized for separation?
Answer:

I^st carbon of Propanone is sp³ hybridisation.
II^nd carbon of Propanone is sp² hybridisation.
III^rd carbon of Propanone is sp³ hybridisation.

iii) Chromatography; Difference in absorption rate of different substances..

Question 5.
1. Give the IUPAC name of the following compound.

2. How can you detect the presence of nitrogen in an organic compound?
3. Arrange the following in the increasing order of stability.
(CH₃)₂CH⁺, CH₃-CH₂⁺, (CH₃)₃C⁺
Answer:
1. 2-chloro, 2-methyl propane

2. Presence of nitrogen can be detected by Lassigne’s test. The sodium fusion extract is boiled with FeSO₄ and then acidified with con.H₂SO₄. The formation of Prussian blue colour conforms the presence of nitrogen.
3.

Question 6.
Predict the product in the following reactions and identify the rules:

Answer:
a) When propene is allowed to react with HBr in the presence of peroxide, 2-Bromo propane is obtained as the minor product. (Peroxide effect, Kharasch effect or Anti MarkownikofFs rule of addition).

Question 7.
a) Draw the structures of the following compounds.
i) 3-hexenoic acid
ii) 2-chloro-2-methyl butanol
iii) 4-nitro-1-pent-l-yne
Answer:

Question 8.
Write the IUPAC names of the following:

Answer:
i) 2,2,4-Trimethyl pentane
ii) 2-Methyl 1-butene
iii) Propyl benzene

Question 9.
Draw the structure of the following molecules.
i) 3, 4-Dimethylhept-3-ene
ii) Neo-pentane
iii) 3-Nitrocyclohexene
Answer:

Question 10.
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Answer:
a) Structural isomers (actually position isomers as well as metamers)
b) geometrical isomers,
c) resonance contributors because they differ in the position of electrons but not atoms:

Question 11.
Expain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and phosphorus.
Answer:
The organic compound is fused with sodium metal to convert these elements (which are present in the covalent form) to ionic form. For example, Sulphur is changed to Na₂S, nitrogen to NaCN and phosphorus to Na₃PO₄ The presence of sulphide ions, cyanide ions and phosphate ions can thus be confirmed by using suitable reagents.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Four Mark Questions and Answers

Question 1.
1) Write the structural formula of
a) 4-Ethyl-1-fluoro-2-nitrobenzene
b) 2,3,6-Trimethyl octane.
c) 1,2-Dibromo benzene.
2) Categorize the following as nucleophile and electrophile
a) HS^–
b) BF₃
c) NO₂⁺
d) C₂H₅O^–
e) (CH₃)₃N
f)NH₂^–
Answer:

Question 2.
1. What is chromatography? Name the different types of chromatography.
2. What is Lassaigne’s test?
Answer:
1. Chromatography is a valuable method for the separations, purification and identification of the constituents of a mixture. Chromatography is classified into 2 types.
a) Adsorption chromatography
b) Partition chromatography

2. Lassigne’s test is used to determine the presence of nitrogen, halogens and sulphur present in the organic compound.

Question 3.
1. Write IUPAC names of the products obtained by addition of HBr to Hex-1-ene.
i) in the absence of peroxide
ii) in the presence of peroxide
2. Complete the following reaction:

Answer:
1.

2. 2HCHO + H₂O₂

Question 4.
1. What is metamerism? Give example for metamers.
2. What are free radicals? How are they formed?
Answer:
1. It is the isomerism which arises due to different alkyl chains on either side of the functional group in the molecule.
e.g. Methoxypropane(CH₃OC₃H₇) and ethoxyethane (C₂H₅OC₂H₅) are metamers.

2. Free radicals are highly reactive species containing unpaired electrons. They are formed by homolytic cleavage of covalent bond. e.g. \(\dot { C } \)H₃

Question 5.
1. How will you prepare butane?
2. Explain Markownikoff’s rule for the addition reaction using a suitable example.
Answer:
1. When an alkyl halide (ethyl chloride) is allowed to react with metalic sodium in presence of dry ether,

2. Markownikoff’s rule of addition:
When a hydrogen hallide is added to an unsymmetrical alkane the halogen atom will goes to the double bond carbon containing lesser number of hydrogen atom, eg:

Question 6.
Identify the isomerism exhibited by the following compounds.

1. CH₃CH₂CH₂OH and (CH₃)₂CHOH
2. CH₃CH₂CHO and CH₃COCH₃
3. CH₃CH2₂CH₂CH₃andCH₃OCH₂CH₂CH₃
4. CH₃(CH₂)₃ CH₃ and (CH₃)₄C

Answer:

1. Position isomerism
2. Functional group isomerism
3. Metamerism
4. Chain isomerism

Question 7.
i) Predict the products of

ii) Classify the following compounds into aromatic and non-aromatic.

(iii) Which of the following compounds will show geometrical isomerism?
a) CH₃ CH=CHCH₃
b) (CH₃)₂₃ C=CH₂
Answer:

Question 8.
Write the IUPAC name

b) Write the structure of the following.
i) 2-Chloro-2-methyl butanol
ii) 4-Nitro-1-pentene
Answer:
a) i) 4-Methyl pentanal
ii) Cyclohexanol

Question 9.
a) Write IUPAC names of the products obtained by addition reactions of HBrto hex-1-ene:
i) In the absence of peroxide.
ii) In the presence of peroxide.
b) How will you convert:
i) Benzene to toluene.
ii) Benzene to nitrobenzene
Answer:

Question 10
a) Explain the term
(i) Inductive effect
(ii) Nucleophile
b) Write the structural formula of the following.
i) 2,5,6-Trimethyloctane
ii) 2,4-Dimethylpentane
c) Suggest the suitable technique for separation of organic compounds given in the data.
i) Aniline-water mixture
ii) Glycerol from spent-lye
Answer:
a) Inductive effect:
If more electronegative atom X linked to a carbon atom then the bonded electron pair will be shifted more towards X. So X acquires a small negative charge and carbon get a small charge. So this carbon atom attracts the a bonded electron pair towards its from nearest carbon atom. Therefore the change is transferred from one carbon to other.

Nucleophile :
Nucleus loving species negatively charge species like \(\overline { OH } \), \(\overline { X } \) etc. are called nucleophiles.
b) 2,5,6-Trimethyloctane

c) i) Fractional distillation.
ii) Distillation under reduced pressure.

Question 11.
Compounds having the same molecular formula exhibit different properties is called isomerism. Explain different types of isomerism with examples.
Answer:
Chain isomerism: Consider the molecular formula C₅H₁₂ the following 3 isomers are possible for this compound.
They are

Position isomerism:
This isomerism arises as a result of the difference in the position of double bond triple bond and functional group.
The following chain isomers are possible for the molecular formula C₄H₈.
CH₂ = CH – CH₂ – CH₃ 1-butene
CH₃ – CH = CH – CH₃ 2-butene

Metamerism:
This isomerism arises as a result of the difference in the alkyl group present on either side of the functional group.
Consider the molecular formula C₄H₁₀O. The following two metamers are possible for this molecular formula.
CH₃ – O – CH₂ -CH₂ -CH₃ Methyl propyl ether
CH₃ – CH₂ – O – CH₂ – CH₃ Diethyl ether

Functional group isomerism:
Consider the molecular formula C₂H₆O the following two functional isomers are possible for this compound.
1) CH₃-CH₂-OH Ethanol
2) CH₃ – O – CH₃ Di methyl ether

Question 12.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer:
Sodium extract is boiled with nitric acid to decompose NaCN and Na₂S, if present,

If the sodium extract is not boiled with nitric acid then NaCN and Na₂S formed will react with AgN03 and hence will interfere with the test as shown below:

Students can Download Chapter 12 Thermodynamics Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 12 Thermodynamics

Plus One Physics Thermodynamics One Mark Questions and Answers

Question 1.
If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature
(a) is doubled
(b) becomes one-fourth
(c) remains constant
(d) is halved
Answer:
(b) becomes one-fourth
According to ideal gas law

Question 2.
The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T) is

where k_B is the Boltzmann constant and N is the number of molecules.
Answer:
c) PV = k_BNT
According to ideal gas equation,
PV = Nk_BT.

Question 3.
Which of the following laws of thermodynamics forms the basis for the definition of temperature?
(a) First law
(b) Zeroth law
(c) Second law
(d) Third’law
Answer:
(b) Zeroth law
The Zeroth law of thermodynamics gives the definition of temperature.

Question 4.
Which of the following is a true statement?
(a) The total entropy of thermally interacting systems is conserved
(b) Carnot engine has 100% efficiency
(c) Total entropy does not change in a reversible process.
(d) Total entropy in an irreversible process can either increase or decrease.
Answer:
(c) Total entropy does not change in a reversible process.

Question 5.
In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas
(a) the temperature will decrease,
(b) the volume will increase
(c) the pressure will remain constant
(d) the temperature will increase
Answer:
(a) From first law of thermodynamics
dQ = dU + dW
dQ = dU (if dW = 0)
Since, dQ < 0
dU < 0
or U_final < U_intial
Hence, temperature will decrease.

Question 6.
An electric fan is switched on in a closed room will the air of the room be cooled?
Answer:
No. Infact speed of air molecules will increase and this results in increase in temperature.

Question 7.
When an iron nail is hammered, if becomes hot. Why?
Answer:
The kinetic energy of hammer gets converted in to heat energy which increases temperature of iron nail.

Question 8.
Identify the thermodynamic process in which temperature of system may increase even when no heat is supplied to the system.
Answer:
Adiabatic process.

Question 9.
Which thermodynamic variable is defined by first law of thermodynamics?
Answer:
Internal energy

Question 10.
The door of an operating refrigerator kept open in a closed room. Will it make the room cool?
Answer:
No. The room will be slightly warmed.

Plus One Physics Thermodynamics Two Mark Questions and Answers

Question 1.
A Carnot engine working between 300K and 600K has a work output of 800J per cycle. Find the amount of energy consumed per cycle

• 800J
• 400J
• 1600J
• 1200J
• 3200J

Answer:
Efficiency of carrots engine

Efficiency also can be written as

∴ input power = 1600 J.

Question 2.
A carnote’s engine is made to work between 200°C and 0°C first and then between 0°C and minus 200°C. Compare the values of efficiencies in the two cases.
Answer:
Case -1:

= 1 – 0.577
= 0.42 = 42%
Case – 2:
η = 1 – \(\frac{73}{473}\)
T2 = -200 = 73k
= 1 – 0.26 = 0.73
= 73%
T₁ = 0 = 273k.

Question 3.
Match the following

Answer:

Question 4.
What is the value of specific heat capacity of gas in

1. Isothermal process
2. Adiabatic process

Answer:

1. Infinite
2. Zero

Question 5.
Is the function of refrigerator against the second law of thermodynamics? Explain.
Answer:
No. The refrigerator transfers heat from the inside space to outer atmosphere, at the expense of external work supplied by the compresser fed by electric supply.

Plus One Physics Thermodynamics Three Mark Questions and Answers

Question 1.
Heat is supplied to a system, but its internal energy does not increase

1. Which process is involved in this case? (1)
2. Obtain an expression for the work done in the above process (2)

Answer:
1. Isothermal process.

2. Work done by adiabatic process
Let an ideal gas undergoes adiabatic charge from (P₁, V₁, T₁) to (P₂, V₂, T₂). The equation for adiabatic charge is
PV^γ = constant = k
ie; P₁V₁^γ = P₂V₂^γ = k _____(a)
The work done by

from equation (a) P₁V₁^γ = P₂V₂^γ = k

Substituting ideal gas equation.

Question 2.
A heat engine is a device which effectively converts heat energy into mechanical energy.

1. State the law which describes this principle. (1)
2. Derive an expression for the efficiency of a Carnot engine. (2)

Answer:
1. Kelvin – Plank statement:
No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of heat into work.

Clausius statement:
No process is possible whose sole result is the transfer of heat from a colder object to hotter object.

2. Carnot’s cycle:
The Carnot cycle consists of two isothermal processes and two adiabatic processes.

Let the working substance in Carnot’s engine be the ideal gas.
Step 1: The gas absorbs heat Q₁ from hot reservoir at T₁ and undergoes isothermal expansion from (P₁, V₁, T₁) to (P₂, V₂, T₁).

Step 2: Gas undergoes adiabatic expansion from (P₂, V₂, T₁) to (P₃, V₃, T₂)

Step 3: The gas release heat Q₂ to cold reservoir at T₂, by isothermal compression from (P₃, V₃, T₂) to (P₄, V₄, T₂).

Step 4: To take gas into initial state, work is done on gas adiabatically [(P₄, V₄, T₂) to (P₁, V₁, T₁)]. Efficiency of Carnot’s engine:

Question 3.
A thermo flask contains coffee. It is violently shaken. Considering the coffee as a system:

1. does its temperature rise?
2. has heat been added to its?
3. has internal energy changed?

Explain your answers.
Answer:

1. Yes
2. Heat is added to coffee
3. Internal energy is changed
4. When we shake, the mechanical energy is added to the liquid contained in a flask.

Question 4.
It is predicted by the meteorologists that global warming will result in the flooding of oceans due to the melting of ice caps on the earth.

1. Name the thermodynamic process involved in the melting of ice.
2. Determine the heat energy required to melt 5kg of ice completely at 0°C. Latent heat of fusion of water is 3336 × 10³ J Kg^-1.
3. During the melting process, what change will occur to its internal energy?

Answer:

1. Isothermal process
2. ∆Q = mL = 5 × 33.36 × 10⁵ = 16.68 × 10⁶J
3. Internal energy increases

Question 5.
Water is heated in an open vessel.

1. This process is
   • isothermal
   • isobaric
   • isochoric
   • adiabatic
2. Which law of thermodynamics is suitable to explain the transfer of heat here?
3. Draw the heat-temperature graph of ice below 0°C heated up to steam above 100°C.

Answer:
1. Isobaric
2. First law of thermodynamics
3.

Question 6.

1. The molar heat capacity of oxygen at constant volume is 20J mol^-1K^-1. What do you mean by molar heat capacity at constant volume (C_V)?
2. The difference between C_P and C_V is always a constant. Give a mathematical proof.

Answer:
1. Molar specific heat at constant volume is the heat required to raise the temperature of I mol substance by IK at constant volume.

2. According to 1^st law of thermodynamics
∆Q = ∆U + PAV
If ∆Q heat is absorbed at constant volume (∆V = 0).

From ideal gas equation for one mole PV = RT. Differentiating w.r.t. temperature (at constant pressure

Equation (4) – Equation (1), we get
C_P – C_V = R.

Question 7.
A heat engine is a device which converts heat energy into work.

1. What is the working substance in an ideal heat engine.
2. A Carnot engine is working between in melting point and steam point. What is its efficiency?

Answer:
1. Ideal gas.

2. T₂ = 0°C = 273K
T₁ = 100°C = 373K
Efficiency η = I – \(\frac{T_{2}}{T_{1}}=I-\frac{273}{373}\)
η = 0.27
η = 27%.

Question 8.
A gas is taken in a cylinder. The walls of the cylinder is insulated from the surroundings

1. The gas in a cylinder is suddenly compressed. Which thermo dynamic process involves in this statement. Explain the thermo dynamic process.
2. If gas is suddenly compressed to 1/4^th of its, original volume. Calculate the rise in temperature. The initial temperature was 27°C and γ = 1.5.

Answer:
1. Adiabatic. A thermo dynamic process in which no heat enters or leaves the system.

2. T. = 27°C = 27 + 273 = 300k

= 300 × 4^{1.5 – 1}
= 300 × 4^1/2
= 600K
rise in temperature = 600K – 300K = 300K.

Question 9.
Raju brought a motor pump from a shop. The efficiency of the motor pump is printed on the label as 60%.

1. The efficiency of a water pump is 60%. What is . meant by this?
2. Can you design an engine of 100% efficiency? Justify your answer.

Answer:
1. Efficiency means that, 60% of the total energy received is converted into useful work.

2. The efficiency of heat engine η = 1 – \(\frac{T_{2}}{T_{1}}\). The efficiency will be 100% if T₂ = OK, both those conditions cannot be attained practically. So art engine can’t have 100% efficiency.

Question 10.
Categorise into reversible and irreversible process

1. Waterfall
2. rusting of iron
3. electrolysis
4. slow compression
5. diffusion of gas
6. melting of ice
7. dissolving NaCl in water
8. flow of heat from hot body to cold body
9. slow compression of spring
10. heat produced by friction.

Answer:
Reversible process → (3), (4), (6) and (9)
Irreversible process → (1), (2), (5), (7), (8) and (10)

Plus One Physics Thermodynamics NCERT Questions and Answers

Question 1.
A geyser heats water flowing at the rate of 3.0 litres per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10⁴ Jg^-1?
Answer:
Volume of water heated = 3.0 litre per minute
mass of water heated, m = 3000 g per minute
increase in temperature, ∆T = 77°C – 27°C = 50°C
specific heat of water, c= 4.2Jg^{-1 °}C^-1
amount of heat used, Q = mc ∆T
or Q = 3000 g min^-1 × 4.2Jg^-1
°C^-1 × 50°C
= 63 × 10⁴J min^-1
rate of combustion of fuel = \(\frac{63 \times 10^{4} \mathrm{Jmin}^{-1}}{4.0 \times 10^{4} \mathrm{Jg}^{-1}}\)
= 15.75gmin^-1.

Question 2.
What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N₂ = 28, R = 8.3 J mol^-1 K^-1).
Answer:
m = 2 × 10³ kg^-1 = 20g,
∆T = 45°C, M = 28
R = 8.3 J mol^-1 K^-1, M = 28
Number of moles, n = \(\frac{m}{M}=\frac{20}{28}=\frac{5}{7}\)
Since nitrogen is diatomic,
∴ C_P = \(\frac{7}{2}\)
R = \(\frac{7}{2}\) × 8.3 J mol^-1K^-1
Now, ∆Q = nC_P ∆T
= \(\frac{5}{2}\) × \(\frac{7}{2}\) × 8.3 × 45J = 933.75J.

Question 3.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas, is compressed to half its original volume?
Answer:
P₂V₂^γ = P₁V₁^γ

Question 4.
A steam engine delivers 5.4 × 10⁸ J of work per minute and services 3.6 × 10⁹J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Answer:
Work done per minute, output = 5.4 × 10⁸J
Heat absorbed per minute, input = 3.6 × 10⁹J
Efficiency, η = \(\frac{5.4 \times 10^{8}}{3.6 \times 10^{9}}\) = 0.15
%η = 0.15 × 100 = 15
Heat energy wasted/ minute = Heat energy absorbed/minute – Useful work done/minute
= 3.6 × 10⁹ – 5.4 × 10⁸
= (3.6 – 0.54) × 10⁹ = 3.06 × 10⁹J.

Question 5.
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joule per second, at what rate is the internal energy increasing?
Answer:
Heat supplied, ∆Q = 100W = 10OJs^-1
Useful work done, ∆W = 75J s^-1
Using first law of thermodynamics
∆Q = ∆U+ ∆W
∆U = ∆Q – ∆W
= 100Js^-1 – 75Js^-1 = 25Js^-1.

Question 6.
A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.
Answer:
T₁ = 36°C = (36 + 273) K = 309 K
T₂ = 9°C = (9 + 273) K = 282 K
Coefficient of performance = \(\frac{T_{2}}{T_{1}-T_{2}}\)

Question 7.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer:
PV = µRT or V = \(\frac{\mu \mathrm{RT}}{\mathrm{P}}\)

= 22.4 × 10^-3 m³ = 22.4 litre.

Students can Download Chapter 11 Thermal Properties of Matter Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Plus One Physics Thermal Properties of Matter One Mark Questions and Answers

Question 1.
In which of the following processes, convection does nottake place primarily?
(a) Sea and land breeze
(b) Boiling of water
(c) Warming of glass of bulb due to filament
(d) Heating air around a furnace
Answer:
(c) Warming of glass of bulb due to filament
In convection process, the heat is transferred by the bodily motion of the heated particles. It is not so in case of warming of glass bulb due to filament heating. In fact, warming of glass bulb is due to radiation.

Question 2.
\(\frac{\text { Watt }}{\text { Kelvin }}\) a unit of
(a) Stefan’s constant
(b) Wien’s constant
(c) Cooling’s constant
(d) Thermal conductance
Answer:
(d) Thermal resistance

Question 3.
For measuring temperatures in the range of 2000°C, we should employ
(a) gas thermometer
(b) platinum-rhodium thermometer
(c) barometer
(d) pyrometer
Answer:
(d) pyrometer

Question 4.
There is a hole in metal disc. What happens to the size of metal disc if the metal disc is heated?
Answer:
The size of hole increases.

Question 5.
Which has more specific heat capacity, water, and sand?
Answer:
Water.

Question 6.
Two solid spheres of the same material have the same radius but one is hollow while the other is solid. Both spheres are heated to same temperature. Then
(a) the solid sphere expands more
(b) the hollow sphere expands more
(c) expansion is same for both
(d) nothing can be solid about their relative expansion if their masses are not given
Answer:
(c) expansion is same for both

Question 7.
The sprinkling of water reduces slightly the temperature of a closed room because
(a) temperature of water is less than that of the room.
(b) specific heat of water is high
(c) water has large latent heat of vaporisation
(d) water is a bad conductor of heat
Answer:
(c) water has large latent heat of vaporisation
When water is sprinkled over a large area, evaporation takes place. As the latent heat of vaporisation is large cooling takes place.

Question 8.
Why specific heat of gas at constant pressure (C_p) is greater than specific heat at constant volume?
Answer:
More heat is required to raise the temperature of gas at constant pressure than at constant volume.

Question 9.
A body is heated. But there is no change in its temperature. Is it possible?
Answer:
Yes. During change of state, there will be no increase in temperature even when heat is supplied.

Question 10.
When boiling water is put in glass tumbler, the tumbler cracks. Why?
Answer:
Glass is poor conductor of heat. So inner and outer surfaces of tumbler suffer uneven expension. Hence it breaks.

Question 11.
A small space is left between two rails on railway track. Why?
Answer:
If no space is left, the rails would bend due to thermal expansion in summer. So small space is left between two rails to allow thermal expansion.

Question 12.
The pendulum of clock is made of invar, Why?
Answer:
Invar has low value of coefficient of linear expansion. So length of pendulum remains almost same in all seasons. (The change in length affects time period of pendulum).

Question 13.
Tea gets cooled, when sugar is added to it. Why?
Answer:
When sugar is added, heat content of tea gets shared with sugar & hence’ temperature decreases.

Question 14.
Ice covered in gunny bag does not melt for a long time. Why?
Answer:
A gunny bag is poor conduct of heat & hence it does not allow external heat to enter.

Question 15.
Why two layers of cloth of equal thickness provide warmer covering than a single layer of cloth of double the thickness?
Answer:
Because air between two layers of clothes is a bad ‘ conductor of heat.

Question 16.
On winter nights, we feel warmer when clouds cover the sky than the sky is clear. Why?
Answer:
The clouds are bad conductor of heat. So heat of earth’s atmosphere is not conducted out.

Plus One Physics Thermal Properties of Matter Two Mark Questions and Answers

Question 1.
1. Two bodies at different temperatures T₁K and T₂K are brought in contact with each other

• Is the resultant temperature be necessarily (T₁ + T₂)/2? If not, Why?
• Should the resultant temperature be between T₁ and T₁ only? If not, Why?

Answer:
1. Two bodies at different temperatures:

• The resultant temperature may not be necessarily \(\left(\frac{T_{1}+T_{2}}{2}\right)\). Because specific heat capacity is different for all substances.
• If heat is not lost to the surroundings, resultant temperature must lie in between T₁ and T₂.

Question 2.
A Patient is admitted to hospital. The temperature of the patient is measured by nurse and is found to be 97.6°F

1. What is meant by temperature?
2. Convert the temperature (97.6°F) is to centigrade

Answer:
1. Temperature is the degree of hotness.

2.

Question 3.
Why iron rims are heated red hot before being put on the cart wheels?
Answer:
The radius of iron rim is smaller than radius of cart wheel. When iron rim is heated, its radius increases due to thermal expansion. After rim has planted on the wheel, iron rim is allowed to cool. Then it fits tightly on the wheel due to thermal contraction.

Question 4.
How woolen clothes helps us in winter against cold?
Answer:
Wool is a heat insulator. Moreover, it contains a layer of dry air in it. This air, which is bad conductor of heat does not allow the body heat to escape & it also does not allow external cold to come in. The rough surface of woollen clothes absorbs more and reflects less heat falling on it.

Plus One Physics Thermal Properties of Matter Three Mark Questions and Answers

Question 1.
A Solid material is supplied with heat at constant rate. The temperature of the material is changing with the heat input as shown in figure.

1. What do the horizontal region AB and CD represent?
2. What does the slope of DE represent?
3. The slope of OA is greater than the slope of BC. What does this indicate?

Answer:

1. Latent heat of fusion, Latent heat of vaporization
2. It indicates that, the material is in vapour state
3. It indicates that latent heat of vaporization of the material is greater than the latent heat of fusion.

Plus One Physics Thermal Properties of Matter Four Mark Questions and Answers

Question 1.
A copper block of mass 2.5kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39Jg^-1 K^-1; latent heat of fusion of water = 335 Jg^-1).
Answer:
Mass, m = 2.5kg
= 2.5 × 10³ g;
Change in temperature, ∆T = 500°C
Specific heat, c = 0.39 Jg^-1K^-1;
Latent heat of fusion, L = 335Jg^-1
If m’ be the mass of ice melted, then m’L = mc ∆T
or m’ × 335 = 2.5 × 10³ × 0.39 × 500 2.5 × 103 × 0.39 × 500
or m’ = \(\frac{2.5 \times 10^{3} \times 0.39 \times 500}{335} \mathrm{g}\)
= 1.5 kg.

Question 2.
Heat from the sun reaches the earth.

1. Write the method of transmission of heat.
2. Thermal conductivity of Aluminium is 205 Jm^-1s^-1deg^-1. What do you mean by this?
3. Birds swell feathers in winter. Why?

Answer:
1. Radiation, conduction, convenction.

2. The coefficient of thermal conductivity of a substance is defind as the quauitity of heat conducted normally persecond through unit area of the substance per unit temperature gradient when the substance attains steady state.

3. By doing so the birds enclose air between the feathers. Air being a poor conductor, prevents the loss of heat from the body of the bird to the cold surroundings.

Question 3.
Two accidents are happened. The first one with water at 100°C and the second one with steam at 100°C.

1. Which is dangerous burn due to water at 100°C, and bum due to steam at 100°C? Why?
2. Latent heat of vapourisation of water to 536 cal/g. Explain the idea of latent heat of vopourisation.
3. Find the heat required to convert 1g of ice at 0°C to steam at 100°C is

Answer:
1. Burn due to steam is more dangerous, because heat content in steam is very high compared to 100°C water.

2. Latent heat of vapourisation is the amount of heat required to change the state of 1 kg water in to vapour.

3. Q = ML + MC∆Q + ML¹

= 716 cal.

Question 4.
When a fluid is heated, the particle rises up.

1. Name the phenomonon behind it.
2. Explain the formation of land breeze and sea breeze.

Answer:
1. Convection

2. During the day, land heats up more quickly than, water in lake (due to high specific heat capacity of water). The air on the surface of earth gets heated, expands, becomes less dense and rises up. The colder air (wind) replaces the space created by hot air.

It creates a sea breeze. At night the land loses its heat very quickly than water. So water remains more warmer at night.

Plus One Physics Thermal Properties of Matter NCERT Questions and Answers

Question 1.
A 10kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.19Jg^-1 K^-1.
Answer:
Power, P = 10 kW
= 10 × 10³W
= 10⁴W
Mass, m = 8 kg; Time, t = 2.5min = 150s
Specific heat, c = 0.91Jg^-1K^-1
= 0.91 × 10³Jkg^-1K^-1
Energy, Q = pt = 10⁴ × 150J = 1.5 × 10⁶J
It is given that 50% of energy is lost to the. surroundings. So, energy absorbed by the block is given by
Q = \(\frac{1}{2}\) × 1.5 × 10⁶J = 0.75 × 10⁶J
But Q = mc∆T
∴ ∆T

Question 2.
A brass boiler has a base area of 0.15m2 and thickness 1.0cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^-1 m^-1K^-1;
Heat of vaporisation of water = 2256 × 10³ J kg^-1.
Answer:
A = 0.15m², d = 10^-2m,
\(\frac{m}{t}=\frac{6}{60}\) kgs^-1 = 0.1kgs^-1,
K = 109J s^-1 m^-1 K^-1, L = 2256 × 10³ Jkg^-1,
θ₁ = ?,
θ₂ = 100°C

Question 3.
Explain why?
(a) A body with large reflectivity is a poor emitter.
(b) A brass tumbler feels much colder than a wooden tray on a chilly day.
(c) An optical pyrometer (for measuring high temperature) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
(d) The earth without its. atmosphere would be inhospitably cold?
(e) Heating systems based on circulation of steam are more efficient in warming a building than heating system based on circulation of hot water.
Answer:
(a) a body whose reflectivity is large would naturally absorb less heat. So, a body with large reflectivity is a poor.

(b) The thermal conductivity of brass is high i.e., brass „ is a good conductor of heat. So, when a brass tumbler is touched, heat quickly flows from human body to tumbler. Consequently, the tumbler appears colder. On the other hand, wood is a bad conductor, so, heat does not flow from the human body to the wooden tray in this case. Thus, it appears comparatively hotter.

(c) Let T the temperature of the hot iron in the furnace. Heat radiated per second per unit area, E = σT⁴

Students can Download Chapter 11 The p Block Elements Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Plus One Chemistry The p Block Elements One Mark Questions and Answers

Question 1.
The aqueous solution of borax is
a) Acidic
b) Alkaline
c) Neutral
d) Amphoteric
Answer:
b) Alkaline

Question 2.
Say TRUE or FALSE.
Boron in aqueous solution forms B³⁺ ion.
Answer:
False

Question 3.
Which of the halide of group 14 does not exist?
a) CF₄
b) Cl₄
c) SiF₄
d) Pbl₄
Answer:
d) Pbl₄

Question 4.
Orthoboric acid, H₃BO₃ is a
a) Protonic acid
b) Arrhenius acid
c) Lewis acid
d) Bronsted-Lowery acid
Answer:
c) Lewis acid

Question 5.
The zeolite used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerization is _________ .
Answer:
ZSM-5

Question 6.
Dry ice is __________ .
Answer:
Solid CO₂

Question 7.
Thermodynamically most stable allotrope of carbon is __________ .
Answer:
Graphite

Question 8.
The alkalimetal used in solar cells is __________ .
Answer:
Cs (Caesium)

Question 9.
Acidity is in the order
Answer:
BBr₃ > BCl₃ > BF₃

Question 10.
AlCl₃ fumes in most air because
Answer:
HCl is formed due to hydrolysis in moist air

Plus One Chemistry The p Block Elements Two Mark Questions and Answers

Question 1.
Silicon belongs to the carbon family. Graphite is an important allotrope of carbon. But Si does not form an analogue of graphite. What are the possible reasons?
Answer:
Silicon atom is much bigger in size than carbon. Si-Si bond energy is less than C-C bond energy and also Si does not form compounds in sp² hybrid state.

Question 2.
Boron is an element with atomic number 5.
a) Write down the electronic configuration of boron.
b) Mention any two uses of boron.
c) Write down some compounds of boron.
Answer:
a) 1s²2s²2p¹
b) To increase the hardness of the steel.
Used as the semiconductor in electronic devices.
c) Diborane, borax, orthoboric acid, boron trifluoride

Question 3.
Answer the questions, with the help of the following:

• Hard solid
• Melting point above 450 K.
• Four allotropic forms are known.
• Low electrical conductivity.
• Mass number! 3

1. Which is the element?
2. Write any two uses of this element.
3.
4. Write any three compounds of this element.
Answer:
1. Boron
2. To increase hardness of steel
As semiconductor in electric devices
3. 2B
4. Diborane, Borax, Boric acid

Question 4.
1. Write any two allotropic forms of carbon.
2. C + ½ O₂ → …………….
3. Write any two uses of carbon monoxide.
4. How is carbon dioxide produced?
5. What is the hardest element/form of an element in the world?
Answer:
1. Diamond, Graphite
2. CO (Carbon monoxide)
3. Reducing agent in metallurgy
Used in the manufacture of methanol
4. Carbon dioxide is formed by the complete combustion of carbon.
C + O₂ → CO₂
5. Diamond

Question 5.
From the compounds of group 14 elements write an appropriate example for each of the following:

Answer:
1-CO
2-SiO₂
3-SnCl₂
4-CCl₄

Question 6.
Diborane has an unusual structure. Justify the statement with figure.
Answer:
In diborane, each Batom uses sp³ hybrid orbitals for bonding. Out of the four sp³ hybrid orbitals on each B atom, one is without an electron. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Question 7.
Match the following:

Answer:
CO – ii,
N₂ – iii,
O₃ – iv,
Boron – i

Question 8.
BCl₃ fumes in moist air.
1. Give reason.
2. Write down a balanced equation that can reveal the answer.
Answer:
1. When water of moist air reacts with boron halide hydrogen chloride is formed which causes fumes,

2. BCl₃ + 3H₂O → H₃BO₃ + 3HCl

Question 9.
‘Generally, non-metal oxides are basic.’
1. Do you agree?
2. What do you meant by oxides?
3. Which are the different types of oxides?
4. Give examples for each type of oxides.
Answer:
1. No. Generally, non metallic oxides are acids.
2. The binary compounds formed by the combination of oxygen with metals or nonmetals are called oxides.
3. Acidic oxides, basic oxide, amphoteric oxide, neutral oxide.
4. Acidic oxide → SO₂, NO₂
Basic oxide → MgO
Amphoteric oxide → Al₂O₃, ZnO
Neutral oxide → CO, N₂O

Question 10.
Match the following:
1. Borane – H₃BO₃
2. Boric acid – Na₂B₄O₇.10H₂O
3. Borax – Amphoteric oxide
4. Al₂O₃ – Boron hydride
Answer:
1. Borane – Boron hydride
2. Boric acid – H₃BO₃
3. Borax – Na₂B₄O₇.10H₂0
4. Al₂O₃ – Amphoteric oxide

Question 11.
How diborane reacts with
1. Oxygen?
2. Water?
Answer:
1. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen releasing an enormous amount of energy.
B₂H₆ + 3O₂ → B₂O₂ + 3H₂O; ∆_rH^Θ -1976 kJ mol^-1

2. Diborane is readily hydrolysed by water to give boric acid.
B₂H₆(g) + 6H₂O(l) → 2B(OH)₃(aq) + 6H₂(g)

Question 12.
1. CCl₄ cannot be hydrolysed. Give reason.
2. Draw the structure of the dimer of AlCl₃.
Answer:
1. Carbon has no d orbitals to accommodate the lone pair of electrons from oxygen atom of H₂0.
2.

Question 13.
1. Draw the structure of boric acid.
2. Starting from borax how will you prepare boric acid? (Write the chemical equation).
Answer:
1.

2. Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃.

Question 14.
CO₂ is a gas but SiO₂ is a solid. Give reason.
Answer:
CO₂ exists as discrete molecules due to the formation of pπ -pπ double bond between carbon and oxygen. But SiO₂ has a three dimensional network structure in which each Si atom is covalently bonded in a tetrahedral manner to four oxygen atoms.

Question 15.
1. What are zeolites?
2. What is ZSM-5?
Answer:
1. Zeolites are aluminosilicates with three-dimensional network structure in which Al atoms replace few Si atoms. Cations such as Na⁺, K⁺ or Ca²⁺ balance the negative charge of aluminosilicate anion.

2. ZSM-5 is a zeolite catalyst used in petrochemical industry to convert alcohols directly into gasoline.

Plus One Chemistry The p Block Elements Three Mark Questions and Answers

Question 1.
Explain the following:
1. Allotropy
2. Coke and Charcoal
Answer:
1. The phenomenon of existence of an element in two or more forms, which have different physical properties but almost similar chemical properties, is called allotropy and the different forms are called allotropes.

2. Coke and Charcoal are amorphous forms of carbon. Coke is formed-by the destructive distillation of coal. It is used as a fuel and also as a reducing agent in metallurgy.

Question 2.
Give justifications.
1. The first ionisation enthalpy of carbon is greater than that of boron, whereas the reverse is correct for the second ionisation enthalpy.
2. Graphite is a better lubricant on moon than that on earth.
Answer:
1. This is because carbon has greater nuclear charge. For second ionisation enthalpy, an electron is to be removed from 2p of carbon while in boron the second electron is in 2s orbital. Removal of a 2s electron is more difficult due to the high penetrating power of 2s orbital.

2. Graphite contains hexagonal sheets of carbon held together by weak van der Waals’ forces and low density. In the moon, the different layers experience less weight due to less gravitational force and the density is still reduced. The lighter layers can easily slide over another to make graphite more lubricating on the moon than on earth.

Question 3.
1. Identify the compound with the following structure:

2. Mention the angles a & b.
3. Write down the corresponding elements in the figure.
1……………..
2……………..
3……………..
4……………..
5……………..
6……………..
Answer:
a) Diborane(B₂H₆).
b) a-97°
b-120°
c) 1-H, 2-H, 3-B, 4-B, 5-H, 6-H

Question 4.
The simplest boron hydride is diborane.
1. Write down the molecular formula of diborane.
2. Distinguish between terminal hydrogen and bridging hydrogen atoms of diboran.
Answer:
1. B₂H₆

2. In diborane, four hydrogen atoms and two boron atoms are present in a single plane. These hydrogen atoms are called terminal atoms. The terminal B-H bonds are regular two centre-two electron bonds. The remaining two hydrogen atoms above and below this plane are called bridging hydrogen atoms. The two bridge bonds (B-H-B) are three centre-two electron bonds.

Question 5.
1. How is orthoboric acid prepared?
2. Account for the acidic nature of orthoboric acid.
Answer:
1. Orthoboric acid is prepared by acidifying an
aqueous solution of borax.
Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃

2. Orthoboric acid is a weak nonobasic acid. It acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)₃ + 2HOH → [B(OH)₃]^– + H₃O⁺

Question 6.
1. How is diborane prepared in the laboratory?
2. BCl₃ is a good Lewis acid. Why?
Answer:
1. Diborane can be conveniently prepared in the laboratory by the oxidation of sodium borohydride with iodine.
2NaBH₄ + l₂ → B₂H₆ + 2Nal + H₂

2. In BCl₃, the central boron atom contains only six electrons. Hence, it has the tendency to accept electrons and acts as a Lewis acid.

Question 7.
1. Name the allotropes of carbon.
2. Carbon monoxide is highly poisonous. Do you agree? Justify.
Answer:
1. Graphite, diamond and fullerene.

2. I agree with this statement. Because, CO has strong and reversible binding with haemoglobin resulting in the formation of carboxyhaemoglobin. This reduces the amount of haemoglobin available in blood for oxygen transport. This causes laboured respiration, muscle weakness and even death.

Question 8.
1. Diamond is hard and non conducting while graphite is soft and conducting. Why?
2. Explain the action of heat on boric acid.
3. What is inorganic benzene? How is it formed?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. Due to this closely packed arrangement, it is hard. Since, there are no free electrons it is an insulator. But in graphite, the carbon atoms are in sp² hybridised state. It is a good conductor due to the presence of delocalised electrons between the layers. Graphite cleaves easily between the layers. Therefore, It is very soft and slippery.

2. On heating, orthoboric acid above 370 K forms metaboric acid, HBO₂ which on further heating yields boric oxide, B₂O₃.

3. B₃N₃H₆ is called Inorganic benzene. It is obtained by heating diborane with ammonia.

Question 9.
1. Explain the difference in properties of diamond and graphite on the basis of their structures.
2. How do you explain the lower atomic radius of gallium as compared to aluminium?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. As a result of this hybridisation, a closely packed arrangement is present in diamond and it explains the hardness of diamond. Due to the absence of electrons, diamond is an insulator.

In graphite, the carbon atoms are in sp² hybridised state. As a result of this hybridisation, graphite has a layered structure with hexagonal rings. Each layer can slide over the other which explains the lubricating property of graphite.

2. This can be explained on the basis of the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic redius of gallium (135 pm) is less than that of aluminium (143 pm).

Question 10.
1. Boron resembles silicon in many of its properties. What is this resemblance generally known as?
2. What is dry ice? What is it used for?
3. What are silicones? How do they differ from silicates?
Answer:
1. Diagonal relationship

2. Carbon dioxide can be obtained as a solid in the form of dry ice, by allowing the liquified CO₂ to expand rapidly.
Dry ice is used as a refrigerant for ice-cream and frozen food.

3. Silicones are a group of organosilicon polymers containing R₂SiO repeating units.
Silicates are minerals with SiO₄^4- as the basic structural unit. In silicates either the discrete unit is present or a number of such units are joined together via corners by sharing 1, 2, 3 or 4 oxygen atoms per silicate units.

Plus One Chemistry The p Block Elements Four Mark Questions and Answers

Question 1.
Match the following:

Answer:
1 – c – f
2 – a – h
3 – b – g
4 – d – e

Question 2.
a) Elements of group 13 are
1) Al, Cr, Cd, Ga, Ti
2) B, C, Si, Ga, Ti
3) B, Al, Ga, Ti, In
4) B, Al, Cr, Ca, Ti
b) Which of the following is not a mineral of boron? (borax, bauxite, colemanite, tincal)
c) In the reaction between NH₃ and BF₃ ammonia acts as
(Lewis base, Lewis acid, brownsted base)
d) Diamond and Graphite are carbon’s
(isotropic forms, allotropic forms, isotopic forms, amorphous form)
Answer:
a) 3) B, Al, Ga, TI, In
b) Bauxite
c) Lewis base
d) allotropic forms

Question 3.
The hydrides of boron are called boranes.
1. How diborane reacts with ammonia?
2. Account for the exceptional hardness of diamond.
Answer:
1. Diborane reacts with ammonia to give B₂H₆.2NH₃
initially, which can be formulated as [BH₂(NH₃)₂]⁺[BH₄]^–. Further heating gives borazine, B₃N₃H₆ known as ‘inorganic benzene’ in view of its ring structure with alternate B-H and N-H groups,

2. In diamond, all the carbon atoms are in sp3 hybridised state. As a result of this hybridisation, a rigid three dimensional network of carbon atom is generated with directional covalent bonds throughout the lattice. It is very difficult to break this network structure.

Plus One Chemistry The p Block Elements NCERT Questions and Answers

Question 1.
Why does boron trifluoride behave as a Lewis acid? (2)
Answer:
The B atom in BF₃ has only 6 electrons in the valence shell and thus needs two more electrons to complete its octet. Therefore, it easily accepts a pair of electrons from nucleophiles such as F^–, NH₃, (C₂H₅)₂O, RCH₂OH etc. and thus behaves as a Lewis acid.

Question 2.
Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon? (2)
Answer:
Due to increase in atomic size and screening effect the force of attraction of the nucleus for the valence electron decreases considerably in Si as compared to C. As a result, there is a phenomenal decrease in ionisation enthalpy from carbon to silicon.

Question 3.
Consider the compounds, BCl₃ and CCl₄. How will they behave with water? (2)
Answer:
The B atom in BCl₃ has only six electrons in the valence shell and hence is an electron-deficient molecule. It easily accepts a pair of electrons donated by water and hence BCl₃ undergoes hydrolysis to form boric acid (H₃BO₃) and HCl.
BCl₃ + 3H₂O → H₃BO₃ + 3HCl

In contrast, C atom in CCl₄ has 8 electrons in the valence shell. Therefore, it is an electron-precise molecule. As a result, it neither accepts nor donates a pair of electrons. In simple words, it does not accept a pair of electrons from H₂O molecule and hence CCl₄ does not undergo hydrolysis in water.

Question 4.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF₃ is buddled through. Give reasons. (3)
Answer:
Anhydrous HF is a covalent compound and is strongly H-bonded. Therefore, it does not give free F^– ions and hence AlF₃ does not dissolve in HF. In contrast, NaF is an ionic compound and hence F^– ions are easily available. As a result, it combines with AlF₃ to form the soluble complex.
3NaF + AlF₃ → Na₃[AlF₃]

On bubbling gaseous BF₃, AlF₃ is precipitated. It is because BF₃ is a stronger Lewis acid than AlF₃. This is attributed to smaller size and higher electronegativity of Boron. As a result of this B has much higher tending to form complexes than Al. Therefore, when BF₃ is added to the above solution, AlF₃ gets precipitated.
Na₃[AlF₆] + 3BF₃ → 3Na[BF₄] + AlF₃(s)

Question 5.
In some of the reactions, thallium resembles aluminium, whereas in others it resembles with group 1 metals. Support this statement by giving some evidences. (3)
Answer:
Aluminium shows a uniform oxidation state of+3 in its compounds. Like aluminium, thallium also shows +3 oxidation state in some of its compounds like TlCl₃, Tl₂O₃, etc. Al is known to form octahedral complexes like [AlF₆]^3-. Similarly, Tl also forms octahedral complexes as [TlF₆]^3-.

Thallium also resembles group 1 metals. Like group 1 metals which show a stable oxidation state of +1 in their compounds Tl, due to inert pair effect, also shows +1 oxidation state in some of its compounds such as Tl₂O, TlCl, TlClO₄, etc. Similarly, like group 1 oxides, Tl₂O is strongly basic.

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

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Plus One Physics Mechanical Properties of Fluids One Mark Questions and Answers

Question 1.
Water is flowing through a very narrow tube. The velocity of water below which the flow remains a streamline flow is known as
(a) relative velocity
(b) terminal velocity
(c) critical velocity
(d) particle velocity
Answer:
(c) critical velocity
Critical velocity is that velocity of liquid flow, upto which the flow of liquid is a streamlined and above which its flow becomes turbulent.

Question 2.
Bernoulli’s equation for steady, non-viscous, imcompressible flow expresses the
(a) conservation of angular momentum
(b) conservation fo density
(c) conservation of momentum
(d) conservation of energy
Answer:
(d) conservation of energy

Question 3.
When cooking oil is heated in a frying pan, the oil moves around in the pan more easily when it is hot. The main reason for this is that with rise in temperature, there is a decrease in
(a) surface tension
(b) viscosity
(c) angle of contact
(d) density
Answer:
(d) density

Question 4.
At what temperature density of air is maximum?
Answer:
4°C

Question 5.
A thin glass plate is lying on a wet marble floor, It is difficult to pull the glass plate because of
(i) surface tension
(ii) Viscosity
(iii) friction
(iv) atmosphere
(v) Gravity
Answer:
(i) Viscosity

Question 6.
Why do clouds float in the sky?
Answer:
Zero terminal velocity

Question 7.
A spinning cricket ball does not follow parabolic path. Why?
Answer:
Due to the magnus effect.

Question 8.
The deep water runs slow. Why?
Answer:
According to equation of continuity velocity is inversly proportional to velocity (AV = constant). Hence deep water runs slow.

Question 9.
Why dust generally settles down in closed room?
Answer:
The dust particles (tiny spheres) acquire terminal velocity as it fall through air. The terminal velocity is directly proportional to square of radius. Hence terminal velocity of dust particle is very small. So they settle down gradually.

Question 10.
Why more viscous oil is used in summer than in winter in scooters?
Answer:
The viscosity decreases with increase in temperature.

Question 11.
Why is sand drier than clay?
Answer:
Capillary action

Question 12.
Why cotton dress is preferred in summer?
Answer:
Cotton dresses have fine pores which act as capillaries for the sweat.

Question 13.
Why oil is poured to calm the sea?
Answer:
When oil is poured in water, the surface tension of water is reduced and water spreads over large area of sea.

Question 14.
How plants draw water from ground?
Answer:
The capillary action.

Question 15.
How do insects run on the surface of water?
Answer:
Because of surface tension, the surface of water behaves like stretched membrane hence it can support weight of small insects.

Question 16.
How ploughing a field helps to retain moisture?
Answer:
When field is ploughed, capillaries are broken and hence water can not rise up and retains moisture.

Question 17.
Hot soup tastes better than cold soup. Why?
Answer:
The surface tension of hot soap is less compared to cold soap. So hot soap spreads larger area.

Plus One Physics Mechanical Properties of Fluids Two Mark Questions and Answers

Question 1.
Remya found that a piece of metal weighs 210 g in air and 180 g when it is immersed in water. Determine the density of the metal piece.
Answer:
Relative density,
R.D = \(\frac{\text { Weight in air }}{\text { Loss of Weight in water }}\)
= \(\frac{210}{30}\) = 7.

Question 2.
Why is hot soup tastier than cold one?
Answer:
When temperature increases, the surface tension of soap decrease. Hence hot soap can enter into tiny pours of taste buds.

Question 3.
Why straws are used to drink soft drinks?
Answer:
When we suck the straw, pressure inside straw becomes lower than atmospheric pressure. This pressure difference cause the soft drink to rise through the straw.

Question 4.
Why new earthen pots keeps water more cool than old earthen pots?
Answer:
The capillaries of old earthen pots will get blocked with passage of time. For new earthen pots, water oozes out through capillaries, gets evaporated at the surface and makes it cool.

Plus One Physics Mechanical Properties of Fluids Three Mark Questions and Answers

Question 1.
Air is blown in between two pith balls suspended freely.

1. What will happen to the balls?
   • They repel each other
   • They attract each other
   • They start oscillating
   • They remain in their initial position They fall on the ground
2. Give your explanation

Answer:

1. They attract each other
2. When air is blown in between two pith balls, the pressure between the balls decreases. Due to this decrease in pressure between the balls, they attract each other.

Question 2.
A child dipped two identical capillary tubes, one in a beaker containing mercury. He observed that water and mercury have risen through the tubes to a certain heights.

1. Name this phenomenon.
2. What difference did he observe in the shape of the meniscus of the two liquids in the tubes?
3. If he plots a graph connecting the radius of the capillary tube and capillary height, what will be the shape of the graph?

Answer:
1. Capillary rise.

2. The shape of the water meniscus in the tube becomes concave upwards. But the shape of mercury measures in the tube become convex upward.

3. h α \(\frac{1}{r}\), this is in the form y α \(\frac{1}{x}\)
Hence when we draw graph between ‘h’ and ‘r’ we get a graph of hyperbola.

Question 3.
Bernoulli’s theorem is a consequence of energy conservation principle. Using this theorem explain the working of atomiser.
Answer:
Atomiser (application of Bernoulli’s theorem)

Atomizer 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.

Question 4.
Surface tension is numerically equal to the surface energy.

1. Difine surface tension.
2. Derive an expression for the rise of liquid in a capillary tube.

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

2. When a drop is split into tiny droplets, the surface area increases. So work has to be done for splitting the drop. Let R be radius of the drop and r the radius of the droplets: R = 1 × 10^-3m
surface area of the drop = 4πR²
= 4π × (1 × 10^-3)²
= 4π × 10^-6m²
Volume of the drop = Volume of 10⁶ droplets

∴ Surface area of million droplets =10⁶ × 4π²
= 10⁶ × 4π(1 × 10^-5)²
= 4π × 10^-4 m²
∴ Increase in surface area = 4π × 10^-4 – 4π × 10^-6
= 3.96π × 10^-4m²
∴ Energy expended = 3.967π × 10^-4 × S
= 3.96π × 10^-4 × 72 × 10^-3J
= 8.95 × 10^-5J.

Question 5.
Match the following

Answer:
1 – C, 2 – f, 3 – e, 4 – a, 5 – b, 6 – d.

Question 6.
Give reasons for the following cases.

1. It is easier to swim in sea water than in river walls.
2. The passangers are advised to remove ink from pen while going up in an aeroplane.

Answer:
1. The density of sea water is more than that of river water due to the presence of salt. Hence sea water offers more upthrust and only a very small portion of human body will be in sea water compared to river water.

2. In ink pen, ink is filled in atmospheric pressure. As we go higher pressure decreases and hence ink will have a tendancy to come out in order to equalise the pressure.

Plus One Physics Mechanical Properties of Fluids Four Mark Questions and Answers

Question 1.
A large tank containing water has a small hole near the bottom of the tank 1.5 m below the surface of water.

1. What is the velocity of the water flowing from the hole?
2. Explain the principle used in deriving the velocity of water flowing from the hole.
3. Where must a second hole to be drilled so that the velocity of water leaving this hole is half of water flowing through the first hole.

Answer:
1. Velocity of water flowing through the hole
u = \( \sqrt{2 g h}\)
= \(\sqrt{2 \times 10 \times 1.5}\) = 5.47m/s.

2. 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)

Mathematically Bernoulli’s theorem can be written as
P + \(\frac{1}{2}\)ρv² + ρgh = constant.

3.

h₂ = \(\frac{1.5}{4}\) m = 0.375 m, from the top side of tank.

Question 2.
Rain drops have an average size of 1 mm when it is formed at the upper atmosphere.

1. Why the velocity of the rain drop is uniform?
2. Derive an expression for the terminal velocity of the drop in terms of coefficient of viscosity of air.
3. If the size of the rain drop become half, then what happens to its terminal speed?

Answer:
1. Due to viscous force acting on the raindrop, it moves with uniform speed.

2. Viscous force, boyancy force and weight of the body
Expression for terminal velocity:
Consider a sphere of radius ‘a’ densitity σ falling through a liquid of density a and viscocity η. 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 viscous force is equal to the weight of the body in the medium, the net force on the body is zero. It moves with a constant velocity called the terminal velocity.
The weight of a body in a medium,

When body has terminal velocity, we can write.

Question 3.

1. Fill in the blanks using the word from the list appended with each statement.
   • Viscosity of gases_____with temperature (increase/decrease)
   • For a fluid in steady flow, the increase in flow speed at a constriction follows from_____ (conservation of mass/Bernoulli’s theorem)
   • The working of a hydraulic lift is based on (Pascal’s Law/ principle of Conservation of Energy)
   • Small insects can walk over the surface of water. It is due to the_____(surface tension of water/viscosity of water)
2. A girl dips a thin capillary tube in water. Water rises through it.
   • Name the phenomenon.
   • How does this rise vary with the diameter of the tube?

Answer:
1. Fill in the blanks :

• increases
• Conservation of mass
• Pascals law
• Surface tension

2. A girl dips a thin capillary tube in water:

• Capillary rise
• h α \(\frac{1}{r}\) ie. when diameter of tube increases, the. capillary rise decreases.

Question 4.
The schematic diagram of a sprayer or atomiser is given below.

1. Name the principle of working of this device from the following:
   • Surface tension
   • Viscosity
   • Bernoulli’s principle
   • Archimedes’ principle
2. Write its mathematical expression.
3. Wings of an aeroplane are curved outwards while flattened inwards. Why?

Answer:
1. Bernollis principle

2. P + \(\frac{1}{2}\)ρv² + ρgh = constant.

3. When the aeroplane moves forward, the air blown in the form of stream lines over the wings of aeroplane is shown figure.

As the upper surface of wing is more curved than its lower surface, the speed of air above the wings is larger than the speed of the air below the wings.

Hence the pressure above the wings becomes less than the pressure below the wings. Due to this pressure difference the aeroplane will get upward force to overcome gravitational force.

Question 5.
During windstorms, roofs of certain houses are blown off without damaging other parts of the houses.

1. Name the theorem which explains this phenomenon.
2. State the theorem.
3. Explain this phenomenon on the basis of this theorem.

Answer:

1. Bernoulli’s theorem
2. For a small amount of liquid in stream line flow, between two points, the total energy is constant.
3. When windstorm blown off, the pressure on the top side of roof decreases. Hence a pressure difference is developed in between roof. Due to this pressure difference, roof of certain houses are blown off.

Question 6.
Two thin evacuated (one end closed) glass take A and B are carefully immersed in a beaker containing mercury such a way that there is no chance to get air in to the tubes. A is stand vertically and B is making an angle θ with the vertical.

1. Is any rise of mercury in the tubes?
2. Is any height difference of mercury levels in tube A and B? Justify your answer.
3. When the doctors are measuring body pressure, it is advisable to lie on a table. Why?

Answer:

1. Yes
2. No. Pressure is same at same level. To get same pressure, height of mercury becomes same.
3. When we lie on the table, the pressure of our body will be same at all points.

Question 7.
A small metal sphere is falling through a caster oil.

1. Name the forces acting on the metal sphere?
2. Which of these forces change? Why?
3. Name the velocity of the sphere when the unbalanced force on it is zero?
4. Write down the expression for this velocity in terms of coefficient of viscosity?

Answer:

1. forces acting on the metal sphere:
   • Weight of the body (mg)
   • Buoyant force or up thrust
   • Viscous force
2. Viscous force. Viscous force is the friction offered by the liquid. It is a self adjusting force.
3. Terminal velocity
4. Terminal velocity, V = \(\frac{2}{9} a^{2}\left(\frac{f-N}{\eta}\right) g\).

Plus One Physics Mechanical Properties of Fluids Five Mark Questions and Answers

Question 1.
A capillary tube when dipped into water, it is commonly observed that water will rise through the tube.

1. Which of the following is responsible for this?
   • Gravitational force
   • Viscous force
   • Nuclear force
   • Surface tension
   • Elastic force
2. Derive an expression for the capillary rise.
3. If the radius of the tube becomes doubled, then what happens to the height of water column in the tube?

Answer:
1. Surface tension.

2.

Consider a capillary tube of radius ‘a’ dipped in a liquid of density ρ and 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
p₁ – p₀ = \(\frac{2 \mathrm{S}}{\mathrm{R}}\)
Where R is the radius of the concave meniscus. The tangent to the meniscus at the point A makes an angle θ with the wall of the tube.
In the right angled triangle ACO

substituting the values of R in the equation (1)

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 = p_i, the pressure over the concave meniscus and pressure at N = p_o + hρg
∴ P_i = P_o + hρg
or P_i – P_o = hρg ……..(3)
From equations (2) and (3), we get

3. We know

The capillary rise decreases to half of the original value.

Question 2.

1. Find the odd one out and justify your answer Atomiser, venturi meter, aeroplane, hydraulic lift
2. Mention one use of venturi meter.
3. Explain the working of the odd one which you have selected in question (a)

Answer:
1. Hydraulic lift – It is based on pascals law.

2. Venturimeter can be used to find velocity of flow of fluid through a pipe.

3.

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.

Using this method we can lift heavy load by applying small force.

Question 3.
When a capillary tube of radius ‘r’ is dipped in water, the water rises through it up to height ‘h’.

1. Which of the following is responsible for the above phenomenon?
   • Viscous force
   • elastic force
   • surface tension
   • gravitational force
   • negative force
2. To what height will water rise in a glass tube with a bore of radius 0.1 mm (take the angle of contact of glass with 0°, surface tension S = 0.0728 N/m)
3. If the length of tube is less that the length of capillary rise, will it overflow. Justify your answer.

Answer:
1. Surface tension

2.

3. The water will never overflow. If the tube is of insufficient length, the radius of curvature of liquid meniscus goes on increasing, making it more and more flat till water is in equilibrium.

Question 4.
A steel ball of radius 1 mm is falling vertically through a tank of oil at 30°C.

1. After some time the ball attains a constant velocity called_____
2. What are the forces acting on the ball and give their directions?
3. Write down the expression for resultant force acting on the ball?)
4. If the density of oil is 2 × 10³kg/m³, density of steel is 8 × 10² Kg/m³ and ‘η’of oil 2NS/m², What will be the constant velocity attained by the ball?

Answer:
1. Terminal velocity.

2. Weight of body (down ward), bouyanant force (up ward), Viscous force (upward).

3. Resultant force = weight of body – buoyant force.

4. Terminal velocity,

Plus One Physics Mechanical Properties of Fluids NCERT Questions and Answers

Question 1.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer:
Force, F = Weight of girl
= mg = 50 × 9.8N = 490N
Radius, r = 0.5 × 10^-2m
Area A = πr² = \(\frac{22}{7}\)(0.5 × 10^-2)² m²
Pressure

Question 2.
Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmospheric pressure.
Answer:
p = hρg, h = \(\frac{p}{\rho g}=\frac{1.01 \times 10^{5}}{984 \times 9.8}\)m = 10.47m.

Question 3.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 of spirit in the other. What is the specific gravity of spirit?

Answer:
Since the mercury columns in the two arms are at the same level,
∴ pressure due to water column = pressure due to spirit column
∴ h_wρ_wg = h_sρ_sg
or h_wρ_w = h_sρ
But h_w = 10 cm,
ρ_w = 1 gcm^-3,
h_s = 12.5cm
∴ 10 × 1 = 12.5 × ρ_s
or ρ_s = \(\frac{10}{12.5}\)gcm^-3
= 0.8cm^-3
∴ Specific gravity of spirit = 0.8.

Question 4.
Figs, (a) and (b) refer to the steady flow of a non-viscous liquid. Which of the two figures is incorrect? Why?

Answer:
Fig (a) is incorrect. This is because at a constriction (ie., where the area of cross-section of the tube is smaller), the flow speed is larger due to mass conservation. Consequently, pressure there is smaller according to Bernoulli’s equation. We assume the fluid to be incompressible.

Question 5.
What is the prssure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^-1Nm^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Excess pressure = \(\frac{2 \sigma}{R}\)

Total pressure = 1.01 × 10⁵ + \(\frac{2 \sigma}{R}\)
= 1.01 × 10⁵ + 310
= 1.0131 × 10⁵Pa
Since data is correct upto three significant figures. We should write total pressure inside the drop as 1.1 × 10⁵Pa.

Question 6.
During blood transfusion, the needle is inserted in a vein where the guage pressure is 2000 Pa, at what height must the blood container be placed so that blood may just enter the vein? Given: density of whole blood = 1.06 × 10³kgm^-3
Answer:
Guage pressure,
p = hρg, h

= 0.19m.

Students can Download Chapter 8 Arrays Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Plus One Arrays One Mark Questions and Answers

Question 1.
From the following which is not true for an array.
(a) It is easy to represent and manipulate array variable
(b) Array uses a compact memory structure
(c) Readability of program will be increased
(d) Array elements are dissimilar elements
Answer:
(d) Array elements are dissimilar elements

Question 2.
Consider the following declaration.
int mark(50) Is it valid? If no give the correct declaration
Answer:
It is not valid. The correct declaration is as follows, int mark[50]. Use square brackets instead of parenthesis

Question 3.
Consider the following declaration.
int mark[200]
The index of the last element is ________.
Answer:
199

Question 4.
Consider the following declaration int mark[200]. The index of the first element is ________.
Answer:
0

Question 5.
Consider the following int age[4] = {15,16,17,18};
From the following which type of initialisation is this.
(a) direct assignment
(b) along with variable declaration
(c) multiple assignment
(d) None of these
Answer:
(b) along with variable declaration

Question 6.
From the following which is used to read and display array elements.
(a) loops
(b) if
(c) switch
(d) if else ladder
Answer:
(a) loops

Question 7.
Write down the corresponding memory consumption in bytes

1. int age[10] = ——-
2. charname[10] = ——-
3. int age[10][10]= ——–

Answer:

1. 2 × 10 = 20 bytbs (2 bytes for one integer)
2. 1 × 10 = 10 (one byte for each character)
3. 2 × 10 ×10 = 200 (2 × (100 elements))

Question 8.
A two dimensional array is having ________ number of subscript
(a) one
(b) two
(c) three
(d) none of these
Answer:
(b) two

Question 9.
Consider the following
int age[10][10];
In this array how many elements will contain.
Answer:
10 × 10 = 100 elements

Question 10.
Consider the following int age[4] = {12, 13, 14};
cout<<age[3];
What will be the output?
(а) 14
(b) 12
(c) 13
(d) 0
Answer:
(d) 0

Question 11.
The elements of 2 dimensional array can be read using _________ loop.
Answer:
nested loop

Question 12.
________ is the process of reading/visiting elements of an array.
Answer:
traversal

Question 13.
Anjali wants to read the 10 marks that already stored in an array and find the total. This process is known as ________.
(a) insertion
(b) deletion
(c) traversal
(d) linear search
Answer:
(c) traversal

Question 14.
Mani wants to check a number from 100 numbers already stored in an array. This process is called ___________.
(a) insertion
(b) deletion
(c) searching
(d) None of these
Answer:
(c) searching

Question 15.
Divide and conquer method used in ____________ search.
Answer:
binary

Question 16.
“Each element of the array is composed with value to be searched from the beginning of the array”. This method is adopted by _____________ search.
Answer:
linear search

Question 17.
____________ search requires a sorted array as input.
Answer:
binary search

Question 18.
___________ searching is slower for larger array.
Answer:
linear searching

Question 19.
____________ is a simple sorting algorithm.
(a) bubble sort
(b) selection sort
(c) linear search
(d) binary search
Answer:
(a) bubble sort

Question 20.
Adjacent elements are checked and inter changed in _____________ sort.
Answer:
bubble sort

Question 21.
The array is divided into sorted part and unsorted part in ____________ sort.
selection.

Question 22.
The elements of an array of size ten are numbered from _____ to _______.
Answer:
0 to 9

Question 23.
Element mark[6] is which element of the array?
(a) The sixth
(b) the seventh
(c) the eighth
(d) impossible to tell
Answer:
(b) the seventh

Question 24.
When a multidimensional array is accessed, each array index is ________.
(a) Separated by column.
(b) Surrounded by brackets and separated by commas.
(c) Separated by commas and surrounded by brackets.
(d) Surrounded by brackets,
Answer:
(d) surrounded by brackets

Question 25.
You have used a 2D array with the Name Mat representing a matrix. Write the C++ expression to access the 3rd element in the 2nd row.
Answer:
mat[1] [2];

Question 26.
Write a C++ statement that defines a string variable called ‘name’ that can hold a string of upto 20 characters.
Answer:
char name[21j;

Question 27.
Given some array declaration. Pick the odd man out.
Float a[+40], int num[0-10], double [50], char name[50], amount[20] of float.
Answer:
char name[50]. It is a valid array decalaration the remaining are not valid.

Question 28.
__________ is a collection of elements with same datatype.
Answer:
Array

Question 29.
int num[10];
The above C++ statement declares an array named num that can store maximum ________ integer numbers. (SEP-2015 (IMP) (1)
(а) 9
(b) 10
(c) N
(d) none of these
Answer:
(b) 10

Question 30.
Declare a two dimensional array to store the elements of a matrix with order 3 × 5. (MARCH-2016) (1)
Answer:
int m[3] [5]; or float m[3] [5]

Question 31.
_________ search method is an example for ‘divide and conquer method’. (SEP-2016) (1)
Answer:
Binary

Question 32.
Read the following C++ statement:
int AR[10]
How many bytes will be allocated for this array? (1)
Answer:
To store an integer 4 bytes needed so AR[10] needs 10 × 4 = 40 bytes

Question 33.
Find the value of score [4] based on the following declaration statement. int score [5] = {988,87,92,79,85}; (MARCH-2017) (1)
Answer:

Hence (158)₁₀ = (9E)₁₆

Plus One Arrays Two Mark Questions and Answers

Question 1.
Whether the statement char text[] = “COMPUTER”; is True/False? Justify.
Answer:
It is a single dimensional array. If the user doesn’t specify the size the operating system allocates the number of characters + one (for null character for text) bytes of memory. So here OS allocates 9 bytes of memory.

Question 2.
Given a word ” Mathematics”. How will you locate this word in a standard dictionary. What are the steps used for this purpose
Answer:
In the standard dictionary the words are arranged in sorted order so binary search method is more suitable to find the word “MATHEMATICS”.

Question 3.
How can you find out the word “MATHEMATICS” from a text book. Which is the most convenient method? Justify your answer.
Answer:
In a text book the words are not arranged in sorted order so linear search method is convenient.

Question 4.
Consider the statement char str[] = “PROGRAM” What will be stored in last location of this array. Justify.
Answer:
The last location is the null character(\0) because each string must be appedend by a null character.

Question 5.
Explain different array operations in detail.
Answer:

1. Traversal: All the elements of an array is visited and processed is called traversal
2. Search: Check whether the given element is present or not
3. Sorting: Arranging elements in an order is called sorting

Question 6.
Read the following C++ statement: int MAT[5][4];

1. How many bytes will be allocated for this array?
2. Suppose MAT[4][4] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the’ sum of all the elements in the array. (MARCH – 2015) (1)

Answer:

Question 7.
Declare an array of size 5 and intitialize it with the numbers 8, 7, 2, 4 and 6. (MARCH-2016) (2)
Answer:
int n[5] = {8, 7, 2, 4, 6};

Question 8.
Predict the output of the following C++ program.
#include <iostream.h>
int main()
{
int array[] = {1, 2, 4, 6,7,5};
for (int n =1; n<=5; n++)
array [n] = array [n – 1 ];
for (n = 0; n <= 5; n++)
cout<< array [n];
return 0;
}
Answer:
1, 1, 1, 1, 1, 1

Question 9.
Suppose M[5][5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (SCERT SAMPLE – II) (2)
Answer:
for (i = 0; i < 5, i++)
for (j = 0; j < 5; j++)
if (i == j)
S = S + M[i][j];

Question 10.

1. Write C++ program for sorting a list of numbers using Bubble Sort Method. (2)
2. Write the different passes of sorting the following numbers using Bubble Sort.
   32, 21, 9, 17, 5

Answer:
1. Example for Bubble sort is given below

2. Bubble sort (ascending order):
Consider the first 2 elements of the numbers and compare it. 32 is greater than 21 then interchange both of them. Then the numbers are
21, 32, 9, 17, 5
Next compare 32 and 9. Here 32 is greater so interchange both of them. Then the numbers are
21, 9, 32, 17, 5
Next compare 32 and 7. Then interchange and the numbers are
21, 9, 17, 32, 5
Next compare 32 and 5. Then interchange and the numbers are
21, 9, 17, 5, 32.
After the first phase the largest number is on the right side. Similarly do the remaining. At last we will get the result as follows.
5, 9, 17, 21, 32

Question 11.
Suppose M[5] [5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (MARCH-2017)
Answer:

Plus One Arrays Three Questions and Answers

Question 1.
Suppose you are given Total mark of 50 students in a class

1. How will you store these values using ordinary variable?
2. Is there any other efficient way to store these values? Give reason.

Answer:
We have to declare 50 variables individually to store total marks of 50 students. It is a laborious work. Hence we use array, it is an efficient way to declare 50 variables. With a single variable name we can store multiple elements.
eg: int mark[50]. Here we can store 50 marks. The first mark is in mark[0], second is in mark[1]…. etc the fiftieth mark is in mark[49],

Question 2.
Four friends are in a queue for watching a movie. How can you locate a particular person from that queue? Write all the steps sequentially.
Answer:
Locate a particular person from a queue can be carried out by using two ways.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Question 3.
Total mark of 50 students in a class are given in an array. A bonus of 10 marks is awarded to all of them. Write the program code for making such a modification.
Answer:

Question 4.
In School Assembly, try to arrange all the students in your class according to the increasing order of their heights separately for male/female students. Smallest one will be in front and tallest one at the last position. Write sequential steps for doing this, in two different methods.
Answer:
Arranging elements from smallest to largest (ascending) or largest to smallest(descending) is called sorting. There are two sorting methods
1. Bubble sort:
It is a simple sorting method. In this sorting considering two adjacent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest(in the case of descending sorting) will be the end of the array. This process continues.

2. Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Question 5.
Using the concept obtained from algorithms in question number (9), write complete program for the two types of sorting.
Answer:

Question 6.
Collect the heights of 12 students from your class in which 7 students are male and others are female students. Suppose these male and female students be seated in two separate benches and you are given a place which is used for sitting these 12 students in linear form. How will you combine and make them sit without mixing male/female students. Write a program for the same.
Answer:

Question 7.
Collect the heights of 12 students from your class in which 7 students are male and others are female students. Suppose you are given the heights according to a sorted order separately for male/female students. How will you combine these group according to the same order in linear form. Write a program for the same.
Answer:

Question 8.
Write the program code for counting the number of vowels from your school name.
Answer:

Question 9.
Write the program code for counting the number of words from the given string.” Directorate of Higher Secondary Examination”
Answer:

Question 10.
Given a word like “ECNALUBMA” Write the program code for arranging it in into a meaningful word.
Answer:

Question 11.
Explain the needs for arrays Array
Answer:
Array is collection of same type of elements. With the same name we can store more elements. The elements are distinguished by using its index or subscript. To store 50 marks of 50 students we have to declare 50 variables, it is a laborious work.

Hence the need for arrays arise. By using array this is very easy as follows int mark[50]. Here the index of first element is 0, then 1, 2, 3, etc upto 49.

Question 12.
Explain different types of arrays.
Answer:
There are two tyes of array
1. Single dimensional:
It contains only one index or subscript. The index starts from 0 and ends with size-1.
eg: int n[50];
char name[10];

2. Multi-dimensional:
It contains more than one index or subscript. The two dimensional array contains two indices, one for rows and another for columns. The row index starts from 0 and end at row size-1 and column index starts at 0 and ends at colunn size-1.
eg: int n[10][10] can store 10 × 10 = 100 elements. The index of the first element is n[0][0] and index of the 100th element is n[9][9],

Question 13.
Write a program to read the 5 marks of a students and display the marks and total
Answer:

Question 14.
Write a program to read 3 marks of 5 students and find the total and display it.
Answer:

Question 15.
Differentiate linear search and binary search
Answer:
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is re¬turned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Question 16.
Differentiate bubble sort and selection sort.
Answer:
1. Bubble sort:
It is a simple sorting method. In this sorting considering two adjascent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest (in the case of descending sorting) will be the end of the array. This process continues.

2. Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Question 17.
Write a program to read a string and a character and find the character by using linear search.
Answer:

Question 18.
Write a program to read marks of 10 students and read a mark and check whether it is in the array or not using binary search
Answer:

Question 19.
Write a program reverse a string without using a function.
Answer:

Question 20.
Write a program to read a string and find the no. of vowels .consonents and special characters.
Answer:

Question 21.
Given a word “COMPUTER”, write a C++ program to reverse the word without using any string functions.
Answer:
#include<iostream>
using namespace std;
int main(()
{
char a[9] = “COMPUTER”;
inti;
for (i=8; i>=0; i- -)
{
cout<<a[i];
}
}

Question 22.
Write a program to accept marks of 10 students and find out the largest and smallest mark from the list.

OR

C++ program to store the scores of 10 batsmen of a school cricket team and find the largest and smallest score.
Answer:

Question 23.
Write the names of two searching methods in arrays. Prepare a chart that shows the comparisons of two searching methods. (MARCH-2015)
Answer:
It is the process of finding the position of the given element.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Question 24.
Write a C++ program to illustrate array traversal. (SAY-2015 (IMP) (3)
Answer:
#include<iostream>
using namespace std;
int main()
{
int i, mark[50];
for(i=0;i<50; i++)
{
cout<<“Enter mark “<<i + 1<<“: cirt>>mark[i];
}
cout<<“\nThe mark of 50 students after adding bonus mark 10 is given below \n”; for(i=0;i<50; i++)
cout<<mark[i] + 10<<endl;
}

Question 25.
Define an array. Also write an algorithm for searching an element in the array using any one method that you are familiar with. (MARCH-2016) (3)
Answer:
An array is a collection of elements with same data type. The index of first element is 0 (zero) and the index of last element is size -1. There are 2 methods linear search and binary search.

Searching: It is the process of finding the position of the given element.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.8.3 Two dimensional (2D) arrays.

Some occasions we have to store 6 different marks of 50 students. For this we use 2D arrays. An array with two subscripts is used, eg. int mark[r][c]; Here r is the row and c is the column.

Question 26.

1. Finding sum of all elements in an array is an example of _________ operation.
2. If 24, 45, 98, 56, 76, 24, 15 are the elements of an array, illustrate the working of selection sort for arranging the elements in descending order. (SCERT SAMPLE -1) (3)

Answer:
1. Traversal

2. In selection sort the array is divided into two parts the sorted part and unsorted part. First smallest elemant in the unsorted part in searched and exchanged with the first elemant now there is 2 parts sorted part and unsorted part. This process continues.

Question 27.
Read the following C++ statement:

1. How many bytes will be allocated for this array?
2. Write algorithm to sort this array using bubble sort method. (SCERT SAMPLE – II) (3)

Answer:
1. To store an integer 4 bytes needed so AR[10] needs 10 × 4 = 40 bytes.

2. Sorting – Arranging elements of an array in an order(ascending or descending).
(a) Bubble sort:
It is a simple sorting method. In this sorting considering two adjascent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest(in the case of descending sorting) will be the end of the array. This process continues.

(b) Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Question 28.
Explain ‘Call by Value’ and ‘Call by Reference’ methods of function calling with the help of a suitable example. (SEP-2016) (3)
Answer:
Two types are call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example

Question 29.
What is an array? Write C++ program to declare and use a single dimensional array for storing the computer science marks of all students in your
class. (SEP-2016) (3)
Answer:
An array is a collection of elements with same data type or with the same name we can store many elements, the first or second or third etc can be distinguished by using the index(subscript). The first element’s index is 0, the second elements index is 1, and so on.

Question 30.
Write an algorithm for arranging elements of an array in ascending order using bubble sort. (MARCH-2017) (3)
Answer:

Plus One Arrays Five Mark Questions and Answers

Question 1.
Write a program to perform matrix addition, multiplication, subtraction.
Answer:

Students can Download Chapter 9 Hydrogen Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Plus One Chemistry Hydrogen One Mark Questions and Answers

Question 1.
In which of the following compounds does hydrogen have an oxidation state of -1?
a) CH₄ b) NH₃ C) HCl d) CaH₂
Answer:
d) CaH₂

Question 2.
The radio active isotope of hydrogen is __________ .
Answer:
Tritium

Question 3.
Temporary hardness of water is due to the presence of
a) MgSO₄
b)Ca(HCO₃)₂
c) CaSO₄
d) NaHCO₃
Answer:
b) Ca(HCO₃)₂

Question 4.
D₂O is used as __________ in nuclear reactors.
Answer:
Moderator

Question 5.
30 volumes of H₂O₂ means
a) 30% H₂O₂ solution
b) 30 cm³ of the solution contains 1 g of H₂O₂
c) 1 cm³ of the solution liberates 30 cm3 of O₂ at STP
d) 30 cm³ of the solution contains 1 mole of H₂O₂
Answer:
c) 1 cm³ of the solution liberates 30 cm3 of O₂ at STP

Question 6.
Name the three isotopes of hydrogen.
Answer:

Question 7.
Dihydrogen is prepared on industrial scale from syngas by
Answer:
Water gas shift reaction

Question 8.
Among the following elements which does not make a hydride is _________
a) Ti
b) Mg
c) Co
d) Pd
Answer:
c)Co

Question 9.
Hydrogen is purified by _________ .
Answer:
Oculusion on pd

Question 10.
H₂O₂ is _________ .
Answer:
Diamagnetic

Question 11.
Ortho & para hydrogen are __________ .
Answer:
Nuclear spin isomers

Plus One Chemistry Hydrogen Two Mark Questions and Answers

Question 1.
Write one method each for the laboratory preparation of dihydrogen from
i) mineral acid
ii) aqueous alkali.

1. Which is the catalyst used for the reaction?
2. Which is the product in this reaction?

Answer:
1.By the reaction of granulated zinc with dilute HCl.
Zn + 2HCl → ZnCl₂ + H₂

2. By the reaction of zinc with aqueous alkali.
Zn + 2NaOH → Na₂ZnO₂ + H₂

Question 2.
Classify the following into those causing temporary hardness and permanent hardness:
[Mg(HCO₃)₂, MgCl₂, CaCO₃, CaSO₄, NaCl, NaHCO₃, Ca(HCO₃)₂]
Answer:
Temporary hardness – Mg (HCO₃)₂, Ca(HCO₃)₂
Permanent hardness – MgCl₂, CaSO₄

Question 3.
The bond angle in water is different from the tetrahedral bond angle.

1. What is the bond angle and shape of water?
2. Justify.

Answer:
1. 104.5°

2. There are three types of repulsions in water. They are: Ip – Ip repulsion, Ip – bp repulsion and bp – bp repulsion. In order to minimize the stronger Ip- Ip repulsion, the bond angle is reduced to 104.5° from 109.5°. Thus, the shape of water is distorted tetrahedral or angular.

Question 4.
Among NH₃, H₂O and HF which would you expect to have highest magnitude of hydrogen bonding? Why?
Answer:
Strength of H-bond depends upon the atomic size and electronegativity of the other atom to which H- atom is covalently bonded. Smaller size and higher electronegativity favour H-bonding. Among N, F and O atoms, F is the smallest and its electronegativity is highest. Hence, HF will have highest magnitude of H-bonding.

Question 5.
1. Soap does not give lather with hard water. Why?
2. What are the disadvantages of hard water?
Answer:
1. Hard water contains Ca⁺, Mg²⁺ions in the form of
their bicarbonates, chlorides or sulphates. Hard water forms scum/precipitate with soap. So soap does not give lather with hard water,

2. Hard water is unsuitable for laundry. It is harmful for boilers because of deposition of salts in the form of scale. This reduces the’ efficiency of the boiler.

Question 6.
Write one example each for the oxidising action of H₂O₂ in acidic medium and basic medium.
Answer:
In acidic medium, H₂O₂ oxidises PbS to PbSO₄.
PbS(s) + 4H₂O₂(aq) → PbSO₄(s) + 4H₂O(l)
In basic medium, H₂O₂ oxidises Fe²⁺ to Fe³⁺.
2Fe²⁺ + H₂O₂ → 2Fe³⁺ + 2OH^–

Question 7.
Write one example each for the reducing action of H₂O₂ in acidic medium and basic medium.
Answer:
In acidic medium H₂O₂ reduces MnO₄^– to Mn²⁺.
2MnO₄^– + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
In basic medium H₂O₂ reduces l₂ to l^–.
l₂ + H₂O₂ + 2OH^– → 2l^– + 2H₂O + O₂

Question 8.
Write two examples for redox reactions involving water.
Answer:
1. Water can be easily reduced to dihydrogen by highly electropositive metals like Na. Here, Na is oxidised to NaOH.
2H₂O(I) + 2Na(s) → 2NaOH(aq) + H₂(g)

2. With F₂, water is oxidised to O₂. Here, F₂ is reduced to F-.
2F₂(g) + 2H₂O(l) → 4H+(aq) + 4F^–(aq) + O₂(g)

Question 9.
Distinguish between

1. Hard and Heavy water
2. Temporary and permanent hardness of water.

Answer:
1. A sample of water said to be hard water if it does not give lather with soap. Heavy water is the oxide of deuterium (D₂O).

2. Temporary hardness is due to the presence of bicarbonate of calcium or magnesium. It can be removed by boiling.

Permanent hardness is due to the presence of chlorides and sulphates of calcium and magnesium. It is not removed by boiling.

Question 10.
Hydrogen combines with elements to give binary compounds known as hydrides.
1. Name the three categories of hydrides.
2. Classify the given hydrides into different categories: NH₃ LiH, TiH, CH₄, NaH
Answer:
1. Ionic hydrides, Covalent hydrides and Metallic/ Interstitial hydrides

2. NaH, LiH – Ionic hydrides
NH₃, CH₄ – Covalent hydrides
TiH – Interstitial hydride

Plus One Chemistry Hydrogen Three Mark Questions and Answers

Question 1.
A chart prepared by a student based on the similarities of hydrogen with alkali metals and halogens is as shown below. Correct the mistakes in it.

Answer:

Question 2.
Prepare a short note on different types of hydrides.
Answer:
Hydrogen reacts with metals or non metals to form binary compounds known as hydrides.
Hydrides are mainly classified into the following 3 types:
1. Ionic or salt like hydrides
These are stoichiometric compounds of dihydrogen formed with most of the s-block elements which are highly electropositive in character. These are crystalline, non-volatile and non-conducting in solid state. But their melts conduct electricity and on electrolysis liberate dihydrogen gas at anode. They react violently with water producing dihydrogen gas. e.g. NaH, KH.

2. Covalent hydrides or Molecular hydrides
These are formed by the action between dihydrogen and non metals (p-block elements). Covalent hydrides are classified into electron-deficient (e.g. B₂H₆), electron-precise (e.g. CH₄) and electron-rich hydrides (e.g. NH₃).

3. Metallic or Non-stoichiometric or Interstitial hydrides
These are formed by the reaction of dihydrogen with many d-block and f-block elements. These hydrides conduct heat and electricity. They are almost always non- stoichiometric, being deficient in hydrogen, e.g.
LaH_2.87,YbH_2.55,VH_0.56 etc.

Question 3.
Consider the chemical equation and fill in the blanks:

1. This method is used for the preparation of ……………
2. The electrode used is …………
3.  ……….. is produced at cathode.

Answer:

1. Hydrogen
2. Platinum
3. Hydrogen

Question 4.
Analyse the equation: 2Na + H₂ → 2NaH

1. In this chemical equation, H₂ reacts with …………
2. Hydrogen reacts with metals to form …………
3. Hydrogen is in ………… oxidation state in NaH.

Answer:

1. Na
2. Metal hydrides
3. -1

Question 5.
a) How is hydrogen of high purity prepared?
b) Dihydrogen is relatively inert at room temperature. Give reason.
c) Write any two uses of hydrogen.
Answer:
a) High purity (>99.95%) dihydrogen is obtained by electrolysing warm aqueous barium hydroxide . solution bewteen nickel electrodes.
b) This is due to high H-H bond enthalpy.
c) 1. In oxy-hydrogen torches.
2. Liquid hydrogen is used as a fuel in rockets.

Question 6.
A sample of cold river water does not easily give lather with soap, but on boiling it does.

1. Evaluate and write the chemical equation involved.
2. In some cases, the water does not give ready lather even if it is boiled. Why?

Answer:
1. If the sample of water has temporary hardness, it is removed during boiling. Here, the bicarbonates of magnesium is precipitated as Mg(OH)₂ and that of calcium is precipitated as CaCO₃.

2. The sample of water might have permanent hardness due to the presence of dissolved chlorides and sulphates of Ca and Mg. This cannot be removed just by boiling.

Question 7.
Hydrogen has three isotopes – Protium, Deuterium and Tritium.
a) Of these which is the radio active one?
b) Name a compound which contains the isotope deuterium.
c) Make a table which shows number of protons, neutrons and electrons in each isotope.
Answer:
a) Tritium
b) Heavy water – D₂O
c)

Question 8.
What are the three types of hydrates? Give examples for each.
Answer:
1. Hydrates with coordinated water, e.g. [Cr(H₂O)₆]Cl₃.
2. Hydrates with interstitial water, e.g. BaCl₂.2H₂O
3. Hydrates with hydrogen bonded water, e.g. CuSO₄.5H₂O.

Question 9.
1. What is meant by amphoteric nature of water?
2. Suggest an example to show this property.
Answer:
1. The amphoteric nature of water means that it has the ability to act as an acid as well as a base.

2. Water acts as an acid with NH₃ and a base with H₂S.

Question 10.
a) What are the advantages of dihydrogen as a fuel?
b) What are the disadvantages of using dihydrogen as a fuel?
c) What is meant by the term ‘hydrogen economy’?
Answer:
a) 1. It is abundantly available in the combined state as water.
2. Use of dihydrogen as fuel provides pollution free atmosphere because its combustion product is only water.
3. Heat of combustion per gram of dihydrogen is more than twice that of jet fuel.

b) 1. Dihydrogen does not occur in free state in nature.
2. Hydrogen gas has explosive flammability which causes problem to its storage and transportation.
3. A cylinder of compressed dihydrogen weighs about 30 times as much as a tank of petrol containing the same amount of energy.

c) Hydrogen economy refers to the use of dihydrogen as an alternative source of energy. The basic principle of hydrogen economy is storage and transportation of energy in the form of dihydrogen instead of fossil fuels or electric power.

Question 11.
What do you mean by temporary and permanent hardness? Explain one chemical method each for removing temporary hardness and permanent hardness.
Answer:
Temporary hardness is due to the presence of bicarbonates of calcium or magnesium. Permanent hardness is due to the presence of chlorides and sulphates of calcium or magnesium ions.
Clark’s method :
This method is used for removing temporary hardness. In this method, calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off.

Treatment with washing soda: This method is used to remove permanent hardness. Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.

Question 12.
a) Write a method for the preparation of H₂O₂.
b) What is 100 volume H₂O₂?
c) Draw the structure of H₂O₂ in gas phase.
Answer:
a) H₂O₂ is prepared by acidifying barium peroxide and removing excess water by evaporation under reduced pressure.
BaO₂.8H₂O(s) + H₂SO₂(aq) → BaSO₄(s) + H₂O₂(aq) + 8H₂O(I)

b) A 30% solution of H₂O₂ is called 100 volume H₂O₂. It means that one mL of 30% H₂O₂ solution will give 100 V of oxygen at STP.

c) H₂O₂ is has non-planar structure as shown below:

Question 13.
lon exchange process is commonly employed for large scale production of soft water.

1. What is the basic principle involved in ion exchange method of softening water?
2. What are inorganic cation exchangers? Give example.

Answer:
1. Adsorption

2. These are complex inorganic salts like sodium- aluminium silicate NaAlSiO₄ which can exchange cations such as Ca²⁺ and Mg²⁺ ions in hard water for Na⁺ ions. e.g. Zeolite.

Question 14.
1. What is heavy water? Mention one use of heavy water.
2. Explain why hydrogen peroxide is not stored in glass vessels.
3. What is calgon? What is its use?
Answer:
1. D₂O. Used as moderator in nuclear reactors.

2. To prevent its decomposition.

3. Calgon is chemically sodium hexametaphosphate (Na₆P₆O₁₈). It is used to remove Ca²⁺ and Mg²⁺ ions of hard water by converting them into soluble complexes.
M²⁺ + Na₄P₆O₁₈^2- → [Na₂MP₆O₁₈]^2- + 2Na⁺ (M = Mg,Ca)

Question 15.
1. Name the oxide of isotope of hydrogen used in nuclear reactor.
2. What are cation exchange resins? What is their role in removing permanent hardness of water?
Answer:
1. Heavy water (D₂O).

2. Cation exchange resins contain large organic molecule with -SO₃H group and are water insoluble. Ion exchange resin (RSO₃H) is changed to RNa by treating with NaCl. The resin exchanges Na⁺ ions with Ca²⁺ and Mg²⁺ ions present in hard water to make the water soft.

Question 16.
1. What is permuitit? How is it useful in removing permanent hardness of water?
2. Compare the structures of water and hydrogen peroxide.
Answer:
1. Hydrated sodium aluminium silicate is called permutit (NaZ). When it is added to hard water it exchanges Ca²⁺ and Mg²⁺ ions with Na⁺ ions.
2NaZ(s) + M²⁺(aq) —> MZ₂(s) + 2Na⁺(aq) M = Ma, Ca)

2. Water molecule has an angular or bent shape. H₂O₂ has a non-planar structure.

Question 17.
1. Give the industrial method of preparation of H₂O₂.
2. How is heavy water prepared?
3. How is pure de-mineralised water obtained?
Answer:
1. Industrially H₂O₂ is prepared by the auto-oxidation of 2-ethylanthraquinol.

2. Heavy water is prepared by exhaustive electrolysis of water or as a byproduct in some fertilizer industries.

3. Pure de-mineralised water is obtained by passing water successively through a cation exchange and an anion exchange resins.

In the cation exchange process, the H⁺ exchanges for Na⁺, Ca²⁺, Mg²⁺ and other cations present in water. This process results in proton release and makes the water acidic.
2RH(s) + M²⁺(aq) \(\rightleftharpoons \) MR₂(s) + 2H⁺(aq)

In the anion exchange process, the OH^– exchanges for anions like Cl^–, HCO₃^–, SO₄^2- etc. present in water.
RNH₃⁺OH^–(S) + X^–(aq) \(\rightleftharpoons \) RNH₃⁺X^–(s) + OH^–(aq)

The OH^– ions, thus liberated neutralise the H⁺ ions set free in the cation exchange process to get pure de-mineralised water.
H⁺(aq) + OH^–(aq) → H₂O(l)

Question 18.
1. Explain why hydrogen peroxide is stored in coloured plastic bottles.
2. Write any two uses of H₂O₂.
Answer:
1. In the presence of metal surfaces or traces of alkali (present in glass conatiners), the decomposition of H₂O₂ (2H₂O₂ → 2H₂O + O₂) is catalysed. It is therefore stored in wax-lined glasses or plastic vessels in dark.

2. 1. As a hair bleach and as a mild disinfectant.
2. In the syntheseis of hydroquinone, tartaric acid and certain food products and pharmaceuticals.

Question 19.
1. A sample of hard water was found to loose its hardness on boiling. Name the type of hardness associated with this sample. 0
2. Write the name and formula of four minerals which cause permanent hardness of water.
3. What are electron deficient hydrides? Whether they behave as Lewis acids or Lewis bases? Why?
Answer:
1. Temporary hardness

2. CaCl₂, MgCl₂, CaSO₄, MgSO₄

3. These are covalent hydrides having too few electrons for writing conventional Lewis structure. They act as Lewis acids because they can accept electrons, e.g. B₂H₆

Plus One Chemistry Hydrogen Four Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2.
Certain samples of water do no produce easy lather with soap.

1. What is this condition of water called?
2. Which are two types of this condition?
3. Suggest two methods to change this condition of water.
4. What are the problems caused by this condition of water?

Answer:

1. Hardwater.
2. Temporary hardness and permanent hardness.
3. By boiling water and by adding Na₂CO₃
4. Wastage of soap and boiler explosion.

Question 3.
Match the following:

Answer:
1) – d)
2) – a)
3) – f)
4) – g)
5) – c)
6) – e)
7) – b)
8) – h)

Question 4.
1. A list of compounds are given below:
H₂O, HCl, CH₄
Construct chemical reactions to show the preparation of H₂ from each of the above compounds.
2. What is syn gas? How is it prepared?
3. What is coal gasification?
4. Explain water gas shift reaction.
Answer:
1.

2. Syn gas is a mixture of CO and H₂. It is prepared by passing steam over red hot coke.

3. The process of production of syngas from coal is called coal gasification.

4. The production of dihydrogen can be increased by reacting CO of syngas mixtures with steam in the presence of iron chromate as catalyst. This is called water gas shift reaction.

The CO₂ is removed by scrubbing with sodium arsenite solution.

Plus One Chemistry Hydrogen NCERT Questions and Answers

Question 1.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? (2)
Answer:
The various isotopes of hydrogen are:

Question 2.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions? (2)
Answer:
Hydrogen atom has only one electron and thus, to achieve stable inert gas configuration of helium, it shares its single electron with electron of another hydrogen atom to form a stable diatomic molecule. The stability of H₂ is further confirmed by the fact, that formation of one mole of gaseous H₂ molecules results in the release of 435.8 kJ of energy.
H(g) + H(g) H₂(g); ∆H = – 435.8 kJ mol^-1

Question 3.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain. (2)
Answer:
In some of the transition metal hydrides, hydrogen is absorbed as H-atoms. Due to the inclusion of H- atoms, the metal lattice expands and thus becomes less stable. Therefore, when such metallic hydride is heated, it decomposes to release hydrogen gas and very finely divided metal. The hydrogen evolved in this manner can be used as a fuel. Thus, transition metals or their alloys can act as sponge and can be used to store and transport hydrogen to be used as a fuel.

Question 4.
What causes the temporary and permanent hardness of water? (2)
Answer:
Temporary hardness is caused by presence of bicarbonates of calcium and magnesium, i.e., Ca(HCO₃)₂ and Mg(HCO₃)₂ in water whereas permanent hardness is caused by presence of soluble chlorides and sulphates of calcium and magnesium, i.e., CaCl₂, CaSO₄, MgCl₂ and MgSO₄ in water.

Question 5.
Write chemical reactions to show amphoteric nature of water. (2)
Answer:
Water is amphoteric in character. It means that it can act as proton donor as well as proton acceptor. When it reacts with acids, (stronger than itself), it behaves as a base. When it reacts with bases (stronger than itself) it acts as an acid.

Question 6.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful? (2)
Answer:
Demineralised or distilled water is not useful for drinking purposes because it does not contain even useful minerals. Therefore, to make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralised or distilled water.

Question 7.
How does H₂O₂ behave as a bleaching agent? (2)
Answer:
The bleaching action of H₂O₂ is due to the nascent oxygen which is liberates on decomposition.
H₂O₂ → H₂O + [O]
The nascent oxygen oxidise the colouring matter to colourless products. Hence, H₂O₂ is used for the bleaching of delicate maerials like ivory, feather, silk, wool, etc.