Category: CBSE Class 11

By going through these CBSE Class 11 Maths Notes Chapter 13 Limits and Derivatives Class 11 Notes, students can recall all the concepts quickly.

Limits and Derivatives Notes Class 11 Maths Chapter 13

Left Hand Limit: The \(\lim _{x \rightarrow a^{-}}\)f(x) is the expected value f(x) at x = a, given the values of fix) near x to the left of a. This value is called the left hand limit of f(x) at a.

Right Hand Limit : The \(\lim _{x \rightarrow a^{+}}\)f(x) is the expected value of f(x) at x – a, given the values of fix) near x to the right of a. ‘ ‘his value is called the right hand limit of f(x) at a.

Limit of a Function: If the right and left hand limits coincide, the common value of the limit of f(x) as x → a is called the limit of a function. It is denoted by \(\lim _{x \rightarrow a}\)f(x).

Algebra of Limits:
(i) Limit of sum of two functions is sum of the limits of the functions, i.e.,
\(\lim _{x \rightarrow a}\) [(x)+g(x)] = \(\lim _{x \rightarrow a}\)f(x) + \(\lim _{x \rightarrow a}\) g(x).

(ii) Limit of difference of two functions is difference of the limits of the functions, i.e.,
\(\lim _{x \rightarrow a}\) [(x)-g(x)] = \(\lim _{x \rightarrow a}\)f(x) – \(\lim _{x \rightarrow a}\) g(x).

(iii) Limit of product of two functions is product of the limits
of the functions i.e.,
\(\lim _{x \rightarrow a}\) [(x)+g(x)] . \(\lim _{x \rightarrow a}\)f(x) . \(\lim _{x \rightarrow a}\) g(x).

(iv) Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non-zero i.e..
\(\lim _{c \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}\)

(v) \(\lim _{x \rightarrow a}\) (c.f)(x) = c\(\lim _{x \rightarrow a}\)f(x)

Limit of Polynomial:
Let f(x) = a₀ + a₁x + a + a₂x² + … + a_nxⁿ be a polynomial function.
Let \(\lim _{x \rightarrow a}\) x^k = a^k.
\(\lim _{x \rightarrow a}\) f(x) = [a₀ + a₁x + a + a₂x² + … + a_nxⁿ]
= a₀ + a₁\(\lim _{x \rightarrow a}\)x + a + a₂\(\lim _{x \rightarrow a}\)x² + … + a_n\(\lim _{x \rightarrow a}\)xⁿ
= a₀ + x₁a + a₂a² + … + a_naⁿ
= f(a).

Limit of Rational Function : A function f is said to be a rational function, if f(x) = \(\frac{g(x)}{h(x)}\), where g(x) and h(x) are polynomials h(x) ≠ 0

However, ifg(a) = 0 and h(a) = 0, i.e., this is of the form 0/0 then factor(s),x-a of g(x) and h(x) are determined and then cancelled out.
Let g(x) = (x-a)p(x)
h(x) = (x-a)q(x)

(ii) For any positive integer n,
\(\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = na^n-1

Theorems:
(i) Let f and g be two real valued functions with the same domain such that f(x) ≤ g(x) for allx in the domain of definition. For some a, if both \(\lim _{x \rightarrow a}\)f(x) and \(\lim _{x \rightarrow a} g(x) exist, then
[latex]\lim _{x \rightarrow a}f(x) ≤ [latex]\lim _{x \rightarrow a}\)g(x)

(ii) Sandwich Theorem : Let f, g and h be real functions
such that f(x) ≤ g(x) ≤ h(x) for all x in the common domain of
definition. For some real number a, if \(\lim _{x \rightarrow a}\)g(x) f(x) = Z, \(\lim _{x \rightarrow a}\)g(x) = h(x) = l, then lim g(x) = l.

Limit of Trigonometric Functions :
(i) \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) = 1
(ii) \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x}\) = 0

Derivative of f(x) at x = a : Suppose f is a real valued of function and a is a point in its domain of definition. The derivative of f at x = a is defined by
\(\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)
provided this limit exists. It is denoted by f'(a).

Derivative of fix) : Suppose f is a real valued function, the function defined by \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) wherever the limit exists is defined to be the derivative of/‘and is denoted by fix). Thus,

f ‘(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

Algebra of Derivative of Functions:
(i) Derivative of sum of two functions is sum of the derivatives of functions:
\(\frac{d}{d x}\)[f(x) + g(x)] = [\(\frac{d}{d x}\)f(x) + \(\frac{d}{d x}\)g(x)]
or (u + v)’ = u’ + v’.

iii) Derivative of difference of two functions is the difference of the derivatives of the functions :
\(\frac{d}{d x}\)[f(x) – g(x)] = [\(\frac{d}{d x}\)f(x) – \(\frac{d}{d x}\)g(x)]
or (u + v)’ = u’ + v’.

(iii) Derivative of product of two functions is given by the product rule, i.e.,
or (u .v)’ = u’v + uv’.

(iv) Derivative of quotient of two functions is given by quotient rule (whenever the denominator is non-zero).

(v) Derivative of λf(x)
\(\frac{d}{d x}\)[λf(x)] = λ \(\frac{d y}{d x}\)f(x)
or (λu)’ = λu’

Some Derivatives:
(i) \(\frac{d}{d x}\)xⁿ = nxn-1
(ii) \(\frac{d}{d x}\)(ax + b)ⁿ = na(ax + b)^n-1
(iii) If f(x) = a₀xⁿ + a₁x^n-1 + a₂x^n-2 + … a_m
then na₀x^n-1 + (n-1)a₁x^n-2 + (n-2)a₂x^n-3+ …….. +a_n-1
(iv) \(\frac{d}{d x}\) (sin x) = cos x.
(v) \(\frac{d}{d x}\) (cos x) = – sin x.
(vi) \(\frac{d}{d x}\) (tan x) = sec² x.
(vii) \(\frac{d}{d x}\) (cosec x) = – cosec x cot x.
(viii) \(\frac{d}{d x}\) (sec x) = sec x tan x.
(ix) \(\frac{d}{d x}\) (cot x) = cosec² x

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 10 Cell Cycle and Cell Division. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 10 Important Extra Questions Cell Cycle and Cell Division

Cell Cycle and Cell Division Important Extra Questions Very Short Answer Type

Question 1.
Who first described meiosis?
Answer:
Strasburger,

Question 2.
What is a genome?
Answer:
It is a full set of DNA instructions or a single set of chromosomes in a cell.

Question 3.
What is meant by the non-disjunction of chromosomes?
Answer:
Non-disjunction means failure in the separation of homologous chromosomes during anaphase.

Question 4.
Why is mitosis an equational division?
Answer:
Mitosis is an equational division because the daughter cells get the same number of chromosomes from the parent.

Question 5.
What is crossing over?
Answer:
The exchange of segments of chromatids of homologous chromosomes during meiosis is called crossing over.

Question 6.
Why is meiosis a reductional division?
Answer:
Meiosis is a reductional division because it reduces the number of chromosomes from diploid number to haploid number in the daughter cells.

Question 7.
What are the two successive divisions in meiosis?
Answer:
The first division is reductional followed by the second equational di¬vision.

Question 8.
Name the two phases of the cell cycle of a somatic cell.
Answer:

1. Interphase and
2. M-phase or mitotic phase

Question 9.
During which part of interphase active synthesis of RNA and proteins take place.
Answer:
G. phase.

Question 10.
What amount of DNA is present in the cell during the G₂ phase?
Answer:
Double the amount of DNA present in the original diploid cell.

Question 11.
What is a kinetochore?
Answer:
A part of the chromosome for the attachment of chromosomal fibers.

Question 12.
Define Eumitosis.
Answer:
Chromosomes are attached to the spindle by their centromere and this type of mitosis is called Eumitosis.

Question 13.
Who gave the term mitosis?
Answer:
W. Flemming.

Question 14.
How many mitotic divisions will be required to produce 128 daughter cells from a single cell?
Answer:
127.

Question 15.
What is the Gj phase of the interphase?
Answer:
It is the first period of growth of the neatly formed undivided cells, during which the cell synthesizes a lot of RNA and proteins.

Cell Cycle and Cell Division Important Extra Questions Short Answer Type

Question 1.
Define cell cycles.
Answer:
The cell cycle is the sequence of events that occur between the formation of a cell and its division into daughter cells.

Question 2.
What do you understand by homologous chromosomes?
Answer:
Homologous chromosomes are pairs of chromosomes that have similar characteristics. They show pairing during meiosis. One chromosome in each pair is inherited from the father and the other one from the mother.

Question 3.
Why is mitosis an equational division?
Answer:
Mitosis is an equational division because the daughter cells have the same number of chromosomes and an equal amount of cytoplasm.

Question 4.
Why is meiosis necessary in sexually reproducing organisms?
Answer:
Meiosis is necessary for sexually reproducing organisms because

1. It maintains the number of chromosomes constant in generation as meiosis is reductional division.
2. It causes variations among the progeny because crossing over takes place during meiosis. This variation is important for evolution.

Question 5.
What is the importance of mitosis?
Answer:
Mitosis is important because

1. It maintains genetic stability through generations.
2. It helps in the growth of multicellular organisms.
3. Many plants and animals multiply by mitosis i.e., asexual repro-duction to regenerate the whole organism.
4. It helps to regenerate lost parts of an animal’s body.
5. It helps in the regeneration of new cells in place of dead and worn-out cells.

Question 6.
What are homologous chromosomes? What happens to homologs during meiosis?
Answer:
Each diploid nucleus has pairs of similar chromosomes called homologous chromosomes. The two homologous chromosomes each derived from one parent during sexual reproduction come together and form pairs during the zygonema of meiosis I. Individuals of a pair are similar in length and in the position of their centromere.

Question 7.
What is the significance of meiosis?
Answer:
Significance of meiosis:

1. Sexual reproduction: Maintains a number of chromosomes constant. Characteristic of a species from generation to generation.
2. Genetic variation: Through crossing over, it produces variations of genetic characters of the progeny essential for evolution.

Question 8.
What do you mean by cell reproduction?
Answer:
Cell reproduction: Reproduction is an essential phenomenon in the continuity of life. New cells arise by the division of the pre-existing cells. It was proposed by Rudolf Virchow.

Reproduction is of two types:

1. sexual and
2. asexual reproduction.

The growth and development of the living being are dependent on the division of cells. The single-celled zygotes by means of cell division develop into an adult having a large number of cells.

Question 9.
What is P-oxidation?
Answer:
Fats are broken down into glycerol and fatty acids during digestion. Glycerol enters the glycolytic pathway at the triose phosphate stage. Fatty acids undergo β oxidation by which two-carbon fragments of acetyl A are split off at a time from the fatty acid chain so that the long fatty acid molecule is shortened by two carbon-carbon atoms at a time. This ultimately results in incomplete oxidation of fatty acids.

Question 10.
Distinguish between Anaphase of mitosis and Anaphase of meiosis I.
Answer:

Anaphase of mitosis	Anaphase of Meiosis I
(i) Centromeres divide into two	(i) Centromeres do not separate the chromosomes.
(ii) Chromatids separate and move towards the opposite direction	(ii) Half number of chromosomes move towards opposite poles.
(iii) Separated chromatids are identical.	(iii) Separated chromosomes are homologous.

Question 11.
Distinguish between combustion and respiration
Answer:

Combustion	Respiration
(i) It is a non-living process.	(i) It is a biological process, taking place in cells of living organisms.
(ii) It oxidizes the substrate releasing the entire energy at once.	(ii) It brings about oxidation of organic compounds releasing energy stepwise.
(iii) The energy released is in the form of heat and sometimes partly as light.	(iii) The chemical energy is either made available to the cell or stored in it as ATR
(iv) Heat is generated in large amounts.	(iv) The heat is generated in small amounts and does not harm the cell.

Question 12.
Distinguish between mitosis and meiosis.
Answer:

Mitosis	Meiosis
1. It takes place in somatic or vegetative cells.	1. It takes place in gametic or reproductive cells.
2. Completes in one sequence or phase.	2. Completes in two sequences or phases.
3. Form two daughter cells that are diploid.	3. Form four haploid daughter cells.
4. Prophase is short and completes in one step.	4. Prophase I is long and complicated. It completes in five steps.
5. Crossing oyer does not take place.	5. Crossing over takes place during prophase.
6. Daughter cells are identical to each other and the parent cells.	6. Daughter cells are not identical to each other and the parent cells show variations.

Cell Cycle and Cell Division Important Extra Questions Long Answer Type

Question 1.
Describe the changes that take place during the prophase and metaphase of mitosis.
Answer:
Following changes take place during prophase

1. Chromosomes become short and thick and sister chromatids are held at the centromere.
2. Nucleus and nuclear envelope disappear.
3. In animal cells, centrioles move to opposite poles.
4. Chromosomes begin to move towards the equatorial plane.

Following changes take place in metaphase:

• Chromosomes lie on the equatorial plate.
• Chromatids become attached by spindle fibers.
• Maximum condensation of chromosomes takes place.

Question 2.
Explain the main steps in aerobic glycolysis.
Answer:
Glycolysis is the breakdown of glucose to pyruvic acid, which includes the following:
(a) Phosphorylation: Transfer of phosphate from ATP to glucose to form glucose – 6 phosphate. One molecule of ATP is consumed enzyme, hexokinase is present.

(b) Isomerisation: There is internal molecular rearrangement to form fructose 6 phosphates. The enzyme is hexose phosphate isomerase.

(c) Second phosphorylation: The fructose – 6-phosphate undergoes phosphorylation to form fructose 1,6 diphosphate. One molecule of ATP is consumed. The enzyme is phosphofructokinase.

(d) Triose phosphates are 3-phosphoglyceraldehyde (PGAL) and dihydroxyacetone phosphate (DHAP). The enzyme phosphorize isomerase maintains the two isomers in equilibrium.

(e) Phosphorylation and oxidative dehydrogenation: PGAL under¬goes simultaneous phosphorylation and oxidative dehydrogenation to form 1,3 diphosphoglyceric acid.

(f) ATP generation: 1,3 diphosphoglyceric acid transfers its phosphate with a high energy bond to ADP to form ATP and 3-phosphoric acid. One molecule of ATP is produced from one triose molecule.

One enzyme is phosphoglyceric kinase Glucose

Question 3.
How cytokinesis is different in an animal and a plant cell?
Answer:
Cytokinesis in plant and animal cells: The separation of daughter nuclei and cytokinesis or cell cleavage maybe two different processes. The first visible changes consist of an appearance of dense material around the microtubules at the equator of the spindle at either mid or late phase then although spindle the fibre tends to disorganize and disappear during telophase, they usually persist and may even increase in number at the equator, frequently intermingled with a row of vesicles and the dense material.

The entire structure is called the midbody. Simultaneously there is a depression on the cell surface a kind of constriction that deepens gradually until reaching the midbody with the completion of the furrowing, the separation of cells is concluded.

The phragmoplast begins to form in the mid anaphase of plant cells. Under the electron microscope, it is possible to observe that the vesicles are of dense material applied together to their surface. The vesicles are derived from Golgi complexes which are found in the regions adjacent to phrag¬moplast which migrate to the equatorial region to be clustered around the microtubules.

Although phragmoplast is initially found as a ring on the periphery of the cell, with time it grows centripetally by the addition of microtubules and partition until it extends across the entire equatorial plane. The vesicles increase in size and just until the two cells are separated by a fairly continuous plasma membrane.

All this time the phragmoplast has been transformed into, cell plate. Thin cytoplasmic connection is plasmoids- data transverse the cell plate and remain in place for communication between the adjacent daughter cells.

The formation of the cell plate also leads to the synthesis of the cell wall. The Golgi-vesicles in phragmoplast is already filled with secretory material consisting mainly of the pectin. The fusion of vesicles results in the combining of the pectin in the extracellular space between the two daughter cells thereby forming the main body of the periphery cell wall.

By going through these CBSE Class 11 Maths Notes Chapter 12 Introduction to three Dimensional Geometry Class 11 Notes, students can recall all the concepts quickly.

Introduction to three Dimensional Geometry Notes Class 11 Maths Chapter 12

Co-ordinates of a Point: The co-ordinates of a point are the distances from the origin of the feet of the perpendiculars from the point on the respective co-ordinate axes.

Distance Formula : The distance between the points (x₁ y₁, z₁) and (x₂, y₂, z₂) is given by \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\). The distance of the point (x, y, z) from the origin is given by \(\sqrt{x^{2}+y^{2}+z^{2}}\)

Section Formulae: .
(i) Section formula for internal division :
If P(x₁ y₁, z₁) and Q(x₂, y₂, z₂) are two points. Let R be a point on the line segment P and Q such that it divides the join of P and Q internally in the ratio m₁ : m₂.

Then, the co-ordinates of R are

(ii) Section formula for external division :
If P(x₁ y₁, z₁) and Q(x₂, y₂, z₂) are two points and let R be a
point on PQ produced dividing it
externally in the ratio m₁ : m₂(m₁ ≠ m₂). Then, the co-ordinates of Rare

Mid-Point : The mid-point of the line segment joining (x₁ y₁, z₁) and (x₂ y₂, z₂) is \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\)

Centroid : Centroid of the triangle whose vertices are (x₁ y₁, z₁), (x₂ y₂, z₂) and (x₃ y₃, z₃) is

By going through these CBSE Class 11 Maths Notes Chapter 11 Conic Sections Class 11 Notes, students can recall all the concepts quickly.

Conic Sections Notes Class 11 Maths Chapter 11

Conic : Let T be a fixed line and F be a fixed point (not on line T).
Let P be any point in the plane of the line T and the point F. Then, the set of all points P, such that | FP | = e | PM |, where M is the foot’ of perpendicular from P on the line l and e > 0 is . a fixed real number is called a conic.

Notes :

1. The fixed point F is called focus of the conic and the fixed line T is called directrix of the conic associated with F.
2. The positive real number e is called eccentricity of the conic.
3. A line through the focus F and perpendicular to the directrix l is called axis of the conic.
4. The point of intersection of the conic and its axis is called
   vertex.
5. A conic with eccentricity ‘e’ is called
   (i) a parabola, if e = 1. (ii) an ellipse, if e < 1. (iii) a hyperbola if e > 1..

Parabola : Let T be a fixed line and F be a fixed point (not on 1). Let P be any point in the plane of line l and the point F. Let M be the foot of perpendicular from P on the line l. Then, the set of all points P such that | FP | = | PM | is called a parabola.

Equation of the Parabola (Standard Form): The equation of the parabola in the standard form is y² = 4ax.

Latus Rectum : The latus rectum of conic is the chord passing through the focus and perpendicular to its axis.

Length of Latus Rectum : The length of the latus rectum of the parabola y² = 4ax is 4a.

Focal Distance : The distance of any point on parabola from the focus is called focal distance of that point.

Focal Chord : Any chord that passes through the focus of the parabola is called a focal chord of the parabola.

The focal distance of a point P(x₁, y₁ of the parabola y² = 4ax is a + x₁

Main Facts about Four Standard Forms of Parabola

Ellipse : Let l be a fixed line, F be a fixed point (not on l). Let P be any point in the plane of the line l and the point F.

Let M be the foot of perpendicular from P on the line l.
Then, the set of all points P such that | FP | = e | PM | (0 < e < 1) is called an ellipse.

Notes : 1. The fixed line is called directrix of the ellipse.
2. The fixed point F is called focus of the ellipse.
3. The positive real number e < 1 is called eccentricity of the ellipse.

Standard form of an Ellipse : The equation of the ellipse in the standard form is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, where b² = a²( 1 – e²) = a² – a²e².
Putting ae = c. Therefore, c² = a² – b².

Note: The ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 has two foci, namely (ae,o) and (- ae, 0) and their respective directrices are x – \(\frac{a}{e}\) = 0 and x + \(\frac{a}{e}\) = 0.

Main Facts about two Standard forms of Ellipse

Notes : 1. The focal distances of any point (x₁, y₁) of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 are a ± ex₁
2. Sum of the focal distances of a point of the ellipse is equal to length of the major axis.

Hyperbola: Let l be a fixed line and F be a fixed point (not on l). Let P be any point in the plane of the line l and the point F. Let M be the foot of perpendicular from P on the line l. Then, the set of all points P such that | FP | = e | PM | (e > 1) is called a hyperbola.

2. Sum of the focal distances of a length of the major axis.

Hyperbola: Let l be a fixed line and F be a fixed point (not on l). Let P be any point in the plane of the line l and the point F. Let M be the foot of perpendicular from P on the line l. Then, the set of all points P such that | FP | = e | PM | (e > 1) is called a hyperbola.

Standard form of Hyperbola : The equation of a hyperbola in the standard form is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where b² = a²(e² – 1). Putting ae -c, b² – c² – a² or c² – a² + b².

Notes: 1. The hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 has two foci namely (- ae, 0) and (ae, 0) with two respective directrices as x + \(\frac{a}{e}\) = 0 and x – \(-\frac{a}{e}\) = 0

2. | ex₁ – a | and | ex₁ + a | are the focal distances of any point (x₁, y₁) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.

Main facts about two standard forms of hyperbola

Circle : The locus of a point which moves in a plane such that its distance from a fixed point, in the plane, is always a constant, is called a circle. The fixed point is called the centre and the constant distance is called the Radius of the circle.

Equation of the Circle : The equation of a circle with centre (h, k) and radius V’ is given by (x – h)² + (y – k)² = r².
If the centre of a circle is origin and radius is ‘r’ then its equation is x² + y² = r².

General Equation of a Circle : The general equation of a circle is x² + y² + 2gx + 2fy + c = 0.

Its centre is (-g,-f) and radius = \(\sqrt{g^{2}+f^{2}-c}\)

Note : 1. If g² + f² – c = 0, then radius = 0.
In this case, x² +y² + 2gx + 2fy + c = 0 represents a circle whose radius is zero. Such a circle is called a Point Circle, [i.e., it represents the point (-g, -f)].

2. If g² + f² – c < 0, then radius is non-real. In this case, x2 +y2 + 2gx + 2fy + c = 0 represents the empty set. 3. If g2 + f2 – c > 0, then radius is real and therefore,
x² + y² + 2gx + 2fy + c = 0 represents a circle whose centre is at the point (- g,- f) and radius = \(\sqrt{g^{2}+f^{2}-c}\). (where g² + f² – c > 0).

Conditions for an Equation to Represent a Circle :

(i) It should be a second degree equation in x and y.
(ii) Coefficient of x² = Coefficient of y² ≠ 0.
(iii) It should not contain any term of the product xy.

Method to find the centre, radius of a circle :

(i) Write the equation of the circle with the coefficient of x² and y² being unity.
(ii) Compare this with x² + y² + 2gx + 2fy + c = 0 and find the value of g,f and c.
(iii) Then, (-g,-f) is the centre and \(\sqrt{g^{2}+f^{2}-c}\) is the radius of the circle.

By going through these CBSE Class 11 Maths Notes Chapter 10 Straight lines Class 11 Notes, students can recall all the concepts quickly.

Straight lines Notes Class 11 Maths Chapter 10

Co-ordinate Geometry: The branch of Mathematics in which geometrical problem are solved through algebra by using the co-ordinate system, is known as co-ordinate geometry or analytical geometry.

Algebrical Representation of a Point: It is done by means of two numbers defined in a particular way called co-ordinates of the point.

Co-ordinate Axes : Let X’OX and Y’OY, be two mutually perpendicular lines taken as axes whose positive directions are shown by arrows on the axes. Let these axes intersect at O. Then, point O is called origin and the axes taken are called co¬ordinate axes.

(i) X’OX is called axis of x or x-axis.
(ii) Y’OY is called axis of y or y-axis.
Since the axes taken are mutually at right angles, they are called rectangular axes.

Quadrants : The co-ordinate axes divide the plane into four parts called quadrants.

XOY, YOX’, XOY’ and Y’OX are called first, second, third and fourth quadrants respectively. The names of the quadrants as I, II, III and IV are fixed once for all.

Co-ordinates : Let P be any point in the plane of axes. From P, draw PL, PM ⊥ on y-axis and x-axis respectively. Then, length LP is called the x-coordinate or the abscissa of point P and MP is called the y-co-ordinate or the ordinate of point P. Point whose abscissa is x and ordinate is y, is called the point (x, y).

Signs of Co-ordinates : The co-ordinates of point lying in the first, second, third and fourth quadrants are as (+, +), (-, +), (-, -) and (+, -) respectively.

Note :

1. Co-ordinates of orgin are (0, 0).
2. Ordinate of every point on the x-axis is zero.
3. Abscissa of every point on the y-axis is zero.

Distance Between Two Points : The distance between two
points P(x₁, y₁) and Q(x₂,y₂) is given by PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Note : 1. When the line PQ is parallel to they-axis, the abscissa of points P and Q will be equal, i.e., x₁ = x₂. Therefore, PQ = |y₂ – y₁|

2. When the line PQ is parallel to the x-axis, the ordinate of points P and Q wll be equal i.e.,y1= y2. Therefore, PQ = |x₂ – x₁ |.

3. The distance of a point P(x, y) from the origin (0, 0) is given
by OP = \(\sqrt{x^{2}+y^{2}}\)

Tests : In questions relating to geometrical figures such as triangle, quadrilateral, etc., take the given vertices in the given order.

(i) For an isosceles triangle : At least two sides are equal.
(ii) For an equilateral triangle : Three sides are equal.
(iii) For a right angled triangle : Sum of the squares of two sides is equal to the square of the third side.
(iv) For collinear points : Sum of the distances between two point-pairs is equal to the distance between the third point pair.
(v) For a square : Four sides are equal, two diagonals are
equal.
(vi) For a rhombus : Four sides are equal and there is no need to prove that two diagonals are unequal as a square is also a rhombus.
(vii) For a rectangle : Opposite sides are equal and two diagonals are equal.
(viii) For a parallelogram: Opposite sides are equal and there is no need to prove that two diagonals are unequal as a rectangle is also a parallelogram.

Internal Division : A point R between the points P and Q on the line segment PQ is said to divide PQ internally in the ratio m₁ : m₂,

Here, in the internal division, sense of line segments PR and RQ is same.
∴ m₁ and m₂ are both positive.

External Division : A point R on the part of the line segment PQ produced or QP produced is said to divide PQ externally in the ratio m₁ : m₂

Hence, if a point R divides PQ externally in ratio m₁ : m₂, then we can say that R divides internally in the ratio – m₁ : m₂ or m₁: – m₂.

Section Formula (Internal division) : The co-ordinates of the point R(x, y) which divides internally the straight line segment joining points P(x₁, y₁) and Q(x₂, y₂) in the ratio m₁: m₂ is given by
\(\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right)\)

How to remember: To find the x-co-ordinate of R, multiply m₁ with the x-coordinate of the point remote from m₁ and m₂ by the x co-ordinate of the point remote from m₂ and add these product, and divide the sum by m₁ + m₂. Similarly, findy.

Note : 1. The co-ordinates of the mid-point of the line segment joining( the points P(x₁ y₁) and Q(x₂, y₂) are given by
\(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

2. If a point R divides the line segment joining the points P(x₁ y₁) and Q(x₂, y₂) in the ratio k : 1, then the co-ordinates of R are given by \(\left(\frac{k x_{2}+x_{1}}{k+1}, \frac{k y_{2}+y_{1}}{k+1}\right)\)

3. If R divides PQ externally in the ratio m₁ : m₂
divides PQ internally in the ratio m₁: -m₂.
∴ The co-ordinates of R are given by
\(\left(\frac{m_{1} x_{2}-m_{2} x_{1}}{m_{1}-m_{2}}, \frac{m_{1} y_{2}-m_{2} y_{1}}{m_{1}-m_{2}}\right)\)

Area of a Triangle : The area of the triangle having vertices
as (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by \(\frac { 1 }{ 2 }\) [(x₁y₂ – x₂y₁ + (x₂y₃ – x₃y₂ + (x₃y₁ – x₁y₃)]
or
\(\frac { 1 }{ 2 }\) [x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)]

How to remember : Result of area can be written down in a simpler way as in the figure given :

Numbers to be multiplied are shown by arrows. All the products with arrow downwards are of positive sign and are added to the products with arrow upwards with negative sign.

Condition of Collinearity of Three Points : Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) will be collinear, if and only if the area of the triangle ABC is zero. That is,
(x₁y₂ – x₂y₁) + (x₂y₃ – x₃y₂) + (x₃y₁ – x₁y₃) = 0.
or x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) = 0.

Slope of a Line : The slope (or gradient) of a line is the tangent of the angle which the part of the line above the x-axis makes with the positive direction of the x-axis.

Note : 1. The slope of the line is independent of the sense of line. Consider sense AB. Then slope = tan θ and on cosidering sense BA, the slope = tan (π + θ) = tan θ. Thus, it is same for both senses of the line.

2. The slope of a line is denoted by m, i.e., m = tan θ, where θ is the angle which the line makes with the positive direction ofx-axis.

3. Slope of a line is not defined, when θ = 90°, as tan 90° is not defined.

4. The slope of a line equally inclined to axes is ± 1.
[∵ tan 45° = 1, tan 135° = – 1]

5. If a line is parallel to the x-axis, θ = 0 and its slope m = tan 0° = 0.

Slope of a Line Joining Two Points : The slope of the line joining points (x₁,y₁) and (x₂ , y₂) is given by \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Condition of Parallelism and Perpendicularity :
(i) The two lines are parallel, if their slopes are equal, i.e., m₁ = m₂.
(ii) The two lines are perpendicular, if the product of their slopes = – 1, i.e., m₁m₂ = -1
Or the slope of one line is the negative reciprocal of the other.

Intercepts : Let a straight line AB meet the axes in A and B. Then,
(i) OA is called the intercept of the line on x-axis or x-intercept.
(ii) OB is called the intercept of the line ony-axis or y-intercept.
(iii) AB is called the portion of the line intercepted between the axes.

Rule for Signs of Intercepts: (i) The intercept on the x-axis is positive, if measured to the right of the origin and negative, if measured to the left.
(ii) The intercept on the y-axis is positive, if it is measured above the origin, and negative, if measured below.

Locus of a point: Locus of a point is the path traced by a moving point when it moves under some given geometrical conditions.

Method to Find Locus of Moving Point:

Step 1 : Suppose the co-ordinates of moving point as (x, y).
Step 2 : Write down the given condition under which the point moves.
Step 3 : Express the geometrical condition in step 2 in terms of x and y, i.e., co-ordinates of the moving point.
Step 4 : Simplify, if required, the result in step 3. Equation so obtained, will be the equation of the locus.

Equation of Straight Lines Parallel to Co-ordinate Axes
(i) Equation of a straight line parallel to x-axis and at a distance of ‘b’ from it is y = b.
(ii) Equation of x-axis is y = 0.
(iii) Equation of a straight line parallel to y-axis and at a distance of ‘a’ from it is x = a.
(iv) Equation of y-axis is x – 0.

A Line Through The Origin : The equation of a straight line passing through the origin and making an angle θ with the positive direction of the x-axis is y = mx, where m = tanθ.

Slope-Intercept Form: (i) The equation of straight line which cuts off a given intercept ‘c’ on they-axis and is inclined at a given angle ‘θ’ to the x-axis is y = mx + c, where m = tan θ.
(ii) The equation of a line whose slope is m and the x-intercept is d, which is y = m(x – d).

Intercept Form : The equation of a straight line which cuts off given intercepts a and b from the axes is \(\frac{x}{b}+\frac{y}{b}\) = 1.

Normal Form : The equation of a straight line in terms of the length of the perpendicular p from the origin upon it and the angle a which this perpendicular makes with the x-axis is x cos α+y sin α = P

Point-Slope Form : The equation of a straight line drawn through a given point (x₁,y₁) making an angle 0 with x-axis is y – y₁ = m(x – x₁), where m = tan θ.

Two-Point Form : The equation of a straight line passing through, two given points (x₁, y₁) and (x₂, y₂) is y – y₁ =\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x-x₁).

Distance Form : The equation of a straight line in the form x – x y – y
\(\frac{x-x_{1}}{\cos \theta}=\frac{y-y_{1}}{\sin \theta}\) = r. where (x₁, y₁) is a point on the line, r the distance between the points (x, y) and (x₁, y₁), θ is the angle which the line makes with the x-axis.

General Equation of a Straight Line : Any equation of the first degree in x and y represents a straight line, such as
Ax + By + C = 0 …(1)

Some Cases :

(i) If A = 0, B ≠ 0, then (1) reduces to By + C = 0
⇒ y = \(-\frac{\mathrm{C}}{\mathrm{B}}\) which is the equation of horizontal line ii.e. parallel to x axis)
(ii) If A = C = 0, B ≠ 0, then (1) reduces to y = 0, which is the equation of x-axis.
(iii) If A ≠ 0, B = 0, then (1) reduces to Ax + C = 0 ⇒ x = \(-\frac{\mathrm{C}}{\mathrm{A}}\) , which is the equation of a vertical line (i.e., parallel to y-axis).
(iv) If B = C = 0, A ≠ 0, then (1) reduces x = 0, which is the equation of y-axis.
(v) If A ≠ 0, B ≠ 0, C = 0, then (1) reduces to Ax + By = 0 ⇒ y = \(-\frac{\mathrm{A}}{\mathrm{B}} x\) , which is the equation of a straight line passing through the origin.

Rule to Find the Intercepts of a Line on the Axes : In the equation of the line. Puty = 0, and find x. This gives the intercept on the x-axis. Again, put x = 0, and findy. This is intercept on the v- axis.

Rule to Reduce the General Equation to the Normal Form:
In the given equation :
Divide both sides by \(\sqrt{(\text { coeff. of } x)^{2}+\left(\text { coeff. of } y^{2}\right)^{2}}\)
Transpose the constant term to R.H.S. and make it positive (by changing the signs throughout, if necessary).

Identical Lines : If equations ax + by + c = C and a’x + b’y + c’ = 0 represent the same straight line, then \(\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}\)

Point of Intersection of Two Lines : Let equation of the lines be
a₁x + b₁y + c₁= 0
and a₂c + b₂y + c₂ = 0

The point of intersection can be obtained by solving these equations by cross-multiplication.

Test for Concurrence of Three Lines : If the equations of three lines are a₁x + b₁y + c₁= 0, a₂c + b₂y + c₂ = 0, a₃x + b₃y + c₃= 0 and if three constants l, m, n can be found such that l(a₁x + b₁y + c₁) + m(a₂x + b₂y + c₂) + n(a₁x + b₃y + c₃ = 0 identically (i.e., = 0, for all values of x and y), then the three straight lines meet in a point.

Angle between Two Lines: (i) The angle θ between two lines y = m₁x + c₁ and y = m₂x + c₂ is given by tan θ = ±\(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) .

Condition of Parallelism : If lines are parallel, then θ, the angle between them = 0.
⇒ tan θ = tan 0° = 0
⇒ \(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\)= 0
⇒ m₁ – m₂ = 0
⇒ m₁ = m₂.
Thus, the two lines are parallel, if their slopes are equal.

Condition of perpendicularity : If lines are perpendicular to each other, then 0, the angle between them = 90°.
⇒ tan θ = tan 90° = ∞
⇒ \(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) = ∞
⇒ The denominator, 1 + m₁m₂ = 0.
⇒ m₁ m₂ = – 1 or m₂ = \(-\frac{1}{m_{1}}\)

Hence, the two lines are perpendicular to each other, if the product of their slopes = – 1, or the slopes of one line is the negative reciprocal of the other.

Note : 1. Angle between the two lines means the acutal angle between the lines Hence, for finding the angle between the two lines, we shall be rejecting the negative values of tan 0.
2. The complete angle formula is used, when the angle between the lines is given.

(ii) The angle θ between the lines Ax + By + C = 0 and A’x + By + C’ = 0 is given by tan θ = .
\(\frac{\mathbf{A}^{\prime} \mathbf{B}-\mathbf{A B}^{\prime}}{\mathbf{A A}^{\prime}+\mathbf{B B}^{\prime}}\)

Condition of parallelism : If the lines are parallel, then θ =
⇒ tan θ = tan 0° = 0.
⇒ \(\frac{\mathrm{A}^{\prime} \mathrm{B}-\mathrm{AB}^{\prime}}{\mathrm{AA}^{\prime}+\mathrm{BB}^{\prime}}\) = 0
⇒ A’B-AB’ = 0
⇒ \(\frac{\mathrm{A}}{\mathrm{A}^{\prime}}=\frac{\mathrm{B}}{\mathrm{B}^{\prime}}\)

Thus, if two lines (equations in the general form) are parallel, then the ratio of the coeff. of x = ratio of the coeff. of y.
Condition of perpendicularity : If lines are perpendicular to each other, then θ = 90°.
tan θ = tan 90° = ∞
⇒ \(\frac{A^{\prime} B-A B^{\prime}}{A A^{\prime}+B B^{\prime}}\) = ∞
⇒ AA’ + BB’ = 0.

Hence, if two lines (equation in the general form) are perpendicular to each other, then the product of the coeffs. of x + product of the coeffs. of y = 0.

Perpendicular Distance Formula : (i) The perpendicular distance of the point (x1, yx) from the line xcos a + ysin a = p is ± (x₁cos α+ y₁sin α – p).
(ii) The perpendicular distance of the point (x₁ y₁) from the line ax + by + c = 0is ± \(\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\)

Distance Between the Parallel Lines: The distance between parallel lines ax + by + c = 0 and ax + by + c is \(\frac{c-c^{\prime}}{\sqrt{a^{2}+b^{2}}}\)

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 17 Breathing and Exchange of Gases. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 17 Important Extra Questions Breathing and Exchange of Gases

Breathing and Exchange of Gases Important Extra Questions Very Short Answer Type

Question 1.
Name the site of the exchange of gases in man.
Answer:
Alveoli in the lungs.

Question 2.
Name the respiratory organs in insects.
Answer:
Tracheae and spiracles.

Question 3.
Name the respiratory pigment of the blood of mammals.
Answer:
Haemoglobin.

Question 4.
What makes the exchange of gases in gills and lungs possible?
Answer:
Thin-walled blood capillaries.

Question 5.
Why are erythrocytes unable to carry out cellular oxidation?
Answer:
They lack mitochondria.

Question 6.
What is the main source of energy?
Answer:
Carbohydrates lipids and proteins.

Question 7.
Why energy obtained from ATP molecules is called biologically useful energy?
Answer:
Because it drives the life processes.

Question 8.
How does the exchange of gases across a respiratory membrane occur?
Answer:
By diffusion and circulatory system.

Question 9.
What holds the energy in the body?
Answer:
Molecules of food hold energy in their chemical bonds.

Question 10.
What is external respiration?
Answer:
It is the intake of oxygen by the blood from air or water to the respiratory organs and the elimination of CO₂.

Question 11.
Name the parts of the human respiratory system in a sequence starting from the nose.
Answer:
External nostrils, nasal cavity, internal nostrils, pharynx, trachea, bronchi, lungs.

Question 12.
What is glottis?
Answer:
The opening of the pharynx into the trachea is called the glottis.

Question 13.
What is breathing?
Answer:
Breathing is one phase of respiration. During breathing, we get oxygen from the atmosphere inside our body.

Question 14.
How breathing is different from circulation give a reason?
Answer:
Breathing is a phase of respiration. It is a physical process.

Question 15.
Which gases take part in breathing?
Answer:
Carbon dioxide, oxygen, nitrogen and mixture.

Question 16.
What is glomerulus?
Answer:
The glomerulus is a lift of blood capillaries in a cup-shaped Bowman’s capsule.

Question 17.
What is tidal volume?
Answer:
It is the volume of air breathed in and out normally.

Question 18.
Why gaseous exchange continues in the lungs even after expiration?
Answer:
Because of the presence of residual volume.

Question 19.
Where does the exchange of gases occur in our body?
Answer:
In lung alveoli and tissue cells.

Question 20.
What is vital capacity in regard to breathing?
Answer:
Vital capacity is the volume of air that can be inspired and expired with maximum efforts.

Question 21.
Name excretory organs of earthworm.
Answer:
Nephridia.

Breathing and Exchange of Gases Important Extra Questions Short Answer Type

Question 1.
Define partial pressure of a gas.
Answer:
It is the pressure exerted in a mixture of gases and is equal to the total pressure of the mixture of gases divided by the percentage of that gas in a mixture. The atmospheric air pressure at sea level is 760mm of Hg. Oxygen forms 35% of the air. The partial pressure of oxygen is 760 × 35/ 100 = 266 mmHg.

Question 2.
How would you differentiate between tidal volume and residual volume?
Answer:

1. Tidal volume is the amount of air inhaled or exhaled in one complete breathing. It is about 500 ml.
2. Residual volume is the volume of air that remained in the lungs after the maximum effort of exhalation. It is about 1500ml.

Question 3.
What is the need for a circulatory system in a bigger animal?
Answer:
In larger animals the deeper cells cannot obtain oxygen directly from the atmosphere simply through the process of diffusion or eliminate CO₂. In such a case, the respiratory system transports the respiratory gases from the respiratory’ surface to the deep-lying tissues.

Question 4.
Why does one experience difficulty at a high altitude?
Answer:
At high altitude, the pressure of air falls and the person cannot get enough oxygen in the lungs for diffusion in the blood. Due to insufficient O_2, the person has difficulty breathing at high altitude. The person feels difficulty such as breathlessness, headache, dizziness, irritability, nausea, vomiting, mental fatigue and a blush (///) on the skin, nails and lips.

Question 5.
What are the conditions essential for effective respiration?
Answer:
Conditions essential for effective respiration:

1. The respiratory surface should be thin and permeable to O₂ and CO₂
2. The rich supply of blood to the respiratory surface.
3. Passage for bringing oxygen to the respiratory surface and removing CO₂ through the same passage.
4. The respiratory surface should be moist.
5. Presence of a circulatory system.
6. Presence of a respiratory pigment to carry out the respiratory gases (CO₂ and O₂)

Question 6.
What is a specialized respiratory surface and what are its advantages?
Answer:
A specialized respiratory surface is thin, moist and highly vascular. It remains in contact with the environment outside the body and tissues inside the body. Diffusion of gases takes place from the respiratory surface between the body and outside the environment. The epidermal capillaries release carbon dioxide and take up oxygen dissolved in the film of surface moisture.

Question 7.
What is respiration?
Answer:
A process of physiochemical change by which environmental oxygen is taken in to oxidise the stored food to release CO₂ water and energy; the energy released is used for doing various life activities whereas CO₂ being foul gas is thrown out from the body. The main source of energy are carbohydrates, lipids and proteins. Respiratory mediums are air and water.

Question 8.
Define the
(a) Inspiratory reserve volume (IRV)
Answer:
Inspiratory reserve volume: It is the amount of air that can be inhaled forcibly after a normal inspiration. It is about 200 – 250 ml.

(b) Expiratory reserve volume (ERV)
Answer:
Expiratory reserve volume: It is the volume of ah, which can be exhaled forcibly after a normal expiration. It is about 1000 – 1500 ml.

(c) Vital capacity of lungs (VC)
Answer:
Vital capacity of lungs: It is the amount of air that one can exhale with maximum effort. It is about 3500-4500 ml.

(d) Residual volume. (RV)
Answer:
Residual volume: The amount of air left in the lungs after forcible expiration is called residual volume. It is about 1500ml.

Question 9.
What is Bronchial Asthma? How it is caused? What are the symptoms of this disease?
Answer:
It is characterised by the spasm of the smooth muscles present in the walls of the bronchiole. It is generally caused due to the hypersensitivity of the bronchiole to the foreign substances present in the air passing through it.

The symptoms of this disease are coughing, difficulty in breathing mainly due to expiration, the mucous membrane starts secreting an excess amount of mucous.

Question 10.
What is the preventive measure of the disease Bronchitis?
Answer:
Bronchitis is caused by cigarette smoking and exposure to air pollutants like carbon monoxide. It can be prevented and cured by avoiding exposure to the cause i.e. smoke, chemicals and pollutants. The underlying infection of the disease is treated with suitable antibiotics Bronchodilator drugs (for widening the constriction of bronchial passage by relaxing the smooth muscles) provide symptomatic relief.

Question 11.
How respiration fulfil the energy requirement of an organism?
Answer:
Respiration is a catabolic process. A catabolic biochemical process of exchange of gases by which atmospheric oxygen is taken in to oxidise the stored food to liberate energy, CO2 and water. Energy set free is used for doing life activities. For oxidation usually, oxygen is used. The energy is released in steps from the continuous breakdown of foodstuffs and is stored in the high energy bonds of ATP molecules. This energy obtained from ATP molecules is termed biologically useful energy as it allows the working of all life processes.

Question 12.
What is chloride shift? Write its significance during respiration.
Answer:
The chloride ions (CI^–) inside RBC combine with potassium ion (K⁺) to form potassium chloride (KCL), whereas hydrogen carbonate ions (HCO^–₃) in the plasma combine with Na’ to form sodium hydrogen carbonate (NaHCO₃) Nearly 70% of carbon dioxide is transported from tissues to the lungs in this form.

In response to chloride ions (CI^–) diffuse from plasma into erythrocytes to maintain the ionic balance. This is called the chloride shift.

Significance: It maintains electrochemical neutrality during respiration.

Question 13.
Write true or false.
(a) Inspiratory reserve volume is the volume of air, which can be inspired in addition to the normal inspiration.
Answer:
False

(b) Vital capacity is a measure of maximum inspiration.
Answer:
true

(c) During the gaseous exchange the gases diffuse from high partial pressure to low partial pressure.
Answer:
true

(d) Carbon dioxide cannot be transported with haemoglobin.
Answer:
true

(e) Earthworm respires through parapodia.
Answer:
false.

Question 14.
What is the role of the carbonic anhydrase enzyme in the transport of gases during respiration?
Answer:
Carbon dioxide produced by the tissues diffuses passively into the bloodstream and passes into the red blood corpuscles where it reacts with water to form carbonic acid (H₂CO₃). This reaction is catalysed by the enzyme, carbonic anhydrase found in the erythrocytes and takes less than one second to complete the process. Immediately after its formation, carbonic acid dissociates into hydrogen (H⁺) and bicarbonate (HCO₃^–) ions. The majority of bicarbonate ions (HCO₃^– ) formed within the erythrocytes diffuse out into the plasma along a concentration gradient. These combine with haemoglobin to form the haemoglobin acid (H.Hb).

Question 15.
What is partial pressure? How does it help in gaseous exchange during respiration?
Answer:
During inspiration and expiration, gases move freely by the process of diffusion. Diffusion of any molecule takes place from high to low concentration. The process of diffusion is directly proportional to the pressure caused by the gas alone. The pressure exerted by an individual gas is called partial pressure. It is represented as PO₂, PCO₂, and PN₂, for oxygen, carbon dioxide and nitrogen respectively.

The inspired air ultimately reaches the alveoli of the lung, which in turn receives the blood supply of the pulmonary circulation. At this stage the oxygen of the inspired air is taken in by the blood and carbon dioxide is released into the alveoli for expiration.

In this way, the gases exchange takes place due to partial pressure.

Question 16.
How does haemoglobin help in the transport of oxygen from the lung to tissues?
Answer:
Blood is the medium for the transport of oxygen from the respiratory organ to the different tissues and carbon dioxide from tissues to the respiratory organs. 97% of the oxygen is transported from the lungs to the tissues in combination with haemoglobin (Hb + O₂ — HbO₂), oxyhaemoglobin and 3% is transported in dissolved condition by the plasma.

Under high partial pressure oxygen easily binds with haemoglobin in the pulmonary capillaries. When this oxygenated blood reaches the different tissues, the partial pressure of oxygen declines and the bonds holding oxygen to haemoglobin become unstable. As a result, oxygen is released from the capillaries.

Question 17.
Write the names of the respiratory organs present in human.
Answer:
The human respiratory system consists of external nares or nostrils, nasal cavity, nasopharynx, larynx, trachea, bronchi, bronchiole and lungs.

Question 18.
How skin of the earthworm helps in respiration?
Answer:
Earthworms exchange O₂ and CO₂ between their looped epidermal blood capillaries and their body surface have a moist film. The epidermal capillaries release carbon dioxide; take up the oxygen dissolved in the film of surface moisture.

Question 19.
Fill in the blanks.
Answer:
(a) 15ml of oxygen is transported per decilitre of blood,
(b) Total lung capacity is 3400 to 4800 ml.
(c) There are 10 pairs of spiracles in the cockroach.
(d) Lung is enclosed by a pleural membrane
(e) Streptococcus bacteria causes pneumonia.

Question 20.
Explain breathing disorders in brief.
Answer:

1. Asthma is caused by an allergic reaction. There is difficulty in breathing.
2. Pneumonia is caused by bacterial infection. There are fever, pain and severe cough.
3. Tuberculosis is an infectious bacterial disease of the lungs and in serious cases, blood may come out while coughing.

Question 21.
With the help of arrow marks show1 the sequence of airflow up to lungs.
Answer:
Air → Nostrils → Nasal cavity → Pharynx → Larynx → Bronchi Bronchioles → Lung, alveoli

Question 22.
In what form O₂ is carried in blood? What happens to it when blood reaches the tissue?
Answer:
O₂ is carried in combination with the haemoglobin of RBCs and forms oxyhaemoglobin.

In tissues, there is the dissociation of oxyhaemoglobin and release of Or It diffuses into the tissue cells where it is used in oxidation.

Breathing and Exchange of Gases Important Extra Questions Long Answer Type

Question 1.
Explain gas transport in the blood.
Answer:
It may be explained in two steps.
(a) Transport of O₂ from lungs to tissues.
(b) Transport of CO₂ from tissues to lungs.

A. Oxygen Transport

1. O₂ is transported in the blood via haemoglobin.
2. O₂ diffuses into RBC and combines with haemoglobin to form oxyhaemoglobin.
3. Oxyhaemoglobin breaks into haemoglobin and oxygen at the tissues, where there are high PCO₂ and PO₂.
4. In the lungs, oxyhaemoglobin is formed due to high PO₂ and low PCO₂.

B. CO₂ Transport: CO₂ is transported in 3 ways with blood.

1. 70% of CO₂ in RBC reacts with H₂O to form H₂CO₃
   
2. The rest 30% CO₂ combining with Hb to form carbon haemoglobin. (HCO₃^– carried by RBC and plasma)
3. Some CO₂ dissolves in plasma on reaching the lungs.
   HCO₃^– + H + H₂CO₃
   H₂CO₃ CO₂ + 2H₂O
   And this CO₂ is expelled out through the lungs.

Question 2.
Name and explain the respiratory organs of the following,
(i) Insect
Answer:
Insect: The integument of insects is thick and highly impermeable to minimise the loss of water through the environment. The exchange of gases cannot take place through the skin covering of these insects. These insects have a highly developed complex system called the tracheal. This mode of respiration is called tracheal respiration.

(ii) Neries
Answer:
Neries: Parapodia is the respiratory oxygen in neries. In this organism respiratory occurs through the skin covering the parapodia (Locomotory organs), which is again very thin, moist, permeable and highly vascular.

(iii) Prawn
Answer:
Prawn: Gills, in the animals like prawns, certain molluscs, fishes, tadpoles, the process of gaseous exchange occur by special respiratory organs called gills. These are richly supplied with blood and readily absorb oxygen found dissolved in water and release CO₂ back into the water.

(iv) Birds
Answer:
Birds: (lungs). In birds and mammals, the skin is impermeable. These have a high metabolic rate and their oxygen requirement is very high. Birds have spongy lungs to have a more extensive respiratory surface. These lungs always remain in the body to keep the respiratory surface moist, which is necessary for the exchange of respiratory gases.

(v) Fishes
Answer:
Oxygen and carbon dioxide dissolves in water, and most fishes exchange dissolved oxygen and carbon dioxide in water by means of the gills.

(iv) Earthworm.
Answer:
Earthworms do not have lungs. They breathe through their skin. Oxygen and carbon dioxide pass through the earthworm’s skin by diffusion

Question 3.
Define the following terms:
(a) Anaerobic respiration,
Answer:
Anaerobic respiration: It is a process that does not involve the use of molecular oxygen. Food is not completely oxidised to CO₂ and water. Less energy is present in anaerobic respiration.

(b) Breathing,
Answer:
Breathing: It is a physical process, which brings in fresh air to the respiratory surface and removes foul impure airs from the outside. It occurs outside the cells and is thus an extracellular process.

(c) Vital capacity,
Answer:
Vital capacity: It is defined as an important measure of pulmonary capacity. It is the maximum amount of air a person can expel from the lungs after first filling the lungs to their maximum extent.

Vital capacity is the sum total of inspiration reserve volume, tidal volume and expiratory reserve volume.
(1 + 1 + VC = IRV = TV/ERV)

(d) Tidal volume,
Answer:
Tidal volume: It is defined as the volume of air normally inspired or expired in one breath without doing any effort. It is about 500 ml in an adult person. It represents the volume of air, which is renewed in the respiratory system during every breathing.

(e) Respiratory centre.
Answer:
Respiratory centre: A number of groups of neurons located bilaterally in the medulla oblongata control the respiratory. These are called respiratory centres. These centres are named the dorsal respiratory group. Ventral respiratory group and pneumatic centre.

Question 4.
Write the role of the diaphragm and its Costals muscles in the breathing process.
Answer:
During breathing, when the lungs contract their volumes decrease resulting in the increase of air pressure in the lungs. Hence, the air is exhaled from the lungs. These two processes are called inspiration and expiration. During normal breathing, the downward and upward movement of the diaphragm takes place. When the diaphragm, contracts, the lower surface of the lung is pulled downward consequently the volume of the lungs increases.

This causes the inhalation of air or inspiration. When the diaphragm relaxes, lungs are compressed and air exhaled, expiration takes place. The demand for extra oxygen is fulfilled by the expansion of the rib cage, during exercise when the rate of breathing increases.

During expiration, high pressure is generated in the lungs and air moves out. The upward movement of the rib cage is caused mainly by the external intercostals muscles present between the ribs along with the assistance of few other adjacent muscles.

Similarly, the downward movement of the rib cage is facilitated by the internal intercostals, external oblique and internal oblique muscles, position of the diaphragm, ribs and sternum during breathing as shown in the diagram

Position of diaphragm, ribs and sternum during breathing

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 15 Plant Growth and Development. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 15 Important Extra Questions Plant Growth and Development

Plant Growth and Development Important Extra Questions Very Short Answer Type

Question 1.
In which phase of the growth curve the growth is maximum?
Answer:
Exponential phase

Question 2.
Write the full form of IAA?
Answer:
Indole acetic acid

Question 3.
Which plant hormone controls the process of apical dominance?
Answer:
Auxin

Question 4.
Which hormone acts as a “stress hormone”?
Answer:
Abscisic acid (ABA)

Question 5.
Name the only gaseous natural plant growth regulator.
Answer:
Ethylene.

Question 6.
What are photo plastic seeds?
Answer:
Photo plastic seeds require light for germination.

Question 7.
What is the exponential period of growth?
Answer:
It is the second phase of maximum growth.

Question 8.
What is vernalization?
Answer:
Vernalization is a promoter of flowering by previous cold treatment.

Question 9.
What is senescence?
Answer:
It is the period between reproductive maturity and the death of a plant.

Question 10.
What will happen if short-day plants are exposed in day lengths in excess of their certain critical photoperiod?
Answer:
These will remain only vegetative.

Question 11.
Name the three stages of cellular growth
Answer:

1. Cell division
2. Cell enlargement
3. Cell maturation

Question 12.
Name the plant hormones concerned with the following:
(i) Elongation of cell
Answer:
Auxins

(ii) Shedding of leaves
Answer:
Abscisic acid

(iii) Breaking of seed dormancy
Answer:
Gibberellins

Question 13.
List four growth regulators of plants
Answer:

1. Auxins
2. Gibberellins
3. Cytokinins
4. Abscisic acid

Question 14.
Mention two functions of auxins
Answer:

1. Promote cell elongation
2. Bring about rooting on stem cutting

Question 15.
Does far-red light help in growth?
Answer:
Far-red light brings about flowering in short-day plants.

Question 16.
Name the chemical which carries stimulus for flowering from leaves.
Answer:
Florigen

Question 17.
Where is auxin synthesized in the plants
Answer:
Meristems, enlarging tissues and young leaves.

Question 18.
Which hormone can be used to prevent abscission?
Answer:
Auxin (growth hormone)

Question 19.
Define photoperiodism.
Answer:
The response of plants in terms of flowering to the relative length of day and night is called photoperiodism.

Question 20.
Define development.
Answer:
When the cells change their shapes and take up specialized functions, it is called development.

Plant Growth and Development Important Extra Questions Short Answer Type

Question 1.
Draw a diagram to show the sigmoid growth curve and write the names of the three phases in it.
Answer:
The rate of growth whether measured as length, area, volume or weight is not uniform. Under ideal conditions when the rate of growth is plotted against time, an S-shaped curve called the sigmoid curve.

1. Lag-phase: Growth is slow in the initial stage.
2. Exponential period: It is the second period of maximum growth.
3. Stationary phase: When the nutrients become limiting growth slows down.

S-shaped or sigmoid population growth curve characteristic of many species when introduced into a variable new environment.

Question 2.
What is vernalization? Give its importance in flowering plants.
Answer:
The term vernalizations are the promotor of flowering by previous cold treatment. In flowering plants, plants that require cold treatment usually behave as biennials. They germinate and grow vegetatively in the first season and produce flowers in the second season. It is now definitely known that by various grafting experiments the growing point is the site that receives the cold stimulus. It is responsible for the productions of a hormone-like substance called remain.

The effect of vernalization can be removed if plants are again treated with high temperatures.

Question 3.
Explain the biological meaning of growth. In what essential ways does plant growth differ from animal growth?
Answer:
Growth is the sum total of various processes that combine to cause an irreversible increase in mass, weight, or volume. The growth is invariably accompanied by differentiation, which is explained by quantitative changes in terms of the structure and functions of the cell. Plant growth differs from animal growth in its unlimited and undefined pattern of growth.

Question 4.
Explain how the method of science operated in the discovery of auxins?
Answer:
The discovery of auxin was the result of an investigation by Darwin (1880) while studying the bending of the coleoptile of phalaris sp (grass) towards the light. He established that the tip of coleoptiles was able to perceive the light stimulus.

The light stimulus was transmitted to the sub-apical region where differential growth caused bending. A hypothesis was formulated that there is a transmitter. Boysen-Jensen (1913) demonstrated experimentally. In (1928) it was finally proved the existence of a chemical transmitter and called the substance Auxin.

Question 5.
Discuss the role of growth regulators in agriculture.
Answer:
Growth regulators play important role in agriculture:

1. Dorminy of seed is broken within a few time
2. The miniature of the plait body is improved.
3. Time of germination becomes less.
4. Some plant growth regulators are IBA, IAA
5. Initiation and promotion of cell division are very useful in tissue culture by growth regulators.

Question 6.
Explain Bolting.
Answer:
Just prior to the reproductive phase in ‘rosette’ plants like cabbage, the internodes elongate enormously causing a marked increase in stem height. This is called Bolting. In natural conditions, bolting requires either long days or cold nights.

Question 7.
Write the functions of Auxin (IAA).
Answer:

1. Auxin promotes elongation and growth of stems and roots and the enlargement of fruits by stimulating cell walls to stretch in more than one direction.
2. Auxin promotes cell division in vascular cambium.
3. Auxin promotes root initiation.
4. It causes the development of callus in tissue cultures.
5. Auxin is also involved in apical dominance and abscission.

Question 8.
What part of the plant perceives the light response?
Answer:
It has been demonstrated that a plant from which all leaves have been removed fails to flower even under the inductive light regime. This has been confirmed from experiments with Xanthium, a short-day plant. Even if one eight of a leaf was exposed to short days, flowering occurred. Even a single leaf exposed to a short day was able to induce flowering when it was grafted onto a plant kept under non-inductive conditions.

Question 9.
Distinguish between phototropism and photoperiodism.
Answer:

Phototropism	Photoperiodism
(i) Phototropism is tropic movement in response to light.	(i) Photoperiodism is a physiological response to the changes in relative lengths of day and night.
(ii) It is caused by the differential growth in the elongation zone.	(ii) It is caused by the replacement of vegetative bud into reproductive bud

Question 10.
Distinguish between Long day plants and Short-day plants.
Answer:

Long day plants	Short day plants
(i) These plants begin flowering when the day length exceeds a critical length	(i) These plants begin flowering when the day length is shorter than a critical length.
(ii) Light period is critical for flowering, i.e., they require darkness below the critical period. hence called short-night plants.	(ii) Long continuous and uninterrupted dark period is critical for flowering, i.e., they require darkness above a critical Level hence called long-night plants.
(iii) Supply of gibberellins does not induce flowering under non-inductive photoperiods	(iii) Supply of gibberellins induces flowering under non-inductive photoperiods.

Question 11.
What do you understand by photoperiodism and vernalization? Describe their significance.
Answer:
Some plants require the exposure of light for a longer period and some requirements for a shorter period than the critical period for flowering.

The plants requiring a longer exposure to light in than critical period are called long-day plants, whereas those requiring light for a shorter period are called short-day plants and the remaining come under the class or neutral or intermediate day plants.

The ability of the plant to detect and respond to the length of the daily period of light or more precisely, the relative length of day and night to which the plant is exposed, is called photoperiodism.

Significance: Photoperiodism plays a decisive role in the flowering process.

Vernalization: The method of inducing early flowering by pretreatment of seeds with a certain low temperature is known as vernalization.

The effect of temperature on the growth of plants, especially for flowering. The plants from the temperate regions require a period of low temperature before flowering takes place.

Significance: Significance of temperature between 1 – 10° C to certain varieties of wheat, rice, millets, and cotton accelerates the growth of seedlings and results in early flowering.

Question 12.
Write an essay on growth regulators in plants.
Answer:
The analysis or growth curves provide evidence of physiological control on growth. Plants produce some specific chemical substances, which are capable of moving from one organ to the other so as to produce their effect on growth. These substances which are active in very small amounts are called plant hormones. They are organic compounds and are capable of influencing physiological activities leading to promotion, inhibition, and modification of growth.

The growth regulatory substances are grouped under five major classes, namely auxins, gibberellins, cytokinins, ethylene, and abscisic acid. The other related growth regulators are jasmonic acid, salicylic acid, and brassinosteroids. Some vitamins also regulate the growth of plants.

Organic substances, other than nutrients, which in low concentration regulate growth, development, and differentiation, are termed, growth regulators. These may either promote or inhibit growth. The growth regulators synthesized are not produced naturally within the plants.

Growth hormones or phytohormones are organic substances, which are synthesized in one part of the plant and transported to another part where these affect a specific physiological process to regulate growth. Growth hormones play a very important role in the growth and development of animals.

Question 13.
How will you measure the rate of growth? Describe an instrument used to measure the increase in height of an angiosperm plant.
Answer:
The growth in length can easily be measured with the help of an ordinary measuring scale at an interval of time. For precise measurement, the equipment named auxanometer or autograph can be used. An auxanometer is used to measure the rate of growth of a plant in terms of shoot length.

A thread is tied to the growing tip of a potted plant and at the other end, weight is tied after passing the thread over a pulley. The needle attached in the center of the pulley will show the deflection, which can be read on the graduated arc to find out the increase in length of the plant.

Measurement of growth by Arc-Aux. nanometer

The growth can also be measured by an increase in weight, both fresh and dry volume of the plant. The increase in the number of cells especially in algae, yeast, and bacteria also gives an idea about the rate of growth.

The measurement of the area of the volume of an organ of the plant will also provide information about the rate of growth.

Question 14.
“Growth is an important phenomenon of living.” Justify this statement with reasons.
Answer:
Growth is one of the most important phenomena of all living organisms. Growth in plants occurs by cell division and cell enlargement, invariably followed by cell differentiation. It is the result of coordination of biophysical increase in size, weight, and volume of an organism or its part. In plants, it is associated with both anabolic and catabolic activities, involve an increase in size.

Thus, growth is a quantitative phenomenon and can be measured in relation to time. Growth in living organisms is intrinsic and differs from extrinsic growth in non-living objects. Plants also show movement due to growth. Hence growth is an important phenomenon in living organisms.

Question 15.
What are the important characteristics of growth? Describe in brief.
Answer:
Growth in plants occurs by cell division and cell differentiation. The cell division generally occurs in apical regions of shoot and root. The meristematic cells present at shoot and root apices are responsible for growth in plants. These cells are also present in vascular bundles of roots and stems of dicot plants. They help in increasing the thickness of stem and root due to secondary growth.

The rate of plant growth is slow in the initial stages and increases rapidly later op. The growth slows down due to the limitation of nutrients. The rate of growth is also called the efficiency index.

Question 16.
What are the different phases of growth? Explain with the help of well-labeled diagrams.
Answer:
Growth in plants is localized in the meristematic region only, i.e., apical, lateral and even, intercalary regions. The growth in length is due to the enlargement and elongation of cells at the apical regions and in thickness due to the activity of lateral and intercalary meristems.

The period of growth is generally divided into three phases, namely, formative, elongation, and maturation.

The formative phase: This phase has constantly dividing cells and restricted to the apical meristem both at the root and shoot tips. The cells of this region are rich in protoplasm with a large nucleus and thin cellulose wall.

Growth regions in a root by parallel line (Different phases of growth)

The phase of elongation: It lies just behind the formative phase and is aimed at the enlargement of cells.

The phase of maturation: Is further behind and here the cells start maturing to obtain a permanent size. These phases are also known as regions. The time interval from the formative phase to the maturation phase is called the grand period of growth.

Plant Growth and Development Important Extra Questions Long Answer Type

Question 1.
What do you understand by senescence? What are the various types of senescence observed in plants? Can growth regulators restart senescence?
Answer:
1. Senescence: Senescence may be defined as the period between reproductive maturity and death of a plant or a plant part. Senescence is accompanied by a reduction in functional capacity and an increase in cellular breakdown and metabolic failures. This process ultimately leads to the complete loss of organisms or plant parts.

2. Types of senescence: In plants, it is of four types:
1. Whole plant senescence: It occurs in plants in which the whole plant dies after seed production e.g., wheat, gram, etc. These plants are annuals and die after seed production. This phenomenon also occurs in some monocarpic plants, which live for several years but flower once. For example, certain bamboos and sago palms.

2. Sequential senescence: In some perennial plants, the tips of the main shoot and branches remain in the meristematic stage. They continue to produce new buds and leaves. The older leaves and lateral organs senescence and die. This type of senescence is called sequential senescence. Example: Mango and Eucalyptus.

Types of plant senescence

3. Shoot senescence: In certain perennial plants, the aerial shoot dies each year after flowering and fruiting. But the underground modified stem and roots survive under unfavorable conditions. These parts give rise to new shoots again next year under favorable conditions. Example: Banana and gladiolus.

4. Simultaneous or synchronous senescence: This occurs in temperate deciduous trees which shed their leaves annually in autumn. Example: Maple and elm.

5. Reduction of senescence: Cytokinins, the growth regulators retard senescence. They prevent the breakdown of proteins and other biomolecules. Instead, they stimulate the rate of synthesis of proteins and their mobilization. Auxins also retard senescence.

Question 2.
Mention any two causes of seed dormancy. Give its significance.
Answer:
Many seeds fail to germinate even when they are provided with favorable conditions. This phase when the seed remains in action is called seed dormancy. This natural barrier for development is gradually overcome with time. Sometimes this dormancy is due to the conditions in the seed itself then it is known as innate dormancy.

It may be due to the following reasons:

1. Impermeability of seed coat: In some plants, the embryo is undifferentiated and unorganized when the seed is shed. It takes time to attain full development before it germinates.
2. Due to immature embryo: In some plants, the embryo is un-differentiated and unorganized when the seed is shed. It takes time to attain full development before it germinates.

Significance of seed dormancy:

1. It enables the seed to be disseminated in time and space.
2. It helps them to germinate when environmental conditions are more favorable.

Advantages of reproduction by seed:

1. The plant is independent of water for sexual reproduction and therefore, better adapted for a land environment.
2. The seed protects the embryo.
3. The seed contains food for the embryo (either or cotyledons or in the endosperm).
4. The seed is usually adapted for dispersal.
5. The seed can remain dormant and survive adverse conditions.
6. The seed is physiologically sensitive to favorable conditions and sometimes must undergo a period of after-ripening so that it will not germinate immediately.
7. The seed is a product of sexual reproduction and, therefore, has the attendant advantages of genetic variation.

Question 3.
What is the difference between:
(a) Nastic and tropic movements.
Answer:
Difference between Nastic and tropic movements:

Tronic Movements	Nastic Movements
(1) These are paratonic directional movements of curvature.	(1) These are paratonic non-directional movements of curvature.
(2) These are growth movements caused due to differential growth only.	(2) These are either growth or turgor movements, caused due to differential growth or changes in turgor.
(3) These occur in radically symmetrical organs like root and stem.	(3) These occur in either symmetric organs or those having bilateral symmetry.

(b) Phototropism and geotropism.
Answer:
Difference between Phototropism and geotropism:

Phototropism	Geotropism
(1) It is the directional movement of curvature caused by light.	(1) It is the unidirectional movement of curvature caused by gravitational pull.
(2) The response depends upon the direction of light.	(2) Different organs show different types of geotropic responses. The stem shows a negative geotropic response.
(3) The stimulus is perceived by the apical meristem, perception is done by yellowish pigment like carotenoids.	(3) The stimulus of gravitational pull is perceived by the root cap in the root and apex of the stem.
(4) It occurs due to differential growth of the organs in the elongation zone.	(4) The geotropic response is caused due to unequal distribution of auxin on two sides of the horizontally growing axis.

Question 4.
What do you understand by the spontaneous and induced movements in plants? Illustrate your answer with suitable diagrams.
Answer:
Spontaneous movements: Plastids in cells may show movements in response to light, the stem of a plant grows upwards against the force of gravity or bends towards light. Spontaneous movement may be at protoplasm or organ or even at the whole plant level. The protoplasmic movements, accomplished by naked protoplasm in unicellular plants, are generally divided into ciliary, amoeboid, cyclosis, and gliding. Spontaneous movement is self-controlled.


Induced movement: The induced (paratonic) movement is the movement of a complete cell or organelle and is influenced by external stimuli. This is also called tactic (taxes or taxis) movement and is common among lower plants. The movement may be due to chemical substances, such as sucrose and malic acid, present in archegonium of ferns and moss, which attract spermatozoids; and mobile bacteria are attached by peptone.

Different types of movement.

Question 5.
Describe the various steps involved in seed germination?
Answer:
Awakening of inactive embryos present inside the seed into a seedling, capable of independent existence, is termed germination.

Two conditions affect the germination of seed:
(a) External conditions: External conditions necessary for germi¬nation are water, oxygen, temperature, and light.
(b) Internal conditions: Sometimes the seeds fail to germinate even if various external conditions are favorable.

Germination is also controlled by certain internal factors which are discussed as:

1. Maturity of the embryo: In certain, plants, the embryo is immature in fully developed seeds.
2. After ripening: In certain, plants, even if the embryo is mature but they do not possess the necessary growth hormones. These germinate only when necessary growth hormones are synthesized.
3. Dormancy period: Certain seeds remain dormant after their shedding.
4. Viability period: The period for which the embryo in seed remains living is termed as viability period. The seeds germinate only within the viability period.
5. Reserve food material: The availability of sufficient reserve food material is essential for germination.
6. Enzymes and growth hormones: Digestive and respiratory enzymes play an essential role in germination. In the absence of activation of certain enzymes, seeds will not germinate. Hormones like auxins, gibberellins, and cytokinins also play important role in germination.

Seed germination includes all the physical and physiological changes that occur in the seed.
(a) Absorption of water: The water enters the seed mainly through the micropyle and imbibed by the reserve food and cell wall material, resulting in swelling of the seed. It causes the rupture of seed coats, allowing the radical to grow into the primary root. Aerobic respiration is essential for seed germination as it makes available the energy needed for the growth of the embryo.

(b) Mobilization of reserve food materials: The food is stored in the endosperm, or in cotyledons. The various enzymes convent this food into soluble substances, which serve as a respiratory substrate, and the energy released during aerobic respiration is used in various metabolic and physiological changes dining embryo growth. The digested food passed towards the embryo through cotyledon.

(c) Development of embryo into seedling: The embryonic cells become metabolically very active. The radicle starts growing and is first to emerge out through ruptured seed coats. It develops into a root that grows downwards into the soil. The root developing from the radicle is termed as the primary root.

After the development of the primary root from the radicle, either the epicotyl or the hypocotyl starts elongation, forming dicotyledonous seeds.

Different stages of germination of seed.

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 14 Respiration in Plants. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 14 Important Extra Questions Respiration in Plants

Respiration in Plants Important Extra Questions Very Short Answer Type

Question 1.
Whether respiration is a catabolic or anabolic process?
Answer:
Catabolic (destructive)

Question 2.
Name the cell organelle where cellular respiration takes place.
Answer:
Mitochondria

Question 3.
Give the chemical equation for aerobic respiration.
Answer:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂ + 686 kcal.

Question 4.
Name the substance that is oxidized during respiration.
Answer:
Glucose

Question 5.
What is fermentation?
Answer:
Respiration by microorganisms without the utilization of oxygen is called fermentation.

Question 6.
What is the main source of energy?
Answer:
Carbohydrates, lipids, and proteins.

Question 7.
In what form the energy released by oxidation is stored in the body?
Answer:
In the high-energy bonds of ATP molecules.

Question 8.
Name two respiratory, mediums for living beings.
Answer:
Air and water.

Question 9.
What is cell respiration?
Answer:
Enzymatic oxidation of food in body cells is known as cell respiration.

Question 10.
What holds the energy in the body?
Answer:
Molecules of food hold energy in their chemical bonds.

Question 11.
Name some industrial products of fermentation.
Answer:
Wine, Alcohol, Vinegar, Bread, etc.

Question 12.
Define anaerobic respiration.
Answer:
Respiration carried out in the absence of oxygen is called anaerobic respiration.

Question 13.
Define RQ is words
Answer:
RQ is the ratio between the amount of CO₂ evolved to the amount of oxygen used.

Question 14.
How is oxygen distributed in the plant body?
Answer:
By diffusion from a cell of the cell.

Question 15.
Name the parts of the plant body that allow the exchange of gases.
Answer:
Lenticels, stomata, and general body surface.

Question 16.
How many ATP molecules are produced in aerobic respira¬tion?
Answer:
36

Question 17.
When a molecule like that of C₆H₁₂O₆ is oxidized completely, what are its end products?
Answer:
CO₂ and water,

Question 18.
Name two Anti-transpirants.
Answer:
Abscisic acid, Phenyl mercuric acetate.

Question 19.
What metal element is involved in the movement of stomata?
Answer:
Potassium

Question 20.
Why is respiration necessary for the plant?
Answer:
Respiration is necessary as it releases the energy which is used for daily activities.

Question 21.
Why is less energy released during anaerobic respiration?
Answer:
During anaerobic respiration, food is incompletely oxidized and so less energy is released.

Question 22.
What is fermentation?
Answer:
Anaerobic respiration by some microbes is called fermentation.

Respiration in Plants Important Extra Questions Short Answer Type

Question 1.
What is respiration?
Answer:
A process of physiochemical change by which environmental oxygen is taken in to oxidize the stored food or release CO₂, water, and energy. The energy released is used for doing various life activities, whereas CO₂ being is used by the plants for their growth and development.

Question 2.
Define aerobic respiration.
Answer:
The process of release of energy through intake of molecular oxygen and release of CO₂ is known as Aerobic respiration.

Question 3.
Define the process of fermentation showing chemical equation.
Answer:
In this process, the carbohydrate is incompletely oxidized into some carbonic compounds such as ethyl alcohol, acetic acid, lactic acid, and CO₂. This process of oxygen being carried out in microbes is known as fermentation.
C₆H₁₂O₆ → 2 C₂H₅OH + Energy (247 kJ).

Question 4.
On what factor the respiratory quotient depends?
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called the respiratory quotient

RQ (the respiratory quotient) depends upon the type of respiratory substrate used during respiration. This is different for different substrates.

Question 5.
What is a citric acid cycle?
Answer:
This is also known as the Tricarboxylic acid cycle. When acetyl CO- A enters into a reaction to form citric acid and how pyruvate is broken down during metabolism is highlighted by the concept of cycles. This series of reactions is known as the citric acid cycle.

Question 6.
What will be the value of RQ when organic acids are used as respiratory substrate?

Answer:
Organic acids contain more oxygen than carbohydrates; therefore the RQ is more than one. Less amount of oxygen is required for their oxidation.

Question 7.
How many types of respiration occur in plants?
Answer:
Depending upon the availability of oxygen, respiration is of two types.

1. Aerobic respiration
2. Anaerobic respiration.

Aerobic respiration: Complete oxidation of organic substances in presence of oxygen takes place.
Anaerobic respiration: This type of respiration takes place in the complete absence of oxygen.

Question 8.
Describe the process of glycolysis and where it occurs.
Answer:
Glycolysis is the first stage of breakdown of glucose and common in all organisms. In anaerobic organisms, it is only the process of respiration. In this process, glucose undergoes partial oxidation. This process occurs in the cytoplasm of the cell.

Question 9.
What is the role of the citric acid cycle?
Answer:

1. Carbon skeletons are obtained for the growth and maintenance of the cell during this cycle.
2. Many intermediate compounds are formed which are used in the synthesis of biomolecules like amino acids, fats, etc.
3. ATP molecules having high energy are generated during this pathway.

Question 10.
What is a compensation point?
Answer:
At a low concentration of CO₂ and non-limiting light intensity, the photosynthetic rate of a given plant will be equal to the total amount of respiration. The atmospheric concentration of CO₂ at which photosynthesis just compensates for respiration is referred to as the CO₂ compensation point.

Question 11.
Write the significance of the citric acid cycle.
Answer:
During this pathway, carbon skeletons are obtained for use in the growth and maintenance of the cell. Many intermediate compounds are formed, which are used in the synthesis of other biomolecules like amino acids, nucleotides, chlorophyll cytochromes, and fats.

Question 12.
Why does anaerobic respiration produce less energy than aerobic respiration?
Answer:
Because it does not involve the use of molecular oxygen. Food is not completely oxidized to CO₂ and waste. So a lot of energy is still retained with the food. Less energy is produced on anaerobic respiration. It yields only about 5% of the energy available in glucose. Then it is a wasteful process or respiration. In anaerobic respiration, yeast metabolizes glucose to ethanol and CO₂ without any use of molecular oxygen.

Question 13.
What is oxidative phosphorylation?
Answer:
Glucose is completely oxidized by the end of the citric acid cycle but the energy is not released unless NADH and FADH are oxidized. Oxygen acts as the final hydrogen acceptor. The whole process by which oxygen effectively allows the formation of ATP molecule by phosphorylation of ADP is called oxidative phosphorylation.

Question 14.
Fill in the blanks with suitable words.
Answer:
(a) Pyruvic acid oxidized into CO₃ and H₂O before entering the citric acid cycle.
(b) The RQ is more than one if the respiratory substrate is oxalic acid.
(c) Glycolysis takes place in crystal.
(d) The acetyl —COA is accepted by Coenzyme A (a sulfur-containing compound).

Question 15.
Explain the effects of temperature on the rate of respiration.
Answer:
Respiration is reduced at very high (above 50°c) and very low (near freezing temp.) temperatures. This is because enzymes can work best between 30°c – 40°c and get inactivated at very high and very’ low tempera¬tures.

Question 16.
How does the exchange of respiratory gases occur in plants?
Answer:
The gaseous exchange takes place through

1. General body surface
2. Lenticels.
3. Stomata present in leaves and young stems. Oxygen becomes transported from cell to cell by diffusion.

Question 17.
Explain RQ significance.
Answer:
RQ value for carbohydrates is 1. It is less than one if proteins are being burnt and more than one if fats are being burnt. So RQ values are important in identifying the kind of substrate used in respiration.

Question 18.
Describe in detail the aerobic oxidation of pyruvic acid.
Answer:
Pyruvic acid generated in the crystal is transported to mitochondria and initiates the second phase of respiration. Before pyruvic acid enters the Kreb’s cycle, one of its three carbon atoms is oxidized to carbon dioxide in the reaction called oxidative decarboxylation.

Pyruvate is first decarboxylated and then oxidized by the enzyme pyruvate dehydrogenase. The combination of the remaining two-carbon acetate unit is readily accepted by a sulfur-containing compound coenzyme A to form acetyl COA. During the process, NAD is reduced to NADH. This process is represented as.

During this process, two molecules of NADH are produced, and thus, it results in a net gain of 6ATP molecules.

(2 NADH + 3 = 6 ATP). 2 molecules of pyruvic acid produced during glycolysis,

Question 19.
Describe the net gain of ATP during respiration.
Answer:
There is a gain of 36 ATP molecules during aerobic respiration of one molecule of glucose. The detail is given in the table.

In most eukaryotic cells, 2 molecules of ATP are required for transporting the NADH produced in glycolysis into mitochondrial for further oxidation. Hence, the net gain of ATP is 36 molecules.

Table ATP molecules produced during respiration.

Question 20.
Define the following:
(a) Respiration
Answer:
Respiration: It is defined as the phenomenon of the release of energy by oxidation of various organic molecules, for cellular use is known as respiration.

(b) Respiratory substrate
Answer:
Respiratory substrate: The compounds that are oxidized during the process of respiration are called respiratory substrates.

(c) Respiratory quotient
Answer:
Respiratory quotient: During respiration oxygen is used and CO₂ is released. The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called respiratory quotient (RQ).

(d) Anaerobic respiration
Answer:
Anaerobic respiration: The type of respiration, in which the carbohydrate is incompletely oxidized into some carbonic compounds in absence of oxygen, is called Anaerobic respiration.

(e) Aerobic respiration
Answer:
Aerobic respiration: It is that process of respiration which leads to complete oxidation of organic compound in the presence of oxygen.

This type of respiration is common in higher organisms.

(f) Fermentation.
Answer:
Fermentation: In anaerobic respiration yeasts metabolize glucose to ethanol and CO₂ without any use of molecular oxygen. This process is called fermentation in yeasts.

Question 21.
Describe the pentose phosphate pathway.
Answer:
Sometimes oxidation of glucose takes place by another pathway, which is called the pentose phosphate pathway (PPP). In the pentose pathway, glucose-6 phosphate (6C) produced during the early stages of glycolysis or the photosynthates produced during photosynthesis are oxidized to give rise to 6-phosphogluconate. This reaction takes place in the enzyme glucose 6phosphale dehydrogenase and generates NAD PH.

The 6-phosphoglucose molecules are further oxidized by the enzyme 6-phosphogluconate dehydrogenase. As a result of this, one molecule of each ribose-5 phosphate, carbon dioxide, and NADPH is produced, which in turn undergoes many changes to produce glycolytic intermediate. These reactions take place in the cytoplasm.
(From glycolysis)

Question 22.
Calculate the efficiency of respiration in the living system.
Answer:
During aerobic respiration, O₂ is consumed and CO₂ is released. The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called respiratory quotient (RQ) or respiratory ratio.

The RQ (respiratory quotient) depends upon the respiratory substrate. When a carbohydrate is used as a substrate.

C₆H₁₂O₆ – 2C₂H₅OH + 2CO + Energy (247 K.J) on complete oxidation. RQ will be

The total energy yield from 38 ATP molecules comes to 1298 kJ. The energy released by one molecule of glucose on complete oxidation is 2870 kJ. Thus, the efficiency is 45%. Much of the energy generated during respiration is released in the form of heat.

Respiration in Plants Important Extra Questions Long Answer Type

Question 1.
Illustrate the mechanism of the electron transport system.
Answer:
The glucose molecule is completely oxidized by the end of the citric acid cycle. But the energy is not released unless NADH and FADH are oxidized through the electron transport system. Here oxidation means the removal of electrons from it.

The metabolic pathway through which the electron passes from one carrier to another is called the electron transport system (ETS) and it is operative in the inner mitochondrial membrane. Electrons from NADH produced in the mitochondrial matrix are oxidized by an NADH dehydrogenase (Complex I) and electrons are then transferred to ubiquinone.

The ubiquinone located within the inner membrane also receives reducing equivalents via FADH, which is generated during the oxidation of succinate, through the activity of the enzyme, succinate dehydrogenase (complex II). The reduced ubiquinone is then oxidized with the transfer of electrons to the cytochrome complex (Complex III).

Cytochrome is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for the transfer of electrons between complex III and complex IV.

(Complex IV) is cytochrome.

When the electrons pass from one carrier to another via complex 1 to IV in the electron transfer chain, they are coupled to ATP synthase (Complex V) for the production of ATP from ADP and inorganic phosphate. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of FADH, produces 2 molecules of ATP.

The electrons are earned by the cytochromes and recombine with their protons before the final stage when the hydrogen atom is accepted by oxygen to form water. Oxygen acts as the final hydrogen acceptor. The whole process by which oxygen allows the production of ATP by phosphorylation of ADP is called oxidative phosphorylation.

Note: There are two routes by which hydrogen from the substrate molecule passes. In route 1.3 ATP molecules are formed for every pair of hydrogen atoms. In route 2, only 2ATP molecules are formed from one pair of hydrogen atoms.

Oxygen acts as the final hydrogen acceptor and forms water.
NAD = nicotinamide adenine dinucleotide.
MN = flavin mononucleotide,
FAD = flavin adenine dinucleotide.

ETC produces 32 ATP molecules per glucose molecule and is the major source of cell energy.

Electron Transport Chain.

Question 2.
Describe the process and role of the citric acid cycle in living organisms.
Answer:
In the process of respiration, the carbohydrates are converted into pyruvic acid through a series of enzymatic reactions. These reactions are known as glycolysis and take place in the cytosol. The pyruvic acid thus formed enters in mitochondria where O₂ and necessary enzymes are available; the pyruvic acid is finally converted into CO₂ and H₂O. This reaction series is known as Krebs Cycle or Citric acid cycle or Tricarboxylic acid (TCA) cycle.

During this cycle, 3 molecules of NAD and one molecule of FAD (Flavin Adenine Dinucleotide) are reduced to produce NADH and FADH respectively. NADH and FADH, so produced during the citric acid cycle are linked with the electron transport system and produce ATP by oxidative phosphorylation, The summary equation for this phase of respiration may therefore be written as follows:


Kreb’s cycle. It follows glycolytic reactions shown in and pyruvate oxidation.

It involves two processes

1. removal of hydrogen and
2. the breaking off of carbon dioxide units one by one.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 1 Some Basic Concepts of Chemistry. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 1 Important Extra Questions Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry Important Extra Questions Very Short Answer Type

Question 1.
What is the number of significant figures in 1050 × 10⁴?
Answer:
Four.

Question 2.
What is the SI unit of length?
Answer:
Metre.

Question 3.
What is the SI unit of density?
Answer:
kg m^-3.

Question 4.
What is the law called which deals with the ratios of volumes of the gaseous reactants and products?
Answer:
Gay Lussac’s law of combining volumes.

Question 5.
What is one a.m.u. or one u?
Answer:
1 a.m.u. or 1 u = \(\frac{1}{12}\)th of the mass of an atom C-12.

Question 6.
Which isotope of carbon is used for getting relative atomic masses?
Answer:
Carbon-12 (C-12).

Question 7.
Write down the empirical formula of benzene.
Answer:
CH.

Question 8.
The mass of human DNA molecules is 1 FG. What is its mass in kg?
Answer:
10^-18 kg

Question 9.
Vanadium metal is added to steel to impart strength. The density of vanadium is 5.96 g/cm³. Express this in the SI unit.
Answer:
5960 kg/m³.

Question 10.
What physical quantities are represented by the following units and what are their most common names?
(i) kg ms²
Answer:
Force (Newton)

(ii) kg m²s².
Answer:
Work (joule)

Question 11.
If in the reaction HgO(s) → Hg(l) + \(\frac{1}{2}\)O₂(g) 100.0 g of HgO on heating in a closed tube gives 92.6 g of Hg, what is the weight of oxygen formed?
Answer:
Weight of oxygen formed = 100 – 92.6 = 7.4 g.

Question 12.
Name three compounds formed by dinitrogen and dioxygen.
Answer:
1. N₂O (Nitrous Oxide)
2. NO (Nitric Oxide)
3. N₂O₅ (Nitrogen pentoxide).

Question 13.
If 700 mL of H₂ at STP contains x molecules of it, how many molecules of O₂ are present in 700 mL of it at the same temperature and pressure?
Answer:
x molecules.

Question 14.
nitrogen combines with dihydrogen according to the reaction.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
What is the ratio in their volumes under similar conditions of temperature and pressure?
Answer:
The ratio in their volumes is 1: 3: 2.

Question 15.
What is the relationship between molecular weight and vapor density of a gas?
Answer:
Molecular weight = 2 × Vapour density.,

Question 16.
What is the mass in gms of 11.2 L of N₂ at STP?
Answer:
22.4 L of N₂ at STP weighs = 28.0 g.
11.2 L of N₂ at STP weighs = \(\frac{28}{22.4}\) × 11.2 = 14.0 gm.

Question 17.
What is the mass of one molecule of sodium chloride?
Answer:
6.023 × 10²³ molecules of NaCl weighs = 58.5 g

I molecule of NaCI weighs = \(\frac{58.5}{6.023 \times 10^{23}}\) = 9.71 × 10^-23 g.

Question 18.
How many total electrons are present in 1.4 g of nitrogen gas?
Answer:
1.4 g of N₂ = \(\frac{14}{28}\) mol = 0.05 and 0.05 × 6.02 × 10²³ molecules
= 3.01 × 10²¹ molecules
= 3.01 × 10²¹ × 14 electrons [ ∵ 1 Molecules of 2 = 14 ē]
= 4.214 × 10²² electrons.

Question 19.
Why atomic masses are the average values?
Answer:
Most of the elements exist in different isotopes. Hence their average values are taken.

Question 20.
What is the percentage composition of Ca in CaCO₃?
Answer:
% of Ca = \(\frac{40 \times 100}{40+12+3 \times 16}\) = 40.0%

Question 21.
What is the mass of 2 moles of CO₂?
Answer:
1 mole of CO₂ = 1 × 12 + 2 × 16 = 44 g.
2 mole of CO₂ = 2 × 44.0 = 88.0 g.

Question 22.
Express the number 45000 in exponential notation to show
(i) two significant figures
Answer:
4.5 × 10⁴

(ii) four significant
Answer:
4.500 × 10⁴.

Question 23.
How many moles of sulphuric acid are present in 1 dm3 of 0.5 M solution?
Answer:
0.5 mole.

Question 24.
Give one example each of a molecule in which empirical formula and molecular formula are
(i) different
(ii) same.
Answer:

Compound	Empirical formula	Molecular formula
(i) Hydrogen peroxide	HO	H2O2
(ii) Copper sulfate	CuSO4	CuSO4.

Question 25.
The radius of an atom is 10-10 m. What will be its value in micro metre?
Answer:
10^-10 m = \(\frac{10^{-10}}{10^{-6}}\) = 10^-4 micrometer.

Question 26.
Define the limiting reagent.
Answer:
The reacting substance gets used up first in the reactions.

Question 27.
What is meant by 1 gram atom of iron?
Answer:
1 g atom of iron means the atomic mass of iron expressed in g i.e., 56 g.

Question 28.
What is the number of oxygen atoms in one mole of CuSO₄.5H₂O?
Answer:
It is moles of oxygen atoms = 9 × 6.02 × 10²³ = 5.418 × 10²⁴ atoms.

Question 29.
What is the SI unit of velocity?
Answer:
ms-1.

Question 30.
A worker reads the resistance of the wire as 20.01 ohm. What may be the actual range of resistance?
Answer:
20.01 ± 0.01 ohm, i.e., 20.00 to 20.02 ohm.

Question 31.
What is AZT? To which use is it put?
Answer:
Azidothymidine used for AIDS victims.

Question 32.
Write down the empirical formula of acetic acid.
Answer:
CH₂O.

Question 33.
What is the advantage of using Molality over Molarity?
Answer:
Molality does not depend upon temperature.

Question 34.
Balance the equation
C₃Hg(g) + O₂(g) → CO₂(g) + H₂O(1)
Answer:
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

Question 35.
Balance the equation
Fe(s) + O₂(g) → Fe₂O₃(s)
Answer:
4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s)

Question 36.
Balance the equation
P₄(s) + O₂(g) → P₄O₁₀(s)
Answer:
P₄(s) + 5 O₂(g) → P₄O₁₀(s)

Question 37.
What is the SI scale of temperature?
Answer:
Kelvin (K).

Question 38.
What is the relation between Empirical formula and molecular formula?
Answer:
Molecular formula = n × Empirical formula
where n = \(\frac{\text { Molar mass }}{\text { empirical formula mass }}\)

Question 39.
What is the mass of NH₃ in 0.5 mol of it?
Answer:
0.5 mol of NH₃ weights 8.5 g.

Question 40.
What is the mass of 3.01 × 10²² g of SO₂?
Answer:
3.2 g.

Question 41.
How many atoms of silver are present in a 0.5 g atom of silver? At the mass of silver = 108.
Answer:
3.01 × 10²³ atoms.

Question 42.
Which of the two is heavier – 2 g of iron or 2 g atom of iron.
Answer:
2 g atom of iron.

Question 43.
How many molecules of CO₂ are present in 1.12 L of it at STP?
Answer:
3.011 × 10²² molecules.

Question 44.
What is the weight of H₂SO₄ in 0.01 M solution of it?
Answer:
0.98 g.

Question 45.
What volume of CO₂ at STP can be obtained on the decomposition of 100.0 g of CaCO₃?
Answer:
22.4 L.

Question 46.
How many molecules of cane sugar (C₁₂H₂₂O₁₁) are present in 34.20 grams of it.
Answer:
6.022 × 10²² molecules.

Question 47.
How many molecules of water are present in a drop of it having a mass of 0.05 g?
Answer:
1.673 × 10²¹ molecules.

Question 48.
Calculate the mass of 0.1 moles of KNO₃.
[At. wt. of K = 39, N = 14, O = 16]
Answer:
1 Mole of KNO-, = 1 × 39 + 1 × 14 + 3 × 16 = 101 g
∴ 0.1 mole of it = 101 × 0.1 = 10.1 g.

Question 49.
What is the molarity of a solution of oxalic acid containing 0.63 g of it in 250 cm³ of the solution?
Answer:
Molarity = 0.02 M.

Question 50.
A solution of NaCl has been prepared by dissolving 5.85 g of it 1 L of water. What is its molality?
Answer:
0.1 m.

Some Basic Concepts of Chemistry Important Extra Questions Short Answer Type

Question 1.
Define Mole. What is its numerical value?
Answer:
A mole is the amount of a substance that contains as many entities (atoms, molecules, or other particles) as there are atoms in exactly 0.012 kg or 12 g of the carbon-12 isotope.

Its numerical value is 6.023 × 10²³.

Question 2.
Define molarity. Is it affected by a change in temperature?
Answer:
The molarity of a solution is defined as the number of moles of the solute present per liter of the solution. It is represented by the symbol M. Its value changes with the change in temperature.

Question 3.
What do you mean by Precision and accuracy?
Answer:
Precision and accuracy: The term precision refers to the closeness of the set of values obtained from identical measurements of a quantity.

Accuracy refers to the closeness of a single measurement to its true value.

Question 4.
Distinguish between fundamental and the derived units.
Answer:
Fundamental units: Fundamental units are those units by which other physical units can be derived. These are mass (M), Length (L), time (T), temperature (°).

Derived units: The units which are obtained by the combination of the fundamental units are called derived units.

Question 5.
Define molality and write its temperature dependence.
Answer:
Molality is defined as the number of g moles of the solute dissolved per kilogram of the solvent.
Molality (m) = \(\frac{\text { Mole of solute }}{\text { Mass of the solvent in } \mathrm{kg}}\)

The molality of the solution does not depend upon the temperature.

Question 6.
Distinguish between an atom and a molecule.
Answer:
Atom: An atom is the smallest particle of an element that takes part in a chemical reaction. It may or may not be capable of independent existence.

Molecule: It is the smallest particle of a substance (element or compound) that is capable of independent existence

Question 7.
Derive the SI unit of Joule (J) in terms of fundamental units.
Answer:
Joule is the SI unit of work or energy

Question 8.
Define the SI unit of energy.
Answer:
Energy = Force × distance through which point of application of force moves.
S.l. unit of energy = SI unit of force × SI unit of distance.
= 1 Newton × 1 metre
= 1 Nm

But 1 N = 1 kg ms^-2
∴ SI unit of energy = 1 kg ms^-2 × 1 m = 1 kg m²s^-2

The unit of energy is joule (J)
Thus 1 Joule (J) = 1 kg m² s^-2.

Question 9.
One volume of a gaseous compound requires 2 volumes of O₂ for combustion and gives 2 volumes of CO₂ and 1 volume of N₂2. Determine the molecular formula of the compound.
Answer:
X + 2O₂ → 2 CO₂ + N₂ (According to problem)
Since the product has 2 atoms of C (in 2 CO₂) and 2 atoms of N (in N₂) these must have come from X. Hence X is C₂N₂.

Question 10.
What is a chemical equation? What are its essential features?
Answer:
The shorthand representation of a chemical change in terms of symbols and formulae is called a chemical equation.

Essential features:

• It should represent a true chemical change.
• It should be balanced.
• It should be molecular.

Question 11.
Write down the significance of a chemical equation.
Answer:
Significance of a chemical equation:

1. It tells about the reactants taking part in the chemical reaction and products formed as a result of chemical change.
2. It tells about the relative number of atoms or- molecules of reactants and products.
3. It tells about the relative weights of reactants and products

Question 12.
How is mole related to
(a) number of atoms/molecules
Answer:
A mole contains 6.022 × 10²³ atoms/molecules.

(b) mass of the substance?
Answer:
A mole of a substance denotes the molecular mass of that substance.

Question 13.
How is mole related to the volume of a gas?
Answer:
A mole of a gas occupies 22.4 L of the gas measured at STP, i. e., 0° C and 1 atmospheric pressure.

Question 14.
Why atomic masses are the average values?
Answer:
Most of the elements exist in different isotopic forms. Chlorine has 2 isotopes with mass numbers 35 and 37 existing in the ratio of 3: 1. Hence the average value is taken.

Question 15.
How many molecules approximately do you expect to be present in a small crystal of sugar which weighs 10 mg?
Answer:
10 mg sugar (C₁₂H₂₂O₁₁) = 0.01 g = \(\frac{0.01}{342}\) mol
= 2.92 × 10^-5 mole
= 2.92 × 10^-5 × 6.02 × 10²³ molecules
= 1.76 × 10¹⁹ molecules.

Question 16.
Two containers of equal capacity A₁ and A₂ contain 10 g of oxygen (O₂) and ozone (O₃) respectively. Which of the two will have greater no. of O-atoms and which will give greater no. of molecules?
Answer:
10 g of O₂ = \(\frac{10}{32}\) mol = \(\frac{10}{32}\) × 6.02 × 10²³ molecules
= 1.88 × 10²³ molecules
= 3.76 × 10²³ atoms.

10 g of O₂ = \(\frac{10}{48}\) mole = \(\frac{10}{48}\) × 6.02 × 10²³ molecules
= 1.254 × 10²³ molecules
= 3.76 × 10²³ atoms

Thus both A₁ and A₂ contain the same no. of atoms, but A₁ contains more numbers of molecules.

Question 17.
What do you mean by the term ‘Formality’? To what type of compounds it is applied?
Answer:
Ionic compounds like Na⁺Cl^– are not molecular. To express the concentration of their solution the term formality is used in place of molarity. Formality is the number of formula weights present in one liter of a solution.

Question 18.
What is meant by a ‘standard solution’?
Answer:
A standard solution is one whose molarity or normality is known. For example 0.1 M NaOH.

Question 19.
Assuming the density of water to be 1 g/cm³, calculate the volume occupied by one molecule of water.
Answer:
1 Mole of H₂O = 18 g = 18 cm³[∵ density of H₂O = 1 g/cm³]
= 6.022 × 10²³ molecules of H₂O
1 Molecule will have a volume
= \(\frac{18}{6.022 \times 10^{23}}\) cm- = 2.989 × 10^-23 cm³.

Question 20.
Define Avogadro’s number. What is its equal to?
Answer:
Avogadro’s number may be defined as the number or number of molecules present in one gram molecule of the substance.
Avogadro’s No. = 6.022 × 10²³.

Some Basic Concepts of Chemistry Important Extra Questions Long Answer Type

Question 1.
State the law of Multiple Proportions. Explain with two examples.
Answer:
The Law of Multiple Proportions states:
“When two elements combine to form two or more than two chemical compounds than the weights of one of elements which combine with a fixed weight of the other, bear a simple ratio to one another.

Examples:
1. Compound of Carbon and Oxygen: C and O combine to form two compounds CO and CO₂.
In CO₂ 12 parts of wt. of C combined with 16 parts by wt. O.
In CO₂ 12 parts of wt. of C combined with 32 parts by wt. of O.
If the weight of C is fixed at 12 parts by wt. then the ratio in the weights of oxygen which combine with the fixed wt. of C (= 12) is 16: 32 or 1: 2.
Thus the weight of oxygen bears a simple ratio of 1: 2 to each other.

2. Compounds of Sulphate (S) and Oxygen (O):
S forms two oxides with O, viz., SO₂ and SO₃
In SO₂, 32 parts of wt. of S combine with 32 parts by wt. of O.
In SO₃, 32 parts of wt. of S combine with 48 parts by wt. of O.
If the wt. of S is fixed at 32 parts, then’ the ratio in the weights of oxygen which combine with the fixed wt. of S is 32: 48 or 2: 3.

Thus the weights of oxygen bear a simple ratio of 2: 3 to each other.

Question 2.
State the law of Constant Composition. Illustrate with two examples.
Answer:
Law of Constant Composition of Definite Proportions states: “A chemical compound is always found, to be made up of the same elements combined together in the same fixed proportion by weight”.

Examples:

1. CO₂ may be prepared in the laboratory as follows:

In all the above examples, CO₂ is made up of the same elements i. e., Carbon (C) and Oxygen (O) combined together in the same fixed proportion by weight of 12: 32 or 3: 8 by weight.

Question 3.
Define empirical formula and molecular formula. How will you establish a relationship between the two? Give examples.
Answer:
The empirical formula of a compound expresses the simplest whole-number ratio of the atoms of the various elements present in one molecule of the compound.

For example, the empirical formula of benzene is CH and that of glucose is CH₂O. This suggests that in the molecule of benzene one atom of Carbon (C) is present for every atom of Hydrogen (H). Similarly in the molecule of glucose (CH₂O), for every one atom of C, there are two atoms of H and one atom of O present in its molecule. Thus, the empirical formula of a compound represents only the atomic ratio of various elements present in its molecule.

The molecular formula of a compound represents the true formula of its molecule. It expresses the actual number of atoms of various elements present in one molecule of a compound. For example, the molecular formula of benzene is C₆H₆ and that of glucose is C₆H₁₂O₆. This suggests that in one molecule of benzene, six atoms of C and 6 atoms of H are present. Similarly, one molecule of glucose (C₆H₁₂O₆) actually contains 6 atoms of C, 12 atoms of H, and 6 atoms of O.

Relation between the empirical and molecular formula
Molecular formula = n × Empirical formula where n is an integer such as 1, 2, 3…
When n = 1; Molecular formula = Empirical formula
When n = 2; Molecular formula = 2 × Empirical formula.
The value of n can be obtained from the relation.
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\)

The molecular mass of a volatile substance can be determined by Victor Meyer’s method or by employing the relation.
Molecular mass = 2 × vapour density .

Empirical formula mass can however be obtained from its empirical formula simply by adding the atomic masses of the various atoms present in it.

Thus the empirical formula mass of glucose CH20
= 1 × 12 + 2 × 1 + 1 × 16 = 30.0 u.

Some Basic Concepts of Chemistry Important Extra Questions Numerical Problems

Question 1.
In the commercial manufacture of nitric acid, how many moles of NO2 produce 7.33 mol HN03 in the reaction
3 NO₂(g) + H₂O(1) → 2HNO₃(aq) + NO(g).
Answer:
2 mols of HNO₃ are produced by 3 mols of NO₂
7.33 mol HNO₃ are produced by \(\frac{3 \times 7.33}{2}\) mol of NO₂
= 10.995 mols.

Question 2.
A sample of NaNO₃ weighing 0.83 g is placed in a 50,0 mL volumetric flask. The flask is then filled with water upon the etched mark. What is the molarity of the solution?
Answer:
Molar mass of NaNO₃ = 23 + 14 + 3 × 16 = 85 g mol^-1

Molarity = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in } L}\)
= \(\frac{0.83 \times 1000}{85 \times 50}\)
= 0.196 M.

Question 3.
Potassium bromide’ KBr contains 32.9% by mass of potassium. If 6.40 g of bromine reacts with 3.60 g of potassium, calculate the number of moles of potassium that combine with bromine to form KBr.
Answer:
Mass percentage of K = 32.9
∴ Mass percentage of Br = 100 – 32.9 = 67.1
Now 67.1 g of bromine combines with 32.9 g of potassium

6.40 g of bromine combines with = \(\frac{32.9}{67.1}\) × 6.40 = 3.14 g of K
Potassium (K) which remains unreacted = 3.60 – 3.14 = 0.46 g

Thus, Br₂ is the limiting reagent.
Mass of K = \(\frac{\text { Mass of } \mathrm{K}}{\text { Atomic mass of } \mathrm{K}}=\frac{3.14 \mathrm{~g}}{39 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.0802 mol.

Question 4.
Calculate the molarity of water in a sample of pure water.
Answer:
1 L of pure water = 1000 cm³
= 1000 g (assuming density = 1 g cm^-3)

No. of moles of H₂O in pure water = \(\frac{1000}{18.0}\) = 55.55 L^-1

Question 5.
How many molecules are there in 10.0 liters of a gas at a pressure of 75 cm at 27°C?
Answer:
PV = nRT

∴ Number of molecules present in 0.4 mol
= 6.023 × 10²³ × 0.4 = 2.409 × 10²³ molecules.

Question 6.
Two acids H₂SO₄ and H₃PO₄ are neutralized separately by the same amount of an alkali when sulfate and dihydrogen orthophosphate are formed respectively. Find the ratio of the masses of H₂SO₄ and H₃PO₄ (P = 31).
Answer:
1 g equivalent of alkali (NaOH) will neutralize 1 g Eqn. of H₂SO₄ and 1 g eq. of H₃PO₄. For the given neutralization reaction.
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
H₃PO₄ + NaOH = Na H₂PO₄ + H₂O

Eq. wt. of H₂SO₄ = \(\frac{\text { Mol. wt. }}{2}=\frac{98}{2}\) = 49;
Eq wt. of H₃PO₄ .
\(\frac{\text { Mol. wt. }}{1}\) = 98

Hence the ratio of masses of H₂SO₄ and HPO₄ = 49: 98 = 1: 2.

Question 7.
What weight of iodine is liberated from a solution, of potassium iodide when 1 liter of Cl₂ gas at 10° C and 750 mm pressure is passed through it?
Answer:
The chemical equation is
2KI + Cl₂ → 2KCl + I₂
22.4 L   2 × 127
at STP = 254 g

(V₁) volume of Cl₂ gas=1 L
V₂ = vol. of gas at
STP =?

(P₁) Pressure = \(\frac{750}{760}\)atm
P₂ = 1 atm

(T₁) Temperature = 10°C + 273 = 283 K
T₂ = 273 K

∴ V₂ = \(\frac{P_{1} V_{1}}{T_{1}^{-}} \times \frac{T_{2}}{P_{2}}=\frac{750 \times 1 \times 273}{760 \times 283 \times 1}\)
= 0.952 L

∴ Vol of Cl₂ g passed at STP = 0.952 L
Now 22.4 L of Cl₂ produces at STP = 254 g of I₂

0.952 L of Cl₂ at STP produces = \(\frac{254}{22.4}\) × 0.952 g of I₂
= 10.78 g of I₂.

Question 8.
A crystalline salt on being rendered anhydrous loses 45.6% of its weight. The percentage composition of the anhydrous salt is Aluminium = 10.50%, Potassium = 15.1% Sulphur = 24.96%, Oxygen = 49.92%. Find the simplest formula of the anhydrous and crystalline salt.
Answer:
Step 1: To calculate the empirical formula of the anhydrous salt


The empirical formula of the anhydrous salt is KAlS2O8

Step 2: Empirical formula mass of anhydrosis salt
= 39 + 27 + 2 × 32 + 8 × 16 = 258 u

Step 3: To calculate the empirical formula mass of the hydrated salt.
Loss of water due to dehydration = 45.6%

∴ The empirical formula mass of anhydrous salt (assuming empirical formula mass of the hydrated sample to be 100) = 100 – 45.6 = 54.4 u

When empirical formula mass.of anhydrous salt is 258 u, that of 100 hydrated is = \(\frac{100}{54.4}\) × 258 = 473.3 u

Step 4: To calculate the no. of water molecules
Total loss in wt. due to dehydration = 473.3 – 258 = 215.3
No. of water molecules = \(\frac{215.3}{18}\) = 11.96 ≈ 12

Step 5: Empirical formula of the hydrated salt = KAlS2O8.12H2O

Question 9.
An organic compound containing C, H, and O gave the following percentage composition: C = 40.687%.
H = 5.085% O = 54.228%
The vapor density of the compound is 59. Calculate the molecular formula of the compound.
Answer:
Calculation of the empirical formula of the compound

∴ The empirical formula of the compound is C2H3O2
Molecular massof compound = 2 × V.D. = 2 × 59 = 118
n = \(\frac{\text { Molecular Mass }}{\text { Empirical formula mass }}\)
= \(\frac{118}{2 \times 12+3 \times 1+2 \times 16}=\frac{118}{59}\) = 2

∴ Molecular formula of the compound
= n × empirical formula = 2 × C₂H₃O₂
= C₄H₆O₄.

Question 10.
Calculate the number of molecules present in 350 cm³ of NH₃ gas at 273 K and 2-atmosphere pressure.
Answer:
Calculate the volume of the gas at STP.
V₁ = 350 cm³
V₂ = ?
T₁ = 273 K
T₂ = 273 K
P₁ = 2 atm.
P₂ = 1 atm

V₂ = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}=\frac{2 \times 350 \times 273}{273 \times 1}\) = 700 cm³

22,400 cm³ of NH₃ at STP contains
= 6.023 × 10²³ molecules of NH₃

700 cm3 of NH₃ at STP will contain
= 1.882 × 10²² molecules.

Question 11.
Calculate the percentage of water of crystallization in the sample of blue vitriol (CuS04.5H20).
Answer:
Mol. wt. of CuSO₄.5H₂O) = 63.5 + 32 + 4 × 16 + 5 × 18 = 249.5
No. of parts bv wt. of H₂O = 5 × 18 = 90
∴ % of H₂O = \(\frac{90}{249.5}\) × 100 = 36.07

Question 12.
Calculate the weight of iron which will be converted into its oxide (Fe₃O₄) by the action of 18 g of steam on it.
Answer:
The chemical equation representing the reaction is

Now 72 g of steam [H₂O(g)] react with 168 g of Iron

∴ 18 g of steam will’reacts with = \(\frac{168}{72}\) × 18 = 42.0 g of iron
Thus, the weight of iron required = 42.0 g

Question 13.
To account for the atomic mass of nitrogen as 14.0067, what should be the ratio of 15N and 14N atoms in natural nitrogen?
[At. mass of 14N = 14.00307 u and 15N = 15.001 u]
Answer:
Let the % of 14N = x
% of 15N = 100 – x
According to the definition of average atomic mass

Question 14.
2.16 g of copper metal, when treated with nitric acid followed by ignition of the nitrate, gave 2.70 g of copper oxide. In another experiment, 1.15 g of copper oxide upon reduction with hydrogen gave 0.92 g of copper. Show that the above data illustrates the law of definite proportions.
Answer:
% of Cu in copper oxide in 1st case = \(\frac{2.16 \times 100}{2.70}\) = 80
% of oxygen = 20%
% of Cu in copper oxide in 2nd case = \(\frac{0.92 \times 100}{1.15}\) = 80
% of oxygen = 20%

Thus, the percentage of copper in copper oxide from both experiments is the same. Hence the above data illustrates the law of definite proportions.

Question 15.
The mass of precious stones is expressed in terms of ‘carat’. Given that 1 carat = 3.168 grains and 1 gram = 15.4 grains, calculate the total mass of a ring in grams and kilograms which contains 0.500 carat diamond and 7.00 gram gold.
Answer:
The unit conversion factors to be used will be

Question 16.
“The star of India” sapphire weighs 563 carats If one carat is equal to 200 mg, what is the weight of the gemstone in grams?
Answer:
The weight of the gemstone in grams will be
= 563 carats × \(\frac{200 \mathrm{mg}}{1 \text { carat }}=\frac{1 \mathrm{~g}}{100 \mathrm{mg}}\)
= 112.6 g

Question 17.
When 4.2 g of NaHCO₃ is added to a solution of acetic acid (CH₃COOH) weighing 10.0 g, it is observed that 2.2 g of CO₂ is released into the atmosphere. The residue left behind is found to weigh 12.0 g. Show that these observations are in agreement with the law of conservation of mass.
Answer:

Total mass of the reactants = 10.0 + 4.2 = 14.2 g
Total mass of the products = 12.0 + 2.2 = 14.2 g

Thus, during the above chemical reactions, the total mass of the reactants is equal to the total mass of the products, i.e., the matter is neither gained nor lost.

This is in keeping with the law of conservation of mass.

Question 18.
Calculate the volume of 0.05 M KMnO₄ solution required to oxidize completely 2.70 g of oxalic acid in acidic solution.
Answer:
The equation representing the chemical change is
2KMnO₄ + 3 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 3 H₂O = 5[0]
H₂C₂O₄ + [O] → 2 CO₂ + H₂O × 5

Step I. To calculate the no. of moles of KMn04 required to completely oxidize 2.70 g of’H₂C₂O₄ in acidic medium:
Molar mass of H₂C₂O₄ = 2 × 1 +2 × 12 + 4 × 16
= 90.0 g mol^-1

∴ No. of moles of H₂C₂O₄ contained in 2.70 g of it = \(\frac{2.70}{90.0}\) = 0.03

5 Moles of H₂C₂O₄ are oxidized by 2 moles of KMnO₄
∴ 0.03 moles of it oxidized by = \(\frac{2}{5}\) × 0.03
= 0.012 mole of KMnO₄

Step II. To calculate the volume of 0.05 M KMnO₄ solution
Now 0.05 mole of KMnO₄ are contained in 1000 cm³ of solution
∴ 0.012 mole of KMnO₄ will be contained in
\(\frac{1000}{0.05}\) × 0.012 = 240 cm³ of solution

Thus, the required volume of 0.05 M KMnO₄ solution = 240 cm³.

Question 19.
4 g carbon was heated with 8 g of sulfur. How much carbon disulfide (CS₂) will be formed when the reaction is complete? What will be its percentage purity?
Answer:

Obviously sulphur will be limiting reactant.

8 g of sulphur will produce CS₂ = \(\frac{76}{64}\) × 8 = 9.5 g

Amount of carbon reacted = \(\frac{12}{64}\) × 8 = 1.5g

Amount of carbon left = 4 – 1.5 = 2.5 g
Total weight of the products = 9.5 + 2.5 = 12 g

% purity of CS₂ in the product = \(\frac{9.5}{12}\) × 100 = 79.2.

Question 20.
A gas mixture of 3.0 liters of propane and butane on complete combustion at 25°C produced 10 liters of CO₂. Find out the composition of the gas mixture.
Answer:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
C₄H₁₀ + 6 \(\frac{1}{2}\)O₂ → 4 CO₂ + 5 H₂O

Suppose volume of propane (C₃H₈) = xL
∴ Volume of butane (C₄H₁₀) = (3 – x)L
1 L of C₃H₈ gives 3 L of CO₂ and 1 L of C₄H₁₀ gives 4 L of CO₂
∴ CO₂ produced = 3x + 4(3 – x) = 12 – x (Given)
12 – x = 10
Hence x = 2 L
∴ Volume of propane = 2 L
and volume of butane = 3 – 2 = 1 L

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 3 Important Extra Questions Classification of Elements and Periodicity in Properties

Classification of Elements and Periodicity in Properties Important Extra Questions Very Short Answer Type

Question 1.
An element is present in the third period of the p-block. It has 5 electrons in its outermost shell. Predict its group. How many unpaired electrons does it have?
Answer:
It belongs to the 15th group (P). It has 3 unpaired electrons.

Question 2.
An element X with Z = 112 has been recently discovered. Predict its electronic configuration and suggest the group in which it is present.
Answer:
[Rn] 5f¹⁴ 6d¹⁰ 7s². It belongs to the 12th group.

Question 3.
The electronic configuration of an element is 1s² 2s² 2p⁶ 3s² 3p⁵. Name the period and the group to which it belongs?
Answer:
Third-period Group 17.

Question 4.
Arrange Cl, Cl^–, Cl⁺ ion in order of increasing size.
Answer:
CP < Cl < CP.

Question 5.
Arrange the following in increasing order of size.
N^3-, Na⁺, F^–, O^2-, Mg²⁺
Answer:
Mg²⁺< Na⁺< F^– < O^2- < N^3-

Question 6.
Give the formula of one species positively charged and one negatively charged that will be isoelectronic with Ne.
Answer:
Na⁺, F^–.

Question 7.
Argon has atomic number 18 and belongs to the 3rd period and 18th group. Predict the group and period for the element having atomic number 19.
Answer:
Group I, Period 4th.

Question 8.
Which of these belong to
(i) the same period and
(ii) the same group.

Element	Atomic Number
A	2
B	10
C	5

Answer:
A and B belong to the same group (18th group)
B and C belong to the same period (II period).

Question 9.
Among the elements of the 3rd period, Na to Al pick out the element.
(i) With the highest first ionization enthalpy.
Answer:
Argon

(ii) With the largest atomic radius.
Answer:
Sodium

(iii) That is the most reactive non-metal.
Answer:
Chlorine

(iv) That is the most reactive metal.
Answer:
Sodium.

Question 10.
Arrange the following ions in the order of increasing size:
Be²⁺, Cl^– , S^2-, Na⁺, Mg²⁺, Br.
Answer:
Be²⁺ < Mg²⁺ < Na⁺ < Cl^– < S^2- < Br^–

Question 11.
Arrange the following elements in the increasing order of metallic character: B, Al, Mg, K.
Answer:
B < Al < Mg < K

Question 12.
Among the elements, Li, K, Ca, S and Kr which one is expected to have the lowest first ionization enthalpy and which one the highest first ionization enthalpy?
Answer:
K has the lowest IE₁
Kr has the highest IE₁

Question 13.
Predict the position of the element in the periodic table satisfying the electronic configuration (n – 1)d¹ ns² for n = 4.
Answer:
(n – 1 )d¹ ns² for n = 4 becomes 3d¹ 4s². It lies in the 4th period and, in the 3rd group.,

Question 14.
Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:
(a) Silicon and Oxygen
Answer:
SiO₂

(b) Aluminium and Bromine
Answer:
AlBr₃

(c) Calcium and Iodine
Answer:
CaI₂

(d) Element 114 and Fluorine
Answer:
Uuq F₄

(e) Element 120 and Oxygen.
Answer
(Ubn) O

Question 15.
Consider the elements N, P, O, and S and arrange them in order of
(a) increasing first ionization enthalpy
Answer:
S < O < P < N

(b) increasing negative electron gain enthalpy
Answer:
N < P < O < S

(c) increasing non-metallic character.
Answer:
P < N < S < O

Question 16.
Among the elements B, Al, C, and Si
(a) Which has the highest IE?
Answer:
C (Carbon)

(b) Which has the highest negative electron gain enthalpy?
Answer:
C (Carbon)

(c) Which has the largest atomic radius?
Answer:
Al

(d) Which has the most metallic character?
Answer:
Al

Question 47.
Arrange the following ions in order of decreasing ionic radii: Li²⁺, He⁺, Be³⁺.
Answer:
He⁺, Li²⁺, Be³⁺ are all isoelectronic ions
∴ He⁺ > Li²⁺ > Be³⁺.

Question 18.
Name the group of elements classified as s-, p-, d-blocks.
Answer:

• s-block consists up of groups 1 and 2. Alkali and alkaline earth metals.
• p-block consists up of groups 13-18.
• d-block consists up of groups 3-12, called transition elements.

Question 19.
Al atom loses electrons successively to form Al⁺, Al²⁺, Al³⁺ ions. Which step will have the highest ionization enthalpy?
Answer:

will have the highest IE.

Question 20.
In terms of electronic configuration, what the elements of a given period and a given group have in common?
Answer:
In a given period, the no. of shells is equal and in a given group, the no. of electrons in the valence shell is the same.

Question 21.
Which of the following elements has the most positive electron gain enthalpy? Fluorine, Nitrogen, Neon.
Answer:
Neon.

Question 22.
Why electron affinities of Be and Mg are positive?
Answer:
They have fully s-orbitals and the additional electron cannot be placed in the much higher energy p-orbitals of the valence shell.

Question 23.
What would be the IUPAC name and symbol for the element with atomic number 120?
Answer:
Unbinillium and its symbol are Ubn.

Question 24.
Why ionization enthalpy of Nitrogen is greater than that of oxygen?
Answer:
Nitrogen has. exactly half-filled orbitals, due to which it is difficult to remove an electron from the valence shell of N atoms.

Question 25.
Give four examples of species that are isoelectronic with Ca²⁺.
Answer:
Ar, K⁺, S^2-, P^3- are isoelectronic with Ca2+.

Question 26.
Comment on the statement, “The electron gain enthalpy of halogens decrease in the order F > Cl > Br > I.”
Answer:
The statement is wrong. The actual order is I > Br > F > Cl.

Question 27.
Arrange the following in order of increasing radii I, I⁺, I^–.
Answer:
I⁺ < I < I^–

Question 28.
Which of the following pairs would have a larger size
(i) K or K⁺
Answer:
K

(ii) Br or Br
Answer:
Br

(iii) O^2- or F^–
Answer:
O^2-

(iv) Na⁺ or K⁺
Answer:
K⁺

Question 29.
Select from each group the one with the smallest radius.
(i)O, O^–, O^2-
Answer:
O

(ii) K⁺, Sr²⁺, Ar
Answer:
Sr²⁺

(iii) Si, P, Cl
Answer:
Cl

Question 30.
Which of the following has the largest and smallest size? Mg, Mg²⁺, Al, Al³⁺.
Answer:
Mg has the largest and Al³⁺ the smallest size.

Question 31.
What is the recommended, IUPAC name and symbol of the element with atomic number 108?
Answer:
Element with At. No. = 108 is Uno. Its recommended name is Unniloctium and its official IUPAC .name is Hassnium.

Question 32.
What is the general outer shell electronic configuration of p-block elements?
Answer:
ns² np^1-6 where n = 2 to 6.

Question 33.
What is the general outer shell electronic configuration of s-block elements?
Answer:
ns1 ~ 2, where n = 2 to 7.

Question 34.
Give the general outer shell electronic configuration of d-block elements?
Answer:
(n – 1 )d^1-10 ns^0-2 where n = 4 to 7.

Question 35.
What is the general outer shell electronic configuration of f-block elements?
Answer:
(n -2)f^0-14 (n – l)d^0-2 ns² where n = 6 – 7.

Question 36.
Which are the three elements among the first three transition series (1st, 2nd, and 3rd) which generally do not show v that properties shown by other members of the series.
Answer:
Zn, Cd, and Hg (Zinc, cadmium, and mercury).

Question 37.
Which is the shortest period in the modern periodic table? Name the elements in it.
Answer:
First period. Hydrogen (H) and helium (He).

Question 38.
What is the number of groups in the d-block?
Answer:
Group numbers 3, 4, 5, 6, 7, 8, 9, 10,11, 12 belong to the d- block

Question 39.
Give the electronic configuration of the first and the last element of the I transition series.
Answer:
Sc (21) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰4s² .
Zn (30) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰ 4s².

Question 40.
Give the names and symbols of two metalloids.
Answer:
Arsenic (As) and antimony (sb).

Question 41.
What do you mean by Vander Waals’ radius?
Answer:
It is defined as one-half the distance between the nuclei of two identical non-bonded isolated atoms or two adjacent identical atoms belonging to two neighboring molecules of an element in the solid state.

Question 42.
Arrange the following ions in order of decreasing ionic radii: Li²⁺, He⁺, Be³⁺.
Answer:
He⁺> Li²⁺> Be³⁺.

Question 43.
Arrange the following in decreasing, order of their van der Waals radii: Cl, H, O, N.
Answer:
Cl > N > O > H. .

Question 44.
What are super heavy elements?
Answer:
Elements with Z > 100 which have high densities are called superheavy elements.

Question 45.
Which transition element has the maximum oxidation state?
Answer:
Osmium (Os) shows an oxidation state of + 8.

Question 46.
Why do the elements of the 2nd period show anomalous properties than the other members of their respective groups?
Answer:

1. small size
2. large charge/radius ratio
3. high electronegativity
4. Absence of d-orbitals.

Question 47.
Which group elements in the periodic table show the maximum electronegativity in their compounds?
Answer:
Halogens belonging to the 17th group.

Question 48.
For each of the following pairs, predict which one has lower first ionization enthalpy?
(i) N or O
Answer:
O

(ii) Na or Na+
Answer:
Na

(iii) Be⁺ or Mg²⁺
Answer:
Be⁺

(iv) I or I^–
Answer:
I^–

Question 49.
Which of the following electronic configurations of the elements do you expect to have a higher value of ionization enthalpy?
(i) 1s² 2s² 2p³ or 1s² 2s² 2p⁴
Answer:
1s² 2s² 2p³

(ii) 1s² or 1s² 2s² 2p⁶
Answer:
1s²

(iii) 3s² or 3s² 3p¹
Answer:
3s²

Question 50.
Arrange the following elements in order of decreasing electron gain enthalpy: B, C, N O.
Answer:
N > B > C > O.

Classification of Elements and Periodicity in Properties Important Extra Questions Short Answer Type

Question 1.
Do elements with high I.E. have high E.A.?
Answer:
Normally is true that the elements with haying high value of I.E. have a high value of E affinity. But however, there are marked exceptions. It is seen that elements, with stable electronic configurations, have very high values of I-Energies as it is difficult to remove electrons as is the case with 15th and 18th group elements but in such case electron cannot be added easily so that is why elements of 15th group have almost zero E.A. and elements of 18th group have got zero E.A. whereas their Ionisation energy values are very high.

Question 2.
What is a periodic classification of elements?
Answer:
By periodic classification of the elements we mean the arrangement of the elements in such a way that the elements with similar physical and chemical properties are grouped together and for this various scientists made contributions but however the contributions made by Mendeleev are of great significance and he gave a periodic table which called as Mendeleev’s Periodic ‘Table which was older and replaced by the long form of the periodic table.

Question 3.
Distinguish between s and p block elements.
Answer:
They can be distinguished as follows: s block elements:

1. They have got the general configuration of the valence shell, ns^1-2.
2. They are all metals.
3. Their compounds are mostly ionic.
4. They are generally strong reducing agents.
5. They mostly impart characteristic color to the flame.
6. They have low ionization energies.
7. They show fixed oxidation states,

p block elements:

1. The valence shell electronic configuration of p block elements in ns² p^1-6.
2. They are mostly non-metals.
3. Their compounds are mostly covalent.
4. They are generally strong oxidizing agents.
5.  Mostly they do not impart color to the flame.
6. They have got a comparatively higher value of I.E.
7. They show variable oxidation states. ,

Question 4.
Explain why ionization enthalpies decrease down a group of the Periodic Table.
Answer:
The decrease in ionization enthalpies down any group is because of the following factors:

1. There is an increase in the number of the main energy shells
2. moving from one element to another.
3. There is also an increase in the magnitude of the screening effect due to the gradual increase in the number of inner electrons.

Question 5.
Why does the first ionization enthalpy increase as we go . from left to right across a given period of the Periodic Table?
Answer:
The value of ionization enthalpy increases with the increase in atomic number across the period.

This is due to the fact that in moving across the period from left to right.,

1. Nuclear charge increases regularly by one unit.
2. The progressive addition of electrons occurs at the same level.
3. Atomic size decreases.

This is due to the gradual increase in nuclear charge and a simultaneous decrease in atomic size the electrons are more and more tightly bound to the nucleus. This results in a gradual increase in ionization energy across the period.

Question 6.
How do atomic radii vary across a period with an atomic number in the periodic table? Explain.
Answer:
Variation of Atomic radii across a period: Atomic radii decrease with the increase in the atomic number in a period. For example, atomic radii decrease from lithium to fluorine in the second period.

In moving from left to right across the period, the nuclear charge increases progressively by one unit but the additional electron goes” to the same principal shell. As a result, the electron cloud is pulled closer to the nucleus by increased effective nuclear charge. This causes a decrease in atomic size.

Variation of the atomic radius with an atomic number across the second period

Question 7.
In each of the following pairs, which species has a larger size? Explain.
(i) Kor K⁺
Answer:
K is larger in size than K⁺.
The electronic configurations of K and K⁺ are:
K: 1s², 2s² 2p⁶, 3s² 3p⁶, 4s¹
K⁺: 1s², 2s² 2p⁶, 3s² 3p⁶.

K⁺ is formed from K when the latter loses its 4s electrons.
K → K⁺ + e^–

As seen from the electronic configurations of K and K⁺, the K⁺ ion has 18 electrons and the K atom has 19 electrons but the nuclear charge in both these species is the same (+ 19). Since the K⁺ ion has 1 electron less than the K atom, the forces of attraction between the nucleus and the electrons are more strong in the K⁺ ion than in the K atom. This results in more inward pulling of electrons towards the nucleus in K+ ion and hence its size decreases.

(ii) Br or Br
Answer:
Br has formed .by the gain of one electron by Br atom. In Br, the nuclear charge is the same as that in the Br atom but the number of electrons has increased. Since the same nuclear charge how acts on the increased number of electrons, the effective nuclear charge per electron decreases in Br”. The electron cloud is held less tightly by the nucleus. This causes an increase in size.

Question 8.
Explain why chlorine has a higher electron affinity than Fluorine? :
Answer:
Cl has a higher electron affinity than F. This is due to the small size of fluorine. As a result of its very small size the inter-electronic repulsions in relatively compact 2p-subshell of a chlorine atom. On comparing chloride ion with fluoride ion we find that electron density per unit volume in fluoride ion (F^–) is more than in chloride (Cl^–) ion. This means that the coming election in the fluorine atom finds less attraction than in the chlorine atom. Consequently, the electron affinity of chlorine is higher than that of fluorine.

Question 9.
Two elements C and D have atomic numbers 36 and 58 respectively. On the basis of electronic configuration predict the following:
(i) The group, period, and block to which each element belongs,
Answer:
The electronic configuration of elements C and D are:
C (At. no 36) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s² 4p⁶.
D (At. no. 58) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s² 4p⁶, 5s² 4d¹⁰ 5p⁶,6s² 5d¹ 4f¹.

The element C belongs to the 18th group, lies in the 4th period, and belongs to the p-block of elements. The element D belongs to lanthanides, lies in the 6th period, and belongs to/-block of elements.

(ii) Are they representative elements?
Answer:
The element C is a noble gas, and D is the inner transition element. They are not representative elements.

Question 10.
Account for the fact that the fourth period has 18 and not 8 elements.
Answer:
The first element of fourth period is potassium (Z = 19) having electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. Thus it starts with the filling of 4s orbital., After the filling of 4s orbital, instead of 4p orbitals, there start the filling 3d orbitals. This is in keeping with the fact that 3d orbitals have fewer energies than 4p orbitals.

Thus 10 elements are built up by the filling of 3d orbitals. After the filling of 3d orbitals, 4p orbitals are filled up and that process is completed at krypton [Kr, Z = 36]. Hence the fourth period consists up of 18 and not 8 elements.

Question 11.
Explain the term ‘valency’ of an element. How does it vary in a period and in a group in the periodic table?
Answer:
The chemical properties of elements depend upon the number of electrons in the outermost shell of an atom. These electrons are called valence electrons and thus determines the valency of the atom (or element).

In representative elements, the valency is generally equal to either n or (8 – n), where n is no. of. valence electrons in the atom.

In a period valence electrons increase from 0 to 8 on moving from left to light. The valency of an element w.r.t. H and Cl increase from 1 to 4 and then decrease to zero. However w.r.t. oxygen, valency increases from 1 to 7 and then becomes zero in noble gases.

In a group, the number of valence electrons remains the same, and therefore all elements in a group exhibit the same valency, e.g., all elements of group I have valency one and those of group 2 have valency two. However, the transition elements exhibit variable valency.

Question 12.
The elements, Z = 107 and Z = 109 have been made recently, element Z = 108 has not yet been made. Indicate the groups in which you will place the above elements.
Answer:
The element with Z = 107, will be placed in the 7th Group, an element with Z = 108 in group 8th and the element with Z = 109 will be placed in group 9th of the periodic table. All three elements will belong to the d-block of the periodic table. (The valence shell electronic configuration will be 7s² 5f¹⁴ 6d⁵ to 7s²5f¹⁴ 6d⁷).

Question 13.
Lanthanides and actinides are placed in separate rows at the bottom of the periodic table. Explain the reason for this arrangement.
Answer:
Lanthanides and actinides are placed in separate rows at the bottom of the periodic table

• to save space
• to keep elements? with similar properties in a single column and
• because their antepenultimate 4/and 5/orbitals are filled respectively.

Question 14.
Which of the following species will have the largest and the smallest size? Mg, Mg²⁺, Al, Al³⁺.
Answer:
Atomic radii decrease from left to right across a period. Cations are always smaller than their parent atoms. Among the isoelectronic species, the one with a larger positive nuclear charge will have a smaller ionic radius.

Hence the largest species is Mg, the smallest one is Al³⁺

Question 15.
The first ionization enthalpy (IE) of the third-period elements Na, Mg, and Si are respectively 496, 737, and 786 kJ mol^-1. Predict whether the first IE value of Al will be more close to 575 or 760 kJ mol^-1 Justify your answer.
Answer:
The first ionization enthalpy (IE) of Al will be more close to 575 kJ mol^-1. The value of Al should be lower than that of Mg because of the effective shielding of 3p electrons from the nucleus by 3s electrons.

Question 16.
Which of the following will have the most negative electron gain enthalpy and which the least?
P, S, Cl, F. Explain your answer.
Answer:
Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down a group. However, adding to electron to 2p orbital leads to greater repulsion than adding an electron to the larger 3p orbital. Hence the element with the most negative electron gain enthalpy is chlorine; the one with the least negative electron gain enthalpy is phosphorus.

Question 17.
Predict the formulae Of compounds which right be formed by the following pairs of elements
(a) Silicon and bromine
Answer:
Silicon (Si) is a group 14 element with a valence of 4; bromine belongs to the group 17 halogen family with a valence of one. Hence the formula of the compound will be SiBr₄.

(b) aluminum and sulfur.
Answer:
Aluminum belongs to group 13 with a valence of 3, sulfur belongs to group 16 with a valence of 2. Hence the formula of the compound would be Al₂S₃.

Question 18.
Show by a chemical reaction with water that Na₂O is a basic oxide and C₁₂O₇ is an acidic oxide.
Answer:
Na₂O reacts with water to form a strong base whereas C₁₂O₇ forms a strong acid.
Na₂O + H₂O → 2NaOH (strong base)
C₁₂O₇ + H₂O → 2HClO₄ (strong acid) ,
Their basic or acidic nature can be qualitatively tested with litmus paper.

Question 19.
Describe the nature of oxides formed by the extreme left elements of the s-blocks and elements on the extreme right Of the p- block.
Answer:
Elements on two extremes of a period easily combine with oxygen to form oxides. The normal oxide formed by the element in the extreme left is the most basic (e.g., Na₂O) whereas that formed by the element on the extreme right is the most acidic (e.g., C₁₂O₇). Oxides of elements in the center are amphoteric (e.g., Al₂O₃ As₂O₃) or neutral (e.g., CO, NO. N₂O).

Question 20.
What are super heavy elements?
Answer:
Elements that have Z > 100 and have high densities are called superheavy elements.

Question 21.
Calculate the energy required to convert all atoms of magnesium to magnesium ions present in 24 mg of magnesium vapors. IE₁ and IE₂ of Mg are 737.76 and 1450.73 kJ mol-1 respectively.
Answer:
Mg (g) + IE₁ → Mg+ (g) + e^– (g)
IE₁ = 737.76 kJ mol^-1

Mg+ (g) + IE₂ → Mg²⁺ + e^– (g)
IE₂ = 1450.73 kJ mol^-1

Total amount of energy needed to convert Mg (g) into ,
Mg²⁺ ion = IE₁ + IE₂

Now 24 mg of Mg = \(\frac{24}{1000}\) g = \(\frac{24}{1000 \times 24}\) mol [1 Mole of Mg = 24 g]
= 10^-3 mole.

Question 22.
The IE₁ and IE₂ of Mg. (g) are 740 and 1450 kJ mol^-1. Calculate the percentage of Mg⁺ (g) and Mg²⁺ (g) if 1 g of Mg (g)
absorbs 50 kJ of energy.
Answer:
No. of moies of Mg vapours present in I g = \(\frac{1}{24}\) = 0.0147

Energy absorbed to convèrt Mg (g) to Mg⁺ (g) = 0.0417 × 740
=30.83 kJ

Energy left unused = 50 – 30.83
= 19.17 kJ
Now 19.17 kJ will be used to e Mg⁺ (g) to Mg²⁺ (g) .
∴ No. of moles of Mg+ (g) converted to Mg2+ (g) = \(\frac{19.17}{1450}\) = 0.132

%age of Mg⁺ (g) = \(\frac{0.0285}{0.0417}\) × 100 = 68.35
and %age of Mg²⁺ (g) = 100 – 68.35 = 31.65

Question 23.
Which of the elements Na, Mg, Si, P would have the greatest difference between the first and the second ionization enthalpies. Briefly explain your answer
Answer:
Among Na, Mg, Si, P, Na is an alkali metal. It has only one electron in the valence shell. Therefore, its IE₁ is low: However, after the removal of the first electron, it acquires a Neon gas configuration i.e., Na+ (1s², 2s² 2p⁶). Therefore its IE₂ is expected to be very high. Consequently, the difference in IE₁ and IE₂ comes to be greatest in the case of Na.

Question 24.
The IE₂ of Mg is higher than that of Na. On the other hand, the IE₂ of Na is much higher than that of Mg. Explain,
Answer:
The first electron in both cases has to be removed from the 3s orbital, but the nuclear charge of Na is less than that of Mg. After the removal of the first electron from Na, the electronic configuration of Na⁺ is 1s², 2s² 2p⁶, i.e., that of noble gas which is very stable and the removal of the 2nd electron is very difficult. In the case of Mg after the removal of the first electron, the electronic configuration of Mg⁺ is 1s², 2s² 2p⁶ 3s. The 2nd electron to be removed is again from 3s orbital which is easier.

Question 25.
The electron gain enthalpy of chlorine is – 349 kJ mol^-1 How much energy in kJ is released when 3.55 g of chlorine is converted completely into Cl^– ion in the gaseous state.
Answer:
According to the definition of electron gain enthalpy
Cl(g) + e^– (g) → Cl^– (g) + 349 kJ mol^-1

The energy released when 1 mole (= 35.5 g) of chlorine atoms change completely with Cl^– (g) = 349 kJ

Question 26.
The amount of energy released when 1 × 10¹⁰ atoms of chlorine in vapor state are converted to Cl^– ions according to the equation.
Cl (g) + e^– → Cl^– (g) is 57.86 × 10^-10 J
Calculate the electron gain enthalpy of the chlorine atom in terms of kJ mol^-1 and eV pet atom.
Answer:
The electron gain enthalpy of chlorine

Question 27.
What is electronegativity? How does it vary along a period and within a group?
Answer:
Electronegativity is defined as the tendency/power/urge of an atom in a molecule to attract the shared pair of electrons v towards itself.

• Variation across a period: Electronegativity generally increases from left to right across a period.
• Variation within a group: It generally decreases in going from top to bottom within a group.

Question 28.
Electronic configuration of the four elements are given below:
Arrange these elements in increasing order of their metallic character. Give reasons for your answer.
(i) [Ar]4s²
Answer:
[Ar]4s² is Calcium metal with At. no. = 20.

(ii) [Ar]3d¹⁰ 4s²
Answer:
[Ar]3d¹⁰ 4s² is Zinc metal with At. no. = 30.

(iii) [Ar]3d¹⁰ 4s² 4p⁶ 5s²
Answer:
[Ar]3d¹⁰ 4s² 4p⁶ 5s² is Strontium metal with At. no. = 38.

(iv) [Arl 3d¹⁰ 4s² 4p⁶ 5s¹
Answer:
[Ar] 3d¹⁰ 4s² 4p⁶, 5s¹ is*Rubidium metal with At. no. = 37.

Alkali metals are the most metallic, followed by alkaline earth metals and transition metals. Among alkali metals – Rubidium (37) is the most metallic. Among alkaline earth metals (Ca, Sr) Sr (Strontium) is more metallic than Calcium (Ca) as the metallic character increases from top to bottom in a group. Zinc – the transition metal is the least metallic. Thus metallic character increases from
Zn < Ca < Sr < Rb or (ii) < (i) < (iii) < (iv)

Question 29.
lthough F is more electronegative than chlorine but the electron gain enthalpy of Cl is more negative than that of F. Why?
Answer:
L. Pauling gave the highest value of 4.0 to F due to its highest electronegative character. Electronegativity decreases from top to bottom in a group and therefore Cl is less electronegative than F. Electron gain enthalpy of F.is less negative than Cl because of the compact size of its atom (2 orbits) as compared to Cl (3 orbits) the mutual electronic repulsion in F is more than that of Cl.

Question 30.
The formulation of F (g) from F (g) is exothermic whereas that of O^2- (g) from O (g) is endothermic. Explain.
Answer:
F (g) + e^– (g) → F (g); ΔH = Negative
Energy is released when an extra electron from outside is added to a neutral isolated gaseous atom of an element.
To convert, O (g) to O^2- (g) two steps are required
(i) O (g) + e^– (g) → O^– (g); ΔH₁ = – 141 kJ mol^-1
(ii) O^– (g) + e^– (g) → O^2- (g); ΔH₂ = + 780 kJ mol^-1

Hence the over all processes endothermic (+ 780 – 141 = + 639 kJ mol^-1) whereas F (g) to F^– (g) is exothermic.

Classification of Elements and Periodicity in Properties Important Extra Questions Long Answer Type

Question 1.
Explain the important general characteristics of groups in the modem periodic table in brief.
Answer:
The elements of a group show the following important similar characteristics.
(0 Electronic configuration. All elements in a particular group have similar outer electronic configuration e.g., all elements of group I’, i.e., alkali metals have ns1 configuration in their valency shell. Similarly, group 2 elements (alkaline Earths) haye ns² outer configuration and halogens (group 17) have ns² np⁵ configuration (where n is the outermost shell).

(it) Valency. The valency of an element depends upon the number of electrons in the outermost shell. So elements of a group show the same valency, e.g., elements of group 1 show + 1 valency and group 2 show + 2 valencies i.e. valency i.e., NaCl > MgCl₂ etc.

(iii) Chemical properties. The chemical properties of the elements are related to the number of electrons in the outermost shell of their atoms. Hence all elements belonging to the same group show similar chemical properties. But the degree of reactivity varies gradually from top to bottom in a group. For example, in group 1 all the elements are highly reactive metals but the degree of reactivity increases from Li to Cs. Similarly, elements of group 17, i.e., halogens: F, Cl, Br, I are all non-metals and they’re- reactivity goes on decreasing from top to bottom.

Question 2.
Explain the electronic configuration in periods in the periodic table. „
Answer:
Each successive period in the periodic table is associated with the filling Up of the next higher principal energy level (n – 1, n – 2, etc.). It can be readily seen that the number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled. The first period starts with the filling of the lowest level (1s) and has thus the two elements – hydrogen (1s¹) and helium (1s²) when the first shell (K) is completed. The second period starts with lithium and the third electron enters the 2s orbital.

The next element, beryllium has four electrons and has the electronic configuration 1s² 2s². Starting from the next element boron, the 2p orbitals are filled with electrons when the L shell is completed’ at neon (2s² 2p⁶). Thus there are 8 elements in the second period. The third period (n = 3) being at sodium, and the added electron enters a 3s orbital. Successive filling of 3s and 3p orbitals give rise to the third period of 8 elements from sodium to argon.

The fourth period (n = 4) starts at potassium with the filling up of 4p of 4s orbital. Before the 4p orbital is filled, the filling up of 3d orbitals becomes energetically favorable and we come across the so-called 3d transition series of elements. The fourth period ends at krypton with the filling up of the 4p orbitals. Altogether we have 18 elements in the fourth period. The fifth period (n = 5) beginning with rubidium is similar to the fourth period and contains the 4d transition series starting at yttrium (Z = 39).

This period ends at xenon with the filling up of the 5p orbitals. The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4/, 5d, and 6p orbitals, in that order. Filling up of the 4/ orbitals being with cerium, (Z = 58) and ends at lutetium (Z = 71) to give the 4/-inner transition series which is called the lanthanide series. The seventh period (n = 7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d, and 7p orbitals and includes most of the man-made radioactive elements.

This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z = 89) gives the 5f-inner transition series known as the actinide series. The 4f and 5f transition series of elements are placed separately in the periodic table to maintain its structure and to preserve the principle of classification by keeping elements with similar properties in a single column.

Question 3.
Explain the variation of valence in the periodic table.
Answer:
Variation of valence in a group as well as across a period in the periodic table occurs as follows:
1. In a group: All elements in a group show the same valency. For example, all alkali metals (group 1) show a valency of 1+. Alkaline earth metals (group 2) show a valency of 2+.

However, the heavier elements of p-block elements (except noble gases) show two valences: one equal to the number of valence electrons or 8-No. of valence electron# and the other two less. For example, thallium (Tl) belongs to group 13. It shows valence of 3+ and 1+.

Lead (Pb) belongs to group 14. If shows valance of 4+ and 2+.
Antimony (Sb) and Bismuth (Bi) belong to group 15. They show valence of 5+ and 3+ being more stable.

This happens due to the non-participation of tie two s-electrons present in the valence shell of these elements. This non-participation of one pair of s-electrons in bonding is called the inert-pair effect.

2. In a period: The number of the valence electrons increases – in going from left to right in a period of the periodic table. Therefore the valency of the elements in a period first increases, and then decreases.

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 11 Mechanical Properties of Fluids. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 11 Important Extra Questions Thermal Properties of Matter

Thermal Properties of Matter Important Extra Questions Very Short Answer Type

Question 1.
The fact that the triple point of a substance is unique is used in modern thermometry. How?
Answer:
In modern thermometry, the triple point of water is a standard fixed point.

Question 2.
Is it possible for a body to have a negative temperature on the Kelvin scale? Why?
Answer:
No. Because absolute zero of temperature is the minimum possible temperature on the Kelvin scale.

Question 3.
(a) Why telephone wires are often given snag?
Answer:
It is done to allow for safe contraction in winter.

(b) The temperature of a gas is increased by 8°C. What is the corresponding change on the Kelvin scale?
Answer:
8 K.

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

Question 5.
Milk is poured into a cup of tea and is mixed with a spoon. Is this an example of a reversible process? Why?
Answer:
No. When milk is poured into tea, some work is done which is not recoverable. So the process is not reversible.

Question 6.
The top of a lake is frozen. Air ¡n contact with it is at -15°C. What do you expect the maximum temperature of water in contact with the lower surface ice? What do you expect the maximum temperature of water at the bottom of the lake?
Answer:
0°C, 4°C.

Question 7.
How does the heat energy from the sun reaches Earth?
Answer:
It reaches by radiation.

Question 8.
Why does not the Earth become as hot as the Sun although it has been receiving heat from the Sun for ages?
Answer:
Earth loses heat by convection and radiation.

Question 9.
Why felt rather than air is employed for thermal insulation?
Answer:
In the air, loss of heat by convection is possible. But convection currents cannot be set up in felt.

Question 10.
What are the three modes of transmission of heat energy from one point to another point?
Answer:
Conduction, Convection and Radiation.

Question 11.
Name suitable thermometers for measuring:
(a) – 80°C,
Answer:
gas thermometer,

(b) 50°C,
Answer:
mercury thermometer,

(c) 700°C,
Answer:
Platinum resistance thermometer,

(d) 1500°C.
Answer:
radiation pyrometer.

Question 12.
Why a thick glass tumbler cracks when boiling liquid is poured into it?
Answer:
Its inner and outer surfaces undergo uneven expansion due to the poor conductivity of glass, hence it cracks.

Question 13.
What is the basic principle of a thermometer?
Answer:
The variation of some physical property of a substance with temperature constitutes the basic principle of the thermometer.

Question 14.
Out of mass, radius and volume of a metal ball, which one suffers maximum and minimum expansion on heating? Why?
Answer:
Volume and radius suffer maximum and minimum expansions respectively as γ = 3α.

Question 15.
The higher and lower fixed points on a thermometer are separated by 160 mm. If the length of the mercury thread above the lower point is 40 mm, then what is the temperature reading?
Answer:
The temperature reading = \(\frac{100 \times 40}{160}\) =25.

Question 16.
Two thermometers are constructed in the same way except that one has a spherical bulb and the other an elongated cylindrical bulb. Which of the two will respond quickly to temperature changes.
Answer:
The thermometer with a cylindrical bulb will respond quickly as the area of the cylindrical bulb is greater than the area of the spherical bulb.

Question 17.
Why a gas is cooled when expanded?
Answer:
Due to a decrease in internal energy, gas is cooled.

Question 18.
Why two layers of cloth of equal thickness provide warmer covering than a single layer of cloth of double thickness?
Answer:
This is because the air enclosed between the two layers of cloth acts as a good heat insulator.

Question 19.
Why snow is a better heat insulator than ice?
Answer:
Snow has air enclosed in it which reduces the chances of loss of heat by convection, hence it is a better heat insulator than ice.

Question 20.
Why water in a metallic pot can be boiled quickly if the bottom of the pot is made black and rough than a highly polished surface?
Answer:
The black and rough surface is a better absorber of heat than a highly polished surface.

Question 21.
Pieces of glass and copper are heated to the same temperature. Why does the piece of copper feel hotter OIL touching?
Answer:
Since copper is a better conductor of he, than gases, so copper transmits heat quickly to the hand and hence it feels hotter on touching.

Question 22.
Why people in the desert wear heavy clothes?
Answer:
Because they protect them from both heat and cold.

Question 23.
Why a wooden table fixed with iron nails become loose after some time?
Answer:
Due to uneven expansion of wood and nail.

Question 24.
Wooden charcoal and a metal piece of the same dimension are heated in the same oven to the same temperature and then removed in the dark. Which one would shine more and why?
Answer:
The charcoal will shine more as it is a good absorber and good emitter, so it will emit more energy.

Question 25.
What is the condition for the difference between the length of a certain brass rod and that of a steel rod to be constant at all temperature?
Answer:
The condition is that the lengths of the rods are inversely proportional to the coefficients of linear expansion of the materials of the rods.

Question 26.
What is the cause of the hotness of a body?
Answer:
The K.E. of the molecules constituting the body causes its hotness.

Question 27.
When are two bodies said to be in thermal equilibrium?
Answer:
Two bodies are said to be in thermal equilibrium when they are at j the same temperature.

Question 28.
What is the value of the temperature coefficient for metals and alloys?
Answer:
Its value is always positive f_c > r metals and alloys.

Question 29.
What is the value of the temperature coefficient for insulators and semi-conductor?
Answer:
It is always negative for insulators and semiconductors.

Question 30.
Are Coefficients of thermal expansion constant for a given solid?
Answer:
No.

Question 31.
Out of a gas and mercury thermometer, which one is more sensitive?
Answer:
A gas thermometer is more sensitive.

Question 32.
At what temperature will wood and iron appear equally hot or equally cold?
Answer:
At the temperature of the human body, both wood and iron will appear equally hot or cold.

Question 33.
A liquid of cubical coefficient of expansion is heated in a vessel having a coefficient of linear expansion \(\frac{\gamma}{3}\). What would be the effect on the level of liquid?
Answer:
No effect.

Question 34.
Is a fur coat itself a source of heat or merely an insulator?
Answer:
It is merely an insulator.

Question 35.
Why fur coat is an insulator?
Answer:
Because it is a bad conductor of heat and prevents the outflow of body heat.

Question 36.
Why does a bullet heat up when it hits a target?
Answer:
It is because its K.E. is converted into heat energy.

Question 37.
How can you measure a temperature of less than 100 K?
Answer:
It can be measured by using a germanium resistance thermometer.

Question 38.
Why water at the bottom of a waterfall is warmer than at the top?
Answer:
It is because P.E. of stored water is converted into heat energy.

Question 39.
What is the S.I. unit of heat capacity?
Answer:
Joule/Kelvin (JK^-1) is the S.I. unit of heat capacity.

Question 40.
Name the substance that contracts on heating?
Answer:
Ice.

Question 41.
Why is a vacuum created between two glass walls of a thermos flask?
Answer:
It is done to prevent the transfer of heat by conduction and convection.

Question 42.
What happens to the K.E. of water which stops after being stirred?
Answer:
It is converted into heat energy.

Question 43.
Why do telephone wires become tight in winter?
Answer:
It becomes tight due to contraction in length with the fall of temperature.

Question 44.
How does the coefficient of cubical expansion of a gas changes with temperature?
Answer:
It decreases with the rise in temperature.

Question 45.
Why do dogs draw out their tongues in summer?
Answer:
They do so to create cooling by evaporation.

Question 46.
How does the boiling point of a liquid changes with temperature?
Answer:
It rises with an increase in temperature.

Question 47.
What is the effect of pressure on the melting point of ice?
Answer:
It is lowered due to the increase in pressure.

Question 48.
What is the thermal capacity of 40 g of copper of specific heat 0.3?
Answer:
It is for 12 cal.

Question 49.
When water is heated from 0° to 10°C, what happens to its volume?
Answer:
It first decreases and then increases.

Question 50.
Why the coefficient of cubical expansion of water is zero at 4°C?
Answer:
This is because the density of water is maximum at 4°C.

Thermal Properties of Matter Important Extra Questions Short Answer Type

Question 1.
Why gas thermometers are more sensitive than mercury thermometers?
Answer:
This is because the coefficient of expansion of a gas is very large as compared to the coefficient of expansion of mercury. For the same temperature change, the gas would undergo a much larger change in volume as compared to mercury.

Question 2.
Why the brake drum of an automobile gets heated up when the automobile moves down a hill at constant speed?
Answer:
Since the speed is constant so there is no change of kinetic energy. The loss in gravitational potential energy is partially the gain in the heat energy of the brake drum.

Question 3.
A solid is heated at a constant rate. The variation of temperature with heat input is shown in the figure here:

(а) What is represented by AB and CD?
Answer:
The portions AB and CD represent a change of state. This is because the supplied heat is unable to change the temperature. While AB represents a change of state from solid to liquid, the CD represents a change of state from liquid to vapour state.

(b) What conclusion would you draw1 if CD = 2AB?
Answer:
It indicates that the latent heat of vaporisation is twice the latent heat of fusion.

(c) What is represented by the slope of DE?
Answer:
Slope of DE represents the reciprocal of the thermal or heat capacity of the substance in vapour state i.e. slope 0f DE = \(\frac{d T}{d Q}=\frac{1}{m C}\)(∴ dQ = mCΔT).

(d) What conclusion would you draw from the fact that the slope of OA is greater than the slope of BC?
Answer:
Specific heat of the substance in the liquid state is greater than that in the solid-state as the slope of OA is more than that of BC i.e. \(\frac{1}{\mathrm{mC}_{1}}\) > \(\frac{1}{\mathrm{mC}_{2}}\) where C₁, C₂ are specific heats mC₁ mC₂ of the material in solid and liquid state respectively.

Question 4.
Define:
(a) Thermal conduction.
Answer:
It h defined as the process of the transfer of heat energy from one part of a solid. to another part at a lower temperature without the actual motion of the molecules. It is also called the conduction of heat.

(b) Coefficient of thermal conductivity of a material.
Answer:
It is defined as the quantity of heat flowing per second across the opposite faces of a unit cube made of that material when the opposite faces are maintained at a temperature difference of 1K or 1°C.

Question 5.
On what factors does the amount of heat flowing from the hot face to the cold face depend? How?
Answer:
If Q is the amount of heat flowing from hot to the cold face, then it is found to be:

1. directly proportional to the cross-sectional area (A) of the face
   i. e. Q ∝ A …(1)
2. directly proportional to the temperature difference between the two faces, i.e. Q ∝ Δθ ….(2)
3. directly proportional to the time t for which the heat flows i.e. Q ∝ t …. (3)
4. inversely proportional to the distance ‘d’ between the two faces
   i.e. Q ∝ \(\frac{1}{Δx}\) …(4)

Combining factors (1) to (4), we get
Q ∝ \(\frac{\mathrm{A} \Delta \theta}{\Delta \mathrm{x}}\)t
or
Q ∝ K A \(\frac{\Delta \theta}{\Delta \mathrm{x}}\)t
where K is the proportionality constant known as the coefficient – of thermal conductivity.

Question 6.
State Newton’s law of cooling and define the cooling curve. What is its importance?
Answer:
Newton’s law of cooling: States that the rate of loss of heat per unit surface area of a body is directly proportional to the temperature difference between the body and the surroundings provided the difference is not too large.

Cooling Curve: It is defined as a graph between the temperature of a body and the time. It is as shown in the figure here.

The slope of the tangent to the curve at any point gives the rate of fall of temperature.

Question 7.
Explain why heat is generated continuously in an electric heater but its temperature becomes constant after some time?
Answer:
When the electric heater is switched on, a stage is quickly reached when the rate at which heat is generated by an electric current becomes equal to the rate at which heat is lost by conduction, convection and radiation and hence a thermal equilibrium is established. Thus temperature becomes constant.

Question 8.
A woollen blanket keeps our body warm. The same blanket if wrapped around ice would keep ice cold. How do you explain this apparent paradox?
Answer:
Wool is a bad conductor of heat. Moreover wool encloses air in it which is a bad conductor. There can also be no loss of heat by convection. The woollen blanket keeps us warm by preventing the heat of the human body to flow outside and hence our body remains warm.

In the case of ice, the heat from outsiders prevented to flow inside and thus ice remains cold.

Question 9.
A liquid is generally heated from below. Why?
Answer:
When a liquid is heated, it becomes rarer due to a decrease in density and it rises up. Liquid from the upper part of the vessel comes down to take its place and thus convection currents are formed. If the liquid is heated at the top, no such convection currents will be formed and only the liquid in the upper part of the vessel will become hot. However, the temperature in the lower part of the vessel will rise slightly due to a small amount of heat conducted by the hot liquid in the upper part of the vessel.

Question 10.
If a drop of waterfalls on a very hot iron, it does not evaporate fora long time. Why?
Answer:
When a drop of waterfalls on a very hot iron, it gets insulated from the hot iron due to the formation of a thin layer of water vapour, which is a bad conductor in nature. It takes quite a long to evaporate as heat is conducted from hot iron to the drop through the layer of water vapour very slowly.

On the other hand, if a drop of waterfalls on an iron which is not very hot, then it comes in direct contact with the iron and evaporates immediately.

Question 11.
On a hot day, a car is left in sunlight with all the windows closed. After some time, it is found that the inside of the car is considerably warmer than the air outside. Explain why?
Answer:
Glass possesses the property of selective absorption of heat radiation. It also transmits about 50% of heat radiation coming from a hot source like the sun and is more or less opaque to the radiation from moderately hot bodies (at about 100°C or so). Due to this, when a car is left in the sun, heat radiation from the sun gets into the car but as the temperature inside the car is moderate, it cannot escape through its windows. Thus glass windows of the car trap the sun rays and because of this, the inside of the car becomes considerably warmer.

Question 12.
It takes longer to boil water with a flame in a satellite in gravitational field-free space, why? How it will be heated?
Answer:
Water boils with flame by the process of convection. Hot lighter particles raise up and heavier particles move down under gravity. In. a gravity-free space in the satellite, the particles cannot move down hence, water can’t be heated by convection.

It will be heated by conduction.

Question 13.
Find γ for polyatomic gas and hence determine its value for a triatomic gas in which the molecules are linearly arranged.
Answer:
The energy of a polyatomic gas having n degrees of freedom is given by
E = n × \(\frac{1}{2}\) kT × N =\(\frac{n}{2}\) RT

In case of a triatomic gas, n = 7
∴ γ = 1 + \(\frac{2}{7}\) = \(\frac{9}{7}\).

Question 14.
Food in a hot case remains warm for a long time during winter, how?
Answer:
The hot case is a double-walled vessel. The space between the walls is evacuated in a good hot case. The food container placed inside the hot case is made of crowning steel, thus neither the outside low-temperature air can enter the container nor the heat from inside can escape through the hot case by conduction or convection. The highly polished shining surface of the food container helps in stopping loss of heat due to radiation. Thus, the heat of the food is preserved for a long time and food remains hot in winter.

Question 15.
You might have seen beggars sleeping on footpaths or in open in winter with their hands and knees pulled inside. Similarly dogs too curl while sleeping in winter. How does such action help anybody?
Answer:
The heat radiated or emitted from a body at a given temperature depends on

1. the temperature difference between the body and the surrounding,
2. area of the body in contact with the surroundings and
3. the nature of the body.

For man and animals in winter (1) and (2) factors remain what they are. So, in order to preserve, their body heat they curl up to reduce the surface area in contact with cold air.

Question 16.
Many people enjoy bathing below Kempty fall in Mussoorie, even though the water is quite cold, explain?
Answer:
When water falls from a height, it loses its potential energy. This is converted into the kinetic energy of the water molecules. It is a known fact from the kinetic theory of gases that an increase in kinetic energy of the substance increases its temperature, hence the temperature of water at the bottom of the fall rises and it does not remain as cold as on the top. Therefore, people taking bath below the Kempty fall do not get frozen, they rather feel warm and enjoy bathing.

Question 17.
Cycle, scooter handles and steering wheels of four-wheelers have plastic, rubber or cotton thread coverings. Why?
Answer:
Cycle and scooter handle, as well as steering wheels of four-wheelers such as car, truck, tempo, are made of metals which are good conductors of heat. In winter when we hold them the body heat is conducted to them being at lower temperature and in summer their temperature is more than the body temperature, transfer heat to the human hands. In either case, the human being feels uncomfortable.

To avoid this heat transfer, those portions from where the driver holds the vehicle are covered with heat insulators such as plastic, rubber or cotton.

Thus, the driver is saved from cold or heat being felt in his hands.

Question 18.
Why metals like copper, iron, brass etc. are good conductors of heat whereas wood, cardboard, ply are not conductors of heat?
Answer:
Heat conductibility in solids apart from temperature depends on the availability of conducting particles i.e. free electrons. In metals like copper, brass and iron-free electrons are available but in insulators like wood, cardboard and ply free electrons are not available, so metal is good conductors of heat.

Question 19.
House in Rajasthan made of stone and lime are cooler than those made of brick and cement why?
Answer:
The thermal conductivity of lime is smaller than that of cement. Even though stone tends to be Rotter than a brick in the day time as much as. does the cement plaster. Similarly, on cold days or cold nights even during summer, the heat of the house is not allowed to escape out. The thickness of stone walls compared to brick walls is yet another factor in keeping the houses, cool or warm in opposite”conditions.

Question 20.
Housing on hills are either made of wood or have wooden lining and walls. Why? Why people in plains where temperature variations are extreme winter and summer do not use-wooden house?
Answer:
On hills, people make wooden houses or put wooden lining, on house walls to insulate the house from the extreme cold of winter because wood being a bad conductor of heat does not allow the heat of house to escape out or exchange low outside temperature with the high temperature inside.

In plains even though wooden houses are helpful in fulfilling the odd weather conditions yet because of other factors such as

• non-availability of plenty of wood and
• chances of catching fire etc. wooden houses are generally not built.

Question 21.
People who own cars know well if they close all glass windows of the car park it in the Sun, it remains very hot inside the car even after sunset. To keep inside the car cooler insulating screens are put inside the car covering, the winds screen and curtains are pulled on the glass of windows, explain.
Answer:
Glass at normal temperature say up to 100° is opaque to heat radiations. But if radiations are coming from a source having a high temperature (surface temperature of sun ~ 6000k), it becomes transparent. Thus heat from the sun enters the car but radiations from the car being at a lower temperature (~ 50°C) cannot escape out so it remains hot, glass is opaque for radiations at such low temperature and the car body is double-walled or lined with insulators also does not allow the inner body heat to escape, Thus inside of the car remains hot.

But when screen and curtains are put the solar radiations are either reflected back or get absorbed and the temperature inside the car does not rise too much. It remains bearable.

Question 22.
What are the various properties of a thermometric substance?
Answer:
Following are the properties of a thermometric substance:

1. They should have a high boiling point and low freezing point to provide for the measurement of a wide range of temperature.
2. It should not be volatile.
3. It should be a variable in abundance.
4. Its rate of expansion should be uniform so that the calibration becomes easier.
5. The coefficient of expansion of the substance should be high so that the thermometer may be sensitive.
6. It should be available in a pure state.
7. It should not stick to the side of the glass tube.
8. It should have good thermal conductivity.

Question 23.
It is generally very cold after the hail storm than during or before it. Why?
Answer:
After a hail storm, the ice balls melt by absorbing a lot of heat energy from the atmosphere, thus reducing its temperature. Hence there is very cold after the hail storm.

Question 24.
Why pendulums made of invar are used in clocks?
Answer:
Invar has a low value of the coefficient of linear expansion, so the length of the pendulum almost remains the same in different seasons and hence gives the correct time.

Question 25.
Should a thermometer bulb have a large heat capacity or a small heat capacity? Why?
Answer:
A thermometer bulb should have a small heat capacity. In case, if it has a large capacity, the temperature of the substance will get lowered due to a large amount of heat absorbed by the thermometer bulb.

Question 26.
Explain why the coefficient of cubical expansion of water is negative between 0°C and 4°C?
Answer:
This is because of the fact that the volume of water decreases between 0°C and 4°C and water contracts within this range of temperature.

Question 27.
Why good conductors of electricity are also good conductors of heat?
Answer:
Both thermal and electrical conductivity depends upon the number of free electrons which are in large number in good conductors of electricity.

Question 28.
How is a gas thermometer better than a mercury thermometer?
Answer:
A gas thermometer is better because it can measure very low and very high temperature while the range of mercury thermometer is limited due to its freezing point – 37°C and boiling point 357°C.

Question 29.
A vertical glass jar is filled with water at 10°C. It has one thermometer at the top and another at the bottom. The central region of the jar is gradually cooled. It is found that the bottom thermometer reads 4°C earlier than the top thermometer and the top thermometer reads 0°C earlier than the bottom thermometer. Why?
Answer:
This is because water expands on cooling from 4°C to 0°C, thus it becomes of highest density at 4°C and then its density goes down. So the bottom thermometer reads 4°C earlier as the temperature of water at the top will be more than 4°C.

Question 30.
Two spheres are m&de of the same metal and have the same mass. One is solid and the other is hollow. They are heated to the same temperature. What is the percentage increase in their diameters?
Answer:
It will be the same We both spheres as the expansion of solid is like photographic enlargement, so \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) = γΔθ.

Question 31.
Nam the effects produced by the heat given to a body.
Answer:
The following are the effects produced by the heat:

1. It changes the temperature of the body.
2. It changes the volume of the body.
3. It changes the state of the body.
4. It changes the physical and chemical properties of the body.
5. It produces light.
6. It generates electricity.

Thermal Properties of Matter Important Extra Questions Long Answer Type

Question 1.
(a) Define the following:
(i) Coefficient of linear expansion (α)
Answer:
α – It is defined as the change in the length per unit original length per unit change in temperature of the material of a solid rod.
i.e. α = \(\frac{\Delta l}{l_{0} \Delta \mathrm{T}}\)

(ii) Coefficient of superficial expansion (β)
Answer:
β – It is defined as the change in surface area per unit original surface area per unit change in the temperature of the material of the solid.
i.e. β = \(\frac{\Delta S}{S_{0} \Delta \mathrm{T}}\)

(iii) Coefficient of cubical expansion (γ)
Answer:
γ – It is defined as the change in volume per unit original volume per unit change in the temperature of the material of the solid.
γ = \(\frac{\Delta V}{V_{0} \Delta \mathrm{T}}\)

(b) Derive the relation α, β and γ.
Answer:
Relation between α and β

Consider a solid of length 1 cm at 0°C. Heat it by 1°C. If l0, lt be its length at 0°C and 1°C respectively, then
l_o = 1 cm
and
l₁ = l₀ + l_o ∝ Δt
= 1 + 1 × α × 1
= (1 + α) (∵ Δt = 1 – 0 = 1°C)

If S_o and S_t be its area at 0°C and 1°C respectively, then
So = l_o²= 1 cm²
S_t = l₁² = (1 + α)² cm²
∴ ΔS = S_t – S_o = (1 + α)² – 1

∴ By definition ,
β = \(\frac{\Delta \mathrm{S}}{\mathrm{S}_{0} \Delta \mathrm{T}}=\frac{(1+\alpha)^{2}-1}{1 \times 1}\)
= 1 + α² + 2α – 1 = α² + 2α
= 2α
(∵ α is very small, so α² is neglected)
or
α = \(\frac{\beta}{2}\) …(1)

Relation between α and γ:
Consider a cube of side 1 cm at 0°C. Heat it by 1°C
∴ l_o = 1 cm
and l₁ = l_o(1 + α Δt) = 1(1 + α × 1)
= (1 + α) cm

If V_o and V_t be its volumes at 0°C and 1°C respectively.
Then V_o = l_o³ = 1 cm³
and V_t = l_t³ = (1 + α)³
= 1 + α³ + 3α(1 + α)
= 1 + α³ + 3α + 3α²
= 1 + 3α
(∵ α is very small so square and higher powers are neglected.)
ΔV change in its volume, then
ΔV = V_t – V_o = 1 + 3α – l = 3α

∴ By definition,

Numerical Problems:

Question 1.
Calculate the increase in the temperature of the water which falls from a height of 100 m. Assume that 90% of the energy due to fall is converted into heat and is retained by water. J = 4.2 J cal^-1.
Answer:
Here, h = 100 m
Let m (kg) = mass of water

∴ Its P.E. at a height h = mgh
Energy of fall retained by water i.e. useful work done is given by,
W = 90% of mgh
= \(\frac{90}{100}\) mgh
= \(\frac{90}{100}\) × m × 9.8 × 100
= 882 m J.

∴ Heat retained, Q = \(\frac{W}{J}=\frac{m \times 882 J}{4.2 J c a l^{-1}}\)
= m × 210 cal …(i)
Specific heat of water C = 10 cal kg^-1 °C^-1

Let Δθ (°C) be the rise in the temperature of water.
∴ Heat gained, Q = mCΔθ
.= m × 10³ × Δθ
= m × Δθ × 10³ cal …. (ii)

∴ From (1) and (ii), we get
m × 210 = m × Δθ × 10³
or
Δθ = \(\frac{210}{10^{3}}\)= 0.21°C.

Question 2.
A clock with a steel pendulum has a time period of 2s at 20°C. If the temperature of the clock rises to 30°C, what will be the gain or loss per day? The coefficient of linear expansion of steel is
1.2 × 10^-5 C^-1
Answer:
Here α = 1.2 × 10^-1  °C^-1
Δt = 30 – 20= 10°C
T = 2s.

Using the relation, Δl = l α Δt, we get
\(\frac{\Delta l}{l}\) = α Δt
= 1.2 × 10^-5 × 10 = 1.2 × 10^-4 …. (i)
∴ T = 2π \(\sqrt{\frac{l}{g}}\) ….(ii)

If T’ be the time period of the pendulum, when l increases by Δl, then

∴ loss in time in one oscillation T’ – T
Hence loss in time in one day is given by

Question 3.
The thermal conductivity of brick is 1.7 W m^-1 K^-1 and that – of cement is 2.9 W m^-1 K^-1. What thickness of cement will have the same insulation as the brick of thickness 20 cm.
Answer:
Here, K_B = 1.7 W m^-1 K^-1
K_C = 2.9 W m^-1 K^-1
dB = 20 cm
dc = ?

We know that the heat flow is given by
Q = KA \(\frac{\Delta \theta}{\mathrm{d}}\) t

For the same insulation by the brick and the cement, Q, A, Δθ and t don’t change
Thus \(\frac{K}{d}\) should be a constant.

Question 4.
Two metal cubes A and B of the same size are arranged as shown in the figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficient of thermal conductivity of A and B are 300 W/m°C and 200 W/m°C respectively. After a steady-state is reached, what will be the temperature of the interface?

Answer:
Let T (°C) be the temperature of the interface =?
Here, K₁ = 300 Wm^-1 °C^-1 for A
K₂ = 200 Wm^-1 °C^-1 for B

∴ Δθ₁ = 100 – T for A
Δθ₂ = T – 0 for B .
x = size of each cube A and B

∴ x₁ = x₂ = x
Let a = area of cross-section of the faces between which there is the flow of heat

If \(\left(\frac{\Delta \mathrm{Q}_{1}}{\Delta \mathrm{t}}\right)_{\mathrm{A}}\) and \(\left(\frac{\Delta \mathrm{Q}_{2}}{\Delta \mathrm{t}}\right)_{\mathrm{B}}\) be the rate of.tlow of heat for A and B respectively, then in steady state,

Question 5.
The heat of combustion of ethane gas is 373 Kcal per mole. Assuming that 50% of heat is lost, how many litres of ethane measured at STP must be burnt to convert 50 kg of water at 10°G to steam at 100°C? One mole of gas occupies 22.4 litres at S.T.P. Latent heat (L) of steam = 2.25 × 106 JK^-1.
Answer:
Here
L = 2.25 × 106 JK^-1
= \(\frac{2.25}{4.2}\) × 106 cal°C^-1

Q = Heat of Combustion
= 373 × 103 Cal/mole

C = 103 J Kg^-1 °C^-1
m = 50 kg
Δθ = 100- 10 = 90°C
V = 22.4 litres

If Q₁ = Total heat energy required to convert 50 kg of water at 10°C to steam at 100°C
Q₁ = mCΔθ + mL
= 5.0 × 1000 × 90 + 50 ×\(\frac{2.25 \times 10^{6}}{4.2}\)
= 4.5 × 106 + 26.79 × 106
= 31.29 × 106 cal

As 50% of heat is lost,
∴ total heat produced = \(\frac{100}{50}\) × 3.129 × 106

Let n = no. of moles of ethane to be burnt, then
n = \(\frac{2 \times 31.29 \times 10^{6}}{373 \times 10^{3}}\) mole

∴ Volume of ethane = nV
= \(\frac{2 \times 31.29 \times 10^{6}}{373 \times 10^{3}}\) × 22.4 litres
= 3758.2 litres.

Question 6.
Calculate the heat of combustion of coal when 10 g of coal on burning raises the temperature of 2 litres of water from 20°C to 55°C.
Answer:
Here, C = 1 cal g^-1 °C^-1
m = 2 × 10³ g
Δθ = 55 – 20 = 35°C
M = 10g

If Q be the heat produced, then
Q = mCΔθ
= 2 × 10³ × 1 × 35
= 70 × 10³ cal

∴ If Q₁ be the heat of combustion, then
Q₁ = \(\frac{Q}{M}=\frac{70 \times 10^{3}}{10}\) = 7000 cal g^-1

Question 7.
Calculate two specific heats of a gas from the following data:
γ = \(\frac{\mathbf{C}_{\mathrm{P}}}{\mathbf{C}_{\mathbf{v}}}\) = 1.51, density of gas at NTP = 1.234 g litre J = 4.2 × 10⁷ erg cal^-1.
Answer:
\(\frac{\mathbf{C}_{\mathrm{P}}}{\mathbf{C}_{\mathbf{v}}}\) = 1.51
P = 1.013 × 10⁶ dyne cm^-2
T= 273K
ρ = 1.234 g litre^-1
m = 1 g

∴ V = \(\frac{\mathrm{m}}{\rho}=\frac{1}{1.234}\) litre
= \(\frac{1}{1.234 \times 10^{-3}}\)

Also we know that for 1g gas,

Question 8.
Specific heats of argon at constant pressure and volume are 0.125 cal g^-1 and 0.075 cal g^-1 respectively. Calculate the density of argon at N.T.P. (J = 4.18 × 10⁷ ergs/cal and normal pressure = 1.01 × 10⁶ dynes cm^-2.)
Answer:
Here, CP = 0.125 cal g^-1
Cv = 0.075 cal g^-1 J
J = 4.18 × 107 ergs cal^-1
P = 1.01 × 106 dyne cm^-2
d = density at NTP = ?
m = 1 g
T = 273 K

Using the relation,
C_p – C_v = \(\frac{r}{J}=\frac{P V}{T J}=\frac{P m}{d T J}\) (∵ V = \(\frac{m}{d}\))
d = \(\frac{P m}{\mathrm{~TJ}\left(\mathrm{C}_{r}-\mathrm{C}_{\mathrm{v}}\right)}\)

Question 9.
A piece of metal weighs 46 g in air. When it is immersed ¡n a liquid of specific gravity 124 at 27°C, it weighs 30g. When the temperature of the liquid is raised to42°C, the metal piece weighs 30.5 g. The specific gravity of the liquid at 42°C ¡s 1.20. Calculate the coefficient of linear expansion of the metal.
Answer:
Here, the Weight of the metal piece at 27°C in air 46 g
Weight of metal piece at 27°C in liquid =30 g
Weight of metal piece at 42°C in liquid = 30.5 g
α =?
Loss in weight of the metal = weight of liqiid displaced = 46 – 30
= 16 g.

The volume of metal at 27°C = Volume of liquid displaced at 27°C
or
V1 = \(\frac{16 g}{\text { specific gravity of liquid }}\)
= \(\frac{16 \mathrm{~g}}{1.24 \mathrm{gcm}^{-3}}\)
= 12.903 cm3

Similarly volume of metal piece at 42°C = V₂ = \(\frac{(46-30.5)}{1.2 \mathrm{gcm}^{-3}}\)
= 12.917 cm³

∴ Coefficient of cubical expansion of the metal

Since γ = 3α
∴ α = \(\frac{1}{3}\) γ = \(\frac{1}{3}\) × 2.41 × 10⁵ °C^-1
= 0.803 × 10⁵ °C^-1

Question 10.
In an industrial process, 10 kg of water per hour is to be heated from 20°C to 80°C. To do so, steam at 150°C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90°C. How many kg of steam is required per hour? Specific heat of steam = 1 Kcal kg^-1 °C^-1 and latent heat of steam = 540 Kcal kg^-1.
Answer:
C = sp. heat of steam
= 1 Kcal kg^-1 °c^-1

L = latent heat of steam
= 540 Kcal kg^-1

Let m (kg) = mass of steam required per hour.

Heat is given by steam first from 150°C to steam at 100°C = mCΔθ
= m × (150 – 100)Kcal = 50 m Kcal.

Then steam changes from steam at 100°C to water at 100°C and gives out heat = mL = 540 m Kcal.

After this water at 100°C gives heat is going to temperature 90°C = m (100 – 90) = 10m Kcal.

Total amount of heat given by the steam = 50 m + 540 m + 10 m = 600 m Kcal.
∴ 600 m K cal = 600 K cal
∴ m = 1 kilogram.

Question 11.
An electric heater is used in a room with a total wall area of 137 m² to maintain a temperature of +20°C inside it when the outside temperature is -10°C. The wall has three layers of different materials. The innermost layer is of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outermost layer is of brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivity of wood, cement and brick are 0. 125,1.5 and 1.0 Wm^-1 °C^-1 respectively.
Answer:
Let the temperature inside the room be θ₁, at the interface of wood and cement be θ₂, at the interface of cement and brick θ₃ respectively. The outside temperature is θ₄.

Here, K₁ = 0.125 Wm^-1 °C, K₂ = 1.5 Wm^-1 °C^-1, K₃ = 1.0 Wm^-1 °C^-1 A = 137m2 , θ₁ = 20°C θ₄ = -10°C , d₁ = 2.5 cm, d₂ = 1.0 cm, d₃ = 25 cm.

∴ \(\frac{0.125\left(20-\theta_{2}\right)}{2.5}=\frac{1.5\left(\theta_{2}-\theta_{3}\right)}{1}\)
or
5(20 – θ₂) = 150(θ₂ – θ₃)
or
20 – θ₂ = 30(θ₂ – θ₃) ,
or
20 = 31θ₂ – 30θ₃

Similarly from

or
20 – θ₂ = \(\frac{4}{5}\) (θ₃ + 10)
or
θ₃ = \(\frac{5}{4}\) (20 – θ₂) – 10 …(2)

∴ From (1) and (2), we get
20 = 3lθ₂ – 30 × (20 – θ₂) + 30 × 10
or
80 = 124θ₂ – 150(20 – θ₂)+ 1200
= 274θ₂ – 3000 + 1200
or
274θ₂ = 18800

∴ θ₂ = \(\frac{940}{137}\) °C

∴ If P be the heater power, then

Question 12.
The opposite faces of a cubical block of iron of cross-section 4 cm2 are kept in contact with steam and melting ice. Calculate the amount of ice melted at the end of 10 minutes if K = 0.2 cal cm^-1 s^-1 °C^-1 for iron. Latent heat of fusion of ice = 80 cal g^-1.
Answer:
Here, L = latent heat of ice
= 80 cal g^-1
A = 4 cm² .
K = 0.2 cal cm^-1 s^-1 °C^-1
Δθ = θ₂ – θ₁ = 100 – 0 = 100°C
t = time = 10 minutes = 10 × 60 = 600s
d = \(\sqrt{4}\) =2 cm

Let m = amount of ice melted = ?
Using the relation, Q = \(\frac{\mathrm{K} \mathrm{A}\left(\theta_{2}-\theta_{1}\right) \mathrm{t}}{\mathrm{d}}\), we get

Q = \(\frac{0.2 \times 4 \times 100 \times 600}{2}\)
Q = 24000 cal …(i)
Also, Q = mL = m. 80 …(ii)

∴ From (i) and (ii), we get
m × 80 = 24000

∴ m = \(\frac{24000}{80}\) = 300 g.

Question 13.
Water is boiled in a rectangular steel tank of thickness 2 cm by a constant temperature furnace. Due to vaporisation, the water level falls at a steady rate of 1 cm in 9 minutes. Calculate the temperature of the furnace. Given the thermal conductivity of steel is 0.2 cal cm^-1 s^-1 °C^-1 and latent heat of steam is 540 cal.
Answer:
d = 2 cm
t = 9 minutes = 9 × 60 = 540 s
K = 0.2 cal cm^-1 s^-1 °C^-1
L = 540 cal.

Let A be the area of the bottom of the steel tank.
θ₁ = temperature of the tank which is in contact with the constant temperature furnace?
fall in level of water = 1cm

∴ volume of water evaporated, V = A × 1 = A cm³
ρ = density of water = 1 g/cm³

Mass of water evaporated, m = ρV – 1 × A = Ag .

Q = heat required to evaporate Ag of water in 9 minute = A × 540 cal.
Also, Q = \(\frac{\mathrm{KA}\left(\theta_{\mathrm{1}}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}}\)
or
540A = \(\frac{0.2 \mathrm{~A} \times\left(\theta_{1}-\theta_{2}\right) 540}{2}\)
or
θ₁ – θ₂ = \(\frac{2}{0.2}\) = 10
Here, θ₂ = 100°C
θ₁ = θ₂ + 10 = 100 + 10= 110 °C.

Question 14.
A circular hole of a radius of 1 cm is drilled in a brass sheet kept at 293K. What will be the diameter of the hole when the sheet is heated to 393K? a for brass = 18 × 10^-6 K^-1.
Answer:
Here, α = 18 × 10^-6 K^-1
ΔT = 393 – 293 K = 100 K
r = radius = 1 cm
∴ D = 2r = diameter = 2 cm
and it acts as the original length l (say).

∴ Let D’ be the new diameter = ?
I f Δl be the increase in length, then using the relation,
Δl = αlΔt, we get
Δl = 18 × 10^-6 × 2 × 100 = 36 × 10^-4 cm

∴ increase in diameter, AD = Al = 36 × 10^-4 cm
∴ D’ = D + ΔD = 2 + 0.0036 = 2.0036cm.

Question 15.
Find the change in the length of the steel bridge of the original length of 200 m in a locality where temperature changes from 243 to 313K. a for steel = 11 × 10^-6 K^-1.
Answer:
Here, T₁ = 243 K
T₂ = 313 K
∴ ΔT = change in temperature
= T₂ – T₁ = 313 – 243 = 70 K
l = 200 m
α = 11 × 10^-6 K^-1

Let Δl = change in the length of the bridge = ?
∴ Δl = lα ΔT
= 200 × 11 × 10^-6 × 70
= 154 × 10^-3 m
= 0.154 m = 15.4 cm.

Question 16.
Equal volumes of copper and mercury have the same thermal capacity. Mercury has a specific heat of0.046 and a density of 13.6 g cm^-3. Calculate the density of copper having sp. heat 0.092.
Answer:
Let C₁ and C₂ be the sp. heat of mercury and copper respectively having densities ρ₁ and ρ₂
Here, ρ₁ = 13.6 g cm^-3
C₁ = 0.046
C₂ = 0.092
∴ ρ₂ =?

Let V = volume of copper and mercury.
If m₁ and m₂ be the mass of mercury and copper, then
m₁ = V × ρ₁
and
m₂ = V × ρ₂
Also let H₁ and H₂ be their respective thermal capacity, so using
Thermal capacity = mass × sp. heat, we get
H₁ = Vρ₁ × C₁
and
H₂ = Vρ₂ × C₂

As H₁ = H₂
Vρ₁C₁ = Vρ₂C₂
or
ρ₂ = ρ₁\(\frac{C_{1}}{C_{2}}=\frac{13.6 \times 0.046}{0.092}\)
= 6.8gcm³.

Question 17.
What would be the final temperature of the mixture when 5 g of ice at -10°C is mixed with 20 g of water at 30°C? Sp. the heat of ice is 0.5 and latent heat of water – 80 cal g^-1.
Answer:
Here, m1 = mass of water = 20 g
θ₁ = temperature of water = 30°C
θ₂ = temperature of ice = -10°C
C₁ = sp. heat of waterr = 1
C₂ = sp, heat of ice.= 0.5
m₂ = mass of ice = 5 g

Let θ = final temperatre of the mixture
L = latent heat of water = 80 cal g^-1

∴ Heat lost by water is given by
H₁ = 20 × 1 × (30 – θ) ….(1)
(∵ H = mCΔθ)

Heat required to raise 5 g of ice at -10°C to 0°C
H₂ = 5 × [0 – (-10)] × 0.5
= 5 × 10 × \(\frac{5}{10}\) = 25 cal ….(2)

Let H₃ = heat required to meet 5 g of ice at 0°C to water at 0°C
H₃ = m₂L
= 5 × 80 = 400 cal ….(3)

Also let H₄ = heat required to raise 5 g of water at 0°C to θ°C
∴ H₄ = 5 × 1 × θ = 5θ

∴ Total heat gained by ice = H₂ + H₃ + H₄
= 25 + 400 + 5θ

∴ Heat lost = Heat gained gives
20(30 – θ) = 425 + 5θ
or
600 – 200 = 425 + 5θ
or
600 – 425 = (20 + 5)θ
or
25θ = 175

∴ θ = \(\frac{175}{25}\) = 7 °C.

Question 18.
A faulty Celcius thermometer has lower and upper fixed points as 1°C and 101°C; it reads a certain temperature as 50°C. What is the correct temperature?
Answer:
Let Correct temp. = x°C
∴ \(\frac{x-0}{100-0}=\frac{50-1}{101-1}\)
or
\(\frac{x}{100}=\frac{49}{100}\)
or
x = 49°C.

Question 19.
At what temperature do the Celsius and Fahrenheit readings have the same numerical value?
Answer:
Let x be the required value in °C and °F.
i.e. t_F = t_c = x°

∴ Using the relation, \(\frac{\mathbf{C}-0}{100}=\frac{\mathrm{F}-32}{180}\), we get

Question 20.
At what temperature do the Kelvin and Fahrenheit readings have the same numerical value?
Answer:
Let x be the required value in °K and °F
t_K = t_F = x

Using the relation,

Question 21.
Express 37°C into Farenheit.
Answer:
Here, t_c = 37°C
t_F = ?
Using the relation, \(\frac{C-0}{100}=\frac{F-32}{180}\) we get
or
F – 32 = \(\frac{9}{5}\) × C
∴ F = \(\frac{9}{5}\)C + 32
= \(\frac{9}{5}\) × 37+32
= 66.6 + 32
= 98.6°F.

Question 22.
The temperature of a body is increased from 100°F to 910°F. What is the temperature range in the Kélvin scale?
Answer:
Here, Δt = (T₂ — T₁)°F
= 910 – 100°F
= 810°F

Using the relation, ΔC = \(\frac{5}{9}\) × ΔF
or
ΔC = \(\frac{5}{9}\) × 810 – 450
∴ ΔC = 450°K

Question 23.
A faulty thermometer has its fixed points marked as 50 and 95°. When it reads 590 what is the corresponding temperature in °C?
Answer:
Let x be the reading ¡n °C.
∴ \(\frac{x-0}{100-0}=\frac{59-5}{95-5}\)
or
\(\frac{x}{100}=\frac{54}{90}=\frac{3}{5}\)
or
x = \(\frac{3}{5}\) × 100 = 60°C.

Question 24.
The temperature coefficient of resistance of a Wire is 0.00125 per °C. At 300 K its resistance is 1 Q. At what temperature will the resistance of the wire be 2Ω?
Answer:
Here, α = 125 × 105 °C^-1
R₃₀₀ = 1Ω

Let R_θ be its resistance at θ°C
∴ R_θ = 2Ω
∴ Δθ = rise in its temp.

∴ Using the relation

or
θ – 300 = 800

∴ θ = 800 + 300 = 1100 K

Question 25.
Calculate the temperature difference in °F equivalent to temperature difference of 25°C.
Answer:
Here, ΔC = 25°C
ΔF =?
Using the relation, ΔF = \(\frac{9}{5}\) × ΔC, we get
ΔF = \(\frac{9}{5}\) × 25
= 45°F.

Value-Based Type:

Question 1.
Venkat having found his mother suffering from fever Venkat took her to the doctor for treatment. While checking the status, the doctor used a thermometer to know the temperature of the body. He kept the thermometer in the mouth of the patient and noted the reading as 102°F. The doctor gave the necessary medicines. After coming home, Venkat asked his mother, who Is a science teacher, why mercury is used in a thermometer when there are so many liquids. Then his mother explained the reason.
(a) Comment upon the values of the mother.
Answer:
Mother has an interest in educating her son and explained that Mercury has got the following properties for being used in their- monitors

1. The expansion of Mercury is fairly regular and uniform.
2. It is opaque and shining, hence can be easily seen through the glass tube.
3. Mercury is a good conductor of heat and has a low thermal capacity,
4. Mercury dose is not wet on the sides of the glass tube in which it is tilled.

(b) A newly designed thermometer has its lower fixed point and upper, the fixed point marked at 5° and 95° respectively. Compute the temperature on this scale corresponding to 50°C.
Answer:
Let θ be the temperature on the scale corresponding to θ = 50°C.
Then \(\frac{\theta-5}{95-5}=\frac{C-0}{100-0}=\frac{C}{100}\)
or
θ = 50°
Thus, the required temperature on the scale of the designed ther¬mometer is 50°.

Question 2.
Raman noticed that his grandfather to be suffering from fever. He took him to the doctor. The doctor gave him some pills. When the pills were used he sweated much, after some time became normal. Rahim enquired the Doctor about how his grandfather become normal.
(a) According to you what values are possessed by Raman?
Answer:
Raman is responsible and he has concern for others, inquisitiveness in gaining knowledge, curiosity

(b) A child running a temperature of 101°Fis given an Antipyrin which causes an increase in the rate of evaporation of the body. If the fever is brought down to 98° F in 20 m, what is the amount of heat lost by the body? The mass of the child is 30 kg.
Answer:
Loss in temperature (Dt) =101°F – 98°F = 3°F = 3 × \(\frac{5}{9}\) °C
= 1.67 °C
Specific heat of water (S) = 1000 Gal. kg^-1 °C^-1
m = 30 kg

∴ Heat lost by the body = ms Δt
= 30kg × 1000 Cal. kg^-1 °C^-1 × 1.67°C
=501000 Cal

Question 3.
Radha and Kavita are two friends of science stream. They were discussing the phenomena of cooling. Radha told that a hot body will lose its temperature more rapidly in a room where the temperature would be less. Kavita did a small activity by putting two j glasses having the same temperature in two different room of different room temperature and note down the temperature after 10 min. She found that Radha’s idea was correct.
(i) What values are exhibited by Kavita?
Answer:
Kavita’s values are:
Curiosity, Interested in the experiment. Habit to check the correctness of an idea.

(ii) Write the law relating to this phenomena.
Solution:
The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study.

According to Newton’s law of cooling, the rate of loss of heat (-dQ/dt) of the body is directly proportional 1o the difference of temperature DT = (T₂– T₁) of the body and the surrounding. The law holds goods only for the small difference in temperature.
i.e – \(\frac{dQ}{dt}\) = K(T₂ – T₁)

-ve sign implies that as time passes, temperature (T) decreases.

Question 4.
Anurag and Akash were discussing heat transfer. Anurag told that houses of concrete roofs get very hot during summer days because the thermal conductivity of concrete (though much smaller than that of metal) is still not small enough. Therefore, people usually prefer to give a layer of earth or foam insulation on the ceiling so that heat transfer is prohibited and keeps the room cooler. But Akash was not convinced by this idea. He told me that it is just to decorate the ceiling.
(i) What values are displayed by Anurag?
Answer:
Application of scientific reason, curiosity, the attitude of group discussion and problem-solving skill.

(ii) Whether Anurag was telling right or wrong?
Answer:
Yes, Anurag was telling right.

(iii) On what factors the amount of heat flowing from the hot face to the cold face depends?
Answer:
Heat flowing Q α A; A = Cross-sectional area
α Δθ ; Δθ = Temperature difference
α t ; t = time
α 1/Δx; Δx = Distance between the two face

On combining the above relations, we get.
i.e Q = \(\frac{\mathrm{KA} \Delta \theta}{\Delta x}\) t
Where K is the proportionality constant known as the coefficient of thermal conductivity