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Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 2 Biological Classification Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 2 Biological Classification

PSEB 11th Class Biology Guide Biological Classification Textbook Questions and Answers

Question 1.
Discuss how classification systems have undergone several changes over a period of time?
Answer:
(i) Linnaeus proposed a two kingdom system of classification with Plantae and Animalia kingdoms was developed that included all plants and animals respectively. But as this system did not distinguish between the eukaryotes and prokaryotes, unicellular and multicellular organisms and photosynthetic (green algae) and non-photosynthetic (fungi) organisms, so scientists found it an inadequate system of classification. Classification systems for the living organisms have hence, undergone several changes over time.

(ii) The two kingdom system of classification was replaced by three kingdom system, then by four and finally by five kingdom system of classification of RH Whittaker (1969).

(iii) The five kingdoms included Monera, Protista, Fungi, Plantae and Animalia. This is the most accepted system of classification of living organisms.

(iv) But, Whittaker has not described viruses and lichens. Then Stanley described viruses, viroids, etc.
Thus, over a period of time, classification systems have undergone several changes.

Question 2.
State two economically important uses of:
(a) Heterotrophic bacteria
(b) Archaebacteria
Answer:
(a) Heterotrophic Bacteria

• Maintain soil fertility by nitrogen fixation, ammonification and nitrification, e.g., Rhizobium bacteria (in the root nodules of legumes).
• The milk products such as butter, cheese, curd, etc., are obtained by the action of bacteria. The milk contains bacterial forms like Streptococcus lacti, Escherichia coli, Lactobacillus lactis and Clostridium sp., etc.

(b) Archaebacteria

• Methanogens are responsible for the production of methane (biogas) from the dung of animals.
• Archaebacteria help in the degradation of waste materials.

Question 3.
What is the nature of cell walls in diatoms?
Answer:
In case of diatoms, the cell wall forms two thin overlapping cells, which fit together as in a soap box. The cell wall is made up of silica. Due to siliceous nature of cell wall, it is known as diatomite or diatomaceous Earth. Diatomaceous Earth is a whitish, highly porous, chemically inert, highly absorbant and fire proof substance.

Question 4.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify?
Answer:
Sometimes, greqn algae such as Chlorella, Scenedesmus and Spirogyra, etc,, grow in excess in water bodies and impart green colour to the water. These are called algal blooms. Red dianoflagellates (Gonyaulax) grow in abundance in sea and impart red colour to the ocean. This looks like red tides. Both due to algal blooms and ‘red tide’ the animal life declines due to toxins and deficiency of oxygen inside water.

Question 5.
How are viroids different from viruses?
Answer:
Viroids different from viruses as follows:

Virus	Viroids
1. These are smaller than bacteria.	Smaller than viruses.
2. Both RNA and DNA present.	Only RNA is present.
3. Protein coat present.	Protein coat absent.
4. Causes diseases like mumps and AIDS.	Causes plant diseases like spindle tuber diseases – potato.

Question 6.
Describe briefly the four major groups of Protozoa.
Answer:
Protozoans are divided into four phyla on the basis of locomotory organelles – Zooflagellata, Sarcodina, Sporozoa and Ciliates.
(i) Zooflagellates: These protozoans possess one to several flagella for locomotion. Zooflagellates are generally uninucleate, occasionally multinucleate.
The body is covered by a firm pellicle. There is also present cyst formation.
Examples: Giardia, Trypanosoma, Leishmania and Trichonympha, etc.

(ii) Sarcodines: These protozoans possess pseudopodia for locomotion. Pseudopodia are of four types, i.e., lobopodia, filopodia, axopodia and reticulopodia. Pseudopodia are also used for engulfing food particles. Sarcodines are mostly free living, found in freshwater, sea water and on damp soil only a few are parasitic. Nutrition is commonly holozoic. Sarcodines are generally uninucleates. Sarcodines are of four types – Amoeboids (i.e.,Amoeba, etc.), radiolarians (i.e., Acanthometra, etc.), foraminiferans (i.e., Elphidium, etc.) and heliozoans (i.e., Actinophrys, etc.).

(iii) Sporozoans: All of them are endoparasites. Locomotoryorganelles (cilia, flagella, pseudopodia, etc.) are absent. Nutrition is parasitic (absorptive), Phogotrophy is rare. The body is covered with an elastic pellicle or cuticle. Nucleus is single. Contractile vacuoles are absent. Life cyle consists of two distinct asexual and sexual phases. They may be passed in one (monogenetic) or two different hosts (digenetic),e.g., Plasmodium, Monocystis, etc.

(iv) Ciliates: These are aquatic, actively moving organisms because of the presence of thousands of cilia. They have a cavity (gullet) that opens to the outside of the cell surface. The coordinated movements of rows of cilia causes the water laden with food to enter into the gullet, e.g., Paramecium.

Question 7.
Plants are autotrophic. Can you think of some plants that are partially heterotrophic?
Answer:
Plants are autotrophs, i.e., they prepare their own food through the process of photosynthesis. But, in nature there are also some other plants which are partjally heterotrophic, i.e., they partially depend upon another organisms for food requirements, e.g.,
(i) Loranthus and Viscum are partial stem parasites which have leathery leaves. They attack several fruit and forest trees and with the help of their haustoria draw sap from the xylem tissue of the host.

(ii) Insectivorous plants have special leaves to trap insects. The trapped insects are killed and digested by proteolytic enzymes secreted by the epidermis of the leaves, e.g., pitcher plant.

(iii) Parasitic plant, e.g., Cuscutta develops haustoria, which penetrate, the vascular bundles of the host plant to absorb water and solutes.

Question 8.
What do the terms phycobiont and mycobiont signify?
Answer:
In case of lichens (t. e., an association of algae and fungi), the algal partner which is capable of carrying out photosynthesis is known as phycobiont, whereas the fungal partner which is heterotrophic in nature is known as mycobiont.

Question 9.
Give a comparative account of the classes of Kingdom Fungi under the following:
(i) Mode of nutrition
(ii) Mode of reproduction

Fungal Class	Mode of Nutrition	Mode of Reproduction
Myxomycetes	Heterotrophic and mostly saprophytic	Asexual and sexual reproduction
Phycomycetes	Mostly parasites	Asexual and sexual methods
Zygomycetes	Mostly saprophytic	Asexual and sexual reproduction
Ascomycetes	Saprophytes or parasites	Asexual and sexual reproduction
Basidiomycetes	Saprophytes or parasites	Asexual and sexual method
Deuteromycetes	Saprophytes or parasites	Only asexual reproduction

Question 10.
What are the characteristic features of Euglenoids?
Answer:
The characteristic features of euglenoids are as follows :

• They occur in freshwater habitats and damp soils.
• A single long flagella present at the anterior end.
• Creeping movements occur by expansion and expansion of their body known as euglenoid movements.
• Mode of nutrition is holophytic, saprobic or holozoic.
• Reserve food material is paramylum.
• Euglenoids are known as plant and animal.
  Plant characters of them are as follows:
  (a) Presence of chloroplasts with chlorophyll.
  (b) Holophytic nutrition
  Animal characters of them are as follows:
  (a) Presence of pellicle, which is made up of proteins and not a cellulose.
  (b) Presence of stigma.
  (c) Presence of contractile vacuole.
  (d) Presence of longitudinal binary fission.
• Under favourable conditions euglenoids multiply by longitudinal binary fission, e.g., Euglena, Phacus, Paranema, etc.

Question 11.
Give a brief account of viruses with respect to their structure ’ and nature of genetic material. Also name four common viral diseases.
Answer:
Viruses are non-cellular, ultramicroscopic, infectious particles. They are made up of envelope, capsid, nucleoid and occasionally one or two enzymes. Viruses possess an outer thin loose covering called envelope. The central portion of nucleoid is surrounded by capsid that is made up of ( smaller sub-units known as capsomeres.

The nucleic acid present in the viruses is known as nucleoid. It is the r infective part of the virus which utilises the host cell machinery. The
genetic material of viruses is of four types –

(i) Double stranded DNA (dsDNA) as found in pox virus, hepatitis-B virus and herpes virus, etc.
(ii) Single stranded DNA (ssDNA) occur in coliphage Φ, coliphage Φ x 174.
(iii) Double stranded RNA (dsRNA) occur in Reo virus,
(iv) Single stranded RNA (ssRNA) occur in TMV virus, polio virus, etc.
Four common viral diseases are – (i) Polio, (ii) AIDS, (iii) Hepatitis-B (iv) Rabies.

Question 12.
Organise a discussion in your class on the topic—Are viruses living or non-living?
Answer:
Viruses are intermediate between living and non-living objects. They resemble non-living objects in:

• Lacking protoplast. ‘
• Ability to get crystallised.
• High specific gravity which is found only in non-living objects.
• Absence of respiration and energy storing system.
• Absence of growth and division.
• Cannot live independent of a living cell.

They resemble living objects in:

• Presence of genetic material (DNA or RNA).
• Property of mutation.
• Irritability.
• Can grow and multiply inside the host cell.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 8 Gravitation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 8 Gravitation

PSEB 11th Class Physics Guide Gravitation Textbook Questions and Answers

Question 1.
Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?
Answer:
(a) No, Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects.

(b) Yes, If the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).

(c) Tindal effect depends inversely upon the cube of the distance while gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

Question 2.
Choose the correct alternative:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula -G Mm(1 /r₂ -1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the centre of the earth.
Solution:
(a) Decreases,
Explanation : Acceleration due to gravity at altitude h is given by the relation:
g_h = \(\left(1-\frac{2 h}{R_{e}}\right) g\)
where, R_e = Radius of the Earth
g = Acceleration due to gravity on the surface of the Earth
It is clear from the given relation that acceleration due to gravity decreases with an increase in height.

(b) Decreases,
Explanation : Acceleration due to gravity at depth d is given by the relation:
g_d = \(\left(1-\frac{d}{R_{e}}\right) g\)
It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.

(c) Mass of the body,
Explanation : Acceleration due to gravity of body of mass m is given by the relation:
g = \(\frac{G M_{e}}{R^{2}}\)
where, G = Universal gravitational constant
M_e = Mass of the Earth
R_e = Radius of the Earth
Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

(d) More,
Explanation : Gravitational potential energy of two points r₂ and r₁ distance away from the centre of the Earth is respectively given by:
V(r₁) = -GmM
V(r₂) = – \(\frac{G m M}{r_{2}}\)
V = V (r₂) – V(r₁) = -GmM\(\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right)\).
Hence, this formula is more accurate than the formula mg (r₂ – r₁).

Question 3.
Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Solution:
Time taken by the Earth to complete one revolution around the Sun,
T_e = 1 year
Orbital radius of the Earth in its orbit, R_e = 1 AU
Time taken by the planet to complete one revolution around the Sun,
T_p = \(\frac{1}{2} T_{e}=\frac{1}{2}\) year
Orbital radius of the planet = R_p
From Kepler’s third law of planetary motion, we can write

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

Question 4.
I₀, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Solution:
Orbital period of I₀, TI₀ = 1.769 days = 1.769 x 24 x 60 x 60 s
Orbital radius of I_I0, R_I0 = 4.22 x 108 m
Satellite I₀ is revolving around the Jupiter Mass of the latter is given by the relation:
M_j = \(\frac{4 \pi^{2} R_{I_{O}}^{3}}{G T_{I_{O}}^{2}}\) ……………………….. (i)

where M_j = Mass of Jupiter
G = Universal gravitational constant Orbital period of the Earth,
T_e = 365.25 days = 365.25 x 24 x 60 x 60 s Orbital radius of the Earth,
Re =1 AU = 1.496 x 10¹¹ m
Mass of Sun is given as:
M_s = \(\frac{4 \pi^{2} R_{e}^{3}}{G T_{e}^{2}}\) ………………………. (ii)

= 1045.04
∴ \(\frac{M_{s}}{M_{j}} \sim 1000\)
\(M_{s} \sim 1000 \times M_{j}\)
Hence, it can be inferred that the mass of Jupiter is about one thousandth that of the Sun.

Question 5.
Let us assume that our only consists of 2.5 x 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic center take to complete one revolution? Take the diameter of the Milky Way to be 10⁵ ly.
Solution:
Number of stars in our galaxy, = 2.5 x 10¹¹
Mass of each star = one solar mass =2 x 10³⁰ kg
Mass of our galaxy,M =2.5 x 10¹¹ x 2 x 10³⁰ = 5 x 10⁴¹ kg
Diameter of Milky Way, d = 10⁵ ly
Radius of Milky Way,r= 5 x 10⁴ ly
∵ 1 ly=946 x 10¹⁵m
∴ r=5 x 10⁴ x 9.46 x 10¹⁵
=4.73 x 10²⁰m

Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:
T = \(\left(\frac{4 \pi^{2} r^{3}}{G M}\right)^{1 / 2}\)

1 year = 365 x 324 x 60 x 60 s
∵ 1 s = \(\frac{1}{365 \times 24 \times 60 \times 60} \) years
∴ 1.12 x 10¹⁶ s = \(\frac{1.12 \times 10^{16}}{365 \times 24 \times 60 \times 60}\) = 3.55 x 10 ⁸ years

Question 6.
Choose the correct alternative:
(a) If the zero of potential energy Is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence Is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.
Solution:
(a) Kinetic energy
Explanation: Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (maybe negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative.

Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.

(b) Less
Explanation: An orbiting satellite acquires certain amount of energy that enables it to revolve around the Earth. This energy is provided by its orbit. It requires relatively lesser energy to move out of the influence of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy.

Question 7.
Does the escape speed of a body from the Earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?
Answer:
(a) No, we know that the escape velocity of the body is given by
V_e = \(\sqrt{\frac{2 G M}{R}} \) , where M and R are mass and radius of Earth. Thus clearly, it does not depend on the grass of the body as V_e is independent of it.

(b) Yes, we know that V_e depends upon the gravitational potential at the point from where the body is launched. Since the gravitational potential depends on the latitude and height of the point, therefore the escape velocity depends on the location of the point from where it is projected. It can also be experienced as:
V_e = \(\sqrt{2 g r}\) . As g has different values at different heights. Therefore, V_e depends upon the height of location.

(c) No, it does not depend on the direction of projection as V_e is independent of the direction of projection.

(d) Yes, it depends on the height of location from where the body is launched as explained in (b).

Question 8.
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant
(a) linear speed,
(b) angular speed,
(c) angular momentum,
(d) kinetic energy,
(e) potential energy,
(f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
(a) No, according to law of conservation of linear momentum L = mvr constant, therefore the comet moves faster when it is close to the sun and moves slower when it is farther away from the sun. Therefore, the speed of the comet does not remain constant.

(b) No, as the linear speed varies, the angular speed also varies. Therefore, angular speed of the comet does not remain constant.

(c) Yes, as no external torque is acting on the comet, therefore, according to law of conservation of angular momentum, the angular momentum of the comet remain constant.

(d) No, kinetic energy of the comet = \(\frac{1}{2}\) mν² As the linear speed of the comet changes as its kinetic energy also changes. Therefore, its KE does not remains constant.

(e) No, potential energy of the comet changes as its kinetic energy changes.

(f) Yes, total energy of a comet remain constant throughout its orbit.

Question 9.
Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem?
Answer:
(b), (c), and (d)
Explanations:
(a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space.

(b) A swollen face is caused generally because of apparent weightlessness in space sense organs such as eyes, ears, nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space.

(c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space.

(d) Space has different orientations. Therefore, orientational problems can affect an astronaut in space.

Question 10.
In the following two exercises, choose the correct answer from among the given ones:
The gravitational intensity at the center of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see figure) (i) a, (ii) b, (iii) c, (iv) 0.

Solution:
(iii) Gravitational potential (V) is constant at all points in a spherical shell.
Hence, the gravitational potential gradient \(\left(\frac{d V}{d r}\right)\) is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.

If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle located at center O will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at center O of the given hemispherical shell has the direction as indicated by arrow c.

Question 11.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow
(i) d,
(ii) e,
(iii) f,
(iv) g.
Solution:
(ii) Gravitational potential (V) is constant at all points in a spherical shell.
Hence, the gravitational potential gradient \(\left(\frac{d V}{d r}\right)\) inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. Ibis indicates that gravitational forces acting at a point in a spherical shell are symmetric. If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point P will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.

Question 12.
A rocket Is fired from the Earth towards the Sun. At what distance from the Earth’s centre is the gravitational force on the rocket zero? Mass of the Sun = 2 x 10³⁰kg, mass of the Earth = 6 x 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 10¹¹ m)
Solution:
Mass of the Sun, M_s = 2 x 10³⁰ kg
Mass of the Earth, M_e = 6 x 10 ²⁴ kg
Orbital radius, r = 1.5 x 10¹¹ m
Mass of the rocket = m

Let x be the distance from the center of the Earth where the gravitational force acting on satellite P becomes zero.
From Newton’s law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as:

= 2.59 x 10⁸ m

Question 13.
How will you ‘weigh the Sun’, that is estimate its mass? The mean orbital radius of the Earth around the Sun is 1.5 x 10⁸ km.
Solution:
Orbital radius of the Earth around the Sun, r = 1.5 x 10⁸ km = 1.5 x 10¹¹ m
Time taken by the Earth to complete one revolution around the Sun,
T = 1 year = 365.25 days = 365.25 x 24 x 60 x 60 s
Universal gravitational constant, G = 6.67 x 10^-11N-m² kg^-2
Thus, mass of the Sun can be calculated using the relation,
M = \(\frac{4 \pi^{2} r^{3}}{G T^{2}}\)
= \(\frac{4 \times(3.14)^{2} \times\left(1.5 \times 10^{11}\right)^{3}}{6.67 \times 10^{-11} \times(365.25 \times 24 \times 60 \times 60)^{2}}\)
= \(\frac{133.24 \times 10^{33}}{6.64 \times 10^{4}}\) = 2.0 x 10³⁰kg
Hence, the mass of the Sun is 2.0 x 10³⁰ kg.

Question 14.
A Saáurn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.5 x 10⁸ km away from the Sun?
Solution:
Distance of the Earth from the Sun, r_e = 1.5x 10⁸ km= 1.5 x 10¹¹ m
Time period of the Earth = T_e
Time period of Saturn, T_s = 29.5 T_e
Distance of Saturn from the Sun = r_s
From Kepler’s third law of planetary motion, we have
T=\(\left(\frac{4 \pi^{2} r^{3}}{G M}\right)^{\frac{1}{2}}\)
For Saturn and Sun, we can write

= 1.43 x 10¹² m

Question 15.
A body weigths 63 N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Solution:
Weight of the body, W =63 N
Acceleration due to gravity at height h from the Earth’s surface is given by the relation:
g’ = \(\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}\)
where, g = Acceleration due to gravity on the Earth’s surface
R_e =Radius of the Earth
For h= \(\frac{R_{e}}{2}\)
g’ = \(\frac{g}{\left(1+\frac{R_{e}}{2 \times R_{e}}\right)^{2}}=\frac{g}{\left(1+\frac{1}{2}\right)^{2}}=\frac{4}{9} g\)

Weight of a body of mass m at height h is given as
W’ = mg’
= m x \(\frac{4}{9}\) g = \(\frac{4}{9}\) x mg
= \(\frac{4}{9}\) W
= \(\frac{4}{9}\) x 63 =28 N

Question 16.
Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the Earth if it weighed 250 N on the surface?
Solution:
Weight of a body of mass m at the Earth’s surface, W = mg = 250 N
Body of mass m is located at depth, d = \(\frac{1}{2}\) R_e
where, R_e = Radius of the Earth
Acceleration due to gravity at depth d is given by the relation
g’ = \(\left(1-\frac{d}{R_{e}}\right) g=\left(1-\frac{R_{e}}{2 \times R_{e}}\right) g=\frac{1}{2} g\)
Weight of the body at depth d,
W’ = mg’
= m x \(\frac{1}{2} g=\frac{1}{2}\) mg = \(\frac{1}{2}\) W
= \(\frac{1}{2}\) x 250 = 125 N

Question 17.
A rocket is fired vertically with a speed of 5 km s^-1 from the Earth’s surface. How far from the Earth does the rocket ‘ go before returning to the earth? Mass of the Earth = 6.0 x 10²⁴ kg; mean radius of the Earth = 6.4 x 10⁶ m;
G = 6.67 x 10 ^-11 N-m² kg^-2.
Solution:
Velocity of the rocket, ν = 5 km/s = 5 x 10³ m/s
Mass of the Earth, M_e = 6.0 x 10²⁴ kg
Radius of the Earth, R_e = 6.4 x 10⁶ m
Height reached by rocket mass, m = h

At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
= \(\frac{1}{2} m v^{2}+\left(\frac{-G M_{e} m}{R_{e}}\right)\)
At highest point h,
ν = 0
and, Potential energy = – \(\frac{G M_{e} m}{R_{e}+h}\)
Total energy of the rocket = 0+ \(\left(-\frac{G M_{e} m}{R_{e}+h}\right)=-\frac{G M_{e} m}{R_{e}+h}\)
From the law of conservation of energy, we have
Total energy of the rocket at the Earth’s surface=Total energy at height h


Height achieved by the rocket with respect to the centre of the Earth = R_e +h
= 6.4 x 10⁶ +1.6 x 10⁶ = 8.0 x 10⁶ m

Question 18.
The escape speed of a projectile on the Earth’s surface is 11.2 km s^-1. A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of the Sun and other planets.
Solution:
Escape velocity of a projectile from the Earth, υ_esc = 11.2 km/s
Projection velocity of the projectile, v_p = 3 v_esc
Mass of the projectile = m
Velocity of the projectile far away from the Earth = v_f
Total energy of the projectile on the Earth = \(\frac{1}{2} m v_{p}^{2}-\frac{1}{2} m v_{\mathrm{esc}}^{2}\)
The gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth = \(\frac{1}{2}\) mv²_f
From the law of conservation of energy, we have

Question 19.
A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out Of the Earth’s gravitational influence? Mass of the satellite = 200 kg mass of the Earth = 6.0 x 10²⁴ kg; radius of the Earth=6.4x 10⁶ m;G=6.67 x 10^-11 N-m² kg^-2.
Solution:
Mass of the Earth, M_e =6.0 x 10²⁴ kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 x 10⁶ m
Universal gravitational constant, G = 6.67 x 10^-11 N-m²kg²
Heightofthesatellite,h =400 km=4 x 10⁵ m=0.4 x 10⁶ m

Total energy of the satellite at height h = \(\frac{1}{2} m v^{2}+\left(\frac{-G M_{e} m}{R_{e}+h}\right)\)
Orbital velocity of the satellite, ν = \(\sqrt{\frac{G M_{e}}{R_{e}+h}}\)
Total energy of height, h = \(\frac{1}{2} m\left(\frac{G M_{e}}{R_{e}+h}\right)-\frac{G M_{e} m}{R_{e}+h}=-\frac{1}{2}\left(\frac{G M_{e} m}{R_{e}+h}\right) \)

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.
Energy required to send the satellite out of its orbit = – (Bound energy)

Question 20.
Two stars each of one solar mass (= 2x 10³⁰ kg) are approaching each other for a head-on collision. When they are a distance 10⁹ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10⁴ km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Solution:
Mass of each star, M = 2 x 10³⁰ kg
Radius of each star, R = 10⁴ km = 10⁷ m
Distance between the stars, r = 10⁹ km = 10¹² m
For negligible speeds, ν = 0 total energy of two stars separated at distance r
= \(\frac{-G M M}{r}+\frac{1}{2} m v^{2}\)
= \(\frac{-G M M}{r}+0\) = 0 ………………………… (i)

Now, consider the case when the stars are about to collide:
The velocity of the stars = ν
Distance between the centers of the stars = 2R
Total kinetic energy of both stars = \(\frac{1}{2} M v^{2}+\frac{1}{2} M v^{2}\) = Mv²
Total potential energy of both stars = \(\frac{-G M M}{2 R}\)
Total energy of the two stars = Mv² – \(\frac{-G M M}{2 R}\) ………………………. (ii)
Using the law of conservation of energy, we can write

Question 21.
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Solution:
The situation is represented in the following figure :

Mass of each sphere, M = 100 kg
Separation between the spheres, r = 1 m
X is the mid-point between the spheres. Gravitational force at point X will be zero. This is because gravitational force exerted by each sphere will act in opposite directions. Gravitational potential at point X

Any object placed at point X will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

Additional Exercises

Question 22.
As you have learnt in the text, a geostationary satellite orbits the Earth at a height of nearly 36,000 km from the surface of the Earth. What is the potential due to Earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the Earth = 6.0 x 10²⁴ kg, radius = 6400 km.
Solution:
Mass of the Earth, M = 6.0 x 10²⁴ kg
Radius of the Earth, R = 6400 km = 6.4 x 10⁶ m
Height of a geostationary satellite from the surface of the Earth,
h = 36000 km = 3.6 x 10⁷m
Gravitational potential energy due to Earth’s gravity at height h,

Question 23.
A star 2.5 times the mass of the Sun and collapsed to a size of 12 km rotates with a speed of 1.2 revs. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the Sun = 2 x 10³⁰ kg).
Solution:
Yes, A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.
Gravitational force, fg = \(\frac{G M m}{R^{2}}\)
where,M =Mass of the star=2.5 x 2 x 10³⁰ = 5 x 10³⁰ kg
m = Mass of the body

R=Radiusofthestar=12 km=1.2 x 10⁴ m
∴ f_g = \(\frac{6.67 \times 10^{-11} \times 5 \times 10^{30} \times m}{\left(1.2 \times 10^{4}\right)^{2}}\)
= 2.31 x 10¹² mN
Centrifugal force, f_c = mrω²

where, ω = Angular speed = 2 πv
ν = Angular frequency = 1.2 rev s^-1
f _c=mR(2πv)²
= m x (1.2 x 10⁴) x 4 x(3.14)² x (1.2)² = 1.7 x 10⁵ mN
Since f_g > f_c, the body will remain stuck to the surface of the star.

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the spaceship = 1000 kg; mass of the Sun = 2 x 10³⁰ kg; mass of mars = 6.4 x 10²³ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 x 10⁸ km;
G = 6.67 x 10^-11 N-m²kg^-2.
Solution:
Given, mass of the Sun, M = 2 x 10³⁰ kg

Mass of the mars, m = 6.4 x 10²³ kg
Mass of spaceship, Δm = 1000 kg
Radius of orbit of the mars, r₀ = 2.28 x 10¹¹
Radius of the mars, r = 3.395 x 10⁶ m
If v is the orbital velocity of mars, then

Since the velocity of the spaceship is the same as that of the mars, Kinetic energy of the spaceship,


Total potential energy of the spaceship,
U = Potential energy of the spaceship due to its being in the gravitational
field of the mars + potential energy of the spaceship due to its being in the gravitational field of the Sun.

Total energy of the spaceship E=K +U = 2.925 x 10¹¹ J – 5.977 x 10¹¹J
= – 3.025 x 10¹¹ J = -3.1 x 10¹¹ J
Negative energy denotes that the spaceship is bound to the solar system. Thus, the energy needed by the spaceship to escape from the solar system = 3.1 x 10¹¹ J.

Question 25.
A rocket is fired ‘vertically’ from file surface of mars with a speed of 2 kms^-1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of Mars before returning to it?
Mass of mars = 6.4x 10²³ kg; radius of mars = 3395 km;
G = 6.67 x 10^-11 N-m² kg².
Solution:
Initial velocity of the rocket, ν = 2 km/s = 2 x 10³ m/s
Mass of Mars, M = 6.4 x 10²³ kg .
Radius of Mars, R = 3395 km = 3.395 x 10⁶ m
Universal gravitational constant, G = 6.67 x 10^-11 N-m² kg^-2
Mass of the rocket = m
Initial kinetic energy of the rocket = \(\frac{1}{2} m v^{2}\)
Initial potential energy of the rocket = \(\frac{-G M m}{R}\)
Total initial energy = \(\frac{1}{2} m v^{2}-\frac{G M m}{R}\)

If 20 % of initial kinetic energy is lost due to martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height.
Total initial energy available = \(\frac{80}{100} \times \frac{1}{2} m v^{2}-\frac{G M m}{R}=0.4 m v^{2}-\frac{G M m}{R} \)
Maximum height reached by the rocket = h
At this height, the velocity and hence, the kinetic energy of the rocket will become zero.
Total energy of the rocket at height h = \(-\frac{G M m}{(R+h)}\)
Applying the law of conservation of energy for the rocket, we can write

= 495 x 10³ m = 495 km

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 1 The Living World Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 1 The Living World

PSEB 11th Class Biology Guide The Living World Textbook Questions and Answers

Question 1.
Why are living organisms classified?
Answer:
Living organisms are classified because:

• there are millions of organisms on the earth, which need a proper system of classification for identification.
• a number of new organisms are discovered each year. They require a particular system to be identified and to find out their correct position in a group.

Question 2.
Why are the classification systems changing every now and then?
Answer:
Evolution is the major factor responsible for the change in classification systems. Since, evolution still continues, so many different species of plants and animals are added in the already existed biodiversity. These newly discovered plant and animal specimens are then identified, classified and named according to the already existing classification systems. Due to evolution, animal and plant species keep on changing, so necessary changes in the already existed classification systems are necessary to place every newly discovered plant and animal in their respective ranks.

Question 3.
What different criteria would you choose to classify people that you meet often?
Answer:
The different scientific criteria to classify people that we meet often would be :
(i) Nomenclature: It is the science of providing distinct and proper names to the organisms. It is the determination of correct name as per established universal practices and rules.

(ii) Classification: It is the arrangement of organisms into categories based on systematic planning. In classification various categories used are class, order, family, genus and species.

(iii) Identification: It is the determination of correct name and place of an organism. Identification is used to tell that a particular species is similar to other organism of known identity. This includes assigning an organism to a particular taxonomic group.
The same criteria can be applied to the people we meet daily. We can identify them will their names classify them according to their living areas, profession, etc.

Question 4.
What do we learn from identification of individuals and populations?
Answer:
Identification of individuals and population categorized it into a species. Each species has unique characteristic features. On the basis of these features, it can be distinguished from other closely related species, e. g.,

Question 5.
Given below is the scientific name of Mango. Identify the correctly written name.
(i) Mangifera Indica
(ii) Mangifera indica.
Answer:
(ii) Mangifera indica (the name of species can never begins with a capital letter).

Question 6.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Answer:
Taxon is a grouping of organisms of any level in hierarchical classification, which is based on some common characteristics, e.g., insects represent a class of phylum – Arthropoda. All the insects possess common characters of three pairs of jointed legs. The term ‘taxon’ was introduced by ICBN in 1956. Examples of taxa are kingdom, phylum or division, class, order, family, genus and species. These taxa form taxonomic hierarchy, e.g., taxa for human :
Phylum – Chordata
Class – Mammalia
Order – Primata
Family – Hominidae
Genus – Homo
Species – sapiens

Question 7.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Answer:
The correct sequence of taxonomical categories is as follows :
Species → Genus → Order → Phylum

Question 8.
Try to collect all the currently accepted meanings for the word ‘species’. Discuss with your teacher the meanings of species in case of higher plants and animals on one hand, and bacteria on the other hand.
Answer:
A group of individual organisms with fundamental similarities is called species. It can be distinguished from other closely related species on the basis of distinct morphological differences.
In case of higher plants and animals, one genus may have one or more than one species, e.g.,Panthera leo (lion) and Panthera tigris (tiger).
In this example, Panthera is genus, which includes leo (lion) and tigris (tiger) as species.
In case of bacteria, different categories are present on the basis of shape. These are spherical, coccus, rod-shaped, comma and spiral-shaped. Thus, meaning of species in case of higher organisms and bacteria are different.

Question 9.
Define and understand the following terms:
(i) Phylum,
(ii) Class,
(iii) Family,
(iv) Order,
(v) Genus.
Answer:
(i) Phylum: Phylum comes next to Kingdom in the taxonomical hierarchy. All broad characteristics of an animal or plant are defined in a phylum. For example all chordates have a notochord and gill at some stage of life cycle. Similarly all arthropods have jointed legs made of chitin.

(ii) Class: The category class includes related orders. It is higher than order and lower than phylum. For example, class – Mammalia has order – Carnivora, Primata, etc.

(iii) Family: It is the category higher than genus and lower than order, which has one or more related genera having some common features. For example, Felidae, Canidae, etc.

(iv) Order: Order further zeroes down on characteristics and includes related genus. For example humans and monkeys belong to the order primates. Both humans and monkeys can use their hands to manipulate objects and can walk on their hind legs.

(v) Genus: It comprises a group of related species which has more characters in common in comparison to species of other genera. We can say that genera are aggregates of closely related species. For example, j potato, tomato and brinjal are three different species but all belong to the genus Solatium. Lion (Panthera leo), leopard (P. pardus) and tiger (P. tigris) with several common features, are all species of the genus Panthera. This genus differs from another genus Felis which includes cats.

Question 10.
How is a key helpful in the identification and classification of an
organism?
Answer:
Key is a device (scheme) of diagnostic alternate (contrasting) characters, which provide an easy means for the identification of unknown organism. The keys are taxonomic literature based on the contrasting characters generally a pair called couplet. Each statement in the key is called a lead. Separate taxonomic keys are required for each taxonomic category such as family, genus and species for identification purposes. Being analytical ‘ in nature, two types of keys are commonly used-indented key and bracketed key.

(i) Indented key provides sequence of choice between two or more statements of characters of species. The user has to make correct choise for identification.

(ii) Bracketed key (1) are used for contrasting characters like indented key but they are not repeated by intervening sub-dividing character and each character is given a number in brackets.

Question 11.
Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.
Answer:

Taxonomical hierarchy is the system of arrangement of taxonomic categories in a descending order depending upon their relative dimensions. It was introduced by Linnaeus (1751) and is therefore, also called Linnaeus hierarchy. Each category, referred to as a unit of classification commonly called as taxon (pi. taxa), e.g., taxonomic categories and hierarchy can be illustrated by a group of organisms, i.e., insects. The common features of insects is ‘three pair of jointed legs’. It means insects are recognisable objects which can be classified, so given a rank or category.
Category further denotes a rank. Each rank or taxon, represents a unit of classification taxonomic studies of all plants and animals led to the development of common categories such as kingdom, phylum or division (for plants), class, order, family, genus and species. All organisms, including those in the plant and animal, kingdoms have ‘species’ as the lowest category.

To place an organism in various categories is to have the knowledge of characters of an individual or group of organism. This helps to identify similarities and dissimilarities among the individual of the same kind of organisms as well as of other kinds of organism. Some organisms with their taxonomical categories are given in following table:

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 21 Neural Control and Coordination Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 21 Neural Control and Coordination

PSEB 11th Class Biology Guide Neural Control and Coordination Textbook Questions and Answers

Question 1.
Briefly describe the structure of the following:
(a) Brain
(b) Eye
(c) Ear
Answer:
(a) Brain
The human brain has the following parts :
(i) Cerebrum
A deep cleft called longitudinal fissure divides the brain/cerebrum into two halves-cerebral hemispheres.
The two cerebral hemispheres are joined together by bundles of densely packed nerve fibers, called corpus callosum.
The outer surface of cerebrum, the cerebral cortex, is called grey matter, due to its greyish appearance; the cell bodies of the neurons are concentrated in this region; it contains motor areas, sensory areas and association areas.
Inner to the cortex is the white matter, that consists of myelinated nerve fibers in the form of nerve fibre tracts.

(ii) Thalamus
Thalamus is the major coordinating centre for sensory and motor signals.

(iii) Hypothalamus
It has centers to control body temperature, hunger, thirst, etc.
It contains several groups of neurosecretory cells, which secrete hormones.

(iv) Limbic System
The inner parts of the cerebral hemispheres and a group of deep structures called amygdala, hippocampus, etc. form a complex structure, called limbic system. Along with the hypothalamus, it is involved in the regulation of sexual behavior, expression of emotions, motivation, etc.

(v) Midbrain
Midbrain is located between the hypothalamus of the forebrain and the pons of the hindbrain. The dorsal portion of the midbrain consists of four small lobes, called corpora quadrigemina. A canal, called cerebral aqueduct passes through the midbrain.

(vi) Hindbrain
It consists of pons, cerebellum and medulla oblongata. The medulla contains centres which control vital functions like respiration, cardiovascular reflexes and gastric secretions. The medulla continues down as the spinal cord.

(b) Eye

• Each eye ball consists of three concentric layers, the outermost sclera, middle choroid and innermost retina.
• The sclera in the front (l/6th) forms the transparent cornea.
• The middle choroid is highly vascular and pigmented, that prevents internally reflected light within the eye; just behind the junction between cornea and sclera, the choroid becomes thicker forming the ciliary body.
• The iris extends from the ciliary body in front of the lens; it controls the dilation or constriction of pupil.
• The lens is suspended from the ciliary body, by suspensory ligaments.
• The anterior chamber of eye is filled with an aqueous clear fluid, aqueous humor and the posterior chamber has a gelatinous material, vitreous humor.
• The retina is composed of three layers of cells; the photoreceptor layer contains rods and cones, the intermediate layer has bipolar neurons, which synapse with retinal ganglion cells and their axons bundle to form optic nerve.
• The photoreceptor cells (rods and cones) contain the light sensitive proteins, called photopigments.
• The point in the retina where the optic nerve leaves the eye and the retinal blood vessels enter the eye is called a blind spot; there are no photoreceptor cells in this region.
• Lateral to blindspot, there is a yellowish pigmented spot, called macula lutea with a central pit called fovea.
• The fovea is the region where only cones are densely packed and it is the point where acuity (resolution) vision is the greatest.

(c) Ear
The ear performs two sensory functions, namely
(a) hearing and

(b) maintenance of body balance.

• Ear consists of three parts: external ear, middle ear and internal ear.
• The external ear consists of the pinna, and external auditory meatus.
• The tympanic membrane separates the middle ear from the external ear.
• The middle ear (tympanic cavity) is an air- filled chamber, which is connected to pharynx by Eustachian tube.
• The middle ear lodges three small bones, the ear ossicles namely, the malleus, incus and stapes.
• The middle ear communicates with the internal ear through the oval window and round window.
• The inner ear is a fluid-filled chamber and called labyrinth; it has two parts, an outer bony labyrinth, inside
• which a membranous labyrinth is floating in the perilymph; the membranous labyrinth is filled with a fluid, called endolymph.
• The labyrinth is divided into two parts, the cochlea and vestibular apparatus.
• Cochlea is the coiled portion of the labyrinth and its membranes, Reissner’s membrane and basilar membrane divide the perilymph-filled bony labyrinth into an upper scala vestibule, middle scala media and a lower scala tympani; scala media is filled with endolymph.
• At the base of the cochlea, scala vestibuli ends at the oval window, while the scala tympani terminates at the round window, that opens to the middle ear.
• Organ of Corti is the structural unit of hearing; it consists of hair cells which are the auditory receptors and is located on the basilar membrane.
• A thin elastic tectorial membrane lies over the row of hair cells.
• The vestibular apparatus is composed of three semicircular canals and an otolith organ or vestibule.
• The otolith organ has two parts namely the utricle and saccule.
• The utricle and saccule also contain a projecting ridge, called macula.
• The crista ampullar and macula are the specific receptors of the vestibular apparatus, for maintaining body balance.

Question 2.
Compare the following:
(a) Central Neural System (CNS) and Peripheral Neural System (PNS)
(b) Resting potential and action potential
(c) Choroid and retina
Answer:
(a) Comparison between Central Neural System (CNS) and Peripheral Neural System (PNS): The CNS includes the brain and the spinal cord and is the site of information processing and control. The PNS comprises of all the nerves of the body associated with the CNS (brain and spinal cord). The nerve fibers of the PNS are of two types :
(i) Afferent fibers, (ii) Efferent fibers

(b) Comparison between Resting Potential and Action Potential:
The electrical potential difference across the resting plasma membrane A is called the resting potential. The electrical potential difference across the plasma membrane at the site A is called the action potential, which is in fact termed as a nerve impulse.

(c) Comparison between Choroid and Retina: The middle layer of eyeball which contains many blood vessels and looks bluish in colour is known as choroid. The choroid layer is thin over the posterior two-thirds of the eyeball, but it becomes thick in the anterior part to form the ciliary body. The ciliary body itself continues forward to form a pigmented and opaque structure called the iris.

Retina is the inner layer of eye ball and it contains three layers of cells from inside to outside, i. e., ganglion cells, bipolar cells and photoreceptor cells. There are two types of photoreceptor cells, namely, rods and cones. These cells contain the light-sensitive proteins called the photopigments.

Question 3.
Explain the following processes:
(a) Polarisation of the membrane of a nerve fibre
(b) Depolarisation of the membrane of a nerve fibre
(c) Conduction of a nerve impulse along a nerve fibre
(d) Transmission \of a nerve impulse across a chemical synapse
Answer:
(a) Polarisation of the Membrane of a Nerve Fibre: During resting condition, the concentration of K+ ions is more inside the axoplasm while the concentration of Na+ ions is more outside the axoplasm. As a result, the potassium ions move faster from inside to outside as compared to sodium ions. Therefore, the membrane becomes positively charged outside and negatively charged inside. This is known as polarisation of membrane or polarised nerve.

(b) Depolarisation of the Membrane of a Nerve Fibre: When an electrical stimulus is given to a nerve fibre, an action potential is generated. The membrane becomes permeable to sodium ions than to potassium ions. This results into positive charge inside and negative charge outside the nerve fibre. Hence, the membrane is said to be depolarised.

(c) Conduction of a Nerve Impulse Along a Nerve Fibre: There are two types of nerve fibers-myelinated and non-myelinated. In myelinated nerve fibre, the action potential is conducted from node to node in jumping manner. This is because the myelinated nerve fibre is coated with myelin sheath.

The myelin sheath is impermeable to ions. As a result, the ionic exchange and depolarization of nerve fiber is not possible along the whole length of nerve fiber. It takes place only at some point, known as nodes of Ranvier, whereas in non-myelinated nerve fiber, the ionic exchange and depolarization of nerve fiber takes place along the whole length of the nerve fiber. Because of this ionic exchange, the depolarised area becomes repolarised and the next polarised area becomes depolarised.

(d) Transmission of a Nerve Impulse Across a Chemical Synapse:
Synapse is a small gap that occurs between the last portion of the axon of one neuron and the dendrite of next neuron. When an impulse reaches at the endplate of axon, vesicles consisting of chemical substances or neurotransmitters, such as acetylcholine, fuse with the plasma membrane.

This chemical moves across the cleft and attaches to chemo-receptors present on the membrane of the dendrite of next neuron. This binding of chemical with chemo-receptors leads to the depolarization of membrane and generates a nerve impulse across nerve fibre. The chemical, acetylcholine, is inactivated by enzyme acetylcholinesterase. The enzyme is present in the postsynaptic membrane of the dendrite. It hydrolyses acetylcholine and this allows the membrane to repolarise.

Question 4.
Draw labeled diagrams of the following:
(a) Neuron
(b) Brain
(c) Eye
(d) Ear
Answer:
(a) Neuron

(b) Brain

(c) Eye

(d) Ear

Question 5.
Write short notes on the following:
(a) Neural coordination
(b) Forebrain
(c) Midbrain
(d) Hindbrain
(e) Retina
(f) Ear ossicles
(g) Cochlea
(h) Organ of Corti
(i) Synapse
Answer:
(a) Neural Coordination: The organized network of point-to-point connections for quick coordination provided by neural system is called neural coordination. The mechanism of neural coordination involves transmission of nerve impulses, impulse conduction across a synapse, and the physiology of reflex action.

(b) Forebrain: The forebrain consists of :
1. Olfactory lobes: The anterior part of the brain is formed by a pair of short club-shaped structures, the olfactory lobes. These are concerned with the sense of smell.

2. Cerebrum: It is the largest and most complex of all the parts of the human brain. A deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres connected by a large bundle of myelinated fibres the corpus callosum. The outer cover of cerebral hemisphere is called cerebral cortex. The cerebral cortex is referred to as the grey matter due to its greyish appearance (as neuron cell bodies are concentrated here).

The cerebral cortex is greatly folded. The upward folds, gyri, alternate with the downward grooves or sulci. Beneath the grey matter, there are millions of medullated nerve fibers, which constitute the inner part of the cerebral hemisphere. The large concentration of medullated nerve fibers gives this tissue an opaque white appearance. Hence, it is called the white matter.

3. Lobes: A very deep and a longitudinal fissure, separates the two cerebral hemispheres. Each cerebral hemisphere of the cerebrum is divided into four lobes, i.e., frontal, parietal, temporal, and occipital lobes.

In each cerebral hemisphere, there are three types of functional areas:
(i) Sensory areas receive impulses from the receptors and motor areas transmit impulses to the effectors.
(ii) Association areas are large regions that are neither clearly sensory nor motor injunction. They interpret the input, store the input and initiate a response in light of similar past experiences. Thus, these areas are responsible for complex functions like memory, learning, reasoning, and other intersensory associations.

(iii) Diencephalon is the posteroventral part of forebrain. Its main parts are as follows :
Epithalamus is a thin membrane of non-neural tissue. It is the posterior segment of the diencephalon. The cerebrum wraps around a structure called thalamus, which is a major coordinating center for sensory and motor signaling. The hypothalamus, that lies at the base of thalamus contains a number of centers, which control body temperature, urge for eating and drinking. It also contains several groups of neurosecretory cells, which secrete hormones called hypothalamic hormones.

(c) Midbrain: The midbrain is located between the thalamus and hypothalamus of the forebrain and pons of the hindbrain. A canal called the cerebral aqueduct passes through, the midbrain.
The dorsal portion of the midbrain mainly consists of two pairs (i.e., four) of rounded swellings (lobes) called corpora qua trigeminal.

(d) Hindbrain: The hindbrain consists of :

(i) Pons: It consists of fiber tracts that interconnect different regions of the brain.

(ii) Cerebellum: It is the second-largest part of the human brain (means little cerebrum). It has very convoluted surface in order to provide the additional space for many more neurons.

(iii) Medulla: It (oblongata) is connected to the spinal cord and contains centers, which control respiration, cardiovascular reflexes, and gastric secretions.

(e) Retina: The inner layer of eyeball is the retina and it contains three layers of cells from inside to outside—ganglion cells, bipolar cells and photoreceptor cells. There are two types of photoreceptor cells namely, rods and cones. These cells contain the light-sensitive proteins called the photopigments.

(f) Ear Ossicles: The middle ear contains three ossicles called malleus, incus and stapes which are attached to one another in a chain-like fashion. The malleus is attached to the tympanic membrane and the stapes is attached to the oval window or the cochlea. The ear ossicles increase the efficiency of transmission of sound waves to the inner ear.

(g) Cochlear: The membranous labyrinth of inner ear is filled with a fluid called endolymph. The coiled portion of the labyrinth is called cochlea. The membranes constituting cochlea, the Meissner’s and basilar, divide the surrounding perilymph-filled bony labyrinth into an upper scale vestibule and a lower scala tympani. The space within cochlea called scala media is filled with endolymph. At the base of the cochlea, the scala vestibule ends at the oval window, while the scala tympani terminates at the round window which opens to the middle ear.

(h) Organ of Corti: The organ of Corti is a structure located on the basilar membrane of inner ear, which contains hair cells that act as auditory receptors. The hair cells are present in rows on the internal side of the organ of Corti. The basal end of the hair cell is in close contact with the afferent nerve fibers. A large number of processes called stereocilia are projected from the apical part of each hair cell. Above the rows of the hair cells is a thin elastic membrane called tectorial membrane.

(i) Synapse: It is a junction between two neurons, where one neuron expands and comes in near contact with another neuron. A synapse is formed by the membranes of a pre-synaptic neuron, and a post-synaptic neuron, which may or may not be separated by a gap called synaptic cleft.

There are two types of synapses-an electrical synapse and a chemical synapse. In electrical synapse, membranes of pre and post-synaptic neurons are is very close proximity field. In chemical synapse, these membranes are separated by a fluid-filled space called synaptic cleft.

Question 6.
Give a brief account of:
(a) Mechanism of synaptic transmission
(b) Mechanism of vision
(c) Mechanism of hearing
Answer:
(a) Mechanism of Synaptic Transmission: Synapse is a junction between two neurons. It is present between the axon terminal of one neuron and the dendrite of next neuron separated by a cleft.

There are two ways of synaptic transmission :
1. Chemical transmission: When a nerve impulse reaches the end plate of axon, it releases a neurotransmitter (acetylcholine) across the synaptic cleft. This chemical is synthesized in cell body of the neuron and is transported to the axon terminal. The acetylcholine diffuses across the cleft and binds to the receptors present on the membrane of next neuron. This causes depolarization of membrane and initiates an action potential.

2. Electrical transmission: In this type of transmission, an electric current is formed in the neuron. This electric current generates an action potential and leads to transmission of nerve impulses across the nerve fiber. This represents a faster method of nerve conduction than the chemical method of transmission.

(b) Mechanism of Vision: Retina is the innermost layer of eye. It contains three layers of cells-inner ganglion cells, middle bipolar cells and outermost photoreceptor cells. A photoreceptor cell is composed of a protein called opsin and an aldehyde of vitamin A called retinal. When light rays are focused on the retina through cornea, it leads to the dissociation of retinal from opsin protein.

This changes the structure of opsin. As the structure of opsin changes, the permeability of membrane changes, generating a potential difference in the cells. This generates an action potential in the ganglionic cells and is transmitted to the visual cortex of the brain via optic nerves. In the cortex region of brain, the impulses are analyzed and image is formed on the retina.

(c) Mechanism of Hearing: The pinna of the external region collects the sound waves and directs it towards ear drum or external auditory canal. These waves strike the tympanic membrane and vibrations are created. Then, these vibrations are transmitted to the oval window, fenestra ovalis, through three ear ossicles, named as malleus, incus, and stapes. These ear ossicles act as lever and transmit the sound waves to internal ear.

These vibrations from fenestra ovalis are transmitted into cochlear fluid. This generates sound waves in the lymph. The formation of waves generates a ripple in the basilar membrane. This movement bends the sensory hair cells present on the organ of corti against tectorial membrane. As a result of this, sound waves are converted into nerve impulses. These impulses are then carried to auditory cortex of brain via auditory nerves. In cerebral cortex of brain, the impulses are analysed and sound is recognised.

Question 7.
Answer briefly:
(a) How do you perceive the colour of an object?
(b) Which part of our body helps us in maintaining the body balance?
(c) How does the eye regulate the amount of light that falls on the retina?
Answer:
(a) The daylight (photopic) vision and colour vision are functions of cones. In the human eye, there are three types of cones which possess their own characteristic photopigments that respond to red, green and blue lights. The sensations of different colours are produced by various combinations of these cones and their photopigments. When these cones are stimulated equally a sensation of white light is produced.

(b) The crista and macula are the specific receptors of the vestibular apparatus of inner ear which are responsible for the maintenance of balance of the body and posture.

(c) The diameter of the pupil is regulated by the muscle fiber of iris. Photoreceptors, rods, and cones regulate the amount of light that falls on the retina.

Question 8.
Explain the following:
(a) Role of Na⁺ in the generation of action potential.
(b) Mechanism of generation of light-induced impulse in the retina.
(c) Mechanism through which a sound produces a nerve impulse in the inner ear.
Answer:
(a) Role of Na⁺ in the Generation of Action Potential: When a stimulus is applied to a nerve, the membrane of the nerve becomes freely permeable to Na ^{+ .} This leads to a rapid influx of Na⁺ followed by the reversal of the polarity at that site, i. e., the outer surface of the membrane becomes negatively charged and the inner side becomes positively charged. The electrical potential difference across plasma membrane at the membrane is called the action potential, which is in fact termed as a nerve impulse. Thus, this shows that Na⁺ ions play an important role in the conduction of nerve impulses.

(b) Mechanism of Generation of Light-induced Impulse in the Retina: Light induces dissociation of the retina from opsin resulting in changes in the structure of the opsin. This causes membrane permeability changes. As a result, potential differences are generated in the photoreceptor cells. This produces a signal that generates action potentials in the ganglion cells through the bipolar cells.

These action potentials (impulses) are transmitted by the optic nerves to the visual corted area of the brain, where the nerve impulses are analysed and the image formed on the retina is recognised.

(c) Mechanism through which a Sound Produces a Nerve Impulse in the Inner Ear: In the inner ear, the vibrations are passed through the oval window on to the fluid of the cochlea, where they generate waves in the lymph.
The waves in the lymphs induce a ripple in the basilar membrane.

These movements of the basilar membrane bend \ the hair cells, pressing them against the tectorial membrane. As a result, nerve impulses are generated in the associated afferent neurons. These impulses are transmitted by the afferent fibres via auditory nerves to the auditory cortex of the brain, where the impulses are analyzed and the sound is recognized.

Question 9.
Differentiate between:
(a) Myelinated and non-myelinated axons
(b) Dendrites and axons
(c) Rods and cones
(d) Thalamus and hypothalamus
(e) Cerebrum and cerebellum
Answer:
(a) Differences between Myelinated and Non-myelinated Axons

Myelinated Axon	Non-myelinated Axon
1. The myelinated nerve fibers are enveloped with Schwann cells, which form a myelin sheath around the axon.	1. Unmyelinatcd nerve fibers are enclosed by a Schwann cell that does not form a myelin sheath around the axon.
2. Myelinated nerve fibres are found in spinal and cranial nerves.	2. They are commonly found in autonomous and the somatic neural systems.

(b) Differences between Dendrite and Axon

Dendrite	Axon
1. These are short fibres which branch repeatedly and project out of the cell body also contain Nissl’s granules	The axon is a long branched fibre, which terminates as a bulb-like structure called. synaptic knob. It possess synaptic vesicles containing chemicals called neurotransmitters.
2. These fibres transmit impulses towards the cell body.	The axons transmit nerve impulses away from the cell body to a synapse.

(c) Differences between Rods and Cones

Rod	Cone
1. The twilight vision is the function of rods.	The daylight vision and colour vision are functions of cones.
2. The rods contain a purplish-red protein called the rhodopsin or visual purple, which contains a derivative of Vitamin-A	In the human eye, there are three types of cones which possess their own characteristic photopigments that respond to red, green and blue lights.

(d) Differences between Thalamus and Hypothalamus

Thalamus	Hypothalamus
1. The cerebrum wraps around a structure called thalamus.	It lies at the base of the thalamus.
2. All types of sensory input passes synapses in the thalamus	It contains neurosecretory cells that secrete hypothalamus hormones.
3. It controls emotional and memory functions.	It regulates, sexual behavior, expression of emotional reactions and motivation.

.(e) Differences between Cerebrum and Cerebellum

Cerebrum	Cerebellum
1. It is the most developed part in brain.	It is the second developed part of brain also called as little cerebrum
2. A deep cleft divides cerebrum into two cerebral hemispheres.	Externally the whole surface contains gyri and sulci.
3. Its functions are intelligence, learning, memory, speech, etc.	It contains centres for coordination and error checking during motor and cognition.

Question 10.
Answer the following:
(a) Which part of the ear determines the pitch of a sound?
(b) Which part of the human brain is the most developed?
(c) Which part of our central neural, system acts as a master clock?
Answer:
(a) Inner ear
(b) Cerebrum
(c) Brain

Question 11.
The region of the vertebrate eye, where the optic nerve passes out of the retina, is called the
(a) fovea
(b) iris
(c) blind spot
(d) optic chiasma
Answer:
(d) Optic Charisma

Question 12.
Distinguish between:
(a) Afferent neurons and efferent neurons.
(b) Impulse conduction in a myelinated nerve fibre and unmyelinated nerve fibre.
(c) Aqueous humour and vitreous humour.
(d) Blind spot and yellow spot.
(e) Cranial nerves and spinal nerves.
Answer:
(a) Differences between Afferent neurons and Efferent neurons

Afferent Neurons	Efferent Neurons
The afferent nerve fibres transmit impulses from tissues/organs to the CNS.	The efferent fibres transmit regulatory impulses from the CNS to the concerned peripheral tissues/organs.

(b) Differences between Myelinated and Non-myelinated Axons

Myelinated Axon	Non-myelinated Axon
1. The myelinated nerve fibers are enveloped with Schwann cells, which form a myelin sheath around the axon.	1. Unmyelinatcd nerve fibers are enclosed by a Schwann cell that does not form a myelin sheath around the axon.
2. Myelinated nerve fibres are found in spinal and cranial nerves.	2. They are commonly found in autonomous and the somatic neural systems.

(c) Differences between Aqueous humour and Vitreous humour

Aqueous Humour	Vitreous Humour
1. It is the space between the cornea and the lens.	The space between the lens and the retina is called the vitreous chamber.
2. It contains a thin watery fluid.	It is filled with a transparent gel.

(d) Differences between Blindspot and Yellow spot

Blind Spot	Yellow Spot
1. Photoreceptor cells are not present in this region.	Yellow spot or macula lutea is located at the posterior pole of the eye lateral to the blind spot. It has a central pit called fovea.
2. The light focuses on that part of the retina is not detected.	The fovea of yellow spot is a thinned-out portion of retina where only the cones are densely packed is the point where visual cavity is greatest.

(e) Differences between Cranial nerves and spinal nerves

Cranial Nerves	Spinal Nerves
1. The cranial nerves originate in the brain and terminate mostly in organs head and upper body.	The spinal nerves originate in the spinal cord and extend to parts of the body below the head.
2. There are 12 pairs of cranial nerves.	There are 31 pairs of spinal nerves.
3. Most of the cranial nerves contain axons and both sensory and motor neurons.	All of the spinal nerves contain axons of both sensory and motor neurons.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 11 Thermal Properties of Matter Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

PSEB 11th Class Physics Guide Thermal Properties of Matter Textbook Questions and Answers

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Solution:
Kelvin and Celsius’s scales are related as
T_C =T_K -273.15
Celsius and Fahrenheit’s scales are related as …(i)
T_F = \(\frac{9}{5}\) T_C +32 ………………………….. (ii)
For neon:
T_K = 24.57K
∴ T_G = 24.57 – 273.15 = -284.58°C
T_F =\(\frac{9}{5}\) T_C +32
= \(\frac{9}{5}\) (-248.58) + 32
=-415.44° F

For carbon dioxide:
T_K = 216.55K
∴ T_C =216.55-273.15 = -56.60°C
T_F =\(\frac{9}{5}\) T_C +32
= \(\frac{9}{5}\) (-56.60) + 32 = -69.88°C

Question 2.
Two absolute scales A and B have triple points of water defined to be 200A and 350B.
What is the relation between T_Aand T_B?
Solution:
Triple point of water on absolute scale A, T₁ = 200 A
Triple point of water on absolute scale B, T₂ = 350 B
Triple point of water on Kelvin scale, T_K = 273.15K
The temperature 273.15K on Kelvein scale is equivalent to 200 A on absolute scale A.
T₁ =T_K
200 A = 273.15 K .
∴ A= \(\frac{273.15}{200}\)

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.
T₂=T_K
350 B = 273.15
∴ B = \(\frac{273.15}{350}\)
T_A is triple point of water on scale A.
T_B is triple point of water on scale B.
∴ \(\frac{273.15}{200} \times T_{A}=\frac{273.15}{350} \times T_{B}\)
T_A = \(\frac{200}{350} T_{B}\)
T_A = \(\frac{4}{7} T_{B}\)
Therefore, the ratioT_A:T_B is given as 4 : 7.

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = R₀[l + α (T-T₀)]
The resistance is 101.6Ω at the triple-point of water 273.16 K, and 165.50 at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4Ω?
Solution:
It is given that:
R = R₀[l + α (T-T₀)] ………………………….(i)
where R₀ and T₀ are the initial resistance and temperature respectively R and T are the final resistance and temperature respectively α is a constant At the triple point of water, T₀ = 273.16 K
Resistance of lead, R₀ =101.6 Ω

At normal melting point of lead, T = 600.5 K
Resistance of lead, R = 165.5 Ω
Substituting these values in equation (i), we get:
R=R₀[l + α (T-T₀)]
165.5 = 101.6[1 + α (600.5-273.16)]
1.629 = 1+α (327.34)
∴ α = \(\frac{0.629}{327.34} \) = 1.92 x10^-3K^-1
For resistance, R₁ = 123.4Ω
R₁ =R₀[l + α (T-T₀)]

where, T is the temperature when the resistance of lead is 123.4Ω
123.4 =101.6[1 +1.92 x 10^-3(T-273.16)]
1.214 =1+1.92 x 10^-3(T- 273.16)
\(\frac{0.214}{1.92 \times 10^{-3}}\) = T -273.16
111.46 = T-273.16
⇒ T =111.46 +273.16
∴ T = 384.62 K

Question 4.
Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) T is related to the temperature t_c on the Celsius scale by
t_c = T – 273.15
Why do we have 273.15 in this relation, and not 273.16?

(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
Solution:
(a) The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

(b) The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.

(c) The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K. Hence, absolute temperature (Kelvin scafe) T, is related to temperature t_c, on Celsius scale as:
t_c =T -273.15

(d) Let T_F be the temperature on Fahrenheit scale and T_K be the temperature on absolute scale. Both the temperatures can be related as
\(\frac{T_{F}-32}{180}=\frac{T_{K}-273.15}{100}\) ………………………….. (i)
Let T_F1 be the temperature on Fahrenheit scale and T_K1 be the temperature on absolute scale. Both the temperatures can be relates as
\(\frac{T_{F 1}-32}{180}=\frac{T_{K 1}-273.15}{100}\) ……………………….. (ii)
It is given that:
T_K1-T_K= 1K

Subtracting equation (i) from equation (ii), we get
\(\frac{T_{F 1}-T_{F}}{180}=\frac{T_{K 1}-T_{K}}{100}\)
\(T_{F 1}-T_{F}=\frac{1 \times 180}{100}=\frac{9}{5}\)
Triple point of water = 273.16 K
∴ Triple point of water on absolute scale = 273.16 x \( \frac{9}{5}\) = 491.69

Question 5.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

Temperature	Pressure thermometer A	Pressure thermometer B
Triple-point of water	1.250 x 105 Pa	0.200 x 105 Pa
Normal melting point of sulphur	1.797 x 105 Pa	0.287 x 105 Pa

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Solution:
Triple point of water, T = 273.16 K.
At this temperature, pressure in thermometer A, P_A = 1.250 x 10⁵ Pa
Let T₁ be the normal melting point of sulphur.
At this temperature, pressure in thermometer A,P₁ = 1.797 x 10⁵ Pa
According to Charles’ law, we have the relation
\(\frac{P_{A}}{T}=\frac{P_{1}}{T_{1}}\)
∴ T₁ = \(\frac{P_{1} T}{P_{A}}=\frac{1.797 \times 10^{5} \times 273.16}{1.250 \times 10^{5}}\)
= 392.69 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.
At triple point 237.16 K, the pressure in thermometer B,
P_B =0.200 x 10⁵ Pa
At temperature T₂, the pressure in thermometer B, P₂ = 0.287 x 10⁵ Pa
According to Charles’ law, we can write the relation
\(\frac{P_{B}}{T}=\frac{P_{2}}{T_{2}}\)
\(\frac{0.200 \times 10^{5}}{273.16}=\frac{0.287 \times 10^{5}}{T_{2}}\)
∴ T₂ = \(\frac{0.287 \times 10^{5}}{0.200 \times 10^{5}} \times 273.16\) = 391.98 K
Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K

(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B. To reduce the discrepancy between the two readings, the experiment should be carried under low-pressure conditions. At low pressure, these gases behave as perfect ideal gases.

Question 6.
A steel tape lm long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cmon a hot day when the temperature is 45°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0°C? Coefficient of linear expansion of steel = 1.20 x 10^-5 K^-1.
Solution:
Length of the steel tape at temperature T = 27°C, l = 1 m = 100 cm
At temperature T₁ = 45°C,
the length of the steel rod, l₁ = 63 cm
Coefficient of linear expansion of steel, α = 1.20 x 10^-5K^-1

Let l₂ be the actual length of the steel rod and l’ be the length of the steel tape at 45°C.
l’ =l+αl(T₁ -T)
∴ l’ = 100 +1.20 x 10^-5x 100(45 -27)
= 100,0216 cm

Hence, the actual length of the steel rod measured by the steel tape at 45° C can be calculated as
l₂ = \(\frac{100.0216}{100} \times 63\) = 63.0136cm
Therefore, the actual length of the rod at 45°C is 63.0136 cm. Its length at 27.0 °C is 63.0 cm

Question 7.
A large steel wheel is to be fitted onto a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm, and the diameter of the central hole in the wheel is 8.69cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume the coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.20 x 10^-5 K^-1.
Solution:
The given temperature, T = 27°C can be written in Kelvin as
27 + 273 =300K
Outer diameter of the steel shaft at T, d₁ = 8.70 cm
Diameter of the central hole in the wheel at T, d₂ = 8.69 cm
Coefficient of linear expansion of steel, a steel = 1.20 x 10^-5 K ^-1
After the shaft is cooled using ‘dry ice’, its temperature becomes T₁.
The wheel will slip on the shaft if the change in diameter,
Δd = 8.69-8.70 =-0.01 cm

Temperature T₁; can be calculated from the relation
Δd = d₁α_steel(T₁ -T)
-0.01 =8.70 x 1.20 x 10^-5(T₁ -300)
(T₁ – 300) = \(\frac{-0.01}{8.70 \times 1.20 \times 10^{-5}}\)
(T₁ -300) = -95.78
∴ T₁ = 300 – 95.78 = 204.22 K
= (204.22-273)°C
= -68.78°C ≈ 69°C
Therefore, the wheel will slip on the shaft when the temperature of the shaft is -69°C.

Question 8.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of copper = 1.70 x 10^-5 K ^-1.
solution:
Initial temperature, T₁ = 27.0°C
Diameter of the hole at T₁, d₁ = 4.24 cm
Final temperature, T₂ = 227°C
Let, diameter of the hole at T₂=d₂
Coefficient of linear expansion of copper, α_Cu = 1.70 x 10^-5 K^-1
For coefficient of superficial expansion β, and change in temperature ΔT, we have the relation:
\(\frac{\text { Change in area }(\Delta \mathrm{A})}{\text { Original area }(\mathrm{A})}=\beta \Delta \mathrm{T}\)

Change in diameter = d₂ – d₁ = 4.2544 – 4.24 = 0.0144 cm
Hence, the diameter increases by 1.44 x 10^-2 cm.

Question 9.
A brass wire 1.8m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of-39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 x 10^-5K^-1; Young’s modulus of brass = 0.91 x 10¹¹ Pa.
Solution:
Initial temperature, T₁ = 27°C
Length of the brass wire at T₁,l = 1.8 m
Final temperature, T₂ = -39 °C
Diameter of the wire, d = 2.0 mm = 2 x 10^-3 m
Let, tension developed in the wire = F
Coefficient of linear expansion of brass, a = 2.0 x 10^-5 K^-1

Young’s modulus of brass, Y = 0.91 x 10¹¹Pa
Young’s modulus is given by the relation ‘
Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\)
ΔL = \(\frac{F \times L}{A \times Y}\) ……………………………….. (i)
where, F = Tension developed in the wire
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation
ΔL = αL(T₂ -T₁) …………………………… (ii)

Equating equations (i) and (ii), we get

= -3.8 x 10² N
(The negative sign indicates that the tension is directed inward.) Hence, the tension developed in the wire is 3.8 x 10² N.

Question 10.
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 x 10^-5 K^-1, steel = 1.2 x 10^-5K^-1).
Solution:
Initial temperature, T₁ = 40°C
Final temperature, T₂ = 250 °C
Change in temperature, ΔT = T₂ – T₁ = 210°C
Length of the brass rod at T₁,l₁ = 50 cm
Diameter of the brass rod at T₂, d₁ = 3.0 mm
Length of the steel rod at T₂,l₂ = 50 cm
Diameter of the steel rod at T₂,d₂ = 3.0 mm
Coefficient of linear expansion of brass, α₁ = 2.0 x 10^-5 K^-1
Coefficient of linear expansion of steel, α₂ = 1.2 x 10^-5 K^-1

For the expansion in the brass rod, we have
\(\frac{\text { Change in length }\left(\Delta l_{1}\right)}{\text { Original length }\left(l_{1}\right)}=\alpha_{1} \Delta T\)
∴ Δl₁ = 50 x (2.1 x 10^-5)X210
= 0.2205cm
For the expansion in the steel rod, we have Change in length
\(\frac{\text { Change in length }\left(\Delta l_{2}\right)}{\text { Original length }\left(l_{2}\right)}=\alpha_{2} \Delta T\)
∴ Δl₂ =50 x (1.2 x 10^-5)x 210
= 0.126 cm

Total change in the lengths of brass and steel,
Δl = Δl₁ + Δl₂
=0.2205 + 0.126
= 0.346 cm
Total change in the length of the combined rod = 0.346 cm Since the rod expands freely from both ends, no thermal stress is developed at the junction.

Question 11.
The coefficient of volume expansion of glycerin is 49 x 10^-5K^-1.
What is the fractional change in its density for a 30°C rise in temperature?
Solution:
Coefficient of volume expansion of glycerin, α_v = 49 x 10^-5 K^-1
Rise in temperature, ΔT = 30°C
Fractional change in its volume = \(\frac{\Delta V}{V}\)

where, m = Mass of glycerine
PT₁ = Initial density at T₁
PT₂ = Final density at T₂
\(\frac{\rho_{T_{1}}-\rho_{T_{2}}}{\rho_{T_{2}}}=\alpha_{\mathrm{V}} \Delta T\)
Where, \(\frac{\rho_{T_{1}}-\rho_{T_{2}}}{\rho_{T_{2}}}\) = Fractional change in the density
∴ Fractional change in the density of glycerin
= 49 x 10^-5 x 30 = 1.47 x 10^-2

Question 12.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 Jg^-1K^-1.
Solution:
Power of the drilling machine, P =10 kW = 10 x 10³ W
Mass of the aluminum block, m = 8.0 kg = 8 x 10³ g
Time for which the machine is used, t = 2.5min = 2.5x 60 = 150 s
Specific heat of aluminium, C = 0.91 J g^-1K^-1
Rise in the temperature of the block after drilling = ΔT

Total energy of the drilling machine = Pt
= 10 x 10³ x 150 = 1.5 x 10⁶ J

It is given that only 50% of the power is useful.

Useful energy, ΔQ = \(\frac{50}{100} \times 1.5 \times 10^{6}\) = 7.5×10⁵J
But ΔQ = mCΔT
∴ ΔT = \(\frac{\Delta Q}{m C}=\frac{7.5 \times 10^{5}}{8 \times 10^{3} \times 0.91}\)
=103°C
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

Question 13.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 Jg^-1K^-1; heat of fusion of water = 335Jg^-1).
Solution:
Mass of the copper block, m = 2.5kg = 2500 g
Rise in the temperature of the copper block, Δθ = 500 °C
Specific heat of copper, C = 0.39 J g^-1 °C^-1
The heat of fusion of water, L = 335 J g^-1

The maximum heat the copper block can lose, Q = mCΔθ
= 2500×0.39×500 =487500J
Let m₁ g be the amount of ice that melts when the copper block is placed on the ice block.

The heat gained by the melted ice, Q = m₁L
∴ m₁ = \(\frac{Q}{L}=\frac{487500}{335}\) =1455.22g
Hence, the maximum amount of ice that can melt is 1.45 kg.

Question 14.
In an experiment on the specific heat of a metal, a 0.20kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40° C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Solution:
Mass of the metal, m = 0.20 kg = 200 g
Initial temperature of the metal, T₁ = 150°C
Final temperature of the metal, T₂ = 40°C
Calorimeter has water equivalent of mass, m’ = 0.025kg = 25g
Volume of water, V = 150 cm³

Mass (M) of water at temperature T = 27°C,
150×1 =150g
Fall in the temperature of the metal,
ΔT =T₁ -T₂ =150-40 =110°C
Specific heat of water, C_w = 4.186 J/g/K
Specific heat of metal = C
Heat lost by the metal, Q = mCΔT …………………………….. (i)

Rise in the temperature of the water and calorimeter system,
ΔT’ = 40 -27 = 13°C
Heat gained by the water and calorimeter system,
ΔQ’ = m₁C_wΔT’
= (M + m’)C_wΔT’ ……………………………… (ii)

Heat lost by the metal = Heat gained by the water and colourimeter system
mCΔT =(M + m’)C_wΔT’
200 xC x 110 = (150+25) x 4.186 x 13
∴ C = \(\frac{175 \times 4.186 \times 13}{110 \times 200} \) = 0.43 Jg^-1K^-1
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.

Gas	Molar specific heat (Cv) (cal mol-1K-1)
Hydrogen	4.87
Nitrogen	4.97
Oxygen	5.02
Nitric oxide	4.99
Carbon monoxide	5.01
Chlorine	6.17

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
Solution:
The gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion). Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion.

Hence, the molar specific heat of diatomic gases is more than that of monatomic gases. If only rotational mode of motion is considered, then the molar specific heat of a diatomic gas = \(\frac{5}{2} R=\frac{5}{2} \times 1.98\) = 4.95 cal mol^-1K^-1 With the exception of chlorine, all the observations in the given table agree with(\(\frac{5}{2} R\)) This is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.

Question 16.
Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium?
(b)What is the effect of decrease of pressure on the fusion and boiling point of CO₂?
(c) What are the critical temperature and pressure for CO₂? What is their significance?
(d) Is CO₂ solid, liquid or gas at
(a) -70°C under 1 atm,
(b) -60°C under 10 atm,
(c) 15°C under 56 atm?
Solution:
The P-T phase diagram for CO₂ is shown in the following figure.

(a) C is the triple point of the CO₂ phase diagram. This means that at the temperature and pressure corresponding to this point (i.e., at -56.6°C and 5.11 atm), the solid, liquid, and vaporous phases of CO₂ co-exist in equilibrium.
(b) The fusion and boiling point of CO₂ decrease with a decrease in pressure.
(c) The critical temperature and critical pressure of CO₂ are 31.1°C and 73 atm respectively.
Even if it is compressed to a pressure greater than 73 atm, CO₂ will not liquefy above the critical temperature.
(d) It can be concluded from the P-T phase diagram of CO₂ that:
(i) CO₂ is gaseous at -70 °C, under 1 atm pressure
(ii) CO₂ is solid at -60 °C, under 10 atm pressure
(iii) CO₂ is liquid at 15°C, under 56 atm pressure.

Question 17.
Answer the following questions based on the P-T phase diagram of CO₂:
(a) CO₂ at 1 atm pressure and temperature -60°C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when C0₂ at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure.
(d) CO₂ is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?
The P-T phase diagram for CO₂ is shown in the following figure:

Solution:
(a) No
Explanation:
At 1 atm pressure and at -60°C, CO₂ lies to the left of -56.6°C (triple point C). Hence, it lies in the region of vaporous and solid phases. Thus, CO₂ condenses into the solid-state directly, without going through the liquid state.
(b) It condenses to solid directly.
Explanation:
At 4 atm pressure, C0₂ lies below 5.11 atm (triple point C). Hence, it lies in the region of vaporous and solid phases. Thus, it condenses into the solid-state directly, without passing through the liquid state.

(c) The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.
Explanation:
When the temperature of a mass of solid CO₂ (at 10 atm pressure and at -65° C) is increased, it changes to the liquid phase and then to the vaporous phase. It forms a line parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.

(d) It departs from ideal gas behaviour as pressure increases.
Explanation:
If CO₂ is heated to 70 °C and compressed isothermally, then it will not exhibit any transition to the liquid state. This is because 70 °C is higher than the critical temperature of CO₂. It will remain in the vapour state but will depart from its ideal behaviour as pressure increases.

Question 18.
A child running a temperature of 101°F is given an antipyrin (i. e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g^-1.
Solution:
Initial temperature of the body of the child, T₁ = 101°F
Final temperature of the body of the child, T₂ = 98°F
Change in temperature, ΔT = \(\left[(101-98) \times \frac{5}{9}\right] \)°c
Time taken to reduce the temperature, t = 20 min
Mass of the child, m = 30 kg = 30 x 10³ g

Specific heat of the human body = Specific heat of water = C
= 1000 cal/kg/°C
Latent heat of evaporation of water, L = 580 cal g^-1
The heat lost by the child is given as:
ΔQ = mCΔT
= 30 x 1000 x (101-98)x \(\frac{5}{9}\) = 50000cal

Let m₁ be the mass of the water evaporated from the child’s body in 20 min.
Loss of heat through water is given by
ΔQ = m₁L
∴ m₁ = \(\frac{\Delta Q}{L}=\frac{50000}{580}\) = 86.2g
∴ Average rate of extra evaporation caused by the drug = \(\frac{m_{1}}{t}\)
= \(\frac{86.2}{20}\) = 4.3 g/mm.

Question 19.
A ‘thermal’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C, and coefficient of thermal conductivity of thermal is 0.01 J s^-1m^-1 K^-1.
[Heat of fusion of water = 335 x 10³ Jkg ^-1 ]
Solution:
Side of the given cubical icebox, s = 30 cm = 0.3 m
Thickness of the icebox, l = 5.0 cm = 0.05 m
Mass of ice kept in the icebox, m = 4 kg
Time gap, t=6h = 6x 60 x 60 s
Outside temperature, T = 45°C

Coefficient of thermal conductivity of thermacole,
K =0.01 Js^-1 m^-1 K^-1
Heat of fusion of water, L = 335 x 10³ J kg^-1
Let m be the total amount of ice that melts in 6 h.
The amount of heat lost by the food,
Q = \(\frac{K A(T-0) t}{l}\)

where, A = Surface area of the box = 6s² =6 x (0.3)² = 0.54 m²
Q = \(\frac{0.01 \times 0.54 \times(45) \times 6 \times 60 \times 60}{0.05}\) = 104976 J
But Q=m’L
∴ m’ = \(\frac{Q}{L}=\frac{104976}{335 \times 10^{3}}\) = 0.313 kg
Mass of ice left = 4-0.313 = 3.687kg .
Hence, the amount of ice remaining after 6 h is 3.687 kg.

Question 20.
A brass boiler has a base area of 0.15m² and a thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js^-1m^-1 K^-1; Heat of vaporisation of water = 2256 x 10³ Jkg^-1.
Solution:
Base area of the boiler, A = 0.15 m²
Thickness of the boiler,l = 1.0 cm = 0.01 m
Roiling rate of water, R = 6.0 kg/min
Mass, m = 6 kg
Time, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s^-1 m^-1 K^-1
Heat of vaporisation, L = 2256 x10³ J kg^-1
The amount of heat flowing into water through the brass base of the boiler is given by
Q = \(\frac{K A\left(T_{1}-T_{2}\right) t}{l}\) ………………………. (i)
where, T₁ = Temperature of the flame in contact with the boiler
T₂ = Boiling point of water = 100 °C
Heat required for boiling the water
Q = mL …………………………………. (ii)

Equating equations (i) and (ii), we get
∴ mL = \(\frac{K A\left(T_{1}-T_{2}\right) t}{l}\)
T₁ -T₂= \(\frac{m L l}{K A t}\)
= \(\frac{6 \times 2256 \times 10^{3} \times 0.01}{109 \times 0.15 \times 60}\) = 137.98°C
T₁ -T₂ = 137.98°C
∴ T₁=137.98+100 = 237.98°C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

Question 21.
Explain why:
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day.
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water
Solution:
(a) A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter.

(b) Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler. Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and one does not feel cool. •
Thus, a brass tumbler feels colder than a wooden tray on a chilly day.

(c) An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of a red hot iron piece kept in the open. Black body radiation equation is given by
E = σ(T⁴ -T₀⁴) where, E – Energy radiation
T = Temperature of optical pyrometer
T₀ = Temperature of open space
σ = Constant
Hence, an increase in the temperature of open space reduces the radiation energy.
When the same piece of iron is placed in a furnace, the radiation energy, E =σT⁴

(d) Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat will be .trapped. All the heat would be radiated back from earth’s surface.

(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g).

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20 C.
Solution:
According to Newton’s law of cooling, the rate of cooling cc difference in temperature.
Here, average of 80 °C and 50 °C, T = \(\frac{T_{1}+T_{2}}{2}=\frac{80+50}{2}\) = 65°C
Temperature of surroundings, T₀ = 20 °C .
∴Difference, ΔT = T – T₀ = 65 – 20 = 45°C
Under these conditions, the body cools 30 °C in time 5 minutes.
∴ 
or \(\frac{30}{5}\) = k x 45 ……………………………….. (i)
The average of 60°C and 30 °C is 45°C which is 25°C (45 – 20) above the room temperature and the body cools by 30 °C (60 – 30) in a time t (say).
∴\(\frac{30}{t}\) = k x 25 ……………………………….. (ii)

where k is same for this situation as for the original.
Dividing eq. (i) from eq. (ii), we get
\(\frac{30 / 5}{30 / t}=\frac{k \times 45}{k \times 25}\)
or \(\frac{t}{5}=\frac{9}{5}\)
or t = 9 min