tenth class physical...

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Tenth Class Physical Sciences2 Marks Questions

1. a) CH3 −− CH2 −− CH2 −− CH3b) CH3 −− CH −− CH3

CH3In both the substances number of carbon and hydrogen atoms are the same. What do you understand by seeing these structures?

A: Substances a and b have the same molecular formula but different structural formula.

So they are structural isomers.

2. Identify the functional group in the below compound and write the IUPAC name of this.

CH3 O

CH3 −− CH −− C −− CH3

A: The functional group present in

CH3 O

CH3 − CH − C − CH3 is Ketone.

The IUPAC name of the compound is 3 - Methyl - 2 - Butan - One.

The characteristic properties of an organic compound depend mainly on an atom or group of atomsin its molecule known as the functional group.

3. How does metals occur in nature? Give examples to any two types of minerals?

A: The earth's crust is the major source of metals.

Sea water also contains some soluble salts such as Sodium Chloride and Magnesium Chloride... etc.

Some metals like gold (Au), Silver (Ag) and Copper (Cu) are available in nature in free state.

The elements or compounds of the metals which occur in the earth crust are called minerals.

The minerals from which the metals are extracted without economical loss are called ores.

a) Bauxite - Al2O3. 2 H2O - is the ore (mineral) for the metal Al.

b) Cinabar - Hgs - is the ore (mineral) for Hg, these are the exampels.

4. Explain thermite process?

A: Thermite process involves the reaction of metal oxides with Aluminium.

When highly reactive metals such as sodium, calcium, aluminium... etc are used as reducing agents,they displace metals of lower reactivity from the compound.

These displacement reactions are highly exothermic.

The amount of heat evolved is so large that the metals produced are in molten state.

850°Ce.g.: TiCl4 + 2 Mg → Ti + 2 MgCl2

850°CTiCl4 + 4 Na → Ti + 4 NaCl

5. What is 22 carat gold? Why is it preferred for making jewellery?

A: 24 carat gold is pure gold. It is very soft.

Soft gold is not suitable for making jewellery.

Pure gold is mixed with either copper or silver to make it relatively hard.

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So 2 parts of silver or copper is mixed with pure gold to make it 22 carat gold.

This 22 carat gold is sufficiently hard to make jewellery.

6. What is the difference between roasting and calcination. Give one example for each.

A:

1 Mark Questions1. Name three metals which occur in free state in nature.

A: Gold (Au), Silver (Ag) and Copper (Cu) are available in nature in free state.

2. How are metals extracted from mineral ores?

A: The extraction of metals from their ores are done in three stages.

a) Concentration or dressing.

b) Extraction of crude metal.

c) Refining or purification of metal.

3. What are the various types of furnaces?

A: 1) Blast furnace 2) Reverberatory furnace and 3) Retort furnace are the three types of furnaces.

4. Why are we advised not to use animal fats for Cooking?

A: Animal fats are recently identified as the cause for heart disease and obesity. So it is advised not to usethem for cooking.

5. What is Ethanoic acid?

A: Ethanoic acid is commonly called as acetic acid.

6. Name the acid present in Vinegar.

A: 5 - 8 % solution of acetic acid in water is called Vinegar.

7. What happens when a small piece of sodium is dropped into ethanol?

A: When a small piece of sodium is dropped into ethanol, sodium ethoxide and hydrogen are formed.

2 C2H5OH + 2 Na → 2 C2H5ONa + H2Ethanol Sodium Ethoxide

Writer: C.V. Sarveswara Sarma

Roasting

1. Roasting is the process in which the oreis heated in the presence of air below itsmelting point.

2. In this process metal oxide is obtainedfrom sulphide ore.

e.g.: 2 ZnS + 3 O2 → 2 ZnO + 2 SO2

Calcination

1. Calcination is the process in which the

ore is heated in the absence of air.

2. In this process metal oxide is obtained

by decomposition of the ore.

e.g,: CaCO3 → CaO + CO2

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Tenth Physical Sciences1. An organic compound 'X' with molecular formula C2H6O undergoes oxidation with alkaline KMnO4 and

forms the compound Y; that has molecular formula C2H4O2.

a) Identify X and Y.

b) Write your observation regarding the product when the compound 'X' is made to react with compound

'Y', which is used as a preservative for pickles. (4 Marks)

A: a) Organic compound X is C2H6O: This is Ethanol.

Organic compound Y is C2H4O2: This is Ethanoic acid.

Ethyl alcohol undergoes oxidation to form Ethanal (Acetaldehyde) and finally Ethanoic acid (Acetic acid).

Alkaline KMnO4CH3CH2OH → CH3CHO → CH3COOHEthanol Heat Ethanol Ethanoic acid

(Acetaldehyde) (Acetic acid)

Here Ethanoic acid (Acetic acid) is used as a preservative for pickles.

b) When X (Ethanol) reacts with Y (Ethanoic acid) in the presence of concentrated H2SO4, ethyl acetate an ester

with fruity odour is formed.

Con.H2SO4

C2H5OH + CH3COOH → CH3COOC2H5 + H2O

Ethanol Acetic acid Ethyl acetate

Here CH3COOH is used as preservative for pickles.

2. What is the action of soap particles on greasy cloth? (4 Marks)

A: Soaps and detergents make oil and dirt present on the cloth come out into water, there by making the cloth clean.

Soap has one polar end and one non-polar end.

The polar end is hydrophilic in nature and this end is attracted towards water.

The non-polar end is hydrophobic, in nature and it is attracted towards grease or oil on the cloth, but not

attracted towards water.

When soap is dissolved in water, its hydrophobic ends attach themselves to dirt and remove it from the cloth.

The hydrophobic end of the soap molecules move towards the dirt or grease particles.

The hydrophobic ends attached to the dirt particle and try to pull out.

The molecules of soap surround the dirt particle at the centre of the cluster and form a spherical structure

called micelle.

These micelles remain suspended in water like particles in a collidal solution.

The various micelles present in water do not come together to form a precipitate as each micelle repells the

other because of the ion- ion- repulsion.

Thus, the dust particles remain trapped in micelles and are easily rinsed away with water.

Hence, soap micelles remove dirt by dissolving in water.

3. Draw the diagram of Frothflotation Process and name the parts. (4 Marks)

A:



Parts:

1) Compressed air

2) Sulphide ore particles

3) Water containing pine oil

4) Gangue

5) Froth bubbles carrying sulphide ore particles.

5

23

4

1

Frothflotation process for the concentration of Sulphideore

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4. We know metals are arranged basing on their reactivity. What is the name given to that arrangement? How

it helps in extraction of metals? (4 Marks)

A: Activity Series: Extraction of the metal from its ore depends on the reactivity of the metal.

To understand the order of reactivity of metals that are very familiar, we study their chemical reactions with

cold water, steam, dilute, strong acids and Cl2 and based on their vigorous activity order in these reactions,

we frame out activity series.

Arrangement of the metals in decreasing order of their reactivity is known as activity series.

Usefulness of activity series in extraction of metals:

Simple chemical reduction methods like heating with C, Co etc., to reduce the ores of these metals depends

mainly on the position of the metal in the activity series.

Extration of metals at the top of the activity series can be done by electrolysis.

Extraction of metals in the middle of the activity series can be extracted by

a) Sulphide ores are converted into oxides by roasting.

b) The metal oxides are then reduced to the corresponding metal by using suitable reducing agents like carbon.

Extraction of metals at the bottom of the acitivity series by heating alone as they are often found in free state.

Extraction of these metals can also be done by displacement from aqueous solutions.

5. How do you appreciate the role of Buckminister fullerence in medicine. (2 Marks)

A: Fullerences are under study for potential medicinal use such as specific antibiotics to target resistant bacteria and

even target certain cancer cells such as melonoma.

This is still under research studies.

I appreciate its usefulness in medicine which may become a good medicine for curing certain cancer in the

time to come.

6. What is the role of esters in everyday life? (2 Marks)

A: Esters are generally volatile liquids with sweet or pleasent smell or fruity smell.

Artificial perfumes are made with esters.

Esters are used as flavouring agents.

Esters are used in making artificial flavours and essences used in ice-creams, sweets, cool drinks... etc.,

Saponification (soap making) is the alkaline hydrolysis of esters.

Writer: C.V.Sarveswara Sarma



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Tenth Physical Sciences4 Marks Questions

1. A house has 3 tube lights, two fans and a television. Each tube light draws40 W. The fan draws 80 W and television draws 60 W. On the average all thetube lights are kept on for five hours, two fans for 12 hours and the television for five hours every day. Find the cost of electric energy used in 30days at the rate of Rs.3.00 per KWH.

A: Power consumed by 3 tube lights in a day = 40 W × 3 × 5 H = 600 WH

Power consumed by 2 fans in a day = 80 W × 2 × 12 H = 1920 WH

Power consumed by 1 television in a day = 60 W × 1 × 5 H = 300 WH

Total power consumed in the house per day = 600 WH + 1920 WH + 300WH

= 2820 Wh

Total power consumed in the house per month(30 days) = 2820 Wh × 30

= 84,600 Wh

84,600Total power consumed in KWH =

1000

= 84.6 KWH

Rate per KWH = Rs.3

Cost of 84.6 KWH electrical energy consumed = Rs.84.6 × 3 = Rs.253.80

2. Draw a neat diagram of an A.C. generator and name the parts.

A:



Parts:

1. Direction of rotation.

2. AB rectangular coil of wire.

3. South pole of a curve shaped permanent magnet.

4. North pole of a curve shaped permanent magnet.

5. Slip rings.

6. Brushes.

5

6

4

3

21

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2 Marks Questions1. How do you appreciate Faraday's Laws, which is the consequence of

conservation of energy.

A: 'Energy can neither be created one destroyed. But it can be converted from the

form to other'. - This is law of conservation of energy.

'When there is a change of magnetic flux linked with a closed coil, a current

is generated in the coil. This induced E.M.F. is equal to the rate of change of

magnetic flux passing through 'it'. - This is Faraday's Law.

To move the magnet in the coil, some work is to be done. This work produces

energy which is converted into electrical energy in the coil.

This is how conservation of energy takes place in electromagnetic induction.

2. State Kirchhoff's Laws.

A: Kirchhoff's Junction Law: At any junction point in a circuit when the current

can divide, the sum of the currents into the junction must equal to the sum of the

currents leaving the junction.

The Loop Law: The algebraic sum of the increases and decreases in potential

difference across various components of a closed circuit loop must be zero.

3. How could we use the principle of electromagnetic induction in the case ofusing ATM Card when its magnetic strip is swiped through a scanner.

A: If the card is moved through a card reader, then a change in magnetic flux is

produced in one direction, which induces e.m.f. or potential.

The current received by the pickup coil goes through signal amplification and

translated into binary code, so that it is read by the computer.

1 Mark Questions

1. The value of magnetic field induction which is uniform is 2T. What is theflux passing through a surface area of 1.5 m2 perpendicular to the field.

A: Magnetic field induction B = 2T, area A = 1.5 m2

Magnetic flux φ = BA = 2T × 1.5 m2 = 3 wb.

2. What is value of 1 KWH in Joules?

A: 1 KWH = 1000 W × 1 hour = 1000 J/S × 60 × 60 Sec.

∴ 1 KWH = 36 × 10 5 Joules.



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3. Silver is a better conductor of electricity than copper. Why do we use copperwire for conduction of electricity?

A: Resistivity of silver = 1.59 × 10−8 Ω m

Resistivity of Copper = 1.68 × 10−8 Ω m

As resistivity of silver is less than copper, its conductivity is more than

copper.

As silver is costly compared to copper, copper is widely used for conduction

of electric current.

4. A wire of length 1 m and radius 0.1 mm has a resistance of 100 ΩΩ . Find theresistivity of the material.

A: Resistance: R = 100 Ω , Length l = 1 m

= 1000 mm

radius of the wire r = 0.1 m.m.

area of cross section A = πr2 = 227

× (0.1) (0.1)

= 3.14 × 0.01 = 0.0314 (mm)2

RA 100 × (.0314)2 Resistivity ρ = =

l 1000

= 0.00314 Ω m




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Tenth Physical Sciences1. Explain the formation of the following molecules using valance bond theory.

a) N2 molecule b) O2 molecule. (4 Marks)

A: a) N2 molecule:

Nitrogen (Z = 7) has electronic configuration: 1s2 2s2 2px1 2p1

y 2pz1

'px' orbital of one 'N' atom overlaps the 'px' orbital of other 'N' atom giving

σ px − px bond along the inter nuclear axis.

The py and pz orbital of one 'N' atom overlap the py and pz orbital of other 'N'

atom laterally, respectively perpendicular to inter-nuclear axis giving

Π py − py and Π pz − pz bonds.

Therefore, N2 molecule has a triple bond between two nitrogen atoms.

b) O2 Molecule:

Oxygen (O) (Z = 8) has electronic configuration: 1s2 2s2 2px1 2p1

y 2pz1

If the 'py' orbital of one 'O' atom overlaps the 'py' orbital of other 'O' atom

along the inter nuclear axis, a sigma py − py (σ py − py) bond is formed.

pz orbital of one 'O' atom overlaps the pz orbital of other 'O' atom laterally,

perpendicular to the inter-nuclear axis giving a Π pz − pz bond.

Thus oxygen molecule has a double bond between two oxygen atoms.



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2. A chemical compound has the following Lewis notation:

a) How many valence electrons does element Y have?

b) What is the valency of the element Y?

c) How many covalent bonds are there in the molecule?

d) Suggest a name for the elements X and Y. (2 Marks)

A: a) Valence electrons of Y are 6.

b) Valency of the element Y is 2.

c) Two covalent bonds are there in the molecule.

d) X is hydrogen and Y is oxygen. The molecule formed is H2O.

3. Why does Sodium always form a cation where as fluorine atom always forman anion? (2 Marks)

A: Sodium atom has low ionisation energy and low electron affinity. It has high

atomic size and low electro negativity.

So it easily looses an electron and hence sodium element always form as a

cation.

Fluorine atom has high ionisation energy and high electron affinity. It has

small atomic size and high electro negativity.

So it easily gains an electron and hence fluorine element always form as an

anion.

4. How does metallic character change when we move

a) Down a group b) Across a period. (2 Marks)

A: a) When we move down a group of the Periodic table the metallic character of

the elements increases.

When we move across a period of the Periodic table the metallic character of

the elements decreases.

5. All alkali metals are solids but hydrogen is a gas with diatomic molecules.How do you justify the inclusion of hydrogen in first group with alkali metals? (2 Marks)

A: The electronic configuration of Hydrogen (Z = 1) is 1s1.

On the basis of electronic configuration it can be placed in first group with

alkali metals.



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It can be placed in 17th group along with halogens as it is shoot by one

electron of the inert gas (He) electronic configuration.

As this is a special case it can be placed in either of the groups or in both

groups.

6. The densities of Calcium (Ca) and Barium (Ba) are 1.55 and 3.51 gm. cm−−3

respectively. Based on Dobereiner's law of triads, can you give the approximate density of Strontium (Sr)? (1 Mark)

A: Density of Ca = 1.55 gm. cm−3

Density of Ba = 3.51 gm. cm−3

3.51 + 1.55Average density =

2

= 2.53 gm.cm−3

But the density of Strontium is = 2.64

The average density of Ca and Ba is approximately equal to the density of

Strontium (Sr).




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1. Without knowing the electronic configuration of atoms of elementsMendeleef still could arrange the elements nearly close to the arrangementsin Modern Periodic table. How can you appreciate this?

A: Mendeleef arranged the elements known at that time in the increasing order of

their atomic weights.

Based on the arrangement of elements in the table he predicted that some

elements were missing and left blank spaces at appropriate places in the table.

Mendeleef believed that some new elements would be discovered definitely.

He predicted the properties of these new additional elements in advance

purely depending on his table.

His predicted properties were almost the same as observed properties of those

elements after their discovery.

The placement of elements in Mendeleef's periodic table helped in correcting

the atomic masses of some elements like, beryllium, indium and gold.

Thus Mendeleef's vision must be appreciated. If we compare the long form

periodic table with Mendeleef's table, we find many elements with their

places unchanged.

Even today his table of elements is followed with few modifications.

There are eight vertical columns in his periodic table called as groups. The

horizontal rows in his periodic table are called periods. There are seven

periods.

This table provided a scientific base for the study of elements in chemistry

which was at that time criticised.

In his honour the 101th element was named as 'Mendelevium'.

2. An element X belongs to 2nd period and 13th group of periodic table. Thenstate a) Number of valence electrons in the element b) The valency of the element c) Is it a metal or non-metal? d) Write the name of the element andits electronic configuration. e) If a halogen Y reacts with it, write the formula of the compound formed.



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A: a) As the element X belongs to 13th group, it will have 3 valence electrons.

b) As the atom of this element looses 3 electrons in a chemical reaction, the

valency of the element is 3.

c) In the periodic tables period the elements which are towards the left are

generally metals. So this element is a metal.

d) The element X is Boron and its electronic configuration is: 1s2 2s2 2p1.

e) We know the valency of halogens is 1. So when Boron reacts with halogens

the formula of the compound is BY3.

3. Collect the information about properties and uses of compounds having lowmelting and boiling points.

A: Covalent compounds have low melting and boiling points.

Properties of covalent compounds: The forces of attractions among covalent

molecules are weak. So the covalent compounds are gases or liquids at room

temperature.

As 'Like dissolve in like' these covalent compounds are soluble in non polar

solvents because of non-polar nature of solvent molecules.

It means covalent compounds are insoluble in water and soluble in organic

solvents.

Covalent compounds have low melting and boiling points.

Covalent compounds do not conduct electricity.

Uses of Covalent Compounds: Our body can produce covalent compounds.

Water is an example for covalent compound. It is used in many ways by us.

Our eatables, sugars, food substances, tea, coffee are all covalent substances.

The air we breathe contains covalent molecules of oxygen, nitrogen.

Many things on earth are covalent compounds except some organic salts.




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Tenth Physical SciencesHuman Eye

1. Derive an expression for the refractive index of the material of a prism.(4 Marks)

A: Observe the ray diagram in the figure (1).

From the ∆OMN we get

d = i1 − r1 + i2 − r2

d = (i1 + i2) − (r1 + r2) .........(1)

From the ∆PMN we have

A = r1 + r2 ......(2)

∴ d = (i1 + i2) − A

∴ A + d = i1 + i2 .........(3)

From Snells law we know

n1 sini = n2 sin r

Using Snells law at M

n1= 1, i = i1

n2 = n, r = r1

∴ sin i1= n sin r1 .........(4)

Similarly at N, n1= n, i = r2, r = i2 and n2 = 1

∴ n sin r2 = sin i2 .........(5)

We know at the angle of minimum

deviation (D)

i1 = i2 and MN is parallel to the base QR

(see fig 2)

Equation (3) becomes:

A + D = i1 + i1 = 2i1

A + D or i1 =

2



P

Q R

Ad

M

i1i2

Nr1 r2

O

r2r1

i1 i1M

A

P

RQ

N

DO

fig - 1

fig - 2

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When i1 = i2 then r1 = r2, So

Equation (2) becomes: A = r1 + r1 = 2 r1

Aor r1 = 2

Substituting the values of i and r1 in Equation (4) we get..

A + D Asin () = n. sin ()2 2

∴ Refractive index of the prism

A + D sin ()2

n = A

sin 2

2. How do you appreciate the working of ciliary muscle in the eye?(4 Marks)

A: The ciliary muscle to which eye lens is attached helps the eye lens to change its

focal length by changing the radii of curvature of the eye lens.

When the eye is focussed on a distant

object, the ciliary muscles are relaxed

so that the focal length of the eye lens

has its maximum value which is equal

to its distance from the retina.

The parallel rays coming into the eye

are then focussed on to the retina and

we see the object clearly.

When the eye is focussed on a closer

object, the ciliary muscles are strained and focal length of eye-lens decreases.

The ciliary muscles adjust the focal length in such a way that the image is

formed on retina and we see the object clearly.

This process of adjusting focal length is called "accommodation".

This is a wonderful arrangement the nature provided in the eye.

If this is not provided we would not have lead the present comfortable life with

regard to our vision.



Optic nerve

Lens

Ciliary muscles

Retina Iris

PupilAqueoushumour

Cornea

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3. Explain briefly the reason for the blue of the sky. (2 Marks)

A: When we look at the sky in a direction perpendicular to the direction of the Sun

rays the sky appears blue.

If our angle of view is changed, the intensity of blue colour also changes.

Molecules of N2 and O2 are the causes for the blue colour of the sky.

The sizes of these molecules are comparable to the wavelength of blue colour

light.

So these molecules act as scattering centres for the blue colour of light. So the

sky appears blue.

4. How can you correct the defect 'hypermetropia'? (2 Marks)

A: Eye lens can form a clear-image on the retina when any object is placed beyond

near point.

To correct the defect of hypermetropia, we need to use a lens which forms an

image of an object beyond near point, when the object is better near point and

least distance of distinct vision.

This is possible only when a double convex lens is used.

5. What is power of lens? Write the unit for power. (1 Mark)

A: The degree of convergence or divergence of light rays that can be achieved by a

lens is expressed in terms of its power.

The reciprocal of focal length is called power of lens.

1 100 Power of lens P = =

f (in m) f (in cm)

This unit of power is dioptre. It is denoted by letter 'D'.

6. Explain the basic process in scattering of light. (1 Mark)

A: Scattering of light is a complex phenomenon.

Atoms or molecules which are exposed to light absorb light energy and emit

some part of the light energy in different directions.

This is the basic process happens in scattering of light.




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Tenth Physical SciencesQuestion And Answers

1. Explain the formation of a mirage? (4 Marks)

A: 1) During a hot summer day, air just above the road is very hot and the air at

higher altitudes is cool.

2) It means that the temperature decreases with height. As a result the density of

air increases with height.

3) We know that the refractive index of air increases with density. Thus the

refractive index of air increases with height.

4) So the cooler air at the top has greater refractive index than the hotter air just

above the road.

5) Light travells faster through the thinner hot air than through the denser cool air

above it.

6) When light from a tree passes through a medium just above the road, whose

refractive index decreases towards the ground, it suffers, refraction and takes

a curved path because of total internal reflection.

7) This appears to the observer as if the ray is reflected from the ground. Hence

we feel the illusion of water being present on the road.

8) The inverted virtual image of the tree is seen by the observer. This is a mirage.

2. If the focal length of a symmetric convergent lens is equal to the radius of curvature of its surface then find its refractive index. (4 Marks)

A: 1) Refractive index of glass = n (say)



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Given focal length of the lens = f = radius of curvature R

2) As the lens is a symmetric convergent lens:

R1 = R, R2 = − R

1 1 1 3) We know: = (n − 1) ( − )f R1 R2

(Lens maker's formula)

4) Substituting the values:

1 1 1 2 = (n − 1) ( + ) = (n − 1) R R R R

1 2 5) ∴ = (n − 1) (or) (n − 1) 2 = 1

R R

1 (or) n − 1 =

2

1 36) ∴ n = 1 + = = 1.5

2 2

7) We know for glass the refractive index n = 1.5

3. A transparent sphere of radius R and refractive index n is kept in air. Atwhat distance from the surface of the sphere should a point object be placedon the principal axis so as to form a real image at the same distance from thesecond surface of the sphere. (2 Marks)

A: 1) As the object and image should be at the same distance from the sphere, the

rays in the sphere should travel parallel to the principal axis as shown in the

figure.

2) Consider the first surface A of the sphere.

a) Object distance u = − x and v = ∞ as the rays after refraction at the first surface travel parallel to the principal axis.



>> >

>>

RO x x I. . .

AC

B

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b) Now n1 = 1 (air) and n2 = n (glass)

n2 n1 (n2 − n1)3) We know the formula: − =

v u Rn 1 (n − 1)

4) Substituting the values: − = ∞ (−x) R

1 (n − 1) R∴ = (or) x =

x R (n − 1)

5) ∴ Object distance from the first surface of the sphere is x = R

(n−1)

.

4. Two converging lenses are to be placed in the path of a parallel rays so thatthe rays remain parallel after passing through both lenses. How should thelenses be arranged? Explain with a neat diagram. (2 Marks)

A: 1) In the figure L1 and L2 are the two converging lenses. P1 and P2 are their optic

centres respectively.

2) F is the focus of the both the lenses.

3) For lens L1 the focal length is P1 F and for the lens L2 the focal length is P2 F.

4) To get the incident and emergent beam of light rays parallel, it is evident that

the lenses should be arranged at a distance equal to P1F + P2F as shown in the

figure.

5. Write the lens maker's formula and explain the terms in it. (2 Marks)

1 1 1A: 1) Lens maker's formula is: = (n − 1) ( − )f R1 R2

2) Here f is the focal length of the lens. n is the refractive index of the material

of the lens and R1 and R2 are the radii of curvature of the two spherical

surfaces of the lens.




>

>

>

> >

> >

>>

>

>

>

>

>

>

>

L1 L2

P2P1 F

P1F = P2F = f (focal length)

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Q: You know that Archimedes burned ships using mirrors.

a) What type of mirrors were used?

b) Which energy was put to use for burning?

c) Mention the principle in this process.

d) Draw a ray diagram showing this principle.

A: a) Archimedes used concave mirrors for burning the ships.

b) Solar energy was put to use for burning.

c) Archimedes focused the image of the Sun on the ships by adjusting the concave mirrors in such a way that their focus lie on the ships.

d) Ray diagram is shown below.

Q: What are your observations about the effects of Oxidation reactions in dailylife. Can you suggest any preventive measures for them?

A: When some metals are exposed to moisture, acids ... etc., they tarnish due to oxidation. This is called Corrosion.

Tarnishing of Silver ware, green coating on Copper are some examples forthis.

Corrosion causes damage to car bodies, bridges, iron railings, ships and to allother objects that are made of metals.

Corrosion can be prevented or at least minimised by shielding the metal surface from Oxygen and moisture.

It can be prevented by painting, oiling, greasing, galvanizing, chrome platingor making alloys.



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When fats and oils are oxidized they become rancid. Their smell and taste

change. Rancidity is an oxidation reaction.

The spoilage of these food items can be prevented by adding preservatives

like Vitamin C and Vitamin E.

Keeping food in air tight containers helps to slow down oxidation process.

2 Marks Questions

Q: You are given a cylindrical vessel with a frictionless movable piston. Wateris taken in the vessel and it is heated by putting weights on the piston.

a) How can you know when the boiling process of water started?

b) If the weights on the piston are increased what happens to the boilingpoint of water?

A: a) The weights fall down when the boiling process starts.

b) There is no change in the boiling point of water, when

the weights on the piston are increased.

This is because, as the piston is a frictionless movable

one, any increase in pressure in water pushes the

piston upwards.

Q: Write the balanced chemical equations for the following reaction.

Aluminium + Copper chloride →→ Aluminium chloride + Copper

A: Step - 1: Unbalanced equation using correct chemical formula.

Al + CuCl2 → AlCl3 + Cu

Step - 2: Observe the number of atoms on both sides.



Element No. of atoms No. of atoms

on L.H.S. on R.H.S.

Aluminium (Al) 1 (in Al) 1 (in AlCl3)

Copper (Cu) 1 (in CuCl2) 1 (in Cu)

Chlorine (Cl) 2 (in CuCl2) 3 (in AlCl3)

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Equate the No. of Cl atoms on both sides

Al + 3 CuCl2 → 2 AlCl3 + Cu

Equate the No. of Al. atoms on both sides

2 Al + 3 CuCl2 → 2 AlCl3 + Cu

Equate the No. of Cu atoms on both sides

2 Al + 3 CuCl2 → 2 AlCl3 + 3 Cu

Step - 3: The coefficients are to be reduced to their smallest whole number

values.

2 Al + 3 CuCl2 → 2 AlCl3 + 3 Cu

Step - 4: Count the number and kinds of atoms on both sides of the equation to

make sure that they are the same. Hence the balanced equation is

2 Al + 3 CuCl2 → 2 AlCl3 + 3 Cu

1 Mark Questions

Q: How much energy is released or absorbed when 1 gm. of water at 0°°Cfreezes to ice at 0°°C?

A: Given: m = 1 gm.

Latent heat of fusion of ice L = 80 Cal /gm (We know)

Formula: Energy released Q = mL Cal.

∴ Substituting the values: Q = 1 × 80 = 80 Cal.

Q: Why is slaked lime used for white washing walls?

A: A solution of slaked lime produced is used for white washing of walls.

Slaked lime (Calcium hydroxide Ca(OH)2) reacts slowly with the Carbon

dioxide in air to form a thin layer of Calcium Carbonate on the walls.

After two to three days of white washing, Calcium Carbonate formed gives a

shiny finish to the walls.

Q: Are the process of evaporation and boiling the same?

A: Boiling of a liquid differs essentially from evaporation.

Evaporation takes place at any temperature while boiling occurs at a definite

temperature called the boiling point.



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Q: Mention the uses of Propane, giving its formula.

A: Propane is a colourless and odourless gas often used as a heating and cooking

fuel.

Its chemical formula is C3H8.

Writer: C.V. Sarveswara Sarma



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Tenth Physical Sciences1. The graph given shows that an ice of 1 kg at −−5°°C is heated till it vaporises

completely.

a) What is the state of ice at C?

b) What does the part DE in the graph represent? Explain.

c) What is the value of Q (heat energy) at E shown in the graph? ( 4 Marks)

A: a) At 'C' ice is at its solid state (Partly)

b) The part DE in the graph indicates the state of vapourization of water. In this

position the heat supplied to water is made use of to change the state of water

to vapour at constant temperature.

c) 1) At E all the water is converted into vapour. The amount of heat taken by

1 kg of water to change its state to vapour is given by Q = mL. Where m = 1

kg = 103 gm and L = 540 Cal / gm.

2) So Q = m × L = 103 × 540 = 54 × 104 Cal.

2. Sri Lakshmi inserted two nails in a beaker which are connected to two wireswith A.C. small bulb. She has taken different solutions like Sodium hydroxide, Potassium hydroxide, Alcohol, Glucose, Hydrochloric acid andSulphuric acid in the beaker one after another. Answer the following questions:

a) Which solutions can make the bulb glow? Why?

b) Categorize these as acids, bases and neutral?

c) Mention what is common in acids and bases? (4 Marks)



100°C100°C

D FE

QCB

A-5°C

0°C

Tem

pera

ture

→

→ Heat

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A: a) 1) Hydrochloric acid, Sulphuric acid, Sodium hydroxide and Potassium

hydroxide solutions can make the bulb glow.

2) These solutions have ions and the moment of these ions in solution helps

for flow of electricity through the solution.

b) 1) Hydrochloric acid, Sulphuric acid are acids.

2) Sodium hydroxide, Potassium hydroxide are bases.

3) Glucose and alcohol are neutral substances.

c) 1) Acids produce H+ ions in solutions

2) Bases produce OH− ions in solutions.

3.

By observing the above pH Scale, answer the following:

a) Which of the body fluid have basic in nature.

b) Is lemon juice a strong acid or weak acid?

c) Which of the above liquid have strong basic character?

d) What is the pH of distilled water. (2 Marks)

A: a) The body fluid blood is basic in nature as its pH is nearly 7.4.

b) The lemon juice is a weak acid as its pH is 2.2.

c) NaOH (Sodium hydroxide) has a strong basic character as its pH value is

nearly 13.6.

d) Distilled water being neutral, its pH is 7.

4. An object is placed at a distance of 10 cm from a convex mirror of focallength 15 cm. Find the position and nature of the image. (2 Marks)

A: 1) Given: Object distance u = −10 cm

Focal length of the convex mirror f = 15 cm

Image distance v = ?



Gas

tric

juic

e

Lem

on ju

ice

Rai

n W

ater

Blo

od

Bak

ing

soda

Milk

of

Mag

nesi

a

Mg(

OH

) 2

NaO

H

Solu

tion

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2) Formula:1f

= 1u +

1v (OR)

1v =

1f

− 1u

3) Substituting the values: 1v =

115

+ 110

2 + 3 5 = = =

1630 30

∴ Image distance v = 6 cm

4) The image formed is erect and virtual.

Magnification m = HiHo

= v

u = 610

= 0.6

∴ The size of the image Hi = 0.6 × Ho = 0.6 times the size of the object.

5. How do you appreciate the role of Toothpaste in preventing the tooth decay.

(1 Mark)

A: 1) Bacteria present in the mouth produce acids by degradation of sugar and food

particles remaining in the mouth.

2) Using tooth pastes, which are generally basic neutralize the excess acid and

prevent tooth decay.

6. Why does water kept in a clay pot remain cool? (1 Mark)

A: 1) A clay pot is porus.

2) Water in the pot comes out of the pot through pores and stay on the outer

surface of the pot.

3) The necessary heat to evaporate for these water particles is taken from the

water inside the pot.

4) The water on the outer surface of the pot thus evaporates cooling the water

inside the pot.

7. Name the mirror used in car headlights? How is it arranged in the car

headlight? (1 Mark)

A: 1) A concave mirror is used in car headlights.

2) When a source of light is placed at the principal focus of a concave mirror, it

produces a strong parallel beam of light.

3) Thus concave mirror is used as a reflector in car (automobile) headlights.



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8. Dry hydrogen chloride gas does not turn blue litmus to red where as hydrochloric acid does. Why? (1 Mark)

A: 1) Dry hydrogen gas does not dissociate into ions in the absence of water. So it

does not turn blue litmus to red.

2) Hydrochloric acid dissociates into H+ and Cl− ions in the presence of water.

3) So Hydrochloric acid turns blue litmus to red.

HCl + H2O → H3O+ + Cl−




tenth class physical...

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