Chapter 10, Space Missions, SSC, SCIENCE PART I, NEW SYLLABUS, FOR BOARD EXAM 2021,

Question 1

Fill in the blanks and explain the statements with reasoning:

a. If the height of the orbit of a satellite from the earth surface is increased, the tangential velocity of the satellite will ……

b. The initial velocity (during launching) of the Managalyaan, must be greater than …………..of the earth.

ANSWER

a. If the height of the orbit of a satellite from the earth surface is increased, the tangential velocity of the satellite will decrease.

Explanation: This can be easily inferred from the formula of critical or tangential velocity.

b. The initial velocity (during launching) of the Managalyaan, must be greater than escape velocity of the earth.

Question 2

State with reasons whether the following sentences are true or false

a. If a spacecraft has to be sent away from the influence of earth’s gravitational field, its velocity must be less than the escape velocity.

b. The escape velocity on the moon is less than that on the earth.

c. A satellite needs a specific velocity to revolve in a specific orbit.

d. If the height of the orbit of a satellite increases, its velocity must also increase.

ANSWER

a. The statement is false.

This is because the minimum velocity with which the spacecraft must be projected so that it escapes the Earth’s gravitational pull is known as escape velocity.

So, the initial velocity of the spacecraft must be greater than or equal to escape velocity of Earth.

b. Escape velocity on a planet is given as:

The statement is True.

Since, acceleration due to gravity on Moon is one-sixth of that on Earth, thus the escape velocity on the Moon is less than that on the Earth.

Hence, the statement is true.

c. The specific velocity: with which the satellite revolves around a planet is known as critical velocity. It is given as

So, it can be observed that the critical velocity of a satellite changes depending on the height of the orbit from the surface of a planet.

Thus, to revolve in a specific orbit, a satellite would require specific velocity.

Hence, the statement is true.

d. The specific velocity with which the satellite revolves around a planet is known as critical velocity. It is given as

So, we see that as the height (h) of the orbit of a satellite increases, it velocity must decrease.

Hence, the statement is false.

Question 3

Answer the following questions:

a. What is meant by an artificial satellite? How are the satellites classified based on their functions?

b. What is meant by the orbit of a satellite? On what basis and how are the orbits of artificial satellites classified?

d. What is meant by satellite launch vehicles? Explain a satellite launch vehicle developed by ISRO with the help of a schematic diagram.

e. Why it is beneficial to use satellite launch vehicles made of more than one stage?

ANSWER:

a. A man made object revolving around a planet in a fixed orbit is known as artificial satellites.

Based on their functions, satellites are classified as following:

Type of satellite	Function of satellite
Weather satellite	Study and prediction of weather
Communication satellite	Establish communication betweendifferent location in the world through use of specific waves
Broadcast satellite	Telecasting of television programs
Navigational satellite	Fix the location of any place on the Earth’s surface in terms of its very precise latitude and longitude
Military Satellite	Collect information for security aspects
Earth Observation Satellite	Study of forests, deserts, oceans, polar ice on the earth’s surface, exploration and management of natural resources, observation and guidance in case of natural calamities like flood and earthquake

b. An orbit is a specific path (elliptical or circular) or trajectory around a planet in which a satellite revolves. ​

Depending on the height of the satellite’s orbit above the Earth’s surface, the satellite orbits are classified as below:

High Earth Orbits (Height from the earth’s surface > 35780 km): If the height of the satellite’s orbit above the earth’s surface is greater than or equal to 35780 km, the orbit is called High earth Orbit

Medium Earth Orbit (Height above the earth’s surface 2000 km to 35780 km): If the height of the satellite orbit above the earth’s surface is in between 2000 km and 35780 km, the orbits are called medium earth orbits.

Low Earth orbit (Height above the earth’s surface: 180 km to 2000 km): If the height of the satellite orbit above the earth’s surface is in between 180 km and 2000 km, the orbits are called Low earth Orbits.

c. The geostationary satellites orbit above the equator.

Thus, these are not useful for studies of polar regions.

d. Satellite launch vehicles are used to place the satellites in their specific orbits.

The structure of polar satellite launch vehicle (PSLV) developed by ISRO is shown below.

It is a vehicle with 4 stages using solid and liquid fuels.

The weight of the vehicle decreases after each stage because of consumption of fuel at that stage and detachment of that stage (i.e. the empty tank) from its body.

Thus, the vehicle moves with higher speed after every stage.

e. The major portion of weight in satellite launch vehicles is contributed by fuel.

Thus, vehicles have to carry a large weight of fuel during their course of journey.

To overcome this problem of launch vehicles carrying heavy load during its entire journey, it is provided with more than one stage.

Because to this, the weight of the vehicle can be reduced step by step, after its launching.

For example, consider a launch vehicle having two stages.

For launching the vehicle, the fuel and engine in the first stage are used.

This imparts a specific velocity to the vehicle and takes it to a certain height.

Once the fuel in this first stage is exhausted, the empty fuel tank and the engine are detached from the main body of the vehicle and fall either into a sea or on an unpopulated land.

As the fuel in the first stage is exhausted, the fuel in the second stage is ignited.

However, the vehicle now contains only one (i.e. the second) stage.

The vehicle can move with higher speed as the weight has reduced. 

Question 4

Complete the following table.

ANSWER

Question 5

Solve the following problems.

a. If the mass of a planet is eight times the mass of the earth and its radius is twice the radius of the earth, what will be the escape velocity for that planet?

Solution

b. How much time a satellite in an orbit at height 35780 km above earth’s surface would take, if the mass of the earth would have been four times its original mass?

Solution

c. If the height of a satellite completing one revolution around the earth in T seconds is h1 meter, then what would be the height of a satellite taking seconds for one revolution?

Solution

ASTER-CLASSES

Maharashtra State Board of Secondary & Higher Secondary Education (MSBSHSE) change 12th syllabus in this year 2020. Maharashtra Board also published 12th new syllabus booksin pdf format. Class 12 or HSC new Syllabus Books pdf Maharashtra Board online download 2020.

In this page we are sharing Maharashtra state board books for 12th 2020 in pdf. Any one can download easily 12th new syllabus 2020-21 from our

Maharashtra State Board conduct 12th exam every year in the month of Feb or March, about more than 10 lakhs of Students enroll in the 12 Class exam, Maharashtra Board also conduct the same Examination in the Month of June for the students who Failed in main examination.

Maharashtra State Board offered courses Arts, Science, Commerce, etc, Students can download 12th syllabus of Maharashtra board 2020 from our website in pdf format.

12th new syllabus 2020 is very important for 12th standard students for self-study. Students can download online free pdf 12th Maharashtra board books..

12th New Syllabus 2020-21 Download

Maharashtra Board HSC Syllabus	Download Below Link
12th std new syllabus	English Yuvakbharati
12th std new syllabus	Marathi Yuvakbharati
12th std new syllabus	Hindi Yuvakbharati
12th std new syllabus	Kannada Yuvakbharati
12th std new syllabus	Urdu Yuvakbharati
12th std new syllabus	Telgu
12th std new syllabus	Sindhi
12th std new syllabus	Sindhi Yuvakbharati
12th std new syllabus	Shiksha Shastra
12th std new syllabus	Secretarial Practice
12th std new syllabus	Sanskrit
12th std new syllabus	Sahakar
12th std new syllabus	Rajyashastra
12th std new syllabus	pustpalan and lekhakarm
12th std new syllabus	Political Science
12th std new syllabus	Physics
12th std new syllabus	Paryavaran shikshan va jal suraksha
12th std new syllabus	Pali Pakaso
12th std new syllabus	Organisation of Commerce and Management
12th std new syllabus	Mathematics & Statistics Commerce Part 1
12th std new syllabus	Mathematics & Statistics Commerce Part 2
12th std new syllabus	Mathematics & Statistics Art & Science Part 1
12th std new syllabus	Mathematics & Statistics Art & Science Part 2
12th std new syllabus	Maharashtri Prakrit
12th std new syllabus	Logic
12th std new syllabus	Information Technology
12th std new syllabus	History
12th std new syllabus	Hindi yuvakbharati
12th std new syllabus	Hidayatul Arabiya
12th std new syllabus	Gulariya Farsi
12th std new syllabus	Gujarati
12th std new syllabus	Environment Education and Water Security
12th std new syllabus	Education
12th std new syllabus	Economics
12th std new syllabus	Co-Operation
12th std new syllabus	Chitnisachi Karypaddhati
12th std new syllabus	Child Development English
12th std new syllabus	Chemistry
12th std new syllabus	Book Keeping & Accountancy
12th std new syllabus	Biology
12th std new syllabus	Bengali
12th std new syllabus	Arthashastra
12th std new syllabus	Ardhamagadhi Prakrit

Maharashtra Board Class 12 Marathi Medium Books

Asterclasses

Maharashtra Board Class 12 Arthshastra Book

Maharashtra Board Class 12 Chitnisachi Karyapadhati Book

Maharashtra Board Class 12 itihas Book

Maharashtra Board Class 12 Paryawaran Shikshan Book

Maharashtra Board Class 12 Pustapalan Aani Lekhakarma Book

Maharashtra Board Class 12 Rajyashastra Book

Maharashtra Board Class 12 Sahakar Book

Asterclasses

Maharashtra Board Class 12 Shikshan Shastra Book

Maharashtra Board Class 12 Wanijya Sanghatan Wa Wyawasthapan Book

Maharashtra Board Class 12 Yuwakbharti Book

Maharashtra Board Class 12 Sanskrit Medium Books

Asterclasses

Maharashtra Board 12th Standard Sanskrit Aalhad Book

Maharashtra Board Class 12 Hindi Medium Books

Asterclasses

Maharashtra Board 12th Standard Hindi Yuwakbharti Book

Maharashtra Board Class 12 Urdu Medium Books

Asterclasses

Maharashtra Board 12th Standard Urdu Yuwakbharti Book

Maharashtra Board Class 12 Gujarati Medium Books

Asterclasses

Maharashtra Board 12th Standard Gujarati Yuwakbharti Book

Maharashtra Board Class 12 Kannada Medium Books

Maharashtra Board 12th Standard Kannada Yuwakbharti Book

Maharashtra Board Class 12 Sindhi Medium Books

Maharashtra Board 12th Standard Sindhi Yuwakbharti Book

Maharashtra Board 12th Standard SindhiDev Yuwakbharti Book

Maharashtra Board Class 12 Bengali Medium Books

Maharashtra Board 12th Standard Bengali Yuwakbharti Book

Maharashtra Board Class 12 Telugu Medium Books

Maharashtra Board 12th Standard Telugu Yuwakbharti Book

Maharashtra Board Class 12 Other Books

Maharashtra Board Class 12 Ardha Magdhi Prakrut

Maharashtra Board Class 12 Gul Hay Farsi

Maharashtra Board Class 12 Hidayaul Arbiyah

Maharashtra Board Class 12 Maharashri Prakrut

Maharashtra Board Class 12 Palippakaso

Science Chapter 6 – Refraction Of Light SSC, SCIENCE PART I, NEW SYLLABUS, FOR BOARD EXAM 2020

Question 1:

Fill in the blanks and Explain the completed sentences.

a . Refractive index depends on the …………. of light.

b. The change in ……………. of light rays while going from one medium to another is called refraction.

ANSWER:

a. Refractive index depends on the velocity of light.

b. The change in direction of light rays while going from one medium to another is called refraction.

---

Question 2:

Prove the following statements.

Question 3:

Mark the correct answer in the following questions.

c. If the refractive index of glass with respect to air is 3/2, what is the refractive index of air with respect to glass?

ANSWER:

2/3

---

A. The correct reason for the twinkling of stars is changing the refractive index of the atmospheric gases.

Light coming from the stars undergoes refraction on entering the Earth’s atmosphere.

This refraction continues until it reaches the Earth’s surface.

This happens because of temperature variation of atmospheric air.

Hence, the atmospheric air has changing refractive index at various altitudes.

In this case, starlight continuously travels from a rarer medium to a denser medium.

Hence, it continuously bends towards the normal.

The refractive index of air medium gradually increases with a decrease in altitude.

The continuous bending of starlight towards the normal results in a slight rise of the apparent position of the star.

---

B. We can see the Sun even when it is a little below the horizon because of refraction of light.

The rays of light from the Sun travel in a straight line until they reach the Earth’s atmosphere. 

The rays of light from the Sun enter obliquely in the Earth’s atmosphere. The light rays coming from the Sun bend because of refraction, and this bending increases further because of the further increase in the refractive index of the successive layers. 

This causes the light rays to bend and we see the Sun early. 

Similarly, at sunset, the apparent position of the Sun is visible to us and not the actual position because of the same bending of light rays effect. 

Thus, due to refraction we see the Sun rise about two minutes before it is actually there and during sunset, we see it for around two minutes more, even though it has already moved from that position.

---

Question 4:

Solve the following examples.

a. If the speed of light in a medium is 1.5 × 108 m/s, what is the absolute refractive index of the medium?

b. If the absolute refractive indices of glass and water are 3/2 and 4/3 respectively, what is the refractive index of glass with respect to water?

ANSWERS

A.

---

B.

---

EXTRA QUESTION

IMPORTANT FOR BOARD EXAM

---

---

Question 1:

Choose the correct option from the bracket and explain the statement giving reason.

(Oxidation, displacement, electrolysis, reduction, zinc, copper, double displacement, decomposition)
a. To prevent rusting, a layer of …….. metal is applied on iron sheets.

b. The conversion of ferrous sulphate to ferric sulphate is …….. reaction.

c. When electric current is passed through acidulated water …….. of water takes place.

d. Addition of an aqueous solution of ZnSO4 to an aqueous solution of BaCl2 is an example of ……. reaction.

ANSWER:

a. To prevent rusting, a layer of zinc metal is applied on iron sheets.

b. The conversion of ferrous sulphate to ferric sulphate is oxidation reaction.

c. When electric current is passed through acidulated water of electrolysis water takes place.

d. Addition of an aqueous solution of ZnSO4 to an aqueous solution of BaCl2 is an example of a double displacement reaction.

Question 2:

Write answers to the following.

a. What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.

b. How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?

c. Explain the term reactant and product giving examples.

d. Explain the types of reaction with reference to oxygen and hydrogen. Illustrate with examples.

e. Explain the similarities and differences in two events, namely adding NaOH to water and adding CaO to water.

ANSWER:

a. The process in which oxidation and reduction occurs simultaneously is called redox reaction. Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion. Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.
For example: If we add stannous chloride solution to the yellow solution of ferric chloride then light green ferrous chloride solution and stannic chloride solution are produced.

Before the reaction, 3Cl atoms were attached to each iron atom. After the reaction, only two chlorine atoms are attached. That is one negative chlorine atom is released. Therefore, reduction of FeCl3 happened or this reactor on the other hand, before reaction two chlorine atoms were attached with each atom of tin (stannum). Due to the above reaction the number of chlorine attached to tin atom increases to four. That is, oxidation of SnCl2has taken place. Therefore, in this reaction oxidation of one substance and reduction of another substance take place simultaneously. This is called redox reaction.

b. The rate of decomposition of hydrogen peroxide can be increased by having the reaction occurs in the presence of iodide ions. The reaction is proceed by two step mechanism:

c. Reactant: The substance which take part in a chemical reaction are called reactants.

Product: the substance which forms as a result of chemical reaction is called products.
The new substance produced as a result of chemical reaction is called products.

For example: When two sodium atoms react with two chlorine atoms(reactants), they give a completely new compound (product) i.e. sodium chloride (two atoms).

d. There are three main types of chemical reactions with reference to oxygen and hydrogen:

1.Combination Reaction:- When two atoms react to form a compound, it is known as combination reaction.

For example: 

2. Decomposition Reaction:- When a compound breaks into simple molecular substances from which it is made up of, it is known as decomposition reaction.

For example: 

3. Oxidation and reduction reaction: –

Oxidation: (i) The addition of oxygen to a substance is called oxidation.(ii) The removal of hydrogen to a substance is called oxidation.
Reduction: (i) The addition of hydrogen to a substance is called reduction.(ii) The removal of oxygen to a substance is called reduction. For example:In the above reaction, copper oxide is changing to Cu. That is, oxygen is being removed from copper oxide. So, copper oxide is being reduced to copper. In the above reaction, H2 is changing into H2O. That is , oxygen is being added to hydrogen. So, hydrogen is being oxidized to water.

e. Chemical equations involved:

Similarities:

1) Both of the equations are exothermic. It means a lot of heat is evolved during the reaction.
2) Both reaction form strong basic solution.

Differences :

1) Sodium Hydroxide is strong base dissociates to form Na⁺ and OH⁻ ion. While Calcium oxide added water to form Calcium Hydroxide which further dissociates.
2) NaOH is a monoacidic base. and CaO is a Di-Acidic base.
3) NaOH, CaO should be added to water gradually with constant stirring.CaO on reacting with water produces a basic solution called Calcium hydroxide which is used for white washing and this reaction is more dangerous as compared to NaOH.

Question 3:

Explain the following terms with examples.

a. Endothermic reaction and exothermic reaction

b. Combination reaction

c. Balanced equation

d. Displacement reaction

ANSWER:

a. Endothermic reaction: Endothermic reactions are chemical reactions in which the reactants absorb heat energy from the surroundings to form products. These reactions lower the temperature of their surrounding area, thereby creating a cooling effect. Physical processes can be endothermic as well – Ice cubes absorb heat energy from their surrounding and melt to form liquid water (no chemical bonds are broken or formed).

Endothermic reactions generally involve the formation of chemical bonds through the absorption of heat from the surroundings. On the other hand, exothermic reactions involve the release of heat energy generated from bond-breakage. The terms ‘endo’ and ‘exo’ have Greek roots, meaning ‘within’ and ‘out’ respectively. As the names suggest, the primary difference between endothermic and exothermic reactions is that the former absorbs heat from the surroundings whereas the latter involves a release of heat.

An exothermic reaction is a chemical reaction that releases energy in the form of heat and light. It is the opposite of an endothermic reaction. Exothermic Reaction means “exo” meaning releases and “thermic” means heat. So the reaction in which there is release of heat with or without light is called an exothermic reaction.
Expressed in a chemical equation: reactants → products + energy.For example: Combustion is an exothermic reaction.The chemical reaction between zinc granules and dilute sulphuric acid is exothermic reaction.

b. Combination reaction: Those reactions in which two or more substances combine to form single substance is called combination reaction For example: Magnesium and oxygen combine, when heated, to form magnesium oxide.

c. Balanced equation:
A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge are the same for both the reactants and the products. In other words, the mass and the charge are balanced on both sides of the reaction.
The balanced equation is:

Both the left and right sides of the equation have 4 Fe, 6 O, and 3 C atoms. When you balance equations, it’s a good idea to check your work by multiplying the subscript of each atom by the coefficient. When no subscript is cited, consider it to be 1. It’s also good practice to cite the state of matter of each reactant. This is listed in parentheses immediately following the compound. For example, the earlier reaction could be written as:
where s indicates a solid and g is a gas state of matter.
d. Displacement reaction:
Displacement reaction is a chemical reaction in which a more reactive element displaces a less reactive element from its compound.

Both metals and nonmetals take part in displacement reactions.

Chemical reactivity of metals is linked with their relative positions in the activity series.

A metal placed higher in the activity series can displace the metal that occupies a lower position from the aqueous solution of its salt.

Question 4:

Give scientific reasons.

a. When the gas formed on heating limestone is passed through freshly prepared lime water, the lime water turns milky.

ANSWER:

Limestone is made up of calcium carbonate.

When calcium carbonate is heated Carbon dioxide is evolved.

Lime water is made up of Calcium hydroxide.

When carbon dioxide is passed through lime water, the Carbon dioxide reacts with calcium hydroxide to form Calcium carbonate which is a white precipitate, then lime water turns milky as there is formation of calcium carbonate.

b. It takes time for pieces of Shahabad tile to disappear in HCl, but its powder disappears rapidly.

ANSWER:

It takes time for pieces of Shahabad tile to disappear in HCl, but its powder disappears rapidly because in pieces of the tile, the surface atoms can only react with HCl but in powdered form all the atoms can react with HCl.

This makes the difference in their reactivity. Powders are simpler substances of the pieces and they take lesser time for the reaction as compared to the whole piece of tile, which is a compound.

For example: powdered salt react and water will dissolve, but salt rocks and water will not dissolve.

c. While preparing dilute sulphuric acid from concentrated sulphuric acid in the laboratory, the concentrated sulphuric acid is added slowly to water with constant stirring.

ANSWER:

While preparing dilute sulphuric acid from concentrated sulphuric acid in the laboratory, the concentrated sulphuric acid is added slowly to water with constant stirring because sulphuric acid (H2SO4) reacts very vigorously with water, it is a highly exothermic reaction.

If you add water to concentrated sulphuric acid, it can boil and bump out due to which you may get a nasty acid burn.

Water is a good absorber of heat, so we add acid to the water, slowly and with constant stirring.

d. It is recommended to use air tight container for storing oil for long time.

ANSWER:

It is recommended to use air tight container for storing oil for a long time in order to avoid the problem of rancidity.

Rancidity is the condition produced by aerial oxidation of fats and oils marked by unpleasant smell and taste.

Question 5:

Observe the following picture a write down the chemical reaction with explanation.

ANSWER:

i. The given picture represents the electrochemical reaction which takes place during the corrosion of iron. (rusting) 
ii. Different regions on the surface of iron behave as anode and cathode. 
iii. In the anode region, Fe is oxidized to Fe²⁺. 

iv. In the cathode region, O2 is reduced to form water. 

V. When Fe²⁺ ions migrate from the anode region, they react with water and further get oxidized to Fe³⁺ ions. 
vi. Fe³⁺ ions form an insoluble hydrated oxide (Fe2O3.H2O), which is deposited as reddish brown layer on the surface. It is called rust. 

---

Question 6:

Identify from the following reaction the reactants that undergo oxidation and reduction.

ANSWER:

a. Fe + S → FeSIn a reaction, Fe is changing to FeS. That means, iron loses electrons to form FeS. Loss of electrons from a substance is called oxidation, so iron undergoes oxidation.
b. 2Ag2O → 4 Ag + O2 ↑In a reaction, silver oxide is changing to silver. That is, oxygen is being removed from silver oxide. Removal of oxygen from a substance is called reduction, so silver oxide undergoes reduction.
c. 2Mg + O2 → 2MgOIn a reaction, magnesium is changing to magnesium oxide. That means, oxygen is being added to magnesium. Addition of oxygen to a substance is called oxidation, so magnesium undergoes oxidation.
d. NiO + H2 → Ni + H2OIn a reaction, Nickel oxide is changing to nickel. That is, oxygen is being removed from nickel oxide. Removal of oxygen from a substance is called reduction, so nickel oxide undergoes reduction.In a reaction, hydrogen is changing to H2O. That is,oxygen is being added to hydrogen. Addition of oxygen to a substance is called oxidation, so hydrogen undergoes oxidation.

Question 7:

Balance the following equation stepwise.

ANSWER:

a. H2S2O7(l) + H2O(l) → H2SO4(l)

Step1. Count the number of each atom in reactant side:H= 4S=2O=8

Step2. Count the number of each atom in product side:H= 2S=1O=4

Step3.Then balance the number of each atom in an equation by multiplying reactant and product side with numeral value:

If we multiply product side by 2, then the number of atoms in product and reactant side gets balance.H2S2O7(l) + H2O(l) → 2H2SO4(l)

b. SO2(g) + H2S(aq) → S(s) + H2O(l)      

Step1. Count the number of each atom in reactant side: H= 2S=2O=2

Step2. Count the number of each atom in product side:H= 2S=1O=2

Step3.Then balance the number of each atom in an equation by multiplying reactant and product side with numeral value:

If we multiply H2S by 2 in the reactant side and S by 3 and H2O by 2 in the product side, then the number of atoms in product and reactant side gets balance.SO2(g) + 2H2S(aq) → 3S(s) + 2H2O(l) 

c. Ag(s) + HCl(aq) → AgCl ↓+ H2 ↑

Step1. Count the number of each atom in reactant side: H= 1Ag=1Cl=1

Step2. Count the number of each atom in product side:H= 2Ag=1Cl=1

Step3.Then balance the number of each atom in an equation by multiplying reactant and product side with numeral value:

If we multiply Ag by 2 and HCl by 2 in the reactant side and AgCl by 2 in the product side, then number of atoms in product and reactant side gets balance.2Ag(s) + 2HCl(aq) → 2AgCl ↓+ H2 ↑

d. NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)

Step1. Count the number of each atom in reactant side: Na= 1H=3O=5S=1

Step2. Count the number of each atom in product side:Na= 2H=2O=5S=1

Step3.Then balance the number of each atom in an equation by multiplying reactant and product side with numeral value:

If we multiply NaOH by 2 in the reactant side and H2O by 2 in the product side, then the number of atoms in product and reactant side gets balance.2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) +2H2O(l)

Question 8:

Identify the endothermic and exothermic reaction.

ANSWER:

Identify the endothermic and exothermic reaction.Heat is released in the product side, as it mentioned in the above reaction. So, It is an exothermic reaction because heat is evolved in exothermic reaction.Heat is given in the product side to break the compound into simpler substances, as it mentioned in the above reaction. So, It is an endothermic reaction because heat is supplied in exothermic reaction.
Heat is released in the product side, as it mentioned in the above reaction. So, It is an exothermic reaction because heat is evolved in exothermic reaction.
Heat is provided in the product side to break the compound into simpler substances, as it mentioned in the above reaction. So, It is an endothermic reaction because heat is supplied in exothermic reaction.

Question 9:

SOLUTION

Science Chapter 2 – Periodic Classification Of Elements SSC, SCIENCE PART I, NEW SYLLABUS, FOR BOARD EXAM 2020,

Question 1:

Rearrange the columns 2 and 3 so as to match with the column 1.

Column 1	Column 2	Column 3
i. Triad ii. Octave iii. Atomic number iv. Period v. Nucleus vi. Electron	a. Lightest and negatively charged particle in all the atoms b. Concentrated mass and positive charge c. Average of the first and the third atomic mass d. Properties of the eighth element similar to the first e. Positive charge on the nucleus f. Sequential change in molecular formulae	1.Mendeleev 2. Thomson 3. Newlands 4. Rutherford 5. Dobereiner 6. Moseley

ANSWER:

Column 1	Column 2	Column 3
i. Triad	a. Average of the first and the third atomic mass	1.Dobereiner
ii. Octave	b. Properties of the eighth element similar to the first	2.Newlands
iii. Atomic number	c. Positive charge on the nucleus	3.Mendeleev
iv. Period	d. Sequential change in molecular formulae	4.Moseley
v. Nucleus	e. Concentrated mass and positive charge	5.Rutherford
vi. Electron	f.  Lightest and negatively charged particle in all the atoms	6.Thomson

Question 2:

Choose the correct option and rewrite the statement.

a. The number of electrons in the outermost shell of alkali metals is …..(i) 1 (ii) 2 (iii) 3 (iv) 7

b. Alkaline earth metals have valency 2. This means that their position in the modern periodic table is in—-(i) Group 2  (ii) Group 16 (iii) Period 2 (iv) d-block

c. Molecular formula of the chloride of an element X is XCl. This compound is a solid having high melting point. Which of the following elements be present in the same group as X.(i) Na  (ii) Mg (iii) Al  (iv) Si

d. In which block of the modern periodic table are the nonmetals found?(i) s-block   (ii) p-block(iii) d-block (iv) f-block

ANSWER:

a. The number of electrons in the outermost shell of alkali metals is 1.

b. Alkaline earth metals have valency 2. This means that their position in the modern periodic table is in group2.

c. Molecular formula of the chloride of an element X is XCl. This compound is a solid having high melting point. An element to be present in the same group as X is Na.

d. In p-block of the modern periodic table are the nonmetals found.

Question 3:

An element has its electron configuration as 2, 8, 2. Now answer the following question.

a. What is the atomic number of this element?

b. What is the group of this element?

c. To which period does this element belong?

d. With which of the following elements would this element resemble? (Atomic numbers are given in the brackets)N(7), Be(4), Ar(18), Cl(17)

ANSWER:

An element has its electron configuration as 2, 8, 2.

a. The atomic number of this element is 12.

b. The group  number of this element is second.

c. This element belongs to third period.

d. This element resembles with Be(2).

Question 4:

Write down the electronic configuration of the following elements from the given atomic numbers.

Answer the following question with explanation.

a. 3Li, 14He, 11Na, 15P Which of these elements belong to be period 3?

b. 1H, 7N, 20Ca, 16S, 4Be, 18Ar Which of these elements belong tot he second group?

c. 7N, 6C, 8O, 5B, 13Al Which is the most electronegative element among these?

d. 4Be, 6C, 8O, 5B, 13Al Which is the most electropositive element among these?

e. 11Na, 15P, 17Cl, 14Si, 12Mg Which of these has largest atoms?

f. 19K, 3Li, 11Na, 4Be Which of these atoms has smallest atomic radius?

g. 13Al, 14Si, 11Na, 12Mg, 16S Which of the above elements has the highest metallic character?

h. 6C, 3Li, 9F, 7N, 😯 Which of the above elements has the highest nonmetallic character?

ANSWER:

a. 3Li, 14He, 11Na, 15P
Electronic configuration of the following elements is: (3Li = 2,1) (14He =2,8,4) (11Na = 2,8,1) (15P = 2,8,5) 14He, 11Na, 15P belong to the third period because according to their electronic configuration, each element contains three shell i.e. K,L,M.

b. 1H, 7N, 20Ca, 16S, 4Be, 18Ar
Electronic configuration of the following elements is:1H = 17N = 2,520Ca = 2,8,8,216S = 2,8,64Be = 2,218Ar = 2,8,820Ca, 4Be belong to second group because these elements have 2 electrons in its outermost shell.

c. 7N, 6C, 8O, 5B, 13Al
Electronic configuration of the following elements is:7N =2,56C = 2,48O = 2,65B = 2,313Al = 2,8,38O is the most electronegative element among these because electronegativity increases as we move from left to right in a period of the periodic table.

d. 4Be, 6C, 8O, 5B, 13Al
Electronic configuration of the following elements is: (4Be = 2,2)  (6C = 2,4) (8O = 2,6) (5B = 2,3) (13Al = 2,8,3) 13Al is the most electropositive element among these because 4Be, 6C, 8O, 5B belong to same period, but 13Al belong to next period. According to the trend, electropositive character of an elements increases as we move from top to bottom in a group of the periodic table. This happens as the tendency of an atom to lose electrons increases due to decrease in nuclear charge and increase in numbers of shell.

e. 11Na, 15P, 17Cl, 14Si, 12Mg
Electronic configuration of the following elements is: (11Na = 2,8,1) (15P = 2,8,5) (17Cl = 2,8,7) (14Si = 2,8,4) (12Mg = 2,8,2) 11Na has largest size among these because according to the trend, atomic radius decreases as we move from left to right in a period of the periodic table. The atomic number of elements increases which means the number of protons and electrons in  the atoms increases. Due to large positive charge on the nucleus, the electrons are pulled closer to the nucleus and the size of atom decreases.

f. 19K, 3Li, 11Na, 4Be
Electronic configuration of the following elements is: .

(19K = 2,8,8,1) (3Li = 2,1) (11Na = 2,8,1) (4Be = 2,2) 4Be has the smallest atomic radius because 19K, 3Li, 11Na are present in the same group 1 but Be is present in group 2. According to the trend, as we move from left to right atomic size of an atoms decreases. Due to large positive charge on the nucleus, the electrons are pulled closer to the nucleus and the size of atom decreases.

g. 13Al, 14Si, 11Na, 12Mg, 16S
Electronic configuration of the following elements is: (3Al = 2,8,3) (14Si = 2,8,4) (11Na = 2,8,1) (12Mg = 2,8,2) (16S = 2,8,6) (11Na) has the highest metallic character because metallic character of an elements decreases as we move from left to right in a modern periodic table. This happens as the tendency of an atom to lose electrons decreases due to gradual increase in the number of protons and nuclear charge.

Question 5:

Write the name and symbol of the element from the description.

a. The atom having the smallest size.

b. The atom having the smallest atomic mass.

c. The most electronegative atom.

d. The noble gas with the smallest atomic radius.

e. The most reactive nonmetal.

ANSWER:

a. The atom having the smallest size = Hydrogen (H)

b. The atom having the smallest atomic mass = Hydrogen (H)

c. The most electronegative atom = Fluorine (F)

d. The noble gas with the smallest atomic radius = Helium (He)

e. The most reactive nonmetal = Fluorine (F)

Question 6:

Write short notes.

a.  Mendeleev’s periodic law.

ANSWER:

According to Mendeleev’s Periodic Law, “Physical and chemical properties of elements are a periodic function of their atomic masses”.

Mendeleev classified elements according to their atomic masses and arranged these elements in increasing order of  their atomic masses.

Mendeleev classified periodic table into horizontal rows and vertical columns.

The horizontal rows are called periods and vertical columns are called groups.

Mendeleev’s Periodic Table contains seven horizontal rows and nine vertical columns.

The elements with similar properties comes into same group.

Mendeleev also left gaps in his periodic table for undiscovered elements like aluminum, silicon and Boron in his periodic table and named them Eka-Aluminium, Eka-silicon and Eka-Boron.

Mendeleev not only predicted the existence of Eka-Aluminium, Eka-silicon and Eka-Boron but also described the general physical properties of these elements.

These elements discovered later and named as Gallium, Germanium and Scandium.

Mendeleev’s periodic table could predict the properties of several elements on the basis of their position in the periodic table.

Mendeleev’s periodic table could accommodate noble gases when they were discovered.

Demerits of Mendeleev’s periodic table:

The position of isotopes could not explained.

Wrong order of atomic masses of some elements could not be explained.

Position of Hydrogen could not be assigned in a periodic table.

b. Structure of the modern periodic table:

ANSWER:

Periodic Table: It is the table of chemical elements arranged in order of atomic number such that elements with similar atomic structure appear in the vertical columns.

The Modern periodic law states that The chemical and physical properties of elements are periodic functions of their atomic numbers.

Modern periodic table is based on modern periodic law. 

Main features:

Groups – There are 18 vertical columns in the periodic table. Each column is called a group. All elements in a group have similar chemical and physical properties because they have the same number of valence electrons.

Periods – In periodic table elements are arranged in a series of rows. Elements of the same period have the same number of electron shells.

Classification of elements:

Group 1 contains alkali metals (Li, Na, K, Rb, Cs and Fr).

The alkaline earth metals are metallic elements found in the group 2 of the periodic table. 

Elements present in groups 3 to 12 in the middle of the periodic table are called transition elements.

In the transition elements, valence electrons are present in more than one shell.

Group 18 on extreme right side position contains noble gases ( He, Ne, Ar, Kr, Xe and Rn ). Their outermost shell contains 8 electrons except He as its outermost shell is K shell and it can hold only 2 electrons. 

Inner transition elements:

1. 14 elements with atomic numbers 58 to 71 (Ce to Lu) are called lanthanides  and they are placed along with the element lanthanum (La), atomic number 57 in the same position (group 3 in period 6) because of very close resemblance between them.

2. 14 elements with atomic numbers 90 to 103 (Th to Lr) are called actinides and they are placed along with the element actinium (Ac), atomic number 89 in the same position (group 3 in period 7) because of very close resemblance between them.

c. Position of isotopes in the Mendeleev’s and the modern periodic table:

ANSWER:

Isotopes: Isotopes are atoms having the same atomic number but different atomic masses.

Therefore, according to Mendeleev’s classification these should be placed at different places depending upon their atomic masses.

For example, hydrogen isotopes with atomic masses 1,2 and 3 should be placed at three places.

However, isotopes have not been given separate places in the periodic table because of their similar properties.

So this was a drawback of Mendeleev’s periodic table as he could not explain the position of isotopes.

Modern periodic table is based upon arrangement of the elements on the basis of their atomic number.

So that, all the isotopes of hydrogen should be placed at the same place depending upon their atomic number.

Question 7:

Write scientific reasons.a. Atomic radius goes on decreasing while going from left to right in a period.

ANSWER:

Atomic radius goes on decreasing while going from left to right in a period.

because the atomic number of the elements increases which means the number of protons and electrons in the atoms increases(the extra electrons being added to the same shell).

Due to large positive charge on the nucleus, the electrons are pulled closer to the nucleus and the size of an atom decreases.

b. Metallic character goes on decreasing while going from left to right in a period.

ANSWER:

Metallic character goes on decreasing while going from left to right in a period because the tendency of atoms of the elements to lose electrons(or gain  electrons) changes in a period.

As we move from left to right in a period, the nuclear charge increases due to gradual increase in the number of protons.

Due to the increase in nuclear charge, the valence electrons are pulled strongly by the nucleus and it becomes difficult for the atoms to lose electrons.

Hence, metallic character decreases.

c.  Atomic radius goes on increasing down a group.

ANSWER:

Atomic radius increases as we move from top to bottom in a group of the periodic table because a new shell of electrons is added to the atoms at every step.

As the number of shells in the atoms increases gradually due to which the size of atoms also increases.

As the size of the atoms increases which leads to increase in atomic radius of an atom.

d.  Elements belonging to the same group have the same valency.

ANSWER:

Elements belong to the same group have the same valency.

because the number of valence electrons in a group is same due to which the tendency of an atom to lose or gain electrons in order to attain nearest noble gas configuration is also same.

e. The third period contains only eight elements even through the electron capacity of the third shell is 18 .

ANSWER:

The third period contains only eight elements even through the electron capacity of the third shell is 18.

because when the other shells get filled and the resultant no of electrons becomes eighteen, it gets added up and settles in the third electron shell and three shells is acquired by fourth period.

Question 8:

Write the names from the description.

a. The period with electrons in the shells K, L and M.

b. The group with valency zero.

c. The family of nonmetals having valency one.

d. The family of metals having valency one.

e. The family of metals having valency two.

f. The metalloids in the second and third periods.

g. Nonmetals in the third period.

h. Two elements having valency 4.

ANSWER:

a. The period with electrons in the shells K, L and M = 3 period

b. The group with valency zero = 18 group

c. The family of nonmetals having valency one = Halogens

d. The family of metals having valency one = Alkali metals

e. The family of metals having valency two = Alkaline earth metals

f. The metalloids in the second and third periods = Boron( second period), Silicon (third period) 

g. Nonmetals in the third period = Sulphur, Chlorine

h. Two elements having valency 4 = Carbon, Silicon

Science Chapter 1 – Gravitation SSC, SCIENCE PART I, NEW SYLLABUS, FOR BOARD EXAM 2020,

Question 1:

Study the entries in the following table and rewrite them putting the connected items in a single row.

I	II	III
MASS	m/s2	Zero at the Centre
WEIGHT	kg	Measure of inertia
ACCELARATION DUE TO GRAVITY	Nm2/kg2	same in the entire universe
GRAVITATIONAL CONSTANT	N	Depends on the height

SOLUTION: –

I	II	III
MASS	kg	Measure of inertia
WEIGHT	N	Zero at the Centre
ACCELARATION DUE TO GRAVITY	m/s2	Depends on the height
GRAVITATIONAL CONSTANT	Nm2/kg2	same in the entire universe

Question 2:

Answer the following questions.

a. What is the difference between mass and weight of an object. Will the mass and weight of an object on the earth be same as their values on Mars? Why?

ANSWER:

^{a. Difference between mass and weight of an object is as follows}

MASS	WEIGHT
Mass is the amount of matter contained in a body.	Weight is the force exerted on a body due to the gravitational pull of another body such as Earth, the sun and the moon.
Mass is an intrinsic property of a body.	Weight is an extrinsic property of a body.
Mass is the measure of inertia.	Weight is the measure of force.
The mass of a body remains the same everywhere in the universe.	The weight of a body depends on the local acceleration due to gravity where it is placed.
The mass of a body cannot be zero.	The weight of a body can be zero.

The mass of an object on the Earth will be same as that on Mars but its weight on both the planets will be different. This is because the weight (W) of an object at a place depends on the acceleration due to gravity of that place i.e. 

b. What  are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force ?

ANSWER:

ANSWER:

Three laws given by Kepler is as follows:

First Law: The orbits of the planets are in the shape of ellipse, having the Sun at one focus.

Second Law: The area swept over per hour by the radius joining the Sun and the planet is the same in all parts of the planet’s orbit.

Third Law: The squares of the periodic times of the planets are proportional to the cubes of their mean distances from the Sun.Newton used Kepler’s third law of planetary motion to arrive at the inverse-square rule. He assumed that the orbits of the planets around the Sun are circular, and not elliptical, and so derived the inverse-square rule for gravitational force using the formula for centripetal force. This is given as:

where, m is the mass of the particle, r is the radius of the circular path of the particle and v is the velocity of the particle. Newton used this formula to determine the force acting on a planet revolving around the Sun. Since the mass m of a planet is constant, equation (i) can be written as:

Now, if the planet takes time T to complete one revolution around the Sun, then its velocity v is given as:

where, r is the radius of the circular orbit of the planet. Or, we can write it as, 

[as the factor 2π is a constant]

On squaring both sides of this equation, we get:

On multiplying and dividing the right-hand side of this relation by r, we get:

According to Kepler’s third law of planetary motion, the factor is constant , Hence, equation (vi) becomes:

On using equation (vii) in equation (ii), we get:

Hence, the gravitational force between the sun and a planet is inversely proportional to the square of the distance between them.

d. A stone thrown vertically upwards with an initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is the same as the time taken to come down.

For vertical upward motion of the stone:

S = hu = uv = 0a = -g
Let t be the time taken by the ball to reach height h. Thus, using second equation of motion, we have

For vertical downward motion of the stone:

S = hu = 0a = g

Let v’ be the velocity of the ball with which it hits the ground. Let t’ be the time taken by the ball to reach the ground. Thus, using the second equation of motion, we have

Hence, from (i) and (ii), we observe that the time taken by the stone to go up is the same as the time taken by it to come down.

e. If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?

ANSWER:

Let the mass of the heavy object be m. Thus, the weight of the object or the pull of the floor on the object is

W = mg

Now, if g becomes twice, the weight of the object or the pull of the floor on the object also becomes twice i.e.

W’ = 2mg = 2W

Thus, because of the doubling of the pull on the object due to the floor, it will become two times more difficult to pull it along the floor.

Question 3:

Explain why the value of g is zero at the centre of the earth.

ANSWER:

At the centre of the Earth, the force due to the upper half of the Earth will cancel the force due to lower half. In the similar manner, force due to any portion of the Earth at the centre will be cancelled due to the portion opposite to it. Thus, the gravitational force at the centre on any body will be 0. Since, from Newton’s law, we know

F = mg Since, mass m of an object can never be 0. Therefore, when F = 0, g has to be 0. Thus, the value of g is zero at the centre of the Earth.

Question 4:

Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be

Question 5:

Solve the following examples.

a. An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?

b. The radius of planet A is half the radius of planet B. If the mass of A is M_A, what must be the mass of B so that the value of g on B is half that of its value on A?

c. The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.

d. An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s² 

e. A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s², calculate its speed on reaching the ground and the height of  the table.

f. The masses of the earth and moon are 6 × 10²⁴ kg and 7.4 × 10²² kg, respectively. The distance between them is 3.8 × 10⁵km. Calculate the gravitational force of attraction between the two? Use G = 6.7 × 10^–11 N m² kg^–2

g. The mass of the earth is 6 × 10²⁴ kg. The distance between the earth and the sun is 1.5 × 10¹¹ m. If the gravitational force between the two is 3.5 × 10²² N, what us the mass of the sun? Use G = 6.7 × 10^–11 N m² kg^–2